Question

Make any reasonable assumptions Q1 (50 pts): Design the T-beam shown below for bending: 4 in 24 in 4 ft Span=24 Use: Driveway and car parking f' = 4,500 psi fy= 60 ksi

Answer 1

fy=60000 psi, fc= 4500 psi, Assumption bw= 12 in , d = 24-3 =21 in, Live load (L.L )= 40 psf ( As per ASCE 2010 for driveway & car parking )) Dead load (D.L) slab = 4 x 150 /12 = 50 psf load of slab on beam ( in udl) = 50 * 4 * 24/24 = 200 lb/ft beam = 12 x ( 24-4 ) x 150 /144 = 250 lb/ft total dead load on beam (Wdl ) = 450 lb/ft = 0.45 kips/ft live load in beam ( in udl) WII= 40 x 4 x 24 /24= 160 lb/ft =0.16 kips /ft Design load = 1.2 *D.L. + 1.6* L.L = 1.2 * 0.45 + 1.6* 0.16 = 0.796 kips/ft Maximum bending moment (Mu ) = w*l*1/8 = 0.796 * 24*24/8 = 57.312 ft-kips Design moment (Mn) = Mu/ = 57.312/0.9 = 63.68 ft- kips Effective flange width a) 1* 24 / 4 = 6 ft 72 in b) bw + 16hf = 12 + 16 x 4 = 76 in c) c/c spacing of beam = 4 ft = 48in USE bf = 48 in Z = assuming lever arm z equal to the larger of 0.9d or d-hf/2) 0.9 x d= 0.9 x 21 = 18.9 in z = d- (hf/2) = 21-(4/2) = 19 in Trial steel area As x fy xz = Mn As = 63.68 x 121(60 x 19 ) = 0.67 sqin Computing values of a and z 0.85 x fc x Ac = As xfy 0.85 x 4.5 x Ac = 0.67 x 60 Ac = 10.51 sq in < 48 x 4 = 192 sq in (N.A in flange ) a=10.51 sqin / 48 = 0.22 in z = d-a/2 = 21-(0.22/2) = 20.89 in Calculating As with this revised z As = 63.68 x 12 /( 60 x 20.89 ) = 0.61 sqin

Computing values of a and z 0.85 x fc x Ac = As xfy 0.85 x 4.5 x Ac = 0.61 x 60 Ac = 9.57 sq in a= 9.57 sqin / 48 = 0.20 in z= d-a/2 = 21- (0.20/ 2 ) = 20.9 in Calculating As with this revised z As = 63.68 x 12 /( 60 x 20.9 ) = 0.61 sqin ( ok close enough to preivious value) As min = 0.0033 x 12 x 21 = 0.83 sqin Checking minimum reinforcing As min = 3 x sqrt (fc) x bwxd/fy = 3 xsqrt ( 4500 ) x 12 x 21 / 60000 =0.84 sqin but not less than As min = 200 x bw xd/fy = 200 x 12x 21/600000 = 0.84 sqin OK Computing values of a and z 0.85 x fc x Ac = As xfy 0.85 x 4.5 x Ac= 0.84 x 60 Ac = 13.18 sq in a= 13.18 sqin / 48 = 0.27 in z= d-a/2 = 21- (0.2712 ) = 20.865 in Computing c, Et and phi c=a/ beta 1 = 0.27 1.825 = 0.33 in Et= (d-c)x 0.003/c = ( 21-0.33 ) x 0.003/0.33 = 0.188 > 0.005 ( so phi = 0.9 as assumed) As reqd = 0.84 sq in use 3 #5 bars (As = 0.91 sq in ) Max S.F Vu= Wux1/2= 0.796 x 24 12 = 9.552 kips Vu at a distance d = 9.552-21*0.796/12=8.159 kips phi x Vc = phi x 2x lemda x sqrt (fc) x 5 xd = 0.75 x 2 x 1 x sqrt ( 4500 ) x 12 x 21 = 25357 lb=25.36 kips Nominal shear rein forecement required Taking #3 Stirrups (Av = 0.11 sqin ) S= Av xfy xd/Vs = 2 x 0.11x 60000 x 21 / (8.159x 10^3 )= 33.97 in Maximum spacing to provide minimum Av stirrups S= Av x fy/{ 0.75 x sqrt ( f'c) b}= 2 x 0.11 x 60000 / {0.75 x sqrt ( 4500 ) x 12}= 21.86 in S= Av xfy / ( 50 x b) = 2 x 0.11x 60000 / ( 50 x 12) = 22 in 4 sqrt (4500) x 12x 21= 67.62 kips > Vu hence max spacing d/4 = 21/4 = 5.25 in Use #3 @ 5 in spacing center to center