When a bar of length I and diameter d is rigidly fixed at the upper end and hanging freely, then the total elongation produced in the bar due to its own weight is

where w = weight per unit volume of the bar.

Concept:

When a solid metal bar is hanged vertically from one of its ends, it tends to

experience elongation due to the weight and deformation that occurs.

The elongation of the bar of weight ‘W’, length ‘L’, diameter ‘D’, and

elasticity ‘E’ due to its self-weight is = \(\frac{{2WL}}{{{D^2}E}}\)

Let’s take a look into the case of a metal bar of weight ‘W’, diameter ‘D’

and length ‘L’ hanging vertically from one of its ends.

The elastic property represented by the Modulus of Elasticity of the metal bar is ‘E’.

Consider an element of length ‘dy’ at a distance ‘y’ from the bottom of the bar being elongated due to the force ‘P’, at section x - x, as shown in the figure.

Weight of the portion below x - x = \(w\) × A × y

Where \(w\) = weight of the bar per unit volume = \(\frac{W}{{AL}}\)

A = area of cross-section of the bar

Now, the weight of the portion P = \(\frac{W}{{AL}} \times \left( {A \times y} \right)\) = \(\frac{{W\; \times \;y}}{L}\)

Change in length of the element ‘dy’ 

= \(\frac{P}{{AE}}\)

= \(\frac{{Wy}}{L} \times \frac{1}{{AE}}\)

Total elongation of the bar = \(\mathop \smallint \limits_0^L \frac{{Wy}}{L} \cdot \frac{1}{{AE}} \cdot dy\)

= \(\frac{{WL}}{{2AE}} = \frac{w AL~\times~L }{2AE}=\frac{w L^2}{2E}\)