Finite Element Analysis - The Basics

Finite ElementAnalysis

Sujith Jose

Introduction

Steps in FiniteElementAnalysis

Finite ElementDiscretization

ElementaryGoverningEquations

Assembling of allelements

Solving theresultingequations

i.IterationMethod

ii.Band MatrixMethod

Example

References

Finite Element Analysis - The Basics

Sujith Jose

University of California, Los Angeles

[email protected]

May 31, 2016


Sujith Jose

Introduction






i.IterationMethod


Example

References

Overview

1 Introduction

2 Steps in Finite Element AnalysisFinite Element DiscretizationElementary Governing EquationsAssembling of all elementsSolving the resulting equations

i.Iteration Methodii.Band Matrix Method

3 Example

4 References


Sujith Jose

Introduction






i.IterationMethod


Example

References

Introduction

Origin in structural analysisMathematical treatment - 1948Applied to Electromagnetic problems - 1968Can handle complex geometriesUsed in almost all engineering disciplines including electrical, aeronautical,biomedical and civil


Sujith Jose

Introduction






i.IterationMethod


Example

References

Steps in Finite Element Analysis

1 Discretize the solution region into elements

2 Derive governing equations for one element

3 Assemble all elements

4 Solving system of equations obtained


Sujith Jose

Introduction






i.IterationMethod


Example

References

Finite Element Discretization

Figure: A typical finite element subdivision of an irregular 2D domain


Sujith Jose

Introduction






i.IterationMethod


Example

References


Consider a single element (triangular or quadrilateral)

Let Ve = Potential at any point (x,y)

Ve 6= 0, inside elementVe = 0, outside element

For triangular element (used here)

Ve = a + bx + cy

For quadrilateral element

Ve = a + bx + cy + dxy


Sujith Jose

Introduction






i.IterationMethod


Example

References


Consider triangular element,

Ve(x , y) = a + bx + cy

Linear variation of potential is the same as assuming that electric field is uniformwithin the element.i.e

~Ee = −∇Ve = −(b ~ax + c ~ay )


Sujith Jose

Introduction






i.IterationMethod


Example

References

Elementary Governing Equations

At any point (x,y), Ve(x , y) = a + bx + cy .We can find potential at any point if we can find values of a, b and c.

Ve1(x1, y1) = a + bx1 + cy1Ve2(x2, y2) = a + bx2 + cy2Ve3(x3, y3) = a + bx3 + cy3Ve1

Ve2

Ve3

=

1 x1 y11 x2 y21 x3 y3

abc

Coefficients a, b and c can be found byinverting matrix

Figure: Typical triangular element


Sujith Jose

Introduction






i.IterationMethod


Example

References


abc

=

1 x1 y11 x2 y21 x3 y3

−1 Ve1

Ve2

Ve3

abc

= 1DET

(x2y3 − x3y2) (x3y1 − x1y3) (x1y2 − x2y1)(y2 − y3) (y3 − y1) (y1 − y2)(x3 − x2) (x1 − x3) (x2 − x1)

Ve1

Ve2

Ve3

Let

DET =

∣∣∣∣∣∣1 x1 y11 x2 y21 x3 y3

∣∣∣∣∣∣ = 2A


Sujith Jose

Introduction






i.IterationMethod


Example

References


abc

= 12A


Ve1

Ve2

Ve3

Ve(x , y) = a + bx + cy =

[1 x y

] abc

Ve(x , y) =[1 x y

]12A


Ve1

Ve2

Ve3

Ve(x , y) = 12A

[α1 α2 α3

] Ve1

Ve2

Ve3

=3∑

i=1αi (x , y)Vei


Sujith Jose

Introduction






i.IterationMethod


Example

References


Potential Ve(x , y) at any point (x,y) within the element (provided the potential atvertices)

Ve(x , y) =3∑

i=1

αi (x , y)Vei

where

α1 =1

2A[(x2y3 − x3y2) + (y2 − y3)x + (x3 − x2)y ] (1)

α2 =1

2A[(x3y1 − x1y3) + (y3 − y1)x + (x1 − x3)y ] (2)

α3 =1

2A[(x1y2 − x2y1) + (y1 − y2)x + (x2 − x1)y ] (3)

αi are called linear interpolation functions or element shape functions


Sujith Jose

Introduction






i.IterationMethod


Example

References

Elementary Governing Equations - Energy term

Energy density = 12εE

2

Energy per unit length

We =1

2

∫ε|~E |2dS =

1

2

∫ε|∇Ve |2dS (4)

Ve =3∑

i=1

αi (x , y)Vei ⇒ ∇Ve =3∑

i=1

Vei∇αi (5)

Substituting

We =1

2

3∑i=1

3∑j=1

εVei |∫∇αi .∇αjdS |Vei (6)


