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Introduction to the Finite Element MethodDr. Mohammad Tawfik

Introduction to the Finite Element Method

Spring 2010


Course Objectives

• The student should be capable of writing simple programs to solve different problems using finite element method.


Assessment

• 10% Assignments (1 per week)• 20% Quizzes (best 2 out of 3)

– Week of 12/11/2006– Week of 20/12/2006– Week of 17/1/2006

• 20% Course Project• 25% Midterm exam (Week of 2/12/2006)• 25% Final exam (starting 3/2/2007)


Fundamental Course Agreement

• Homework is sent in electronic format (No hardcopies are accepted)

• Computer programs have to written in MATLAB or Mathematica script

• No late homework is accepted• No excuses are accepted for missing a

quiz• Best two out of three quizzes are counted


References

• J.N. Reddy, “An Introduction to the Finite Element Method” 3rd ed., McGraw Hill, ISBN 007-124473-5

• D.V. Hutton, “Fundamentals of Finite Element Analysis” 1st ed., McGraw Hill, ISBN 007-121857-2

• K. Bathe, “Finite Element Procedures,” Prentice Hall, 1996. (in library)

• T. Hughes, “The finite Element Method: Linear Static and Dynamic Finite Element analysis,” Dover Publications, 2000. (in library)


Numerical Solution of Boundary Value Problems

Weighted Residual Methods


Objectives

• In this section we will be introduced to the general classification of approximate methods

• Special attention will be paid for the weighted residual method

• Derivation of a system of linear equations to approximate the solution of an ODE will be presented using different techniques


Why Approximate?

• Ignorance

• Readily Available Packages

• Need to Develop New Techniques

• Good use of your computer!

• In general, the problem does not have an analytical solution!


Classification of Approximate Solutions of D.E.’s

• Discrete Coordinate Method– Finite difference Methods– Stepwise integration methods

• Euler method• Runge-Kutta methods• Etc…

• Distributed Coordinate Method


Distributed Coordinate Methods

• Weighted Residual Methods– Interior Residual

• Collocation• Galrekin• Finite Element

– Boundary Residual• Boundary Element Method

• Stationary Functional Methods– Reyligh-Ritz methods– Finite Element method


Basic Concepts

• A linear differential equation may be written in the form:

xgxfL

• Where L(.) is a linear differential operator.• An approximate solution maybe of the form:

n

iii xaxf

1


Basic Concepts• Applying the differential operator on the approximate

solution, you get:

01

1

xgxLa

xgxaLxgxfL

n

iii

n

iii

xRxgxLan

iii

1

Residue


Handling the Residue

• The weighted residual methods are all based on minimizing the value of the residue.

• Since the residue can not be zero over the whole domain, different techniques were introduced.


Collocation Method

• The idea behind the collocation method is similar to that behind the buttons of your shirt!

• Assume a solution, then force the residue to be zero at the collocation points


Collocation Method

0jxR

0

1

j

n

ijii

j

xFxLa

xR


Example Problem


The bar tensile problem

02

2

xFx

uEA


Bar application

02

2

xFx

uEA

n

iii xaxu

1

xRxFdx

xdaEA

n

i

ii

12

2Applying the collocation method

01

2

2

j

n

i

jii xF

dx

xdaEA


In Matrix Form

nnnnnn

n

n

xF

xF

xF

a

a

a

kkk

kkk

kkk

2

1

2

1

21

22212

12111

...

...

...

Solve the above system for the “generalized coordinates” ai to get the solution for u(x)

jxx

iij dx

xdEAk

2

2


Notes on the trial functions

• They should be at least twice differentiable!

• They should satisfy all boundary conditions!

• Those are called the “Admissibility Conditions”.


Using Admissible Functions

• For a constant forcing function, F(x)=f

• The strain at the free end of the bar should be zero (slope of displacement is zero). We may use:

l

xSinx

2


Using the function into the DE:

• Since we only have one term in the series, we will select one collocation point!

