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ollegr [7]
3 years ago
6

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
Engineering
1 answer:
MrRissso [65]3 years ago
7 0

Answer:

Option D

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

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A 1.5-m-long aluminum rod must not stretch more than 1 mm and the normal stress must not exceed 40 MPa when the rod is subjected
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We use the larger value of diameter = 9.77 mm

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