Question 1 of 6: Soil Phase Relationships

Question 1: Soil Phase Relationships (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) bulk unit weight $\gamma$; (b) dry density $\rho_d$; (c) degree of saturation $S$; (d) saturated unit weight $\gamma_{sat}$; (e) porosity $n$.

Approach. Use the measured mass and volume to get the water mass directly, find the solids and void volumes from $G_s$, and combine into every requested phase ratio.

1. Part (a) — bulk unit weight. The whole moist sample over its total volume: $$\rho=\frac{M_t}{V}=\frac{1.8}{0.001}=1800\ \text{kg/m}^3,\qquad \gamma=\rho g=1800\times9.81/1000=\boxed{17.66\ \text{kN/m}^3}.$$
2. Volumes of solids, water and voids. $V_s=\dfrac{M_s}{G_s\rho_w}=\dfrac{1.7}{2.5\times1000}=6.80\times10^{-4}\ \text{m}^3$ (680 cm³); mass of water $M_w=M_t-M_s=0.1$ kg, so $V_w=1.00\times10^{-4}\ \text{m}^3$ (100 cm³); void volume $V_v=V-V_s=3.20\times10^{-4}\ \text{m}^3$ (320 cm³).
3. Part (b) — dry density. $\rho_d=\dfrac{M_s}{V}=\dfrac{1.7}{0.001}=\boxed{1700\ \text{kg/m}^3}$ ($\gamma_d=1700\times9.81/1000=16.68\ \text{kN/m}^3$).
4. Part (e) — porosity and void ratio. $n=\dfrac{V_v}{V}=\dfrac{320}{1000}=\boxed{32.0\%}$, and equivalently $e=\dfrac{V_v}{V_s}=\dfrac{320}{680}=0.4706$ (check: $n=e/(1+e)=0.4706/1.4706=0.320$✓).
5. Part (c) — degree of saturation. $S=\dfrac{V_w}{V_v}=\dfrac{100}{320}=\boxed{31.25\%}.$
6. Part (d) — saturated unit weight. Flood the same void space to $S=100\%$: total saturated mass $M_{sat}=M_s+V_v\rho_w=1.7+0.32=2.02$ kg, so $$\gamma_{sat}=\frac{M_{sat}}{V}g=\frac{2.02}{0.001}\times9.81/1000=\boxed{19.82\ \text{kN/m}^3}.$$