Steel Buildings in Europe

Title A.2 Worked Example – Simply supported beam with intermediate lateral restraints 6 of 7 4 – 69 c0  c,Rd M / M y,Ed = 0,50 244,97 307,15  = 0,627 f  = 0,583 ≤ c0  c,Rd M / M y,Ed = 0,627 OK Combination 2 k c = 1 L c = 5,00 m f  = 4,57 93,9 1 500   = 1,165 c0  c,Rd M / M y,Ed = 0,50 57,66 307,15  = 2,663 f  = 1,165 ≤ c0  c,Rd M / M y,Ed = 2,663 OK Shear Resistance In the absence of torsion, 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 EN 1993-1-1 § 6.2.6 (3) A v = 8446 – 2 × 180 × 13,5 + (8,6 + 2 × 21) × 13,5 = 4269 mm 2 579,21 kN 1000 1,0 (235 3) 4269 ( 3) M0 v,z y pl,Rd      f A V EN 1993-1-1 § 6.2.6 (2) V Ed / V pl,Rd = 65,33 / 579,21 = 0,113 < 1 OK Note that the verification to shear buckling is not required when : h w / t w ≤ 72  /  The value  may conservatively be taken as 1,0 h w / t w = (400 – 2 × 13,5) / 8,6 = 43,37 < 72 × 1 / 1,0 = 72 EN 1993-1-1 § 6.2.6 (6) Note: No interaction between moment and shear has to be considered, since the maximum moment is obtained at mid-span and the maximum shear force is obtained at supports. Generally for combined bending and shear see EN 1993-1-1, § 6.2.8. 2.6. Serviceability Limit State verification 2.6.1. Actions on the beam Characteristic combination: G k + Q s = 2,45+ 3,60 = 6,05 kN/m EN 1990 § 6.5.3 § A1.4.2