Civil Engineering Reference
In-Depth Information
==
(
)
=
025
.
kinin
/
288
wL
FEP
36
k
23
2
2
(
)
2
wL
2
025
.
kinin
/
288
FEM
=
=
=
1728
k n
−
23
12
12
==
(
)
=
025
.
kinin
/
288
wL
FEP
36
k
32
2
2
(
)
=−
2
0
.
25
kinin
/
288
wL
2
FEM
=−
=−
1728
k n
−
32
12
12
The 18 k lateral load is placed directly on joint 2 in the
y
direction.
P
P
M
P
P
M
P
P
M
P
0
0
0
18
36
172
1
x
1
y
1
z
2
x
−
2
y
−
8
−
[
]
[]
=
T
=
2
z
PPM
&
FEPM R
=
g
g
m
0
36
1728
0
0
0
3
x
−
3
y
3
z
4
P
M
x
4
y
4
z
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 1 and 4. The resulting matrix is just joints 2
and 3. This is shown in Table 4.6 along with the reduced load. The solution
for the deformations will be in inches and radians.
∆
∆
0 3040
0 0207
0 0034
0 2852
0 0270
0 0010
.
.
.
.
.
.
2
x
−
−
2
y
q
−1
2
z
=
=
∆
KP
=
g
g
g
∆
∆
3
x
−
3
y
q
3
z
