Square footing size in clay, 750 kN

Given: square footing, $Q = 750$ kN, clay with $c = 35$ kN/m², $\gamma = 17$ kN/m³, FS = 3. The depth is not given.

Assumptions: the clay is purely cohesive ($\phi = 0$), so Terzaghi's factors are $N_c = 5.7$, $N_q = 1$, $N_\gamma = 0$. Because the footing is backfilled, the weight of footing plus backfill is taken as about equal to the soil it replaces, so the net load intensity on the soil is $Q/B^2$ and the net ultimate capacity is used.

$$q_u = 1.3cN_c + \gamma D_f N_q \Rightarrow q_{nu} = q_u - \gamma D_f = 1.3cN_c$$ $$\begin{aligned} q_{nu} &= 1.3 \times 35 \times 5.7 = 259.35\ \text{kN/m}^2 \\ q_{ns} &= \frac{259.35}{3} = 86.45\ \text{kN/m}^2 \\ B^2 &= \frac{750}{86.45} = 8.68\ \text{m}^2 \Rightarrow B = 2.95\ \text{m} \end{aligned}$$

Since the depth cancels out ($N_q = 1$), the answer does not depend on $D_f$.

Answer: provide a square footing of 3.0 m × 3.0 m (net pressure $750/9 = 83.3$ kN/m² < 86.45 kN/m²; FS $= 259.35/83.3 = 3.11$).