Robot Buckling length/critical load calculation

[Anonymous]

Hi, quite new to Robot,

just out of interest does Robot have the capability to analysis a sub-frame and work out critcial buckling load or buckling length?

if so is there a tutorial on this at all?

thanks,
Rich

[Rafacascudo]

Yes , just change the analysis type to Buckling analysis

[Artur.Kosakowski]

You may want to check: http://forums.autodesk.com/t5/Robot-Structural-Analysis/Critical-buckling-length/m-p/3841687

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[Anonymous]

Hi, I have setup a simple test columns for the buckling length check.

1) 3m cantilever column with axial load and a fully fixed base and 254UC73.

2) 3m propped cantilever with fully fixed base (prop is a pin with sliding in Z direction) and 254UC73.

the results are not what I expect for buckling length and critical load capacity. 

1) the critical load = 2213kN (which is slightly higher than what i expected) but the buckling lengths are out by a large margin.

for the condition defined the buckling length should be maximum 1.4L which is 4.2m not the length calculated.

2) the critical load = 26705kN (which greatly exceed the maximum sectional capacity of the steel) but the buckling length isn't correct either as indicated 2.92m which should be approx. 0.7L.

is there something wrong in the way I have setup the model?

1) input the column section at 0,0,0 and specified the length being 3m

2) add fixed base support to the bottom of the column (Node 1)

3) Added load case - DL1

4) Defined load - Nodal Load @ Node 2 - top of column - 100kN (downwards)

5) Changed Analysis Type for DL1 to buckling

6) Calculated and opened results->advance->critical load

[Artur.Kosakowski]

Could you attach the rtd file and your hand calculation?

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.

[Anonymous]

The calculation for the cantilever and the propped cantilever buckiling load is based on Blue Book approach

http://tsbluebook.steel-sci.org/EN/Browsers/Main.htm

Case 1) Le = 1.4L = 4.2m; extrapolate between Le of 4m and 5m gives Ncr = Nb,z rd = 2090kN

Case 2) Le = 0.7L = 2.1m; similarly to the above; Ncr = 2922kN;

Maximum sectional capacity 254UC73 S355 = A x Wpl,y = 3305kN

please see the two models http://www.filemail.com/d/gjdsmtbrvikbrgy.

thanks for any assistances provided.

[Artur.Kosakowski · Accepted solution]

I may be wrong but my assumption is that you have 2L for a cantilver and 0.7L for the other scheme.

E.g. http://www.mitcalc.com/doc/buckling/help/en/buckling.htm

As it has already been stated on this forum you need to divide the bar for better accuracy of the results. In my case I divided the bars in 5 pieces.

1). 2 * 3 = 6 m

2). 0.7 * 3 = 2.1m

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.

[Anonymous]

Many thanks,
I take it the critical force is irrelevant in this respect?

[Artur.Kosakowski · Accepted solution]

Let me write this again. The aim of the buckling analysis in Robot is to determine if the applied load will result in the lost of the overall stability of a structure or any of its part. The main result is the critical coefficient which indicates the value of the critical load that would cause this effect (being the applied load multiplied by the critical coefficient). This analysis is not intended to determine buckling lengths or critical forces in elements of a model.

See: http://forums.autodesk.com/t5/Robot-Structural-Analysis/Buckling-Analysis/m-p/3280269

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.

[Anonymous]

much appreciated!

[v.brigot]

Hi, could you give please what is the cr value authorized according to Eurocodes? We see often on forums a minimum values of 10?

Many thanks.