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 10⁴ kg

m_f = mass of fuel = 35000 lbs = 35000 lbs (1 kg/2.21 lbs ) = 1.6 x 10⁴ kg

Take off weight is given as

W_t = (m_a + m_f ) g = ((5.4 x 10⁴) + (1.6 x 10⁴ )) (9.81) = 6.87 x 10⁵ 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 10⁴) + (1.6 x 10⁴ ) + (158.4)) (9.81)

W' = 6.8 x 10⁵ N

4.

E = 200 x 10⁹ Pa

stress = 120 x 10⁶ Pa

strain = ?

Using the equation

E = stress/strain

so strain = stress/E

strain = (120 x 10⁶)/( 200 x 10⁹) = 0.0006