FEM Introduction Mohammed Bakkesin

8/16/2019 FEM Introduction Mohammed Bakkesin

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Introduction to the Finite Element Method

Dr. Mohammad Tawfik

Introduction to the Finite

Element Method

Spring 2010


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Course Objectives

• The student should be capable of writing

simple programs to solve different

problems using finite element method.


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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)


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Fundamental Course Agreement

• Homework is sent in electronic format (Nohardcopies 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


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References

• J.N. Reddy, “An Introduction to the FiniteElement Method” 3rd ed., McGraw Hill, ISBN007-124473-5

• D.V. Hutton, “Fundamentals of Finite Element

Analysis” 1st ed., McGraw Hill, ISBN 007-121857-2

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

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


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Numerical Solution of

Boundary Value Problems

Weighted Residual Methods


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Objectives

• In this section we will be introduced to thegeneral classification of approximatemethods

• Special attention will be paid for theweighted residual method

• Derivation of a system of linear equations

to approximate the solution of an ODE willbe presented using different techniques


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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!


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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


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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


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Basic Concepts

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

x g x f L

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

n

i

ii xa x f 1


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Basic Concepts

• Applying the differential operator on the approximatesolution, you get:

01

1

x g x La

x g xa L x g x f L

n

i

ii

n

i

ii

x R x g x Lan

i

ii 1

Residue


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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.


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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


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Collocation Method

0 j x R

01

j

n

i

jii

j

x F x La

x R


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Example Problem


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The bar tensile problem

02

2

x F

x

u EA


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Bar application

02

2

x F

x

u EA

n

i

ii xa xu1

x R x F

dx

xd a EA

n

i

i

i

12

2

Applying the collocation method

01

2

2

j

n

i

ji

i x F dx

xd a EA


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In Matrix Form

nnnnnn

n

n

x F

x F

x F

a

a

a

k k k

k k k

k k k

2

1

2

1

21

22212

12111

...

...

...

Solve the above system for the “generalized

coordinates” ai to get the solution for u(x)

j x x

iij

dx

xd EAk

2

2


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Notes on the trial functions

• They should be at least twice

differentiable!

• They should satisfy all boundary

conditions!

• Those are called the “ Admissibility

Conditions”.


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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

xSin x

2


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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

l EA

dx

xd EA

22

2

2

2

f aSinl

EA

1

2

42


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Solving:

• Then, the approximate

solution for this problem is:

• Which gives the maximum

displacement to be:

• And maximum strain to be:

EA f l

EA

f l

Sinl EA

f a2

2

2

21 57.0

24

42

l

xSin EA

f l xu 257.0

2

5.057.02

exact EA

f l l u

0.19.00 exact EA

lf u x


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The Subdomain Method (free

reading)

• The idea behind the

subdomain method is

to force the integral

of the residue to beequal to zero on an

subinterval of the

domain


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The Subdomain Method

01

j

j

x

x

dx x R

011

1

j

j

j

j

x

x

n

i

x

x

ii dx x g dx x La


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Bar application

02

2

x F

x

u EA

n

i

ii xa xu1

x R x F

dx

xd a EA

n

i

i

i

12

2

Applying the subdomain method

11

1

2

2 j

j

j

j

x

x

n

i

x

x

ii dx x F dx

dx

xd a EA


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In Matrix Form

11

2

2 j

j

j

j

x

x

i

x

x

i dx x F adxdx

xd EA




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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 performedover the whole domain!!!

• Actually, it turned out to be a VERY

GOOD idea


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The Galerkin Method

0 Domain

j dx x x R

01

Domain

j

n

i Domain

i ji dx x g xdx x L xa


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In Matrix Form

Domain

ji

Domain

i j dx x F xadx

dx

xd x EA

2

2




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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

x

Sin x 2


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Substituting with the approximate

solution:

Domain

j

n

i Domain

i ji dx x F xdx

dx

xd xa EA

1

2

2

l

l

fdxl

xSin

dxl

xSin

l

xSina

l EA

0

0

1

2

2

222

l l a

l EA

2

22 1

2

EA

fl l

EA

f a

2

3

2

1 52.016


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Substituting with the approximate

solution: (Int. by Parts)

Domain

j

n

i Domain

i ji dx x F xdx

dx

xd xa EA

1

2

2

l l a

l EA

2

22 1

2

EA

fl l

EA

f a

2

3

2

1 52.016

Domain

i j

l

i j

Domain

i j

dxdx

xd

dx

xd

dx

xd x

dx

dx

xd x

0

2

2

Zero!


