Civil Engineering Reference
In-Depth Information
From Table 4.4, the variable load provides 37.2/60.9
=
61% of the shear
force. Hence, the shear force at each pin joint is
V
Ed
=
246
×
0.61
×
0.767
+
246
×
0.39
=
115
+
96
=
211 kN
(5.38)
It is assumed that the steel section for the column will be from the
203
×
203 UC serial size, so the eccentricity at the pin joint is 0.203/2
+
0.1
0.20 m, and the major-axis bending moment applied to the column
at each loaded floor level is
=
M
Ed
=
211
×
0.2
=
42.2 kN m
(5.39)
The bending moment in length 0-1 is determined mainly by the load
from level 1, so factor
α
n
is taken as 1.0 at that level, giving the applied
moments and shears shown in Fig. 5.13(a). The bending moments in the
lower part of the column, found by moment distribution, are shown in
Fig. 5.13(b).
For minor-axis bending, all the loading is permanent, and equal on the
two sides of the column, so bending moment arises only from the initial
bow of the member.
Including the permanent load from Equation 5.11, the axial load for
length 0-1 is:
N
Ed
=
9
×
97
+
8
×
211
+
246
=
2807 kN
(5.40)
5.7.2
Properties of the cross-section, and y-axis slenderness
A cross-section for the column must now be assumed, and is shown
in Fig. 5.14. Applying the usual partial factors to the properties of the
Figure 5.14
Assumed cross-section for external column length 0-1
