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

Question 1 of 6: Phase Relationships of an Undisturbed Soil Sample

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 1: Phase Relationships of an Undisturbed Soil Sample (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. The core sample and mass data below.

Given data

Quantity	Value
Core diameter, $D$	100 mm
Core length, $L$	250 mm
Empty tube mass	100 g
Tube + moist soil mass	3490 g
Oven-dry soil mass, $M_d$	3098 g
Water displaced by solids, $V_s$	1178 mL

Find. The bulk (moist) density, dry unit weight, degree of saturation, and saturated unit weight of the soil.

Approach. Compute the sample's total volume from the core dimensions, get the moist and dry masses from the tube readings, take the water-displacement volume as the volume of solids $V_s$, then work through the standard phase-relationship definitions.

1. Total (bulk) volume of the core. $$V = \dfrac{\pi}{4}D^2 L = \dfrac{\pi}{4}(100\text{ mm})^2(250\text{ mm}) = 1{,}963{,}495\text{ mm}^3 = 1963.5\text{ cm}^3$$
2. Moist and dry masses. Moist soil mass $M = 3490 - 100 = 3390$ g. Dry mass $M_d = 3098$ g (already free of the tube), so pore-water mass $M_w = 3390-3098 = 292$ g.
3. (a) Bulk density and unit weight. $$\rho_{bulk} = \dfrac{M}{V} = \dfrac{3390\text{ g}}{1963.5\text{ cm}^3} = 1.727\text{ g/cm}^3 = 1727\text{ kg/m}^3$$ $$\gamma_{bulk} = \rho_{bulk}\, g = 1.727\times 9.81 = \boxed{16.94\text{ kN/m}^3}$$
4. (b) Dry unit weight. Same volume, dry mass only: $$\rho_d = \dfrac{M_d}{V} = \dfrac{3098}{1963.5} = 1.578\text{ g/cm}^3 \quad\Rightarrow\quad \gamma_d = 1.578\times 9.81 = \boxed{15.48\text{ kN/m}^3}$$
5. (c) Degree of saturation. The water-displacement test gives the volume of solids directly ($V_s = 1178$ cm³, since loosely-poured grains displace their own solid volume). Void volume and water volume follow: $$V_v = V - V_s = 1963.5 - 1178 = 785.5\text{ cm}^3, \qquad V_w = \dfrac{M_w}{\rho_w} = \dfrac{292\text{ g}}{1\text{ g/cm}^3} = 292\text{ cm}^3$$ $$S_r = \dfrac{V_w}{V_v} = \dfrac{292}{785.5} = \boxed{37.2\%}$$ (a quick internal check: $G_s = M_d/(V_s\rho_w) = 3098/1178 = 2.63$ — a physically realistic specific gravity, confirming the displacement reading is self-consistent.)
6. (d) Saturated unit weight. Fill the same voids completely with water: $$M_{sat} = M_d + V_v\rho_w = 3098 + 785.5 = 3883.5\text{ g} \quad\Rightarrow\quad \rho_{sat} = \dfrac{3883.5}{1963.5} = 1.978\text{ g/cm}^3$$ $$\gamma_{sat} = 1.978\times 9.81 = \boxed{19.40\text{ kN/m}^3}$$

Final results — Question 1

Quantity	Value
(a) Bulk unit weight, $\gamma_{bulk}$	16.94 kN/m³
(b) Dry unit weight, $\gamma_d$	15.48 kN/m³
(c) Degree of saturation, $S_r$	37.2%
(d) Saturated unit weight, $\gamma_{sat}$	19.40 kN/m³

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