Question

A rod of uniform cross-sectional area A and length L has a weight W. It is suspended vertically from a fixed support. If the material of the rod is homogeneous and its modulus of elasticity is I. then determine the total elongation produced in the rod due to its own weight.

Answer 1

In this case, different parts of the rod does not elongate to the same extent. The element closer to the support elongates more as the stress is higher with respect to the elements closer to the free end of the rod. To determine total elongation of the rod, let us consider a small element of differential length dx at a distance x from the free end of the rod, as shown in the Fig. 5.19.

The stress at the position of this element is produced by the weight of the rod of length x lying below it. i.e., (W/L) x.

Therefore, stress at this section

$\sigma =\frac{\left(W/L\right)x}{A}=\frac{Wx}{AL}$

The elongation $d\delta$ produced in the differential element $dx$ is $d\delta =\frac{\sigma }{Y}dx=\frac{W}{YAL}xdx$

Thus, total elongation produced in the rod can be calculated as

$\delta =\int d\delta$

$\delta =\frac{W}{YAL}{\int }_0^{L}xdx=\frac{W}{YAL}{\left[\frac{x^2}{2}\right]}_0^L$

$or\sigma =\frac{\left(W/2\right)L}{YA}$