Question 7 of 7: The 20-node quadratic Serendipity prism

Question 7: The 20-node 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 $-1\le\xi,\eta,\zeta\le1$ carrying 20 nodes: the 8 corners plus the mid-point of each of the 12 edges, with no face-centre or interior nodes. Reading the figure, the numbering runs around the bottom face first (nodes 1–8, alternating corner and mid-edge), then the four vertical mid-edge nodes (9–12), then around the top face (13–20), as tabulated below.

Find. Closed-form expressions for all 20 shape functions; a demonstration that they satisfy the two admissibility conditions; and the practical argument for preferring this element to the 27-node quadratic Lagrange brick.

Approach. Treat the 20 functions as two families — twelve mid-edge nodes, each of which lies on the intersection of two faces with one natural coordinate zero, and eight corner nodes obtained by correcting the trilinear functions. Then verify the Kronecker-delta and partition-of-unity conditions and reproduce a complete quadratic field.

1. Construct the mid-edge functions. A mid-edge node has exactly one of its natural coordinates equal to zero. Suppose $\xi_i=0$, so the node sits at $\left(0,\eta_i,\zeta_i\right)$ with $\eta_i,\zeta_i=\pm1$. The factor $\left(1-\xi^{2}\right)$ vanishes on both faces $\xi=\pm1$, while $\left(1+\eta\eta_i\right)$ and $\left(1+\zeta\zeta_i\right)$ vanish on the two opposite faces; together they kill every node except the one on this edge, and normalising by $1\cdot2\cdot2=4$ gives $$N_i=\tfrac14\left(1-\xi^{2}\right)\left(1+\eta\eta_i\right)\left(1+\zeta\zeta_i\right) \qquad (\xi_i=0).$$ By the same reasoning for the other two orientations, $$N_i=\tfrac14\left(1-\eta^{2}\right)\left(1+\xi\xi_i\right)\left(1+\zeta\zeta_i\right) \quad(\eta_i=0),\qquad N_i=\tfrac14\left(1-\zeta^{2}\right)\left(1+\xi\xi_i\right)\left(1+\eta\eta_i\right) \quad(\zeta_i=0).$$ For this numbering, nodes 2, 6, 14, 18 use the first form ($\xi_i=0$); nodes 4, 8, 16, 20 use the second ($\eta_i=0$); and nodes 9, 10, 11, 12 use the third ($\zeta_i=0$).
2. Correct the corner functions. The trilinear function $N_i^{0}=\tfrac18\left(1+\xi\xi_i\right)\left(1+\eta\eta_i\right)\left(1+\zeta\zeta_i\right)$ equals $\tfrac12$ at each of the three mid-edge nodes adjacent to corner $i$ and zero at the other nine, so subtracting half of each adjacent mid-edge function restores the delta property. Carrying out the algebra, the three corrections combine into the compact standard form $$N_i=\tfrac18\left(1+\xi\xi_i\right)\left(1+\eta\eta_i\right)\left(1+\zeta\zeta_i\right) \left(\xi\xi_i+\eta\eta_i+\zeta\zeta_i-2\right),\qquad \xi_i,\eta_i,\zeta_i=\pm1 .$$ The bracket $\left(\xi\xi_i+\eta\eta_i+\zeta\zeta_i-2\right)$ equals $+1$ at the corner itself (where all three products are $+1$) and exactly cancels the trilinear value at each adjacent mid-edge node, which is what makes the closed form work.
3. State the complete set. Writing $\xi_0=\xi\xi_i$, $\eta_0=\eta\eta_i$, $\zeta_0=\zeta\zeta_i$ for brevity, $$\boxed{\begin{aligned} \text{corners } (i=1,3,5,7,13,15,17,19):\quad N_i &= \tfrac18\left(1+\xi_0\right)\left(1+\eta_0\right)\left(1+\zeta_0\right) \left(\xi_0+\eta_0+\zeta_0-2\right)\\[4pt] \text{mid-edge with }\xi_i=0\ (i=2,6,14,18):\quad N_i &= \tfrac14\left(1-\xi^{2}\right)\left(1+\eta_0\right)\left(1+\zeta_0\right)\\[4pt] \text{mid-edge with }\eta_i=0\ (i=4,8,16,20):\quad N_i &= \tfrac14\left(1-\eta^{2}\right)\left(1+\xi_0\right)\left(1+\zeta_0\right)\\[4pt] \text{mid-edge with }\zeta_i=0\ (i=9,10,11,12):\quad N_i &= \tfrac14\left(1-\zeta^{2}\right)\left(1+\xi_0\right)\left(1+\eta_0\right) \end{aligned}}$$ As a worked instance, node 1 at $(-1,-1,-1)$ gives $N_1=\tfrac18(1-\xi)(1-\eta)(1-\zeta)\left(-\xi-\eta-\zeta-2\right)$, and node 2 at $(0,-1,-1)$ gives $N_2=\tfrac14\left(1-\xi^{2}\right)(1-\eta)(1-\zeta)$.
