If the specimen is loaded until it is stressed to 65 ksi, determine the approximate amount of elastic recovery after it is unloaded. Express your answer as a length. Express your answer to three significant figures and include the appropriate units.

ER = 0.008273 in

Explanation:

Given:

- Length of the specimen L = 2 in

- The diameter of specimen D = 0.5 in

- Specimen is loaded until it is stressed = 65 ksi

Find:

- Determine the approximate amount of elastic recovery after it is unloaded.

Solution:

- From diagram we can see the linear part of the curve we can determine the Elastic Modulus E as follows:

E = stress / strain

E = 44 / 0.0028

E = 15714.28 ksi

- Compute the Elastic strain for the loading condition:

strain = loaded stress / E

strain = 65 / 15714.28

strain = 0.0041364

- Compute elastic recovery:

ER = strain*L

ER = 0.0041364*2

ER = 0.008273 in