Soil Physical Properties: 18-Geol-A6 Soil Mechanics, December 2017, Q2

Question 2: Soil Physical Properties (15 marks)

(2.1) Compaction curve interpretation and phase relations (part of 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. Standard Proctor test points (water content $w$, dry unit weight $\gamma_d$): $(7.5\%,18.4)$, $(9.0\%,19.6)$, $(10.0\%,18.8)$, $(11.5\%,19.4)$, $(13.0\%,18.5)$ kN/m³; assume $G_s=2.70$ (standard mineral value, consistent with the zero-air-void reference curves $S=90/95/100\%$ printed on the source chart).

Find. (a) axis labels; (b) interpretation of the curve, $w_{opt}$ and $\gamma_{d,max}$; (c) at $w=7\%$: void ratio $e$, degree of saturation $S$, total unit weight $\gamma$.

Approach. Fit a smooth curve through the five plotted points (the raw digitized points are not perfectly monotonic — typical of hand-plotted lab data) to read the peak and interpolate $\gamma_d$ at $w=7\%$, then apply the three-phase relations.

Check: $G_s=2.70$ is assumed (not stated for this sub-question) — it is the value given later in Q2.2 for the same borrow soil and matches the $S=90/95/100\%$ zero-air-void curves drawn for reference on the source chart. The five raw plotted points are not perfectly monotonic (measurement scatter typical of a hand-plotted Proctor test); a least-squares parabola through all five points is used as the smooth compaction curve for reading the peak and interpolating at $w=7\%$.

Fig. Q2-1 — Standard Proctor compaction curve, fitted to the five plotted test points.

(a) Axes. Horizontal axis: water content $w$ (%); vertical axis: dry unit weight $\gamma_d$ (kN/m³).
(b.i) Interpretation of the curve. $\gamma_d$ rises with $w$ on the "dry side" as added water lubricates particle rearrangement into a denser packing, reaches a single peak, then falls on the "wet side" as further water displaces solids and pushes the state toward (but never onto) the zero-air-void curve for the given $G_s$.
(b.ii) Optimum water content and maximum dry unit weight. The least-squares parabola through the five plotted points peaks at $\boxed{w_{opt}\approx10.3\%}$, $\boxed{\gamma_{d,max}\approx19.3\text{ kN/m}^3}$ (consistent with the source figure's own visually-read peak near $w\approx9\%$, $\gamma_d\approx19.6\text{ kN/m}^3$, given the scatter in the plotted points).
(c) Dry unit weight at $w=7\%$. Reading the fitted curve at $w=7\%$ (a short extrapolation just left of the lowest plotted point, $7.5\%$): $\gamma_d=\boxed{18.16\text{ kN/m}^3}$.
(c) i) Void ratio. $\gamma_d=\dfrac{G_s\gamma_w}{1+e}\ \Rightarrow\ e=\dfrac{G_s\gamma_w}{\gamma_d}-1=\dfrac{2.70\times9.81}{18.16}-1=\boxed{0.458}$.
(c) ii) Degree of saturation. $S=\dfrac{wG_s}{e}=\dfrac{0.07\times2.70}{0.458}=\boxed{41.2\%}$.
(c) iii) Total unit weight. $\gamma=\gamma_d(1+w)=18.16\times1.07=\boxed{19.43\text{ kN/m}^3}$.

Quantity	Value
Optimum water content, $w_{opt}$	≈10.3%
Maximum dry unit weight, $\gamma_{d,max}$	≈19.3 kN/m³
$\gamma_d$ at $w=7\%$	18.16 kN/m³
Void ratio, $e$ (at $w=7\%$)	0.458
Degree of saturation, $S$ (at $w=7\%$)	41.2%
Total unit weight, $\gamma$ (at $w=7\%$)	19.43 kN/m³

(2.2) Embankment borrow-pit volumes (part of 15 marks)

Given. Target embankment: $\gamma_{d,target}=17.5\text{ kN/m}^3$, $w_{target}=19\%$. Borrow pit: $\gamma_{borrow}=15\text{ kN/m}^3$, $w_{borrow}=5.5\%$, $G_s=2.7$.

Find. (a) Borrow volume per m³ of embankment. (b) Water to add per m³ of embankment.

Approach. The mass of dry solids is conserved during hauling and compaction, so equate the dry weight delivered from the borrow pit to the dry weight required in place; any shortfall in water content is made up by adding water.

Borrow dry unit weight. $\gamma_{d,borrow}=\dfrac{\gamma_{borrow}}{1+w_{borrow}}=\dfrac{15}{1.055}=\boxed{14.22\text{ kN/m}^3}$.
Borrow volume for 1 m³ embankment. Dry weight of solids needed per m³ of finished embankment $=\gamma_{d,target}\times1=17.5\text{ kN}$. $V_{borrow}=\dfrac{17.5}{\gamma_{d,borrow}}=\dfrac{17.5}{14.22}=\boxed{1.231\text{ m}^3}$.
Water carried from the borrow pit. $W_{w,initial}=(\gamma_{borrow}-\gamma_{d,borrow})\times V_{borrow}=(15-14.22)\times1.231=0.963\text{ kN}$.
Water required in place. $W_{w,target}=\gamma_{d,target}\times w_{target}=17.5\times0.19=3.325\text{ kN}$.
Water to add. $\Delta W_w=W_{w,target}-W_{w,initial}=3.325-0.963=\boxed{2.36\text{ kN/m}^3\text{ embankment}}$, i.e. $\approx\boxed{241\text{ kg (241 L)}}$ of water per m³.

Quantity	Value
Borrow dry unit weight	14.22 kN/m³
Borrow volume per m³ embankment	1.231 m³
Water to add per m³ embankment	2.36 kN (≈241 kg)

Concept & Theory

Topic. Standard Proctor compaction and three-phase weight-volume relations, with dry-mass conservation in earthwork.

Key relations. $\gamma_d=G_s\gamma_w/(1+e)$; $S=wG_s/e$; $\gamma=\gamma_d(1+w)$; dry mass of solids is invariant under hauling/compaction (only void volume and water content change).

Why this works. The Proctor curve peak balances two competing effects of added water — particle lubrication (densifying) versus pore-water volume displacing solids (loosening); compaction changes only the void ratio of a fixed mass of solids, so tracking dry weight bridges the borrow-pit and in-place states.

Common pitfall. Equating borrow volume directly to embankment volume (they differ because void ratio changes on compaction) instead of conserving dry weight; and assuming $\gamma_d$ from a raw hand-plotted point instead of reading the smooth fitted curve at the required water content.

Source. Das, Principles of Geotechnical Engineering, 9th ed., Ch. 3 & 6 (phase relations; compaction).

Question 2 of 6: Soil Physical Properties

Question 2: Soil Physical Properties (15 marks)

(2.1) Compaction curve interpretation and phase relations (part of 15 marks)

(2.2) Embankment borrow-pit volumes (part of 15 marks)