FEM Basics Presentation

8/11/2019 FEM Basics Presentation

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Indian Institute of Technology Bombay

Basics of FEM

Prof. S. V. Kulkarni

Department of Electrical EngineeringIndian Institute of Technology Bombay


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Outline

1 Introduction

2 FEM ProcedureDiscretisation of the DomainApproximation of the SolutionAssembly of the SystemBoundary Conditions and Solution of the Final System

3 Properties of FEM Matrices


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Introduction

Introduction

Computational Electromagnetics:Finite difference methodFinite element method (FEM)Boundary element method (BEM)Method of moments (MoM)Meshless methods: Particle-in-cell, Petrov Galerkin etc.

Finite Element Analysis:Variational ProcedureGalerkins Method


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Introduction

Illustrative Example: Parallel Plate Capacitor

P :

2 = 0

|y = 1, 0< x < 1 = 10

|y = 0, 0< x < 1 = 0on

(1)

x

y

(0, 0)

(1, 1)(0, 1)

(1, 0)

=10

=0


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FEM Procedure

Variational Procedure

Minimisation or Maximisation of a Functional function offunctions

Functional for the example problem:

E = 12

||2 d (2)

Physical Signicance of functional: Energy of the system

For a single dielectric system, the energy minimisation andthe corresponding solution is not inuenced by and hence itcan be dropped from the expression


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FEM Procedure

Discretisation of the Domain

Overview of Discretisation

FEM Problem domaindivided into a large number ofsmall elements/sub-domains

Type of element:Geometry of problemShape of element:2D Triangular, Rectangularand 3D Cubic, Tetrahedral,

PrismaticNodal versus Vector

Size of element: specic togeometry requirements

y )3

(x, 3

y )2(x, 2

y )1(x, 1

1e

2e

3e

1

2

3

I di I i f T h l B b


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FEM Procedure


Element Size Illustration: Arbitrary Boundary

Indian Instit te of Technolog Bomba


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FEM Procedure


Example Problems Domain Discretisation

1 2

3

1 1

1

1

1 1

1

2

2

2

22

2

3

3

33

33

32

1

2

3

4

5

6 8

7

1 2 3

7 8 9

4 65



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FEM Procedure

Approximation of the Solution

Overview of Solution Approximation

Analytical solution may not exist for many problems

Piece-wise linear polynomial approximation: Polynomialsolution over each elementExamples:

3D Elemental: e = a + bx + cy + dz + exy + gyz + hxz + ixyz 2D Elemental: e = a + bx + cy + dxy Global approximation: =

e

e (3)

Rayleigh-Ritz method (variational procedure)


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gy y

FEM Procedure

Assembly of the System

2D Linear Assembly (2 of 2)

and,

= Area of the elemental triangular element = 1

2

1 x 1 y 11 x 2 y 21 x 3 y 3

(9)

N i (x , y ) has the property that,

N i (x j , y j ) = ij (10)

where,

ij =1 i = j 0 i j

(11)



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gy y

FEM Procedure


Functional Approximation (1 of 3)

Using equations 3 and 7 in equation 2, we can express thefunctional in terms of the (unknown) nodal potentials as:

E = e

12 e

3

i = 1 N i (x , y )

e i

2

de

(12)

E = 12 e

e

3

i = 1

{N i (x , y )}e i 2de (13)

E = 12 e e N 1(x , y )e 1 + N 2(x , y )e 2 + N 3(x , y )e 3

2de

(14)



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FEM Procedure



For any vector a , a a = |a |2

E = 12

e

e

N 1(x , y )e 1 + N 2(x , y )e 2 + N 3(x , y )

e 3

N 1(x , y )e 1 + N 2(x , y )e 2 + N 3(x , y )

e 3 d

e

(15)

E = 12 e

3

i = 1

3

j = 1 e e i N i (x , y ) N j (x , y )e j de (16)where e is the elemental domain



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FEM Procedure



E = 12 e

3

i = 1

3

j = 1

e i e

N i (x , y ) N j (x , y )de e j (17)

a e ij e N i N j de (18)

A e 3

i = 1

3

j = 1

a e ij =a e 11 a

e 12 a

e 13

a e 21 a e 22 a

e 23

a e 31 a

e 32 a

e 33

(19)

A e is referred to as the elemental stiffness matrix. The elementalenergy can, thus, be represented as,

E e = e TA e e (20)

where e = e 1 e 2 e 3T

and E =e

E e .



