18-Env-A3 Geotechnical and Hydrogeological Engineering · December 2017
Question 5 of 6: Primary Consolidation of an Overconsolidated Clay
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2017 — 04-Env-A3 / Geotechnical & Hydrogeological Engineering. 3 hours duration; open book exam, any non-communicating calculator permitted. FIVE (5) questions constitute a complete exam paper (the first five as they appear in the answer book are marked, 20 marks each, 100 marks total); all six printed questions are solved below for completeness.
Reference texts. Braja M. Das, Principles of Geotechnical Engineering (9th ed.) — weight–volume relations, compaction, seepage/flow nets, and consolidation chapters; Craig & Knappett, Craig's Soil Mechanics (8th ed.) — cross-reference for flow-net theory and finite-difference seepage; Freeze & Cherry, Groundwater (1979) — Darcy's law, the Thiem confined-flow equation, and radial travel time.
Question 5: Primary Consolidation of an Overconsolidated Clay (20 marks)
Find. (a) ultimate primary settlement; (b) time for 95% primary consolidation.
Approach. Compute the in-situ effective stress at the clay's mid-height (water table at the top of the clay, since it rests on dry sand), compare it with $\sigma_c'$ to size the recompression and virgin-compression legs of the settlement, then use the standard $T_v$–$U$ correlation with the drainage path length set by double drainage (permeable ground surface above, permeable sand below).
Fig. Q5 — 1 m saturated clay on dry sand, loaded by a uniform surface surcharge.
In-situ effective stress. Saturated unit weight from the phase relation, $\gamma_{sat}=\dfrac{(G_s+e_0)\gamma_w}{1+e_0}=\dfrac{(2.65+5.0)\times9.81}{6.0}=12.51\ \text{kN/m}^3$. The water table sits at the top of the clay (it rests on DRY sand), so at mid-height ($z=0.5\ \text{m}$) the total stress is $\sigma_{v0}=\gamma_{sat}\times0.5=6.25\ \text{kN/m}^2$ and the pore pressure is $u_0=\gamma_w\times0.5=4.91\ \text{kN/m}^2$, giving
$$\sigma_0'=\sigma_{v0}-u_0=6.25-4.91=1.35\ \text{kN/m}^2.$$
Since $\sigma_0'=1.35<\sigma_c'=20\ \text{kN/m}^2$, the clay is heavily overconsolidated ($OCR=20/1.35=14.8$).
Part (a) — two-stage settlement. The final effective stress $\sigma_f'=\sigma_0'+\Delta\sigma=1.35+200=201.35\ \text{kN/m}^2$ exceeds $\sigma_c'$, so consolidation proceeds first along the recompression line up to $\sigma_c'$, then along the virgin-compression line from $\sigma_c'$ to $\sigma_f'$:
$$S=\frac{H_0}{1+e_0}\Big[C_r\log_{10}\Big(\frac{\sigma_c'}{\sigma_0'}\Big)+C_c\log_{10}\Big(\frac{\sigma_f'}{\sigma_c'}\Big)\Big]
=\frac{1.0}{6.0}\Big[0.2\log_{10}(14.83)+0.5\log_{10}(10.07)\Big]=\boxed{122.6\ \text{mm}}.$$
Part (b) — time for 95% consolidation. With the ground surface open (permeable) above and permeable sand below, drainage is double-sided: $H_{dr}=H_0/2=0.5\ \text{m}=50\ \text{cm}$. At $U=95\%$, $T_v=1.781-0.933\log_{10}(100-95)=1.129$, so
$$t_{95}=\frac{T_v\,H_{dr}^2}{c_v}=\frac{1.129\times50^2}{0.001}=2.822\times10^{6}\ \text{s}=\boxed{32.7\ \text{days}}.$$
Check: the very low $\sigma_0'=1.35\ \text{kPa}$ follows directly from the unusually high given $e_0=5.0$ (a very soft clay, consistent with Question 1's own high-void-ratio sample); both the settlement and the $OCR$ are solved literally from the stated data.
