Question

Why is the percent elongation in 2 inches greater than in 8 inches?

Answer 1

Elongation in a bar is

`\epsilon = (PL)/(\pi r^2 E)`

Where,

P = Applied force

L = Lengthof the specified rod

A = Cross-sectional area
`(\pi r^2)`

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.