Civil Engineering Reference
In-Depth Information
Table 4.5
Global Stiffness Matrix
1
2
3
4
5
6
7
2.944
×
10
6
2.944
×
10
6
1
19,627.2
−
19,627.2
0
0
0
2.944
×
10
6
5.888
×
10
8
−
2.944
×
10
6
2.944
×
10
8
2
0
0
0
−
2.944
×
10
6
2.944
×
10
6
3
−
19,627.2
66,350.4
0
−
19,627.2
−
27,096
2.944
×
10
6
2.944
×
10
8
11.78
×
10
8
−
2.944
×
10
6
2.944
×
10
8
4
0
0
−
2.944
×
10
6
−
2.944
×
10
6
5
0
0
−
19,627.2
19,627.2
0
2.944
×
10
6
2.944
×
10
8
−
2.944
×
10
6
5.889
×
10
8
6
0
0
0
7
0
0
−
27,096
0
0
0
27,096
so the element stiffness matrices are (per Equation 4.48)
12
1,800
−
12
1,800
k
(1)
=
k
(2)
=
1,800
360,000
−
1,800
180,000
1,635
.
6
−
12
−
1,800
12
−
1,800
1,800
180,000
−
1,800
360,000
while for element 3,
78
.
54(69)(10
3
)
200
AE
L
=
=
27096 N/mm
so the stiffness matrix for element 3 is
k
(3)
=
27, 096
1
−
1
−
11
Assembling the global stiffness matrix per the displacement correspondence table (noting
that we use a “short-cut” for element 3, since the stiffness of the element in the global
X
direction is meaningless), we obtain the results in Table 4.5. The constraint conditions are
U
1
=
U
7
=
0
and the applied force vector is
F
1
M
1
F
2
M
2
F
3
M
3
F
4
R
1
0
0
0
=
−
10,000
0
R
4
where we use
R
to indicate a reaction force. If we apply the constraint conditions and
solve the resulting
5
×
5
system of equations, we obtain the results
1
=
9
.
3638 (10
−
4
) rad
v
2
=−
0
.
73811 mm
2
=−
0
.
0092538 rad
v
3
=−
5
.
5523 mm
3
=−
0
.
019444 rad
