Question

For a clayey sand c’ = 10kPa, ?’ = 32 and ? = 18kN/m 3 , a 2m x 2m footing is placed at depth of 0.5 m. The unit weight of concrete is 23 kN/m3 . What is the maximum axial column load allowed on the footing with factor of safety of 3? (b) If the column load is 200mm eccentric to one of the neutral axis, calculate the maximum compressive stress applied to the soil. What will be the factor of safety of the compressive stress against the bearing capacity calculated?

Answer 1

Step 1:

Given data:

C=10 Kpa

$\psi$ =32

$\gamma$ =18KN/m³

Square footing=2mx2m

D_f=0.5m

F.O.S=3

Step 2:

Calculate the utlimate bearing capacity of the soil

N_q=e^3.14Tan $\phi$ Tan²(45+ $\phi$ /2)

N_q=e^3.14Tan32Tan²(45+32/2)

N_q=23.04

N_p=0.1054exp(9.6x32x3.14/180)

N_p=22.134

S_q=1+B/LTan32

S_q=1+2/2Tan32

S_q=1.66

S_p=1-0.4B/L

S_p=1-0.4=0.6

d_p=1

d_q=1+2Tan $\phi$ (1-sin $\phi$ )²Arctan(D_f/B)

d_q=1+2Tan32(1-sin32)²arctan(0.5/2x3.14/180)

d_q=1.068

Q_u= $\gamma$ D_f(N_q-1)s_qd_q+0.5 $\gamma$ BN_ps_pd_p

Q_u=18x0.5x(23.04-1)x1.66x1.068+0.5x18x22.134x0.6x1.0

Q_u=471.18Kpa

Step 3:

Calculate the safe bearing capacity and the load acting:

Safe bearing capacity=Q_u/F+ $\gamma$ D_f

=471.18/3+18(0.5)

=166.06kpa

Load=166.06x2x2=664.24KN

The allowable load=664.24KN

(b) e=200mm

e=0.2m

Load=664.24KN

Maximum stress=P/BL(1+6e/B)

=664.24/4(1+6(0.2)/2)

=265.696Kpa

Reduced footing size=2-2(0.2)=1.6m

Calculate the shape and depth factors

S_q=1+B/LTan32

S_q=1+1.6/2Tan32

S_q=1.528

S_p=1-0.4B/L

S_p=1-0.4(1.5/2)=0.7

d_p=1

d_q=1+2Tan $\phi$ (1-sin $\phi$ )²Arctan(D_f/B)

d_q=1+2Tan32(1-sin32)²arctan(0.5/1.5x3.14/180)

d_q=1.078

Q_u= $\gamma$ D_f(N_q-1)s_qd_q+0.5 $\gamma$ BN_ps_pd_p

Q_u=18x0.5x(23.04-1)x1.528x1.078+0.5x18x22.134x0.7x1.0

Q_u=466.177Kpa

F.O.S=Q_u/( $\sigma$ _a- $\gamma$ D_f)

F.O.S=466.177/(265.696-0.5x18)

F.O.S=1.75