A uniform heavy rod of weight W, cross-sectional area A and initial length I is suspended from a rigid support. f is the Young's modulus of the material of the wire. If no Iateral contraction is taken into account, the elongation of the rod will be

Let the whole length of the rod be divided into a large number of small length elements. Consider a small length dx of the rod at a   distance x from the fixed rod. The length of the rod below this part will be (L -x). The tension in the rod at the element will be equal to the weight of the rod below it, i.e.
,$\mathrm{T}=\left(\frac{\mathrm{W}}{\mathrm{L}}\right)(\mathrm{L}-\mathrm{X})$
Elongation in the element 
$=\frac{\text{ original length }\times \text{ stress }}{\text{ Young's modulus }}=\frac{\mathrm{Tdx}}{\mathrm{AY}}$
$=\left\{\frac{\mathrm{W}}{\mathrm{L}}(\mathrm{L}-\mathrm{x})\right\}\times \frac{\mathrm{dx}}{\mathrm{AY}}$
Total elongation $={\int }_0^{\mathrm{L}} \frac{\mathrm{W}(\mathrm{L}-\mathrm{x})\mathrm{dx}}{\mathrm{LAY}}$
$=\frac{\mathrm{W}}{\mathrm{LAY}}{\left[\mathrm{LX}-\frac{{\mathrm{X}}^2}{2}\right]}_0^{\mathrm{L}}=\frac{\mathrm{WL}}{2\mathrm{AY}}$