Question

Base course aggregate has a target dry density of 119.7 lb/cu ft in place. It will be laid down and compacted in a rectangular street repair area of 2000 ft * 48 ft * 6 in. The aggregate in the stockpile contains 3.1% moisture. If the required compaction is 95% of the target, how many tons of aggregate will be needed?

Answer 1

total weight of aggregate = 5627528 lbs = 2814 tons

Step-by-step explanation:

given data

dry density = 119.7 lb/cu ft

area = 2000 ft × 48 ft × 6 in

aggregate = 3.1%

required compaction = 95%

solution

we get here volume of space to be filled with aggregate that is

volume = 2000 × 48 × 0.5 = 48000 ft³

when here space fill with aggregate of density is

density = 0.95 × 119.7 = 113.72 lb/ft³

and

dry weight of this aggregate will be is

dry weight = 48000 × 113.72 = 5458320 lbs

and

we consider take percent moisture by weigh so that there weight of moisture in aggregate is express as

weight of moisture = 0.031 × 5458320 = 169208 lbs

and

total weight of aggregate will be

total weight of aggregate = 5458320 + 169208

total weight of aggregate = 5627528 lbs = 2814 tons