# Elongation of Rod under its own weight

Consider a rod having mass M , Length L and Area of cross section A .

Weight of rod = M g

Now consider an element at a distance x from the bottom of rod , tension at a distance x from bottom is

$\displaystyle T = \frac{M g}{L}x$

Elongation in element ‘ dx ‘ is

$\displaystyle = \frac{T . dx}{A Y }$ ; Where Y = Young’s modulus of elasticity .

Total Elongation in the rod is

$\displaystyle = \int_{0}^{L}\frac{T dx}{A Y}$

$\displaystyle = \int_{0}^{L}\frac{M g x dx}{L A Y}$

$\displaystyle = \frac{M g}{L A Y} \int_{0}^{L} x dx$

$\displaystyle = \frac{M g}{L A Y} [\frac{x^2}{2}]_{0}^{L}$

$\displaystyle = \frac{M g}{L A Y} (\frac{L^2}{2} )$

$\displaystyle = \frac{M g L}{2 A Y}$

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