1.

Derive an expression for elastic potential energy per unit volume stored for the wire is 1/2x stress x strain.

Answer»

Solution :Poisson.s ratio: The ratio fo lateral STRAIN to ongitudinal is known as Possion.s ratio.
In case of cylindrical rod if it has length l and radius r, after stretching, its radius DECREASES and length increases
`THEREFORE (Delta r)/( r ) =- (mu Delta l )/( l ) ""...(1)`
Volume `V = pi r ^(2) l ""...(2)`
FOr small change
`(Delta V)/(V) = 2 (Deltar )/(r) + (Delta l )/(l)`
From equation (1),
`(Delta V)/(V) = - 2mu (Delta l)/(l) + (Delta l)/(l) ""...(3)`
`therefore (Delta V)/(V) = (Delta l)/(l) (1- 2mu ) = epsi _(1) (1- 2mu ) ""...(4)`
Equation (4) shows that , for `Delta V gt 0` value of `mu` can not be increases from `0.5.`
This equation is true for any cross-section of rod.


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