Question

relationship between Strain energy and shear stress ​

Answer 1

ANSWER :-

Consider a wire of length L and area of cross-section A subjected to a deforming force F along the wire. Let l be the increase in the length of the wire. Hence , change in area of cross-section is neglected.

Young's modulus of the wire is

$Y = \frac{FL}{Al} \\ \\ F = \frac{YAl}{L} - - - -( 1)$

The work done in increasing the wire by 'dl' is 'dW'

dW = F.dl = F × dl × cos0°

dW = Fdl--------(2)

From equation (1) and (2)

$dW = \frac{AYldl}{L}$

The amount of work done in increasing the length of wire from 0 to l is W

W = $\displaystyle \int\limits^l_0{dW} \dl$

W = $\displaystyle \int\limits^l_0{\frac{YAl}{L} }\dl$

W = $\frac{YAl}{L}$$\displaystyle \int\limits^l_0 {l }\dl$

W = $\frac{YA}{L}$ $\frac{l^2}{2}$

W = $\frac{YAl^2}{2L}$

W = $\frac{1}{2}$Fl

W = $\frac{1}{2}$$\frac{F}{A}$$\frac{e}{L}$AL

W = Stress energy = U

U = $\frac{1}{2}$ Stress × Strain × Volume