Civil Engineering Reference
In-Depth Information
Using the EC3 less conservative method with
β
=
0.75,
λ
LT
,0
=
0.4 (Clause
6.3.2.3) and
α
LT
=
0.34 (Tables 6.5 and 6.3), and equation 6.27,
Φ
LT
=
0.5
{
1
+
0.34
(
0.826
−
0.4
)
+
0.75
×
0.826
2
}=
0.828
Using equation 6.26,
M
b
,
Rd
=
132.8
/
{
0.828
+
√
(
0.828
2
−
0.75
×
0.826
2
)
}
kNm
=
106.6kNm.
The design moment resistance is further increased by using equations 6.29, 6.30
and 6.28 to find
f
=
1
−
0.5
×
(
1
−
1
/
√
1.75
)
{
1
−
2
×
(
0.826
−
0.8
)
2
}=
0.878, and
M
b
,
Rd
,
mod
=
106.6
/
0.878
=
121.4kNm
<
122.5kNm
=
M
Ed
and the design moment resistance is just inadequate.
6.15.3 Example 3 - checking a cantilever
Problem
. The 8.0 m long 457
×
191 UB 82 cantilever of S275 steel shown in
Figure 6.31c has the section properties shown in Figure 6.30b.The cantilever has
lateral, torsional, and warping restraints at the support, is free at the tip, and has
afactoredupwardsdesignuniformlydistributedloadof12kN/m(whichincludes
an allowance for self-weight) acting at the top flange. Check the adequacy of the
cantilever.
Design bending moment.
M
Ed
=
12
×
8
2
/
2
=
384kNm.
Section resistance.
t
f
=
16.0 mm,
f
y
=
275 N/mm
2
EN10025-2
ε
=
√
(
235
/
275
)
=
0.924
T5.2
c
f
/(
t
f
ε)
=
(
191.3
/
2
−
9.9
/
2
−
10.2
)/(
16.0
×
0.924
)
T5.2
=
5.44
<
9 and the flange is Class 1.
T5.2
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