Question

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

Answer 1

Answer:- given diagram below
4 in 12417 Span=24 4 Ft Fc= 4,500 PSI G stoksi given data- ky = –60,00opsi, Fe=4500 psi Assumption, BW=12 in, d=24-3= 21 in. therefore, Live load (LoL)= 40p5f (as peE ASCE 2010 for driveway and cat parking) Dead road (OL) slab = 4 X150 = Sopsf 12 Load of slab on beam cinud!)=50x4X24 24 = 2001blft Beam =12* (24-4)X150 144 = 250lblft

Total dead load on beam (wd))=4 50161 Ft = 0:45 Kipsiff Live load in beam cin udl) WII = 40X4X4 24 = 160lb/ft=0:16 kipSifh Design love = 12 XO-L+1.6XL.L =12 X 0.45 +1.6X0:16 = 0.796 kips1 Ft maximum bend ending moment (mu) = w XIXI 8 = 0.796X24X24 8 =57.312ft kips Design moment (mn)=MU = 57.312 o 9 - 63.687+-kips Effective flange width oft 1 Хая 4 372 in b bw +johf = 12 +16X4= 76 in © CIC clc spacing of beam = 471 - 48 in USE b F = 48 in

assuming leveɛ arm z equal to the larger of ogd ord-chf/2) 2= 0.gxd=0.9x21=1:in z = d - Ch112) = 21-(412) = 19 in therefore, Trial Steel area AS X fyX ZEMIN xfy X 2= AS = 63.68X12 (COX19) 0.67 squin computing value of g and 2 0 -85x8 X 60 = ASA. ASXFO 0.85 X 4.5 X AC = 0.67 X 60 AC=10.Si se in 248X4 =192 sq in (NA in frange) a = 10:51 sain 148 =0.22in d - 2/2 = 21-10-2212)= 20-89 in z =

calculating as with this reviseal 2. AS = 63.68X12 1160X20-89) = 0.61 Sain therefore, computing value of a and z 0.85 x F'C XACS Asxty 0.85X4.5 XACS 0.61 X 60 AC = g.57 sqin a=9.57 sqin 48 2 = = 0.20 in d-a/2 = 21- (0.2012) 2ogin - as with this revised z Calculating AS=63.68 X 12 60X20.9 =0.61 sin As min = 0.0033 X 12 X21 = 0.83 sq.in minimum checking seinforcing As min = 3x8qrt ( f'c) X bWxd = 3x sqrt (4500)X12X21160,000 = 0.84 sq in alty

values of a and z computing O-85x'c 10 cá nsx Asxfy : 8 5 XS K0 cc 5:8AX 6 6 AC = 13.18 Sq.in a = 13:18 sqin 148 0.27 in z d-a/2 = 21-00-27) = 20.865 in therefore, compe C, Et and phi imputing C = al betal=0:27_= 0:3318 825 Ets (4-6)*0.00310 = (21-0.33) X0.003 033 = 0-18870.005 ( 50 phizo.9 as assiemed) As required = 0.84 Sqin used 3 #s bacs (Asso:gisq.in) max 5.6 Vu= W0X1/2 = 0·796x24 = 9:552 kips

Vu at a distance de 9.552- 21X0.796112 = 8.159 kips Pnixve=phix 2 x lemda x sqrter'c) xbxd = 0.75X2X1 xsqet (4500) X 12 X2) = 25357 16 - 25 36 Kips Nominal sheas sein Forcement required # 3 Stirreps (Av=011 sqin ) S = AVX Fyxd I v5 = 2x0:11X60,000 X 21 (8-159X163) = 33.97 in maximum spacing to provide minimum Av Stirrups s= AV 0.75 x59et (HC) Xby 2x0.11 X60,000/ {0.75 x sqet(4500)x123 AVX F xfyl { = 21.86 in - Av xfy / ( 50xb ) = 2X0.11860,000| ( 50x12) 22 in Hence max spacing 214-2114-5.25in use # 3 @ Sin spacing center to center.