18-Env-A3 Geotechnical and Hydrogeological Engineering · December 2019

Question 1 of 6: Phase Relationships from Moist Weight and Volume

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

Notes on this paper

National Exams December 2019 — 18-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 from Moist Weight and Volume (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 moist sample volume, weight, moisture content and specific gravity below.

Given data
Quantity	Value
Sample volume, $V$	$5.66\times10^{-3}\text{ m}^3$
Moist weight, $W$	$102\times10^{-3}\text{ kN}$
Moisture content, $w$	11%
Specific gravity of solids, $G_s$	2.7

Find. Void ratio, porosity, degree of saturation, moisture content and buoyant unit weight, all in the soil's natural (as-sampled) state.

Approach. Get the moist unit weight directly from the sample's own weight and volume, strip off the water content to get the dry unit weight, then work through the standard phase-relationship definitions to reach every other quantity.

Moist and dry unit weight. $$\gamma_{moist} = \dfrac{W}{V} = \dfrac{102\times10^{-3}\text{ kN}}{5.66\times10^{-3}\text{ m}^3} = 18.02\text{ kN/m}^3$$ Removing the pore water via the moisture content, $$\gamma_d = \dfrac{\gamma_{moist}}{1+w} = \dfrac{18.02}{1.11} = \boxed{16.24\text{ kN/m}^3}$$
(a) Void ratio. From $\gamma_d = G_s\gamma_w/(1+e)$: $$e = \dfrac{G_s\gamma_w}{\gamma_d} - 1 = \dfrac{2.7\times9.81}{16.24} - 1 = \boxed{0.631}$$
(b) Porosity. $$n = \dfrac{e}{1+e} = \dfrac{0.631}{1.631} = \boxed{38.7\%}$$
(c) Degree of saturation. $$S_r = \dfrac{wG_s}{e} = \dfrac{0.11\times2.7}{0.631} = \boxed{47.0\%}$$
(d) Moisture content. The moisture content was directly measured in the laboratory ($w=11\%$), and it also closes the phase-relationship loop: rearranging $S_r=wG_s/e$ gives $w=S_re/G_s = (0.470)(0.631)/2.7 = 0.110$, which reproduces the given value $\boxed{w = 11\%}$ — a self-consistency check that the void ratio and degree of saturation above are correct.
(e) Buoyant (submerged) unit weight. First the saturated unit weight (all voids filled with water at this same $e$): $$\gamma_{sat} = \dfrac{G_s+e}{1+e}\gamma_w = \dfrac{2.7+0.631}{1.631}\times9.81 = 20.03\text{ kN/m}^3$$ $$\gamma_b = \gamma_{sat} - \gamma_w = 20.03 - 9.81 = \boxed{10.22\text{ kN/m}^3}$$

Final results — Question 1
Quantity	Value
(a) Void ratio, $e$	0.631
(b) Porosity, $n$	38.7%
(c) Degree of saturation, $S_r$	47.0%
(d) Moisture content, $w$	11.0%
(e) Buoyant unit weight, $\gamma_b$	10.22 kN/m³

Topic: Soil phase relationships (weight–volume relations).

Key relations: $\gamma=W/V$; $\gamma_d=\gamma_{moist}/(1+w)$; $\gamma_d=G_s\gamma_w/(1+e)$; $S_r=wG_s/e$; $\gamma_{sat}=(G_s+e)\gamma_w/(1+e)$.

Why this works: A soil is a three-phase system whose weight–volume relations are all tied together through $G_s$, $e$ and $w$; once the moist unit weight and moisture content fix the dry unit weight, every other index property (void ratio, porosity, saturation, the saturated and buoyant limits) follows algebraically with no further lab measurements needed.

Common pitfall: Using $\gamma_{moist}$ instead of $\gamma_d$ in the $G_s\gamma_w/(1+e)$ relation, or forgetting that $S_r$ must be computed from $w$, $G_s$ and $e$ together rather than assumed from the moisture content alone.

Source: Das, Principles of Geotechnical Engineering, Ch. 2 (Weight–Volume Relationships).

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