---
title: "Shape functions, eight-noded rectangular element"
subject: "Computational Techniques in Civil Engineering"
chapter: 4
key: "semester_8/computational_techniques_in_civil_engineering:shape-function-eight-noded-rectangle"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/shape-function-eight-noded-rectangle
site: BCT Engineering
---

# Shape functions, eight-noded rectangular element

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

- Asked in: 2078 Kartik, 6 marks; 2075 Bhadra, 8 marks (asked 2 times)
- Video: [Serendipty Element (8-Noded Rectangular element)](https://www.youtube.com/watch?v=R8FT20M-doo) — Dr. Anwesh Virkunwar
- 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/shape-function-eight-noded-rectangle
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4

## Question

Derive the shape functions for the eight-noded rectangular element given in the figure.

[Figure: eight-noded 2-D rectangular element in natural coordinates $\xi$, $\eta$ with origin at the centre and half-sides of 1 (or $a$ and $b$); corner nodes 1, 2, 3, 4 taken anticlockwise from the bottom-left corner, and mid-side nodes 5 (bottom), 6 (right), 7 (top), 8 (left)]

## Short answer

For an eight-noded rectangle in natural coordinates (xi,eta), mid-side shape functions are N5=0.5(1-xi^2)(1-eta), N6=0.5(1+xi)(1-eta^2), N7=0.5(1-xi^2)(1+eta), N8=0.5(1-xi)(1-eta^2). Corner shape functions are Ni=0.25(1+xi*xi_i)(1+eta*eta_i)(xi*xi_i+eta*eta_i-1).

## Answer

#### 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. Which is the shape function of mid-side node 5 at (0,-1)?
   A. (1/4)(1-xi)(1-eta)
   B. (1/2)(1-xi^2)(1-eta)
   C. (1/2)(1+xi)(1-eta^2)
   D. -(1/4)(1-xi)(1-eta)(1+xi+eta)
   Answer: B. N5 is a parabola in xi along the bottom side, vanishing at xi = +-1 and at eta = 1; C = 1/2 gives N5 = 1 at node 5. (Key point: Mid-side shape function)

2. In N1 = C(1-xi)(1-eta)(1+xi+eta), what is the purpose of the factor (1+xi+eta)?
   A. It makes N1 equal 1 at node 1
   B. It makes N1 vanish at mid-side nodes 5 and 8, which lie on the line 1+xi+eta = 0
   C. It makes N1 vanish at node 3
   D. It ensures the sum of shape functions is zero
   Answer: B. The factors (1-xi) and (1-eta) kill the other nodes on xi = 1 and eta = 1; nodes 5 and 8 lie on 1+xi+eta = 0. (Key point: Corner function factor)

3. What is C for N1 and why?
   A. C = 1/4, since N1(-1,-1) = 1
   B. C = -1/4, since at node 1 the factors give 2 x 2 x (-1) = -4
   C. C = 1/2, as for mid-side nodes
   D. C = -1, since the factor 1+xi+eta is -1
   Answer: B. N1(-1,-1) = C(2)(2)(-1) = -4C = 1, so C = -1/4. (Key point: Constant for corner node)

4. At the centre (0,0), what do the corner and mid-side functions contribute and what is the total?
   A. Corners +1, mid-sides 0, total 1
   B. Corners -1, mid-sides +2, total 1
   C. Corners 0, mid-sides 1, total 1
   D. Corners -2, mid-sides +3, total 1
   Answer: B. Each corner gives -1/4 (four give -1) and each mid-side gives 1/2 (four give 2); the sum is 1. (Key point: Partition of unity check)

## Related in Computational Techniques in Civil Engineering

- [Isoparametric formulation; eight-noded shape functions](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Aisoparametric-eight-noded) — Computational Techniques in Civil Engineering, chapter 4
- [Shape functions, six-node triangular element](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Asix-node-triangle-shape-function) — Computational Techniques in Civil Engineering, chapter 4
- [Shape functions, ten-node triangular element](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Ashape-function-ten-node-triangle) — Computational Techniques in Civil Engineering, chapter 4, asked 2×

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