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  • Q1:* Minimum thickness of slab Exterior Bay → t= L/24 = 15x12/24 = 7.5 in Interior Bay → t= 1/28 = 8.5x12/28= 3.6 in d=h-cover-0.5 (diam) = 7.5-0.75 -0.25 = 6.5" Weight & slab = (7.5/12) x 150 pcf = 93.75 psf Total Dead load = 93.75 + 70 163.75 psf * 15 * fo 8.5 8.5 Factored load: Wu= 1₁2DL + 1.6LL = 1.2 (163.75) + 1.6 (130) = 404.5 psf ~ 405 psf = 0.405 ksf * Moment & Shear SAP using IS/nYO Determine the moment and shear using the ACI coefficient Method: M=CM W₁12 and V = C, W₁1/2 Vmax at C= 1.15 (0.354)(11/2) = 2.24Kips -ve Mmax at C = -(0.354)(11)²/10 = 4.3ft-kip at B = +(0.354)(11)²/14 = 3.06ft-kip +ve Mmax ● CM. CV. A 1/24/14 %0 X₁ 1.0 1.15 1.0 E 12 D 4 110 D 411 E ¼/16 CVE 313 Dr. Farid Abed/nL₂ Table 1: Dimensions and distributed loading for each group Live Load psf Group L₁ No. (ft) 123456789 Bedroom L3 L2 (ft) (ft) 5000 60 5000 60 14 17 13 13 15 17 15 15 16 14 15 16 14 14 4500 60 4000 60 16 4000 60 13 14 17 4500 60 15 16 17 4500 60 16 16 17 4000 60 15 15 16 4500 60 4555FHO65 fc (ksi) (psi) 3L2/4 7766ON6 Bathroom 120 120 120 120 130 130 130 130 110 Kitchen Super Imposed Dead Load psf 80 85 75 70 90 80 70 60 90 LivingSee Answer
  • Q2: Can be done by groups of 2. If so please only one of the members upload the solution in UBLearns and put the names of all members in the first page of the report. Instructions for the assignment submission: 1. For the structure, report the displacements, reactions and member forces. 2. Using the obtained reactions, check the global equilibrium by hand calculations 3. Show a deformed shape of the structure. 180 in Provide all the above in a neat, short and well-organized report and keep the details in appendix. 180 in Problem No. 1 Consider the structure shown below with material and section properties as indicated on the figure. Calculate the forces in all truss members, and the shear and bending diagram of the beam. TOXOXO CIE 323 STRUCTURAL ANALYSIS E = 29000 ksi I = 3000 in 3A A SAP2000 ASSIGNMENT 3A 240 in 3A 20kips A 3A 240 in + 20kips I, 36A 360 inSee Answer

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