Civil Engineering Reference
In-Depth Information
Section resistance.
t
f
=
12.5 mm,
f
y
=
275 N
/
mm
2
EN10025-2
ε
=
(
235
/
275
)
=
0.924
T5.2
c
f
/(
t
f
ε)
=
(
204.3
/
2
−
7.9
/
2
−
10.2
)/(
12.5
×
0.924
)
=
7.62
<
9
T5.2
and the flange is Class 1. T5.2
The worst case for the web classification is when the web is fully plastic in
compression, for which
c
w
/(
t
w
ε)
=
(
206.2
−
2
×
12.5
−
2
×
10.2
)/(
7.9
×
0.924
)
=
22.0
<
33
T5.2
and the web is Class 1.
T5.2
Thus the section is Class 1.
γ
M
0
=
1.0 6.1
N
c
,
Rd
=
66.3
×
10
2
×
275
/
1.0 N
=
1823 kN
=
N
pl
,
Rd
6.2.4(2)
0.25
N
pl
,
Rd
=
0.25
×
1826
=
455.8 kN
>
137.2 kN
=
N
Ed
, 6.2.9.1(4)
0.5
h
w
t
w
f
y
=
0.5
(
206.2
−
2
×
12.5
)
×
7.9
×
275 N
=
196.8 kN 6.2.9.1(4)
>
137.2 kN
=
N
Ed
,
and so no allowance need be made for the effect of axial force on the plastic
resistance moment.
6.2.9.1(4)
M
N
,
Rd
=
M
pl
,
Rd
=
567
×
10
3
×
275
/
1.0 Nmm
=
155.9 kNm
>
144.8 kNm
=
M
Ed
6.2.9.1(5)
and the section resistance is adequate.
6.2.9.1(2)
Member resistance.
Becausethememberiscontinuouslybracedout-of-plane,beamlateralbuckling
and column minor axis and torsional buckling need not be considered.
Thus
M
z
,
Ed
=
0,
χ
LT
=
1.0, and
λ
0
=
0,
C
mLT
=
1.0,
a
LT
=
0, and
b
LT
=
0
TA.1
L
cr
=
5000
5.2.2.(7b)
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