Shape functions of eight-node serendipity element

Each shape function must equal 1 at its own node and 0 at the other seven nodes, and the sum must be 1.

Mid-side node 5 (0, -1)

$N_5$ must vanish at the other two nodes of the edge $\eta = -1$ ($\xi = \pm1$) and on the edge $\eta = +1$, and at nodes 6 and 8 ($\xi = \pm1$). Use a quadratic in $\xi$ times a linear function in $\eta$:

$$N_5 = c\,(1 - \xi^2)(1 - \eta)$$

At node 5, $N_5 = c(1)(2) = 1$, so $c = \dfrac{1}{2}$:

$$N_5 = \tfrac{1}{2}(1 - \xi^2)(1 - \eta)$$

Corner node 1 (-1, -1)

Start with the bilinear function $\tfrac{1}{4}(1 - \xi)(1 - \eta)$. It is 1 at node 1 and 0 at nodes 2, 3, 4, but it is non-zero at mid-side nodes 5 and 8 (value $\tfrac{1}{2}$ each). Subtract $\tfrac{1}{2}N_5$ and $\tfrac{1}{2}N_8$:

$$N_1 = \tfrac{1}{4}(1 - \xi)(1 - \eta) - \tfrac{1}{2}N_5 - \tfrac{1}{2}N_8$$

With $N_8 = \tfrac{1}{2}(1 - \xi)(1 - \eta^2)$, simplifying gives

$$N_1 = \tfrac{1}{4}(1 - \xi)(1 - \eta)(-\xi - \eta - 1)$$

Check: at node 1 ($\xi = \eta = -1$): $\tfrac{1}{4}(2)(2)(1 + 1 - 1) = 1$. At node 2 ($\xi = 1$): factor $(1 - \xi) = 0$. At node 5 ($\xi = 0$, $\eta = -1$): factor $(1 - \eta)(-\xi - \eta - 1) = 2(0) = 0$. At node 8 similarly zero.

General formulae ($\xi_i$, $\eta_i$ = nodal values)

• Corner nodes ($i = 1$ to 4):

$$N_i = \tfrac{1}{4}(1 + \xi\xi_i)(1 + \eta\eta_i)(\xi\xi_i + \eta\eta_i - 1)$$

• Mid-side nodes with $\xi_i = 0$ (nodes 5, 7):

$$N_i = \tfrac{1}{2}(1 - \xi^2)(1 + \eta\eta_i)$$

• Mid-side nodes with $\eta_i = 0$ (nodes 6, 8):

$$N_i = \tfrac{1}{2}(1 + \xi\xi_i)(1 - \eta^2)$$

The edge shape functions are quadratic along each edge, so adjacent elements share three nodes per edge and compatibility is maintained.