Problem 2. Frame Model in SAP2000 Friday, May 15, 2020 8:30 AM The moment resisting frame below has A992 Grade 50 (Fy = 50 ksi) steel W12 ×106 columns and W21 x 68 beams. The frame resists both lateral (wind) and gravity loads (dead and live) as shown. The loads are unfactored but live loads have been reduced as allowed. 7.5kip 7.5kip 7.5kip 7.5kip 3.75kip 3.75kip 3.75kip 3.75kip - - 1 - - Į 35 kip 35 kip 12 ft 15kip 15kip 15kip 15kip 7.5kip 7.5kip 7.5kip 25 kip I I 7.5kip 25 kip - 10 ft (TYP) 10 ft (TYP) 12 ft 30 ft 30 ft 30 ft Wind Dead Live a) Model the frame in SAP2000. b) Analyze the frame for the loadings shown and for the consequential ASCE 7-16 load combinations. Load Combos: · 1.2D + 1.6L > LC1 . 1.2D + W + 0.5L - LC2 . 0.9D + W ->LC3 c) Determine the maximum factored moment for the beam supporting the first floor. 1) -171.81k 2) -537.81K - Controls ( Mu ) 3) _ 490.5 1 Lab 7 -Beam Plastic Moment, LRFD, SAP2000 Page 4 d) Lookup the plastic section modulus for that beam and determine the factored moment capacity. Is the beam safe? W21x68 _> Zx=160 in3 QMp = & Fy Zx = (0.9) (50 ksi) (160 in3) = 7200 k-in = 600 kft > 537.8 k-ft e) Determine the maximum factored axial force and moment in the column on the right. LC1 LC2 LC3 Mu (k-ft) -76.9 498.1 488.7 Pu (kips) -90 -109.7 -84.9 f) Look up the cross-sectional area and elastic section modulus for that column. Check to see if the column is yielding under the combined factored axial force and moment from above. W 12×106 : A= 31.2 in2 Sx = 145 in 3 IK 109.7k F= M + 2 = 145 in 3 498. 1 * 12 + 31 . 2 in 2 = 44,7 ksi < Fy= 50 ksi The member is NOT YIELDING 44.7 - 89% utilization 50 Lab 7 -Beam Plastic Moment, LRFD, SAP2000 Page 5 g) Determine the range of maximum factored axial forces and moments that the column base plates should be designed for. Right column see part c) Left column : LC1 LC2 LC3 Mu (k-ft) 76.9 448.7 458.1 Pu (k) -90 -20.8 3.9 h) For design wind loads, IBC suggests that the relative lateral displacement of any story should not exceed h/175 (h is in inches) to avoid possible damage to nonstructural components. Was that serviceability criteria met? 2. 4" 1.17 11 Z > x 4/175= 175 12' # 12 = 0.823 " Do_1=1.17"'> 0.823" No Good X DI-2 = 2.4- 1.17 = 1.23"'> 0.823 No Good X