Finite Element Method (3): 2D FEM

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


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


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


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


6

Example: 2D Potential Equation with Drichlet Boundary Conditions

I,

Y

X

100V

75V 50V

0V

4,4

0,0 10cm

10cm


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


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


9

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


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


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=𝜕


𝜕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


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


13

2D Rectangular Finite Elements Using Galerkin’s Method (3)

The integral 𝐼2 can be evaluated using Green’s Theory (Next Lecture):

𝐼2 = 𝜕


𝜕𝑥+𝜕

𝜕𝑦𝑁 𝑇𝜕𝑉

𝜕𝑦𝑑𝑥𝑑𝑦

𝐴

Finite Element Method (3): 2D FEM

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