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FEM Procedure


A e Example Calculations (1 of 2)

For example,a e 11 =

e

N 1 N 1 de (21)

where, from equation 8,

N 1 = 12

(x 2 y 3 x 3y 2) + ( y 2 y 3)x + ( x 3 x 2)y

= 12

[(y 2 y 3)x + ( x 3 x 2)y ] (22)

a e 11 = N 1 N 1 e de = 1

4 2(y 2 y 3)2 + ( x 3 x 2)2

(23)

= 1

4(y 2 y 3)2 + ( x 3 x 2)2 (24)



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FEM Procedure


A e Example Calculations (2 of 2)

Similarly, we have,

a e 12 = N 1 N 2 = 14 [(y 2 y 3)(y 3 y 1) + ( x 3 x 2)(x 1 x 3)]

(25)

a e 13 = N 1 N 3 = 14

[(y 2 y 3)(y 1 y 2) + ( x 3 x 2)(x 2 x 1)]

(26)


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FEM Procedure


Assembly of Global Stiffness Matrix (1 of 3)

The global stiffness matrix is of the form:

A =

A11 A12 A13 A19A21 A22 A23 A29A31 A32 A33 A39

... . . .

...

A91 A92 A93 A99

(28)



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FEM Procedure



1 2

3

1 1

1

1

1 1

1

2

2

2

22

2

3

3

33

33

32

1

2

3

4

5

6 8

7

1 2 3

7 8 9

65

Example elements:

A11 = a e 111 + a e 211

A12 = a e 212 = A21A

1i = A

i 1 = 0 i = 3, 6 , 7 , 8 , 9

A22 = a e 222 + a e 311 + a

e 411

A23 = a e 412 = A32...

A55 = a e 122 + a e 233 + a

e 333

+ a e 622 + a e 711 + a

e 811

A51 = a e 121 + a e 231 = A15

and so on

Indian Institute of Technology BombayFEM P d


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FEM Procedure



Thus, the global stiffness matrix is:

A =

1 2 3 91 a e 111 + a

e 211 a

e 212 0 0

2 a e 221 a e 222 + a

e 311 + a

e 411 a

e 412 0

3 0 a e 421 a e 422 0

... ... . . . ...

9 0 0 0 a e 722 + a e 833

(29)

Indian Institute of Technology BombayFEM Procedure


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FEM Procedure


Approximate Functional Expression

Thus, equation 17 for the functional can now be expressed as,

E = 12

TA (30)

where, A is the matrix in equation 29 and is a vector of all nodalpotential values,

= 1 2 3 4 5 6 7 8 9T

(31)



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FEM Procedure

Boundary Conditions and Solution of the Final System

Gradient of Functional is Zero

E = 0 (32)

E 1E

2...E 9

= 0 (33)

A11 1 + A12 2 + A13 3 + + A19 9 = 0A21 1 + A22 2 + A23 3 + + A29 9 = 0

...A91 1 + A92 2 + A93 3 + + A99 9 = 0

(34)

A =

0 (35)



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FEM Procedure


Incoporating Boundary Conditions: Method 1

The boundary conditions are:

1 = 2 = 3 = 07 = 8 = 9 = 10 (36)

This species six of the nine unknowns in equation 34.

The remaining three unknowns can be determined by

substituting these six values into any three of the nineequations in ( 34 ) and solving the resulting set of linearequations.



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FEM Procedure



Equation 34 now becomes the reduced order system as shownbelow:

A44 4 + A45 5 + A46 6 = 10 (A47 + A48 + A49 )A54 4 + A55 5 + A56 6 = 10 (A57 + A58 + A59 )A64 4 + A65 5 + A66 6 = 10 (A67 + A68 + A69 )

(37)

A 3 3

Symmetric3 1 = b 3 1 (38)



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Alternatively, the boundary conditions can be applied withoutreducing the size of A as follows:

1 = 02 = 03 = 0

A41 1 + A42 2 + A43 3 + + A49 9 = 0A51 1 + A52 2 + A53 3 + + A59 9 = 0A61 1 + A62 2 + A63 3 + + A69 9 = 0

7 = 108 = 109 = 10

(39)

A 9 9

Unsymmetric

9 1 = b 9 1 (40)



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Linear System Solution

Matrix A of the order of 10 2 : Direct inversion

Matrix A of the order of 10 3 to 10 4 : Direct solution techniques

such as Gaussian elimination, LU decomposition or LDUdecomposition

For linear systems of the order of 10 5 or higher:Computationally efcient to use only iterative solutiontechniques such as steepest descent method or the conjugategradient method. Further, may require the use ofpreconditioners.

Indian Institute of Technology BombayProperties of FEM Matrices


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Important Properties of FEM Induced Matrices ( A )

1 Matrix A is (generally) a large sparse matrix with a sparsityfactor of less than 1 %. The sparsity factor is dened as thetotal number of non-zero elements in A divided by the totalnumber of elements in A expressed as a percentage.

2 Matrix A is positive denite (that is, x T A x > 0 x ) and hence,diagonally dominant. The positive deniteness is a directconsequence of the physical nature of the laws governing anelectromagnetic system.

3 The condition number of A deteriorates as the size of Aincreases. This requires the use of special techniques in thesolution of the large linear system in order to ensure that thesolutions obtained are not spurious.

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FEM Basics Presentation

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