The site earthwork in Phase III of Four Hills Landfill project included a 3.0 ft deep cut across an entire 2.5-acre site. Soil was excavated from within the proposed Phase III footprint. The average unit weight of this soil is 118 lb/ft3, and the average moisture content is 9.6%. It also has a maximum dry unit weight of 122 lb/ft3 and an optimum moisture content of 11.1%, based on the modified Proctor test. The excavated soil will be placed on a nearby site and compacted to an average relative compaction of 93%. Compute the volume of fill that will be produced and express your answer in cubic yards.

11470.4 cubic yards

Explanation:

first we calculate the volume of the site;

V1 = Area × depth

V1 = 2.5 acre ×  (43560 ft² / 1 acre )×3ft

V1 = 326700 ft

next we is the relative compaction

RC = [γd(field) / γdmax(laboratory)] × 100

so we substitute

93 =  [γd(field) / 122 lb/ft³)] × 100

γd(field) = 113.46 lb/ft³

then the dry unit weight of the site

γday1 = γavg / ( 1 + w)

= 118 lb/ft³ / ( 1 + (9.6/100))

= 118 lb/ft³ /1.096

= 107.664 lb/ft³

finally we find the fill volume of the site

V2/V1 = γd / γd(field)

we substitute

V2/326700 = 107.664 / 113.46

V2 = 310010.83 ft³

we convert to cubic yards

= 310010.83 ft³ × (0.037 cubic yard / 1 ft³)

= 11470.4 cubic yards

answer:

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explanation: