Question

The natural material in a borrow pit has a mass unit weight of 110.0 pcf and a water content of 6%, and the specific gravity of the soil solids is 2.63. The specifications require that the soil be compacted to a dry unit weight of 122 lb. per cf and the water content be held to 5.5%.

A. How many cubic yards of borrow are required to construct an embankment having a net section volume of 245,000 cy?
B. How many gallons of water must be added per cubic yard of borrow material assuming no loss by evaporation?

Answer 1

A. 288,030.91 cy

B. The amount of water that must be removed from the natural material is 483541.04254 gallons of water

Step-by-step explanation:

The natural material in the barrow properties are;

The mass unit weight, γ = 110.0 pcf

The water content, w = 6%

The specific gravity of the soil solids,
`G_s` = 2.63

The desired dry unit weight,
`\gamma _d` = 122 pcf

The water content, w₁ = 5.5 %

The net section volume,
`V_T` = 245,000 cy = 6,615,000 ft³

A.
`\gamma _d` =
`W_s`/
`V_T`

∴
`W_s` =
`V_T` ×
`\gamma _d` = 6,615,000 ft³ × 122 lb/ft³ = 807030000 lbs

w = (
`W_w`/
`W_s`) × 100

∴
`W_w` = (w/100) ×
`W_s` = (6/100) × 807030000 lbs = 48421800 lbs

The weight of solids

W =
`W_s` +
`W_w` = 807030000 lbs + 48421800 lbs = 855451800 lbs

V = W/γ = 855451800 lbs/(110.0 lb/ft.³) = 7776834.54545 ft.³ = 288,030.91 cy

V = 288,030.91 cy

The amount of cubic yards of borrow required = 288,030.91 cy

B. The volume of water in the required soil is found as follows;

`W_(w1)` = (w₁/100) ×
`W_s` = (5.5/100) × 807030000 lbs = 44386650 lbs

The amount of water that must be added =
`W_(w1)` -
`W_w` = 44386650 lbs - 48421800 lbs = -4,035,150 lbs

Therefore, 4,035,150 lbs of water must be removed

The density of water, ρ = 8.345 lbs/gal

Therefore, V = 4,035,150 lbs/(8.345 lbs/gal) = 483541.04254 gal of water must be removed from the natural material