Steel Buildings in Europe

Title A.3 Worked Example – Simply supported, secondary composite beam 9 of 10 4 – 79 Assume the spacing of the bars s f = 250 mm and there is no contribution from the profiled steel sheeting: A sf ≥     435 1,0 2,05 62 250 73,05 mm 2 Take 10 mm diameter bars (78,5 mm 2 ) at 250 mm cross-centres extending over the effective concrete breadth. 3.7. Serviceability Limit State verification 3.7.1. SLS Combination EN 1990 § 6.5.3 G k + Q k = 9,80 + 7,50 = 17,30 kN/m Deflection due to G k + Q k : 384 ) 5 ( y 4 E I w G Q L   I y is calculated for the equivalent section, by calculating an effective equivalent steel area of the concrete effective area: 100 NA 263 295 Figure A.5 Equivalent steel section used for the calculation of A and I y EN 1994-1-1 § 5.4.2.2 b equ = b eff / n 0 n 0 is the modular ratio for primary effects ( Q k ) = E a / E cm = 210000 / 33000 = 6,36  b equ = 1,875 / 6,36 = 0,295 m Using the parallel axis theorem the second moment of area is: I y = 24 540 cm 4