Steel Buildings in Europe

Title A.2 Worked Example – Simply supported beam with intermediate lateral restraints 4 of 7 4 – 67 Shear force diagram V V Ed Maximum shear force at supports : Combination 1 V Ed = 0,5 × 8,71 × 15 = 65,33 kN Combination 2 V Ed = 0,5 × -2,05 × 15=  15,38 kN 2.5.3. Section classification The parameter  is derived from the yield strength: 1,0 235 235 235 y    f  EN 1993-1-1 Table 5.2 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  = 9 Flange class 1 Internal compression part: web under pure bending c = h – 2 t f – 2 r = 400 – 2 × 13,5 – 2 × 21 = 331 mm c / t w = 331 / 8,6 = 38,49 < 72  = 72 Web class 1 The class of the cross-section is the highest class (i.e. the least favourable) between the flange and the web. Therefore, the overall section is Class 1. For Class 1 sections, the ULS verifications should be based on the plastic resistance of the cross-section. EN 1993-1-1 Table 5.2 2.5.4. Section resistance Cross-sectional moment resistance The design cross-sectional resistance is: M c,Rd = W pl,y f y /  M0 = (1307 × 235 / 1,0)  10 -3 = M c.Rd = 307,15 kNm EN 1993-1-1 § 6.2.5 Combination 1 M y,Ed / M c,Rd = 244,97 / 307,15= 0,798 < 1,0 OK Combination 2 M y,Ed / M c,Rd = 57,66 / 307,15= 0,188 < 1,0 OK Lateral-torsional buckling verification using the simplified assessment methods for beams with restraints in buildings: In buildings, members with discrete lateral restraint to the compression flange are not susceptible to lateral-torsional buckling if the length L c between restraints or the resulting equivalent compression flange slenderness f  satisfies: EN 1993-1-1 § 6.3.2.4 (1)B