Local buckling of thin-plate elements - The Behaviour and Design of Steel Structures to EC3 - page 145

and the ineffective width of the web is

b c − b e 1 − b e 2 = 744.0 − 209.8 − 314.6 = 219.6 mm. EC3-1-5T4.1
A eff = ( 1540 − 2 × 20 − 219.6 ) × 10 + 2 × 400 × 20
= 28804 mm 2
28804 × z c = ( 2 × 400 × 20 + ( 1540 − 2 × 20 ) × 10 ) × ( 1540 / 2 )
− 219.6 × 10 × ( 1540 − 20 − 6 − 209.8 − 219.6 / 2 )
z c = 737.6 mm
I eff = ( 400 × 20 ) × ( 1540 − 10 − 737.6 ) 2
+ ( 400 × 20 ) × ( 737.6 − 10 ) 2
+ ( 1540 − 2 × 20 ) 3 × 10 / 12 + ( 1540 − 2 × 20 )
× 10 × ( 1540 / 2 − 737.6 ) 2
− 219.6 3 × 10 / 12 − 219.6
× 10 × ( 1540 − 20 − 6 − 209.8 − 219.6 / 2 − 737.6 ) 2 mm 4
= 11.62 × 10 9 mm 4
W eff = 11.62 × 10 9 /( 1540 − 737.6 ) = 14.48 × 10 6 mm 3
M c , Rd = 14.48 × 10 6 × 345 / 1.0 Nmm = 4996 kNm.

4.9.5Example - shear buckling resistance of an
unstiffened plate girder web

Problem. Determine the shear buckling resistance of the unstiffened plate girder
web of S355 steel shown in Figure 4.31d.
Solution.

t w = 10 mm, f yw = 355 N / mm 2 EN10025-2

ε = √ ( 235 / 355 ) = 0.814 T5.2

η = 1.2 EC3-1-5 5.1(2)

h w = 1540 − 2 × 20 = 1500 mm. EC3-1-5 F5.1

η h w /( t w ε) = 1.2 × 1500 /( 10 × 0.814 )
= 221.2 > 72 and so the web is slender.

EC3-1-5 5.1(2)

a / h w =∞ / h w =∞ , k τ st = 0

EC3-1-5A.3(1)

k τ = 5.34 EC3-1-5A.3(1)