Steel Buildings in Europe

Title A.1 Simply supported, laterally unrestrained beam 4 of 7 4 – 60 1.5.5. Section resistance Cross-sectional moment resistance The design cross-sectional resistance is: M c,Rd = W pl,y f y /  M0 = (804,3 × 235 / 1,0)  10 -3 = 189,01 kNm The section must verify that M y,Ed / M c,Rd < 1,0 M y,Ed / M c,Rd = 90,48 / 189,01 = 0,479 < 1,0 OK EN 1993-1-1 § 6.2.5 Lateral-torsional buckling resistance To determine the design lateral-torsional buckling resistance, the reduction factor for lateral-torsional buckling must be determined. The following calculation determines this factor by means of the elastic critical moment. Elastic critical moment The critical moment may be calculated from the following expression:                       2 g 2 2 g z 2 t 2 z w 2 w 2 z 2 1 cr ( ) C z C z E I k L G I I I k k k L M C E I   where: E is Young’s modulus: E = 210000 N/mm 2 G is the shear modulus: G = 80770 N/mm 2 L is the span: L = 5,70 m SN003 [4] In the expression for M cr , the following simplifications can be made: k = 1 since the compression flange is free to rotate about the weak axis of the cross-section, k w = 1 since warping is not prevented at the ends of the beam. z g is the distance from the loading point to the shear centre: z g = h / 2 = +165 mm ( z g is positive when the loads act towards the shear centre) The C 1 and C 2 coefficients depend on the bending moment diagram. For a uniformly distributed load and k = 1: C 1 = 1,127 C 2 = 0,454 SN003 [4] Therefore:   502,75 kN 10 (5700) 10 788,1 210000 3 2 4 2 2 z 2          k L E I 74,91 mm 165 0,454 2 g     C z