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Explain/Clarify Planar Shear Elastomer Specimen Test?

Explain/Clarify Planar Shear Elastomer Specimen Test?

Explain/Clarify Planar Shear Elastomer Specimen Test?

(OP)
I am reading some articles from Axel Products re: tests for specifying elastomer properties for FEA.  I am confused about the planar shear test.  This is the test that looks like a very wide tensile specimen.  My ASCII art below shows the specimen and the axis orientation.

     ^ Y-Axis
     |
-------------
|           |    ---> X-Axis
-------------

Z-Axis is perpendicular to thickness.
Assumption 1:  Thin specimen===>  SigmaZ=0
Assumption 2:  Wide specimen===>  Strain X=0 near middle

Hookes Law reduces to:

SigmaX = SigmaY*Poisson's ratio

For example, SigmaX=.5, SigmaY=1(Applied load), SigmaZ=0.  If I do a Mohr's Circle for three dimensions it looks like I should get a max shear stress if I rotate 45 deg about the X-axis.  So the max shear is across the thickness of the sample correct?  Just trying to make sure I understand this.  I made a simple Pro/Mechanica model and verified most of this, but I am a little fuzzy on the max shear and what value to expect.  In my example above, should I expect a max shear of .5 at an angle of 45 deg in the YZ plane (ie: rotated 45 deg about the X-Axis)?

Thank You

RE: Explain/Clarify Planar Shear Elastomer Specimen Test?

(OP)
Anyone willing to give me their interpretation?

RE: Explain/Clarify Planar Shear Elastomer Specimen Test?

Hi cnuk;

I've been browsing around on Axel's web site; their 10th paper on the site (under technical downloads) is a long paper by MSC Software, that discusses all basic tests needed in an FEA. On page 50, they explain the idea behind the planar tension test; does that help?

I think there are two basic assumptions to keep in mind. The first is, the specimen needs to be AT LEAST ten times as wide (X-axis in your example) as it is tall (Y-axis). Only then do you really have a pure shear state of strain in the center of the specimen. The other thing is, this is a rubber test, and like most rubber tests, the assumption that the material is incompressible (Poisson's Ratio = 0.5).

Keeping those two things in mind, you can see, first of all, that the pure shear state of strain only happens at low strains. When Axel stretches a 150mm x 10mm (X x Y) specimen to 100% strain (so that the shape is 150mm x 20mm), we're no longer exactly in pure shear. The other thing to realize is, this strain state is not universal throughout the specimen. So the essential part of their test set-up is that they measure strain ON the specimen, in the CENTER only (that's the part where the rubber can only get thinner, and not move sideways).

That said, I think it's an excellent, fairly non-complicated testing solution for your FEA (when combined with uniaxial tension and compression data). Yes, I agree that the math behind it can be a bit complicated, but from a testing point of view, it's a great experiment.

Hope this helps...

Ron

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