Answer Happy

Question 4. (50 marks) Determine the maximum uniformly distributed imposed action (live load) wo which can be applied to the top flange of the entire beam shown in Figure 2. The beam is a 610UB125 of Grade 350 steel. The shear force diagram for the beam under the nominal uniformly distributed permanent action of wg = 3 kN is shown in the figure. Both the imposed and permanent actions are applied at the level of the top flange. The beam is fully restrained against lateral buckling at the three supports A, B and C, and there is no lateral rotational restraint along the beam in the two spans. The self-weight of the beam is included in the value of wg, and the bearing capacity of the web can be considered as adequate. (Hint: Both limit states of flexural and shear strength need to be checked). flexible gusset plate w* = 1.2 wgł 1.5w T F -detail-1 -detail-1 F Å 0000 0000 detail-1 4 m 4m 4 m et 4 m Wg= 3 kN/m B Oooo 8 m 8 m m 7 9 KN A с Figure 2 - Two-span continuous beam