Question

A uniform bar of length of ${}^{\prime }{L}^{\prime }$ and cross sectional area of 'A' is subjected to a tensile load of ${}^{\prime }{F}^{\prime }{.}^{\prime }{Y}^{\prime }$ be the Young Modulus and ${}^{\prime }{\sigma }^{\prime }$ be the Poisson's ratio. Then volumetric strain is:

• A. $\frac{F}{AY}(1-\sigma )$
• B. $\frac{F}{AY}(2-\sigma )$
• C. $\frac{F}{AY}(1-2\sigma )$
• D. $\frac{F}{AY}\sigma$

Answer 1

Answer: (C)

$\frac{F}{AY}(1-2\sigma )$

Let increase in length be $\Delta l$

Let decrease in length be $\Delta r$

Final volume $=\pi (r-\Delta r{)}^2.(l+\Delta l)$

$=\pi (r^2-2r\Delta r+\Delta r^2)(l+\Delta l)$

$=\pi r^{2}l+\pi r^2\Delta l-2\pi r\Delta rl$ (The rest of the products of $\Delta rand$\Delta l$ can be neglected})

$\Delta v=$initial volume-final volume

$\Delta v=\pi r^2\Delta l-2\pi r\Delta rl$

Volumetric strain $=\frac{\Delta v}{v}$

$\frac{\Delta v}{v}=\frac{\pi r^2\Delta l}{\pi r^{2}l}-\frac{2\pi r\Delta rl}{\pi r^{2}l}$

$=\frac{\Delta v}{v}=\frac{\Delta l}{l}(1-\frac{2\Delta r}{r}\frac{l}{\Delta l})$

$\frac{\Delta v}{v}=\frac{F}{AY}(1-2\sigma )$