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1. An aluminum alloy rod 1.5 m long required a force of 60,000 N to reach its yield strength of 280 MPa. E = 7.0 x 104 MPa. T

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Answer #1

1. a. Yield strength = Force /Area

280 N/mm2 = 60000/(22/28 d^2)

Diameter, d = 16.51 mm = 1.651 cm

2. Total strain = Total Stress/E

Total strain = 490/(7 x 10^4) = 0.007

3. Final length after load is released, Lf = Li + DL

DL = Li x Total Strain = 1.5m x 0.007 = 0.0105m

Lf= 1.5 + 0.0105 = 1.5105 m

4. Maximum elastic strain, MES = Elastic stress/ Youngs Modulus(E) = [60000/((3.142/4)*(16.51^2))]/ (7 x 10^4)

MES= 0.004

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