Question 2 of 6: Phase relationships of a compacted sample

Question 2: Phase relationships of 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 compacted specimen with the following measured properties.

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

Approach. Convert mass to weight for the unit weights, then split the phase diagram into solids/water/air by mass and volume ($M_s = M/(1+w)$, $V_s = M_s/G_s\rho_w$) and read $e$, $n$, $S$ off the volumes.

1. Bulk unit weight. $\gamma = \dfrac{Mg}{V} = \dfrac{12.5 \times 9.81}{0.006}\ \text{N/m}^3 = 20{,}437\ \text{N/m}^3$, so $\boxed{\gamma = 20.4\ \text{kN/m}^3}$.
2. Dry unit weight. Removing the water, $\gamma_d = \dfrac{\gamma}{1+w} = \dfrac{20.4}{1.10} = 18.6\ \text{kN/m}^3$.
3. Mass and volume of solids. $M_s = \dfrac{M}{1+w} = \dfrac{12.5}{1.10} = 11.36\ \text{kg}$, hence water mass $M_w = 1.14\ \text{kg}$. Volume of solids $V_s = \dfrac{M_s}{G_s\,\rho_w} = \dfrac{11.36}{2.67 \times 1000} = 4.256\times 10^{-3}\ \text{m}^3$.
4. Void and water volumes. $V_v = V - V_s = 0.006 - 0.004256 = 1.744\times 10^{-3}\ \text{m}^3$; $V_w = M_w/\rho_w = 1.14\times 10^{-3}\ \text{m}^3$.
5. Void ratio and porosity. $e = \dfrac{V_v}{V_s} = \dfrac{1.744}{4.256} = 0.41$; $n = \dfrac{V_v}{V} = \dfrac{1.744}{6.00} = 0.291 = 29.1\%$.
6. Degree of saturation. $S = \dfrac{V_w}{V_v} = \dfrac{1.14}{1.744} = 0.652 = 65.2\%$, confirmed by the identity $Se = wG_s \Rightarrow S = \dfrac{0.10\times 2.67}{0.41} = 0.652$.

The specimen is only about two-thirds saturated, consistent with a soil compacted dry of optimum where entrapped air remains in the voids.