Question 1 of 6: Weight–Volume Relations of a Soil

Question 1: Weight–Volume Relations of a Soil (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.

Find. (a) void ratio $e$; (b) porosity $n$; (c) degree of saturation $S$; (d) moisture content $w$; (e) buoyant unit weight $\gamma_b$.

Approach. Back out the void ratio from $\gamma_d=G_s\gamma_w/(1+e)$, then read porosity and moisture content directly off it, use $wG_s=Se$ for saturation, and finally compute the saturated unit weight at the SAME void ratio to get the buoyant (submerged) unit weight.

1. Part (a) — void ratio from the dry unit weight. Since $\gamma_d=\dfrac{G_s\gamma_w}{1+e}$, $$e=\frac{G_s\gamma_w}{\gamma_d}-1=\frac{2.65\times9.81}{15.1}-1=\boxed{0.722}.$$
2. Part (b) — porosity. $$n=\frac{e}{1+e}=\frac{0.722}{1.722}=\boxed{41.9\%}.$$
3. Part (d) — moisture content. With $\gamma=\gamma_d(1+w)$, $$w=\frac{\gamma}{\gamma_d}-1=\frac{17.0}{15.1}-1=\boxed{12.6\%}.$$
4. Part (c) — degree of saturation. From $wG_s=Se$, $$S=\frac{wG_s}{e}=\frac{0.126\times2.65}{0.722}=\boxed{46.2\%}.$$ Only 46% of the void space is filled with water, so the soil is well above the water table — the "natural" state here is moist, not saturated.
5. Part (e) — buoyant unit weight. The buoyant (submerged) unit weight is a property of the soil skeleton once fully saturated at this SAME void ratio (i.e. if this soil were below the water table), $\gamma_b=\gamma_{sat}-\gamma_w$, which simplifies to $$\gamma_b=\frac{(G_s-1)\gamma_w}{1+e}=\frac{(2.65-1)\times9.81}{1.722}=\boxed{9.40\ \text{kN/m}^3}.$$