Derive the formula for the elongation due to self weight. as $\delta 1=\frac{w L}{2 A E}$.

Mass density=$\frac{Mass}{Volume}$

Weight density=$\frac{weight}{volume}$

Now, $Wx=\gamma*volume$

$Wx=\gamma*A*x$

where, $A=\frac{\pi*d^2}{4}$

$d(\delta l) =\frac{Wx.dx}{AE}$

$\delta l=\int \frac{Wx.dx}{AE}$

Now, integrating from O to L

$\delta l=\int_{0}^{L} \gamma *A*x {dx}{AE}$

$\delta l=\frac{\gamma}{E}. \frac{x^2}{2}_{0}^{L}$

$\delta l=\frac{\gamma L^2}{2E}$

Hence, elongation due to self weight is half of the elongation caused due to external load equal to weight of bar applied at the end.

$W={\gamma AL}$

$\gamma=\frac{W}{AL}$

$\delta l=\frac{W}{AL}* \frac{L^2}{2E}$

$\delta l=\frac{WL}{2AE}$