Civil Engineering Reference
In-Depth Information
Usingthevaluesforthecriticalsectioninequation6.25,
λ
LT
=
√
(
2535
/
2885
)
=
0.937
h
/
b
=
(
960
+
2
×
20
)/
300
=
3.33
>
2.
Using the EC3 simple general method with
β
=
1.0 and
λ
LT
,0
=
0.2
(Clause 6.3.2.2) and
α
LT
=
0.34 (Tables 6.4 and 6.3) and equation 6.27,
Φ
LT
=
0.5
{
1
+
0.34
(
0.937
−
0.2
)
+
0.937
2
}=
1.065
Using equation 6.26,
M
b
,
Rd
=
2535
/
{
1.065
+
√
(
1.065
2
−
0.937
2
)
}
kNm
=
1615kNm
=
Q
b
,
Rd
/
2
×
3.0
Q
b
,
Rd
=
1615
×
2
/
3.0
=
1077kN.
6.16 Unworked examples
6.16.1 Example 8 - checking a continuous beam
Acontinuous533
×
210UB92beamofS275steelhastwoequalspansof7.0m.A
uniformlydistributeddesignload
q
isappliedalongthecentroidalaxis.Determine
the maximum value of
q
.
6.16.2 Example 9 - checking an overhanging beam
The overhanging beam shown in Figure 6.34a is prevented from twisting and
deflecting laterally at its end and supports. The supported segment is a 250
×
150
×
10RHSofS275steelandisrigidlyconnectedtotheoverhangingsegment
which is a 254
×
146 UB 37 of S275 steel. Determine the maximum design load
Q
that can be applied at the end of the UB.
Lateral deflection and twist
prevented at end and at supports
Lateral deflection and twist
prevented at supports and load points
254 ×146 UB 37 250×150 ×10 RHS
Q
kN
1.5
Q
kN
Q
5
Q
kNm
7.0m 14.0m 4.0m 6.0m 8.0m
(a) Overhanging beam (b) Braced beam
Figure 6.34
Unworked examples 9 and 10.
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