Respuesta :
Answer:
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
