Answered

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Consider a mild steel specimen with yield strength of 43.5 ksi and Young's modulus of 29,000 ksi. It is stretched up to a point where the strain in the specimen is 0.2% (or 0.002). If the specimen is unloaded (i.e. load reduces to zero), the residual strain (or permanent set) is: 0.05% 0.1% 0% 0.2%

Sagot :

Answer:

0.05%

Explanation:

From the question, we have;

The yield strength of the mild steel, [tex]\sigma _c[/tex] = 43.5 ksi

Young's modulus of elasticity, ∈ = 29,000 ksi

The total strain, [tex]\epsilon _c[/tex] = 0.2% = 0.002

The inelatic strain [tex]\epsilon_c^{in}[/tex] is given as follows;

[tex]\epsilon_c^{in}[/tex] = [tex]\epsilon _c[/tex] - [tex]\sigma _c[/tex]/∈

Therefore, we have;

[tex]\epsilon_c^{in}[/tex] = 0.002 - 43.5/(29,000) = 0.0005

Therefore, the inelastic strain, [tex]\epsilon_c^{in}[/tex] = 0.0005 = 0.05%

Taking the inelastic strain as the residual strain, we have;

The residual strain = 0.05%

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