• The midpoint is a reasonable choice!

l

xSin

lEA

dx

xdEA

22

2

2

2

faSinl

EA

1

2

42


Solving:

• Then, the approximate solution for this problem is:

• Which gives the maximum displacement to be:

• And maximum strain to be:

EA

fl

EA

fl

SinlEA

fa

2

2

2

21 57.024

42

l

xSin

EA

flxu

257.0

2

5.057.02

exactEA

fllu

0.19.00 exactEA

lfux


The Subdomain Method (free reading)

• The idea behind the subdomain method is to force the integral of the residue to be equal to zero on an subinterval of the domain


The Subdomain Method

01

j

j

x

x

dxxR

011

1

j

j

j

j

x

x

n

i

x

x

ii dxxgdxxLa


Bar application

02

2

xFx

uEA

n

iii xaxu

1

xRxFdx

xdaEA

n

i

ii

12

2Applying the subdomain method

11

12

2 j

j

j

j

x

x

n

i

x

x

ii dxxFdx

dx

xdaEA


In Matrix Form

11

2

2 j

j

j

j

x

x

i

x

x

i dxxFadxdx

xdEA



The Galerkin Method

• Galerkin suggested that the residue should be multiplied by a weighting function that is a part of the suggested solution then the integration is performed over the whole domain!!!

• Actually, it turned out to be a VERY GOOD idea


The Galerkin Method

0Domain

j dxxxR

01

Domain

j

n

i Domain

iji dxxgxdxxLxa


Bar application

02

2

xFx

uEA

n

iii xaxu

1

xRxFdx

xdaEA

n

i

ii

12

2Applying Galerkin method

Domain

j

n

i Domain

iji dxxFxdx

dx

xdxaEA

12

2


In Matrix Form

Domain

ji

Domain

ij dxxFxadx

dx

xdxEA

2

2



Same conditions on the functions are applied

• They should be at least twice differentiable!

• They should satisfy all boundary conditions!

• Let’s use the same function as in the collocation method:

l

xSinx

2


Substituting with the approximate solution:

Domain

j

n

i Domain

iji dxxFxdx

dx

xdxaEA

12

2

l

l

fdxl

xSin

dxl

xSin

l

xSina

lEA

0

0

1

2

2

222

ll

al

EA2

22 1

2

EA

fll

EA

fa

2

3

2

1 52.016


Substituting with the approximate solution: (Int. by Parts)

Domain

j

n

i Domain

iji dxxFxdx

dx

xdxaEA

12

2

ll

al

EA2

22 1

2

EA

fll

EA

fa

2

3

2

1 52.016

Domain

ijl

ij

Domain

ij

dxdx

xd

dx

xd

dx

xdx

dxdx

xdx

0

2

2

Zero!


What did we gain?

• The functions are required to be less differentiable

• Not all boundary conditions need to be satisfied

• The matrix became symmetric!


Summary

• We may solve differential equations using a series of functions with different weights.

• When those functions are used, Residue appears in the differential equation

• The weights of the functions may be determined to minimize the residue by different techniques

• One very important technique is the Galerkin method.


NOTE

• Next Sunday 5/11 (No lecture)

• Following week 12/11, Quiz #1 will be held covering all the material up-to this lecture

• Homework #1 is due next week (Electronic submission of report and code is mandatory.


Report Should Include …

• Cover page

• Introduction section indicating the procedure you used with the equations as implemented in your code

• Results section

• Observations and Conclusions if any according to the output of your program.


Homework #1

• Solve the beam bending problem, for beam displacement, for a simply supported beam with a load placed at the center of the beam using– Collocation Method– Subdomain Method– Galerkin Method

• Use three term Sin series that satisfies all BC’s

• Write a program that produces the results for n-term solution.

)(4

4

xFdx

wd

0)()0(

0)()0(

2

2

2

2

dx

lwd

dx

wd

lww


Exact Solution

12/110

3

15

7

412

2/1060

13

12)(

23

3

xxxx

xxx

xw


The Finite Element Method

2nd order DE’s in 1-D


Objectives

• Understand the basic steps of the finite element analysis

• Apply the finite element method to second order differential equations in 1-D


The Mathematical Model

• Solve:

• Subject to:

Lx

fcudx

dua

dx

d

0

0

00 ,0 Qdx

duauu

Lx


Step #1: Discretization

• At this step, we divide the domain into elements.

• The elements are connected at nodes.

• All properties of the domain are defined at those nodes.


Step #2: Element Equations

• Let’s concentrate our attention to a single element.

• The same DE applies on the element level, hence, we may follow the procedure for weighted residual methods on the element level!

21

0

xxx

fcudx

dua

dx

d

21

2211

21

,

,,

Qdx

duaQ

dx

dua

uxuuxu

xxxx


Polynomial Approximation

• Now, we may propose an approximate solution for the primary variable, u(x), within that element.

• The simplest proposition would be a polynomial!



