Poisson's ratio and change in volume [duplicate]

The Poisson's ratio $\nu$ is always less than $0.5$.

A traction force ($\Delta L >0$) can cause an increase in volume, while a compression force ($\Delta L <0$) can only decrease the volume.

$$\Delta V \approx V \frac{\Delta L}{L}(1-2\nu)$$

I'm totally ok with these facts but I don't understand which one follows the other.

Is $\nu<0.5$ a consequence of an empirical observation that $\Delta V<0$ with traction, or viceversa?

• May 9, 2016 at 20:23
• Rubber has Poisson coefficient equal to 0.5 (as all incompressible materials).
– user130529
Oct 16, 2016 at 21:30
• I answered this a while back for linear, isotropic materials. The answer requires a bit of algebra to get to, but shows very clearly why the Poisson ratio is bounded between -1 and 0.5. physics.stackexchange.com/q/99077 Oct 16, 2016 at 21:56
• Possible duplicate of Limits of Poisson's ratio in isotropic solid Apr 16, 2017 at 15:40

Yes, $ν$ being always less than 0.5 would be a consequence of $ΔV$ always being negative with traction.