Shape Function of 3D Simple Element 2-Finite Element Method-Assignment Solution, Exercises of Mathematical Methods for Numerical Analysis and Optimization

This assignment solution was submitted to Amar Sharma for Finite Element Method course at Aligarh Muslim University. It includes: Shape, Function, Simple, Element, Equations, Putting, Values, Determine, Three-dimensional

Typology: Exercises

2011/2012

Uploaded on 07/08/2012

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Problem # 3.16:
Determine the shape function of a 3-D simple element:
ɸ = a1+a2x+a3y+a4z
ɳT= { 1 x y z }
[N] = [Ni Nj Nk Nl]
[NT] = { Ni Nj Nk Nl}
Ni
[N] = Nj
Nk
Nl
Ni
ʃNNtds = ʃʃʃvNNTdV = ʃʃʃv Nj [Ni Nj Nk Nl] dV
Nk
Nl
ʃʃʃv(Ni
2 Nj
2 Nk
2 Nl
2)dV
As Ni=L1
Nj = L2
Nk = L3
Nl = L4
ʃʃʃv L1
2 L2
2 L3
2 L4
2dV
According to Equation 3.81
ʃʃʃv L1
2 L2
2 L3
2 L4
2dV
= α!β!ϒ!δ!*6V/(α+β+ϒ+δ+3)!
Putting the above values we have 2!2!2!2!*6V/(2+2+2+2+3)!
Ans = 96V/11!
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Problem # 3.16:

Determine the shape function of a 3-D simple element: ɸ = a 1 +a 2 x+a 3 y+a 4 z ɳT= { 1 x y z } [N] = [Ni Nj Nk Nl] [NT] = { Ni Nj Nk Nl} Ni [N] = Nj Nk Nl Ni ʃNNtds = ʃʃʃvNNTdV = ʃʃʃv Nj [Ni Nj Nk Nl] dV Nk Nl ʃʃʃv(Ni^2 Nj^2 Nk^2 Nl^2 )dV As Ni=L 1 Nj = L 2 Nk = L 3 Nl = L 4 ʃʃʃv L 12 L 22 L 32 L 42 dV According to Equation 3. ʃʃʃv L 12 L 22 L 32 L 42 dV = α!β!ϒ!δ!6V/(α+β+ϒ+δ+3)! Putting the above values we have 2!2!2!2!6V/(2+2+2+2+3)!

Ans = 96V/11!

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