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Solution of the system of equations
KV
=
P
gives
−
0
.
15482228
(
U
2
)
(
2
)
V
=
(13)
0
.
01214497
Determination of the Member Stress Resultants
From these two nodal values, the displacements and rotations at the element edges can be
determined.
The forces and moments for the elements can be obtained using
p
i
k
i
v
i
p
i
0
=
−
(14)
For element 1, from (13)
0
.
0
(
u
X
1
)
.
(θ
)
0
0
Y
1
v
1
=
(15)
−
0
.
1542228
(
u
X
2
)
0
.
01214497
(θ
)
Y
2
From (3) and (15)
V
1
F
X
1
199
.
535
M
1
M
Y
1
−
1012
.
15
k
1
v
1
V
2
F
X
2
M
2
M
Y
2
=
(16)
−
.
199
535
−
756
.
968
Introduce (4)
V
1
F
X
1
+
=
199
.
535
36
.
00
235
.
535
M
1
M
Y
1
−
1012
.
15
−
48
.
00
−
1060
.
152
V
2
F
X
2
(17)
−
199
.
535
36
.
00
−
163
.
535
M
2
M
Y
2
−
756
.
968
48
.
00
−
708
.
968
k
1
v
1
p
10
p
1
−
=
For element 2
0
.
0
(w
2
)(w
Z
2
)
(θ
2
)(θ
Y
2
)
(w
3
)(w
Z
3
)
v
2
=
0
.
01214497
0
(18)
.
0
From (5) and (18)
V
2
F
Z
2
M
2
M
Y
2
V
3
F
Z
3
−
63
.
3968
k
2
v
2
=
633
.
9675
(19)
63
.
3968
+
=
V
2
F
Z
2
M
2
M
Y
2
V
3
F
Z
3
−
63
.
3968
−
37
.
5
−
100
.
897
633
.
9675
75
.
0
708
.
968
(20)
63
.
3968
−
22
.
5
40
.
897
k
2
v
2
p
20
p
2
−
=
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