Steel Buildings in Europe

Title A.4 Worked Example – Simply supported, primary composite beam 7 of 13 4 – 87 Reduction factor For rolled sections, the reduction factor for lateral–torsional buckling is calculated from: EN 1993-1-1 § 6.3.2.3 (1)          2 LT LT LT 2 LT 2 LT LT LT 1 1.0 but 1         where :     0,5 1 LT LT,0 LT LT LT 2            LT is the imperfection factor for LTB. When applying the method for rolled profiles, the LTB curve has to be selected from Table 6.5: For h a / b = 400 / 180 = 2,22 > 2 Curve ‘c’ (  LT = 0,49) EN 1993-1-1 Table 6.5 Table 6.3 0,4 LT,0   and  = 0,75  Note: The values of LT,0  and  may be given in the National Annex. The recommended values are 0,4 and 0,75 respectively.  We obtain :     0,884 (0,853) 0,4 0,75 0,5 1 0,49 0,853 2 LT         and : 0,730 (0,853) 0,75 (0,884) 0,884 1 2 2 LT       Then, we verify :  LT = 0,730 < 1,0 but :  LT = 0,730 < 2 LT 1 /  = 1,374 So :  LT = 0,730 Design buckling resistance moment M b,Rd =  LT W pl,y f y /  M1 M b,Rd = (0,730 × 1307000 × 355 / 1,0) × 10 -6 = 338,7 kNm M y,Ed / M b,Rd = 269,2 / 338,7 = 0,795 < 1,0 OK EN 1993-1-1 § 6.3.2.1 Shear Resistance The shear plastic resistance depends on the shear area, which is given by: A v = A – 2 b t f + ( t w + 2 r ) t f A v = 8446 – 2 × 180 × 13,5 + (8,6 + 2 × 21) × 13,5 = 4269 mm 2 EN 1993-1-1 § 6.2.6 (3) Shear plastic resistance 874,97 kN 1,0 (355 / 3) 10 4269 ( / 3) 3 M0 v y pl,Rd        A f V EN 1993-1-1 § 6.2.6 (2) V Ed / V pl,Rd = 90,73 / 874,97 = 0,104 < 1,0 OK