Strip footing, change in capacity with water rise

Given: strip footing, $B = 1.5$ m, $D_f = 1.2$ m, $c = 0$, $N_q = 81.3$, $N_\gamma = 100.4$, $G = 2.70$, $\gamma_d = 16$ kN/m³, $\gamma_w = 9.81$ kN/m³. For a strip footing $q_u = \gamma D_f N_q + 0.5\gamma B N_\gamma$ (the $cN_c$ term is zero).

Case 1: dry sand

$$\begin{aligned} q &= 16 \times 1.2 = 19.2\ \text{kN/m}^2 \\ q_{u1} &= 19.2 \times 81.3 + 0.5 \times 16 \times 1.5 \times 100.4 \\ &= 1560.96 + 1204.8 = 2765.76\ \text{kN/m}^2 \end{aligned}$$

Case 2: water table at 0.5 m below ground level

The sand below the water table is saturated. Find its saturated unit weight from $\gamma_d = \dfrac{G\gamma_w}{1+e}$:

$$\begin{aligned} 1 + e &= \frac{2.70 \times 9.81}{16} = 1.655 \Rightarrow e = 0.655 \\ \gamma_{sat} &= \frac{(G+e)\gamma_w}{1+e} = \frac{(2.70 + 0.655) \times 9.81}{1.655} = 19.88\ \text{kN/m}^3 \\ \gamma' &= 19.88 - 9.81 = 10.07\ \text{kN/m}^3 \end{aligned}$$

The water table is above the base, so the soil from 0.5 m to 1.2 m is submerged (0.7 m thick), and the soil below the base is also submerged, so $\gamma'$ is used in the $N_\gamma$ term:

$$\begin{aligned} q &= 16 \times 0.5 + 10.07 \times 0.7 = 15.05\ \text{kN/m}^2 \\ q_{u2} &= 15.05 \times 81.3 + 0.5 \times 10.07 \times 1.5 \times 100.4 \\ &= 1223.7 + 758.6 = 1982.3\ \text{kN/m}^2 \end{aligned}$$

Change in capacity

$$\Delta q_u = 2765.76 - 1982.29 = 783.5\ \text{kN/m}^2$$

Answer: the ultimate bearing capacity falls from 2765.8 kN/m² to 1982.3 kN/m², a reduction of about 783 kN/m² (28.3%).