---
title: "Isoparametric formulation; eight-noded shape functions"
subject: "Computational Techniques in Civil Engineering"
chapter: 4
key: "semester_8/computational_techniques_in_civil_engineering:isoparametric-eight-noded"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/isoparametric-eight-noded
site: BCT Engineering
---

# Isoparametric formulation; eight-noded shape functions

Computational Techniques in Civil Engineering · Chapter 4: Finite element method

- Asked in: 2074 Bhadra, 1+5 marks
- Video: [Isoparametric Elements in Finite Element Method Quadratic Elements in FEM](https://www.youtube.com/watch?v=PhedVyx_G8o) — Mahesh Gadwantikar
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/isoparametric-eight-noded
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4

## Question

What is isoparametric formulation? Obtain shape functions $N_i$ for the eight-noded rectangular element shown in the figure.

[Figure: eight-noded rectangle in natural coordinates $\xi$-$\eta$, origin at centre, sides of length 1 on each side of the origin; corner nodes 1 (bottom-left), 2 (bottom-right), 3 (top-right), 4 (top-left); mid-side nodes 5 (bottom), 6 (right), 7 (top), 8 (left)]

## Short answer

Isoparametric formulation uses the same shape functions for coordinates and displacements on the reference square. For the 8-node serendipity element: corners Ni=(1/4)(1+xi*xi_i)(1+eta*eta_i)(xi*xi_i+eta*eta_i-1); mid-sides N5=(1/2)(1-xi^2)(1-eta), N6=(1/2)(1+xi)(1-eta^2), N7=(1/2)(1-xi^2)(1+eta), N8=(1/2)(1-xi)(1-eta^2).

## Answer

#### Isoparametric formulation
In an isoparametric element the **same shape functions** are used to describe both the geometry (coordinates) and the displacement field:
$$
x=\sum N_i(\xi,\eta)\,x_i,\qquad y=\sum N_i(\xi,\eta)\,y_i,\qquad u=\sum N_i(\xi,\eta)\,u_i,\qquad v=\sum N_i(\xi,\eta)\,v_i
$$
The element is defined in natural coordinates $(\xi,\eta)$ on a square $[-1,1]^2$ and mapped to the real (possibly curved-sided) element. Integrals are evaluated with the Jacobian $|J|$ by Gauss quadrature: $[k]=t\int\int[B]^T[D][B]\,|J|\,d\xi\,d\eta$.

#### Shape functions: natural coordinates and nodes
Use $\xi=x/a$, $\eta=y/b$ so the element is the square $-1\le\xi,\eta\le1$ (for half-sides $a,b$; if the figure has unit half-sides, $\xi=x$, $\eta=y$). Nodes:

| Node | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| $(\xi,\eta)$ | $(-1,-1)$ | $(1,-1)$ | $(1,1)$ | $(-1,1)$ | $(0,-1)$ | $(1,0)$ | $(0,1)$ | $(-1,0)$ |

```text
   4-----7-----3
   |           |
   8     +     6      eta ^
   |           |          |
   1-----5-----2          +--> xi
```

#### Assumed displacement
Eight nodal values need an 8-term polynomial (the serendipity family: complete quadratic plus two cubic terms):
$$
u=\alpha_1+\alpha_2\xi+\alpha_3\eta+\alpha_4\xi^2+\alpha_5\xi\eta+\alpha_6\eta^2+\alpha_7\xi^2\eta+\alpha_8\xi\eta^2
$$
Writing it at the 8 nodes gives $\{u\}=[C]\{\alpha\}$ and $N=[1\ \xi\ \eta\ \dots][C]^{-1}$. The same result is obtained more quickly by building each $N_i$ as a product of straight-line factors that vanish at all the other nodes, as below.

