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
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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i asked this question before and i got an answer, but i have a different answer that is different from who solved it, can you recheck ?
![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe26036de-a789-41c6-8275-0b28efa071c7%2F7a0cb91b-f2fb-4bed-ad85-432c61441796%2Fte2180l_processed.png&w=3840&q=75)
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe26036de-a789-41c6-8275-0b28efa071c7%2F7a0cb91b-f2fb-4bed-ad85-432c61441796%2F5w70905_processed.png&w=3840&q=75)
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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