Question 2 of 7: Density, Porosity and Void Ratio of an Undisturbed Soil Cylinder

Question 2: Density, Porosity and Void Ratio of an Undisturbed Soil Cylinder (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. Compute the cylinder volume from its geometry, get the wet soil mass by difference, split wet mass into solids and water using the wet-basis moisture content, then apply the standard weight-volume relationships (bulk/dry density, porosity, void ratio). For part (c), hold the solid mass fixed and fill the existing void space entirely with water.

1. Cylinder volume and wet soil mass. $$V = \frac{\pi}{4}D^2L = \frac{\pi}{4}(0.50)^2(1.0) = 0.1964\ \text{m}^3, \qquad m_{\text{wet}} = 445 - 37.5 = 407.5\ \text{kg}$$
2. a) Bulk (wet) and dry density. Bulk density uses the full wet mass; dry density strips out the 15% wet-basis water fraction first: $$\rho_{\text{wet}} = \frac{m_{\text{wet}}}{V} = \frac{407.5}{0.1964} = \boxed{2075\ \text{kg/m}^3\ (2.08\ \text{g/cm}^3)}$$ $$m_{\text{dry}} = m_{\text{wet}}(1-w) = 407.5(0.85) = 346.4\ \text{kg}, \qquad \rho_{\text{dry}} = \frac{m_{\text{dry}}}{V} = \frac{346.4}{0.1964} = \boxed{1764\ \text{kg/m}^3\ (1.76\ \text{g/cm}^3)}$$
3. b) Porosity and void ratio. Porosity compares the dry density actually achieved against the density the solids alone would give if packed with zero voids: $$n = 1 - \frac{\rho_{\text{dry}}}{G_s} = 1 - \frac{1764}{2650} = \boxed{0.334\ (33.4\%)}, \qquad e = \frac{n}{1-n} = \frac{0.334}{0.666} = \boxed{0.50}$$
4. c) Masses at full saturation. The solid mass (346.4 kg) is fixed by the grains present; only the void space can gain water. Solid volume and void volume follow from G_s and the cylinder volume already found: $$V_s = \frac{m_{\text{dry}}}{G_s} = \frac{346.4}{2650} = 0.1307\ \text{m}^3, \qquad V_v = V - V_s = 0.1964 - 0.1307 = 0.0656\ \text{m}^3\ (65.6\ \text{L})$$ Filling every void with water (S = 100%) gives the saturated water mass, and adding it to the fixed solid mass gives the saturated soil mass and, with the cylinder, the total mass: $$m_{w,\text{sat}} = V_v\,\rho_w = (0.0656)(1000) = 65.6\ \text{kg}$$ $$m_{\text{soil,sat}} = m_{\text{dry}} + m_{w,\text{sat}} = 346.4 + 65.6 = \boxed{412.0\ \text{kg}}$$ $$m_{\text{total}} = m_{\text{soil,sat}} + m_{\text{cyl}} = 412.0 + 37.5 = \boxed{449.5\ \text{kg}}$$