A 5x5 square foundation is constructed 35 ft below the groun

A 5’x5’ square foundation is constructed 3.5 ft below the ground surface. The water table elevation is 2.75 ft below ground surface. Given dry=106 pcf, ’=28 deg, c’=95 psf, above water table w=5%, below water table w=20%. Using the Terzaghi bearing capacity equation and a safety factor of 3, calculate the allowable gross vertical force in units of [kip] the foundation can resist.

Solution

Solution:-

Given

Size of footing = 5’ *5’

C’ = 95 psf

y_dry = 106 pcf

y_dry =   y/(1 + w)

y = 106 * (1+ 0.05) = 111.3 pcf

y\' = 106 * (1 + 0.2) = 127.2 pcf

D_f = 3.5 feet

D_w = 2.75 feet

’ = 28 degree

N_c = 32   ,   N_q = 18 ,   N_y = 16

For square footing

Q_u = 1.2 C’ N_c + ( y’ D_f + (y - y’ ) * D_w ) * N_q + 0.4 * y’ B N_y

Q_u = 1.2 * 95 * 32 + ( 127.2 *3.5 + ( 111.3 – 127.2 )* 2.75 )* 18 + 0.4 * 127.2 *5 * 16

Q_u = 14944.95 psf

Allowable Q_a   = 14944.95/(factor of safety)

                  Q_a = 14944.95/3 = 4981.65 psf

Allowable gross vertical force (F ) = Q_a * (5 *5)

= 4981.65 * 25

Allowable gross vertical force(F) = 124.541 kip Answer