Saved Bookmarks
| 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. |
|