Answer:
strain = 0.0011575
Explanation:
Given:
- The diameter of specimen d = 0.02 m
- Applied uni-axial force P = 40,000 N
- Modulus of Elasticity E = 110 GPa
Find:
If the deformation is totally elastic, what is the strain experienced by the specimen?
Solution:
- First we need to calculate the stress in the specimen using the relation:
stress = Force / Area
stress = 40,000*4 / pi*d^2
stress = 160,000 / pi*0.02^2
stress = 127.323 MPa
- Now compute Elastic strain from the definition of Elastic modulus of a material:
E = stress / strain
strain = stress / E
- Plug in the values:
strain = 127.323 / 110*10^3
strain = 0.0011575