A cross beam in a highway bridge experiences a stress of 14 ksi due to the dead weight of the bridge structure. When a fully loa
1 answer:
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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
Answer:
critical clearing angle = 70.3°
Explanation:
Generator operating at = 50 Hz
power delivered = 1 pu
power transferable when there is a fault = 0.5 pu
power transferable before there is a fault = 2.0 pu
power transferable after fault clearance = 1.5 pu
using equal area criterion to determine the critical clearing angle
Attached is the power angle curve diagram and the remaining part of the solution.
The power angle curve is given as
= Pmax sinβ
therefore : 2sinβo = Pm
2sinβo = 1
sinβo = 0.5 pu
βo = ⁰
also ; 1.5sinβ1 = 1
sinβ1 = 1/1.5
β1 = = 41.81⁰
∴ βmax = 180 - 41.81 = 138.19⁰
attached is the remaining solution
The critical clearing angle = ≈ 70.3⁰
