Question

A circular section of material is tested. The original specimen is 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. Determine the modulus of elasticity and the proportional limit.

Answer 1

Modulus of Elasticity = 100 GPa

Proportional limit = 0.15 GPa

Step-by-step 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:

`A = (\pi d^(2) )/(4) \\A = (\pi 0.013^(2) )/(4)\\A = 0.000133 m`

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

Answer 2

modulus of elasticity = 100.45 Gpa,

proportional limit = 150.68 N/mm^2.

Step-by-step 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.