finite element method (3): 2d fem
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Finite Element Method (3): 2D FEM
Lecture 12-13 Dr. Amr Bayoumi
Fall 2014 Advanced Engineering Mathematics (EC760)
Arab Academy for Science and Technology - Cairo
Dr. Amr Bayoumi- Spring 2015 Lec. 13 EC760 Advanced Engineering Mathematics
2
Outline
• 2D using Triangular Elements
• 2D using Rectangular Elements
Dr. Amr Bayoumi- Spring 2015 Lec. 13 EC760 Advanced Engineering Mathematics
3
References
• S. Chapra and R. Canale, “Numerical Method’s for Engineers”, McGraw-Hill, 5th Ed., 2006
• S. Moaveni, “Finite Element Analysis, Theory and Application with Ansys”, Pearson Prentice Hall, 3rd Ed., 2008
• E. Thompson, “Introduction to the Finite Element Method: Theory, Programming, and Applications”, Wiley, 2005
Dr. Amr Bayoumi- Spring 2015 Lec. 13 EC760 Advanced Engineering Mathematics
4
Triangular Mesh
𝑢 𝑥, 𝑦 = 𝑎0 + 𝑎1,1𝑥 + 𝑎1,2𝑦
1 𝑥1 𝑦11 𝑥2 𝑦21 𝑥3 𝑦3
𝑎0𝑎1,1𝑎1,2=𝑢1𝑢2𝑢3
Find a0, a11, a12 (Cramer’s Rule, LU, GE,…)
𝑢 = 𝑁1𝑢1 +𝑁2𝑢2 +𝑁3𝑢3 Ae=Area of triangular element=(1/2) Det(A)
• 𝑁1 =1
2𝐴𝑒[𝑥2𝑦3− 𝑥3𝑦2) + (𝑦2− 𝑦3)𝑥 + (𝑥3 − 𝑥2)𝑦]
• 𝑁2 =1
2𝐴𝑒[(𝑥3𝑦1− 𝑥1𝑦3) + (𝑦3− 𝑦1)𝑥 + (𝑥1 − 𝑥3)𝑦]
• 𝑁3 =1
2𝐴𝑒[(𝑥1𝑦2− 𝑥2𝑦1) + (𝑦1− 𝑦2)𝑥 + (𝑥2 − 𝑥1)𝑦]
x
y
1
2
3 u1
u2
u3
Dr. Amr Bayoumi- Spring 2015 Lec. 13 EC760 Advanced Engineering Mathematics
5
Rectangular Mesh (Local Coordinates)
𝑢 𝑥, 𝑦 = 𝑎0 + 𝑎1𝑥 + 𝑎2𝑦 + 𝑎3𝑥𝑦
1 𝑥1 𝑦1 𝑥1𝑦11 𝑥2 𝑦2 𝑥2𝑦211
𝑥3𝑥4
𝑦3𝑥4
𝑥3𝑦3𝑥4𝑦4
𝑎0𝑎1𝑎2𝑎3
=
𝑢1𝑢2𝑢3𝑢4
𝑢 = 𝑁1𝑢1 +𝑁2𝑢2 +𝑁3𝑢3 + 𝑁4𝑢4
𝑁1 = 1 −𝑥
𝑙 1 −
𝑦
𝑤,
𝑁2 =𝑥
𝑙 1 −
𝑦
𝑤,
𝑁3 =𝑦
𝑤1 −𝑥
𝑙 , 𝑁4 =
𝑥𝑦
𝑙𝑤
X
Y 4 3
1 2
u1 u2
u4 u3
𝑥
𝑦
𝑙
𝑤
Dr. Amr Bayoumi- Spring 2015 Lec. 13 EC760 Advanced Engineering Mathematics
6
Example: 2D Potential Equation with Drichlet Boundary Conditions
I,
Y
X
100V
75V 50V
0V
4,4
0,0 10cm
10cm
Dr. Amr Bayoumi- Spring 2015 Lec. 13 EC760 Advanced Engineering Mathematics
7
Example: 2D Potential Equation with Drichlet Boundary Conditions
Element Equations:
𝜕2𝑉(𝑥, 𝑦)
𝜕𝑥2 + 𝜕2𝑉(𝑥, 𝑦)
𝜕𝑦2 = −𝑓 𝑥
𝑓 𝑥 = 0
𝜕𝑉
𝜕𝑥= −𝐸𝑥
𝜕𝑉
𝜕𝑦= −𝐸𝑦
X
Y 4 3
1 2
