Shape functions (defined in natural coordinates)
$N_{1} = \frac{1}{4}(1 – \xi)(1 – \eta)$
$N_{2} = \frac{1}{4}(1 + \xi)(1 – \eta)$
$N_{3} = \frac{1}{4}(1 + \xi)(1 + \eta)$
$N_{4} = \frac{1}{4}(1 – \xi)(1 + \eta)$
The above shape functions are in fact the Lagrangian interpolation
Note that we define the element as a square of size $[-1, 1] \times [-1, 1]$ in the natural space, thus the natural coordinates of nodes are given by
Node 1: $\xi_{1} = -1$, $\eta_{1}= -1$
Node 2: $\xi_{2} = 1$, $\eta_{2} = -1$
Node 3: $\xi_{3} = 1$, $\eta_{3} = 1$
Node 4: $\xi_{4} = -1$, $\eta_{3} = 1$
Substitute the nodal natural coordinates into the shape functions, we have the following so-called “Kronecker delta” property
$N_{i} (\xi_{i}, \eta_{i}) = 1$, and $N_{i} (\xi_{j}, \eta_{j}) = 1$, with $j \neq i$
Approximation of coordinates of a point within the element
$x(\xi, \eta) = N_{1} (\xi, \eta) x_{1} + N_{2} (\xi, \eta) x_{2} + N_{3} (\xi, \eta) x_{3} + N_{4} (\xi, \eta) x_{4}$
$y(\xi, \eta) = N_{1} (\xi, \eta) y_{1} + N_{2} (\xi, \eta) y_{2} + N_{3} (\xi, \eta) y_{3} + N_{4} (\xi, \eta) y_{4}$
where $x_{1}$, $x_{2}$, $x_{3}$, $x_{4}$ (and $y_{1}$, $y_{2}$, $y_{3}$, $y_{4}$) are the coordinates of the four nodes of the element.
Approximation of displacement (for the shake of brevity, only the horizontal component $u$ is written below. The approximation of vertical component of displacement is similar)
$u(\xi, \eta) = N_{1} (\xi, \eta) \hat{u}_{1} + N_{2} (\xi, \eta) \hat{u}_{2} + N_{3} (\xi, \eta) \hat{u}_{3} + N_{4} (\xi, \eta) \hat{u}_{4}$
where $\hat{u}_{1}$, $\hat{u}_{2}$, $\hat{u}_{3}$, $\hat{u}_{4}$ (and $\hat{u}_{1}$, $\hat{u}_{2}$, $\hat{u}_{3}$, $\hat{u}_{4}$) are the nodal parameters for displacement. Thanks to the Kronecker delta property, these values are equal to the nodal values of displacement.
For numerical integration, see post “2D Gaussian quadrature“
Jacobian matrix (necessary when we need to transform the derivative with respect to natural coordinates into the derivative with respect to physical coordinates
Derivative of shape function $N_{1]$ with respect to physical coordinates
$\frac{\partial N_{1} } {\partial x} = \frac{\partial N_{1} }{ \partial \xi } \frac{\partial \xi}{\partial x} + \frac{\partial N_{1} } {\partial \eta } \frac{\partial \eta} {\partial x}$
$\frac{\partial N_{1} } {\partial y} = \frac{\partial N_{1} }{ \partial \xi } \frac{\partial \xi}{\partial y} + \frac{\partial N_{1} } {\partial \eta } \frac{\partial \eta} {\partial y}$
Rewrite the formulation in matrix form
$ \begin{bmatrix}
\frac{\partial N_{1}}{\partial \xi} \\
\frac{\partial N_{1}}{\partial \eta}
\end{bmatrix} = \begin{bmatrix}
\frac{\partial x} {\partial \xi} & \frac{\partial y} {\partial \xi} \\
\frac{\partial x} {\partial \eta} & \frac{\partial y} {\partial \eta}
\end{bmatrix} \begin{bmatrix}
\frac{\partial N_{1}} {\partial x} \\
\frac{\partial N_{1}} {\partial y}
\end{bmatrix}
$
Note that the above relation holds for all the four shape functions. The Jacobian matrix is given by
$ \mathbf{J} = \begin{bmatrix}
\frac{\partial x} {\partial \xi} & \frac{\partial y} {\partial \xi} \\
\frac{\partial x} {\partial \eta} & \frac{\partial y} {\partial \eta}
\end{bmatrix}$
The derivative of shape functions with respect to natural coordinates can be calculated without difficulty. And so does the derivative of x (and y) with respect to $\xi$ and $\eta$. Hence, the derivative of shape functions with respect to physical coordinates can be calculated.