Local buckling of thin-plate elements - The Behaviour and Design of Steel Structures to EC3 - page 150

4.9.10 Example 10 - shear and bending of a Class 2 beam

Problem. Determine the design resistance of the 356 × 171 UB 45 of S355
steel shown in Figure 4.31b at a point where the design moment is M Ed =
230kNm.

Solution.
AsinSection4.9.2, f y = 355N/mm 2 , ε = 0.814, η = 1.2,theflangeisClass2
and the web is Class 1, the section is Class 2, and W pl = 775 cm 3 .

A v = 57.3 × 10 2 − 2 × 171.1 × 9.7 + ( 7.0 + 2 × 10.2 ) × 9.7
= 2676 mm 2

6.2.6(3)
η = 1.2 EC3-1-5 5.1(2)
η h w t w = 1.2 × ( 351.4 − 2 × 9.7 ) × 7.0 = 2789 mm 2 > 2676 mm 2
6.2.6(3)

V pl , Rd = 2789 × ( 355 / √ 3 )/ 1.0 N = 571.6 kN 6.2.6(2)

η h w /( t w ε) = 1.2 × ( 351.4 − 2 × 9.7 )/( 7.0 × 0.814 ) = 70.0 < 72
EC3-1-5 5.1(2)

so that shear buckling need not be considered.
Using a reduced bending resistance corresponding to M V , Rd = M Ed = 230
kNm, then

( 1 − ρ) × 355 × 775 × 10 3 = 230 × 10 6 , so that 6.2.8(3)

ρ = 0.164

Now ρ = ( 2 V Ed / V pl , Rd − 1 ) 2 , which leads to 6.2.8(3)

V c , M , Rd / V pl , Rd =[ √ ( 0.164 ) + 1 ] / 2 = 0.702, so that
V c , M , Rd = 0.702 × 571.6 = 401.5 kN.

4.9.11 Example 11 - shear and bending of a stiffened
plate girder

Problem. DeterminetheshearresistanceofthestiffenedplatewebgirderofSection
4.9.6 shown and shown in Figure 4.31d at the point where the design moment is
M Ed = 4000kNm.

Solution.
As previously, f yf = 345 N/mm 2 , ε f = 0.825, and the flanges are Class 3
(Section 4.9.4), f yw = 345 N/mm 2 and ε w = 0.814 (Section 4.9.5), and V bw , Rd =
1733kNm (Section 4.9.6).