Yards of Dirt

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Yards of Dirt

Postby diannek » Fri Jul 28, 2006 4:46 am

How would i figure how many yards of dirt would be needed to build a 12 foot wide road, 45 feet long,eight feet high with 3 to 1 slopes? How wide would the base be?
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Postby Dirtman » Fri Jul 28, 2006 1:04 pm

Hi Dianne:

A 3:1 slope means simply that for every foot of vertical rise, the slope will extend three feet laterally (horizontally) from the base of the rise. Since you have a proposed elevation differential of eight feet, the base will extend 24 feet (8*3) on each side.

A 12 feet wide road bed with the base extending 24 feet on each side is 60 feet wide at the bottom, IF the sub-grade of the proposed 8' high road bed is on level ground to begin with. Typically it isn't or you wouldn't be needing to build a road.

To calculate the cubic yards, use the average end area method to determine the square footage of each end.

The road cross section is an isosceles trapezoid so add the width of the top of the road and the base together. Divide the answer by 2 and multiply by 8 (the height of the road cross-section). ((12+60)/2)*8 = (72/2)*8 = 36*8 = 288 square feet. Normally, you calculate each end and divide by 2 to get an average but in this case, each end is the same so we’ll save the averaging.

The length of the road is 45 feet so multiply 288 square feet (from above) x 45, which equals 12,960 cubic feet. Divide by 27 to get COMPACTED cubic yards. 12,960/27 = 480 CCY (compacted cubic yards).

If you have the material imported, bear in mind that material loaded in a dump truck is loose, not compacted. Material loaded in a dump truck is called loose cubic yards. Depending on the material, you’ll need to add 12% to 50% to the loose volume to obtain enough material.

If you don’t understand, write back and I’ll explain it more clearly. To save yourself a lot of aggravation, money and possible disaster, write back if you’re not familiar with this type of work.

George
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