Civil Engineering Reference
In-Depth Information
The global deformations can be found from the global stiffness Equa-
tion 4.36 (step 4). The rows and columns corresponding to the support
constraint degrees of freedom must be deleted prior to the solution. This
would be all three motions at 4 and the X and Z motions at joint 5. The
resulting matrix is shown in Table 5.9 along with the reduced load. The
solution for the deformations will be in inches and radians.
∆
∆
0 07505
2 45212
0 00004
0 06168
.
1
x
−
.
1
y
q
.
.
2 25618
0 00261
0 03701
2 25254
0 00449
0 02122
1
z
∆
∆
2
x
−
.
.
.
.
.
.
−1
2
y
=
=
∆
KP
=
g
g
g
q
−
−
−
−
−
2
z
∆
∆
3
x
3
y
q
q
3
z
5
z
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 5.10 shows the
appropriate stiffness terms and deformations needed to find the reactions.
Since there are fixed-end forces and moments at support joint 4, they must
Table 5.9.
Example 5.3 Member stiffness
K
G
-FEPM
2191
531
0
0
0
556
0
0
−
1389
−
802
−
531
531
400
−
2894
0
0
−
40
−
2894
−
531
−
360
−
833
−
9
0
−
2894
277778
0
2894
138889
0
0
0
0
216
0
0
2358
0
6510
0
6510
0
0
−
1389
−
136
0
2894
0
2132
1852
0
0
0
−
40
−
2083
−
24
0
−
2894
138889
6510
1852
861111
−
6510
0
208333
0
384
−
802
−
531
0
−
136
0
−
6510
2326
531
−
7066
0
0
−
531
−
360
0
0
−
2083
0
531
2483
−
2060
−
2894
0
556
−
833
0
6510
0
208333
−
7066
−
2060
867788 138889
0
0
0
0
0
0
0
0
−
2894
138889 277778
0
