Footing 2 m × 3 m, eccentric in X and Y

Given: $B = 2$ m (X-direction), $L = 3$ m (Y-direction), $D_f = 2$ m, $\phi = 35^\circ$, $c = 25$ kPa, $e_x = 0.2$ m, $e_y = 0.3$ m, WT 0.8 m below GL, $\gamma = 16.5$ kN/m³ (above WT), $\gamma_{sat} = 19.6$ kN/m³, $N_c = 57.8$, $N_q = 41.4$, $N_\gamma = 42.4$, FOS = 3, $\gamma_w = 9.81$ kN/m³.

Method: Meyerhof's effective area for double eccentricity, with Terzaghi's equation for a rectangular footing.

Effective dimensions

$$B' = 2 - 2(0.2) = 1.6\ \text{m},\quad L' = 3 - 2(0.3) = 2.4\ \text{m},\quad A' = 3.84\ \text{m}^2$$

Overburden pressure

The water table is above the base, so $\gamma' = 19.6 - 9.81 = 9.79$ kN/m³.

$$q = 16.5 \times 0.8 + 9.79 \times 1.2 = 13.2 + 11.75 = 24.95\ \text{kN/m}^2$$

Ultimate bearing capacity

$$\begin{aligned} q_u &= \left(1 + 0.3\tfrac{B'}{L'}\right)cN_c + qN_q + 0.5\left(1 - 0.2\tfrac{B'}{L'}\right)\gamma' B' N_\gamma \\ &= (1.2)(25)(57.8) + 24.95 \times 41.4 + 0.5(0.8667)(9.79)(1.6)(42.4) \\ &= 1734.0 + 1032.8 + 287.8 \\ &= 3054.6\ \text{kN/m}^2 \end{aligned}$$

Allowable load

$$Q_{all} = \frac{q_u}{FOS} A' = \frac{3054.6}{3} \times 3.84 = 3909.9\ \text{kN}$$

Answer: allowable load $\approx 3910$ kN (allowable pressure $q_u/3 = 1018$ kN/m² on the effective area; gross basis).