Footing 2 m × 3 m, chart factors, eccentricity

Given: $B = 2$ m, $L = 3$ m, $D_f = 1.5$ m, $e = 0.26$ m along the width, $c = 20$ kN/m², $\phi = 35^\circ$, $\gamma = 18$ kN/m³ (taken as the saturated unit weight), water table 0.5 m below GL, Terzaghi's theory.

Bearing capacity factors for $\phi = 35^\circ$ (Terzaghi chart, general shear): $N_c = 57.8$, $N_q = 41.4$, $N_\gamma = 42.4$. Factor of safety assumed $F = 3$, $\gamma_w = 9.81$ kN/m³.

Effective dimensions

$B' = B - 2e = 2 - 2(0.26) = 1.48$ m; $L = 3$ m.

Overburden and submerged weight

The water table is above the base, so the soil below the base is submerged: $\gamma' = 18 - 9.81 = 8.19$ kN/m³.

$$q = 18 \times 0.5 + 8.19 \times 1.0 = 17.19\ \text{kN/m}^2$$

Ultimate bearing capacity (rectangular footing)

$$\begin{aligned} q_u &= \left(1 + 0.3\tfrac{B'}{L}\right)cN_c + qN_q + 0.5\left(1 - 0.2\tfrac{B'}{L}\right)\gamma' B' N_\gamma \\ &= (1.148)(20)(57.8) + 17.19 \times 41.4 + 0.5(0.9013)(8.19)(1.48)(42.4) \\ &= 1327.1 + 711.7 + 231.6 \\ &= 2270.4\ \text{kN/m}^2 \end{aligned}$$

Safe bearing capacity

$$\begin{aligned} q_{nu} &= q_u - q = 2270.4 - 17.19 = 2253.2\ \text{kN/m}^2 \\ q_{ns} &= \frac{q_{nu}}{3} + q = 751.1 + 17.19 = 768.2\ \text{kN/m}^2 \end{aligned}$$

Safe load on the footing $= q_{ns} \times B'L = 768.2 \times 1.48 \times 3 = 3411\ \text{kN}$.

Answer: safe bearing capacity $q_s \approx 768$ kN/m² (net ultimate 2253 kN/m², FS = 3), corresponding to a safe load of about 3410 kN.