I have a simple shaft with a radial load applied to one end and a fixed constraint on the opposite end. When I run the analysis on it, I am confused about what it reports for the "Reaction Force Magnitude". Since I only have one radial load (500 lbs in the Y-direction), I would expect the reaction force to be equal and opposite to this, but it is not even close.
Attached is my ipt (Inv 2020) and a pdf of the report I generated from the results.
TIA,
Kirk
Just ran the same sample part thru Nastran In-CAD. It reports what I expect on the reaction. Not sure what I have set wrong in Inventor FEA>
Hi Arthur,
I believe there was a bug somewhere. Inventor FEA is multi-threaded. Probably there was a threading error or memory corruption (which I suffer a lot myself). If you find a way to reproduce it, please let us know.
Many thanks!
@johnsonshiue wrote:
Hi Arthur,
I believe there was a bug somewhere. Inventor FEA is multi-threaded. Probably there was a threading error or memory corruption (which I suffer a lot myself). If you find a way to reproduce it, please let us know.
Many thanks!
I actually saw this on a more complicated problem. Rather than posting it, I created this basic example to show the same problem. Its rather easy to recreate.
Results from Inv are much different than from Nastran In-CAD. The Nastran In-CAD results look more like what it should be. The value I roughly calculated is 7643 psi. I just don't know what I am missing with the Inventor FEA model.
Hi Kirk,
I think you need to specify the exact direction of the load. I got the same result as yours if no direction is specify. Though it looks like there is only one direction, it is actually around the cylinder, so the upper portion shrink in size. If I specify the direction of the force, the result is similar to NASTRAN. Please share the file, if the result is still different.
Many thanks!
I could have sworn I applied a vector load in the Y-direction. Guess not. It gives the expect results now.
Thanks,
Kirk
