Soil Physical Properties: 18-Geol-A6 Soil Mechanics, May 2015, Q2

Question 2: Soil Physical Properties (15 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. Part 1: $e=0.75$, $w=22\%$, $G_s=2.66$. Part 2: wet+container $=397.6$ g, container $=258.7$ g, $w_0=6.3\%$, target $\Delta w=+3.4\%$. Part 3: target $\gamma_d=18\text{ kN/m}^3$ at $w=7\%$; borrow pit $\gamma=17\text{ kN/m}^3$, $w=5\%$, $G_s=2.7$.

Find. (1a–e) porosity, moist/dry unit weight, saturation, water to add. (2) grams of water to add. (3a–b) borrow volume and water per m³ of embankment.

Approach. Use the standard phase-relation identities ($n=e/(1+e)$, $\gamma=G_s\gamma_w(1+w)/(1+e)$, $Se=wG_s$) for Part 1; a mass balance on dry solids and water for Part 2; and conservation of dry-solid mass between borrow pit and placed embankment for Part 3.

Part 1(a) — porosity. $$n=\frac{e}{1+e}=\frac{0.75}{1.75}=\boxed{42.9\%}$$
Part 1(b) — moist unit weight. $$\gamma_{moist}=\frac{G_s\gamma_w(1+w)}{1+e}=\frac{2.66(9.81)(1.22)}{1.75}=\boxed{18.19\text{ kN/m}^3}$$
Part 1(c) — dry unit weight. $$\gamma_{dry}=\frac{G_s\gamma_w}{1+e}=\frac{2.66(9.81)}{1.75}=\boxed{14.91\text{ kN/m}^3}\quad\left(\text{check: }\gamma_{moist}/(1+w)=18.19/1.22=14.91\ \checkmark\right)$$
Part 1(d) — degree of saturation, via $Se=wG_s$. $$S=\frac{wG_s}{e}=\frac{0.22(2.66)}{0.75}=\boxed{78.0\%}$$
Part 1(e) — water to add for full saturation of 10 m³. Voids occupy $n\times V=0.4286\times10=4.286\text{ m}^3$. At $S=78.0\%$ the voids currently hold $0.780\times4.286=3.343\text{ m}^3$ of water; at $S=100\%$ they would hold all $4.286\text{ m}^3$. $$\Delta V_w=4.286-3.343=0.943\text{ m}^3\ \Rightarrow\ \Delta m_w=\boxed{942\text{ kg}}$$
Part 2 — water to raise $w$ by 3.4 points. Wet soil mass $=397.6-258.7=138.9$ g. Dry mass $M_s=138.9/1.063=130.67$ g, so the current water mass is $M_{w0}=138.9-130.67=8.23$ g. At the target $w=6.3+3.4=9.7\%$, the required water mass is $M_{w,\text{new}}=130.67(0.097)=12.68$ g. $$\Delta m_w=12.68-8.23=\boxed{4.44\text{ g}}$$
Part 3(a) — borrow-pit volume per m³ of embankment. Borrow-pit dry unit weight: $\gamma_{d,borrow}=\gamma/(1+w)=17/1.05=16.19\text{ kN/m}^3$. Dry solids are conserved between borrow pit and embankment, so $$V_{borrow}\,\gamma_{d,borrow}=V_{emb}\,\gamma_{d,target}\ \Rightarrow\ V_{borrow}=\frac{1.0(18)}{16.19}=\boxed{1.112\text{ m}^3\text{ per m}^3\text{ embankment}}$$
Part 3(b) — water required per m³ of embankment. Water carried in the borrow volume moved: $W_{w,borrow}=V_{borrow}(\gamma-\gamma_{d,borrow})=1.112(17-16.19)=0.900\text{ kN}$. Water needed in the placed embankment: $W_{w,target}=\gamma_{d,target}(1.0)(0.07)=1.26\text{ kN}$. $$\Delta W_w=1.26-0.900=\boxed{0.36\text{ kN per m}^3}\ \left(\approx37\text{ kg or }37\text{ L per m}^3\right)$$

Quantity	Value
Porosity $n$	42.9%
Moist unit weight $\gamma_{moist}$	18.19 kN/m³
Dry unit weight $\gamma_{dry}$	14.91 kN/m³
Degree of saturation $S$	78.0%
Water to add (10 m³, full saturation)	942 kg
Water to add (Part 2 sample)	4.44 g
Borrow-pit volume (Part 3a)	1.112 m³ per m³ embankment
Water required (Part 3b)	0.36 kN/m³ (≈37 kg/m³)

Concept & Theory

Topic. Soil phase relations: void ratio, porosity, unit weight and degree of saturation link the three-phase (solid/water/air) model, and mass conservation of dry solids ties a borrow source to a compacted fill.

Key relations. $n=e/(1+e)$; $\gamma=G_s\gamma_w(1+w)/(1+e)$; $\gamma_d=\gamma/(1+w)=G_s\gamma_w/(1+e)$; $Se=wG_s$.

Why this works. The three-phase identities all fall out of defining $e$ and $w$ as ratios of a common solid-volume/solid-mass basis; $Se=wG_s$ is simply the volume of water divided by the volume of voids, expressed through $w$ and $G_s$ instead of directly. For borrow-to-fill problems, dry-solid mass (not total mass or volume) is the conserved quantity, because compaction changes void ratio and water content but not the mass of mineral grains moved.

Common pitfall. Using total (moist) unit weight instead of dry unit weight when balancing a borrow-to-embankment volume — the borrow soil's water content is not the same as the target compacted water content, so only the dry-solid mass balance gives the right conversion factor.

Source. Das, Principles of Geotechnical Engineering, Ch. 3 (Weight–Volume Relationships).

Question 2 of 6: Soil Physical Properties

Question 2: Soil Physical Properties (15 marks)