Question

Johnson & Johnson has been awarded a contract to build a 4-seat outhouse for a rich man. The exterior dimensions will be 10 ft by 40 ft. The pit must be 10-ft deep so as not to fill up too quickly. The excavation has a bottom dimension of 20 ft by 50 ft and a top dimension of 60 ft by 90 ft with the cast-in-place concrete pit walls also being 10 ft by 40 ft. The borrow material has a dry unit weight of 97.0 pcf and a water content of 5%. The specifications call for the backfill to have a dry unit weight of 118 pcf at a water content of 7%. How may cubic yards of borrow material are required? How many gallons of water are required? The specific gravity of the soil solids is 2.67.

Answer 1

Hi,

Solution as follows:

1. Excavation Quantity:

Excavation dimensions provided in the question are as follows:

Bottom Size of pit = 20 ft x 50 ft

Top Size of pit = 60 ft x 90 ft

Depth of pit = 10 ft

Pit is in the shape of a Prismoid. Hence the volume of excavation shall be calculated by below formula:

Where,

h = height

A₁ and A₂ = Bottom and top areas

Hence volume of excavation,

Since 1 cubic yard = 27 cubic feet,

2. Borrow Quantity (CCY):

Earth shall be filled outside the walls. Hence the volume of fill can be calculated by excavation quantity less volume of walls (out-to-out)

Volume of Walls = 10 x 40 x 10 =4,000 cubic feet

Since 1 cubic yard = 27 cubic feet,

Volume of fill = 1077.01 – 148.15 = 928.86 cubic yards

Since the fill need to be compacted after fill; the quantity calculated above will be compacted cubic yards (CCY).

3. Swell factor:

Dry unit weight of soil required (Compacted Density) = 118 pcf

Dry unit weight of borrow soil (Bank density) = 97 pcf

4. Required Volume of earth:

Bank volume of Earth required (BCY) = CCY x Swell factor = 928.86 x 1.216 = 1129.49 bank cubic yards

5. Required Volume of Water:

Water content in borrow soil = 5%

Water content required in compacted soil = 7%

Additional water content required = 7% – 5% = 2%

Weight of solids = Dry density x Volume

Weight of solids in borrowed earth = 118 x 928.86 = 109,605.48 lbs

Unit weight of water = 8.34 lbs per gallon

Hence they will required 1129.49 BCY of borrow earth and 262.84 gallons of water for the job.

Hope the same is clear.

Keep Chegging

Best Regards

Chegg Expert Team