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What did we gain?

• The functions are required to be less

differentiable

• Not all boundary conditions need to be

satisfied

• The matrix became symmetric!


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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.


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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.


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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.


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Homework #1

• Solve the beam bendingproblem, for beamdisplacement, for a simplysupported beam with a loadplaced at the center of the

beam using – Collocation Method

– Subdomain Method

– Galerkin Method

• Use three term Sin series thatsatisfies all BC’s

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

)(4

4

x F dx

wd

0)()0(

0)()0(

2

2

2

2

dx

l wd

dx

wd

l ww


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Exact Solution

12/110

3

15

7

412

2/1060

13

12)(

23

3

x

x x x

x

x x

xw


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The Finite Element Method

2nd order DE’s in 1-D


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Objectives

• Understand the basic steps of the finite

element analysis

• Apply the finite element method to second

order differential equations in 1-D


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The Mathematical Model

• Solve:

• Subject to:

L x

f cudx

dua

dx

d

0

0

00 ,0 Qdxdu

auu L x


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Step #1: Discretization

• At this step, we divide

the domain into

elements.

• The elements areconnected at nodes.

• All properties of the

domain are defined at

those nodes.


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Step #2: Element Equations

• Let’s concentrate ourattention to a singleelement.

• The same DE applieson the element level,hence, we may followthe procedure forweighted residual

methods on theelement level!

21

0

x x x

f cu

dx

dua

dx

d

21

2211

21

,

,,

QdxduaQ

dxdua

u xuu xu

x x x x


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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!


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• Interpolating the values

of displacement

knowing the nodal

displacements, we maywrite:

01 b xb xu

01111 b xbu xu

212

11

12

2 u x x x xu

x x x x xu

02122 b xbu xu


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eu xu

uuu

u x x

x x

u x x

x x

xu

2

1

212211

2

12

1

1

12

2

St #2 El t E ti


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(cont’d)

• Assuming constant

domain properties:

• Applying the

Galerkin method:

21

2

2

0

x x x

f cudxud a

022

Domain

jii jii

j dx f xu x xcudx

xd xa

St #2 El t E ti


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(cont’d)

• Note that:

• And:

ee hdx

xd

hdx

xd 1

,

1 21

Domain

i j

x

x

i

j

Domain

i j

dxdx

xd

dx

xd a

dx

xd xa

dxdx

xd xa

2

1

2

2

St #2 El t E ti


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(cont’d)

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

• Which gives:

01

2

1

2

1

2

2

2

x

x eee

dxh

x x f u

h

x xc

h

a

023

1

ee

e

fhuchha

St #2 El t E ti


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(cont’d)

• Repeating for all terms:

• The above equation is called the element

equat ion .

1

1

221

12

611

11

2

1 ee

e

fh

u

uch

h

a


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What happens for adjacent

elements?


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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).


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Finite Element Procedure

1. Connecting Elements

2. Boundary Conditions3. Solving Equations


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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


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Recall

• In the previous lecture, we obtained the

element equation that relates the element

degrees of freedom to the externally

applied f ields

• 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

uu

k k k k


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Two –Element example

1

2

1

1

1

2

1

1

1

4

1

3

1

2

1

1

f f

uu

k k k k

2

2

2

1

2

2

2

1

2

4

2

3

2

2

2

1

f

f

u

u

k k

k k

3

2

1

3

2

1

3

2

1

2

4

2

3

2

2

2

1

1

4

1

3

1

2

1

1

0

0

QQ

Q

f f

f

uu

u

k k k k k k

k k


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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 differentialequation for constant properties an externally

applied force:

02

2

x F x

u EA

02

1

2

x

x

i ji j

e

dx f udx

d

dx

d

h

EA


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Performing Integration:

1

1

211

11

2

1 e

e

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

i ji j

e

dx f udx

d

dx

d

h

EA


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Two –Element bar example

1

2

1

1

1

2

1

1

11

11

f f

uu

h EA

e

2

2

2

1

2

2

2

1

11

11

f

f

u

u

h

EA

e

0

0

1

2

1

2110

121

011

3

2

1 R

fh

uu

u

h

EA e

e


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Applying Boundary Conditions


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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

R fh

u

uh

EA e

e


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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


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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

4

2

3

0

110

121

011 R fh

fh

fh e

e

e


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Secondary Variables

• Using the first equation, we get:

• Which is the exact value of the reactionforce.

