Question

A new embankment, when completed will occupy a net volume of 257,000cy. The borrow material that will be used to construct this fill is stiff clay. In its "bank" condition, the borrow material has a moist unit weight of 129pcf, a water content of, 16.5% and an in place void ratio of 0.620. The embankment will be constructed in layers of 8 inch depth, loose measure then compacted to a dry unit weight of 114pcf at a moisture content of 18.3%. Trucks with a 35,000 pound capacity will be used for transport a) Compute the required volume of borrow pit excavation. b) Determine how many trucks will be required to complete the job. c) Determine how many gallons of water will need to be added to each truck. (1gallon of water is 8.36 pounds)

Answer 1

Okay, here are the steps to solve this problem:

a) Compute the required volume of borrow pit excavation:

Given: Net embankment volume = 257,000 cy

Unit weight of borrow material (loose) = 129 pcf

unit weight of compacted embankment = 114 pcf

Step 1) Convert 257,000 cy to cubic ft: 257,000 cy * 27 ft^3/cy = 6,989,000 ft^3

Step 2) Compute loose volume required: 6,989,000 ft^3 / (129 pcf - 114 pcf) = 123,890,000 ft^3 (or 287,480 cy)

Therefore, the required volume of borrow pit excavation is 287,480 cy.

b) Determine how many trucks will be required to complete the job:

Given: Truck capacity = 35,000 lbs

Unit weight of borrow material (compacted) = 114 pcf = 1,752 lbs/cy

Step 1) Convert 287,480 cy to tons: 287,480 cy * 1 ton / 27 cu yd = 10,626 tons

Step 2) Number of trucks = 10,626 tons / 35,000 lbs per truck = 304 trucks

Therefore, 304 trucks will be required to complete the job.

c) Determine how many gallons of water is needed to add to each truck:

Given: Moisture content required = 18.3%

Moisture content of borrow material = 16.5%

Unit weight of borrow material (compacted) = 1,752 lbs/cy

Step 1) Additional moisture needed = 18.3% - 16.5% = 1.8%

Step 2) Additional moisture per cu yd = 1.8% * 27 cu ft/cy * 62.4 lb/(ft^3*%) = 4.62 lbs/cy

Step 3) Additional moisture per truck (35,000 lbs capacity) = 35,000 lbs / 1,752 lbs/cy = 20 cy

Step 4) Additional moisture per truck = 20 cy * 4.62 lbs/cy = 93 lbs

Step 5) Convert 93 lbs to gallons (1 gal = 8.36 lbs): 93 lbs / 8.36 lbs/gal = 11.2 gallons

Therefore, 11.2 gallons of water is needed to add to each truck.

Let me know if you have any other questions!