FEM: Element Equations

2nd order DE’s in 1-D

Mohammad Tawfik #WikiCourses

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Introduction to the Finite

Element Method





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

1

212211

2

12

11

12

2





(cont’d)

• Assuming constant

domain properties:

• Applying the

Galerkin method:

21

2

2

0

xxx

fcudx

uda

02

2

Domain

jiijii

j dxfxuxxcudx

xdxa





(cont’d)

• Note that:

• And:

ee hdx

xd

hdx

xd 1,

1 21

Domain

ij

x

x

ij

Domain

ij

dxdx

xd

dx

xda

dx

xdxa

dxdx

xdxa

2

1

2

2





(cont’d)

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

• Which gives:

012

1

21

2

2

2

x

x eee

dxh

xxfu

h

xxc

ha

023

1

ee

e

fhu

ch

h

a





(cont’d)

• 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?




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

1

1

1

2

1

1

1

4

1

3

1

2

1

1

f

f

u

u

kk

kk

2

2

2

1

2

2

2

1

2

4

2

3

2

2

2

1

f

f

u

u

kk

kk

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

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

xF

x

uEA

02

1

2

x

x

ijij

e

dxfudx

d

dx

d

h

EA




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

ijij

e

dxfudx

d

dx

d

h

EA




Two–Element bar example

1

2

1

1

1

2

1

1

11

11

f

f

u

u

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 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.

Education

FEM: Element Equations