Question 4 of 6: Consolidation

Question 4: Consolidation (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.

(a) Settlement of the overconsolidated clay

Given. $H_0=7\text{ m}$; $C_c=0.32$; $C_r=0.065$; $e_0=0.864$; $\sigma'_p=310\text{ kPa}$; $\sigma'_{vo}=126\text{ kPa}$; applied pressure $\Delta\sigma=285\text{ kPa}$.

Find. (i) Settlement given the clay's actual (overconsolidated) stress history. (ii) Settlement if the clay were instead normally consolidated (never preconsolidated beyond the existing overburden).

Approach. Compare the final vertical effective stress $\sigma'_{vf}=\sigma'_{vo}+\Delta\sigma$ against $\sigma'_p$ to decide whether the stress path stays on the recompression line or crosses onto the virgin compression line.

1. Final effective stress. $$\sigma'_{vf}=\sigma'_{vo}+\Delta\sigma=126+285=\boxed{411\text{ kPa}}$$ Since $\sigma'_{vf}=411>\sigma'_p=310\text{ kPa}$, the stress path recompresses from 126 to 310 kPa, then follows the virgin compression line from 310 to 411 kPa — both terms of the combined settlement formula are needed.
2. Part (i) — settlement using the actual stress history. With $H_0/(1+e_0)=7/1.864=3.755\text{ m}$: $$\Delta H=C_r\frac{H_0}{1+e_0}\log\frac{\sigma'_p}{\sigma'_{vo}}+C_c\frac{H_0}{1+e_0}\log\frac{\sigma'_{vf}}{\sigma'_p}$$ $$\Delta H=0.065(3.755)\log\frac{310}{126}+0.32(3.755)\log\frac{411}{310}=0.0954+0.1473=\boxed{0.243\text{ m}\ (24.3\text{ cm})}$$
3. Part (ii) — settlement if never preconsolidated above the overburden. With no preconsolidation "memory" ($\sigma'_p=\sigma'_{vo}$), the entire stress increase from 126 to 411 kPa follows the steeper virgin compression line only: $$\Delta H=C_c\frac{H_0}{1+e_0}\log\frac{\sigma'_{vf}}{\sigma'_{vo}}=0.32(3.755)\log\frac{411}{126}=\boxed{0.617\text{ m}\ (61.7\text{ cm})}$$ This is about 2.5× the overconsolidated-clay settlement of part (i), illustrating why preconsolidation (from prior loading, erosion, or desiccation) makes a clay stratum substantially stiffer against a new load than a normally consolidated clay at the same void ratio.

(b) Time rate of settlement: single vs. double drainage

Given. Design estimate: 10 cm settlement at some elapsed time under double drainage, with 50 cm ultimate settlement; $c_v=2.41\times10^{-4}\text{ cm}^2/\text{s}$; clay thickness $H=7\text{ m}$.

Find. (i) How the double-to-single drainage change affects the ultimate settlement magnitude. (ii)–(iii) Time to reach 10 cm of settlement under single and double drainage, computed directly from $c_v$.

Approach. The degree of consolidation at 10 cm of settlement is $U=\Delta H(t)/\Delta H_{\text{ult}}=10/50=20\%$, independent of drainage condition (ultimate settlement is set by the stress change and compressibility alone, not by how the water escapes). The time to reach that $U$ is then found from the time factor $T=c_vt/H_{dr}^2$, using $H_{dr}=H/2$ for double drainage and $H_{dr}=H$ for single drainage.

1. Part (i) — effect on total settlement. The ultimate settlement $\Delta H_{\text{ult}}$ depends only on $C_c$, $C_r$, $e_0$ and the stress change (the same equations as part (a)) — it does not appear anywhere in $T=c_vt/H_{dr}^2$. Removing the bottom drainage boundary therefore does not change the total (ultimate) settlement, which remains $\boxed{50\text{ cm}}$ either way; it only slows the rate at which that settlement is reached, because the drainage path length $H_{dr}$ doubles (from $H/2$ to $H$) when the bottom boundary is sealed.
2. Degree of consolidation at the 10 cm point. $$U=\frac{\Delta H(t)}{\Delta H_{\text{ult}}}=\frac{10}{50}=\boxed{20\%}$$ Since $U<60\%$, the low-$U$ time-factor relation applies: $$T=\frac{\pi}{4}\left(\frac{U}{100}\right)^2=\frac{\pi}{4}(0.20)^2=\boxed{0.0314}$$
3. Part (iii) — time under double drainage. $H_{dr}=H/2=3.5\text{ m}=350\text{ cm}$. $$t=\frac{TH_{dr}^2}{c_v}=\frac{0.0314(350)^2}{2.41\times10^{-4}}=1.60\times10^{7}\text{ s}=\boxed{0.51\text{ yr}\ (\approx185\text{ days})}$$
4. Part (ii) — time under single drainage. $H_{dr}=H=7\text{ m}=700\text{ cm}$ — the same $T$ (same $U$) but a drainage path twice as long: $$t=\frac{TH_{dr}^2}{c_v}=\frac{0.0314(700)^2}{2.41\times10^{-4}}=6.39\times10^{7}\text{ s}=\boxed{2.03\text{ yr}\ (\approx739\text{ days})}$$ Since $t\propto H_{dr}^2$ at fixed $U$, doubling the drainage path exactly quadruples the time: $t_{\text{single}}/t_{\text{double}}=(H/(H/2))^2=4$, consistent with $2.03/0.51\approx4$.