Steel Buildings in Europe

Title A.4 Worked Example – Simply supported, primary composite beam 12 of 13 4 – 92 To prevent crushing of the compression struts in the concrete flange, the following condition should be satisfied: f f cd Ed cos sin    f v  with   / 250 0,6 1 ck f    and  f = 45° 0,5 4,5 1,5 25 250 0,6 1 25 Ed            v N/mm 2 OK The following inequality should be satisfied for the transverse reinforcement : A sf f yd / s f ≥ v Ed h f / cot θ f where f yd = 500 / 1,15 = 435 N/mm 2 Assume the spacing of the bars s f = 200 mm and there is no contribution from the profiled steel sheeting A sf ≥     435 1,0 2,85 82 200 107,4 mm 2 Take 12 mm diameter bars (113 mm 2 ) at 200 mm spacing. 4.4. Serviceability Limit State verifications The deflection due to G + Q is calculated as: Q Q G G G 4 F E I w a L F E I a L E I q L w y 2 2 y 2 2 y 4 24 4 ) (3 24 ) (3 384 5 a a        And the total deflection is: w = w G + w Q 4.4.1. Construction stage SLS Combination during the construction stage F G + F Q = 49,28 + 13,5 = 62,78 kN EN 1990 § 6.5.3 q G = 0,65 kN/m Deflection during the construction stage I y is the second moment of area of the steel beam. 49280 10 23130 210000 24 ) 3000 4 (3 9000 3000 10 23130 210000 384 9000 5 0,65 4 2 2 4 4               - G w 1,1 26,2 27,3 mm    G w 7,2 mm 13500 10 23130 210000 24 ) - 4 3000 (3 9000 3000 4 2 2 Q          w  w = w G + w Q = 27,3 + 7,2 = 34,5 mm The deflection under ( G + Q ) is L /261