Question 2 of 6: Phase Relations for a Compacted Sample

Question 2: Phase Relations for a Compacted Sample (10 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. A single compacted specimen whose total mass, total volume, water content and specific gravity of solids are all known, from which the complete phase diagram can be reconstructed.

Find. The bulk unit weight $\gamma$, the dry unit weight $\gamma_d$, the degree of saturation $S$, the void ratio $e$ and the porosity $n$.

Approach. Get $\gamma$ from mass and volume, strip the water out with $\gamma_d = \gamma/(1+w)$, back-figure $e$ from the solid-phase identity $\gamma_d = G_s\gamma_w/(1+e)$, then read $n$ and $S$ off the standard relations $n = e/(1+e)$ and $Se = wG_s$.

1. Bulk density and bulk unit weight. The bulk (total, moist) unit weight is the weight of everything in the specimen divided by the volume it occupies: $$\rho = \frac{m}{V} = \frac{12.5}{0.006} = 2083.3\ \text{kg/m}^3 \qquad \gamma = \rho\,g = 2083.3 \times 9.81 = \boxed{20.44\ \text{kN/m}^3}$$ This is the unit weight that would be used directly in an overburden calculation above the water table.
2. Dry unit weight. Water content is defined on the mass of solids, $w = m_w/m_s$, so the solids alone carry the fraction $1/(1+w)$ of the total weight while occupying the same total volume: $$\gamma_d = \frac{\gamma}{1+w} = \frac{20.44}{1.10} = \boxed{18.58\ \text{kN/m}^3}$$ This is the value a field density test is compared against in a compaction specification.
3. Void ratio from the solid phase. Writing the dry unit weight in terms of the solid phase, $\gamma_d = W_s/V = G_s\gamma_w V_s / (V_s + V_v)$, which rearranges to the formula-sheet relation $\gamma_d = G_s\gamma_w/(1+e)$. Solving for the void ratio, $$e = \frac{G_s\gamma_w}{\gamma_d} - 1 = \frac{2.67 \times 9.81}{18.58} - 1 = \frac{26.19}{18.58} - 1 = \boxed{0.410}$$ A void ratio of 0.41 is a well-compacted granular or sandy fill; a loose sand would be nearer 0.7 and a soft clay above 1.0.
4. Porosity. Porosity and void ratio measure the same voids against different denominators — $e$ against the solids, $n$ against the total: $$n = \frac{e}{1+e} = \frac{0.410}{1.410} = 0.291 = \boxed{29.1\ \%}$$
5. Degree of saturation. The identity $Se = wG_s$ on the formula sheet follows directly from the definitions of $S$, $e$, $w$ and $G_s$, so $$S = \frac{wG_s}{e} = \frac{0.10 \times 2.67}{0.410} = 0.652 = \boxed{65.2\ \%}$$ The specimen is well short of saturation, which is exactly what a soil compacted dry of optimum looks like — consistent with the fabric discussion in Question 3.
6. Cross-check by rebuilding the phase diagram. As an independent verification, work forward from masses instead of backward from unit weights. Mass of solids $m_s = 12.5/1.10 = 11.364$ kg, so $V_s = 11.364/(2.67 \times 1000) = 0.004256\ \text{m}^3$ and $V_v = 0.006 - 0.004256 = 0.001744\ \text{m}^3$. Then $e = V_v/V_s = 0.410$ and $n = V_v/V = 0.291$, both reproducing the values above. The mass of water is $m_w = 12.5 - 11.364 = 1.136$ kg, giving $V_w = 0.001136\ \text{m}^3$ and $S = V_w/V_v = 0.652$. Every route agrees.