Civil Engineering Reference
In-Depth Information
Table 4.6.
Example 4.20 Global joint stiffness
K
G
P-FEPM
1056
0
4720
-1007
0
0
18
0
1525
2098
0
-15
2098
-36
4720
2098
1006944
0
-2098
201389
-1728
-1007
0
0
1056
0
4720
0
0
-15
-2098
0
1525
-2098
-36
0
2098
201389
4720
-2098
1006944
1728
The reactions at the supports can be found using the solution of the global
deformation with Equation 4.37 (step 5). Only the terms in the rows cor-
responding to the restrained degrees of freedom and in the columns of the
unrestrained degrees of freedom need to be included. Table 4.7 shows the
appropriate stiffness terms and deformations needed to find the reactions.
Since there were no fixed-end forces and moments at the support joints,
the solution is complete. The reaction forces are in kips and inches (k-in).
[]
=
−
[
]
[]
T
−
PK PM FEPM R
∆
&
g
g
g
m
[]
=
PK
∆
g
g
Table 4.7.
Example 4.20 Global joint stiffness
K
G2
P
D
G
-49
0
-4720
0
0
0
0.3040
0.92
0
-1510
0
0
0
0
-0.0207
31.22
4720
0
302083
0
0
0
-0.0034
419.18
0
0
0
-49
0
-4720
0.2852
-18.92
0
0
0
0
-1510
0
-0.0270
40.78
0
0
0
4720
0
302083
0.0010
1659.74
The final step is finding the member forces for each of the members using
Equation 4.35 (step 6). The member force will be in kips and inches.
The local member stiffness matrix and the rotation matrix were shown
in step 1 and are omitted here. The sign convention for the X-Y system
applies when interpreting the final-end forces and moments.
[
]
=
[
]
[]
+
[
]
PM KR FEPM
&
∆
m
m
g
m
[
]
=
[
]
[]
PM K
&
a
∆
m
m
g