• Interpolating the values of displacement knowing the nodal displacements, we may write: 01 bxbxu

01111 bxbuxu 2

12

11

12

2 uxx

xxu

xx

xxxu

02122 bxbuxu



euxu

uuu

uxx

xxu

xx

xxxu

2

1212211

212

11

12

2


Step #2: Element Equations (cont’d)

• Assuming constant domain properties:

• Applying the Galerkin method:

21

2

2

0

xxx

fcudx

uda

02

2

Domain

jiijii

j dxfxuxxcudx

xdxa



• Note that:

• And:

ee hdx

xd

hdx

xd 1,

1 21

Domain

ijx

x

ij

Domain

ij

dxdx

xd

dx

xda

dx

xdxa

dxdx

xdxa

2

1

2

2



• For i=j=1: (and ignoring boundary terms)

• Which gives:

012

1

21

2

22

x

x eee

dxh

xxfu

h

xxc

ha

023 1

ee

e

fhu

ch

h

a



• Repeating for all terms:

• The above equation is called the element equation.

1

1

221

12

611

11

2

1 ee

e

fh

u

uch

h

a


What happens for adjacent elements?


Homework #2

• Derive the element equation without ignoring the boundary terms.

• What are differences in the element equation.

• The solution should be handed using the same report format (use equation editor to write your report).


Finite Element Procedure

1. Connecting Elements

2. Boundary Conditions

3. Solving Equations


Objectives

• Learn how the finite element model for the whole domain is assembled

• Learn how to apply boundary conditions

• Solving the system of linear equations


Recall

• In the previous lecture, we obtained the element equation that relates the element degrees of freedom to the externally applied fields

• Which maybe written:

1

1

221

12

611

11

2

1 ee

e

fh

u

uch

h

a

2

1

2

1

43

21

f

f

u

u

kk

kk


Two–Element example

1

2

11

12

11

14

13

12

11

f

f

u

u

kk

kk

2

2

21

22

21

24

23

22

21

f

f

u

u

kk

kk

3

2

1

3

2

1

3

2

1

24

23

22

21

14

13

12

11

0

0

Q

Q

Q

f

f

f

u

u

u

kk

kkkk

kk


Illustration: Bar application

1. Discretization: Divide the bar into N number of elements. The length of each element will be (L/N)

2. Derive the element equation from the differential equation for constant properties an externally applied force:

02

2

xFx

uEA

02

1

2

x

x

ijij

e

dxfudx

d

dx

d

h

EA


Performing Integration:

1

1

211

11

2

1 ee

e

e

fh

u

u

h

EA

Note that if the integration is evaluated from 0 to he, where he is the element length, the same results will be obtained.

02

1

2

x

x

ijij

e

dxfudx

d

dx

d

h

EA


Two–Element bar example

1

2

11

12

11

11

11

f

f

u

u

h

EA

e

2

2

21

22

21

11

11

f

f

u

u

h

EA

e

0

0

1

2

1

2110

121

011

3

2

1 Rfh

u

u

u

h

EA e

e


Applying Boundary Conditions


Applying BC’s

• For the bar with fixed left side and free right side, we may force the value of the left-displacement to be equal to zero:

0

0

1

2

1

2

0

110

121

011

3

2

Rfh

u

uh

EA e

e


Solving

• Removing the first row and column of the system of equations:

• Solving:

1

2

211

12

3

2 e

e

fh

u

u

h

EA

4

3

2

2

3

2

EA

fh

u

ue


Secondary Variables

• Using the values of the displacements obtained, we may get the value of the reaction force:

0

0

1

2

1

2

2

42

30

110

121

011 Rfh

fh

fh e

e

e


Secondary Variables

• Using the first equation, we get:

• Which is the exact value of the reaction force.

Rfhfh ee 22

3

efhR 2


Summary

• In this lecture, we learned how to assemble the global matrices of the finite element model; how to apply the boundary conditions, and solve the system of equations obtained.

• And finally, how to obtain the secondary variables.


Homework #3

• Problems #3.9 & 3.13 from the text book

• Write down a computer code that solves the problem for N elements.


Bars and Trusses


Objectives


Bar Example (Ex. 4.5.2, p. 187)

• Consider the bar shown in the above figure.• It is composed of two different parts. One steel tapered

part, and uniform Aluminum part.• Calculate the displacement field using finite element

method.


Bar Example

• The bar may be represented by two elements.

