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Intro to FEM (continued)

Examples of using FEM to solve a problem and comparison with exact solution:

We consider a problem already discussed in the previous class:

g

x

Solution by exact method:

( )xLxE

gu =

2

( )xLg

22 =

0 L

u

2L

2

gL

2

gL

L

2

L0

Now we will solve the same problem by FEM:

In FEM, the displacement is discretized as and the governing equation is

reduced to algebraic equation:

=

=n

k

kk xwuxu1

)()(

FKU = or jn

k

kjk FuK ==1

where , (see notes of( ) ( )=L

kjjk xxwxwEK 0''

d ( ) ( ) xxwgxxwfFL

j

L

jj dd 00 ==

1

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previous lecture).

FEM nodes:

j

1

jw

0 L1j 1+j

( )

[ ]

[ ]

= ++

+

otherwise,0

,,

,,

1

1

1

1

1

1

jj

jj

j

jj

jj

j

j xxx

xx

xx

xxxxx

xx

xw

Case study 1: In the simplest possible case, we choose only one node, i.e. 1=n . In this case, wehave one node and 2 elements.

2/L

( ) ( )xwuxu11

=

( )

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1111 FuK =

( )L

Ex

LExxwEK

LL

4d2

d0

2

0

2'111 =

==

( )2

d0

11

gLxxwgF

L ==

E

gL

K

Fu

8

2

11

11

==

0 L

u

2L

( )xLx

E

gu =

2

Exact solution: ( )E

gLLL

L

E

gLxu

82222

2

=

==

FEM solution for stress:

( ) ( ) ( )

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x

pointsnintegratio

Remarks:

1. ( )elementFEM

x = (Stress calculated using FEM corresponding to an average of stress over

an element).

2. Stress is not well defined at the nodal points (taking on different values depending on which

side of the node). It is better to evaluate stress at integration points.

Case study 2: With slightly more sophistication and still within the possibility of doing calculation

by hand, we can choose two nodes 2=n .

3

L

3

2L

( ) ( ) ( )xwuxwuxu 2211 +=

( )

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( )

( )

( )

( )

=

LL

LLL

LL

xw

,32,0

32,3,3

3,0,3

'1 , ( )

( )

( )

( )

=

LLL

LLL

L

xw

,32,

3

32,3,3

3,0,0

'2

1w 2w

=

2221

1211

KK

KKK

( )L

Ex

LExxwEK

LL 6d

3d

32

0

2

0

2'111 =

==

( )L

Ex

LExxwEK

L

L

L 6d

3d

3

2

0

2'222 =

==

( ) ( ) 2132

30

'2

'112

3d

33d K

L

Ex

LLExxwxwEK

L

L

L

==

==

=

21

123

L

EK

(Comment: The stiffness matrix is highly banded and sparse, and in some cases can be easily

determined. If for the present problem, we can extract the above calculations todetermine

100=n

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=

21

121

121

12

OOOnCK

0

0

)

Nodal forces:3

21

gLFF

==

FKU =

=

=

=

1

1

91

1

21

12

3

1

9

1

1

321

123

22

2

1

2

1

E

gL

E

gL

u

u

gL

u

u

L

E

E

gLuu

9

2

21

==

Exact solution: ( )E

gLLL

L

E

gLxu

93323

2

=

==

( ) EgLL

LL

E

gLxu 93

2

3

2

232

2

=

==

6

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0 L

u( )xLx

E

gu =

2

FEM solution for stress:

( ) ( ) ( ) ( )xwEuxwEuxEux '22'11

' +==

xL0 1 2

2gL

3gL

pointsnintegratio

Brief overview of FEM in 2D and 3D (read course notes Section 1.2 on the web):

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Node number (integer);

Nodal displacement (vector);

(can also include other physical quantities e.g. temperature, pressure, etc.)

Elements: associated with a number of nodes;

Element connectivity;

Type of elements:

Linear elements: Quadratic elements:

3 nodal triangle (linear) 6 nodal triangle

( ) yaxaayxw 321, ++= ( )

26

254

321 4,

yaxaxya

yaxaayxw

+++

++=

8