Dear all,
I use the cladding feature froms its addition in robot and is very usefull.
Today I was designing a rectangular slab with in the cernter a circular hole. I added a rectangular cladding and then a circle: you see on the screen that the circle forms an hole on the cladding. after the calculation if I look the the load ridistribution, i see that the hole is not considered! if I check the reaction the give the same result!
is this a bug or my procedure is worng? PLease find attached the example file.
Solved! Go to Solution.
Solved by Artur.Kosakowski. Go to Solution.
Hello,
You didn't add the RSA-file to your question so I hope I did understand your question correctly.
I think you must ad the whole in your cladding as well.
greeting,
Henk Boelens
The distribution algorithm is not capable of correct detection of openings that stretch up to the edges of a cladding itself. You should either create two claddings and no opening or reduce its size.
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hey,
I am having problems with load distribution on the inclined roof, I added a load on the cladding but it is not distributed on the pannels. I put the load distibution in the right direction as seen in the attachement. It is working for the cladding on the columns.
Regards,
Gabriel Habib
hey,
thanks for the reply but I figured it out. the cladding were not placed correctly at the nodes.
Regards,
Gabriel Habib
I dont know why the Cladding Load distribution is hidden. Please help me out
Q2 .Can some one help me to find a way to distribute the load equally on for sides?
Equally on four sides not 45 degree.
Hi @Anonymous
The cladding load distribution option is based on the principle of assigning load from the tributary area which 'belongs' to given 'supporting' element. The kind of the load you described is also available in RSA. It is applicable to bar elements and doesn't require any cladding definition:
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Is RSA not able to distribute load of a simple cladding with a uniform planar load correctly? Or is there anything I am doing wrong in this matter.
I tried adding RSA file but it won't let do that.
Thanks in advance.
Hi @kmmdD66ZQ
Load distribution procedure assigns cladding area to defined beams.
In your case.
Then loads applied to assigned area are transferred to the corresponding beam.
I am aware of the way of distribution, but is it possible to distribute the load correctly without needing to calculate the distribution by hand and applying it as a lineload.
25% of the load should be placed on beam 2 in your and my case.
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