R fh fh ee

22

3

e fh R 2


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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 ofequations obtained.

• And finally, how to obtain the secondary

variables.


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Homework #3

• Problems #3.9 & 3.13 from the text book

• Write down a computer code that solves

the problem for N elements.


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Bars and Trusses


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Objectives


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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 elementmethod.


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Bar Example

• The bar may be represented by two

elements.

• The stiffness matrices of the two elements

may be obtained using the followingintegration:

2

1

2

122

22

21

2

1

11

11 x

x

ee

ee

x

x

e dx

hh

hh

x EAdxdx

d

dx

d

dx

d dx

d

x EA K


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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:

1111

120

10

11

11

120

10 7

2

7 2

1

x

x Al

dx K

11

11

96

10.75.4

11

11

96

5.05.1

96

10.3 7

2

7 2

1

x

x

Fe dx x

K


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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

10

5

5

3

2

1

4

R

u

u

u


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Bar Example

• Solving, we get:

• For the secondary

variables:

inu

u

181.0

061.0

3

2

lb R 30000


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Reading Task

• Please read and understand examples,4.5.1 & 4.5.3.


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Trusses

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

• The Truss bars are generally oriented in

the plain.


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Trusses

• Now, the problem lies in thetransformation of the local displacements

of the bar, which are always in the

direction of the bar, to the global degreesof freedom that are generally oriented in

the plain.

E i f M i


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Equation of Motion

0

0

0000

01010000

0101

2

1

2

2

1

1

F

F

v

u

v

u

h

EA

T f ti M t i


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Transformation Matrix

DOF d Transforme DOF Local v

u

v

u

CosSin

SinCos

CosSin

SinCos

v

u

v

u

2

2

1

1

2

2

1

1

00

00

00

00

DOF

d Transforme DOF Local T

Th E i f M i B


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The Equation of Motion Becomes

• Substituting into theFEM:

• Transforming the

forces:

• Finally:

F T K

F T T K T T T

F K

R ll


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Recall

T K T K T

CosSin

SinCos

CosSin

SinCos

T

00

00

00

00

Where:

0000

0101

0000

0101

h

EA K

Element Stiffness Matrix in Global


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Coordinates

CosSinSinCos

CosSin

SinCos

CosSin

SinCos

CosSin

SinCos

h

EA K

00

00

00

00

0000

0101

0000

0101

00

00

00

00



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Coordinates

22

22

22

22

22

12

2

1

22

12

2

1

22

1

22

1

22

12

2

1

SinSinSinSin

SinCosSinCos

SinSinSinSin

SinCosSinCos

h

EA K

E l 4 6 1 196 201


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Example: 4.6.1 pp. 196-201

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

El t E ti


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Element Equations

0000

0101

0000

0101

1

L

EA K

1010

0000

1010

0000

2

L

EA K

3536.03536.03536.03536.03536.03536.03536.03536.0

3536.03536.03536.03536.0

3536.03536.03536.03536.0

3

L

EA K

A bl P d


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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 EA K

Gl b l F V t


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Global Force Vector

P

P

F F

F

F

F

F

F

F

F F

F y

x

y

x

y

x

y

x

y

x

2

2

2

1

1

3

3

2

2

1

1

Remember!

NO distributed loadis applied to a truss

B d C diti


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Boundary Conditions

02211

V U V U

Remove the corresponding rows and columns

P

P

V

U

L

EA

23536.13536.0

3536.03536.0

3

3

Continue! (as before)

R lt


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Dr. Mohammad Tawfik

Results

EA

PLV

EA

PLU

3 ,828.5

33

P F F

P F P F

y x

y x

3 ,0

, ,

22

11

Postcomp tation


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Postcomputation

e

e

e

ee

A

P

A

P 21

e

e

ee

e

e

u

u

L

E A

P

P

2

1

2

1

11

11

2

2

1

1

2

2

1

1

00

00

00

00

v

u

vu

CosSin

SinCos

CosSinSinCos

v

u

vu

Postcomputation


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Postcomputation

A

P

A

P 2 ,

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

Summary


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Summary

• In this lecture we learned how to apply thefinite element modeling technique to bar

problems with general orientation in a

plain.

