Civil Engineering Reference
In-Depth Information
0 0246
0 0057
0 0595
0 0022
0 0602
0 0261
.
−
427 −
320 0000
−
.
−
320
−
240 0
−
566 00
.
.
.
[]
=
=
PK
∆
g
g
−
427
320 0000
320
−
240 000566
−
−
.
867
775
12 33
23 75
.
−
.
.
.
=
The final step is finding the member forces for each of the members using
Equation 4.35 (step 6). Since there are no applied forces on the mem-
bers, the fixed-end forces and moments can be omitted. Note that the local
member stiffness matrix is used here and not the global matrix. The mem-
ber force will be in kips. If the
i
i-end is positive, the member is in compres-
sion and if it is negative the member is in tension.
[
]
=
[
]
[]
+
[
]
PM KR FEPM
&
∆
m
m
g
m
[
]
=
[
]
[]
PM K
&
b
∆
m
m
g
For member 1, the deformations at joints 1 and 4 are used.
555 6
.
0 00
0100
1000
000
0
0
0 0595
0 0022
−
124
0
124
0
.
0
−
555600
.
−
=
−
555 6
.
0
55560
.
1
0010
.
.
.
0
0 00
−
For member 2, the deformations at joints 3 and 5 are used.
555 6
.
0 00
0100
1000
000
0
0
0 0602
0 0261
14 5
0
14 5
0
.
0
−
555600
.
−
=
−
555 6
.
0
55560
.
1
0010
.
−
.
0
0 00
−
−
.
