Question

2. Design a square spread footing for the following conditions. Service dead load is 350 kips, service live load is 275 kips. Soil density is 130 lb/ftº. Allowable soil pressure is 4500 psf. Column is 18 in. square. Geotechnical analyses have determined the width to be 13 ft. fe' = 3500 psi normal weight concrete, and fy= 60 ksi. The base of the footing is placed 5 ft below the ground surface. Note: you don't need to design the bar development length, and the connection between column and footing

Answer 1

18 in Given Dead load = 350 kips Live Load = 275 kips Soil density = 130 16 (ft3 Allowable Soil pressure = 4500 psf Column width = 13 ft fc = 3500 psi fy= 60ksi base of footing = sf factored ctored load (pu) 1.2(04) + 1-c(24) (1.2*350) + (1-6 X275) 420+ yub sbok weight of footing = dx soil density Where Assume da 20" Z 20 X 130 1.66x Bo f = 216.67 -psfl weight of soil fill on top of footing (ws) (ws)= (5 - 20 X 130 (2 5-1.66)x136 433. 33 psf We 31 pof Area = Total load effective Sil pressure

2 850+ 275 3.85 625 3.85 A 162.33ft² (24) Net pressure - Area 860 effective Soil 162.33 5:06 ft pressure at bottom of footing pe = Allowable forl pressure) - {weight of footing] u500 216.67 - 433.33 pe= le = 3.85 ksf effective depth by two way shear Ø Vez= & Ves d d bo Vuz According to ACI Code Øvci = 8uxWfc bod CBS2) (2 + 4 & 5fd body (B72) (4&d to) Ife bod Pu -- qu (Ltd.) for square column øvere Vuz ou x fc, bo d = Pu - qu (Cador bo= ulcedi) 16 x x Tfc ((td) di = le-au (Ctd) 2 0.75 % ux $3.500 (18+d) di = 860000 - 35.14 (18+d.)?

3 147. US (18+ di) di = 860000 - 35.14 ( 324 +0,736 d.) 3 194.680, + 177.48d, 2 = 860000 - 11385.36 - 35.14d, 2 - 1265.oud, de 20 3 21a-ssdir + 44 59.72d, - 848614.64co da 9.83 N 10" & vc3= duz & las d3 +2} & Ifle body = Pu- qu (Ct dz) ~ on solving we get d₂ = 5.43" d- h-cover-Ź (bar diameter) 놀 (0.875) Cok) Maximum Bending moment (Mu) Mu Ź (qu) ( - ) 2 xB e 1 / 2 x 5.068 23 - 19 ) *x! 0.5 x 5.06 x 6.254 13 1087.42 k-ft 12 k-ft! reinforcement d = 16.56 > 10 11 Mu Primary Mu bdr = 1087.42 X 12 X 1000 Ra 13 x 12 x 16.54 1304 goyo 2580-24 5057.29 psi S. 0.85 fc ( fy - 2 Ru 0x0.85x fcl - 085x3.5 |- 2x5057.28 به 60 0.9 x 0.8584

f = 0.0143 As = fbd = 0.0143 X 13 X1 2 x 16.54 - 36.89 = 37 in 2 20 # 7 bars @ qu in both corection Spacing = (13 x12) 18-1 156-6 17 150 17 8.823 29 th Bearing foot N₂ = Ni A s 2N, Nyfø (0.85% fc xA1) A) 0.65 0.85 X 3.5 X 18 X 18 Ñ, - 6 26.531 No=626.53 x 13 x 13 x144 52(626.5) 188 18 5429.9 5 125 3.06 Cok)

Answer 2

ANSWER:------------

18 in Given Dead load = 350 kips Live Load = 275 kips Soil density = 130 16 (ft3 Allowable Soil pressure = 4500 psf Column width = 13 ft fc = 3500 psi fy= 60ksi base of footing = sf factored ctored load (pu) 1.2(04) + 1-c(24) (1.2*350) + (1-6 X275) 420+ yub sbok weight of footing = dx soil density Where Assume da 20" Z 20 X 130 1.66x Bo f = 216.67 -psfl weight of soil fill on top of footing (ws) (ws)= (5 - 20 X 130 (2 5-1.66)x136 433. 33 psf We 31 pof Area = Total load effective Sil pressure

2 850+ 275 3.85 625 3.85 A 162.33ft² (24) Net pressure - Area 860 effective Soil 162.33 5:06 ft pressure at bottom of footing pe = Allowable forl pressure) - {weight of footing] u500 216.67 - 433.33 pe= le = 3.85 ksf effective depth by two way shear Ø Vez= & Ves d d bo Vuz According to ACI Code Øvci = 8uxWfc bod CBS2) (2 + 4 & 5fd body (B72) (4&d to) Ife bod Pu -- qu (Ltd.) for square column øvere Vuz ou x fc, bo d = Pu - qu (Cador bo= ulcedi) 16 x x Tfc ((td) di = le-au (Ctd) 2 0.75 % ux $3.500 (18+d) di = 860000 - 35.14 (18+d.)?

3 147. US (18+ di) di = 860000 - 35.14 ( 324 +0,736 d.) 3 194.680, + 177.48d, 2 = 860000 - 11385.36 - 35.14d, 2 - 1265.oud, de 20 3 21a-ssdir + 44 59.72d, - 848614.64co da 9.83 N 10" & vc3= duz & las d3 +2} & Ifle body = Pu- qu (Ct dz) ~ on solving we get d₂ = 5.43" d- h-cover-Ź (bar diameter) 놀 (0.875) Cok) Maximum Bending moment (Mu) Mu Ź (qu) ( - ) 2 xB e 1 / 2 x 5.068 23 - 19 ) *x! 0.5 x 5.06 x 6.254 13 1087.42 k-ft 12 k-ft! reinforcement d = 16.56 > 10 11 Mu Primary Mu bdr = 1087.42 X 12 X 1000 Ra 13 x 12 x 16.54 1304 goyo 2580-24 5057.29 psi S. 0.85 fc ( fy - 2 Ru 0x0.85x fcl - 085x3.5 |- 2x5057.28 به 60 0.9 x 0.8584

f = 0.0143 As = fbd = 0.0143 X 13 X1 2 x 16.54 - 36.89 = 37 in 2 20 # 7 bars @ qu in both corection Spacing = (13 x12) 18-1 156-6 17 150 17 8.823 29 th Bearing foot N₂ = Ni A s 2N, Nyfø (0.85% fc xA1) A) 0.65 0.85 X 3.5 X 18 X 18 Ñ, - 6 26.531 No=626.53 x 13 x 13 x144 52(626.5) 188 18 5429.9 5 125 3.06 Cok)