Steel Buildings in Europe

Title A.3 Worked Example – Simply supported, secondary composite beam 5 of 10 4 – 75 Internal compression part c = h – 2 t f – 2 r = 270 – 2 × 10,2 – 2 × 15 = 219,6 mm c / t w = 219,6 / 6,6 = 33,3 < 72  = 58,3 Web Class 1 The class of the cross-section is the highest class (i.e. the least favourable) between the flange and the web. So the ULS verifications should be based on the plastic resistance of the cross-section since the Class is 1. 3.6.2. Effective width of concrete flange At mid-span, the total effective width may be determined by:    ei 0 eff,1 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, b ei = L e / 8 but ≤ b i = 3,0 m b eff,1 = 0 + 7,5 / 8 = 0,9375 m  b eff = 2 × 0,9375 = 1,875 m < 3,0 m EN 1994-1-1 Figure 5.1 At the ends, the total effective width is determined by:    i ei 0 eff,0 b b b  EN 1994-1-1 Figure 5.1 where:  i = (0,55 + 0,025 L e / b ei ) but ≤ 1,0 = (0,55 + 0,025 × 7,5 / 0,9375) = 0,75 b eff,0 = 0 + 0,75 × 7,5 / 8 = 0,703 m  b eff = 2 × 0,703 = 1,406 m < 3,0 m 3.6.3. Design shear resistance of a headed stud The shear resistance of each stud may be determined by:           V ck cm 2 V 2 u t Rd 0,29 / 4 ; Min 0,8     d f E f d P k h sc / d = 100 / 19 = 5,26 > 4, so  = 1 EN 1994-1-1 § 6.6.3.1 Reduction factor (k t ) For sheeting with ribs transverse to the supporting beam, the reduction factor for shear resistance is calculated by:           1 0,7 p sc p 0 r t h h h b n k but ≤ k tmax for profiled sheeting with holes. EN 1994-1-1 § 6.6.4.2 Table 6.2