• The stiffness matrices of the two elements may be obtained using the following integration:

2

1

2

122

2221

2

1

11

11x

x

ee

ee

x

x

e dx

hh

hhxEAdx

dx

d

dx

d

dx

ddx

d

xEAK


Bar Example

• For the Aluminum bar: E=107 psi, and A=1 in2. we get:

• For the Steel bar: E=38107 psi, and A=(1.5-0.5x/96) in2. we get:

11

11

120

10

11

11

120

10 7

2

7 2

1

x

x

Al dxK

11

11

96

10.75.4

11

11

96

5.05.1

96

10.3 7

2

7 2

1

x

x

Fe dxx

K


Bar Example

• Assembling the Stiffness matrix and utilizing the external forces, we get:

• The boundary conditions may be applied and the system of equations solved.

0

0

10

10.2

0

33.833.80

33.88.575.49

05.495.49

105

5

3

2

14

R

u

u

u


Bar Example

• Solving, we get:

• For the secondary variables:

inu

u

181.0

061.0

3

2

lbR 30000


Reading Task

• Please read and understand examples, 4.5.1 & 4.5.3.


Trusses

• A truss is a set of bars that are connected at frictionless joints.

• The Truss bars are generally oriented in the plain.


Trusses

• Now, the problem lies in the transformation of the local displacements of the bar, which are always in the direction of the bar, to the global degrees of freedom that are generally oriented in the plain.


Equation of Motion

0

0

0000

0101

0000

0101

2

1

2

2

1

1

F

F

v

u

v

u

h

EA


Transformation Matrix

DOFdTransforme

DOFLocalv

u

v

u

CosSin

SinCos

CosSin

SinCos

v

u

v

u

2

2

1

1

2

2

1

1

00

00

00

00

DOF

dTransformeDOFLocal T


The Equation of Motion Becomes

• Substituting into the FEM:

• Transforming the forces:

• Finally:

FTK

FTTKT TT

FK


Recall

TKTK T

CosSin

SinCos

CosSin

SinCos

T

00

00

00

00

Where:

0000

0101

0000

0101

h

EAK


Element Stiffness Matrix in Global Coordinates

CosSin

SinCos

CosSin

SinCos

CosSin

SinCos

CosSin

SinCos

h

EAK

00

00

00

00

0000

0101

0000

0101

00

00

00

00


Element Stiffness Matrix in Global Coordinates

22

22

22

22

22

12

2

1

22

12

2

1

22

12

2

1

22

12

2

1

SinSinSinSin

SinCosSinCos

SinSinSinSin

SinCosSinCos

h

EAK


Example: 4.6.1 pp. 196-201

• Use the finite element analysis to find the displacements of node C.


Element Equations

0000

0101

0000

0101

1

L

EAK

1010

0000

1010

0000

2

L

EAK

3536.03536.03536.03536.0

3536.03536.03536.03536.0

3536.03536.03536.03536.0

3536.03536.03536.03536.0

3

L

EAK


Assembly Procedure

3536.13536.0103536.03536.0

3536.03536.0003536.03536.0

101000

000101

3536.03536.0003536.03536.0

3536.03536.0013536.03536.1

L

EAK


Global Force Vector

P

P

F

F

F

F

F

F

F

F

F

F

Fy

x

y

x

y

x

y

x

y

x

2

2

2

1

1

3

3

2

2

1

1

Remember!

NO distributed load is applied to a truss


Boundary Conditions

02211 VUVU

Remove the corresponding rows and columns

P

P

V

U

L

EA

23536.13536.0

3536.03536.0

3

3

Continue! (as before)


Results

EA

PLV

EA

PLU

3 ,828.5 33

PFF

PFPF

yx

yx

3 ,0

, ,

22

11


Postcomputation

e

e

e

e

e

A

P

A

P 21

e

e

eee

e

u

uL

EA

P

P

2

1

2

1

11

11

2

2

1

1

2

2

1

1

00

00

00

00

v

u

v

u

CosSin

SinCos

CosSin

SinCos

v

u

v

u


Postcomputation

A

P

A

P2 ,

3 ,0 )3()2()1(


Summary

• In this lecture we learned how to apply the finite element modeling technique to bar problems with general orientation in a plain.


