# Range of poissons ratio [duplicate]

I know the range of poisson's ratio is -1 to 0.5 but how do you arrive at this expression? I am a 11th grade student and I am not too familiar with advanced physics

• It's an experimentally determined range. Apr 23, 2014 at 15:56
• Kyle, this is incorrect. See the link in my answer. Apr 23, 2014 at 18:23

The answer is a bit lengthy, but can be arrived at using arguments about elastic strain energy. Here is a very detailed explanation:

Limits of Poisson's ratio in isotropic solid

This was written at a graduate mechanical engineering level, so I'll simplify it here.

Imagine that there exists a function $\psi$ that describes how much energy is contained in a solid per unit volume. This quantity is a function of material properties and deformation. For a linear elastic, isotropic solid, the material properties are Young's modulus (E), and the Poisson ratio ($\nu$).

One of the assumptions of the theory of elasticity is that the elastic energy $\psi$ is a function that is strictly increasing for all conceivable deformations. The details of this assumption are in my other answer (the link), but it turns out that $\nu$ can only be in the interval

$$-1 < \nu < \frac{1}{2}$$

I hope this clears up your question at an appropriate level. Let me know in the comments if not!

• I wonder if there are real life examples of matrieals with negative $\nu$. Great answer BTW. Apr 23, 2014 at 21:13
• Thanks! What you're thinking of are called "auxetic" materials. This property tends to correspond to highly (micro)structured materials. One classical example is a non-convex hexagon. Here's a link: en.wikipedia.org/wiki/Auxetics Apr 24, 2014 at 15:26
• There are quite a few auxetic metamaterials known to us. Monolayer graphene for example at finite temperature exhibits a universal negative poisson ratio of about -1/3 (so do all thermally fluctuating tethered membranes). One can easily construct an auxetic structure even out of paper. Look up Miura-Ori. Sep 30, 2014 at 2:04

Poisson's ratio can be expressed in terms of Young's modulus $E$, Bulk modulus $K$ and Shear modulus $G$. I hope you know how do arrive at these relationships;

$$G=\frac{E}{2(1+\nu)}$$

$$K=\frac{E}{3(1-2\nu)}$$

$E, K$ and $G$ and all NOT less than zero.

From the equation of $G$ you can see that it's the value of $v=-1$ which would make the denominator $0$ and $G$ very large. That's the limit for $G$.

And from the equation of $K$ it's the value of $v=0.5$ that makes the denominator $0$ and $K$ very large. The limit for $K$.

Therefore $v$ can only fall between $-1$ and $0.5$.