Shape functions for quadrilateral element

Element and coordinates

A four-node quadrilateral has nodes 1 to 4 anticlockwise and 8 DOFs $(u_i,v_i)$. Shape functions are derived for the parent square in natural coordinates $(\xi,\eta)$, $-1\le\xi,\eta\le1$, with nodes

$$1:(-1,-1),\quad 2:(1,-1),\quad 3:(1,1),\quad 4:(-1,1)$$

   4-----------3          eta
   |           |           ^
   |     +     |           |
   |           |           +--> xi
   1-----------2

Assumed displacement

A bilinear polynomial (4 terms, one for each node) is used for each displacement component:

$$u=\alpha_1+\alpha_2\xi+\alpha_3\eta+\alpha_4\xi\eta$$

Applying the nodal values:

$$\begin{Bmatrix}u_1\\u_2\\u_3\\u_4\end{Bmatrix}=\begin{bmatrix}1&-1&-1&1\\1&1&-1&-1\\1&1&1&1\\1&-1&1&-1\end{bmatrix}\begin{Bmatrix}\alpha_1\\\alpha_2\\\alpha_3\\\alpha_4\end{Bmatrix}$$

Inverting:

$$\begin{Bmatrix}\alpha_1\\\alpha_2\\\alpha_3\\\alpha_4\end{Bmatrix}=\frac14\begin{bmatrix}1&1&1&1\\-1&1&1&-1\\-1&-1&1&1\\1&-1&1&-1\end{bmatrix}\begin{Bmatrix}u_1\\u_2\\u_3\\u_4\end{Bmatrix}$$

Substituting into $u=[1\ \xi\ \eta\ \xi\eta]\{\alpha\}$ and collecting coefficients of the nodal values:

$$\begin{aligned} N_1&=\tfrac14(1-\xi-\eta+\xi\eta)=\tfrac14(1-\xi)(1-\eta)\\ N_2&=\tfrac14(1+\xi-\eta-\xi\eta)=\tfrac14(1+\xi)(1-\eta)\\ N_3&=\tfrac14(1+\xi+\eta+\xi\eta)=\tfrac14(1+\xi)(1+\eta)\\ N_4&=\tfrac14(1-\xi+\eta-\xi\eta)=\tfrac14(1-\xi)(1+\eta) \end{aligned}$$

General form:

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

Properties

• $N_i=1$ at node $i$ and $0$ at the other three nodes.
• $\sum N_i=1$ (rigid-body motion), and $\sum N_ix_i=x$, so a linear field is reproduced exactly.
• Along each side the variation is linear, so neighbouring elements are compatible.

Use for a general quadrilateral

For an arbitrary quadrilateral the same $N_i$ are used as isoparametric functions: $x=\sum N_ix_i$, $y=\sum N_iy_i$, $u=\sum N_iu_i$. The derivatives with respect to $x,y$ are obtained with the Jacobian:

$$[J]=\begin{bmatrix}\partial x/\partial\xi&\partial y/\partial\xi\\\partial x/\partial\eta&\partial y/\partial\eta\end{bmatrix},\qquad \begin{Bmatrix}\partial N/\partial x\\\partial N/\partial y\end{Bmatrix}=[J]^{-1}\begin{Bmatrix}\partial N/\partial\xi\\\partial N/\partial\eta\end{Bmatrix}$$

For a rectangle of sides $2a\times2b$ centred at the origin, $\xi=x/a$ and $\eta=y/b$.