Steel Buildings in Europe

Title A.4 Worked Example – Simply supported, primary composite beam 6 of 13 4 – 86 Outstand flange : flange under uniform compression c = ( b – t w – 2 r ) / 2 = (180 – 8,6 – 2 × 21)/2 = 64,7 mm c / t f = 64,7 / 13,5 = 4,79 ≤ 9  = 7,29 Flange Class 1 EN 1993-1-1 Table 5.2 (sheet 2 of 3) Internal compression part c = h a – 2 t f – 2 r = 400 – 2 × 13,5 – 2 × 21 = 331 mm c / t w = 331 / 8,6 = 38,5 < 72  = 58,3 Web Class 1 EN 1993-1-1 Table 5.2 (sheet 1 of 3) The class of the cross-section is the highest class ( i.e. the least favourable) of the flange and the web. In this case the overall section is Class 1. For Class 1 sections, the ULS verifications should be based on the plastic resistance of the cross-section . 4.3.4. Construction stage Cross-sectional moment resistance The design bending resistance of a cross-section is given by: M c,Rd = M pl,Rd = W pl,y f y /  M0 = (1307 × 355 / 1,0) / 1000 EN 1993-1-1 § 6.2.5 M c,Rd = 463,98 kNm M y,Ed / M c,Rd = 269,2 / 463,98 = 0,58 < 1 OK Reduction factor for lateral–torsional buckling To determine the design buckling resistance of a laterally unrestrained beam, the reduction factor for lateral–torsional buckling must be determined. The restraint provided by the steel sheet is in this case quite small and it is neglected. The following calculation determines this factor by a simplified method for lateral–torsional buckling. This method avoids calculating the elastic critical moment. Non-dimensional slenderness The non-dimensional slenderness may be obtained from the simplified method for steel grade S355: 0,853 89 300/3,95 89 / z LT    L i  SN002 [4] For rolled profiles, 0,4 LT,0   Note: The value of LT,0  may be given in the National Annex. The recommended value is 0,4. So LT,0 LT 0,853     = 0,4 EN 1993-1-1, § 6.3.2.3(1)