Question

Reiforced concrate

22: (50%) Design for the maximum sold slab supported on and maximum shear of a continouse one-way negative moment on beams spared at 1uft centers. The width of the beam is 12 in. The uniform dead land of 120 psf (including self weight of slab) and live road of 100 psf. use ACI 318 code moment and shear coefficents for continouse beams and slab. Pr = 4ksi, fo= 60ksi, h=6in, d=506 in, try #3 bals. slab carries a

Answer 1

Given na sapancing and location : : 14" Loading Dead Load = 120 psf 1 : : 130 Live load = 100 psf Consider unit with of slab b = 1 ft WDL = 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"

ACI 318 - 2014 6.5.1 Simplified method of analysis for non - prestressed continuous beams and one-way solid slabs Negative moment coefficent At interior face of exterior support support is spandrel beam .......... ..1/24 At other faces of interior supports .......1/11 At exterior face of first interior support 1/10 Hence maximum moment will be at exterior face of first interior support Moment=w. * Ln2 * coefficient *Ln? 10 0.304 * 132 = 5.1376 k-ft 10 W V II

Shear coefficent 6.5.4 Vu due to gravity loads 1.15* wu * Ln Exterior face of first interior support 2 Face of all other supports wu* Ln/2 1.15* wu* Ln Maxium negative shear = 2 0.304*13 Maximum negative shear Vu = 1.15* 2 = 2.272 kips

Design of Longitudinal rebar Mu Mnrequired = 0 = 0.9 for tension controlled member 5.1376 Mn - required -= 5.708 k-ft 0.9 = 68.50 k - in Given fc= 4 ksi fy = 60ksi d=5.06 in Required rebar ratio 0.85*fc 2* Mnrequired p= *(1 fy 0.85* fc*b* d? 0.85* 4 2*68.50 P= *(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 V

Spacing of #3 bars Area of one #3 bars A =0.11 in Area of one #3 bars A =0.11 in? A*12 0.11*12 = 5.65" As 0.2335 hence provide # 3 bars at s = 5.5" S S(min) ACI 318 specify minimum area of steel as bw*d An *max(3* Vfc ,200) fy bw = b=12" max(3* fc ,200) = max(3* /4000 ,200) max(189.73 ,200) = 200 5.06*12* 200 A = 0.2024 in? 60,000 As>A = S(min) S(min) hence ok

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

check for shear

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* fc b=l'=12" where 1 = 1 for normal weght concrete Ve=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 ØVn = 5692.09 lb Vy = 2720 lb Hence Vn>VU Safe in shear