Homework #5

• Problem 4.27, – Due 13/12/2006 before 9:00am

• Problem 4.44,– Due 20/12/2006 before 9:00am


Announcements

• Compensation Tutorial for E15:– Next Sunday 17/12/2006 3rd Period in H6

• Next Lecture:– Wednesday 20/12/2006 3rd Period in H6

• Next Quiz:– Wednesday 20/12/2006 3rd Period in H6 – (This Lecture is included)


Term Projects

• A problem has got to be solved using the finite element method

• A report is going to be presented by each group presenting the problem and its solution


The Report should contain:

• Cover page– Project Title– Names of team members

• Table of contents• Introduction and literature survey

– Introduction to the problem– Historical background and relevance of the problem– Papers and books that presented the problem– Latest achievements in the problem



• The finite element derivation– Governing equation– Derivation of the element matrices

• Using Glerkin method• Application of Symbolic manipulator to derive the

matrix equations will be appreciated

– Solution procedure



• The numerical results and verification– Program results– Verification of results compared to published results– Parametric study

• Discussion– Observations of the results– Further work that may be performed with the problem– Future developments of the model

• References


Evaluation

• Report (50%)

• Code (30%)– Structured: Functions built, easily modified– Readability: Organization, remarks– Length: The shorter the better

• Results (20%)


Projects

• Heat transfer in a 2-D heat sink

• 2-D flow around a blunt body in a wind tunnel

• Vibration characteristics of a pipe with internal fluid flow

• Panel flutter of a beam

• Rotating Timoshenko beam/blade


Heat transfer in a 2-D heat sink

• The heat sink will have heat flowing from one side

• Convection transfer on the surfaces

• Different boundary conditions on the other three sides

• Plot contours of temperature distribution with different boundary conditions


2-D flow around a blunt body in a wind tunnel

• Potential flow in a duct

• Rectangular body with different Dimensions

• Study the effect of the body size on the flow speed on both sides

• Plot contours of potential function, pressure, and velocity potential


Vibration characteristics of a pipe with internal fluid flow

• Study the change of the natural frequencies with the flow speed under different boundary conditions and fluid density

• Indicate the flow speeds at which instabilities occur


Panel flutter of a beam

• A fixed-fixed beam is subjected to flow over its surface

• Plot the effect of the flow speed on the natural frequencies of the beam

• Indicate the speed at which instability occurs


Rotating Timoshenko beam/blade

• Rotating beams undergo centrifugal tension that results in the change of its natural frequencies

• Study the effect of rotation speed on the beam natural frequencies and frequency response to excitations at the root


Teams

• 2-3 Students teams

• Names and selected projects should be submitted before 4PM on Thursday 21/12/2006


Work Progress• A report should be submitted By 4PM every Wednesday• 27/12/2006

– The report should contain a preliminary literature survey– Problem statement– Governing equations

• 10/1/2007– The report should contain a deeper literature survey– The preliminary derivations of the finite element model

• 17/1/2007– A more mature version of the report should be presented– Preliminary results of the code– List of the program script should be included

• 24/1/2007– Final version of the report should be presented together with the code


Beams and Frames


Beams and Frames

• Beams are the most-used structural elements.

• Many real structures may be approximated as beam elements

• Two main beam theories:– Euler-Bernoulli beam theory– Timoshenko beam theory


Euler-Bernoulli Beam Theory

• The main assumption in the Euler-Bernoulli beam theory is that the beam’s thickness is too small compared to the beam length

• That assumption resulted in that the sheer deformation of the beam may be neglected without much error in the analysis


Governing Equation

• The equation governing the deformation of and E-B beam under transverse loading may be written in the form:

)(2

2

2

2

xFdx

wdxEI

dx

d


The Thin-Beam Elements

• The thin beam element has a special feature, namely, the two degrees of freedom at each node are related.


Beam Interpolation Function

axHxw

34

2321)( xaxaxaaxw

axxxxw 321

axxaxHadx

xdH

dx

xdwx

23210



aHww 00 1

aT

a

a

a

a

lH

lH

H

H

w

w

w

w

x

x

4

3

2

1

2

2

1

1

0

0

'

'

aHww x 0'0' 1

alHwlw 2

alHwlw x 2''



4

3

2

1

2

32

2

2

1

1

3210

1

0010

0001

'

'

a

a

a

a

ll

lll

w

w

w

w

2

2

1

1

2323

22

4

3

2

1

'

'

1212

13230010

0001

w

w

w

w

llll

llll

a

a

a

a



ee wxNwTxHaxHxw 1

ewTa 1

4

1iii wxNxw



2

32

3

3

2

2

3

32

3

3

2

2

23

2

231

l

x

l

xl

x

l

xl

x

l

xx

l

x

l

x

xNxN T


Interpolation Functions

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

X

N(x

)

