24-Bld-A6 Geotechnical Materials and Analysis · Undated paper

Question 2 of 7

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07-BLD-A6 Geotechnical Materials and Analysis, National Examinations (printed exam date May 2019). 3 hours, closed book, 100 marks. Section A (Q1–Q3) is compulsory (40 marks); Section B directs "answer any three of Q4–Q7" (60 marks), but for completeness this solution answers all four.

Reference texts: B.M. Das, Principles of Geotechnical Engineering, 9th ed.; B.M. Das, Principles of Foundation Engineering, 9th ed.

Question 2 (10 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.

Quantity	Symbol	Value
Total volume	$V$	5660 cm³ = 5.660×10⁻³ m³
Total mass	$M$	10.4 kg
Moisture content	$w$	10%
Specific gravity of solids	$G_s$	2.7

Find. $\rho_{moist}$, $\rho_{dry}$, porosity $n$, degree of saturation $S$, and $V_w$.

Three-phase diagram for the moist soil sample: volume split (left) into air, water and solids; mass split (right) into (negligible) air, water and solids — drawn to the proportions computed below.

Approach. Split the given total mass into solids and water using $w$, convert the solids mass to a solids volume via $G_s$, and back out every remaining phase quantity from $V=V_s+V_w+V_a$.

Moist (bulk) density. $\rho_{moist}=\dfrac{M}{V}=\dfrac{10.4\text{ kg}}{5.660\times10^{-3}\text{ m}^3}=\boxed{1837\text{ kg/m}^3}$.
Split mass into solids and water. $M_s=\dfrac{M}{1+w}=\dfrac{10.4}{1.10}=9.4545\text{ kg}$, so $M_w=M-M_s=0.9455\text{ kg}$.
Dry density. $\rho_{dry}=\dfrac{M_s}{V}=\dfrac{9.4545}{5.660\times10^{-3}}=\boxed{1670\text{ kg/m}^3}$ (equivalently $\rho_{moist}/(1+w)$).
Volume of solids and voids. $V_s=\dfrac{M_s}{G_s\rho_w}=\dfrac{9.4545}{2.7\times1000}=3.502\times10^{-3}\text{ m}^3$. Then $V_v=V-V_s=5.660\times10^{-3}-3.502\times10^{-3}=2.158\times10^{-3}\text{ m}^3$.
Porosity. $n=\dfrac{V_v}{V}=\dfrac{2.158\times10^{-3}}{5.660\times10^{-3}}=\boxed{38.1\%}$.
Volume of water. $V_w=\dfrac{M_w}{\rho_w}=\dfrac{0.9455\text{ kg}}{1000\text{ kg/m}^3}=\boxed{9.455\times10^{-4}\text{ m}^3}$ (945 cm³).
Degree of saturation. $S=\dfrac{V_w}{V_v}=\dfrac{9.455\times10^{-4}}{2.158\times10^{-3}}=\boxed{43.8\%}$. Cross-check: $wG_s=Se \Rightarrow 0.10\times2.7=0.270$, and $S\,e=0.438\times0.616=0.270$ — consistent.

Quantity	Result
Moist density, $\rho_{moist}$	1837 kg/m³
Dry density, $\rho_{dry}$	1670 kg/m³
Porosity, $n$	38.1%
Degree of saturation, $S$	43.8%
Volume of water, $V_w$	9.46×10⁻⁴ m³ (945 cm³)

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

Key relations: $\rho=M/V$; $M_s=M/(1+w)$; $V_s=M_s/(G_s\rho_w)$; $n=V_v/V$; $S=V_w/V_v$; $wG_s=Se$.

Why this works: every phase quantity (mass or volume of air, water, solids) can be derived from just four independent given values ($V$, $M$, $w$, $G_s$) because the phase diagram has only two independent ratios once total mass and volume are fixed — $w$ splits the mass, $G_s$ converts solids mass to solids volume, and everything else follows from $V=V_a+V_w+V_s$.

Common pitfall: using $\rho_w=1000\text{ kg/m}^3$ inconsistently with cm³-based volumes (always convert to SI first); confusing $n$ (void volume over total volume) with $e$ (void volume over solids volume only).

Source: Das, Principles of Geotechnical Engineering, 9th ed., Ch. 2.

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