Question 2 of 6: Soil Physical Properties

Question 2: Soil Physical Properties (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. Sample diameter $d=50\text{ mm}$, height $h=100\text{ mm}$; degree of saturation $S=85\%$; dry density $\rho_d=1.65\text{ g/cm}^3$; specific gravity $G_s=2.65$; pore fluid densities $\rho_{w,1}=1.0\text{ g/cm}^3$ (distilled) and $\rho_{w,2}=1.2\text{ g/cm}^3$ (saline).

Find. (a) Mass of pore fluid and mass of dry soil for each of the two identical-volume samples (parts i–ii).

Approach. The mould volume and dry density fix the mass of solids and the void ratio (using $G_s$ against the standard reference density of water, since $G_s$ is a fixed material property independent of which fluid saturates the pores). The void volume and the given saturation then fix the volume of pore fluid; only the pore-fluid mass differs between the two samples, through $\rho_{w,1}$ vs $\rho_{w,2}$.

1. Mould volume. $$V=\frac{\pi}{4}d^2h=\frac{\pi}{4}(5\text{ cm})^2(10\text{ cm})=\boxed{196.3\text{ cm}^3}$$
2. Void ratio and mass of dry soil. $G_s$ is referenced to standard water density ($1.0\text{ g/cm}^3$), so $$e=\frac{G_s\rho_{w,\text{std}}}{\rho_d}-1=\frac{2.65\times1.0}{1.65}-1=\boxed{0.606}$$ Dry density is mass of solids per total volume regardless of pore fluid, so the mass of dry soil is identical for both samples: $$M_d=\rho_d V=1.65\times196.3=\boxed{324.0\text{ g}}\quad\text{(both samples)}$$
3. Volume of voids and volume of pore fluid. $$V_v=\frac{e}{1+e}V=\frac{0.606}{1.606}(196.3)=\boxed{74.1\text{ cm}^3}$$ $$V_f=S\,V_v=0.85(74.1)=\boxed{63.0\text{ cm}^3}\quad\text{(both samples — same }e\text{ and }S\text{)}$$
4. Mass of pore fluid — distilled water. $$M_{f,1}=V_f\rho_{w,1}=63.0\times1.0=\boxed{63.0\text{ g}}$$
5. Mass of pore fluid — saline water. Same pore volume, denser fluid: $$M_{f,2}=V_f\rho_{w,2}=63.0\times1.2=\boxed{75.6\text{ g}}$$