Consolidation test, 90% time, k, Cc

Assumptions: the specimen is drained at top and bottom (two porous stones), so $H_{dr}$ is half the average specimen thickness during the test. Average thickness $= 20 - 2.5/2 = 18.75$ mm, so $H_{dr} = 9.375$ mm. $\gamma_w = 9.81$ kN/m³.

(i) Time for 90% consolidation

Since $t \propto T_v$ for the same specimen:

$$t_{90} = t_{50}\,\frac{T_{90}}{T_{50}} = 154 \times \frac{0.848}{0.197} = 663\ \text{s} \approx 11\ \text{min}$$

(ii) Coefficient of permeability

$$c_v = \frac{T_{50}H_{dr}^2}{t_{50}} = \frac{0.197 (9.375)^2}{154} = 0.1124\ \text{mm}^2/\text{s} = 1.124\times10^{-7}\ \text{m}^2/\text{s}$$

Volume compressibility, with strain $= 2.5/20 = 0.125$ and $\Delta\sigma' = 120-60 = 60$ kPa:

$$m_v = \frac{0.125}{60} = 2.083\times10^{-3}\ \text{m}^2/\text{kN}$$ $$k = c_v m_v \gamma_w = (1.124\times10^{-7})(2.083\times10^{-3})(9.81) = 2.30\times10^{-9}\ \text{m/s}$$

(iii) Compression index

Change in void ratio:

$$\Delta e = \frac{\Delta H}{H_0}(1+e_0) = 0.125 \times 1.65 = 0.206$$ $$C_c = \frac{\Delta e}{\log_{10}(\sigma_1'/\sigma_0')} = \frac{0.206}{\log_{10}(120/60)} = \frac{0.206}{0.301} = 0.685$$

Answer: (i) $t_{90} \approx 663$ s (11 min); (ii) $k \approx 2.3\times10^{-9}$ m/s; (iii) $C_c \approx 0.685$.