Answer to Question #102628 in Mechanics | Relativity for Airee

Question

Consider the state of stress in a bar subject to compression in the axial direction. Lateral expansion is restrained to half the amount it would ordinarily be if the lateral faces were load free. Find the effective modulus of elasticity.

Answer 1

As per the given question,

Let the initial length of the bar =L

Initial radius of the bar is r

A is the cross sectional area of the bar.

The load across the axis=F

After the compression, radius becomes R and length becomes L'

We know that Stress$(\sigma)=\dfrac{F}{A}$

Strain$(\epsilon)=\dfrac{\triangle l}{L}$

As per the hook's law, When we will apply the force axially on the bar, then it will compress.

stress$\propto$ strain

$\sigma \propto \epsilon$

So,

Now, to remove the proportionality, we will multiply by a constant.

$\sigma=Y\epsilon$

$Y=\dfrac{\sigma}{\epsilon}$

Now,

$Y=\dfrac{\dfrac{F}{A}}{\dfrac{\triangle l}{L}}=\dfrac{FL}{A\triangle l}$

Here Y represents the modulus of elasticity.