topic Re: Linear loads in Robot Structural Analysis Forum

Linear loads

Anonymous — Wed, 24 Aug 2016 14:44:35 GMT

Hi,

Is there a way to display a mesh indepedent linear load (kN/m) acting at the bottom and/or top of (f.ex) walls in Robot? I.e. from adjacent slabs etc. If the walls is simply supported, is there a way to display which load that acts on the wall for a manual control?

And in the case shown in the picture attached, which what is the maximum linear load (kN/m) at the bottom/top of the wall?

Thanks

Re: Linear loads

Rafacascudo — Thu, 25 Aug 2016 01:27:44 GMT

I guess you are asking about linear reactions instead of linear loads.
Linear reactions can be displayed through Results/diagrams on bars , Reactions tab. Choose average or integral value to be displayed

Re: Linear loads

Anonymous — Thu, 25 Aug 2016 05:10:46 GMT

Thanks. But if you have a simply supported wall, is there a way to display the linear load acting at the top of the wall? Almost like the "forces generated automatically" for bars?

Re: Linear loads

Artur.Kosakowski — Thu, 25 Aug 2016 08:30:35 GMT

No, there isn't.

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.

Re: Linear loads

Anonymous — Thu, 25 Aug 2016 21:05:29 GMT

Is there a way to use panel cuts, and how?

Re: Linear loads

Artur.Kosakowski — Fri, 26 Aug 2016 07:13:04 GMT

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.

Re: Linear loads

Anonymous — Fri, 26 Aug 2016 08:16:37 GMT

Thank you. These results is mesh dependent? What is the correlation between the mesh and the results?

Re: Linear loads

Artur.Kosakowski — Fri, 26 Aug 2016 08:36:21 GMT

Yes, it is as any other type of results for panels.

See:

http://forums.autodesk.com/t5/robot-structural-analysis/finite-element-mesh-size/m-p/3752226

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.