an introduction to the finite element analysis · st.petersburg 2005 agenda part iii finite element...

An Introduction to the Finite Element Analysis

Joint Advanced Student School

St.Petersburg 2005

Agenda

PART I

Introduction and Basic Concepts

1.0 Computational Methods1.1 Idealization

1.2 Discretization

1.3 Solution

2.0 The Finite Elements Method2.1 FEM Notation

2.2 Element Types

3.0 Mechanichal Approach3.1 The Problem Setup

3.2 Strain Energy

3.3 External Energy

3.4 The Potential Energy Functional


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Agenda

PART II

Mathematical Formulation

4.0 The Mathematics Behind the Method

4.1 Weighted Residual Methods

4.2 Approxiamting Functions

4.3 The Residual

4.4 Galerkin’s Method

4.5 The Weak Form

4.6 Solution Space

4.7 Linear System of Equations

4.8 Connection to the physical system


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Agenda

PART III

Finite Element Discretization

5.0 Finite Element Discretization

4.1 The Trial Basis

4.2 Matrix Form of the Problem

4.3 Element Stiffness Matrix

4.4 Element Mass Matrix

4.5 External Work Integral

4.6 Assembling


6.0 References

7.0 Question and Answers

PART I

Introduction and Basic Concepts


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1.0 Computational Methods


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1.1 Idealization

• Mathematical Models• “A model is a symbolic device built to

simulate and predict aspects of behavior of a system.”

• Abstraction of physical reality

• Implicit vs. Explicit Modelling• Implicit modelling consists of using existent

pieces of abstraction and fitting them into the particular situation (e.g. Using general purpose FEM programs)

• Explicit modelling consists of building the model from scratch


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1.2 Dicretization

1. Finite Difference Discretization

• The solution is discretized

• Stability Problems

• Loss of physical meaning

2. Finite Element Discretization

• The problem is discretized

• Physical meaning is conserved on elements

• Interpretation and Control is easier


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1.3 Solution

1. Linear System Solution Algorithms

• Gaussian Elimination

• Fast Fourier Transform

• Relaxation Techniques

2. Error Estimation and Convergence

Analysis


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2.0 Finite Element Method

• Two interpretations

1. Physical Interpretation:The continous physical model is divided into finite pieces called elements and laws of nature are applied on the generic element. The results are then recombined to represent the continuum.

2. Mathematical Interpretation:The differetional equation reppresenting the system is converted into a variational form, which is approximated by the linear combination of a finite set of trial functions.


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2.1 FEM Notation

Elements are defined by the following

properties:

1. Dimensionality

2. Nodal Points

3. Geometry

4. Degrees of Freedom

5. Nodal Forces

(Non homogeneous RHS of the DE)


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2.2 Element Types


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3.0 Mechanical Approach

• Simple mechanical problem

• Introduction of basic mechanical concepts

• Introduction of governing equations

• Mechanical concepts used in mathematical

derivation


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3.1 The Problem Setup


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• Hooke’s Law:

where

• Strain Energy Density:

3.2 Strain Energy

dx

dux )(

)()( xEx

)()(2

1xx


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• Integrating over the Volume of the Bar:

• All quantities may depend on x.

3.2 Strain Energy (cont’d)

dxEAuuU

dxuEAudxpdVU

L

LL

V

0

00

''2

1

')'(2

1

2

1

2

1


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3.3 External Energy

• Due to applied external loads

1. The distributed load q(x)

2. The point end load P. This can be

included in q.

• External Energy:

L

dxquW0


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3.4 The Total Potential

Energy Functional

• The unknown strain Function u is found by

minimizing the TPE functional described below:

)]([)]([)]([ xuWxuUxu

or

WU

PART II

Mathematical Formulation


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4.0 Historical Background

• Hrennikof and McHenry formulated a 2D structural problem as an assembly of bars and beams

• Courant used a variational formulation to approximate PDE’s by linear interpolation over triangular elements

• Turner wrote a seminal paper on how to solve one and two dimensional problems using structural elements or triangular and rectangular elements of continuum.


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The class of differential equations containing also the one

dimensional bar described above can be described as

follows :

.0)1()0(

)1(10),()())((][

uu

xxquxzdx

duxp

dx

duL


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It follows that:

Multiplying this by a weight function v and integrating over

the whole domain we obtain:

For the inner product to exist v must be “square integrable”

Therefore:

Equation (2) is called variational form

0][ quL


)2(0)][,()][(1

0 quLvdxvquL

)1,0(2Lv


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4.2 Approximating Function

• We can replace u and v in the formula with their approximation function i.e.

