Civil Engineering Reference
In-Depth Information
c
w
/(
t
w
ε)
=
(
460.2
−
2
×
16.0
−
2
×
10.2
)/(
9.9
×
0.924
)
T5.2
=
44.6
<
72 and the web is Class 1.
T5.2
M
c
,
Rd
=
275
×
1831
×
10
3
Nmm
6.2.5
=
503.5kNm
>
384kNm
=
M
Ed
and the section resistance is adequate.
Elastic buckling moment.
Theupwardsloadatthetopflangeisequivalenttoadownwardsloadatthebottom
flange, so that
z
Q
=
460.0
/
2
=
230.0 mm. Using equation 6.8,
210000
×
1871
×
10
4
81000
×
69.2
×
10
4
ε
=
230.0
8000
=
0.241
Using equation 6.7,
K
=
√
{
(π
2
×
210000
×
0.922
×
10
12
)/(
81000
×
69.2
×
10
4
×
8000
2
)
}
=
0.730
Using equation 6.59,
qL
3
2
√
EI
z
GI
t
=
27
1.4
(
0.241
−
0.1
)
1
+
{
1
+
1.4
2
(
0.241
−
0.1
)
2
}
1.3
(
0.241
−
0.1
)
+
10
(
0.730
−
2
)
1
+
{
1
+
1.3
2
(
0.241
−
0.1
)
2
}
=
17.23
M
cr
=
qL
2
/
2
=
17.23
×
√
(
210000
×
1871
×
10
4
×
81000
×
69.2
×
10
4
)/
8000 Nmm
=
1011kNm.
(Using the computer program PRFELB [18] leads to
M
cr
=
1051 kNm.)
Member resistance.
Using equation 6.25,
λ
LT
=
√
(
503.5
/
1011
)
=
0.706
h
/
b
=
460.0
/
191.3
=
2.40
>
2
Using the EC3 simple general method with
β
=
1.0,
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
(
0.706
−
0.2
)
+
1.0
×
0.706
2
}=
0.835
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