Circular footing on stiff clay, water table

Given: circular footing on stiff clay, $q_{uc} = 250$ kN/m², so $c = q_{uc}/2 = 125$ kN/m² and $\phi = 0$. $D_f = 2$ m, $Q = 700$ kN, FS = 2.5, $\gamma = 20$ kN/m³.

For $\phi = 0$ (Terzaghi): $N_c = 5.7$, $N_q = 1$, $N_\gamma = 0$. Circular footing: $q_u = 1.3cN_c + \gamma D_f N_q$.

(a) Diameter (water table deep)

$$\begin{aligned} q_u &= 1.3 \times 125 \times 5.7 + 20 \times 2 = 926.25 + 40 = 966.25\ \text{kN/m}^2 \\ q_{nu} &= q_u - \gamma D_f = 926.25\ \text{kN/m}^2 \\ q_{ns} &= \frac{926.25}{2.5} = 370.5\ \text{kN/m}^2 \\ q_s &= q_{ns} + \gamma D_f = 370.5 + 40 = 410.5\ \text{kN/m}^2 \end{aligned}$$ $$\begin{aligned} A &= \frac{700}{410.5} = 1.705\ \text{m}^2 \\ d &= \sqrt{\frac{4 \times 1.705}{\pi}} = 1.47\ \text{m} \end{aligned}$$

Provide a diameter of 1.5 m (capacity $410.5 \times 1.767 = 725$ kN > 700 kN).

(b) Water table at ground level

$\gamma' = 20 - 9.81 = 10.19$ kN/m³, so the overburden becomes $q' = 10.19 \times 2 = 20.38$ kN/m². The $N_\gamma$ term is zero, so only the surcharge changes.

$$\begin{aligned} q_u' &= 926.25 + 20.38 = 946.63\ \text{kN/m}^2 \\ q_{nu}' &= 946.63 - 20.38 = 926.25\ \text{kN/m}^2 \\ q_s' &= \frac{926.25}{2.5} + 20.38 = 390.88\ \text{kN/m}^2 \end{aligned}$$

Answer: diameter $\approx 1.5$ m. If the water table rises to the ground level, the ultimate and safe bearing capacities fall by 19.62 kN/m² each, while the net ultimate capacity is unchanged (clay with $\phi = 0$ is affected only through the overburden).