Answer:
The elongation in the bar equals 2.1 millimeters.
Explanation:
We know from Hooke's Law
[tex]\sigma =E\times \epsilon[/tex]
Where
[tex]\sigma [/tex] is stress in the material
[tex]\epsilon [/tex] is strain in the material
'E' is the young's modulus of the material
Now we know that
[tex]stress =\frac{Force}{Area}[/tex]
Applying values we get
[tex]\sigma =\frac{33000}{\frac{\pi }{4}\times (0.02)^{2}}=105.042\times 10^{6}kg/m^{2}[/tex]
Applying the values in the Hooke's relation we obtain
[tex]\epsilon =\frac{\sigma }{E}\\\\\epsilon =\frac{105.042\times 10^{6}}{2\times 10^{10}}=0.0052521[/tex]
Now by definition of strain we have
[tex]\epsilon =\frac{\Delta L}{L_{o}}\\\\\Delta L=0.0052521\times 0.4\times 10^{3}=2.1mm[/tex]
