Question 4 of 6: Consolidation Settlement of a Clay Layer

Question 4: Consolidation Settlement of a Clay Layer (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) ultimate primary consolidation settlement $S_c$; (b) time $t$ for 25% of primary consolidation.

Approach. The clay is normally consolidated, so the entire stress increase from the new fill drives virgin ($C_c$) compression — $C_r$ is not used. Take the effective stress change at the clay's mid-height as representative of the whole layer, then find $t$ from the time factor $T_v$ for $U=25\%$ using the correct drainage path length.

1. Part (a) — effective stress at clay mid-depth, before and after the fill. Assume the water table sits at the original ground surface (the top of the clay, before the sand was placed), so the clay is saturated throughout and the sand fill sits entirely above the water table. The clay's submerged unit weight is $\gamma'=\gamma_{clay}-\gamma_w=15-9.81=5.19\ \text{kN/m}^3$, so at mid-depth (5 m below the clay's top, i.e. below the water table): $$\sigma_0'=\gamma'\times\frac{H_c}{2}=5.19\times5=\boxed{25.95\ \text{kPa}}.$$ The fill sits above the water table, so its full weight is transmitted as surcharge (no buoyancy): $$\Delta\sigma'=\gamma_{sand}H_{sand}=17\times7=\boxed{119.0\ \text{kPa}},\qquad \sigma_1'=\sigma_0'+\Delta\sigma'=144.95\ \text{kPa}.$$
2. Apply the normally-consolidated settlement equation. Since $\sigma_1'>\sigma_0'=\sigma_p'$ throughout (NC clay, no recompression range), only $C_c$ is used: $$S_c=\frac{C_c H_c}{1+e_0}\log_{10}\!\left(\frac{\sigma_1'}{\sigma_0'}\right)=\frac{0.45\times10}{1+1.0}\log_{10}\!\left(\frac{144.95}{25.95}\right)=2.25\times0.747=\boxed{1.68\ \text{m}}.$$
3. Part (b) — drainage path and time factor. The clay rests on impermeable shale (no drainage downward) and is capped by the pervious sand fill (free drainage upward), so this is SINGLE drainage and the drainage path is the full clay thickness, $H_{dr}=10\ \text{m}=1000\ \text{cm}$. For $U\le60\%$, $T_v=(\pi/4)U^2$: $$T_v=\frac{\pi}{4}(0.25)^2=0.0491.$$
4. Solve for time. $t=T_v H_{dr}^2/C_v$: $$t=\frac{0.0491\times(1000)^2}{0.003}=1.636\times10^7\ \text{s}=\boxed{189\ \text{days}\ (\approx0.52\ \text{yr})}.$$