Question 1 of 6: Compaction — Porosity, Saturation and Unit Weights

Question 1: Compaction — Porosity, Saturation and Unit Weights (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. Compute the bulk density directly from the mold, strip out the moisture to get the dry unit weight, then use an assumed $G_s$ to back out the void ratio and complete the remaining phase relations.

1. Part (c) — bulk (moist) density. The as-compacted density uses the whole mass over the mold volume: $$\rho=\frac{M}{V}=\frac{1.850}{0.001}=\boxed{1850\ \text{kg/m}^3}\quad\left(\gamma=\rho g=1850\times9.81/1000=18.15\ \text{kN/m}^3\right).$$
2. Part (d) — dry unit weight. Strip the pore water using $w=M_w/M_s$: $$M_s=\frac{M}{1+w}=\frac{1.850}{1.12}=1.6518\ \text{kg},\qquad \rho_d=\frac{M_s}{V}=1651.8\ \text{kg/m}^3,$$ $$\gamma_d=\rho_d g=1651.8\times9.81/1000=\boxed{16.20\ \text{kN/m}^3}.$$
3. Part (a) — void ratio and porosity. With $G_s=2.70$ assumed, inverting $\gamma_d=\dfrac{G_s}{1+e}\gamma_w$ gives $$e=\frac{G_s\,\gamma_w}{\gamma_d}-1=\frac{2.70\times9.81}{16.20}-1=0.6346,\qquad n=\frac{e}{1+e}=\frac{0.6346}{1.6346}=\boxed{38.8\%}.$$
4. Part (b) — degree of saturation. $S=\dfrac{w\,G_s}{e}$: $$S=\frac{0.12\times2.70}{0.6346}=\boxed{51.1\%}.$$
5. Part (e) — saturated unit weight. Flooding the same void ratio to $S=100\%$: $$\gamma_{sat}=\frac{G_s+e}{1+e}\,\gamma_w=\frac{2.70+0.6346}{1.6346}\times9.81=\boxed{20.01\ \text{kN/m}^3}.$$