Sujith Jose

Introduction






i.IterationMethod


Example

References

Elementary Governing Equations - Energy term

Let coupling term between nodes i and j be C(e)ij =

∫∇αi .∇αjdS

We =1

2

3∑i=1

3∑j=1

εVei |∫∇αi .∇αjdS |Vei =

1

2

3∑i=1

3∑j=1

εVei |C(e)ij |Vej (7)

Writing in matrix form, energy per unit length is

We =1

2ε[Ve ]T [C (e)][Ve ] (8)

where [Ve ] =

Ve1

Ve2

Ve3

and [C (e)] =

C(e)11 C

(e)12 C

(e)13

C(e)21 C

(e)22 C

(e)23

C(e)31 C

(e)32 C

(e)33

called element coefficient

matrix or stiffness matrix


Sujith Jose

Introduction






i.IterationMethod


Example

References

Assembling of all elements

The energy associated with all the N elements in the solution region

W =N∑

e=1

We =1

2ε[V ]T [C ][V ] (9)

where

[V ] =

V1

V2

.

.Vn

(10)

n is the number of nodes[C ] is called the over-all or global coefficient matrix which is the assemblage ofindividual element coefficient matrices.


Sujith Jose

Introduction






i.IterationMethod


Example

References

Global coefficient matrix - an example

Consider a 3 element finite element mesh.5 nodes give a 5x5 global coefficientmatrix.

[C ] =

C11 C12 C13 C14 C15

C21 C22 C23 C24 C25

C31 C32 C33 C34 C35

C41 C42 C43 C44 C45

C51 C52 C53 C54 C55

Figure: Assembly of three elements


Sujith Jose

Introduction






i.IterationMethod


Example

References


Cij is the coupling term betweenglobal nodes i and j.Cij =

∫∇αi .∇αjdS

Contribution to Cij comes from allelements containing nodes i and j .

Write global coefficient elements interms of contributing elementcoefficient elements



Sujith Jose

Introduction






i.IterationMethod


Example

References


Elements 1 and 2 have node 1 in commonC11 = C

(1)11 + C

(2)11

Node 2 belongs to element 1 only

C22 = C(1)33

Node 4 belongs to elements 1, 2 and 3

C44 = C(1)22 + C

(2)33 + C

(3)33

Nodes 1 and 4 belong simultaneously toelements 1 and 2C14 = C

(1)12 + C

(2)13

No coupling between nodes 2 and 3C23 = C32 = 0



Sujith Jose

Introduction






i.IterationMethod


Example

References


The global coefficient matrix

Symmetric (Cij = Cji )

Sparse

Singular

[C ] =

C

(1)11 + C

(2)11 C

(1)13 C

(2)12 C

(1)12 + C

(2)13 0

C(1)31 C

(1)33 0 C

(1)32 0

C(2)21 0 C

(2)22 + C

(3)11 C

(2)23 + C

(3)13 C

(3)13

C(1)21 + C

(2)31 C

(1)23 C

(2)32 + C

(3)31 C

(1)22 + C

(2)33 + C

(3)33 C

(3)32

0 0 C(3)21 C

(3)23 C

(3)22


Sujith Jose

Introduction






i.IterationMethod


Example

References


Energy associated with assemblage of 3 elements

W =1

2ε[V ]T [C ][V ]

[C ] =

C

(1)11 + C

(2)11 C

(1)13 C

(2)12 C

(1)12 + C

(2)13 0

C(1)31 C

(1)33 0 C

(1)32 0

C(2)21 0 C

(2)22 + C

(3)11 C

(2)23 + C

(3)13 C

(3)13

C(1)21 + C

(2)31 C

(1)23 C

(2)32 + C

(3)31 C

(1)22 + C

(2)33 + C

(3)33 C

(3)32

0 0 C(3)21 C

(3)23 C

(3)22

, [V ] =

V1

V2

V3

V4

V5


Sujith Jose

Introduction






i.IterationMethod


Example

References

Solving the resulting equations

Laplace’s (or Poisson’s ) equation is satisfied when the total energy in thesolution region is minimum

Hence, ∂W∂V1

= ∂W∂V2

= ... = ∂W∂Vn

= 0


Sujith Jose

Introduction






i.IterationMethod


Example

References

Solving the resulting equations

For example,

∂W

∂V1= 0⇒ 0 = 2V1C11+V2C12+V3C13+V4C14+V5C15+V2C21+V3C31+V4C41+V5C51

Or0 = V1C11 + V2C12 + V3C13 + V4C14 + V5C15

In general, ∂W∂Vk

= 0 leads to

0 =n∑

i=1

ViCki

where n is the number of nodes in the mesh.Writing for all nodes k = 1, 2, ..., n→ set of simultaneous equations.From these equations, V1,V2, ..,Vn can be found.


Sujith Jose

Introduction






i.IterationMethod


Example

References

Solving the resulting equationsIteration method

Suppose node 0 is connected to m nodes.0 = V0C00 +V1C01 +V2C02 + ...+VmC0m

or

V0 = − 1

C00

m∑k=1

VkC0k

V0 can be calculated if the potentials atnodes connected to 0 are known.

