18-Env-A3 Geotechnical and Hydrogeological Engineering · May 2018

Question 4 of 6: Primary Consolidation of a Clay Layer Beneath a New Sand Fill

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams May 2018 — 04-Env-A3, Geotechnical and Hydrogeological Engineering — 3 hours, open book. All SIX questions are answered below (the exam marks only the first five as submitted; all six are solved here as a complete study resource). Marking scheme: each question 20 marks, equal value.

Reference texts: Das, Principles of Geotechnical Engineering, 9th ed.; Craig & Knappett, Craig's Soil Mechanics, 8th ed.; Freeze & Cherry, Groundwater (1979).

Question 4: Primary Consolidation of a Clay Layer Beneath a New Sand Fill (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.

Given data

Quantity	Value
Clay thickness, $H$	2 m
Sand fill thickness	20 m
Sand unit weight, $\gamma_{sand}$	15 kN/m³
Clay unit weight, $\gamma_{clay}$	10 kN/m³
Initial void ratio, $e_0$	2.0
Compression index, $C_c$	0.40
Recompression index, $C_r$	0.10
Coefficient of consolidation, $C_v$	0.002 cm²/s

Find. The ultimate primary settlement of the clay and the time for 50% of that settlement to occur.

Approach. The clay is normally consolidated, so the entire stress increase from the new fill follows the virgin compression line ($C_c$ only — $C_r$ is not used here, since there is no over-consolidation range to recompress through). Consolidation time comes from Terzaghi's theory using the single drainage path set by the impermeable shale below.

1. (a) Effective stress before and after the fill. Before the fill, the clay's own self-weight to its mid-depth is the only overburden (and, being normally consolidated, this IS its preconsolidation pressure): $$\sigma_0' = \gamma_{clay}\left(\dfrac{H}{2}\right) = 10(1.0) = 10\text{ kPa}$$ The 20 m sand fill is wide compared to the clay's depth, so it adds a uniform stress increment at the clay: $$\Delta\sigma' = \gamma_{sand}(20) = 15(20) = 300\text{ kPa} \quad\Rightarrow\quad \sigma_f' = \sigma_0'+\Delta\sigma' = 310\text{ kPa}$$
2. Ultimate primary settlement. $$S_c = \dfrac{C_c H}{1+e_0}\log_{10}\left(\dfrac{\sigma_f'}{\sigma_0'}\right) = \dfrac{0.40(2)}{1+2.0}\log_{10}\left(\dfrac{310}{10}\right) = 0.2667\times 1.491 = \boxed{0.398\text{ m} \approx 398\text{ mm}}$$
3. (b) Drainage path and time factor. The clay sits directly on impermeable shale, so drainage occurs only upward into the (permeable) sand — single drainage, $H_{dr}=H=2\text{ m}=200\text{ cm}$. The time factor for 50% average consolidation is $$T_{50} = \dfrac{\pi}{4}\left(\dfrac{U}{100}\right)^2 = \dfrac{\pi}{4}(0.50)^2 = 0.196$$
4. Time for 50% consolidation. $$t_{50} = \dfrac{T_{50}H_{dr}^2}{C_v} = \dfrac{0.196(200)^2}{0.002} = 3.93\times10^6\text{ s} = \boxed{45.5\text{ days}\ (\approx 0.125\text{ yr})}$$

Final results — Question 4

Quantity	Value
(a) Ultimate primary settlement, $S_c$	398 mm
(b) Time for 50% consolidation, $t_{50}$	45.5 days

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