Answered

IDNLearn.com makes it easy to find the right answers to your questions. Our experts are ready to provide in-depth answers and practical solutions to any questions you may have.

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%

We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Your questions deserve precise answers. Thank you for visiting IDNLearn.com, and see you again soon for more helpful information.