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A material has Poisson's ratio 0.5. If a uniform rod of it suffers a longitudinal strain of $2\times 10^{-3}$. What is the percentage increase in volume.

Solution

$V=\pi r^2L\implies\Delta V=\Delta(\pi r^2L)=\pi r^2(\Delta L)+2\pi rL(\Delta r)$

. . .

Can you tell me what happened?

Note- $L$ is increasing by $\Delta L$and $r$ is increasing by $\Delta r$ (which is negative).

Community
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The Cryptic Cat
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  • One thing I want to point out is that I am pretty sure that should not be equality -- this is an approximation based on the fact that the change in volume is assumed to be very small. – Rellek Aug 14 '16 at 02:47
  • what does " it suffers a longitudinal strain of $2\times 10^{−3}$" mean? How is it relatede to$\Delta L$ ? – miracle173 Aug 14 '16 at 20:11

2 Answers2

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$$V_{\text{new}}=V +\Delta V \\= \pi\, \left(r+\Delta r \right)^2 \, \left(L+\Delta L \right)\\= \pi\,\left(\Delta r \right)^2\, \Delta L+2\,\pi\,r\, \Delta r \, \Delta L+\pi\,r^2\, \Delta L+ \pi\,\left(\Delta r \right)^2\,L+2\,\pi\,r\, \Delta r \,L+ \pi\,r^2\,L \\ \approx \pi\,r^2\,L+2\,\pi\,r\,\Delta r \,L+ \pi\, r ^2\,\Delta L$$

The last approximation holds if we can ignore terms containing poducts of $\Delta r$ and $\Delta L$ of degree 2 or higher. $$\Delta V= V_{\text{new}}-V \approx 2\,\pi\,r\,\Delta r \,L+ \pi\,r ^2\,\Delta L$$

miracle173
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If you replace that $\Delta$ with $d$ (the one we use for differentation), you will understand that the result is simply the application of product rule of differentiation

Amritansh Singhal
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