# Poisson's ratio

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?

• 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. – wwarriner Dec 17 '15 at 18:08
• @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 – Air Dec 17 '15 at 18:18
• 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. – wwarriner Dec 17 '15 at 18:22
• @starrise Perfect illustration for the OP of why you show all your work when you ask someone to check it. :) – Air Dec 17 '15 at 18:23
• The conclusion I came to is that the original problem is nonsensical. Neither answer is "correct" because the problem is inconsistent. – wwarriner Dec 18 '15 at 0:00