Primary settlement, sand over clay, 110 kPa

Assumptions: no water content is given, so the sand above the water table (top 1.5 m) is taken as dry and the sand below it as saturated. The clay is normally consolidated, $\gamma_w = 9.81\ \text{kN/m}^3$, and settlement is computed at the middle of the clay layer.

Step 1: Unit weights from $\gamma_d = \dfrac{G_s\gamma_w}{1+e}$ and $\gamma_{sat} = \dfrac{(G_s+e)\gamma_w}{1+e}$

$$\begin{aligned} \gamma_{d,sand} &= \frac{2.62(9.81)}{1.98} = 12.98\ \text{kN/m}^3 \\ \gamma_{sat,sand} &= \frac{(2.62+0.98)(9.81)}{1.98} = 17.84\ \text{kN/m}^3 \\ \gamma_{sat,clay} &= \frac{(2.70+0.62)(9.81)}{1.62} = 20.10\ \text{kN/m}^3 \end{aligned}$$

Step 2: Compression index

$$C_c = 0.0099(w_L-10) = 0.0099(50-10) = 0.396$$

Step 3: Effective stress at mid-clay (depth $3.5+1.75 = 5.25$ m)

$$\begin{aligned} \sigma_0' &= 12.98(1.5) + (17.84-9.81)(2.0) + (20.10-9.81)(1.75) \\ &= 19.47 + 16.06 + 18.01 = 53.54\ \text{kPa} \end{aligned}$$

Step 4: Settlement, with $\Delta\sigma = 110$ kPa

$$\begin{aligned} S_c &= \frac{C_c H}{1+e_0}\log_{10}\frac{\sigma_0'+\Delta\sigma}{\sigma_0'} \\ &= \frac{0.396(3.5)}{1.62}\log_{10}\frac{53.54+110}{53.54} \\ &= 0.8556 \times 0.4850 = 0.415\ \text{m} \end{aligned}$$

Answer: primary settlement of the clay layer $\approx 0.415\ \text{m} \approx 415\ \text{mm}$.

(If the sand above the water table were taken as saturated by capillarity, $\sigma_0' = 60.8$ kPa and the settlement would be about 384 mm.)