• The functions fj and yj are of our choice and are meant to be suitable to the particular problem. For example the choice of sine and cosine functions satisfy boundary conditions hence it could be a good choice.

N

j

jj

N

j

jj

xdxVxv

xcxUxu

1

1

)()()(

)()()(

y

f


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4.2 Approximating Function

• U is called trial function and V is called test function

• As the differential operator L[u] is second order

• Therefore we can see U as element of a finite-diemnsional

subspace of the infinite-dimensional function space C2(0,1)

• The same way

)1,0()1,0( 22 CUCu

)1,0()1,0( 2CSU N

)1,0()1,0(ˆ 2LSV N


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4.3 The Residual

• Replacing v and u with respectively V and U (2) becomes

• r(x) is called the residual (as the name of the method suggests)

• The vanishing inner product shows that the residual is orthogonal to

all functions V in the test space.

qULxr

SVrV N

][)(

ˆ,0),(


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• Substituting into

and exchanging summations and integrals we obtain

• As the inner product equation is satisfied for all choices of V in SN the above equation has to be valid for all choices of dj

which implies that

4.3 The Residual

N

j

jj xdxV1

)()( y 0)][,( qULV

NjdqULd j

N

j

jj ,...,2,1,0)][,(1

y

NjqULj ,...,2,10)][,( y


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• One obvious choice of would be taking it equal to

• This Choice leads to the Galerkin’s Method

• This form of the problem is called the strong form of the

problem. Because the so chosen test space has more

continuity than necessary.

• Therefore it is worthwile for this and other reasons to convert

the problem into a more symmetrical form

• This can be acheived by integrating by parts the initial strong

form of the problem.

4.4 Galerkin’s Method

jyjf

NjqULj ,...,2,10)][,( f


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• Let us remember the initial form of the problem

• Integrating by parts

4.5 The Weak Form

dxqzupuvquLv

1

0

])''([)][,(

.0)1()0(

10),()())((][

uu

xxquxzdx

duxp

dx

duL

0')''(])''([1

0

1

0

1

0

vpudxvqvzupuvdxqzupuv


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4.5 The Weak Form

• The problem can be rewritten as

where

• The integration by parts eliminated the second derivatives

from the problem making it possible less continouity than the

previous form. This is why this form is called weak form of the

problem.

• A(v,u) is called Strain Energy.

0),(),( qvuvA

dxvzupuvuvA )''(),(

1

0


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4.6 Solution Space

• Now that derivative of v comes into the picture v needs to have more

continoutiy than those in L2. As we want to keep symmetry its

appropriate to choose functions that produce bounded values of

• As p and z are necessarily smooth functions the following restriction is

sufficient

• Functions obeying this rule belong to the so called Sobolev Space

and they are denoted by H1. We require v and u to satisfy boundary

conditions so we denote the resulting space as

dxzuupuuA ))'((),(

1

0

22

1

0

1

0

22)'(

H

dxuu


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• The solution now takes the form

• Substituting the approximate solutions obtained earlier in the

more general WRM we obtain

• More explicitly substituting U and V (remember we chose

them to have the same base) and swapping summations and

integrals we obtain


1

0),(),( HvqvuvA

N

N

SVqVUVA

HSVU

0

1

00

),(),(

,

N

k

jkjk NjqAc1

,...,2,1,),(),( fff


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4.8 Connection to the Physical System

Mechanical Formulation Mathematical Formulation

dxEAuuUL

0

''2

1

L

dxquW0

0 WU

dxvzupuvuvA )''(),(

1

0

0),(),( qvuvA

),( qv

PART III

Finite Element Discretization


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• Let us take the initial value problem with constant coefficients

• As a first step let us divide the domain in N subintervals with

the following mesh

• Each subinterval is called finite element.

5.0 Finite Element Discretization

Njxx jj :1),,( 1

1...0 10 Nxxx

.0)1()0(

0,

10),(''

uu

zp

xxqzupu


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• Next we select as a basis the so called “hat function”.

