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Base Elements' Geometry

Carlos Adir edited this page Dec 14, 2020 · 4 revisions

Elements in 2D

For 2D, we have in total 4 elements:

T3 Element T6 Element
Q4 Element Q8 Element

And so, we have that for

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And we can have all the interpolation functions, using the point as the form (s, t).

Element T3

i Pi Ni dNi/ds dNi/dt
1 (0, 0) 1-s-t -1 -1
2 (1, 0) s 1 0
3 (0, 1) t 0 1

Element T6

i Pi Ni dNi/ds dNi/dt
1 (0, 0) (s+t-1)(2s+2t-1) 4(s+t)-3 4(s+t)-3
2 (0.5, 0) -4s(s+t-1) 4(1-2s-t) -4s
3 (1, 0) s(2s - 1) 4s - 1 0
4 (0.5, 0.5) 4st 4t 4s
5 (0, 1) t(2t - 1) 0 4t - 1
6 (0, 0.5) -4t(s+t-1) -4t 4(1-s-2t)

Element Q4

i Pi Ni dNi/ds dNi/dt
1 (-1, -1) (s-1)(t-1)/4 (t-1)/4 (s-1)
2 (1, -1) (s+1)(1-t)/4 (1-t)/4 -(s+1)
3 (1, 1) (s+1)(t+1)/4 (t+1)/4 (s+1)
4 (-1, 1) (1-s)(t+1)/4 -(t+1)/4 (1-s)

Element Q8

i Pi Ni dNi/ds dNi/dt
1 (-1, -1) (1-s)(t-1)(s+t+1)/4 (2s+t)(1-t)/4 -(s-1)(s+2t)/4
2 (0, -1) (s^2-1)(t-1)/2 s(t-1) (s^2-1)/2
3 (1, -1) (s+1)(1-t)(s-t-1)/4 (t-2s)(t-1)/4 (-s+2t)(s+1)/4
4 (-1, 0) (s+1)(1-t^2)/2 (1-t^2)/2 -t(s+1)
5 (1, 0) (s+1)(t+1)(s+t-1)/4 (2s+t)(t+1)/4 (s+1)(s+2t)/4
6 (-1, 1) (1-s^2)(t+1)/2 -s(t+1) (1-s^2)/2
7 (0, 1) (s-1)(t+1)(s-t+1)/4 (2s-t)(t+1)/4 (s-1)(s-2t)/4
8 (1, 1) (s-1)(t^2-1)/2 (t^2-1)/2 t(s-1)
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