Rectangular footing 3 m × 4 m, water table cases

Given: $B = 3$ m, $L = 4$ m, $D_f = 1.5$ m, $\phi = 34^\circ$, $c = 0$, $\gamma = 20$ kN/m³ (above WT), $\gamma_{sat} = 21.5$ kN/m³, $N_q = 29$, $N_\gamma = 41$, FS = 3, $\gamma_w = 9.81$ kN/m³. As the excavation is backfilled, the overburden pressure $q$ acts in full.

Terzaghi's rectangular footing ($c = 0$): $q_u = qN_q + 0.5\left(1 - 0.2\dfrac{B}{L}\right)\gamma B N_\gamma$, with $1 - 0.2(0.75) = 0.85$. Safe load $= \dfrac{q_u}{3}\times BL$ (gross).

(i) WT at 1 m below GL (0.5 m above the base)

Sand below the base is submerged: $\gamma' = 21.5 - 9.81 = 11.69$ kN/m³.

$$\begin{aligned} q &= 20(1.0) + 11.69(0.5) = 25.85\ \text{kN/m}^2 \\ q_u &= 25.85 \times 29 + 0.5(0.85)(11.69)(3)(41) \\ &= 749.5 + 611.1 = 1360.6\ \text{kN/m}^2 \\ Q_{safe} &= \frac{1360.6}{3}\times 12 = 5442\ \text{kN} \end{aligned}$$

(ii) WT at 1 m below the base

$Z_{w2} = 1$ m $< B = 3$ m, so $R_{w2} = 0.5(1 + 1/3) = 0.667$. The overburden is dry: $q = 20 \times 1.5 = 30$ kN/m².

$$\begin{aligned} q_u &= 30 \times 29 + 0.5(0.85)(20)(3)(41)(0.667) \\ &= 870.0 + 1045.5 \times 0.667 = 870.0 + 697.0 \\ &= 1567.0\ \text{kN/m}^2 \\ Q_{safe} &= \frac{1567.0}{3}\times 12 = 6268\ \text{kN} \end{aligned}$$

Answer: (i) safe load $\approx 5440$ kN; (ii) safe load $\approx 6270$ kN. The higher water table reduces the safe load by about 13%.