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.
Answer:
Modulus of Elasticity = 100 GPa
Proportional limit = 0.15 GPa
Explanation:
Axial Load = 20 kN = 20000 N
Original length, L₀ = 200 mm = 0.2 m
diameter, d = 13 mm = 0.013 m
Elongation, ΔL = 0.3 mm = 0.0003 m
Area of the material:
[tex]A = \frac{\pi d^{2} }{4} \\A = \frac{\pi 0.013^{2} }{4}\\A = 0.000133 m[/tex]
Stress = Load / Area
Stress = 20000 / 0.000133
Stress = 150375940 N/m²
Stress = Proportional limit = 0.15 GPa
Modulus of Elasticity = Stress/Strain
Strain = ΔL / L₀
Strain = 0.0003 / 0.2
Strain = 0.0015
Modulus of Elasticity = 0.15 / 0.0015
Modulus of Elasticity = 100 GPa
