Question 2 of 6: Earthwork — Hauling and Compaction-Water Trucking

Question 2: Earthwork — Hauling and Compaction-Water Trucking (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) number of dump trucks required to haul the cut; (b) volume of compaction water and number of tanker trucks required.

Approach. Convert the bank cut volume to loose (bulked) volume for hauling and divide by one truck's total 7-day carrying capacity to size the soil fleet; separately use the fill's target dry unit weight and volume to get the dry mass placed, scale by the moisture deficit to get the water mass/volume, and size the tanker fleet the same way.

1. Part (a) — loose volume to haul. Trucking is sized on loose volume: $V_{loose}=V_{cut}(1+0.25)=100{,}000\times1.25=125{,}000\ \text{m}^3$.
2. Part (a) — hauling capacity per truck. Working time $=7\times16\ \text{h}=112\ \text{h}=6720\ \text{min}$; trips per truck $=6720/30=224$; capacity per truck over the job $=224\times40=8960\ \text{m}^3$.
3. Part (a) — fleet size. $$N_{soil}=\frac{V_{loose}}{224\times40}=\frac{125{,}000}{8960}=13.95\ \Rightarrow\ \boxed{14\ \text{trucks}}\ (\text{round up to cover the full volume}).$$
4. Part (b) — dry mass placed in the fill. Field dry unit weight $\gamma_{d,field}=RC\times\gamma_{d,max}=0.95\times19=18.05\ \text{kN/m}^3$. The dry mass of soil solids the fill contains (solids mass is conserved through hauling/wetting/compaction) is $$M_s=\frac{\gamma_{d,field}V_{fill}}{g}=\frac{18.05\times80{,}000}{9.81\times10^{-3}}=1.472\times10^{8}\ \text{kg}\approx147{,}200\ \text{tonnes}.$$
5. Part (b) — water required. Raising the moisture content from 5% to 12% adds $\Delta w=0.07$ of the dry mass in water: $$M_w=M_s\,\Delta w=1.472\times10^{8}\times0.07=1.030\times10^{7}\ \text{kg}\ \Rightarrow\ V_w=\boxed{10{,}304\ \text{m}^3}\ (\rho_w=1000\ \text{kg/m}^3).$$
6. Part (b) — tanker fleet. Trips per tanker $=6720/60=112$; capacity per tanker over the job $=112\times20=2240\ \text{m}^3$. $$N_{water}=\frac{10{,}304}{2240}=4.60\ \Rightarrow\ \boxed{5\ \text{tanker trucks}}.$$