Question

A uniform heavy rod of weight W, cross-sectional area A and length L is hanging from a fixed support. Young modulus of the material of the rod is Y. Neglect the lateral contraction. Find the elongation of the rod.

• A. $\frac{wL}{Ay}$
• B. $\frac{wL}{\left(2Ay\right)}$
• C. $\frac{2wL}{Ay}$
• D. $\frac{3wL}{2Ay}$

Answer 1

The correct option is B

$\frac{wL}{\left(2Ay\right)}$

Consider a small length dx of the rod at a distance x from the fixed end. The part below this small element has length L - x. The tension T of the rod at the element equals the weight of the rod below it.

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

Elongation in the element is given by

Elongation = original length x stress/Y
=$\frac{Tdx}{AY}=\frac{(L-x)Wdx}{LAY},$
The total elongation=${\int }_0^L\frac{(L-x)Wdx}{LAY}$
$\frac{W}{LAY}{(L_x-\frac{x^2}{2})}_0^L=\frac{WL}{2AY}$