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Chapter 2 – The Basic FEA Procedure
k
2
= A
2
E/L
2
= 2.8125
×
10
7
lb./in.
k
3
= A
3
E/L
3
= 7.95
×
10
6
lb./in.
Element Stiffness Equations
[
K
(1)
] = 43.295
×
10
7
1 -1
-1 1
Similarly,
[
K
(2)
] = 28.125
×
10
6
1 -1
-1 1
[
K
(3)
] = 7.9500
×
10
6
1 -1
-1 1
Global stiffness matrix is
43.295 -43.295 0
0
[K
g
] = -43.295 43.295+28.125 -28.125 0
×
10
6
0 -28.125 28.125+7.95 -7.95
0
0
-7.95 7.95
Now the global structural equations can be written as,
43.295 -43.295 0 0 u
1
F1
10
6
×
-43.295 71.42 -28.125 0
u
2
= F2
0 -28.125 36.075 -7.95
u
3
F3
0
0 -7.95 7.95
u
4
F4
ME 273 Lecture Notes © by R. B. Agarwal
Finite Element Analysis
2
-21
Chapter 2 – The Basic FEA Procedure
Applying the boundary conditions: u
1
= 0, and F
1
= F
2
= F
3
= 0, F
4
= 5000 lb., results in
the reduced matrix,
71.42 -28.125 0 u
2
0
10
6
×
-28.125 36.075 -7.95 u
3
= 0
0 -7.95 7.95 u
4
5000
Solving we get,
u
2
0.0012
u
3
= 0.0029 in.
u
4
0.0092
ME 273 Lecture Notes © by R. B. Agarwal
Finite Element Analysis
2
-22