---
title: "Jacobian of a four-node isoparametric element"
subject: "Finite Element Method"
chapter: 7
key: "semester_8/finite_element_method:isoparametric-jacobian-quadrilateral"
source: https://www.bctengineer.com/q/semester_8/finite_element_method/isoparametric-jacobian-quadrilateral
site: BCT Engineering
---

# Jacobian of a four-node isoparametric element

Finite Element Method · Chapter 7: Higher order Elements

- Asked in: year not given, 8 marks
- Video: [Finite Element Method- Unit 4 (Lecture 5) Isoparametric Formulation, Jacobian Matrix](https://www.youtube.com/watch?v=6JKG2EvvlfA) — Dr. A. S. Sayyad
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/finite_element_method/isoparametric-jacobian-quadrilateral
- Whole chapter: https://www.bctengineer.com/md/semester_8/finite_element_method/chapter-7

## Question

Explain parametric mapping and the isoparametric concept. A four-node quadrilateral element has nodes 1 (0, 0), 2 (4, 0), 3 (5, 3) and 4 (1, 3), coordinates in mm, numbered anticlockwise. Using the bilinear shape functions in $(\xi, \eta)$, find at the point $\xi = \eta = 0.5$: (a) the global coordinates, (b) the Jacobian matrix and its determinant, (c) the derivatives $\partial N_i/\partial x$ and $\partial N_i/\partial y$.

## Short answer

Global coordinates: (3.75, 2.25) mm. Jacobian = [2 0; 0.5 1.5], det J = 3.0. Derivatives ∂N/∂x: -0.0625, 0.0625, 0.1875, -0.1875; ∂N/∂y: -0.0625, -0.2708, 0.1875, 0.1458 (nodes 1–4).

## Answer

#### 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:
$$
\frac{\partial N_i}{\partial\xi} = \tfrac{1}{4}\xi_i(1 + \eta\eta_i),\qquad \frac{\partial N_i}{\partial\eta} = \tfrac{1}{4}\eta_i(1 + \xi\xi_i)
$$

| Node | $\partial N/\partial\xi$ | $\partial N/\partial\eta$ |
|---|---|---|
| 1 | $-0.125$ | $-0.125$ |
| 2 | $0.125$ | $-0.375$ |
| 3 | $0.375$ | $0.375$ |
| 4 | $-0.375$ | $0.125$ |

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

(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

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

| Node | $\partial N/\partial x$ | $\partial N/\partial y$ |
|---|---|---|
| 1 | $-0.0625$ | $-0.0625$ |
| 2 | $0.0625$ | $-0.2708$ |
| 3 | $0.1875$ | $0.1875$ |
| 4 | $-0.1875$ | $0.1458$ |

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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the isoparametric concept?
   A. The same shape functions are used for geometry and for displacement
   B. Geometry is linear while displacement is quadratic
   C. Only the geometry is interpolated
   D. Displacement uses higher order than geometry
   Answer: A. x = sum Ni xi and u = sum Ni ui, with equal order for both, mapped from a parent square. (Key point: Isoparametric concept)

2. At xi = eta = 0.5, what are the shape functions and the global point?
   A. N = 0.0625 each; (x, y) = (0.6, 0.4) mm
   B. N = 0.25 each; (x, y) = (2.5, 1.5) mm
   C. N = 0.5625, 0.1875, 0.0625, 0.1875; (x, y) = (1.5, 0.75) mm
   D. N = 0.0625, 0.1875, 0.5625, 0.1875; (x, y) = (3.75, 2.25) mm
   Answer: D. Ni = (1/4)(1 + xi xi_i)(1 + eta eta_i), and then x = 3.75 and y = 2.25 mm by summing Ni xi and Ni yi. (Key point: Global coordinates)

3. What are [J] and det J for this element?
   A. [J] = [[4, 0],[1, 3]], det J = 12
   B. [J] = [[1.5, 0],[0.5, 2]], det J = 3
   C. [J] = [[2, 0],[0.5, 1.5]], det J = 3.0
   D. [J] = [[2, 0.5],[0, 1.5]], det J = 3.0
   Answer: C. J11 = sum dNi/dxi xi = 2, J12 = 0, J21 = 0.5, J22 = 1.5; det = 3. Area 12 = 4 x det J since the parent area is 4. (Key point: Jacobian)

4. How are global derivatives dN/dx and dN/dy obtained?
   A. Multiply [J] by the natural derivatives
   B. Use the natural derivatives directly
   C. Multiply [J]^-1 by the natural derivatives; each column sums to zero
   D. Divide natural derivatives by det J only
   Answer: C. The chain rule gives {dN/dxi, dN/deta} = [J]{dN/dx, dN/dy}, so the inverse is needed; derivatives sum to zero because the sum of Ni is 1. (Key point: Global derivatives)

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