CSLB General Building (B)Site Work, Foundations and ConcreteHard
A geotechnical technician performs a sand cone test on compacted fill at a foundation site and determines the in-place dry density of the soil is 112 pounds per cubic foot. The laboratory Proctor test established a maximum dry density of 120 pounds per cubic foot for this soil. If the specifications require 90 percent relative compaction, does the fill pass?
- ANo, because 112/120 = 93.3%, which exceeds the required range and indicates over-compaction
- BYes, because the dry density alone always satisfies compaction requirements
- CNo, because 112/120 = 83.3%, which is below the 90% minimum requirement
- DYes, because 112/120 = 93.3%, which exceeds the 90% minimum requirement
Show answer & explanationAnswer & explanation
Correct answer: D. Yes, because 112/120 = 93.3%, which exceeds the 90% minimum requirement
Relative compaction is calculated as field dry density divided by maximum Proctor dry density: 112 ÷ 120 = 0.933, or 93.3%. Since this exceeds the 90% minimum specified, the fill passes; there is no such thing as 'over-compaction' failure at this level.
Why the other options are wrong
- A. Over-compaction is not a real failure criterion in this context; 93.3% simply exceeds the minimum.
- B. Relative compaction, not raw dry density alone, is compared to the spec requirement.
- C. The math is incorrect; 112/120 is 93.3%, not 83.3%.
Sand Cone Test / Relative Compaction
The sand cone test measures in-place soil density in the field, which is then divided by the laboratory maximum Proctor dry density to calculate percent relative compaction.
- Relative compaction = field dry density ÷ max Proctor density
- Common specification minimum is 90% relative compaction
- Sand cone test is a common field density testing method
Memory trick: Divide field density by lab max — the percentage tells the compaction story.