5.1 The Trial Basis

otherwise

xxxxx

xx

xxxxx

xx

x jj

jj

j

jj

jj

j

j

0

)( 1

1

1

1

1

1

f


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• With the basis in the previous slide we construct our

approximate solution U(x)

• It is interesting to note that the coefficients correspond to the

values of U at the interior nodes

5.1 The Trial Basis


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• The problem at this point can be easily solved using the

previously derived Galerkin’s Method

• A little more work is needed to convert this problem into

matrix notation

5.1 The Trial Basis

N

k

jkjk NjqAc1

,...,2,1,),(),( fff


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• Restricting U over the typical finite element we can write

• Which in turn can be written as

in the same way


],[)]()([)(

)(][)( 1

1

1

1

1 jj

j

j

jj

j

j

jj xxxc

cxx

x

xccxU

ff

f

f

],[)()()( 111 jjjjjj xxxxcxcxU ff

],[)]()([)(

)(][)( 1

1

1

1

1 jj

j

j

jj

j

j

jj xxxd

dxx

x

xddxV

ff

f

f


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• Taking the derivative

• Derivative of V is analogus


1

1

1

1 ],[]/1/1[/1

/1][)('

jjj

jj

j

j

jj

j

j

jj

xxh

xxxc

chh

h

hccxU

],[]/1/1[/1

/1][)(' 1

1

1 jj

j

j

jj

j

j

jj xxxd

dhh

h

hddxV


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• The variational formula can be elementwise defined as

follows:


j

j

j

j

j

j

x

x

x

x

M

j

x

x

S

j

M

j

S

jj

N

j

jj

dxVqqV

dxzVUUVA

dxUpVUVA

UVAUVAUVA

qVUVA

1

1

1

),(

),(

''),(

),(),(),(

0]),(),([1


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• Substituting U,V,U’ and V’ into these formulae we obtain

4.3 The Element Stiffnes Matrix

11

11

][11

11][),(

]/1/1[/1

/1][),(

1

1

1

21

1

1

1

1

j

j

j

j

jjj

j

jx

xj

jj

S

j

j

jx

xjj

j

j

jj

S

j

h

pK

c

cKdd

c

cdx

h

pddUVA

c

cdxhh

h

hpddUVA

j

j

j

j


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4.4 The Element Mass Matrix

• The same way

21

12

6

][),(

][][),(

1

1

1

1

1

11

j

j

j

j

jjj

S

j

j

jx

xjj

j

j

jj

M

j

zhM

c

cMddUVA

c

cdxzddUVA

j

j

fff

f


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• The external work integral cannot be evaluated for every

function q(x)

• We can consider a linear interpolant of q(x) for simplicity.

• Substituting and evaluating the integral

4.5 External Work

Integral

j

j

x

xdxVqqV

1

),(

],[),()()( 111 jjjjjj xxxxqxqxq ff

jj

jjj

j

jjj

j

j

jj

x

xj

j

jjj

qq

qqhl

ldddxq

qddqV

j

j

2

2

6

],[],[],[),(

1

1

1

1

1

1

11

fff

f

Element load vector


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• Now the task is to assemble the elements into the whole

system in fact we have to sum each integral over all the

elements

• For doing so we can extend the dimension of each element

matrix to N and then put the 2x2 matrix at the appropriate

position inside it

• With all matrices and vectors having the same dimension the

summation looks like

4.6 Assembling

N

j

S

jA1

KcdT

21

121

.........

121

121

12

h

pK

1

2

1

1

2

1

......

NN d

d

d

d

c

c

c

c


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• Doing the same for the Mass Matrix and for the Load Vector

4.6 Assembling

N

j

M

jA1

McdT

41

141

.........

141

141

14

6

zhM

N

j

jqV1

),( ldT

NNN qqq

qqq

qqq

h

12

321

210

4

...

4

4

6l


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• Substituting this Matrix form of the expressions in

we obtain the following set of linear equations

• This has to be satisfied for all choices of d therefore


0lcMKdT ])[(

N

j

jj qVUVA1

0]),(),([

0lcMK )(


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References

• Carlos Felippahttp://caswww.colorado.edu/courses.d/IFEM.d/IFEM.Ch06.d/IFEM.Ch06.pdf

• Joseph E Flaherty,Amos Eaton Professorhttp://www.cs.rpi.edu/~flaherje/FEM/fem1.ps

• Gilbert Strang, George J. Fix

An Analysis of the Finite Element Method

Prentice-Hall,1973

http://caswww.colorado.edu/courses.d/IFEM.d/IFEM.Ch06.d/IFEM.Ch06.pdf

http://www.cs.rpi.edu/~flaherje/FEM/fem1.ps

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