Answered

Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

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%

Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We hope this was helpful. Please come back whenever you need more information or answers to your queries. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.