Question 1 of 6: Proctor Compaction — Water Addition and Borrow Volume

Question 1: Proctor Compaction — Water Addition and Borrow Volume (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) additional weight of water per unit volume of compacted fill needed to raise the borrow soil to $w_{opt}$; (b) number of borrow-pit truckloads for the embankment.

Approach. Use the field dry unit weight target ($95\%$ of $\gamma_{d,max}$) as the "currency" of solids per unit volume: compare the water carried at the borrow moisture content against the water needed at $w_{opt}$ for that same mass of solids (part a), then use conservation of dry solids between the borrow pit's natural state and the compacted fill to size the required haul volume (part b).

1. Part (a) — dry unit weight required in the field. The fill must reach 95% of the maximum Proctor dry unit weight: $$\gamma_{d,field}=0.95\,\gamma_{d,max}=0.95(19)=\boxed{18.05\ \text{kN/m}^3}.$$ This is also the weight of solids, $W_s$, contained in each cubic metre of compacted fill (by definition of dry unit weight).
2. Water carried per m³ of fill at each moisture content. Since $w=W_w/W_s$, the water weight per unit volume at a given $w$ is $W_w=w\,\gamma_{d,field}$: $$W_{w,bp}=w_{bp}\,\gamma_{d,field}=(0.082)(18.05)=1.480\ \text{kN/m}^3,$$ $$W_{w,opt}=w_{opt}\,\gamma_{d,field}=(0.115)(18.05)=2.076\ \text{kN/m}^3.$$
3. Additional water required. The extra water that must be added to each cubic metre of PLACED fill to raise it from the borrow moisture to the optimum moisture is the difference: $$\Delta W_w=W_{w,opt}-W_{w,bp}=2.076-1.480=\boxed{0.596\ \text{kN/m}^3}.$$
4. Part (b) — borrow volume needed via conservation of dry solids. The borrow pit's own dry unit weight (before any water is added or compaction is applied) is $$\gamma_{d,bp}=\frac{\gamma_{bp}}{1+w_{bp}}=\frac{17.2}{1.082}=15.90\ \text{kN/m}^3.$$ The total weight of solids needed for the whole embankment is fixed by the compacted target, $W_{s,total}=\gamma_{d,field}\,V_{fill}=(18.05)(100{,}000)=1{,}805{,}000\ \text{kN}$. The SAME solids, before compaction, occupy a larger natural (borrow-pit) volume: $$V_{borrow}=\frac{W_{s,total}}{\gamma_{d,bp}}=\frac{1{,}805{,}000}{15.90}=\boxed{113{,}547\ \text{m}^3}.$$
5. Truckloads. At 10 m³ per truck (borrow-state volume, before any water addition or compaction), $$N_{trucks}=\frac{V_{borrow}}{10}=\frac{113{,}547}{10}=11{,}354.7\ \Rightarrow\ \boxed{11{,}355\ \text{truckloads}}\ \text{(rounded up).}$$