Question

How many trips would one rubber-tired Herrywampus have to make to backfill a space with a geometrical volume of 5400 cubic yard? The maximum capacity of the machine is 30 cubic yard (heaped), or 40 tons. The material is to be compacted with a shrinkage of 25% (relative to bank measure) and has a swell factor of 20% (relative to bank measure). The material weighs 3,000 lb/cu yd (bank). Assume that the machine carries its maximum load on each trip. Check by both weight and volume limitations

Answer 1

Answer:

If analyzed by volume capacity, more trips are needed to fill the space, thus the required trips are 288

Explanation:

a) By volume.

The shrinkage factor is:

$\frac{5400cu-yd}{1-0.25} =7200cu-yd$

The volume at loose is:

$V_{loose} =V_{bank} (1+swell-factor)=7200(1+0.2)=8640cu-yd$

If the Herrywampus has a capacity of 30 cubic yard:

$\frac{8640cu-yd}{30cu-yd/trip} =288trip$

b) By weight

The swell factor in terms of percent swell is equal to:

$pounds-per-cubic-yard-loose=\frac{pounds-per-cubic-yard-bank}{\frac{percent-swell}{100}+1 }$

$pounds-per-cubic-yard-loose=\frac{3000}{\frac{20}{100} +1} =2500lb/cu-yd$

The weight of backfill is:

$8640cu-yd*2500\frac{lb}{cu-yd} *\frac{1ton}{2000lb} =10800ton$

The Herrywampus has a capacity of 40 ton:

$\frac{10800}{40ton/trip} =270trip$

If analyzed by volume capacity, more trips are needed to fill the space, thus the required trips are 288