Question

The single-story unbraced frame shown below is subjected to dead load, roof live load, and wind load Figure 1 shows the results of a first-order analysis relative to the columns of the frame. The axial load and end moment (also equal to the maximum moment in the column) are given separately for the different load cases (i.e., dead load, roof live load, and lateral wind load). All vertical loads are symmetrically placed and contribute only to the Mnt moments (i.e., the frame behaves as a braced frame when only vertical loads are applied). The lateral load produces Mi moments Use A992 steel and select a W14 shape for the columns. Bending shown in the figures is about the strong axis. Assume K,-1 for both braced and unbraced conditions. Kr = 1 for the braced conditions, and Kr-2 for the unbraced conditions. Each column is laterally braced at the top and at the bottom (i.e., Lb-18 ft). Use the second-order axial forces and moments based on the moment amplification method. Assume Bi 1 and B2-1.14. Use LRFD (50 points). Points will be given as follows: a) Pu and Mu from the controlling load combination (with separate values for Pnt, Mnt, Ph and Mi) including second-order effects: 15 points b) Trial section: 10 points c Selection of an appropriate cross-section satisfying AISC specifications with an interaction coefficient obtained from Equations H-la/H1-lb of the Steel Manual 20.8: 25 points EXTRA CREDIT #1: Calculate the amplification coefficients Bi and B2 for the designed beam-column by assuming a drift index Δι! = 1/400 based on the service wind load (i.e., H= 3.6 kips). Use the effective length method and compute the amplification coefficient BI using EI* EI. Calculate B2 by assuming that story = 2Pn and RMー0.85. (10 extra credit points) EXTRA CREDIT #2: Check if the selected column is also the most efficient among the W14 sections (10 extra credit points)

Answer 1

Single store unbraced frame is subjected to dead load, root line and wind m_1t = 0.5 times 32 = 16 & t - kips The load combination four which is given by 1.20 - 1.0 w + 0.52 P_net = (1.2 times 19) + (1.0 times (-11)) + (0.5 times 33) = 28.3 kips P_it = 1.0 times 1.0 = 1.4 kips M_1t = 1.2 times 79 + 1.0 times - 46 + 0.5 times 130 = 113.8 & t - kips m_1t = 1.0 times 32 = 32 & t - kips where the load combination 3 will given using p_nt = 70.1 kips, P_1t = 0.7 kips, m_nt = 279.8 & t-kips and m_1t = 16.0 & t-kips Assuming B_1 = 1.0 Pestorey = R_m = HL/delta H = R_m H/delta H/L = 0.85 times 36/1/4000 = 1224 B_2 = 1/1 - alpha Pestory/Pestory = 1/1 - alpha P_mt/Pestory B_2 = 1/1 - alpha P story/Pestory = 1/1 - alpha P_mt/pestory \r\n B_2 = 1/1 - 1 times 701/1224 = 1.06075 m_4 = B_1 M_net + B_2 M_1t = (1 times 279.8) + (1.06075 times 16) = 296.772 & t-kips P_4 = P_nt + B_2 P_1t = 70.1 times (1.06075 times 0.7) = 70.842 & t - kips.