Strip footing FOS, three water table cases

Given: strip footing $B = 2$ m, $D_f = 1.2$ m, $c = 0$ (cohesionless), $N_q = 41$, $N_\gamma = 42$, applied pressure $q = 400$ kN/m², $\gamma = 16.8$ kN/m³, $\gamma_{sat} = 19.5$ kN/m³, $\gamma_w = 9.81$ kN/m³. FOS $= q_u/q$. Strip footing: $q_u = \gamma D_f N_q + 0.5\gamma B N_\gamma$.

(i) WT at 4 m below GL

The water table is $4 - 1.2 = 2.8$ m below the base, greater than $B = 2$ m, so it has no effect.

$$\begin{aligned} q_u &= 16.8 \times 1.2 \times 41 + 0.5 \times 16.8 \times 2 \times 42 \\ &= 826.6 + 705.6 = 1532.2\ \text{kN/m}^2 \\ FOS &= 1532.2/400 = 3.83 \end{aligned}$$

(ii) WT at 2.5 m below GL

$Z_{w2} = 2.5 - 1.2 = 1.3$ m below the base ($< B$), so $R_{w2} = 0.5(1 + 1.3/2) = 0.825$. The surcharge is unchanged ($R_{w1} = 1$).

$$\begin{aligned} q_u &= 826.6 + 705.6 \times 0.825 = 826.6 + 582.1 = 1408.7\ \text{kN/m}^2 \\ FOS &= 1408.7/400 = 3.52 \end{aligned}$$

(iii) WT at 0.5 m below GL

The water table is above the base. $\gamma' = 19.5 - 9.81 = 9.69$ kN/m³.

$$\begin{aligned} q &= 16.8 \times 0.5 + 9.69 \times 0.7 = 15.18\ \text{kN/m}^2 \\ q_u &= 15.18 \times 41 + 0.5 \times 9.69 \times 2 \times 42 \\ &= 622.5 + 407.0 = 1029.5\ \text{kN/m}^2 \\ FOS &= 1029.5/400 = 2.57 \end{aligned}$$

Answer: FOS = 3.83 (WT at 4 m), 3.52 (WT at 2.5 m) and 2.57 (WT at 0.5 m). A rising water table lowers the factor of safety.