Question 4 of 6: Index Properties of a Cylindrical Soil Sample

Question 4: Index Properties of a Cylindrical Soil Sample (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.

Approach. Work from the sample's total volume (from its cylindrical dimensions) and the two masses (moist and oven-dry) to get the mass of water, the volume of solids, and hence every index property in turn.

1. Total volume and water content. $$\begin{aligned} V&=\frac{\pi}{4}D^2L=\frac{\pi}{4}(7.5)^2(15)=662.68\ \text{cm}^3\\ M_w&=1400-1196=204\ \text{g} \end{aligned}$$ d) Water content: $$w=\frac{M_w}{M_s}=\frac{204}{1196}=\boxed{17.06\%}$$
2. a) Void ratio. With ρ_w = 1 g/cm³, the volume of solids follows from the assumed G_s: $$\begin{aligned} V_s&=\frac{M_s}{G_s\rho_w}=\frac{1196}{2.65}=451.32\ \text{cm}^3\\ V_v&=V-V_s=662.68-451.32=211.36\ \text{cm}^3 \end{aligned}$$ $$e=\frac{V_v}{V_s}=\frac{211.36}{451.32}=\boxed{0.468}$$
3. c) Dry unit weight. $$\gamma_d=\frac{M_s}{V}\gamma_w=\frac{1196}{662.68}(9.81)=\boxed{17.71\ \text{kN/m}^3}$$
4. b) Zero-air-voids unit weight. The theoretical dry unit weight the soil would have at this same water content if every void were water-filled (S = 100%, zero air): $$\gamma_{zav}=\frac{G_s\gamma_w}{1+wG_s}=\frac{2.65(9.81)}{1+0.1706(2.65)}=\boxed{17.90\ \text{kN/m}^3}$$ Since γ_zav (17.90) is only marginally above the actual γ_d (17.71), the sample is nearly saturated as retrieved — consistent with S = V_w/V_v = 204/211.36 = 96.5%, an agricultural field soil sample taken shortly after wetting or from below the water table.