It sounds logical untill I tried some thought experiments. So I must be doing something wrong. Or I am missing a fundamental aspect.

What I thought of was:

-lets say you have an orthogonal parallelepiped which is pulled in tension by a stress-strain test. (in the long axis or z-direction).

  • the lateral sides are 5 mm and 4 mm and the axial length is 7 mm.

-It has been elongated by the test untill the specimen had an axial length of 11 mm and hypothetically the 4 mm lateral side did not change.

So the original volume was 4x5x7= 140 mm^3 and since there is no net volume change the final length of the lateral side that did change has to be: 140= a x 11 x 4 -> a= 35/11 mm

now if you calculate the poisson's ratio: v= - Ex/Ez

-> Ex = (35/11 -5)/5= -4/11 and Ez= (11-7)/7= 4/7

v= 4/11 : 4/7 = 7/11 = 0.64 > 0.5

How I understood it: There is no volume change, so when you elongate one direction the other two will compensate this by compressing. And this poisson ratio of 1/2 is in the extreme case that one direction does not compress and so the other side has to take all the compression. Like in the situation that you elongate the axial side to twice it's orginal length, then in order to have same volume the two other directions have to become each 1/4 of their orginal length or one direction has to become 1/2 it's orginal length. This will indeed then give poisson ratio of 1/4 and 1/2. So I don't get why with other numerical examples I don't get this same 1/4 and 1/2 ratio.

  • $\begingroup$ Poisson's ratio depends on the material behavior, for isotropic materials on the compression and shear modulus. You can not derive the Poisson's ratio from the given boundary conditions (hypotherically the 4mm lateral side did not change). $\endgroup$ Commented Jan 24, 2016 at 21:28

1 Answer 1


First poisons ratio for isotropic materials applies equally between each pair of dimensions. That means that if the axial length is stretched $1\%$ both of the lateral dimensions would reduce by $\nu 1\%$. I don't know where you're getting the 0.25 number from.

Second, this approximation only works when the strain is small. Lets say we have an orthogonal parallelepiped with dimensions $x$, $y$, and $z$. If we apply a stress $\sigma_x$ streching the x direction and let the y and z dimensions be stress free. Then we'll have:

$$\epsilon_x=\frac{\sigma_x}{E}$$ $$\epsilon_y=-\nu\frac{\sigma_x}{E}$$ $$\epsilon_z=-\nu\frac{\sigma_x}{E}$$

So the new dimensions would be:

$$x'=x(1+\epsilon_x)$$ $$y'=y(1-\nu\epsilon_x)$$ $$z'=z(1-\nu\epsilon_x)$$

And the new volume would be:

$$V'=x'\,y'\,z'= x\,y\,z\, (1+\epsilon_x)(1-\nu\epsilon_x)^2$$

So the ratio of the new volume to the old volume would be:


To get the increase in volume, you can subtract the 1. Then there are three more terms, but since $\epsilon_x$ is assumed to be small then the square and cube terms should be dominated by the linear term. Thus those terms are usually neglected and you're left with $(1-2\nu)\epsilon_x$. If you have an incompressible solid then this value should always be zero, so then $\nu$ must equal $0.5$

Let us look at a more general example:


Is really saying:

$$\begin{split}1=(1+\epsilon_x)(1+\epsilon_y)(1+\epsilon_z)=&1+ \\ & \epsilon_x+\epsilon_y+\epsilon_z + \\ & \epsilon_x \epsilon_y + \epsilon_y \epsilon_z + \epsilon_x \epsilon_z + \\ & \epsilon_x \epsilon_y \epsilon_z \end{split}$$

but again with a assumption that the strains are small, the product of multiple strains should be really small, so this equation simplifies to:


Now the general equations for stain in isotropic materials are:

$$\epsilon_x=\frac1{E}(\sigma_x-\nu(\sigma_y+\sigma_z))$$ $$\epsilon_y=\frac1{E}(\sigma_y-\nu(\sigma_x+\sigma_z))$$ $$\epsilon_z=\frac1{E}(\sigma_z-\nu(\sigma_x+\sigma_y))$$

We can sum these all together to get:


So we see that one again to conserve volume $\nu=\frac12$

Now why didn't this work in your example?

When you applied the stretching both other directions should have contracted. In order for the second direction to not contract, that direction would also have have to be pulled. This pulling on the second direction would further contract the third direction, but it would also contract the first direction, so in order for that to stay at it's original displacement you'd have to pull even harder.

Then you specified that $\nu=-\frac{\epsilon_z}{\epsilon_x}$ but this is only true if there is only stress in the x direction which is not the case for your scenario (because there must have been a stress applied to keep the 4cm side at 4cm).

Finally the strain you applied was too large. Even if you let the two lateral dimensions strain freely, you'd get a Poisson's ratio not equal to .5 just because the small strain approximation would no longer be valid:

$$\epsilon_x=\frac{11}{7}-1\approx 0.57$$

to conserve volume:

$$\epsilon_y=\epsilon_z=\sqrt{\frac{7}{11}}-1 \approx -0.20$$

Which is quite a bit less than half.


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