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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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