Settlement with preconsolidation 125 kPa

Given: clay 8 m to 14 m ($H = 6$ m), mid-depth at 11 m; $\sigma_c' = 125$ kPa, $C_c = 0.36$, $C_r = 0.06$, $\Delta\sigma = 50$ kPa, $\gamma_w = 10$ kN/m³.

Assumption: the initial void ratio of the clay is not given. Taking $G_s = 2.7$ and $\gamma_{sat} = 20$ kN/m³, $\frac{(2.7+e_0)(10)}{1+e_0}=20$ gives $e_0 = 0.70$.

 0 - 2 m   sand  gamma = 16
 2 - 8 m   sand  gsat  = 18   (WT at 2 m)
 8 - 14 m  clay  gsat  = 20   <- mid at 11 m

Step 1: Initial effective stress at 11 m

$$\sigma_0' = 16(2) + (18-10)(6) + (20-10)(3) = 32 + 48 + 30 = 110\ \text{kPa}$$

Step 2: State of the clay. $\sigma_0' = 110 < \sigma_c' = 125$, so the clay is overconsolidated. Final stress $\sigma_0'+\Delta\sigma = 160 > \sigma_c'$, so the stress path crosses the preconsolidation pressure and both $C_r$ and $C_c$ are needed.

Step 3: Settlement

$$\begin{aligned} S_c &= \frac{H}{1+e_0}\left[C_r\log_{10}\frac{\sigma_c'}{\sigma_0'} + C_c\log_{10}\frac{\sigma_0'+\Delta\sigma}{\sigma_c'}\right] \\ &= \frac{6000}{1.70}\left[0.06\log_{10}\frac{125}{110} + 0.36\log_{10}\frac{160}{125}\right] \\ &= 3529\,[0.00330 + 0.03859] = 148\ \text{mm} \end{aligned}$$

Answer: primary consolidation settlement $\approx 148$ mm (for $e_0 = 0.70$).