Shape functions of four-node rectangular element

   4 (0,b)  o-----------o  3 (a,b)
            |           |
            |           |  b
            |           |
   1 (0,0)  o-----------o  2 (a,0)
                  a

Displacement function

Four nodes per scalar field need four terms. Taking terms symmetrically from Pascal's triangle gives the bilinear polynomial

$$u(x, y) = \alpha_1 + \alpha_2x + \alpha_3y + \alpha_4xy$$

(The $xy$ term is chosen; $x^2$ or $y^2$ alone would make the function unsymmetrical.)

Apply the four nodal conditions $u(0,0) = u_1$, $u(a,0) = u_2$, $u(a,b) = u_3$, $u(0,b) = u_4$:

$$\begin{aligned} \alpha_1 &= u_1 \\ \alpha_2 &= \frac{u_2 - u_1}{a} \\ \alpha_3 &= \frac{u_4 - u_1}{b} \\ \alpha_4 &= \frac{u_1 - u_2 + u_3 - u_4}{ab} \end{aligned}$$

Substituting and collecting the coefficients of $u_i$ gives $u = \sum N_iu_i$ with

$$\begin{aligned} N_1 &= \left(1 - \frac{x}{a}\right)\left(1 - \frac{y}{b}\right), & N_2 &= \frac{x}{a}\left(1 - \frac{y}{b}\right) \\ N_3 &= \frac{xy}{ab}, & N_4 &= \left(1 - \frac{x}{a}\right)\frac{y}{b} \end{aligned}$$

The same functions interpolate $v$, i.e. $v = \beta_1 + \beta_2x + \beta_3y + \beta_4xy = \sum N_iv_i$.

Properties verified

• At node 3 ($x = a$, $y = b$): $N_3 = 1$, $N_1 = N_2 = N_4 = 0$. The same holds at the other nodes.
• Sum: $N_1 + N_2 + N_3 + N_4 = \left(1 - \frac{x}{a}\right)\left(1 - \frac{y}{b}\right) + \frac{x}{a}\left(1 - \frac{y}{b}\right) + \frac{xy}{ab} + \left(1 - \frac{x}{a}\right)\frac{y}{b} = 1$.
• Along an edge, e.g. $y = 0$, only $N_1$ and $N_2$ are non-zero and they are linear in $x$, so the displacement is continuous between neighbouring elements.

Strain variation

$$\varepsilon_x = \frac{\partial u}{\partial x} = \alpha_2 + \alpha_4y,\qquad \varepsilon_y = \frac{\partial v}{\partial y} = \beta_3 + \beta_4x$$

$\varepsilon_x$ varies linearly with $y$ and $\varepsilon_y$ varies linearly with $x$, so the element represents bending better than the CST. A drawback is that it can only be used for rectangles with sides parallel to the axes; the (isoparametric) quadrilateral element removes this limitation.