Final settlement and settlement at 0.5 year

Data from the previous profile: clay thickness $H = 6$ m (8 m to 14 m), $e_0 = 2.4$, $C_c = 1.2$, $c_v = 3.25\times10^{-7}$ m²/s. At the clay centre: $\sigma_0' = 117$ kPa; final $\sigma' = 117 + 15 = 132$ kPa. The clay is assumed normally consolidated.

Final primary settlement

$$\begin{aligned} S_c &= \frac{C_c H}{1+e_0}\log_{10}\frac{\sigma_0'+\Delta\sigma}{\sigma_0'} \\ &= \frac{1.2 \times 6}{3.4}\log_{10}\frac{132}{117} = 2.1176 \times 0.05238 = 0.111\ \text{m} \end{aligned}$$

$S_c \approx 111$ mm.

Settlement after 0.5 year

Sand lies above and below the clay, so drainage is two-way: $H_{dr} = 6/2 = 3$ m.

$$t = 0.5 \times 365 \times 86400 = 1.577\times10^{7}\ \text{s}$$ $$T_v = \frac{c_v t}{H_{dr}^2} = \frac{3.25\times10^{-7}\times1.577\times10^{7}}{3^2} = 0.569$$

From the given table, $T_v = 0.569$ corresponds to $U = 80\%$.

$$S_t = U \times S_c = 0.80 \times 111 = 88.8\ \text{mm}$$

Answer: final settlement $\approx 111$ mm; settlement after 0.5 year $\approx 89$ mm (U = 80%).