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FIGURE 11.25
Element 3 forces.
FIGURE 11.26
Determination of longitudinal forces.
For element 3 (Fig. 11.25)
−
(
u
X
3
)
−
w
3
)
0
.
15482228
0
v
3
=
(21)
.
0
(
u
X
4
)
−
w
4
)
From (7) and (21)
F
X
3
−
V
3
81
.
035
k
3
v
3
p
3
=
V
4
=−
(22)
F
X
4
−
−
81
.
035
Whereas the transverse forces
V
and the moments
M
can be obtained using the stiffness
relations of the element, the longitudinal forces have to be determined by direct nodal
equilibrium. See Fig. 11.26. Sign Convention 1 will be utilized.
F
X
F
X
V
3
=
=
0:
N
2
−
−
V
2
=
=
0:
N
3
+
P
X
2
0
0
N
2
=
.
−
.
=−
.
N
3
=−
.
82
5
163
535
81
035
81
035
F
Z
F
Z
(23)
0:
N
2
+
V
2
+
0:
N
3
−
V
3
+
=
P
Z
2
=
0
=
P
Z
3
=
0
N
2
=−
N
3
=
1050
−
100
.
897
=−
1151
.
897
40
.
887
−
2100
=−
2059
.
113
Iteration with Improved Values of N
In the next step of the calculation procedure of Table 11.5 the results obtained by the initial
analysis with second order effects have to be compared with the results of the linear (first
order) analysis of the frame. The forces
N
i
are given in the summary of results of Table 11.7.
For element 1 we obtain from the linear analysis the value of
N
=−
.
1116
4 kN resulting
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