Steel Buildings in Europe

Title A.3 Worked Example – Simply supported, secondary composite beam 8 of 10 4 – 78 3.6.6. Shear Resistance The shear plastic resistance depends on the shear area of the steel beam, 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 = 4595 – 2 × 135 × 10,2 + (6,6 + 2 × 15) × 10,2 = 2214 mm 2 Shear plastic resistance 453 8 kN 10 1 0 / 3) (355 2214 / 3) ( 3 M0 v y pl,Rd , , A f V -      EN 1994-1-1 § 6.2.2.2 V Ed / V pl,Rd = 91,80 / 453,8 = 0,202 < 1,0 OK Verification to shear buckling is not required when: h w / t w ≤ 72  /   may be conservatively taken as 1,0 h w / t w = (270 – 2 × 10,2) / 6,6 = 37,8 < 72 × 0,81 / 1,0 = 58,3 OK EN 1993-1-1 § 6.2.6 (6) 3.6.7. Longitudinal Shear Resistance of the Slab The plastic longitudinal shear stresses is given by : h x v F   f d Ed  EN 1992-1-1 § 6.2.4 Figure 6.7 where:  x = 7,5 / 2 = 3,75 m The value of  x is half the distance between the section where the moment is zero and the section where the moment is a maximum, so there are two areas for the longitudinal shear resistance of the slab.  F d = N c / 2 = 951,56 / 2 = 475,8 kN h f = h - h p = 120 – 58 = 62 mm      3750 62 475,8 10 3 f d Ed h x v F   2,05 N/mm 2 To prevent crushing of the compression struts in the concrete flange, the following condition should be satisfied: f f cd Ed cos sin    f v  with   / 250 0,6 1 ck f    and  f = 45° 0,5 4,5 1,5 25 250 0,6 1 25 Ed            v N/mm 2 OK The following inequality should be satisfied for the transverse reinforcement: A sf f yd / s f ≥ v Ed h f / cot θ f where f yd = 500 / 1,15 = 435 N/mm 2