Results: Buckling

[Anonymous]

Hi, I am a new RSA user and I decided to start with a simple column buckling case. I selected a simple W 4x13 bar simply supported as shown in file attached. According to Euler theory, the elastic critical load is126,8 kN, however in my RSA model the critical coefficient yields in a very diferent result .i.e. 11,5 kN. Could someone help me to know what I am doing wrong?

[Artur.Kosakowski]

1).  1.15e1 * 10 = 115 rather than 11.5

2) Delete excessive loads: selfweight and bar force

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.

[Rafal.Gaweda]

See also http://forums.autodesk.com/t5/Robot-Structural-Analysis/Results-Buckling/m-p/3406121/highlight/true#...

[Anonymous]

Ups! you right, thanks a lot. I didn't realize that RSA notation. By the way, after that I tried a shorter bar (see file attached). I cut the bar length to the half and the critical load was 507 kN. That it means that RSA still calculates the critical load using Euler approach. However, in this case the slender coeff. is almost 1 and the Euler approach is not longer valid. I expected that RSA uses another approach since when the slenderness < 2 critical loads are lower than those obtained by Euler. Do you know if RSA alllows to calculate critical load in a range of slenderness where at buckling portions of the columns are not longer elastic?  

[Rafal.Gaweda]

http://forums.autodesk.com/t5/Robot-Structural-Analysis/Length-of-buckling/m-p/4342757#M15569

[Anonymous]

Thank you. However, I had already read those posts and I couldn't find a direct link with my question. According to the file attached it seems that RSA applies the Euler approach and doesn't take into account non elastic column behavior. Is that true?

[Rafal.Gaweda]

True.