Answer:
hello your question has some missing information attached to the answer is the missing component
Answer : αaxial,p = -6.034 ksi ( compressive )
αbend,p = 19.648 ksi ( tensile )
Explanation:
αaxial, p = equation 1
αbend, p = equation 2
P = load = 35 kips
A = area of column = 5.8
d = column cross section depth = 9.5 in
= 55.0
Hence equation 1 becomes
αaxial,p = -35 / 5.8 = - 6.034 ksi ( compressive )
equation 2 becomes
αbend, p = = + 19.648 ksi ( tensile )
Answer: power required to maintain level flight=82.20hp
Explanation:
Given
Area = 200 ft^2
Weight = 2000 lb
Cl( Lift coefficient)= 0.39
Cd( Drag coefficient) = 0.06
The density ρ of air at standard atmospheric pressure = 2.38 X 10^-3 slugs/ft^3
For Equilibrium to be maintained during flight conditions, the lift force must be balanced by the weight of the aircraft such that
Lift force = Weight of aircraft
(1/2)ρAU²Cl= W
1/2X 2.38 X 10^-3 X 200 X U² X 0.39 = 2000
U²= 2000 X 2 / 2.38 X 10^-3 X 200 X 0.39
U=
Velocity, U= 146.7892ft/s
Drag force of the velocity can be deduced from the formulae
Cd= Drag force(D) /1/2 ρU²A
Drag force=1/2 ρU²ACd
D=1/2 x (2.38 X 10^-3 slugs/ft^3) x (146.7892ft/s)² x 200 ft^2 x 0.06
D=307.69
Drag force= 308lb
power required to maintain level flight is given as
P = Drag force x Velocity = D x U
=308lb X 146.7892ft/s
=45,211.0736lb.ft/s
Changing to hp we have that
1 Horsepower, hp = 550 ft lbf/s
??=45,211.0736lb.ft/s
45,211.0736lb.ft/s/ 550 ft lbf/s= 82.20hp
Answer:
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