N1

N2

N3

N4


Beam Stiffness Matrix

• The governing equation is:

• Using the series solution

)(2

2

2

2

xFdx

wdxEI

dx

d

4

1iii wxNxw



• The governing equation becomes

• Applying Galerkin method:

)()(4

12

2

2

2

xRxFwdx

NdxEI

dx

d

ii

i

ee l

ji

ii

l

j dxNxFwdx

NdxEI

dx

ddxNxR

0

4

12

2

2

2

0

)()(



• Using integration by parts, twice, and ignoring the boundary terms, we get:

• In matrix form:

0)(0

4

12

2

2

2

el

ji

iji dxNxFw

dx

Nd

dx

NdxEI

ee l

xxe

l

xxxx dxNxFwdxNNxEI00

)(


Use of Symbolic Manipulator

Beam Example


Optional Homework #6

• Derive the expression for the interpolation function for a beam in terms of nodal displacements and slopes.

• Try to use a symbolic manipulator to generate the expressions.

)(4

4

2

2

xFdx

wdEI

dt

wdA


Two Dimensional Elements


2-D Elements

• In this section, we will be introduced to two dimensional elements with single degree of freedom per node.

• Detailed attention will be paid to rectangular elements.


For the 2-D BV Problem

• Let’s consider a problem with a single dependent variable

• We may set one degree of freedom to each node; say fi.

• Further, let’s only consider a rectangular element that is aligned with the physical coordinates


A Rectangular Element

• For the approximation of a general function f(x,y) over the element you need a 2-D interpolation function

xyayaxaayxf 4321,


Let’s follow the same procedure!


2-D Interpolation Function

ayxHyxf ,, xyayaxaayxf 4321),(

aHff 0,00,0 1

aT

a

a

a

a

bH

baH

aH

H

f

f

f

f

4

3

2

1

4

3

2

1

0

,

0,

0,0

aaHfaf ,00, 2 abaHfbaf ,, 3 abHfbf ,0,0 4


2-D Interpolation Function

4

3

2

1

4

3

2

1

001

1

001

0001

a

a

a

a

b

abba

a

f

f

f

f

4

3

2

1

4

3

2

1

1111

100

1

0011

0001

f

f

f

f

abababab

bb

aa

a

a

a

a


2-D Interpolation Function efyxNayxHyxf ,,,

ab

xy

b

yab

xyab

xy

a

xab

xy

b

y

a

x

yxNyxN T

1

,,


How does this look like?


2-D Interpolation Functions

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.3

0.6

0.9

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

N1

x

y

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.3

0.6

0.9

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

N2

x

y


2-D Interpolation Functions

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.3

0.6

0.9

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

N3

x

y

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.3

0.6

0.9

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

N4

x

y


Example: Laplace Equation

02

02

2

2

2

yx

ei

ii yxNyxN ,,4

1


Example: Laplace Equation

ei

ii yxNyxN ,,4

1

0 e

Area

yyxx dANNNN

Applying the Galerkin method and integrating by parts, the element equation becomes


The Element Equaiton

0

222

222

222

222

6

1

22222222

22222222

22222222

22222222

e

babababa

babababa

babababa

babababa

ab


The Logistic Problem!


The Logistic Problem

• In the 2-D problems, the numbering scheme, usually, is not as straight forward as the 1-D problem


1-D Example

• Element #1 is associated with nodes 1&2• Element #2 is associated with nodes 2&3, etc…


2-D Example


For Element #5

Local Node NumberGlobal Node Number

15

26

39

48


Contribution of element #5 to global matrix

123456789101112

1

2

3

4

51,11,21,41,3

62,12,22,42,3

7

84,14,24,44,3

93,13,23,43,3

10

11

12


A Solution for the Logistics’ Problem

• One solution of the logistic problem is to keep a record of elements and the mapping of the local numbering scheme to the global numbering scheme in a table!


Elements Register: Global Numbering

Element Number

Node Number

1234

11254

24587

3781110

42365

55698

6891211


Algorithm for Assembling Global Matrix

1. Create a square matrix “A”; N*N (N=Number of nodes)

2. For the ith element3. Get the element matrix “B”4. For the jth node5. Get its global number k6. For the mth node7. Get its global number n

8. Let Akn=Akn+Bjm

9. Repeat for all m10. Repeat for all j11. Repeat for all i

Element Number

Node Number

1234

11254

24587

3781110

42365

55698

6891211

123456789101112

1

2

3

4

5

6

7

8

9

10

11

12

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