Hi,
I think I have seen the answer on this forum before, but can't find it 😞
I want to know how to define a non-linear release that is compression only. Robot seems to make it tension compression as shown. Any advice?
Solved! Go to Solution.
Solved by Pawel.Pulak. Go to Solution.
Hi Tony,
the possible ways (starting from the simplest one) are:
1/ compression-only members - in this case it will be truss members (pinned), transferring only compression:
2/ unidirectional releases - more detailed description and examples what should be selected (Ux+ or Ux-) in case of beginning or end of bar can be found in Help
3/ you can also use releases with non-linear model as shown in your post - but in such case it will be necessary to define different stiffness (very high in one and zero in opposite direction) for positive and negative semi-axis
---------------------------------------------
If this post answers your question please click "Accept as Solution". It will help everyone to find answer more quickly!
Regards,
Ahh yes. I already understand the two ways you mention. But to make it non linear and compression only, can you please elaborate on how to achieve this. Sorry to be lazy and just ask, I havn't had time to fully investigate by myself.
Tony
Hi Tony,
attached file containing very simple model of compression-only vertical member which supports horizontal cantilever. It is repeated 5 times and model in various ways :
1/ bar 2 - modeled by compression-only truss member
2/ bars 4 and 8 - modeled by unidirectional releases defined on the beginning (bar 4) or on the end (bar 8) of bar
3/ bars 6 and 10 - modeled by non-linear releases defined on the beginning (bar 6) or on the end (bar 10) of bar
It contains 2 load cases with vertical load on the end of cantilever downwards and upwards - see the screen capture below:
Non-linear model of release is defined as zero stiffness in one direction (positive semi-axis) and "rigid with linear hardening" with very high Flim force in opposite direction (negative semi-axis) - see the screen capture below:
---------------------------------------------
If this post answers your question please click "Accept as Solution". It will help everyone to find answer more quickly!
Regards,
Can't find what you're looking for? Ask the community or share your knowledge.