Question

A contractor plans to use a wheeled loader to move 5,000 BCY of soil 500 feet using a 50-minute working hour. The soil has a swell of 15%. The loader has a bucket capacity of 2.5 cubic yards. The soil has a fill factor of 1.2 and a weight of 2,500 lbs per LCY. The loader has an STL of 50% of 15,000 lbs. The loader will haul in one direction loaded at 2.5 miles per hour and return at 3.5 miles per hour. The total cost of the tractor is $55.00 per hour.  It has a fixed time of .35 minutes. Remember that the loader will be moving the dirt in its loose state. How many days will it take to move all of the dirt working 8 hours per day?

Answer 1

Solution :

Given in the Question,
Productivity = (Load Volume) (Operational Efficiency) / Cycle Time,
LV = (BC)(FF),
OE = minutes worked per hour,
CT = FT + VT,
VT = HT and RT each = D/(R*88),
LV IS BC/FF,
= 2.5 CUBIC.Y/1.2,
= 2.08 CUBIC YARDS,
OE IS M= TOTAL WORK FOR DAY IS 3.5/2.5 = 1.4 .
AND CT IS FT = VT WHERE (0.35+50) MIN IS 50.35 MIN
FINALLY PRODUCTIVITY IS LV(OE)/CT
= 2.08 X (1.4/50.35)
= 0.058 M3/MIN
NOTE :
CT IS CYCLE TIME
OE IS OPERATIONAL EFFICIENCY
BC IS BUCKET CAPACITY
FC IS FILL FACTOR
LV IS LOAD VOLUME.