#### Mid-side node 5 $(0,-1)$
$N_5$ must vanish on the sides $\xi=\pm1$ and $\eta=1$, and have a parabola along the bottom side: $N_5=C(1-\xi^2)(1-\eta)$. At node 5, $C(1)(2)=1\Rightarrow C=\tfrac12$. Likewise for nodes 6, 7, 8:
$$
\begin{aligned}
N_5&=\tfrac12(1-\xi^2)(1-\eta)\\
N_6&=\tfrac12(1+\xi)(1-\eta^2)\\
N_7&=\tfrac12(1-\xi^2)(1+\eta)\\
N_8&=\tfrac12(1-\xi)(1-\eta^2)
\end{aligned}
$$

#### Corner node 1 $(-1,-1)$
$N_1$ must vanish on the sides $\xi=1$ and $\eta=1$ (nodes 2, 6, 3, 7, 4) and at the mid-side nodes 5 and 8, which lie on the line $1+\xi+\eta=0$: $N_1=C(1-\xi)(1-\eta)(1+\xi+\eta)$. At node 1: $C(2)(2)(-1)=1\Rightarrow C=-\tfrac14$. Hence
$$
N_1=-\tfrac14(1-\xi)(1-\eta)(1+\xi+\eta)
$$
and for the other corners the general form is
$$
\boxed{N_i=\tfrac14(1+\xi\xi_i)(1+\eta\eta_i)(\xi\xi_i+\eta\eta_i-1)},\quad i=1,2,3,4
$$
that is
$$
\begin{aligned}
N_2&=\tfrac14(1+\xi)(1-\eta)(\xi-\eta-1)\\
N_3&=\tfrac14(1+\xi)(1+\eta)(\xi+\eta-1)\\
N_4&=\tfrac14(1-\xi)(1+\eta)(-\xi+\eta-1)
\end{aligned}
$$

#### Checks
At the centre $(0,0)$: corners give $4\times\tfrac14(-1)=-1$, mid-sides give $4\times\tfrac12=2$; the sum is $1$. In general $\sum N_i=1$ and each $N_i=1$ at its own node and $0$ at the other seven. Each side carries a quadratic variation through three nodes, so adjacent elements are compatible.

## Check yourself

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

1. What makes an element isoparametric?
   A. Geometry uses linear functions while displacement uses quadratic ones
   B. The same shape functions describe both the geometry and the displacement field
   C. Only the displacement is mapped to natural coordinates
   D. All sides must be straight and equal
   Answer: B. In isoparametric elements x = sum N_i x_i and u = sum N_i u_i use the same N_i. (Key point: Definition of isoparametric)

2. Why is the Jacobian |J| needed in [k] = t double-integral [B]^T [D] [B] |J| dxi deta?
   A. To convert integration from natural coordinates to the real (possibly distorted) element, usually done by Gauss quadrature
   B. To apply boundary conditions
   C. To make D symmetric
   D. To normalise shape functions to unity
   Answer: A. The element is mapped from the square [-1,1]^2 to the real shape, so area and derivatives change through |J|. (Key point: Role of the Jacobian)

3. What is the shape function of mid-side node 5 at (0,-1)?
   A. N5 = 1/4(1+xi)(1-eta)
   B. N5 = 1/2(1 - xi^2)(1 - eta)
   C. N5 = 1/2(1 + xi)(1 - eta^2)
   D. N5 = (1 - xi^2)(1 - eta^2)
   Answer: B. It must vanish on xi = +-1 and eta = 1, be parabolic along the bottom side, and equal 1 at node 5: C(1)(2) = 1 gives C = 1/2. (Key point: Mid-side shape function)

4. What is N1 for the corner node (-1,-1)?
   A. 1/4(1-xi)(1-eta)
   B. -1/4(1-xi)(1-eta)(1+xi+eta)
   C. 1/4(1+xi)(1+eta)(xi+eta-1)
   D. 1/2(1-xi)(1-eta)(1+xi+eta)
   Answer: B. The factor (1+xi+eta) is zero at nodes 5 and 8, and C(2)(2)(-1) = 1 gives C = -1/4. (Key point: Corner shape function)

## Related in Computational Techniques in Civil Engineering

- [Shape functions, eight-noded rectangular element](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Ashape-function-eight-noded-rectangle) — Computational Techniques in Civil Engineering, chapter 4, asked 2×
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## Related in other subjects

- [Shape functions of eight-node serendipity element](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Aeight-node-serendipity-shape-functions) — Finite Element Method, chapter 7
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