Question

The soil borrow material to be used to construct a highway embankment has a mass unit weight of 107.0 lb/cf and a water content of 7.3%, and the specific gravity of the soil solids is 2.62. The specifications require that the soils be place in the fill so that the dry unit weight is 113 lb/cf and the water content be held at 6%. How many cubic yards of borrow are required to construct an embankment have a 440,000-cy net section volume? How many gallons of water must be added per cubic yard of borrow material assuming no loss evaporation.466,103 bcy borrow-4.72 gal per bcy borrow534,953 bcy borrow-5.49 gal per bcy borrow517,116 bcy borrow-3.89 gal per bcy borrow498,594 bcy borrow-4.20 gal per bcy borrow

Answer 1

Option D

Step-by-step explanation:

Given information

Bulk unit weight of 107.0 lb/cf

Water content of 7.3%,=0.073

Specific gravity of the soil solids is 2.62

Specifications

Dry unit weight is 113 lb/cf

Water content is 6%.

Volume of embankment is 440,000-cy

Borrow material

`Dry_(unit,weight)=\frac {bulk_(unit,weight)}{1+water_(content)}=\frac {107}{1+0.073}= 99.72041 lb/cf`

Embankment

Considering that the volume of embankment is inversely proportional to the dry unit weight

`\frac {V_(embankment)}{V_(borrow)}=\frac {Dry_(borrow)}{Dry_(embankment)}`

Therefore,
`V_(borrow)=V_(embankment) *\frac {Dry_(embarkement)}{Dry_(borrow)}`

`V_(borrow)=440,000-cy*\frac {113 lb/cf }{99.72041 lb/cf }= 498594-cy`

Therefore, volume of borrow material is 498594-cy

(b)

The weight of water in embankment is found by multiplying the moisture content and dry unit weight.

Assuming that all the specifications are achieved, weight of water in embankment=0.06*113=6.78 lb/cf

Since
`1 yd^(3)= 27 ft^(3)`

The embankment requires water of 6.78*27*440000= 80546400 lb

Borrow materials’ water will also be 0.073*99.72041=7.27959 lb/cf

Borrow material requires water of 7.27959*27*498594=97998120 lb

Extra water between borrow material and embankment=97998120 lb-80546400 lb=17451720 lb

`Unit_(weight)=\frac {17451720}{498594}=35.00186 lb`

1 gallon is approximately
`8.35 yd^(3)` hence

`\frac {35.00186 lb/yd^(3)}{8.35}=4.19184 gallons/yd^(3)`

That's approximately 4.2 gallons