Question

Base course aggregate has a target dry density of 119.7 Ib/cu ft in place. It will be laid down and compacted in a rectangular street repair area of 2000 ft by 48 ft by 6 in. The aggregate in the stockpile contains 3.1 percent 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:

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

volume = 2000 × 48 × 0.5

volume = 48000 ft³

now filling space with aggregate of the density that is

density = 0.95 × 119.7

density = 113.72 lb/ft³

and dry weight of this aggregate is

dry weight = 48000 × 113.72

dry weight = 5458320 lbs

we consider here percent moisture is by weigh

so weight of moisture in aggregate will be

weight of moisture = 0.031 × 5458320

weight of moisture = 169208 lbs

so here total weight of aggregate is

total weight of aggregate = 5458320 + 169208

total weight of aggregate = 5627528 lbs = 2814 tons