V1 V2
V4 V3
𝑥
𝑦
𝑙
𝑤
Dr. Amr Bayoumi- Spring 2015 Lec. 13 EC760 Advanced Engineering Mathematics
8
Derivatives in 2D: 𝜕𝑽
𝜕𝑥
𝑉 = 𝑁1𝑣1 +𝑁2𝑣2 +𝑁3𝑣3 +𝑁4𝑣4 = 𝑁𝑇 𝑉
𝜕𝑉
𝜕𝑥=𝜕 𝑁 𝑇
𝜕𝑥𝑉 =
𝜕
𝜕𝑥𝑁1 𝑁2 𝑁3 𝑁4 𝑉
𝜕𝑁1𝜕𝑥=𝜕
𝜕𝑥1 −𝑥
𝑙1 −𝑦
𝑤= 1 −
𝑦
𝑤 −1
𝑙 =𝑦 − 𝑤
𝑤𝑙
𝜕𝑁2𝜕𝑥=𝜕
𝜕𝑥
𝑥
𝑙 1 −𝑦
𝑤= 1 −
𝑦
𝑤 1
𝑙 =𝑤 − 𝑦
𝑤𝑙
𝜕𝑁3𝜕𝑥=𝜕
𝜕𝑥
𝑦
𝑤1 −𝑥
𝑙 =−𝑦
𝑤𝑙,𝜕𝑁4𝜕𝑥=𝜕
𝜕𝑥
𝑥𝑦
𝑙𝑤=𝑦
𝑤𝑙
𝜕𝑉
𝜕𝑥=1
𝑤𝑙(𝑦 − 𝑤) (𝑤 − 𝑦) (−𝑦) (𝑦) 𝑉
Dr. Amr Bayoumi- Spring 2015 Lec. 13 EC760 Advanced Engineering Mathematics
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Derivatives in 2D: 𝜕𝑽
𝜕𝒚
𝜕𝑁1𝜕𝑦=𝜕
𝜕𝑦1 −𝑥
𝑙1 −𝑦
𝑤= 1 −
𝑥
𝑙 −1
𝑤 =𝑥 − 𝑙
𝑤𝑙
𝜕𝑁2𝜕𝑦=𝜕
𝜕𝑦
𝑥
𝑙 1 −𝑦
𝑤=−𝑥
𝑤𝑙
𝜕𝑁3𝜕𝑦=𝜕
𝜕𝑦
𝑦
𝑤1 −𝑥
𝑙 =𝑙 − 𝑥
𝑤𝑙,𝜕𝑁4𝜕𝑥=𝜕
𝜕𝑥
𝑥𝑦
𝑙𝑤=𝑥
𝑤𝑙
𝜕𝑉
𝜕𝑦=1
𝑤𝑙(𝑥 − 𝑙) (𝑙 − 𝑥) (−𝑥) (𝑥) 𝑉
Dr. Amr Bayoumi- Spring 2015 Lec. 13 EC760 Advanced Engineering Mathematics
10
Residual Equations in 2D Rectangular F.E. Using Galerkin’s Method
𝑉 = 𝑁1𝑣1 +𝑁2𝑣2 +𝑁3𝑣3 +𝑁4𝑣4 = 𝑁𝑇 𝑉
𝜕2𝑉(𝑥, 𝑦)
𝜕𝑥2 + 𝜕2𝑉(𝑥, 𝑦)
𝜕𝑦2 = 0
𝑅 = 𝑅𝑖4
𝑖=1= 𝑁𝑖
𝜕2𝑉(𝑥, 𝑦)
𝜕𝑥2 + 𝜕2𝑉(𝑥, 𝑦)
𝜕𝑦2𝑑𝐴
𝐴
4
𝑖=1
= 𝑁 𝑇𝜕2𝑉(𝑥, 𝑦)
𝜕𝑥2 + 𝜕2𝑉(𝑥, 𝑦)
𝜕𝑦2𝑑𝑥𝑑𝑦
𝐴
= 0
Where: 𝑁 𝑇 = 𝑁1 𝑁2 𝑁3 𝑁4
Dr. Amr Bayoumi- Spring 2015 Lec. 13 EC760 Advanced Engineering Mathematics
11
Green’s Theory in 2D Finite Elements Using Galerkin’s Method (2)
Use: 𝜕
𝜕𝑥𝑁 𝑇𝜕𝑉
𝜕𝑥= [𝑁]𝑇
𝜕2𝑉
𝜕𝑥2+𝜕 𝑁 𝑇
𝜕𝑥
𝜕𝑉
𝜕𝑥
→ [𝑁]𝑇𝜕2𝑉
𝜕𝑥2=𝜕
𝜕𝑥𝑁 𝑇𝜕𝑉
𝜕𝑥−𝜕 𝑁 𝑇
𝜕𝑥
𝜕𝑉
𝜕𝑥
Similarly:
[𝑁]𝑇𝜕2𝑉
𝜕𝑦2=𝜕
𝜕𝑦𝑁 𝑇𝜕𝑉
𝜕𝑦−𝜕 𝑁 𝑇
𝜕𝑦
𝜕𝑉
𝜕𝑦
By substituting: 𝑅 = 𝐼1 + 𝐼2
Dr. Amr Bayoumi- Spring 2015 Lec. 13 EC760 Advanced Engineering Mathematics
12
2D Rectangular Finite Elements Using Galerkin’s Method (2)
The integral 𝐼1 can be easily evaluated using derivatives of 𝑁 𝑇
𝐼1 = −𝜕 𝑁 𝑇
𝜕𝑥
𝜕𝑉
𝜕𝑥 − 𝜕 𝑁 𝑇
𝜕𝑦
𝜕𝑉
𝜕𝑦𝑑𝑥𝑑𝑦
𝑏
𝑎
𝑑
𝑐
𝜕 𝑁 𝑇
𝜕𝑥=1
𝑤𝑙(𝑦 − 𝑤) (𝑤 − 𝑦) (−𝑦) (𝑦) 𝑉
𝜕 𝑁 𝑇
𝜕𝑦=1
𝑤𝑙(𝑥 − 𝑙) (𝑙 − 𝑥) (−𝑥) (𝑥)
𝜕𝑉
𝜕𝑥=𝜕 𝑁 𝑇
𝜕𝑥𝑉 ,
𝜕𝑉
𝜕𝑦=𝜕 𝑁 𝑇
𝜕𝑦𝑉
Dr. Amr Bayoumi- Spring 2015 Lec. 13 EC760 Advanced Engineering Mathematics
13
2D Rectangular Finite Elements Using Galerkin’s Method (3)
The integral 𝐼2 can be evaluated using Green’s Theory (Next Lecture):
𝐼2 = 𝜕
𝜕𝑥𝑁 𝑇𝜕𝑉
𝜕𝑥+𝜕
𝜕𝑦𝑁 𝑇𝜕𝑉
𝜕𝑦𝑑𝑥𝑑𝑦
𝐴
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