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EN0175 09 / 12 / 06
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
=
==
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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