Question 1 of 6: Compaction — Weight–Volume Relations

Question 1: Compaction — Weight–Volume Relations (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.

Find. (a) porosity $n$; (b) degree of saturation $S$; (c) bulk (moist) density $\rho$; (d) dry unit weight $\gamma_d$; (e) saturated unit weight $\gamma_{sat}$.

Approach. Split the moist mass into dry-solids and water masses using $w=M_w/M_s$, convert the dry mass to a solids volume via $G_s$, then read every weight–volume ratio off the resulting phase diagram.

1. Part (a)/(b) — split the compacted mass and find the phase volumes. With $w=M_w/M_s$, $$M_s=\frac{M}{1+w}=\frac{1.820}{1.07}=1.70094\ \text{kg}, \qquad M_w=M-M_s=0.11907\ \text{kg}.$$ The solids volume follows from $G_s=M_s/(\rho_w V_s)$: $$V_s=\frac{M_s}{G_s\,\rho_w}=\frac{1.70094}{2.6\times1000}=6.5421\times10^{-4}\ \text{m}^3=654.2\ \text{cm}^3.$$ The void volume is the remainder of the 1000 cm³ mold: $V_v=V-V_s=1000-654.2=345.8\ \text{cm}^3$, so $$n=\frac{V_v}{V}=\frac{345.8}{1000}=\boxed{34.6\%}.$$
2. Degree of saturation. The water occupies $V_w=M_w/\rho_w=119.1\ \text{cm}^3$ (since $\rho_w=1\ \text{g/cm}^3$), and $$S=\frac{V_w}{V_v}=\frac{119.1}{345.8}=\boxed{34.4\%}.$$
3. Part (c) — bulk density. The as-compacted (moist) density uses the total mass over the mold volume: $$\rho=\frac{M}{V}=\frac{1.820\ \text{kg}}{0.001\ \text{m}^3}=\boxed{1820\ \text{kg/m}^3}.$$
4. Part (d) — dry unit weight. The dry density is $\rho_d=M_s/V=1700.9\ \text{kg/m}^3$, so $$\gamma_d=\rho_d\,g=1700.9\times9.81/1000=\boxed{16.69\ \text{kN/m}^3}.$$
5. Part (e) — saturated unit weight. Compaction does not change the void ratio $e=V_v/V_s=345.8/654.2=0.5286$, so filling every void with water ($S=100\%$) gives $$\gamma_{sat}=\frac{G_s+e}{1+e}\,\gamma_w=\frac{2.6+0.5286}{1.5286}\times9.81=\boxed{20.08\ \text{kN/m}^3}.$$