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In-s [12.5K]
3 years ago
11

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 p

roportional limit, the specimen elongates by 0.3 mm. The total axial load is 20 kN. Determine the modulus of elasticity and the proportional limit.
Engineering
2 answers:
n200080 [17]3 years ago
7 0

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.

Anastaziya [24]3 years ago
3 0

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:

A = \frac{\pi d^{2} }{4} \\A = \frac{\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

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