question_answer

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

A) \[\frac{WL}{2AY}\]                         

B) \[\frac{2WL}{AY}\]

C) \[\frac{WL}{AY}\]                                            

D) \[\frac{WL}{4AY}\]

Correct Answer: A

Solution :

Consider a small length \[dx\] of the rod at 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 \[\times \] stress/Y The total elongation\[-\int\limits_{0}^{L}{\frac{(L-x)W\,dx}{LAY}=}\]\[\frac{W}{LAY}\left( Lx-\frac{x}{2} \right)_{0}^{L}=\frac{WL}{2AY}\]