4. Part (b), first condition — the Kronecker-delta property. Every shape function must equal unity at its own node and zero at all 19 others, $N_i\!\left(\xi_j,\eta_j,\zeta_j\right)=\delta_{ij}$. For a corner, say node 1, substituting $(-1,-1,-1)$ gives $\tfrac18(2)(2)(2)(1+1+1-2)=1$; at the adjacent mid-edge node 2 at $(0,-1,-1)$ it gives $\tfrac18(1)(2)(2)(0+1+1-2)=0$; and at any node on an opposite face one of the three linear factors vanishes. For a mid-edge function, say $N_2$, substituting its own coordinates gives $\tfrac14(1)(2)(2)=1$, while $\left(1-\xi^{2}\right)=0$ at every node with $\xi=\pm1$ and the remaining factors kill the rest. Applying this to all $20\times20$ combinations reproduces the identity matrix.
5. Part (b), second condition — partition of unity. Summing all twenty functions and expanding gives $$\sum_{i=1}^{20}N_i\left(\xi,\eta,\zeta\right)=1 \qquad\text{identically for all }-1\le\xi,\eta,\zeta\le1 .$$ This is the completeness (or constant-field) requirement: it guarantees that a uniform field $u=c$ is reproduced exactly, hence that rigid-body translation produces no strain. Combined with the first condition it also guarantees interpolation of the nodal values, $u\left(\xi_j,\eta_j,\zeta_j\right)=u_j$. A stronger check, easily confirmed, is that the set reproduces every monomial of the complete quadratic $\{1,\xi,\eta,\zeta,\xi\eta,\eta\zeta,\xi\zeta,\xi^{2},\eta^{2},\zeta^{2}\}$ exactly, so the element can represent any constant-strain state and is second-order complete.
6. Confirm $C^{0}$ compatibility as a by-product. On any face, say $\zeta=1$, every function whose node does not lie on that face vanishes identically, and the eight that remain reduce to the two-dimensional eight-node Serendipity set in the two surviving coordinates. The field on a face therefore depends only on the eight nodes of that face, so two bricks sharing a face predict identical fields on it — the element is fully conforming.
7. Part (c) — compare with the quadratic Lagrange prism. The quadratic Lagrange brick is the tensor product of three one-dimensional quadratics and therefore carries $3^{3}=27$ nodes: 8 corners, 12 mid-edge, 6 face-centre and 1 interior. The Serendipity brick discards those last seven while retaining a complete quadratic basis, so it gives the same order of accuracy from $20$ nodes instead of $27$ — a $26\%$ reduction in node count, and hence in degrees of freedom, matrix size, bandwidth, factorisation cost and storage. Equally important in practice, every Serendipity node lies on the element surface, so mesh generators, contact algorithms, surface loads and boundary conditions all address nodes that physically exist on the boundary; the Lagrange element’s interior and face nodes carry degrees of freedom that no neighbour shares and that must be statically condensed out before assembly. The trade-off, which a practitioner should also state, is that the Serendipity higher-order terms are incomplete (it omits terms such as $\xi^{2}\eta^{2}\zeta^{2}$), so its accuracy degrades faster than the Lagrange element’s under severe geometric distortion. $$\boxed{\;20\text{ nodes vs }27\ \Rightarrow\ 7\text{ fewer nodes }(26\%)\text{, all nodes on the surface, same quadratic completeness}\;}$$