Cut/Fill Balance With Rejected Material
Mass grading for a building pad. The plan takes 4,800 BCY of cut from an on-site borrow area and uses it as fill; the pad needs 3,600 CCY of compacted fill. Lab results on the silty clay: 28% swell, 12% shrinkage at the specified 95% Standard Proctor. Trucks are 18 LCY struck capacity, and company policy is to load no more than 90% of struck. Partway through the cut the operator exposes roughly 900 BCY in the northeast corner that is too wet to compact; the geotech rejects it and it has to leave the site. The superintendent says not to worry: 'There's 1,200 CY of extra dirt in that cut, we're fine.' The pad area is 150 ft x 200 ft with an average fill of 3.25 ft.
1. Is the superintendent right? Check the balance in consistent units.
Read the worked answer
Usable cut 4,800 − 900 = 3,900 BCY of acceptable material.
Convert to compacted Shrinkage of 12% means compacted volume = bank × (1 − 0.12): 3,900 × 0.88 = 3,432 CCY.
Requirement 3,600 CCY needed. Shortfall = 3,600 − 3,432 = 168 CCY.
Import in bank measure 168 ÷ 0.88 = 191 BCY of import is required.
Where the superintendent went wrong He is comparing 1,200 BCY of raw cut against a compacted fill number. Bank, loose, and compacted yards are three different quantities; 1,200 BCY of cut only yields about 1,056 CCY in the fill. Once 900 BCY goes in the waste pile, the job is short, not long.
Also worth flagging: the rejected material cannot be quietly buried in the embankment to make the numbers work. Wet, over-moisture clay in a structural fill will settle and fail density tests.
2. How many truckloads are needed to waste the 900 BCY, and how many would it take at struck capacity?
Read the worked answer
Loose volume to haul Swell of 28%: 900 × 1.28 = 1,152 LCY.
Payload under policy 18 LCY struck × 0.90 = 16.2 LCY per load.
Load count 1,152 ÷ 16.2 = 71.1 → round up to 72 loads (you cannot ship 0.1 of a load).
If loaded to struck 1,152 ÷ 18 = 64 loads exactly.
The difference Eight extra loads, plus the extra cycle time and scale tickets, purely from the 90% loading policy. That policy exists for a reason - spillage on the haul road, axle weight limits, and material blowing off - so the correct answer is to plan 72 loads, not to override the policy to save eight trucks.
Note the cut/fill answer does not change either way: swell and shrinkage describe volume, not the amount of material that exists.
3. The fill is 3.25 ft thick on average. With 8-inch loose lifts, how many lifts are required, and what would you tell the superintendent about the wet material?
Read the worked answer
Loose-to-compacted ratio Loose volume = bank × 1.28. Compacted volume = bank × 0.88. Ratio = 1.28 ÷ 0.88 = 1.4545 loose yards per compacted yard.
Compacted thickness of one 8-inch loose lift 8 in ÷ 1.4545 = 5.5 in = 0.458 ft.
Number of lifts 3.25 ft ÷ 0.458 ft = 7.1 → 8 lifts to cover the thickness with full lifts (7 full lifts gets you 3.21 ft; the last lift is thin, so plan on 8 with the final one trimmed).
On the wet material Options in order of preference:
- Stockpile and aerate/disc it to dry toward optimum moisture, then retest - it is only 900 BCY.
- Stabilize it (lime or fly ash) if the geotech approves and there is time for the mellowing period.
- Use it in a non-structural area such as landscaping or a fill slope.
- Waste it - and then own the 72 loads plus 191 BCY of import.
What you do not do is push it into the embankment and compact over it. It will pass no density test, and the failure shows up two years later as a settled slab.