Homework #5


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Homework #5

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

• Problem 4.44,

– Due 20/12/2006 before 9:00am

Announcements


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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


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Term Projects

• A problem has got to be solved using thefinite element method

• A report is going to be presented by each

group presenting the problem and itssolution

The Report should contain:


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• 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



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• 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



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• 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


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Evaluation

• Report (50%)

• Code (30%)

– Structured: Functions built, easily modified

– Readability: Organization, remarks

– Length: The shorter the better

• Results (20%)

Projects


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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 withinternal fluid flow

• Panel flutter of a beam

• Rotating Timoshenko beam/blade

Heat transfer in a 2 D heat sink


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Heat transfer in a 2-D heat sink

• The heat sink will have heat flowing fromone side

• Convection transfer on the surfaces

• Different boundary conditions on the otherthree sides

• Plot contours of temperature distribution

with different boundary conditions

2-D flow around a blunt body in a


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wind tunnel

• Potential flow in a duct

• Rectangular body with different

Dimensions

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

• Plot contours of potential function,

pressure, and velocity potential

Vibration characteristics of a pipe


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with internal fluid flow

• Study the change of the naturalfrequencies with the flow speed under

different boundary conditions and fluid

density• Indicate the flow speeds at which

instabilities occur

Panel flutter of a beam


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Panel flutter of a beam

• A fixed-fixed beam is subjected to flowover 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


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Rotating Timoshenko beam/blade

• Rotating beams undergo centrifugaltension that results in the change of its

natural frequencies

• Study the effect of rotation speed on thebeam natural frequencies and frequency

response to excitations at the root

Teams


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Teams

• 2-3 Students teams

• Names and selected projects should be

submitted before 4PM on Thursday

21/12/2006

Work Progress


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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


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Beams and Frames

Beams and Frames


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Beams and Frames

• Beams are the most-used structuralelements.

• Many real structures may be approximated

as beam elements• Two main beam theories:

– Euler-Bernoulli beam theory

– Timoshenko beam theory

Euler-Bernoulli Beam Theory


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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 theanalysis

Governing Equation


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Governing Equation

• The equation governing the deformation ofand E-B beam under transverse loading

may be written in the form:

)(22

2

2

x F dxwd x EI

dxd

The Thin-Beam Elements


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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


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a x H xw

34

2321)( xa xa xaa xw

a x x x xw 321

a x xa x H adx

xdH

dx

xdw

x

2

3210



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a H ww 001

aT

a

a

aa

l H

l H

H H

w

w

ww

x

x

4

3

2

1

2

2

1

1

0

0

'

'

a H ww x 0'0' 1 al H wl w

2

al H wl w x 2''



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Dr. Mohammad Tawfik


4

3

2

1

2

32

2

2

1

1

3210

1

0010

0001

'

'

a

a

a

a

l l

l l l

w

w

w

w

2

2

1

1

2323

22

4

3

2

1

'

'

1212

1323

0010

0001

w

w

w

w

l l l l

l l l l

a

a

a

a



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Dr. Mohammad Tawfik


ee w x N wT x H a x H xw 1

ewT a 1

4

1i

ii w x N xw



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Dr. Mohammad Tawfik


2

32

3

3

2

2

3

32

3

3

2

2

23

2

231

l

x

l

x

l

x

l

x l

x

l

x x

l

x

l

x

x N x N T

Interpolation Functions


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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


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• The governing equation is:

• Using the series solution

)(2

2

2

2

x F dx

wd x EI

dx

d

4

1i

ii w x N xw



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• The governing equation becomes

• Applying Galerkin method:

