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Why rubber is incompressible material? I know its Poisson's ratio is nearing to 0.5. So I don't understand physically, what it means by 0.5 Poisson's ratio and incompressibility. When I tried searching it, I found that rubber (or similar polymers) conserve volume after deformation and so they are incompressible. But same is the case with steel (Poisson's ratio around 0.3), it conserves volume after deformation. So can someone explain this?

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    $\begingroup$ There are no incompressible materials. For weak pressure most materials are incompressible "enough" for many applications. Rubber can be dense in the sense that it doesn't have cavities like polymer foams, so it's may be as incompressible in the same sense as many fluids are "incompressible" for their intended technical applications. $\endgroup$ – CuriousOne Oct 3 '14 at 19:42
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Why rubber is incompressible material?

While an incompressible material must have a Poisson's ratio of ½, that rubber has a Poisson's ratio close to ½ does not mean rubber is incompressible. In fact, there is no such thing as a truly incompressible material.

That rubber has a Poisson's ratio of ½ merely means that rubber is in some sense a bit like a liquid. Rubber is in fact quite compressible. Apply pressure in all three dimensions and rubber will shrink in volume.

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This 18 month old answer has received no attention until as of late. The two other answers commit the logical fallacy of affirming the consequent.

An incompressible substance (which does not exist) resists compression from any and all directions, including compression applied simultaneously from multiple directions. Poisson's ratio looks at the behavior of a substance when the substance is compressed one dimensionally, with behavior in other two dimensions being unconstrained. While incompressibility necessarily entails a Poisson's ratio of exactly ½, a substance with a Poisson's ratio of exactly ½ does not necessarily mean the substance is incompressible.

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MartinG's answer gives the math, but try to imagine what happens when you squeeze a material. Poisson's ratio tells you how much of the contraction in one direction elongates the other directions. So if you squish a square horizontally by a distance $\delta$, the top and bottom surfaces will both move away from the center of mass by a distance $\nu\delta$.

So as $\nu \rightarrow 0.5$, the displacement in the vertical directions is exactly equal to the displacement in the horizontal direction. This means volume is conserved which is the definition of incompressible.

And note -- materials can only be approximated as incompressible over the length and time scales of interest. The value of $\nu$ will never be exactly 0.5. And even as you get very close to it, extremely high strain rates may cause volume change in the material anyway. So at best we can only say rubber is approximately incompressible.

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  • $\begingroup$ MartinG's answer gives the math that says that an incompressible substance must necessarily have a Poisson's ratio of ½. The and vice versa in that answer is incorrect. $\endgroup$ – David Hammen Apr 27 '16 at 13:58
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Incompressibility implies Poisson's ratio $\nu = 1/2$, and vice versa.

A cylinder of length $L$ and radius $r$ has volume $$ V = \pi r^2 L $$ For an incompressible material (constant $V$) differentiation gives $$ dV = 2\pi rL dr + \pi r^2 dL = 0 \\\nu \equiv -\frac{L}{r}\frac{dr}{dL} = \frac{1}{2} $$ Thus a material with $\nu = 0.3$ cannot be incompressible.

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    $\begingroup$ -1 for the "and vice versa". While incompressibility does indeed require a Poisson's ratio of ½, the reverse (a Poisson's ratio of ½ means a substance is incompressible) is not necessarily true. You are affirming the consequent, which is a logical fallacy. $\endgroup$ – David Hammen Apr 27 '16 at 13:53
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Bulk modulus of elasticity is given by : E/3(1-2v) where E= strain (or linear deformation and in your case compressibility) v= Poisson ratio ( for rubber it is 0.5) and bulk modulus of elasticity basically tells us the capability of compressing of a material. hence inputting the values we result in a zero denominator making the B.M.O.E infinite. which makes it impossible AF to compress :D

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Assuming a cube of side l with normal stress in one direction, its final volume after deformation would be: $$V=(l+d1)*(l-d2)^2=l(1+d1/l)*l^2(1-d2/l)^2$$ $$ d1/l=\frac {\sigma}{E}=\epsilon $$ $$ d2/l=\frac {\nu \sigma}{E} $$ $$\nu =0.5$$ $$d2/l=\frac { \sigma}{2E}$$ $$V=l^3(1+\epsilon)(1-\epsilon /2)^2$$ $$V=l^3(1-\epsilon+\epsilon^2/4+\epsilon-\epsilon^2+\epsilon^3/4)$$ Ignoring the $\epsilon $ of higher order we obtain: $$V=l^3(1-\epsilon+\epsilon)=l^3$$ $$\Delta V=l^3-l^3=0$$ So the volume does not change, thus it is incompressible.

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Conserved volume means volume before and after any deformation must be equal (like in a rolling operation, forging operation, etc). In this situation the Poisson ratio becomes 0.5. Rubber behaves like incompressible deformation; that is, if we stretch rubber its length increases and width decreases proportionally, so its volume remains same. As a result, it has a Poisson ratio of 0.5.

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