Civil Engineering Reference
In-Depth Information
Section Classification
:
0
:
5
ε = ð
235/f
y
Þ
=
0
:
92
Then check the compactness of the section:
d/t
w
= 248.6/7.1 = 37.5
<
72
×
0.92
<
×
…
c/t
f
= 7.0
10
0.92
The section is Class 1 (Compact section)
Moment Resistance
:
There is a full restraint to top flange; so there is no effect of lateral-torsional
buckling.
Resistance moment M
pl.Rd
= W
pl.x
f
y
/
γ
M.0
= 157 kNm
>
154.97
Shear Resistance:
Applied shear V
d
= 123.97 kN
Shear area A
v
= 2567 mm
2
Resistance V
pl.Rd
=2567
×
0.275/(1.732
×
[1.1]) = 370 kN
>
123.97
Steel Beam: Design Resistance at 20
°
C
Design loading in fire
:
M
f
:
d
= η
f
M
d
Combination factor
ψ
1.1
= 0.5
G
k,1
/Q
k
= 2.0
Reduction factor
η
f
= 0.46
M
f
:
d
=
0
:
46
×
154
:
97
=
71
:
25 kNm
Design resistance at 20
C, using fire safety factors
:
For a Class 1 beam with uniform temperature distribution,
Resistance moment at temperature
°
θ
is M
fi.
.Rd
= k
y.
(
γ
M.1
/
γ
M.fi
)M
Rd
θ
θ
Strength reduction factor for 20
°
C: k
y.20
= 1.0
γ
M.1
= [1.1] and
γ
M.f
= [1.0]
Resistance moment for strength at 20
°
CisM
Rd
= 157 kNm
×
×
M
f.20.Rd
= 1.0
([1.1]/[1.0])
157 = 172.7 kNm
M
fi
:
t
:
Rd
=
k
1
k
2
k
1
= [1.0] for beam exposed to fire from all sides
k
2
= 1.0 (note that k
2
is equal to 0.85 at the support for statically
indeterminate beam)
M
f.t.Rd
= 172.7/([1.0]
M
fi
:
q
:
Rd
=
×
1.0) = 172.7 kNm
Critical Temperature of the Beam
:
Degree of utilization
μ
0
= 71.25/172.7 = 0.41
Critical temperature of beam
θ
cr
= 616
°
C
7.7.5 Steel Column: Strength Design
This example is a steel column with section HEB 180 under normal force equal
to 991.8 kN and with a height of 3.5 m.
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