Question

Designs for the maximum negative moment and maximum shear of a continouse oneway solid slab supported or bears spaced at inft on centers. The width of the beam is 12 m. The uniform dead load of 120 psf (including self-weight of slab) and live road of 100 psf. use ACI 318 code moment and shear couffrents for continouse beams and slab. Pc = Hksi, for boksi, n=6in, d=506 in, try #3 bals. slab carries a

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Answer 1

Loading Dead Load = 120 psf Live load = 100 psf Consider unit with of slab b=1 ft W DL = 120*1 = 120 plf WLL = 100*1=100 plf ASCE 7 LRFD Factored load wu=1.2* * W DL +1.6* WLL =1.2* 120+1.6*100 = 304 plf = 0.304 k/ft

Center to center spacing of beam = 14" Width of beam = 12" Clear spacing Ln = 14-1=13" 0.304 * 132 Maximum negative moment Mn 10 = 5.1376 k-ft 0.304 * 13 Maximum negative shear Vu = 1.15* 2 = 2.272 kips

Design of Longitudinal rebar Mu Mnrequired= 0 For tension controlled member o=0.9 5.1376 Mn-required= = 5.708 k - ft 0.9 = 68.50k-in Given fc = 4 ksi fy = 60ksi d=5.06 in h=6" Required rebar ratio 0.85* fc 2* Mnrequired р *(1-1 fy 0.85* fc*b*d? 0.85* 4 2* 68.50 р (1-1 60 0.85* 4*12*5.062 solving p=0.003846 *

Required Area of rebar As = 0.003846*12*5.06 =0.2335 in? / unit width Spacing of #3 bars(s) Area of one #3 bars =0.11 in Area of one #3 bars A = 0.11 in? Area of one #3 bars * 12 0.11*12 = 5.65" As 0.2335 'hence provide # 3 bars at s = 5.5" S= c As(min) ACI318 specify minimum area of steelas bw* d *max(3* _fc ,200) fy bw = b = 12" max(3* _fc ,200) = max(3* /4000 ,200) = max(189.73 ,200) = 200 5.06*12* 200 = 0.2024 in? 60,000 As>A S(min) hence ok As(min)

Maximum spacing as per ACI 318-2014 Min(3h or 18”)=-min(3*6 or 18”)=18” Provided spacing=5.5" 5.5"<18" Hence Satisfied Required spacing Main steel # 3 @5.5” spacing

Design Shear capacity Øn QVn = pvc Vc is the Shear resistance of Concrete The ACI 318 provide formula for computing shear resistance offered by concrete Vc = 2*2*b* d* Vfc b=l'=12" where 2 = 1 for normal weght concrete Vc=2*1*12*5.06* V4000 = 7680.531b Strength reduction factor for shear design is 0 =0.75 Vc =0.75* 7680.53 = 5760.404 lb Vn = 5692.09 lb V = 2720 lb Hence Vn>VU Safe in shear