Steel Buildings in Europe

Title A.4 Worked Example – Simply supported, primary composite beam 11 of 13 4 – 91 M Rd + - N c =  N c,f = 1403 kN N a = 2201 kN h n h p 797 kN Figure A.12 Plastic stress distribution on the beam The position of the plastic neutral axis is: h n = 388 mm Therefore, the design bending resistance of the composite cross-section is: M Rd = 738 kNm M y,Ed / M Rd = 465,6 / 738 = 0,63 < 1,0 OK Shear Resistance The plastic shear resistance is the same as for steel beam alone. 874 97 pl,Rd , V  kN EN 1994-1-1 § 6.2.2.2 V Ed / V pl,Rd = 156,20 / 874,97 = 0,18 < 1,0 OK Interaction between bending moment and shear force If V Ed < V pl,Rd / 2 then the shear force may be neglected. So, V Ed = 156,20 kN < V pl,Rd / 2 = 874,97 / 2 = 437,50 kN OK EN 1993-1-1 § 6.2.8 (2) 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 = 9,0 / 3 = 3,0 m The value for  x is the distance between the restraint and the point load. Therefore there are three areas for the longitudinal shear resistance.  F d = N c / 2 = 1403 / 2 = 701,5 kN h f = h - h p = 140 – 58 = 82 mm      3000 82 701,5 10 3 Ed h x F   f d v 2,85 N/mm 2