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
1 answer:
Answer:
A. 288,030.91 cy
B. The amount of water that must be removed from the natural material is 483541.04254 gallons of water
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, = 2.63
The desired dry unit weight, = 122 pcf
The water content, w₁ = 5.5 %
The net section volume, = 245,000 cy = 6,615,000 ft³
A. =
/
∴ =
×
= 6,615,000 ft³ × 122 lb/ft³ = 807030000 lbs
w = (/
) × 100
∴ = (w/100) ×
= (6/100) × 807030000 lbs = 48421800 lbs
The weight of solids
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;
= (w₁/100) ×
= (5.5/100) × 807030000 lbs = 44386650 lbs
The amount of water that must be added = -
= 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
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Answer:
The shear stress will be 80 MPa
Explanation:
Here we have;
τ = (T·r)/J
For rectangular tube, we have;
Average shear stress given as follows;
Where;
= 100 mm × 200 mm = 20000 mm² = 0.02 m²
t = Thickness of the shaft in question = 2 mm = 0.002 m
T = Applied torque
Therefore, 50 MPa = T/(2×0.002×0.02)
T = 50 MPa × 0.00008 m³ = 4000 N·m
Where the dimension is 50 mm × 250 mm, which is 0.05 m × 0.25 m
Therefore, = 0.05 m × 0.25 m = 0.0125 m².
Therefore, from the following average shear stress formula, we have;
Plugging in then values, gives;
The shear stress will be 80,000,000 Pa or 80 MPa.