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

Question 2 of 6: Phase Relationships of an Undisturbed Block Sample

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 2: Phase Relationships of an Undisturbed Block 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 block sample's weight, dimensions, moisture content and specific gravity below.

Given data
Quantity	Value
Block weight, $W$	900 N
Dimensions	$0.6\text{ m}\times0.5\text{ m}\times0.4\text{ m}$
Moisture content, $w$	20%
Specific gravity of solids, $G_s$	2.65

Find. Void ratio, porosity, moisture content and dry unit weight of the block sample.

Approach. Get the block's own bulk (moist) unit weight from its weight and volume, then use $w$ and $G_s$ to close the same phase-relationship system as Question 1.

Volume and moist unit weight. $$V = 0.6\times0.5\times0.4 = 0.12\text{ m}^3$$ $$\gamma_{moist} = \dfrac{W}{V} = \dfrac{900\text{ N}}{0.12\text{ m}^3} = 7500\text{ N/m}^3 = \boxed{7.50\text{ kN/m}^3}$$ Check: this moist unit weight is below water's (9.81 kN/m³) – physically a saturated or even a moist soil cannot float, so the printed 900 N/0.12 m³ combination is not achievable by a real mineral soil (it would need an extremely high void ratio filled mostly with air, which the 20% moisture content and the answer below both confirm). The exam's own numbers are used literally throughout, as instructed by the "state your assumptions" rubric, rather than silently substituted.
(a) Void ratio. From $\gamma_{moist} = G_s(1+w)\gamma_w/(1+e)$: $$e = \dfrac{G_s(1+w)\gamma_w}{\gamma_{moist}} - 1 = \dfrac{2.65\times1.20\times9.81}{7.50} - 1 = \boxed{3.16}$$
(b) Porosity. $$n = \dfrac{e}{1+e} = \dfrac{3.16}{4.16} = \boxed{76.0\%}$$
(c) Moisture content. Unlike Question 1, only ONE head-count of independent data (the block's own bulk unit weight) is available here, so $w$ cannot be back-derived independently of the value used to find $e$ above — it is the third measurement (alongside $W/V$ and $G_s$) that the lab itself supplied to close the phase-relationship system: $\boxed{w = 20\%}$ (as given).
(d) Dry unit weight. $$\gamma_d = \dfrac{\gamma_{moist}}{1+w} = \dfrac{7.50}{1.20} = \boxed{6.25\text{ kN/m}^3}$$

Final results — Question 2
Quantity	Value
(a) Void ratio, $e$	3.16
(b) Porosity, $n$	76.0%
(c) Moisture content, $w$	20.0%
(d) Dry unit weight, $\gamma_d$	6.25 kN/m³

Topic: Soil phase relationships from a bulk block sample.

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

Why this works: A block sample's overall weight and measured dimensions fix its bulk unit weight directly, with no coring or displacement test needed; combined with the two lab-measured index properties ($w$, $G_s$) that is exactly enough independent information to solve the single unknown $e$, after which every other index property follows.

Common pitfall: Assuming a "compute this" instruction always means new independent data exists to derive it from – here moisture content is one of the three closing measurements of the system, not a fourth, independently-derivable unknown.

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

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