Isoparametric formulation; eight-noded shape functions

Isoparametric formulation

In an isoparametric element the same shape functions are used to describe both the geometry (coordinates) and the displacement field:

$$x=\sum N_i(\xi,\eta)\,x_i,\qquad y=\sum N_i(\xi,\eta)\,y_i,\qquad u=\sum N_i(\xi,\eta)\,u_i,\qquad v=\sum N_i(\xi,\eta)\,v_i$$

The element is defined in natural coordinates $(\xi,\eta)$ on a square $[-1,1]^2$ and mapped to the real (possibly curved-sided) element. Integrals are evaluated with the Jacobian $|J|$ by Gauss quadrature: $[k]=t\int\int[B]^T[D][B]\,|J|\,d\xi\,d\eta$.

Shape functions: natural coordinates and nodes

Use $\xi=x/a$, $\eta=y/b$ so the element is the square $-1\le\xi,\eta\le1$ (for half-sides $a,b$; if the figure has unit half-sides, $\xi=x$, $\eta=y$). Nodes:

   4-----7-----3
   |           |
   8     +     6      eta ^
   |           |          |
   1-----5-----2          +--> xi

Assumed displacement

Eight nodal values need an 8-term polynomial (the serendipity family: complete quadratic plus two cubic terms):

$$u=\alpha_1+\alpha_2\xi+\alpha_3\eta+\alpha_4\xi^2+\alpha_5\xi\eta+\alpha_6\eta^2+\alpha_7\xi^2\eta+\alpha_8\xi\eta^2$$

Writing it at the 8 nodes gives $\{u\}=[C]\{\alpha\}$ and $N=[1\ \xi\ \eta\ \dots][C]^{-1}$. The same result is obtained more quickly by building each $N_i$ as a product of straight-line factors that vanish at all the other nodes, as below.

Mid-side node 5 $(0,-1)$

$N_5$ must vanish on the sides $\xi=\pm1$ and $\eta=1$, and have a parabola along the bottom side: $N_5=C(1-\xi^2)(1-\eta)$. At node 5, $C(1)(2)=1\Rightarrow C=\tfrac12$. Likewise for nodes 6, 7, 8:

$$\begin{aligned} N_5&=\tfrac12(1-\xi^2)(1-\eta)\\ N_6&=\tfrac12(1+\xi)(1-\eta^2)\\ N_7&=\tfrac12(1-\xi^2)(1+\eta)\\ N_8&=\tfrac12(1-\xi)(1-\eta^2) \end{aligned}$$

Corner node 1 $(-1,-1)$

$N_1$ must vanish on the sides $\xi=1$ and $\eta=1$ (nodes 2, 6, 3, 7, 4) and at the mid-side nodes 5 and 8, which lie on the line $1+\xi+\eta=0$: $N_1=C(1-\xi)(1-\eta)(1+\xi+\eta)$. At node 1: $C(2)(2)(-1)=1\Rightarrow C=-\tfrac14$. Hence

$$N_1=-\tfrac14(1-\xi)(1-\eta)(1+\xi+\eta)$$

and for the other corners the general form is

$$\boxed{N_i=\tfrac14(1+\xi\xi_i)(1+\eta\eta_i)(\xi\xi_i+\eta\eta_i-1)},\quad i=1,2,3,4$$

that is

$$\begin{aligned} N_2&=\tfrac14(1+\xi)(1-\eta)(\xi-\eta-1)\\ N_3&=\tfrac14(1+\xi)(1+\eta)(\xi+\eta-1)\\ N_4&=\tfrac14(1-\xi)(1+\eta)(-\xi+\eta-1) \end{aligned}$$

Checks

At the centre $(0,0)$: corners give $4\times\tfrac14(-1)=-1$, mid-sides give $4\times\tfrac12=2$; the sum is $1$. In general $\sum N_i=1$ and each $N_i=1$ at its own node and $0$ at the other seven. Each side carries a quadratic variation through three nodes, so adjacent elements are compatible.