Weight–Volume Relations of a Soil Sample: 18-Env-A3 Geotechnical and Hydrogeological Engineering, December 2016, Q1

Question 1: Weight–Volume Relations of a 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. A single moist soil specimen with the mass and volume below.

Given data
Quantity	Symbol	Value
Moist (as-sampled) mass	$M$	30.6 kg
Total volume	$V$	0.0183 m³
Oven-dry mass	$M_s$	27.2 kg
Specific gravity of solids	$G_s$	2.65

Find. (a) bulk density $\rho$; (b) dry density $\rho_d$; (c) moisture content $w$; (d) saturated density $\rho_{sat}$; (e) void ratio $e$.

Figure — three-phase weight–volume block diagram for the sample (air, water, solids).

Approach. Bulk and dry density come straight from mass/volume; the water lost on drying gives the moisture content directly; converting the dry mass to a solids volume $V_s=M_s/(G_s\rho_w)$ isolates the voids volume $V_v=V-V_s$, which fixes both the void ratio and the saturated density (voids assumed fully water-filled).

Part (a) — bulk (moist) density. $$\rho=\frac{M}{V}=\frac{30.6}{0.0183}=\boxed{1672\ \text{kg/m}^3}\ (1.672\ \text{Mg/m}^3).$$
Part (b) — dry density. $$\rho_d=\frac{M_s}{V}=\frac{27.2}{0.0183}=\boxed{1486\ \text{kg/m}^3}.$$
Part (c) — moisture content. The water driven off in the oven is $M_w=M-M_s=30.6-27.2=3.4$ kg, so $$w=\frac{M_w}{M_s}=\frac{3.4}{27.2}=\boxed{12.5\%}.$$
Part (e) — void ratio (needed before part d). The solids occupy $$V_s=\frac{M_s}{G_s\rho_w}=\frac{27.2}{2.65\times1000}=0.01026\ \text{m}^3,$$ so the voids occupy $V_v=V-V_s=0.0183-0.01026=0.00804\ \text{m}^3$, giving $$e=\frac{V_v}{V_s}=\frac{0.00804}{0.01026}=\boxed{0.783}.$$
Part (d) — saturated density. Filling every void with water adds $V_v\rho_w$ to the dry mass without changing $V$, so $$\rho_{sat}=\frac{M_s+V_v\rho_w}{V}=\frac{27.2+0.00804\times1000}{0.0183}=\boxed{1925\ \text{kg/m}^3}.$$ As a check, the sample's as-found degree of saturation is only $S=(M_w/\rho_w)/V_v\times100=42\%$, confirming it was sampled well short of saturation — consistent with $\rho_{sat}>\rho$.

Quantity	Value
(a) Bulk density, $\rho$	1672 kg/m³
(b) Dry density, $\rho_d$	1486 kg/m³
(c) Moisture content, $w$	12.5%
(d) Saturated density, $\rho_{sat}$	1925 kg/m³
(e) Void ratio, $e$	0.783

Concept & Theory

Topic. The three-phase (solid–water–air) weight–volume diagram and the identities that connect its densities.

Key relations. $\rho=M/V$; $\rho_d=M_s/V$; $w=M_w/M_s$; $V_s=M_s/(G_s\rho_w)$; $e=V_v/V_s$; $\rho_{sat}=(M_s+V_v\rho_w)/V$.

Why this works. Only two independent measurements (mass and volume, before and after drying) are needed because $G_s$ pins down how much of the total volume is solid; everything else — voids, void ratio, and the saturated end-state — follows from simple volume bookkeeping once $V_s$ is known.

Common pitfall. Confusing "bulk density" with "saturated density" — the as-sampled soil here is only 42% saturated, so $\rho\ne\rho_{sat}$; saturated density must be built up from the void ratio, not read off the sample as tested.

Source. Das, Principles of Geotechnical Engineering, ch. 1 (weight–volume relationships).

Question 1 of 6: Weight–Volume Relations of a Soil Sample

Question 1: Weight–Volume Relations of a Soil Sample (20 marks)