Answer:
modulus of elasticity = 100.45 Gpa,
proportional limit = 150.68 N/mm^2.
Explanation:
We are given the following parameters or data in the question as;
=> "The original specimen = 200 mm long and has a diameter of 13 mm."
=> "When loaded to its proportional limit, the specimen elongates by 0.3 mm."
=> " The total axial load is 20 kN"
Step one: Calculate the area
Area = π/ 4 × c^2.
Area = π/ 4 × 13^2 = 132.73 mm^2.
Step two: determine the stress induced.
stress induced = load/ area= 20 × 1000/132.73 = 150.68 N/mm^2.
Step three: determine the strain rate:
The strain rate = change in length/original length = 0.3/ 200 = 0.0015.
Step four: determine the modulus of elasticity.
modulus of elasticity = stress/strain = 150.68/0.0015 = 100453.33 N/mm^2 = 100.45 Gpa.
Step five: determine the proportional limit.
proportional limit = 20 × 1000/132.73 = 150.68 N/mm^2.
