Question

1. An aircraft weights 120,000 lbs., it has 35,000 lbs. of fuel added. What will be its takeoff weight? If there is 350 lbs. of excess weight what will be its new weight?

Hint 1 kg = 2.21 lbs. Gravity = 9.81 N/m

(up to 5 points)

2. Why is the fracture value less than the UTS value shown on a stress-strain graph?

(up to 5 points)

3. Use the table below that gives the yield point for three materials. What applied load

would there be at yield for each when the area is a diameter of 20 mm? Yield is in MPa.

(up to 30 points)

Aluminum 240 (Yield MPa)

Steel 600 (Yield MPa)

Steel – hardened 950 (Yield MPa)

4. A material has an E value is 200 GPa, what is the strain, if the stress is 120MPa?

Hint E = stress/strain and think about the units.

(up to 10 points)

Answer #1

1.

m_{a} = mass of the aircraft = 120,000 lbs = 120,000 lbs
(1 kg/2.21 lbs) = 5.4 x 10^{4} kg

m_{f} = mass of fuel = 35000 lbs = 35000 lbs (1 kg/2.21
lbs ) = 1.6 x 10^{4} kg

Take off weight is given as

W_{t} = (m_{a} + m_{f} ) g = ((5.4 x
10^{4}) + (1.6 x 10^{4} )) (9.81) = 6.87 x
10^{5} N

m_{e} = excess mass = 350 lbs = 350 lbs (1 kg/2.21 lbs)
= 158.4 kg

New weight is given as

W' = (m_{a} + m_{f} + m_{e}) g = ((5.4 x
10^{4}) + (1.6 x 10^{4} ) + (158.4)) (9.81)

W' = 6.8 x 10^{5} N

4.

E = 200 x 10^{9} Pa

stress = 120 x 10^{6} Pa

strain = ?

Using the equation

E = stress/strain

so strain = stress/E

strain = (120 x 10^{6})/( 200 x 10^{9}) =
0.0006

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 2 minutes ago

asked 3 minutes ago

asked 4 minutes ago

asked 7 minutes ago

asked 7 minutes ago

asked 7 minutes ago

asked 8 minutes ago

asked 18 minutes ago

asked 24 minutes ago

asked 27 minutes ago

asked 37 minutes ago

asked 37 minutes ago