Thursday, August 25, 2016

elasticity - Compressible material with Poisson ratio of 0.5


According to comments and an answer to this question, it is claimed that:



  1. Incompressible materials require the Poisson ratio to be 0.5.


  2. A poisson ratio of 0.5 does not imply that the material is incompressible.


It is thus implied that the statement "A material has a Poisson ratio of 0.5 if and only if the material is incompressible" is false, and that causality cannot be reversed. However, the maths seem reversible, and I have not been able to find any compressible material with a Poisson ratio of 0.5, whether that be in practice or theory.


Note: arguably only theory matters; the hypothesis is academic: no real material exists with a Poisson ratio of exactly 0.5.


Can above second statement be proven with theoretical evidence (e.g. by presenting a theoretical material with a Poisson ratio of exactly 0.5 that does change volume for small stress/strain)? Or, alternatively, can the second statement be disproven mathematically (e.g. by proving reversibility of causality)?



Answer



Idealized liquids have a Poisson's ratio of exactly 1/2 and are not incompressible (because no material is incompressible). No solid can have a Poisson's ratio of exactly 1/2.


To explain: If you apply a deviatoric (i.e., not hydrostatic; the 1-D version is shear) load on a liquid, it will deform without resistance and without a change in volume, implying that $\nu=1/2$. One way to see this is to use one of the elasticity relations (which assume a homogeneous isotropic linear elastic material): $G=\frac{3K(1-2\nu)}{2(1+\nu)}$. All stable materials have a positive bulk modulus $K$ (i.e., all stable materials compress to some degree under hydrostatic stress); thus, setting $\nu=1/2$ implies a shear modulus $G$ of zero, which corresponds to a fluid.


Let's look at a couple of the elasticity relations from the Poisson's ratio side: $\nu=\frac{3K-E}{6K}$, where $E$ is the Young's elastic modulus, and $\nu=\frac{3K-2G}{2(3K+G)}$. Thus, elastomers have a Poisson's ratio of nearly 1/2 because their shear and Young's elastic moduli are much smaller than their bulk moduli. When you shear or pull on rubber, for example, it's relatively easy to unkink and uncoil its long polymer chains to obtain shearing or uniaxial deformation. When you apply pressure from all sides, however, you're essentially trying to push C atoms closer to C atoms, which is not easy.


Note that this review (Greaves et al., "Poisson’s ratio and modern materials", Nat Materials 10 2011, DOI: 10.1038/NMAT3134) defines $\nu=\frac{3K-2G}{2(3K+G)}=0$ for gases for unexplained reasons, possibly because they are idealizing gases as exhibiting $K=0$.



No comments:

Post a Comment

classical mechanics - Moment of a force about a given axis (Torque) - Scalar or vectorial?

I am studying Statics and saw that: The moment of a force about a given axis (or Torque) is defined by the equation: $M_X = (\vec r \times \...