* 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 Living