Question

negative moment Q2:-(50%) Design for the maximum negative and maximum shear of a continouse one-way sond slab supported on beams spaced at in ft on centers. The width of the beam is 12 in. The uniform dead lood of 12opsf (including self weight of slab) and live read of 100 psf. use ACI_318 code moment and shear coefficents for continouse beams and slab. Pr = Hksi fy= boksi, h=6in, d = 5:06 in, try #3 bals. slab carries a

Answer 1

ACI 318 6.5 — 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 w *Ln 10 Ln is the Clear span 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* In Maxium negative shear = 2 1

Given Center to center spacing of beam = 14" Width of beam = 12" Clear spacing Ln=14-1 = 13" Loading Dead Load = 120 psf Live load = 100 psf Consider unit with of slab b=1 ft WDL = 120* 1 = 120plf WLL = 100* 1 = 100 plf ASCE 7 LRFD Factored load wu =1.2* w +1.6*WLL =1.2* 120+1.6*100 = 304 plf = 0.304 k/ft 0.304 * 132 Maximum negative moment Mn = 5.1376 k-ft 10 0.304 * 13 Maximum negative shear Vu` = 1.15* = 2.272 kips 2 DL

= Design of main rebar Mu Mn - required = 0 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 0.85* fc 2* Mnrequired p= fy 0.85* fc*b*d? 0.85* 4 2*68.50 60 0.85* 4*12*5.062 solving p=0.003846 As = 0.003846*12*5.06 = 0.2335 in? / unit width C*(1 1 P= (1-1- -- )

for #3 bars Area of one bar A = 0.11 in 2 Spacing of #3 bars A*12 0.11*12 S= = 5.65" As 0.2335 hence provide # 3 bars at s = 5.5" ACI 318 specify minimum area of steel as bw*d A max(3* Vfc ,200) S(min) fy bw = b = 12" max(3* Vfc ,200 )= max(3* 4000 ,200) = max(189.73 ,200) = 200 5.06*12*200 A = 0.2024 in? S(min) 60,000 As>A S(min) hence ok

Design Shear capacity n = ØVc ØVc 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=1'=12" where 1 = 1 for normal weght concrete Vc = 2*1*12*5.06* 4000 = 7680.531b Strength reduction factor for shear design is O=0.75 Vc =0.75* 7680.53 = 5760.404 lb ØVn = 5692.09 lb Vu = 2720 lb Hence ØVn > V Safe in shear =

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