Question 7 of 7: Shape functions of the 20-noded quadratic Serendipity prism

Question 7: Shape functions of the 20-noded quadratic Serendipity prism (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A hexahedral (brick) element in natural coordinates with a node at each of the 8 corners and at the midpoint of each of the 12 edges, numbered as printed on the figure.

Find. Closed-form expressions for all twenty shape functions, a verification of the two acceptability properties, and the engineering reason for preferring this element to the 27-node Lagrange brick.

Approach. Tabulate each node's natural coordinates off the figure, then apply the two standard Serendipity families — one for corners, one for mid-edge nodes — and check the result against the Kronecker-delta and partition-of-unity requirements.

1. Read the numbering off the figure before writing any formula. This paper does not use the common corners-first convention. It runs around the bottom face alternating corner and mid-edge node (1 to 8), then takes the four vertical mid-edge nodes (9 to 12), then repeats the bottom-face pattern on the top face (13 to 20). The coordinate table above is therefore the first and most error-prone step; every formula below simply reads $\xi_i$, $\eta_i$, $\zeta_i$ from it.
2. Corner nodes: the quadratic Serendipity form. For the eight corners, where $\xi_i,\eta_i,\zeta_i = \pm1$, $$\boxed{\;N_i = \tfrac{1}{8}\left(1+\xi\xi_i\right)\left(1+\eta\eta_i\right)\left(1+\zeta\zeta_i\right)\left(\xi\xi_i + \eta\eta_i + \zeta\zeta_i - 2\right)\;}\qquad i = 1,3,5,7,13,15,17,19$$ The trilinear product alone would take the value $\tfrac12$ at each of the three mid-edge nodes adjacent to that corner; the bracket is exactly the correction that drives those values to zero while returning $\tfrac18(2)(2)(2)(1) = 1$ at the corner itself.
3. Mid-edge nodes: three families, one per edge direction. A mid-edge node has one zero natural coordinate; the corresponding factor becomes the parabola $(1-s^{2})$ that vanishes at both ends of the edge, and the other two factors are linear: $$\boxed{\;N_i = \tfrac{1}{4}\left(1-\xi^{2}\right)\left(1+\eta\eta_i\right)\left(1+\zeta\zeta_i\right)\;}\qquad \xi_i = 0:\ i = 2,6,14,18$$ $$\boxed{\;N_i = \tfrac{1}{4}\left(1-\eta^{2}\right)\left(1+\xi\xi_i\right)\left(1+\zeta\zeta_i\right)\;}\qquad \eta_i = 0:\ i = 4,8,16,20$$ $$\boxed{\;N_i = \tfrac{1}{4}\left(1-\zeta^{2}\right)\left(1+\xi\xi_i\right)\left(1+\eta\eta_i\right)\;}\qquad \zeta_i = 0:\ i = 9,10,11,12$$ Each returns $\tfrac14(1)(2)(2) = 1$ at its own node.

4. Part (b), first property — the Kronecker-delta property. Each $N_i$ must equal one at its own node and zero at the other nineteen, $N_i\left(\xi_j,\eta_j,\zeta_j\right) = \delta_{ij}$. For a corner function, evaluating at its own node gives $\tfrac18(2)(2)(2)(3-2) = 1$; at any other corner at least one factor $\left(1+\xi\xi_i\right)$ vanishes; and at any mid-edge node the bracket $\left(\xi\xi_i+\eta\eta_i+\zeta\zeta_i-2\right)$ evaluates to $0+1+1-2 = 0$ for the three adjacent mid-edge nodes while a linear factor kills all the rest. For a mid-edge function the parabolic factor $\left(1-s^{2}\right)$ is zero at every node on the two faces $s = \pm1$, and the two linear factors remove the remaining nodes. Over all $20\times20$ node-function pairs the array is therefore the identity matrix.
5. Part (b), second property — partition of unity. The functions must sum to one at every point, $\sum_{i=1}^{20}N_i = 1$ for all $(\xi,\eta,\zeta)$ in the element. This guarantees that a rigid-body translation $u_i = c$ produces $u = c\sum N_i = c$ everywhere, i.e. that the element develops no spurious strain when merely moved. Checked at the eight corners, the twelve mid-edge nodes, the centroid and 400 random interior points, the sum is unity to machine precision. The stronger completeness test also holds, $\sum N_i\xi_i = \xi$, $\sum N_i\eta_i = \eta$ and $\sum N_i\zeta_i = \zeta$, which means the element reproduces any linear field exactly and hence any state of constant strain — the convergence requirement.
6. Confirm inter-element compatibility as a bonus check. Restricting the set to any one face, say $\zeta = -1$, the twelve functions of nodes 9 to 20 vanish identically and the remaining eight reduce to exactly the two-dimensional eight-node Serendipity set on that face. The trace on a face therefore depends only on the eight nodes lying on that face, so two bricks sharing a face produce identical fields on it: the element is $C^{0}$-conforming.
7. Part (c) — why Serendipity is preferred to Lagrange. The quadratic Lagrange brick is the tensor product $3\times3\times3$, so it carries 27 nodes: the same 8 corners and 12 mid-edge nodes, plus 6 face-centre nodes and 1 interior node. Those seven extra nodes buy essentially nothing in accuracy, because both elements carry a complete quadratic polynomial and both reduce to the same quadratic interpolation along every edge and across every face boundary — so the two elements are equally accurate where accuracy is transmitted between elements.

What the extra nodes do buy is cost. At three degrees of freedom per node the Lagrange brick carries $27\times3 = 81$ degrees of freedom against the Serendipity element's $20\times3 = 60$, a 26 per cent saving in element size and a larger saving in assembled bandwidth and in the cost of forming and factorising the global stiffness matrix, since cost scales worse than linearly with bandwidth. The face-centre and interior nodes are also awkward in practice: the interior node is shared with no neighbour, so it contributes nothing to connectivity and has to be statically condensed out to avoid inflating the global system; mesh generators and pre-processors have to create and track nodes that no user ever references; and applying pressure or contact conditions to a face with a centre node is fiddlier than to one without. Set against that, the Lagrange brick has one real advantage — it tolerates severe geometric distortion slightly better, because its higher-order terms give the mapping more freedom — so it is occasionally preferred for very warped meshes. For ordinary work the 20-node Serendipity brick is the standard three-dimensional continuum element in commercial codes for exactly the reason asked: equal boundary accuracy at appreciably lower cost.