A 5ft by 5ft square footing is located 4 ft below the ground

A 5-ft by 5-ft square footing is located 4 ft below the ground surface. The footing is subjected to an eccentric load of 75 kips. The subsoil consists of a thick deposit of cohesive soil with qu=4.0 kips/ft2 and =130 lb/ft3. The water table is at a great depth, and its effect on bearing capacity can be ignored. Calculate the factor of safety.

Solution

Solution:-

Given

size of footing = 5 x5 feet

depth of footing (D) = 4 feet

y = 135 lb/feet³

let e =0.64 feet

effective width(B’) = B – 2e = 5 – 2 * 0.64 = 3.72 feet

For =0

Taking N_y = 0 , N_c = 5.14 , N_q = 1

q_u   = CN_c *s_c *d_c*i_c + q N_q*s_q*d_q*i_q

s_c =1 + (B/L)(N_q/N_c)

     = 1 + (3.72/5) *(1/5.14) = 0.1447

s_q = 1 + (B’/L)tan

      = 1 + (3.72/5) *tan0 = 1

d_c = 1 + 0.4 *(D_f/B) = 1 + 0.4 *4/5 = 1.32 feet

d_q = 1 + 2tan(0) (1 – sin 0)²(D_f/B) =1

q_u = CN_cs_cd_ci_c + qN_qs_qd_qi_q

         = 4*10³ *5.14*1.32 + (4*130)*1*1*1 = 27659.2 lb/feet²

q_max = Q/(BL) * (1 + 6e/B)

             = 75000/(5*5) * (1 + 6 * 0.64/5) = 5304 lb/feet²

Now

q_nu = 27659.2 – 130*4

       = 27139.2 lb/feet²

q_max = (27139.2/F + 130*4)

5304 = 27139.2/F +520

Factor of safety(F) = 5.67   Answer