Steel Buildings in Europe

Title A.3 Worked Example – Simply supported, secondary composite beam 7 of 10 4 – 77 3.6.5. Verification of bending resistance Minimum degree of shear connection The minimum degree of shear connection for a steel section with equal flanges is given by:   e y min 1- 355 0 75 0 03 , - , L f           with L e ≤ 25 L e is the distance in sagging bending between points of zero bending moment in metres, in this example: L e = 7,5 m   min = 1 – (355 / 355) (0,75 – 0,03 × 7,50) = 0,475  = 0,578 >  min = 0,475 OK EN 1994-1-1 § 6.6.1.2 Plastic Resistance Moment at mid span The design value of the normal force in the structural steel section is: N pl,a = A a f y /  M0 = 4595 × 355 × 10 -3 / 1,0 = 1631 kN EN 1994-1-1 § 6.6.1.2 and § 6.2.1.3  N pl,a = 1631 kN > N c = 952 kN For ductile shear connectors and a Class 1 steel cross-section, the bending resistance, M Rd , of the critical cross-section of the beam (at mid span) is calculated by means of rigid-plastic theory except that a reduced value of the compressive force in the concrete flange N c is used instead of N c,f . The plastic stress distribution is shows in Figure A.4. M Rd + - N c =  N c,f = 952 kN N a = 1291 kN h n h p 339 kN Figure A.4 Plastic stress distribution The position of the neutral axis is: h n = 263 mm Therefore the design resistance for bending of the composite cross-section is: M Rd = 301,7 kNm So, M y,Ed / M Rd = 172,2 / 301,7 = 0,57 < 1 OK