5 m clay, Cc 0.25, 65% consolidation

Assumptions: the layer is drained at top and bottom (double drainage), so $H_{dr} = 2.5$ m; $\gamma_w = 9.81$ kN/m³. $0.15\ \text{N/mm}^2 = 150$ kPa and $0.20\ \text{N/mm}^2 = 200$ kPa.

Step 1: Void ratio at 200 kPa

$$\Delta e = 0.25\log_{10}\frac{200}{150} = 0.0312, \qquad e_2 = 1.9 - 0.0312 = 1.869$$

Step 2: Settlement

$$S = \frac{\Delta e}{1+e_0}H = \frac{0.0312}{1+1.9}\times 5000 = 53.9\ \text{mm}$$

Step 3: Coefficient of consolidation

$$m_v = \frac{\Delta e}{(1+e_0)\Delta\sigma'} = \frac{0.0312}{2.9 \times 50} = 2.154\times10^{-4}\ \text{m}^2/\text{kN}$$

With $k = 3.2\times10^{-3}$ mm/s $= 3.2\times10^{-6}$ m/s:

$$c_v = \frac{k}{m_v \gamma_w} = \frac{3.2\times10^{-6}}{2.154\times10^{-4}\times9.81} = 1.514\times10^{-3}\ \text{m}^2/\text{s}$$

Step 4: Time for 65% consolidation

Since $U > 60\%$, use

$$T_v = 1.781 - 0.933\log_{10}(100-65) = 1.781 - 0.933(1.544) = 0.340$$ $$t_{65} = \frac{T_v H_{dr}^2}{c_v} = \frac{0.340 \times 2.5^2}{1.514\times10^{-3}} = 1405\ \text{s} \approx 23.4\ \text{min}$$

Answer: $e_2 = 1.869$; settlement $= 53.9$ mm; $t_{65} \approx 1405$ s (about 23 min). For single drainage ($H_{dr}=5$ m) the time would be about 94 min.