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2. A medium dense sand is proposed to support a square foundation having a width and length of 5 feet as shown below. The bottom of the footing is 2.5 feet below the ground surface. The water table is 4 feet below the bottom of the footing. Using a factor of safety of 3, what is the allowable bearing capacity of the proposed footing? Gs = 2.6 e = 0.5 ø = 35° 5 ft d = 4 ft

2. A medium dense sand is proposed to support a square foundation having a width and length of 5 feet as shown below. The bottom of the footing is 2.5 feet below the ground surface. The water table is 4 feet below the bottom of the footing. Using a factor of safety of 3, what is the allowable bearing capacity of the proposed footing? Gs = 2.6 e = 0.5 ø = 35° 5 ft d = 4 ft

Fundamentals of Geotechnical Engineering (MindTap Course List)
Fundamentals of Geotechnical Engineering (MindTap Course List)
5th Edition
ISBN:9781305635180
Author:Braja M. Das, Nagaratnam Sivakugan
Publisher:Braja M. Das, Nagaratnam Sivakugan
Chapter16: Shallow Foundations-bearing Capacity
Section: Chapter Questions
Problem 16.6P: The applied load on a shallow square foundation makes an angle of 15 with the vertical. Given: B =...
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2. A medium dense sand is proposed to support a square foundation having a width and length of 5 feet as shown below. The bottom of the footing is 2.5 feet below the ground surface. The water table is 4 feet below the bottom of the footing. Using a factor of safety of 3, what is the allowable bearing capacity of the proposed footing? Gs= 2.6 e = 0.5 Ø = 35° 5 ft d = 4 ft
Transcribed Image Text:2. A medium dense sand is proposed to support a square foundation having a width and length of 5 feet as shown below. The bottom of the footing is 2.5 feet below the ground surface. The water table is 4 feet below the bottom of the footing. Using a factor of safety of 3, what is the allowable bearing capacity of the proposed footing? Gs= 2.6 e = 0.5 Ø = 35° 5 ft d = 4 ft
Olution: 2u- unconfinad compessive stength YE120 pof Lu =12 tof Df =3ft :- 4A x 6ft c undraind ehear shength Here <g loral whear fal lure is AFqumd. B<4f = tan ( tant) = tan tand) 10 ft = 467° itsf=2000 psf c= 0-4x2000=B00psf :- d= 0:4 tsf For soil below the baue of footra to D= f, sguralent densty = 120psf * uthimate beruig copaidy ob footing (RectngelSo), 2u = CNC + YPN2 +('-02)BYNr whore, No Ng & Nr core the fun dios of . whou (ST/4 - (3-4.0%) tan 407 Ng = e 2 cas(45 + % 2 cos²(45°+ tpe7 Ng 1-217 = 1329 Ne = (Ng-1) otE 229-1) cot 4.0€ = 3.86) tand Kpr wheu Kpr=I+ Sind (-Sing Nr= tan 4. 07 185 NSE O-0082 New subeftute 18 abore , 20=(1+03x4)x 800x 3061 + (120xBX1329) +(1-0ax4)x4x12oro.00 82 3552.12 + 478-44 + 77/2 %3D 2= 4032.33 psf
Transcribed Image Text:Olution: 2u- unconfinad compessive stength YE120 pof Lu =12 tof Df =3ft :- 4A x 6ft c undraind ehear shength Here <g loral whear fal lure is AFqumd. B<4f = tan ( tant) = tan tand) 10 ft = 467° itsf=2000 psf c= 0-4x2000=B00psf :- d= 0:4 tsf For soil below the baue of footra to D= f, sguralent densty = 120psf * uthimate beruig copaidy ob footing (RectngelSo), 2u = CNC + YPN2 +('-02)BYNr whore, No Ng & Nr core the fun dios of . whou (ST/4 - (3-4.0%) tan 407 Ng = e 2 cas(45 + % 2 cos²(45°+ tpe7 Ng 1-217 = 1329 Ne = (Ng-1) otE 229-1) cot 4.0€ = 3.86) tand Kpr wheu Kpr=I+ Sind (-Sing Nr= tan 4. 07 185 NSE O-0082 New subeftute 18 abore , 20=(1+03x4)x 800x 3061 + (120xBX1329) +(1-0ax4)x4x12oro.00 82 3552.12 + 478-44 + 77/2 %3D 2= 4032.33 psf
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