Circular footing depth, 2000 kN

Given: circular footing $B = 2.5$ m, gross load $Q = 2000$ kN, $\phi = 30^\circ$, $c = 0$, $\gamma = 19$ kN/m³, FS = 3, $N_q = 22.5$, $N_\gamma = 19.7$.

Terzaghi's equation for a circular footing: $q_u = 1.3cN_c + \gamma D_f N_q + 0.3\gamma B N_\gamma$. With $c = 0$ the first term is zero.

Required ultimate capacity

$$\begin{aligned} A &= \frac{\pi}{4}(2.5)^2 = 4.909\ \text{m}^2 \\ q_{applied} &= \frac{2000}{4.909} = 407.4\ \text{kN/m}^2 \\ q_u &= FS \times q_{applied} = 3 \times 407.4 = 1222.3\ \text{kN/m}^2 \end{aligned}$$

Solve for depth

$$\begin{aligned} 0.3\gamma BN_\gamma &= 0.3 \times 19 \times 2.5 \times 19.7 = 280.7\ \text{kN/m}^2 \\ \gamma D_fN_q &= 19 \times 22.5 \times D_f = 427.5\,D_f \\ 1222.3 &= 427.5\,D_f + 280.7 \\ D_f &= \frac{941.6}{427.5} = 2.20\ \text{m} \end{aligned}$$

Answer: the footing should be placed at a depth of about 2.2 m (gross capacity basis; $D_f/B = 0.88 < 1$, so the shallow-footing equation is valid).