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I am trying to find Poisson's Ratio, given the following information:

A cylindrical specimen of some metal alloy 52 mm in diameter is stressed elastically in tension. A force of 43,354 N produces a reduction in specimen diameter of 0.0031 mm. The elastic modulus of the material is 821 GPa.

I calculated the lateral strain as $0.0031\ \mathrm{mm}/52\ \mathrm{mm} = 5.96\times 10^{-5}$.

Using $\sigma = F/A$, I calculated the axial strain as the stress divided by the elastic modulus. Since $E = 821\ \mathrm{GPa}$ and the stress is $\sigma = 43,354\ \mathrm{N}/(\pi(\frac{52}{2}\ \mathrm{mm})^2)= 0.02041\ \mathrm{GPa}$, the axial strain is $\epsilon_a = 2.486 \times 10^{-5}$.

Dividing the lateral strain by the axial I get the answer $\nu = -2.397$, which seems too big (in module) to be right. I know that the algebra is correct, so the problem must be somewhere else. But where?

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    $\begingroup$ Using area reduction to compute longitudinal strain, since $-\Delta A/A \approx \Delta L/L$, bypasses calculation of stress and gives a Poisson ratio of $~0.5$. This is reasonable if volume is conserved. If you then use the area reduction based longitudinal strain to back calculate stress, you get $98.5\ \textrm{MPa}$ and a force of $210\ \textrm{kN}$. Since the results are inconsistent for different calculations of the same value, I'd conclude the original question has an incorrect value somewhere. Or I am missing something really simple. $\endgroup$ – wwarriner Dec 17 '15 at 18:08
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    $\begingroup$ @starrise I worked the problem with the given values and got 2.4e-3. I'm not familiar with the area reduction simplification but if the problem statement isn't based on realistic values, the assumptions that allow such simplifications may not hold. i.stack.imgur.com/vNn4U.png $\endgroup$ – Air Dec 17 '15 at 18:18
  • $\begingroup$ GPa is 10^9 N/m^2, or alternately 10^6 kN/m^2. Subbing that in changes your answer to 2.4 which is what I get (and except for negative sign), what OP got. $\endgroup$ – wwarriner Dec 17 '15 at 18:22
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    $\begingroup$ @starrise Perfect illustration for the OP of why you show all your work when you ask someone to check it. :) $\endgroup$ – Air Dec 17 '15 at 18:23
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    $\begingroup$ The conclusion I came to is that the original problem is nonsensical. Neither answer is "correct" because the problem is inconsistent. $\endgroup$ – wwarriner Dec 18 '15 at 0:00
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A couple of notable points:

  1. Metals rarely exceed a few hundred GPa Modulus - see the wikipedia page for several listed values. My hunch is this is a textbook error, and that a different modulus would give the correct answer. For example, 82.1 GPa would lead to an axial strain of 2.486×10−4, leading to a Poisson's ratio of 0.2397, which makes sense. The 82.1 GPa is in between Aluminum and Brass on the table - a realistic value for a metal.
  2. Some objects really do have a super high, or even negative modulus - but these are usually composite materials. The textbook lists a range between 0 - 1/2 as acceptable values for Poisson's ratios of engineering, non-composite materials (See Mechanics of Materials, Section 2.13). This makes 0.2397 a better answer.
  3. All scientific values should have a decimal point. A given value of 821 GPa does not list an accuracy. 821. (or 821.0) GPa states that the accuracy is +/- 0.5 GPa by having a decimal point. This is part of significant figures. This highlights that the 821 GPa is a textbook error on an additional level.
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  • $\begingroup$ Theoretically, modulus should fall in the range of -1 to 0.5. At 0.5 volume is conserved and bulk modulus is infinite. Hydrostatic pressure has no effect on a material with 0.5 volume. Materials with poisson ratio greater than 0.5 would have negative bulk modulus, and would expand in hydrostatic compression. At -1, shear modulus is infinite and the material would be perfectly rigid under shear stress. Below -1 the material has negative shear modulus and would strain in the opposite direction of stress. See this. $\endgroup$ – wwarriner Dec 21 '15 at 20:02
  • $\begingroup$ Interesting stuff. I listed my reference, though I obtained it in Sophomore Year of College. I figured more advanced stuff had come out, having had experience with a negative ratio between then and now, but it is nice to hear an explanation. $\endgroup$ – Mark Dec 21 '15 at 20:05
  • $\begingroup$ Agree on the first two points but not the third. 821 has three significant figures. The decimal point is only required if it has more than three, or for whole numbers ending with one or more zeroes where the zeroes are to be counted as significant figures. I'm sure some follow the convention you describe but the majority of my college texts did not - including the good ones. $\endgroup$ – Air Jan 21 '16 at 18:34
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The method you used can be modified a bit, by using True stresses and True strains instead of engineering stresses and strains. Poisson's ratio is based on conservation of volume of an object. Engineering stresses and strains do not conserve volume whereas True stresses and True strains conserve volume.

The relation between True stress and True strain is different from the engineering quantities. Using that relation may solve your problem.

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