Civil Engineering Reference
In-Depth Information
τ
cr
=
5.34
×
190000
×
(
10
/
1500
)
2
=
45.1 N
/
mm
2
EC3-1-5A.1(2)
λ
w
=
0.76
×
√
(
355
/
45.1
)
=
2.132
>
1.08
EC3-1-5 5.3(3)
Assuming that there is a non-rigid end post, then
χ
w
=
0.83
/
2.132
=
0.389
EC3-1-5T5.1
Neglecting any contribution from the flanges,
V
b
,
Rd
=
V
bw
,
Rd
=
0.389
×
355
×
1500
×
10
√
3
×
1.0
N
=
1196 kN. EC3-1-5 2(1)
4.9.6 Example 6 - shear buckling resistance of a
stiffened plate girder web
Problem.
DeterminetheshearbucklingresistanceoftheplategirderwebofS355
steel shown in Figure 4.31d if intermediate transverse stiffeners are placed at
1800 mm spacing.
Solution.
t
w
=
10 mm,
f
yw
=
355 N
/
mm
2
EN10025-2
ε
=
√
(
235
/
355
)
=
0.814
T5.2
h
w
=
1540
−
2
×
20
=
1500 mm.
a
/
h
w
=
1800
/
1500
=
1.20,
k
τ
st
=
0
EC3-1-5A.3(1)
k
τ
=
5.34
+
4.00
/
1.20
2
=
8.12
EC3-1-5A.3(1)
τ
cr
=
8.12
×
190000
×
(
10
/
1500
)
2
=
68.6 N
/
mm
2
EC3-1-5A.1(2)
Assuming that there is a rigid end post, then
λ
w
=
0.76
×
√
(
355
/
68.6
)
=
1.73
>
1.08
EC3-1-5 5.3(3)
χ
w
=
1.37
/(
0.7
+
1.73
)
=
0.564
EC3-1-5T5.1
Neglecting any contribution from the flanges,
V
b
,
Rd
=
V
bw
,
Rd
=
0.564
×
355
×
1500
×
10
√
3
×
1.0
N
=
1734 kN.
EC3-1-5 5.2(1)
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