Why is the default pinned release fixed in RX?

[Anonymous]

From a structural point of view, a pinned fixation is supposed to resist vertical and horizontal forced but not moments, so shouldn't the default be to uncheck RX in the program? 

I'm only asking because I have the type 3 instability and after watching the webinar and reading many forum questions, a popular solution seems to be to uncheck RX, which I believe is a realistic way of modeling it.

Just wanted to understand the reasoning behind this.

Thank you.

[Rafacascudo]

Rx is the torsion moment bar release. If you release it on both bar ends , then the bar can spin freely around its axis.

[Anonymous]

Thanks for your reply Rafa.

I understand that, but it seems to be the only way to stop getting the type 3 instability error and I'm not sure if modeling it that way (unchecking RX) will produce wrong results. 

Issue is, I have purlins that are all pinned to secondary beams and these secondary beams are in turn pinned to main beams which are pinned to columns, hence the type 3 instability. 

My question is, will unchecking RX produce wrong results? 

[Rafacascudo]

  RX "fixed" in one of the ends is just mandatory also in real world. If both ends are released for torsion , a small horizontal excentricity of the vertical loading would be enough to make the beam spin freely.

Try  making a model fully fixed for torsion and the other fixed in just one of the ends and check the difference. Probably neglegible

[mustafahesenow]

Hi guys,

For me even I uncheck Rx in both member ends because its impossible to be free to rotate in both ends. which is more realistic. 

[Artur.Kosakowski · Accepted solution]

Hi @Anonymous

In case you not happy with fully released RX (as in fact any type of connection unless deliberately made not to give some torsional 'restrain') and filly fixed one you can decide on something in between using the partial releases instead. In this way you should be able to 'avoid' instabilities and reduce torsion in a bar when excessive by adjusting the value of RX 'stiffness' assigned to a bar.

If one or more of these posts answered your question, please click Accept as Solution on the posts that helped you so others in the community can find them easily.

[Anonymous]

Thank you all for your suggestions.

I ended up not modeling the purlins altogether and that fixed the model in terms of the instability warning. I believe it might have been caused by the large amount of computations that would need to be processed by Robot.

In any case, I do believe the best way to have modeled it would have been to increase the stiffness in the RX direction as there was a significant difference in reactions when I completely released the RX fixity in both directions