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.

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:

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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