---
title: "Area coordinates and temperature at a point"
subject: "Finite Element Method"
chapter: 5
key: "semester_8/finite_element_method:cst-area-coordinates-interpolation"
source: https://www.bctengineer.com/q/semester_8/finite_element_method/cst-area-coordinates-interpolation
site: BCT Engineering
---

# Area coordinates and temperature at a point

Finite Element Method · Chapter 5: Interpolation Functions

- Asked in: year not given, 6 marks
- Video: [CST Element | Shape function numerical](https://www.youtube.com/watch?v=BjRx3qEjtEE) — Nivrutti Patil
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/finite_element_method/cst-area-coordinates-interpolation
- Whole chapter: https://www.bctengineer.com/md/semester_8/finite_element_method/chapter-5

## Question

A three-node triangular element has nodes 1 (10, 10), 2 (60, 20) and 3 (30, 50), coordinates in mm. (a) Find its area. (b) Find the shape functions at the point P (30, 25) and check that they sum to 1. (c) If the nodal temperatures are 100 °C, 60 °C and 80 °C, find the temperature at P.

## Short answer

Area A=900 mm^2. At P(30,25), shape functions are N1=0.4167, N2=0.2778, N3=0.3056, summing to 1. Temperature at P is 0.4167(100)+0.2778(60)+0.3056(80)=82.78 °C.

## Answer

#### (a) Area

$$
2A = x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 10(20 - 50) + 60(50 - 10) + 30(10 - 20)
$$
$$
2A = -300 + 2400 - 300 = 1800 \Rightarrow A = 900\ \text{mm}^2
$$

The positive value shows the nodes are numbered anticlockwise.

#### (b) Shape functions

Constants:

| $i$ | $a_i$ | $b_i$ | $c_i$ |
|---|---|---|---|
| 1 | $x_2y_3 - x_3y_2 = 3000 - 600 = 2400$ | $y_2 - y_3 = -30$ | $x_3 - x_2 = -30$ |
| 2 | $x_3y_1 - x_1y_3 = 300 - 500 = -200$ | $y_3 - y_1 = 40$ | $x_1 - x_3 = -20$ |
| 3 | $x_1y_2 - x_2y_1 = 200 - 600 = -400$ | $y_1 - y_2 = -10$ | $x_2 - x_1 = 50$ |

At P (30, 25), $N_i = \dfrac{a_i + b_i(30) + c_i(25)}{1800}$:
$$
\begin{aligned}
N_1 &= \frac{2400 - 900 - 750}{1800} = \frac{750}{1800} = 0.4167 \\
N_2 &= \frac{-200 + 1200 - 500}{1800} = \frac{500}{1800} = 0.2778 \\
N_3 &= \frac{-400 - 300 + 1250}{1800} = \frac{550}{1800} = 0.3056
\end{aligned}
$$

Sum $= 0.4167 + 0.2778 + 0.3056 = 1.0000$, as required.

Check on location: $x = \sum N_ix_i = 0.4167(10) + 0.2778(60) + 0.3056(30) = 30.0$ and $y = 0.4167(10) + 0.2778(20) + 0.3056(50) = 25.0$.

#### (c) Temperature at P

$$
T_P = 0.4167(100) + 0.2778(60) + 0.3056(80) = 41.67 + 16.67 + 24.44 = 82.78\ ^\circ\text{C}
$$

**Answer:** $A = 900\ \text{mm}^2$; $N = (0.417,\ 0.278,\ 0.306)$; $T_P = 82.8\ ^\circ$C.

## Check yourself

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

1. For nodes 1 (10,10), 2 (60,20), 3 (30,50), what is the area of the triangle?
   A. 900 mm^2
   B. 1800 mm^2
   C. 450 mm^2
   D. 2400 mm^2
   Answer: A. 2A = 10(20-50) + 60(50-10) + 30(10-20) = 1800, so A = 900 mm^2. The positive value means anticlockwise numbering. (Key point: Area formula)

2. What are the shape functions N1, N2, N3 at P (30, 25)?
   A. 0.333, 0.333, 0.333
   B. 0.500, 0.250, 0.250
   C. 0.417, 0.278, 0.306
   D. 0.306, 0.278, 0.417
   Answer: C. Ni = (ai + bi x + ci y)/1800 gives 750/1800, 500/1800 and 550/1800, which sum to 1. (Key point: Shape functions at P)

3. How is the temperature at P found and what is it?
   A. Simple average of the three temperatures = 80 °C
   B. Sum of nodal temperatures = 240 °C
   C. T = T1 N1 only = 41.7 °C
   D. T = N1T1 + N2T2 + N3T3 = 82.8 °C
   Answer: D. The shape functions weight the nodal temperatures: 0.4167(100) + 0.2778(60) + 0.3056(80) = 82.78 °C. (Key point: Interpolation)

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