Rigid links in custom directions & degrees of freedom
rigid bars from nodes (304, 21 and 306) to nodes (47, 8 and 46) respectively
is not really perpendicular to de panel edges . But if this is what you need , you could model rigid bars between these nodes and release the correspondent uy or uz local directions on these rigid bars accordingly
Rafael Medeiros
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Hi @Anonymous
As a rigid links directions are related to the global coordinate system in case you need to follow e.g. a skew beam it is better to replace it with a rigid bar rotated (gamma angle) as per direction of a beam and define releases at its end where the slave node of the rigid link would be. Mind that for other that Skyline and Frontal solver this is done anyway (use of RLink elements to model rigid links).
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Artur Kosakowski
Due to the fact that the edges (14_Edge(5) and 14_Edge(6)) are not in the same direction and the slab is continuous, the effect you need ,described in your 1st post (free to move independently in a direction perpendicular to these edges), is impossible to achieve.
Any link (with releases or through custom section properties)you create for one of these edges will not "help" or allow the other edge to move freely.
In the direction below it is possible , but that´s not what you really described in your 1st post
Rafael Medeiros
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Accepted solution
Just for the record , I am not official support !!!
Better way to compare displacements of similar models is to look at "nodal displacements" table . If you had checked this table you would have seen that nodes 5 and 323 from bars 1 and 50 have has weird displacements on Y direction for a load totally on direction X.
Then , It happenned that the mistake was with you and it was a pretty basic one. Never trust 2 digits numbers!!!! Always use at least 4 digits for node coordinates!!!!
Due to that , bar 54 is perfectly modelled on Y direction . But bars 1 and 50 are not ,as you can see below.
That was the reason for the enormous and weird displacements on Y direction on bar top nodes of bars 1 and 50. The force applied on X direction was not exactly perpendicular to these 2 bars
Correcting that , you get
That said , The best way to get what you need is to model very rigid bars and assign bar releases for the directions,rotation you want ... released
Rafael Medeiros
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How did you copy node 322? By mouse snapping or writing 0 , 0.25 , 0 on the vector field?
Rafael Medeiros
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well , in 16 years ,it is the 1st time I see an error like this. These kind of mistake happens when you set the vetor through mouse snaps. It´s very easy to snap to some of these below as they are already activated by default.
I would say that it is nearly impossible to be a RSA bug.
Can you reproduce this error again?
Maybe official support has other ideas of how this happenned
Rafael Medeiros
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Hi @Anonymous
As the definition of released d.o.fs in rigid links refers to the global coordinate system in situations when you are working in other planes the best option seems to be replacing them with rigid bars with gamma angle defined in such a way that their local axes correspond to the 'directions' you want to release and define releases in their 'slave' nodes.
For the crashes: have you sent any crash reports?
For the snapping: make sure you switched off the display of the grid when you indicate the translation vector.
I hope this helps.
If you find your post answered press the Accept as Solution button please. This will help other users to find solutions much faster. Thank you.
Artur Kosakowski