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Valentin [98]
3 years ago
15

For the same cross-sectional area, which column provides the higher buckling load: a circular bar or a circular tube?

Engineering
1 answer:
juin [17]3 years ago
3 0

Answer:

Circular tube

Explanation:

Now for better understanding lets take an example

Lets take

Diameter of solid bar= 4\sqrt{2} cm

Outer diameter of tube =6 cm

Inner diameter of tube=2 cm

So from we can say that both tubes have equal cross sectional area.

We know that buckling load is given as P = \dfrac{\pi ^2EI}{L_e^2}      

If area moment of inertia(I) is high then buckling load will be high.

We know that  area moment of inertia(I)

For circular tube I = \dfrac{\pi }{64}(D_o^4-D_i^4)

For circular bar I = \dfrac{\pi }{64}D^4  

Now by putting the values

    For circular tube I=62.83 cm^4

  For circular bar I=50.26 cm^4

So we can say that for same cross sectional area the  area moment of inertia(I) is high for tube as compare to bar.So buckling load  will be higher in tube as compare to bar.

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A light aircraft with a wing area of 200 ft^2 and a weight of 2000 lb has a lift coefficient of 0.39 and a drag coefficient of 0
Gnoma [55]

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=\sqrt{21,547.08}

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

6 0
3 years ago
A cylindrical metal specimen having an original diameter of 12.8 mm and gauge length of 50.80 mm is pulled in tension until frac
Sedaia [141]

Answer:

%Reduction in area = 73.41%

%Reduction in elongation = 42.20%

Explanation:

Given

Original diameter = 12.8 mm

Gauge length = 50.80mm

Diameter at the point of fracture = 6.60 mm (0.260 in.)

Fractured gauge length = 72.14 mm.

%Reduction in Area is given as:

((do/2)² - (d1/2)²)/(do/2)²

Calculating percent reduction in area

do = 12.8mm, d1 = 6.6mm

So,

%RA = ((12.8/2)² - 6.6/2)²)/(12.8/2)²

%RA = 0.734130859375

%RA = 73.41%

Calculating percent reduction in elongation

%Reduction in elongation is given as:

((do) - (d1))/(d1)

do = 72.14mm, d1 = 50.80mm

So,

%RA = ((72.24) - (50.80))/(50.80)

%RA = 0.422047244094488

%RA = 42.20%

3 0
3 years ago
Why is it important to keep hand tools free of oil and grease?
Bezzdna [24]

Answer:

keeps it from rusting

Explanation:

5 0
3 years ago
Whats the best used for cable -stayed bridge
nalin [4]

Answer:

a cable -stayed bridge has, one or more towers,from which cable support the bridge deck.

7 0
3 years ago
A rectangular workpiece has the following original dimensions: 2a=100mm, h=25mm, and width=20mm. The metal has a strengh coeffic
Elena-2011 [213]

Answer:

See attachment for detailed answer.

Explanation:

Download pdf
4 0
3 years ago
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