)()(4

1

2

2

2

2

x R x F wdx

N d x EI

dx

d

i

i

i

ee l

j

i

ii

l

j dx N x F wdx N d x EI

dxd dx N x R

0

4

12

2

2

2

0

)()(



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

• In matrix form: 0)(0

4

12

2

2

2

el

ji

i

ji

dx N x F wdx

N d

dx

N d

x EI

ee l

xx

e

l

xx xx dx N x F wdx N N x EI 00

)(


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Dr. Mohammad Tawfik

Use of Symbolic Manipulator

Beam Example

Optional Homework #6

http://localhost/var/www/apps/conversion/tmp/scratch_4/BeamProgram.pdfhttp://localhost/var/www/apps/conversion/tmp/scratch_4/BeamProgram.pdf


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Optional Homework #6

• Derive the expression for the interpolationfunction for a beam in terms of nodal

displacements and slopes.

• Try to use a symbolic manipulator togenerate the expressions.

)(4

4

2

2

x F dxwd EI

dt wd A


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Two Dimensional Elements

2-D Elements


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• In this section, we will be introduced to twodimensional elements with single degree

of freedom per node.

• Detailed attention will be paid torectangular elements.

For the 2-D BV Problem


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• Let’s consider a problem with a singledependent variable

• We may set one degree of freedom to

each node; say f i.• Further, let’s only consider a rectangular

element that is aligned with the physical

coordinates

A Rectangular Element


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g

• For the approximationof a general function

f(x,y) over the element

you need a 2-D

interpolation function

xya ya xaa y x f 4321,


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Let’s follow the same

procedure!

2-D Interpolation Function


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p

a y x H y x f ,, xya ya xaa y xf 4321),(

a H f f 0,00,0 1

aT

aa

a

a

b H ba H

a H

H

f f

f

f

4

3

2

1

4

3

2

1

0,

0,

0,0

aa H f a f ,00, 2 aba H f ba f ,, 3 ab H f b f ,0,0 4



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p

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

00

11

0001

f

f

f f

abababab

bb

aa

a

a

aa



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p

e f y x N a y x H y x f ,,,

ab xy

b y

ab

xyab

xy

a

xab

xy

b

y

a

x

y x N y x N T

1

,,


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How does this look like?

2-D Interpolation Functions


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p

0 0.1 0.2 0.3 0.4 0.50.6 0.7 0.8 0.9 1

0

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 0.50.6 0.7 0.8 0.9 1

0

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


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p

0 0.1 0.2 0.3 0.4 0.50.6 0.7 0.8 0.9 1

0

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 0.50.6 0.7 0.8 0.9 1

0

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


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p p q

02

02

2

2

2

y x

ei

ii y x N y x N ,,4

1

Example: Laplace Equation


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Dr. Mohammad Tawfik

p p q

ei

ii y x N y x N ,,4

1

0 e

Area

y y x x dA N N N N

Applying the Galerkin method and integrating by parts,

the element equation becomes

The Element Equaiton


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Dr. Mohammad Tawfik

0222

222

222

222

6

1

22222222

22222222

22222222

22222222

e

babababa

babababa

babababa

babababa

ab


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The Logistic Problem!

The Logistic Problem


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• In the 2-D problems, the numberingscheme, usually, is not as straight forward

as the 1-D problem

1-D Example


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

2-D Example


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2-D Example


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For Element #5


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Global Node Number Local Node Number

51

62

93

84

Contribution of element #5 to global

matrix


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matrix121110987654321

1

2

3

4

1,31,41,21,15

2,32,42,22,16

7

4,34,44,24,18

3,33,43,23,19

10

11

12

A Solution for the Logistics’

Problem


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Problem

• One solution of the logistic problem is tokeep a record of elements and the

mapping of the local numbering scheme to

the global numbering scheme in a table!

Elements Register: Global

Numbering


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Numbering

Node Number ElementNumber 4321

45211

785421011873

56324

896551112986

Algorithm for Assembling Global

Matrix


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Matrix

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

2. For the ith element

3. Get the element matrix “B”

4. For the jth node

5. Get its global number k6. For the mth node

7. Get its global number n

8. Let Akn=Akn+B jm9. Repeat for all m

10. Repeat for all j11. Repeat for all i

Node Number Element

Number 4321

45211

78542

1011873

56324

89655

1112986

121110987654321

1

2

3

4

5

6

7

8

9

10

11

12

FEM Introduction Mohammed Bakkesin

Documents