Question

A uniform heavy rod of length L, weight W and cross – sectional area A is hanging from a fixed

support. Young’s modulus of the material is Y. Find the elongation of the rod.

• A. $\frac{WL}{AY}$
• B. $\frac{WL}{2AY}$
• C. $\frac{WL}{4AY}$
• D. $\frac{WL}{3AY}$

Answer 1

Answer: (B)

$\frac{WL}{2AY}$

Answer 2

weight acts at COM.

effective length = $\frac{L}{2}$. Hence $\frac{\Delta l}{L/2} = \frac{W}{AY}$

or ∆l = $\frac{WL}{2AY}$

Alternative method

T = (L - x)$\frac{W}{L}$; elongation = $\frac{Tdx}{AY}$

= $\frac{(L - x)Wdx}{LAY}$; Total elongation = $\frac{W}{LAY}\int_{0}^{L}{(L - x)dx}$

=$\frac{WL}{2AY}$.