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. What is the productivity of the loader in LCY per hour?

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)

Answer 1

ACCODING TO GIVEN DATA :

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

= 2.5CUBIC.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 50.35MIN

FINALLY PRODUCTIVITY IS LV(OE)/CT

2.08X1.4/50.35 =0.058M3/MIN .

HERE NOTE : CT IS CYCLE TIME

OE IS OPERATIONAL EFFICIENCY

BC IS BUCKET CAPACITY

FC IS FILL FACTOR

LV IS LOAD VOLUME