Understanding incompressibility (of rubber or viscoelastic material) Literature gives a lot of explanation why rubber is incompressible. However, I still need some thinking to understand physical behavior of rubber or any such material. 
Often, incompressibility is tied to Poison's ratio ($\nu$) -> 0.5. At Poisson's ratio ->0.5, ratio of bulk modulus ($K$) to shear modulus ($G$), which can be given below, tends to infinity.
$$
 \lim_{\nu \to 0.5} \frac{K}{G}=\lim_{\nu \to 0.5}\frac{2(1+\nu)}{3(1-2\nu)} \rightarrow \infty
$$
However, I feel this is only math. I think following:
If we take water in cylinder and compress it with a piston (assuming no gap to 'leak' water), will water be compressed? I read that water is incompressible. Now, if I take rubber instead, will rubber be compressed? Or with the magical property of incompressibility, rubber will become so stiff that even if I apply tons of force to piston, rubber in the cylinder will not change its volume? 
I would appreciate insights into this. Thanks.
 A: Nothing is incompressible, but most liquids and solids have a very low compressibility i.e. a very high bulk modulus.
The reason for this is that in liquids and solids the atoms/molecules are in contact with each other. To squeeze them closer together you need to deform the bonds in molecules and/or the electron distribution around atoms. Both processes take a lot of energy so the force required is high.
I'm using a rather vague definition of in contact here. Atoms and molecules don't have sharp edges - they are fuzzy objects where the electron density falls off continuously with distance. However there will be an equilibrium distance where the attraction due to Van der Waals or dipolar forces is balanced out by the repulsion due to overlap of the electron clouds. It's trying to push atoms closer together than this equilibrium distance that takes a lot of energy and therefore requires a lot of force.
A: From http://imechanica.org/node/10589 - makes sense to me. 
by Zhigang Suo on Tue, 2011-07-19 07:02.
A rubber is a network of polymer chains.  Each polymer chain consists of many monomers.  The polymer chains are crosslinked by covalent bonds.  The covalent bonds give the solid-like behavior of the rubber.  If these crosslinks are removed, the rubber becomes a polymer melt, and is a liquid. 
Thus, a rubber is very similar to a liquid at the level of monomers.  Like a liquid, the polymers are densely packed, so that the rubber is difficult to change volume.  Also like a liquid, the polymers can move relative to one another, so that the rubber is easy to change shape.
Given this molecular picture, it is clear that the rubber is much easier to change shape than change volume.  The shear modulus is much smaller than the bulk modulus.  In modeling, we often neglect the change in volume, and focus on change in shape.  That is, we assume that the rubber is incompressible. 
This idealization of incompressibility is not always suitable.  For example, an incompressible material will not support longitudinal wave.  But we know rubber can support longitudinal wave.  The speed of the longitudinal wave is much larger than the speed of the shear wave. 
A: One does not have to go into fine details to answer to your question.
Sure when pressures are very high nothing is fully incompressible. Neither water nor rubber.
If you put water in a piston a reasonable pressure will give a negligible compression.
It will be the same thing for rubber, because in a piston, its section is fixed (I assume the shape of the rubber fits exactly that of the piston) and reducing its length would reduce its volume and rubber is very hard to compress, like water. It has nothing to do with elasticity : the rubber in the cylinder is just regular rubber.
Where does elasticity enter? When you free the sides of your rubber.
Take a horizontal cylinder made of rubber (with height small compared to its radius, otherwise it will bend and spoil the demonstration) and push on it from the top. In this situation you will find that you can easily reduce its height ! It is elastic !
But what about the volume ? Since you let the sides of your cylinder free, with nothing to keep it from spreading, well, it will spread.  By how much ? Well the linear increase of radius will be equal to the vertical decrease multiplies by the Poisson's ratio $\nu$ !
Thus the area increase will be twice that.
And the volume decrease will be proportional to ($1-2\nu$) the vertical decrease.
To be perfectly incompressible, the rubber would just need a Poisson's ratio 1/2. In fact, $\nu$ is not exactly 1/2, but you see why an elastic substance can easily respond to a one dimensional pressure, if $\nu$ is close enough to 1/2, by spreading in the two other directions, while being almost incompressible, if its extension in the two other directions is "blocked".
