Handout4-PPT
15
30/10/2014 1 HANDOUT 4 Isoparametric FE Formulation of 2D Elements Let e e a H u T i e u u u u ] ,..., , , [ 3 2 1 a , 0 i u V
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Transcript of Handout4-PPT
30/10/2014
1
HANDOUT 4
Isoparametric FE Formulation of 2D Elements
Let
eeaHu Ti
e uuuu ],...,,,[ 321a
,0 iuV
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Quadrilateral element
44332211),( xhxhxhxhsrx
44332211),( yhyhyhyhsry
44332211),( xxxxx uhuhuhuhsru
44332211),( yyyyy uhuhuhuhsru
e
y
xe
sru
srusr Hau
),(
),(),(
4321
4321
2
1
0000
0000
hhhh
hhhh
H
HH
Tyxyxyxyxe uuuuuuuu 44332211 ,,,,,,,a
In matrix format
)1)(1(4
11 srh
)1)(1(4
12 srh
)1)(1(4
13 srh
)1)(1(4
14 srh
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J is the determinant ofthe Jacobian .
Exercise
Calculate the Jacobian matrix for the above mentioned element
44332211),( xhxhxhxhsrx
44332211),( yhyhyhyhsry
)1)(1(4
11 srh )1)(1(
4
12 srh
)1)(1(4
13 srh )1)(1(
4
14 srh
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Calculating matrix B (1)
eee aBε
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Calculating matrix B (2)
44332211),( xxxxx uhuhuhuhsru
44332211),( yyyyy uhuhuhuhsru
Calculating matrix B (3)
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Calculating matrix B (4)
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Numerical Integration (Quadrature)
1
1 1 2 2
1
( ) ( ) ( ) ..... ( )n nf r dr f r f r f r
n=2e.g., 1
1
( ) 1.0 (0.5773502692) 1.0 ( 0.5773502692)f r dr f f
1
1
( ) 0.5555555556 (0.7745966692) 0.5555555556 ( 0.7745966692)
0.8888888889 (0)
f r dr f f
f
n=3
Gauss – Legendre integration in one dimension
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n
i
n
jjiji srfdsdrsrf
1 1
1
1
1
1
),(),(
Exercise
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Example
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dss
y
s
xd
22 )()(
dsd 5
1
1
520
55
)1(2
10
0)1(2
100
00
00
00
)1(2
10
0)1(2
1
dss
s
s
s
s
eF
