Steel Buildings in Europe

Title A.4 Worked Example – Simply supported, primary composite beam 8 of 13 4 – 88 Note that the verification to shear buckling is not required when : h w / t w ≤ 72  /  The relevant value of  is :  = 1,2 h w / t w = (400 – 2 × 13,5) / 8,6 = 43 < 72 × 0,81 / 1,2 = 48,6 EN 1993-1-1 § 6.2.6 (6) EN 1993-1-5 § 5.1 (2) Interaction between bending moment and shear force If V Ed < V pl,Rd / 2 then the shear force may be neglected. So, V Ed = 90,73 kN < V pl,Rd / 2 = 874,97 / 2 = 437,50 kN OK EN 1993-1-1 § 6.2.8 (2) 4.3.5. Final stage Effective width of concrete flange The effective width is constant between 0,25 L and 0,75 L , where L is the span length. From L /4 to the closest support, the effective width decreases linearly. The concentrated loads are located between 0,25 L and 0,75 L . EN 1994-1-1 § 5.4.1.2 The total effective width is determined by: (Figure 5.1) ∑ eff,1 ei 0 b b   b b 0 is the distance between the centres of the outstand shear connectors, in this case b 0 = 0 b ei is the value of the effective width of the concrete flange on each side of the web and taken as b ei = L e / 8 but ≤ b i = 3,0 m b eff,1 = 0 + 9,0 / 8 = 1,125 m, then b eff = 2 × 1,125 = 2,25 m < 3,0 m Design shear resistance of a headed stud The shear resistance should be determined by:           V cm V u l Rd Min     d f E f d P k ck 0,29 ; 0,8 / 4 2 2 EN 1994-1-1 § 6.6.3.1 h sc / d = 100 / 19 = 5,26 > 4, so  = 1 Reduction factor (k l ) For sheeting with ribs transverse to the supporting beam, the reduction factor for shear resistance is calculated by: EN 1994-1-1 § 6.6.4.1           1 0,6 p sc p 0 l h h h b k but ≤ 1