Transverse stress has wrong sign

[s.mich]

I was running an analysis on a part in which I was looking for the transverse tress caused by the Poisson's effect on the surface of the part. This particular part is in tension, so I was expecting a negative value for this stress but the FEA results is showing a positive stress despite showing deformation in compression.

I have recreated a simple dog-bone shape to illustrate my point. The part is being pulled in the X direction by a force, so the stress in the Y direction would be what I was looking for. I am now wondering if I am misunderstanding what X-Solid stress and Y-solid stress etc is plotting in the post-processor. Is the best way to just look at the longitudinal stress and manually calculate with Poisson's ratio?

Thanks

[John_Holtz · Accepted solution]

Hi @s.mich 

I think your understanding of the X-Normal and Y-Normal stress is correct. Positive values are tension, negative values are compression.

I think your expectation of what the Y-Normal stress should be is incorrect. Because there is no load in the Y direction, the Y-Normal stress "should be" zero. The Y-Normal strain is negative due to Poisson's ratio, but the stress would be 0 in an ideal situation (shown in the block on the left side below). Because the stress in the top/bottom part of the dog bone is lower than in the middle, the ends reduce the amount of negative strain in the middle portion. This creates the positive stress instead of a value of 0 or negative.

Here is my example showing a uniform stress state (left image) and a nonuniform stress state (right image). The transverse stress is shown. (Sorry. My orientation is different than your model. 😞 I should have used the same drawing plane so that the stresses would match! My Z direction corresponds to your Y direction.)

(Left) Uniform stress. Z-Normal is 0. (Right) Nonuniform stress. Z-Normal stress is positive.

[s.mich]

I see. So going back to the stress tensor and our infinitely small cube from our mechanics text books, since there's no force that's transferred in the transverse side, that stress is 0. I had assumed that it would calculate the strain via e_long = -Poisson's*e_lat, then output stresses accordingly per element. In the dog bone, I guess it might be introducing some lateral load due to bending or something which was making the stress non-zero? 

In my actual part, the stress in the lateral direction by pure coincidence was almost exactly the (longitudinal stress) * (Poisson's ratio) which may have been misleading to me thinking it wasn't recognizing the stress direction rather than it shouldn't be outputting lateral stress at all.

Taking your comment about the strain being negative, I edited my analysis to record and output strain and it seems to be working much better thank you.