Jacobian of a four-node isoparametric element

Parametric mapping and isoparametric element

An arbitrary quadrilateral in the $x$-$y$ plane is mapped from a square parent element with natural coordinates $-1 \le \xi, \eta \le 1$. In the isoparametric element, the same shape functions are used for the geometry and for the displacement:

$$x = \sum N_ix_i,\quad y = \sum N_iy_i,\quad u = \sum N_iu_i,\quad v = \sum N_iv_i$$

Using the same order for geometry and displacement is the isoparametric concept (if the geometry uses lower-order functions than the displacement, the element is subparametric; if higher, superparametric). Derivatives need the chain rule, so the Jacobian matrix is required:

$$\begin{Bmatrix} \partial N/\partial\xi \\ \partial N/\partial\eta \end{Bmatrix} = [J]\begin{Bmatrix} \partial N/\partial x \\ \partial N/\partial y \end{Bmatrix},\qquad [J] = \begin{bmatrix} \partial x/\partial\xi & \partial y/\partial\xi \\ \partial x/\partial\eta & \partial y/\partial\eta \end{bmatrix},\quad dA = \det[J]\,d\xi\,d\eta$$

Shape functions at $\xi = \eta = 0.5$

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

$$N_1 = \tfrac{1}{4}(0.5)(0.5) = 0.0625,\ N_2 = 0.1875,\ N_3 = 0.5625,\ N_4 = 0.1875$$

(a) Global coordinates

$$\begin{aligned} x &= 0.0625(0) + 0.1875(4) + 0.5625(5) + 0.1875(1) = 3.75\ \text{mm} \\ y &= 0 + 0 + 0.5625(3) + 0.1875(3) = 2.25\ \text{mm} \end{aligned}$$

(b) Jacobian

Derivatives with respect to the natural coordinates:

$$\frac{\partial N_i}{\partial\xi} = \tfrac{1}{4}\xi_i(1 + \eta\eta_i),\qquad \frac{\partial N_i}{\partial\eta} = \tfrac{1}{4}\eta_i(1 + \xi\xi_i)$$ $$\begin{aligned} J_{11} &= \sum \frac{\partial N_i}{\partial\xi}x_i = 0.125(4) + 0.375(5) - 0.375(1) = 2.0 \\ J_{12} &= \sum \frac{\partial N_i}{\partial\xi}y_i = 0.375(3) - 0.375(3) = 0 \\ J_{21} &= \sum \frac{\partial N_i}{\partial\eta}x_i = -0.375(4) + 0.375(5) + 0.125(1) = 0.5 \\ J_{22} &= \sum \frac{\partial N_i}{\partial\eta}y_i = 0.375(3) + 0.125(3) = 1.5 \end{aligned}$$ $$[J] = \begin{bmatrix} 2.0 & 0 \\ 0.5 & 1.5 \end{bmatrix},\qquad \det[J] = 3.0$$

(The element is a parallelogram of area $4 \times 3 = 12$ mm$^2$, which equals $4\det J$ because the parent square has area 4. For a parallelogram, $J$ is constant.)

(c) Global derivatives

$$[J]^{-1} = \frac{1}{3}\begin{bmatrix} 1.5 & 0 \\ -0.5 & 2.0 \end{bmatrix}$$

Each column sums to zero, as it must because $\sum N_i = 1$.

Answer: $(x, y) = (3.75,\ 2.25)$ mm; $[J] = \begin{bmatrix} 2 & 0 \\ 0.5 & 1.5 \end{bmatrix}$, $\det J = 3.0$; derivatives as tabulated.