A uniform heavy rod of weight W, cross sectional area A and length L is hung from a fixed support. Young’s modulus of the material of the rod is Y. If lateral contraction is neglected, the elongation of the rod under its own weight is

The correct option is WL/2AY.

To find the elongation of the rod under its own weight, we need to consider that the tension (or stretching force) is not uniform throughout the length of the rod. Instead, the tension varies from zero at the bottom free end to a maximum value equal to the total weight of the rod at the top support.

Let us consider a small element of the rod of length dx at a distance x from the lower (free) end of the rod.

The weight of the portion of the rod below this element acts as the tension T(x) at this position. Since the rod is uniform, the weight per unit length is W/L. Therefore, the tension T(x) is given by:

$T\left(x\right)=\frac{W}{L}x$

According to Hooke's Law, the elongation dl of this small element of length dx is:

$dl=\frac{T\left(x\right)dx}{AY}$

Substituting the expression for tension T(x) into the equation:

$dl=\frac{Wxdx}{AYL}$

To find the total elongation of the rod, we integrate dl over the entire length of the rod from x = 0 to x = L:

$\Delta L={\int }_0^L\frac{Wx}{AYL}dx$

Taking the constant terms out of the integral:

$\Delta L=\frac{W}{AYL}{\int }_0^{L}xdx$

Performing the integration:

$\Delta L=\frac{W}{AYL}{[\frac{x^2}{2}}_0^L$

Applying the limits:

$\Delta L=\frac{W}{AYL}·\frac{L^2}{2}$

Simplifying the expression gives the final total elongation:

$\Delta L=\frac{WL}{2AY}$

Thus, the elongation of the rod under its own weight is WL/2AY.