Civil Engineering Reference
In-Depth Information
The collapse load factors calculated for the other possible mechanisms of
•
lower beam
(α
p
=
2
×
(
155.9
+
85.0
+
246.4
)/(
80
×
6
)
=
2.030
)
•
lower-storey sway
(α
p
=
2
×
155.9
/(
30
×
5
)
=
2.079
)
•
upper-storey sway
(α
p
=
2
×
(
84.2
+
85.0
)/(
10
×
5
)
=
6.768
)
•
combined beam sway
α
p
=
2
×
(
84.2
+
84.2
+
246.4
+
85.0
+
155.9
)
{
(
40
+
80
)
×
6
+
(
20
×
5
+
10
×
10
)
}
=
1.425
are all greater than 1.403, and so
α
p
=
1.403.
8.5.11 Example 11 - elastic member design
Problem
.Checktheadequacyofthelower-storeycolumn67(ofS275steel)ofthe
unbracedframeshowninFigure8.4.Thecolumnisfullybracedagainstdeflection
out of the plane of the frame and twisting.
Design actions.
The first-order actions are
N
67
=
136.7 kN and
M
76
=
119.3 kNm (Figure 8.4)
and the amplified moment is 141.8 kNm (Section 8.5.8).
H
Ed
=
10
+
20
=
30 kN and
V
Ed
=
(
20
+
40
+
20
)
+
(
40
+
80
+
40
)
=
240 kN.
H
Ed
/
V
Ed
=
30
/
240
=
0.125
<
0.15,
5.3.2(4)B
and so the effects of sway imperfections should be included.
φ
0
=
1
/
200 5.3.2(3)
α
h
=
2
/
√
10
=
0.632
<
2
/
3 and so
α
h
=
2
/
3 5.3.2(3)
m
=
1 (one of the columns will have less than the average force), 5.3.2(3)
α
m
=
√
{
0.5
(
1
+
1
/
1
)
}=
1.0, and
5.3.2(3)
φ
=
(
1
/
200
)
×
(
2
/
3
)
×
1.0
=
0.00333
5.3.2(3)
The increased horizontal forces are
5.3.2(7)
10
+
0.00333
×
80
=
10.27 kN for the upper storey, and
20
+
0.00333
×
240
=
20.80 kN for the lower storey.
If the frame is reanalysed for these increased forces, then the axial force will
increase from 136.7 to
N
Ed
=
137.2 kN, and the moment from 119.3
×
1.188
=
141.8 kNm to
M
y
,
Ed
=
121.9
×
1.188
=
144.8 kNm.
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