Elongation in a bar is
[tex]\epsilon = \frac{PL}{\pi r^2 E}[/tex]
Where,
P = Applied force
L = Lengthof the specified rod
A = Cross-sectional area [tex](\pi r^2)[/tex]
E = Modulus of Elasticity
Performing a quick analysis we can realize that the larger the radius, the lower the elongation percentage. The radius is inversely proportional to the percentage of elongation. For this reason in a 2in bar the change will be GREATER than that of an 8in bar.