Figure: Node 0 connected to m other nodes


Sujith Jose

Introduction






i.IterationMethod


Example

References

Solving the resulting equationsIteration method

Free nodes - Nodes whose potential are unknown

Fixed nodes - Nodes where the potential V is prescribed or known

Iteration process:1 Set free node potential initial value equal to

1 Zero2 Or average potential of fixed nodes Vave = 1

2 (Vmin + Vmax ), where Vmin andVmax are the minimum and maximum values of V at the fixed nodes.

2 Calculate value for free node using V0 = − 1C00

m∑k=1

VkC0k

3 Use these as fixed node potential for next iteration

4 Repeat until change between subsequent iterations is negligible.


Sujith Jose

Introduction






i.IterationMethod


Example

References

Solving the resulting equationsBand Matrix Method

If all free nodes (f) are numbered first and fixed/prescribed nodes (p) last,W = 1

2ε[Ve ]T [C (e)][Ve ] can be written as

W =1

2ε[Vf Vp

] [Cff Cfp

Cpf Cpp

] [Vf

Vp

]Differentiating wrt Vf , [

Cff Cfp

] [Vf

Vp

]= 0


Sujith Jose

Introduction






i.IterationMethod


Example

References

Solving the resulting equationsBand Matrix Method

[Cff Cfp

] [Vf

Vp

]= 0⇒ [Cff ][Vf ] = −[Cfp][Vp]

This equation can be written as

[A][V ] = [B]

or[V ] = [A]−1[B]

where[V ] = [Vf ], [A] = [Cff ], [B] = −[Cfp][Vp]

Thus, we can solve for [V ] using matrix techniques.


Sujith Jose

Introduction






i.IterationMethod


Example

References

Review of steps

1 Discretize the solution region into elements: Ve = a + bx + cy

2 Derive governing equations for one element: Ve(x , y) =3∑

i=1αi (x , y)Vei

3 Assemble all elements: W = 12ε[V ]T [C ][V ]

4 Solving system of equations obtained: [Cff ][Vf ] = −[Cfp][Vp]


Sujith Jose

Introduction






i.IterationMethod


Example

References

Example - Potential on a 2D surface

Voltage at nodes 1 and 3 are known. Can we find potential at any point within themesh using FEM ?

Using x1, x2, x3, x4, y1, y2, y3 and y4,element

[C (1)] =

1.236 −0.7786 −0.4571−0.7786 0.6929 0.0857−0.4571 0.0857 0.3714

[C (2)] =

0.5571 −0.4571 0.1−0.4571 0.8238 −0.3667−0.1 0.3667 0.4667

Figure: Two element mesh


Sujith Jose

Introduction






i.IterationMethod


Example

References


Using band matrix method,

[Cff ][Vf ] = −[Cfp][Vp][C22 C24

C42 C44

] [V2

V4

]=

[C21 C23

C41 C43

] [V1

V3

]

1 0 0 00 1.25 0 −0.01430 0 1 01 −0.0143 0 0.8381

V1

V2

V3

V4

=

0

3.70810.0

4.438



Sujith Jose

Introduction






i.IterationMethod


Example

References


We get V1 = 0,V2 = 3.708,V3 = 10 andV4 = 4.438

Now voltage at any point inside each element canbe found using linear interpolation functions

Ve(x , y) =3∑

i=1

αi (x , y)Vei



Sujith Jose

Introduction






i.IterationMethod


Example

References

References

Matthew Sadiku (1989)

A Simple Introduction to Finite Element Analysis of Electromagnetic Problems

IEEE Transactions on Education 32(2), 85 - 93.

Jianming Jin (2002)

The Finite Element Method in Electromagnetics

Second Edition


Sujith Jose

Introduction






i.IterationMethod


Example

References

Questions?


Sujith Jose

Introduction






i.IterationMethod


Example

References

Appendix - Boundary value problems

A boundary value problem can be defined by a governing differential equation in adomain Ω:

Lφ = f

together with boundary conditions on the boundary that encloses the domain.Approximate solutions to boundary value problems can be found using Ritz orGalerkin’s method.


Sujith Jose

Introduction






i.IterationMethod


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Appendix - Ritz method

Boundary value problem is formulated in terms of a variational expressioncalled functional.

Minimum of this functional corresponds to the governing differential equationunder the given boundary conditions.

Approximate solution is then obtained by minimizing the functional withrespect to variables that define a certain approximation to the solution.


Sujith Jose

Introduction






i.IterationMethod


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Appendix - Galerkin’s method

This method is one of the weighted residual methods i.e. seek the solution byweighting the residual of the differential equation.

Assume that φ is an approximate solution to boundary value problem. Then,nonzero residual

r = Lφ− f 6= 0

The best approximation for φ will be the one that reduces residual r to leastvalue at all points of Ω.

Ri =

∫wi rdΩ = 0

where Ri denote weighted residual integrals and wi are chosen weightingfunctions.

Engineering

Finite Element Analysis - The Basics