Manopulo Pres

An Introduction to the Finite Element

Analysis

Presented byNiko Manopulo

AgendaPART IIntroduction and Basic Concepts

1.0 Computational Methods1.1 Idealization1.2 Discretization1.3 Solution

2.0 The Finite Elements Method2.1 FEM Notation2.2 Element Types

3.0 Mechanichal Approach3.1 The Problem Setup3.2 Strain Energy3.3 External Energy3.4 The Potential Energy Functional

Agenda

PART IIMathematical Formulation

4.0 The Mathematics Behind the Method 4.1 Weighted Residual Methods

4.2 Approxiamting Functions4.3 The Residual4.4 Galerkin’s Method4.5 The Weak Form4.6 Solution Space4.7 Linear System of Equations4.8 Connection to the physical system

Agenda

PART IIIFinite Element Discretization

5.0 Finite Element Discretization 4.1 The Trial Basis

4.2 Matrix Form of the Problem4.3 Element Stiffness Matrix4.4 Element Mass Matrix4.5 External Work Integral4.6 Assembling4.7 Linear System of Equations

6.0 References7.0 Question and Answers

PART IIntroduction and Basic Concepts

1.0 Computational Methods

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

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

1.3 Solution

1. Linear System Solution Algorithms• Gaussian Elimination• Fast Fourier Transform• Relaxation Techniques

2. Error Estimation and Convergence Analysis

2.0 Finite Element Method

• Two interpretations1. 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.

2.1 FEM Notation

Elements are defined by the following properties:

1. Dimensionality2. Nodal Points3. Geometry4. Degrees of Freedom5. Nodal Forces (Non homogeneous RHS of the DE)

2.2 Element Types

3.0 Mechanical Approach

• Simple mechanical problem• Introduction of basic mechanical concepts• Introduction of governing equations• Mechanical concepts used in mathematical

derivation

3.1 The Problem Setup

• Hooke’s Law:

where

• Strain Energy Density:

3.2 Strain Energy

dxdux )(

)()( xEx

)()(21 xx

• 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

''21

')'(21

21

21

3.3 External Energy

• Due to applied external loads1. The distributed load q(x)2. The point end load P. This can be

included in q.

• External Energy:

L

dxquW0

3.4 The Total Potential Energy Functional

• The unknown strain Function u is found by minimizing the TPE functional described below:

)]([)]([)]([ xuWxuUxuor

WU

PART IIMathematical Formulation

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.

4.1 Weighted Residual Methods

The class of differential equations containing also the one dimensional bar described above can be described as follows :

.0)1()0(

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

uu

xxquxzdxduxp

dxduL

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

4.1 Weighted Residual Methods

)1,0(2Lv

4.2 Approximating Function

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

• The functions jandj 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

jjj

N

jjj

xdxVxv

xcxUxu

1

1

)()()(

)()()(

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

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.

qULxrSVrV N

][)(

ˆ,0),(

• 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

jjj xdxV

1

)()( 0)][,( qULV

NjdqULd j

N

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

1

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

4.4 Galerkin’s Method

• 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.

jj

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

• 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

xxquxzdxduxp

dxduL

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

0

1

0

1

0

vpudxvqvzupuvdxqzupuv

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

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

10

1

0

22)'(

H

dxuu

• 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

4.7 Linear System of Equations

N

kjkjk NjqAc

1

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

10),(),( HvqvuvA

N

N

SVqVUVA

HSVU

0

100

),(),(

,

4.8 Connection to the Physical System

Mechanical Formulation Mathematical Formulation

dxEAuuUL

0

''21

L

dxquW0

0 WU

dxvzupuvuvA )''(),(1

0

0),(),( qvuvA

),( qv

PART IIIFinite Element Discretization

• 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

.0)1()0(0,

10),(''

uuzp

xxqzupu

Njxx jj :1),,( 1

1...0 10 Nxxx

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

5.1 The Trial Basis

otherwise

xxxxxxx

xxxxxxx

x jjjj

j

jjjj

j

j

0

)( 11

1

11

1

• 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

• 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

kjkjk NjqAc

1

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

• Restricting U over the typical finite element we can write

• Which in turn can be written as

in the same way

5.2 Matrix Form of the Problem

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

][)( 11

11

1 jjj

jjj

j

jjj xxx

cc

xxxx

ccxU

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

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

][)( 11

11

1 jjj

jjj

j

jjj xxx

dd

xxxx

ddxV

• Taking the derivative

• Derivative of V is analogus


1

11

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

][)('

jjj

jjj

jjj

j

jjj

xxh

xxxc

chh

hh

ccxU

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

][)(' 11

1 jjj

jjj

j

jjj xxx

dd

hhhh

ddxV

• The variational formula can be elementwise defined as follows:


j

j

j

j

j

j

x

x

x

x

Mj

x

x

Sj

Mj

Sjj

N

jjj

dxVqqV

dxzVUUVA

dxUpVUVA

UVAUVAUVA

qVUVA

1

1

1

),(

),(

''),(

),(),(),(

0]),(),([1

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

4.3 The Element Stiffnes Matrix

1111

][1111

][),(

]/1/1[/1/1

][),(

11

121

11

1

1

jj

j

jjjj

j

jx

xj

jjSj

j

jx

x jjj

jjj

Sj

hpK

cc

Kddc

cdx

hpddUVA

cc

dxhhhh

pddUVA

j

j

j

j

4.4 The Element Mass Matrix

2112

6

][),(

][][),(

11

11

11

1

jj

j

jjjj

Sj

j

jx

x jjj

jjj

Mj

zhM

cc

MddUVA

cc

dxzddUVA j

j

• The same way

• 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

jj

jjjj

jjjj

jjj

x

xj

jjjj

qqqqh

l

ldddxq

qddqV j

j

22

6

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

1

1

11

11

11

Element load vector

• 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

SjA

1

KcdT

21121

.........121

12112

hpK

1

2

1

1

2

1

......

NN d

dd

d

c

cc

c

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

4.6 Assembling

N

j

MjA

1

McdT

41141

.........141

14114

6zhM

NNN qqq

qqqqqq

h

12

321

210

4...

44

6l

N

jjqV

1

),( ldT

• 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

4.7 Linear System of Equations

0lcMKdT ])[(

N

jjj qVUVA

1

0]),(),([

0lcMK )(

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