Question

show tht the strain energy per unit volume =1/2*(stress)²/y when a wire is stretched​

Answer 1

To Show:

$\sf{Strain\:Energy\:per\:unit\:volume=\dfrac{1}{2}\dfrac{(stress)^{2}}{Y}}$

where Y is Young's Modulus

Proof:

Work done in stretching a wire by a distance x on applying force F is

$\longrightarrow\mathrm{W=\dfrac{1}{2}\times F\times x}$

On multiplying and dividing lA in R.H.S.,we get

$\longrightarrow\mathrm{W=\dfrac{1}{2}\times F\times x\times \dfrac{lA}{lA}}$

where

l is length of wire

A is area of cross-section of wire

Now,

$\longrightarrow\mathrm{W=\dfrac{1}{2}\times \dfrac{F}{A}\times \dfrac{x}{l} \times lA}$

We know that,

$\longrightarrow\bf{Stress=\dfrac{Force}{Area}=\dfrac{F}{A}}$

$\longrightarrow\bf{Strain=\dfrac{Change\:in\:length}{Actual\:length}=\dfrac{x}{l}}$

$\longrightarrow\bf{Volume=Length\times Area=lA}$

So,

$\longrightarrow\mathrm{W=\dfrac{1}{2}\times stress\times strain\times volume}$

Now,

Energy stored in unit volume(V) of a stretched wire is

$\longrightarrow\mathrm{U=\dfrac{W}{V} =\dfrac{1}{2}\times\dfrac{stress \times strain\times\cancel{volume}}{\cancel{volume}}}$

$\longrightarrow\mathrm{U=\dfrac{1}{2}\times stress \times strain}-----(1)$

Now again, we know that

$\longrightarrow\bf{Y=\dfrac{longitudinal\:stress}{longitudinal\:strain}}$

where Y is Young's Modulus of elasticity

So, on multiplying and dividing stress in R.H.S of (1), we get

$\longrightarrow\mathrm{U=\dfrac{1}{2}\times stress \times strain\times\dfrac{stress}{stress}}$

$\longrightarrow\mathrm{U=\dfrac{1}{2}\times (stress)^{2}\times\dfrac{1}{\dfrac{stress}{strain}}}$

$\longrightarrow\mathrm{U=\dfrac{1}{2}\times (stress)^{2}\times\dfrac{1}{Y}}$

$\longrightarrow\mathrm{\pink{U=\dfrac{1}{2}\times\dfrac{ (stress)^{2}}{Y}}}$

Hence Proved