The Poisson ratio of the material equals 0,33. If I load a cube or a cylinder that is fixed at one end with compression force in ANSYS I can confirm that the ratio is 0,33 by dividing the lateral by longitudinal strain.

But when I load a geometry that is hollow with changing cross section the lateral/longitudinal ratios across the geometry change in unreasonable intervals that make no sense to me, because I was convinced that the ratio is always constant and does not depend on cross-section or shape. My question is, what does the ratio depend on and how? If somebody could direct me to some useful literature, it would be greatly appreciated.

  • 1
    $\begingroup$ For a hollow geometry with changing cross section, the stress is not uniaxial anymore, this is probably why you are not measuring the same ratio along the rod. $\endgroup$
    – user190081
    Nov 5, 2018 at 16:38
  • $\begingroup$ The Poisson ratio remains constant but the stress varies even with pure axial load, thus the deformation varies non-linearly from point to point. $\endgroup$
    – r13
    Sep 24, 2021 at 0:18
  • $\begingroup$ @user190081 I wonder if there's a way to get ANSYS to output the local stress and strain tensors at each point in the material, so OP can verify that they're related by an elastic modulus tensor built from the same Young's modulus and Poisson's ratio everywhere. $\endgroup$ Jan 12 at 12:02

3 Answers 3


The Poisson ratio is defined for a sufficiently uniform material (i.e., one for which the continuum assumption reasonably holds). If your material is not uniform, then what you're obtaining is some type of effective Poisson ratio, which could be just about anything and doesn't hold broad meaning. This is covered in most textbooks on elasticity or mechanics of materials.

  • $\begingroup$ Material I use is homogeneous. $\endgroup$
    – breza
    Nov 8, 2018 at 13:43

For simplicity lets assume you have a cylinder of length 20 cm which is made up of two short 10 cm cylinders, both hollow and have a 1 cm diameter hole drilled out of them, one has the cross section surface are of 2 cm^2 and the other thinner one has a cross section surface 1 cm.

Now we apply P = 2kN to the top of the cylinder.

the stress on the thick part will be

$$ \sigma = 2\, \text{kN}/ 2\, \text{cm}^2 = \ 1\, \text{kN/cm}^2$$ but for the thinner part $$ \sigma = 2\, \text{kN}/ 1\, \text{cm}^2 = 2\,\text{kN/cm}^2. $$

So the thin section vertically strains twice and so horizontally its strain is two times more than the thick part.

Same idea applies to random changes in cross section of a bar, and even more complication arises when these changes have secondary stress concentration and distort the surface of strain plane along the length of the test sample.

  • $\begingroup$ If the horizontal and vertical components of strain say increase by factor 2, does that mean that the ratio lateral/longitudinal stays the same? Is it possible that, in case of the complicated random cross-sections in the structure and axial loading, the lateral strain is greater than the longitudianal strain? $\endgroup$
    – breza
    Nov 6, 2018 at 12:00

First of all a bit of background, the Poisson ratio is calculated from $$v=-\frac{\epsilon_y}{\epsilon_x}$$

To use the above equation, you need to use a uniaxial loading (axial or compressive). Because at different loads you might have different Poisson ratio usually what happens is that in the elastic region you create a plot of $-\epsilon_y$ vs. $\epsilon_x$ and you can use regression analysis to estimate the coefficient.

Regarding the example you present, I presume that one case of a hollow box with a changing thickness has the following shape.

box with changing cross-section

In this example there are two things that are important:

  1. The Poisson Ratio in the literature can be found as a material or structure property.

The former case (material property), is the most traditional and is measured with the above mentioned formula as $-\frac{\epsilon_y}{\epsilon_x}$ for a uniaxial stress state. In that case, you are getting the averaged property of an infinitesimal cube of the material. (Because each piece of the material is assumed to be in the same state is should exhibit similar behaviour, thus averaging). Essentially, its the same as stacking 1000 pieces of A4 paper in order to estimate the thickness of one sheet of A4 paper with a hand held measure).

The latter case (structure property) is most commonly used for auxetic structures. Sometimes these are referenced as auxetic materials. Again what you are measuring is an averaged property of the structure (the material behaviour is a totally different thing). As such you should have a repeated pattern across the structure. Obviously, the stresses and strains on a small cube of material at different points of the structure differ significantly, and only in a macroscopic and (very loose) sense can be viewed as stresses and strains. In that sense the Poisson ratio is applied to this case of "materials".

  1. Loading axially a structure with changing cross-section (exception is if its symmetric), introduces shear stresses.

Imagine cutting image 1, in for pieces (resulting in a trapezoidal-ish shape). If you apply a force on that geometry, then the material will also shear.

enter image description here

As a result, you can't use the Poisson formula for this scenario. Not unless you have a repeating structure (like a honeycomb or a auxetic structure).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.