Question 1 of 6: Index Properties from a Moist Soil Sample

Question 1: Index Properties from a Moist 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.

Given.

Find. (a) $\gamma_d$; (b) $\gamma_{zav}$; (c) $e$; (d) $S$; (e) $V_w$.

Approach. Get the moist unit weight directly from $W/V$, strip out the water fraction with $w$ to get $\gamma_d$, then use the standard $G_s$–$e$–$\gamma_d$ identity to back out every other index property.

1. Moist and dry unit weight. $$\gamma=\frac{W}{V}=\frac{54.3\ \text{N}}{2.83\times10^{-3}\ \text{m}^3}=19{,}187\ \text{N/m}^3=19.19\ \text{kN/m}^3,$$ $$\gamma_d=\frac{\gamma}{1+w}=\frac{19.19}{1.12}=\boxed{17.13\ \text{kN/m}^3}.$$
2. (b) Zero-air-voids dry unit weight — the theoretical maximum dry unit weight at this water content if all air were expelled ($S=100\%$): $$\gamma_{zav}=\frac{G_s\,\gamma_w}{1+w\,G_s}=\frac{2.72\times9.81}{1+0.12\times2.72}=\frac{26.68}{1.3264}=\boxed{20.12\ \text{kN/m}^3}.$$
3. (c) Void ratio from the $G_s$–$e$–$\gamma_d$ identity $\gamma_d=\dfrac{G_s\,\gamma_w}{1+e}$: $$e=\frac{G_s\,\gamma_w}{\gamma_d}-1=\frac{2.72\times9.81}{17.13}-1=1.5576-1=\boxed{0.558}.$$
4. (d) Degree of saturation from $S\,e=w\,G_s$: $$S=\frac{w\,G_s}{e}=\frac{0.12\times2.72}{0.558}=\boxed{58.5\%}.$$
5. (e) Volume of water — split the moist weight into solids and water using $w=W_w/W_s$: $$W_s=\frac{W}{1+w}=\frac{54.3}{1.12}=48.482\ \text{N},\qquad W_w=W-W_s=5.818\ \text{N},$$ $$V_w=\frac{W_w}{\gamma_w}=\frac{5.818\ \text{N}}{9810\ \text{N/m}^3}=\boxed{5.93\times10^{-4}\ \text{m}^3\ (593\ \text{cm}^3)}.$$