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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