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Area coordinates and temperature at a point

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.

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.

(a) Area

2A=x1(y2−y3)+x2(y3−y1)+x3(y1−y2)=10(20−50)+60(50−10)+30(10−20)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⇒A=900 mm22A = -300 + 2400 - 300 = 1800 \Rightarrow A = 900\ \text{mm}^2

The positive value shows the nodes are numbered anticlockwise.

(b) Shape functions

Constants:

iiaia_ibib_icic_i
1x2y3−x3y2=3000−600=2400x_2y_3 - x_3y_2 = 3000 - 600 = 2400y2−y3=−30y_2 - y_3 = -30x3−x2=−30x_3 - x_2 = -30
2x3y1−x1y3=300−500=−200x_3y_1 - x_1y_3 = 300 - 500 = -200y3−y1=40y_3 - y_1 = 40x1−x3=−20x_1 - x_3 = -20
3x1y2−x2y1=200−600=−400x_1y_2 - x_2y_1 = 200 - 600 = -400y1−y2=−10y_1 - y_2 = -10x2−x1=50x_2 - x_1 = 50

At P (30, 25), Ni=ai+bi(30)+ci(25)1800N_i = \dfrac{a_i + b_i(30) + c_i(25)}{1800}:

N1=2400−900−7501800=7501800=0.4167N2=−200+1200−5001800=5001800=0.2778N3=−400−300+12501800=5501800=0.3056\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= 0.4167 + 0.2778 + 0.3056 = 1.0000, as required.

Check on location: x=∑Nixi=0.4167(10)+0.2778(60)+0.3056(30)=30.0x = \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)=25.0y = 0.4167(10) + 0.2778(20) + 0.3056(50) = 25.0.

(c) Temperature at P

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

Answer: A=900 mm2A = 900\ \text{mm}^2; N=(0.417, 0.278, 0.306)N = (0.417,\ 0.278,\ 0.306); TP=82.8 ∘T_P = 82.8\ ^\circC.

✓ Answer checked by Claude Opus 5.5 (high effort)

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