Answer:
The elastic deformation is 0.00131.
Explanation:
Given that,
Force F = 44500 N
Cross section [tex]A =15.2mm\times19.1\ mm[/tex]
We Calculate the stress
Using formula of stress
[tex]\sigma=\dfrac{F}{A}[/tex]
Where, F = force
A = area of cross section
Put the value into the formula
[tex]\sigma=\dfrac{44500}{15.2\times10^{-3}\times19.1\times10^{-3}}[/tex]
[tex]\sigma=153.27\times10^{6}\ N/m^2[/tex]
We need to calculate the strain
Using formula of strain
[tex]Y=\dfrac{\sigma}{\epsilon}[/tex]
[tex]epsilon=\dfrac{\sigma}{Y}[/tex]
Where,
[tex]\sigma[/tex]=stress
Y = young modulus of copper
Put the value into the formula
[tex]\epsilon=\dfrac{153.27\times10^{6}}{117\times10^{9}}[/tex]
[tex]\epsilon =0.00131[/tex]
Hence, The elastic deformation is 0.00131.
