Earthwork Hauling and Dam Weight: 18-Env-A3 Geotechnical and Hydrogeological Engineering, December 2018, Q2

Question 2: Earthwork Hauling and Dam Weight (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.

Given data
Quantity	Symbol	Value
Cut volume (bank measure)	$V_{cut}$	12,500 m³
Fill volume (compacted)	$V_{fill}$	8,500 m³
Truck capacity per load	—	30 m³
Bulking (swell) factor	—	35%
Round-trip cycle distance	—	80 km at 40 km/h
Fixed load/dump time	—	15 min at each end
Work schedule	—	2×8 h shifts/day, 7 days
Relative compaction	$RC$	95% of modified-Proctor $\gamma_{d,max}$
Modified-Proctor max dry unit weight	$\gamma_{d,max}$	21 kN/m³
Optimum moisture content	$w_{opt}$	8%

Find. (a) the number of truck loads required to haul the cut, and the fleet size that clears the schedule; (b) the total weight of the completed dam.

Approach. (a) Bulk the bank cut volume for hauling, divide by truck capacity for total loads, then use the cycle time and shift schedule to size the fleet. (b) Apply the relative compaction to get field dry unit weight, add the optimum moisture to get field wet unit weight, and multiply by the compacted fill volume.

Part (a) — loose (bulked) volume. The 12,500 m³ bank cut swells by the bulking factor once excavated and loaded: $$V_{loose}=V_{cut}(1+0.35)=12{,}500\times1.35=16{,}875\ \text{m}^3.$$
Total truck loads. $$N_{loads}=\frac{V_{loose}}{30}=\frac{16{,}875}{30}=562.5\ \Rightarrow\ \boxed{563\ \text{loads}}\ \text{(rounded up).}$$
Cycle time per truck. Travel time $=80/40=2.0$ h; plus 15 min at each end (load + dump) $=0.5$ h fixed: $$t_{cycle}=2.0+0.5=2.5\ \text{h}.$$
Loads per truck over the job. Each truck works $2\times8=16$ h/day, so $16/2.5=6.4$ loads/truck/day, and over 7 days: $$6.4\times7=44.8\ \text{loads/truck}.$$
Fleet size. Substituting, prose connects the required total to what one truck can deliver: $$N_{trucks}=\frac{563}{44.8}=12.57\ \Rightarrow\ \boxed{13\ \text{trucks}}\ \text{needed to complete the haul in 7 days.}$$
Part (b) — field unit weight of the fill. The dam is compacted to 95% of the modified-Proctor maximum, at the optimum moisture: $$\gamma_{d,field}=RC\times\gamma_{d,max}=0.95\times21=19.95\ \text{kN/m}^3,$$ $$\gamma_{wet,field}=\gamma_{d,field}(1+w_{opt})=19.95\times1.08=21.55\ \text{kN/m}^3.$$
Total weight of the dam. Multiplying by the compacted fill volume of the dam: $$W=\gamma_{wet,field}\times V_{fill}=21.55\times8{,}500=\boxed{183{,}140\ \text{kN}}\ (\approx183.1\ \text{MN},\ \approx18{,}670\ \text{tonnes}).$$

Check: "how many truck loads will be required" is read literally as the total number of loaded trips (563); since the question supplies full cycle-time and schedule data, the fleet size needed to physically deliver those 563 loads within the 7-day/2-shift window (13 trucks) is reported alongside it as the practically decisive number — both use every given quantity. Part (b) uses the fill volume (compacted dam volume), not the cut volume, since that is the material actually placed in the finished structure. The cut area's 2% natural moisture content is background context (it describes how much water the contractor must add to reach the fill's 8% optimum) but is not itself an input to either requested quantity — a common exam pattern of supplying one more data point than a given sub-question strictly needs.

Quantity	Value
(a) Total truck loads required	563
(a) Truck fleet size (7-day schedule)	13 trucks
(b) Field wet unit weight	21.55 kN/m³
(b) Total weight of the dam	183,140 kN (≈18,670 t)

Concept & Theory

Topic. Earthwork mass-haul sizing and compacted-fill weight.

Key relations. $V_{loose}=V_{bank}(1+\text{bulking})$; loads $=V_{loose}/\text{capacity}$; $t_{cycle}=t_{travel}+t_{fixed}$; $\gamma_{d,field}=RC\cdot\gamma_{d,max}$; $\gamma_{wet}=\gamma_d(1+w)$.

Why this works. Bulking accounts for the loose, air-entrained state material takes on once excavated and loaded — trucks are sized on that loose volume, not the bank volume. Fleet sizing is then a simple productivity balance: total loads needed divided by loads one truck can deliver in the available working time. Compacted unit weight follows directly from the Proctor curve once the target relative compaction and moisture are fixed.

Common pitfall. Sizing trucks on the bank volume instead of the bulked loose volume (undercounts loads by the bulking factor), and using the cut's natural moisture content instead of the fill's optimum moisture when computing the constructed dam's wet unit weight.

Source. Das, Principles of Geotechnical Engineering (9th ed.), Ch. 6 (compaction, relative compaction, Proctor tests).

Question 2 of 6: Earthwork Hauling and Dam Weight

Question 2: Earthwork Hauling and Dam Weight (20 marks)