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.

fem basics presentation

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