Civil Engineering Reference
In-Depth Information
Using equation 6.26,
M
b
,
Rd
=
140.7
/
{
0.748
+
√
(
0.748
2
−
0.510
2
)
}
kNm
=
108.6kNm.
Section resistance (flange in tension).
Using Table 4.2 of EN1993-1-5 [60] and the ratio of the web outstand tip
compression to tension of
ψ
=
σ
2
/σ
1
=−
(
66.0
−
19.6
/
2
−
12.7
)/(
306.0
−
19.6
/
2
−
66.0
)
=−
0.189,
k
σ
=
0.57
−
0.21
×
(
−
0.189
)
+
0.07
×
(
−
0.189
)
2
=
0.612, and
21
√
k
σ
=
21
×
√
(
0.612
)
=
16.4
c
w
/(
t
w
ε)
=
(
306.0
−
19.6
−
12.7
)/(
11.9
×
0.942
)
T5.2
=
24.4
>
16.4
=
21
√
k
σ
and the web is Class 4.
T5.2
ThesectionmaybetreatedasClass3iftheeffectivevalueof
ε
isincreasedsothat
the Class 3 limit is just satisfied, in which case
ε
=
c
w
/
t
w
21
√
k
σ
=
(
306.0
−
19.6
−
12.7
)/
11.9
=
1.400
5.5.2(9)
16.4
in which case the maximum design compressive stress is
σ
com
,
Rd
=
f
y
/γ
M
0
ε
2
=
265
/
1.0
1.400
2
=
135 N/mm
2
5.5.2(9)
and so the section resistance is
M
c
,
Rd
=
135
×
299
×
10
3
Nmm
=
40.4 kNm.
6.2.5
(IftheeffectivesectionmethodofClause4.3(4)ofEN1993-1-5[60]isused,then
an increased section resistance of
M
c
,
Rd
=
W
el
,
y
,
min
f
y
=
53.2kNm is obtained.)
Elastic buckling (flange in tension).
Using Figure 6.27 leads to
β
y
=−
215.6mm.
Using equation 6.76 leads to
M
cr
=
188.3kNm.
(Using the computer program PRFELB [18] leads to
M
cr
=
188.3kNm.)
Member resistance (flange in tension).
Using equation 6.25,
λ
LT
=
√
(
40.4
/
188.3
)
=
0.463
Using the EC3 simple general method as before and equation 6.27,
Φ
LT
=
0.5
{
1
+
0.76
(
0.463
−
0.2
)
+
1.0
×
0.463
2
}=
0.708
Using equation 6.26,
M
b
,
Rd
=
40.4
/
{
0.708
+
√
(
0.708
2
−
0.463
2
)
}
kNm
=
32.6kNm.
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