BUCKLING CHECK ANALYSIS-STAAD PRO

Question

Please help me..i m still newly user with this software..and i m already out of idea how to solve it..can anyone help me how can i do my buckling analysis of my concrete tube length 36 meter and diameter size of my conc.tube is 600 mm.. 
 when i m going to do the analysis..i m stuck..whether to use perform analysis or perform buckling analysis?....i need to know how its buckle condition and the value of bending moment of concrete tube... 
 here i attached the Staad Pro editor that contains my model to make it clearly to everyone...Can anyone help me out.. 
 
 STAAD SPACE START JOB INFORMATION ENGINEER DATE 22-Jan-14 END JOB INFORMATION INPUT WIDTH 79 UNIT METER KN JOINT COORDINATES 1 0 0 0; 2 0 -1.5 0; 3 0 -3 0; 4 0 -4.5 0; 5 0 -6 0; 6 0 -7.5 0; 7 0 -9 0; 8 0 -10.5 0; 9 0 -12 0; 10 0 -13.5 0; 11 0 -15 0; 12 0 -16.5 0; 13 0 -18 0; 14 0 -19.5 0; 15 0 -21 0; 16 0 -22.5 0; 17 0 -24 0; 18 0 -24.5 0; 19 0 -25.5 0; 20 0 -27 0; 21 0 -28.5 0; 22 0 -30 0; 23 0 -31.5 0; 24 0 -33 0; 25 0 -34.5 0; 26 0 -36 0; MEMBER INCIDENCES 1 1 2; 2 2 3; 3 3 4; 4 4 5; 5 5 6; 6 6 7; 7 7 8; 8 8 9; 9 9 10; 10 10 11; 11 11 12; 12 12 13; 13 13 14; 14 14 15; 15 15 16; 16 16 17; 17 17 18; 18 18 19; 19 19 20; 20 20 21; 21 21 22; 22 22 23; 23 23 24; 24 24 25; 25 25 26; DEFINE MATERIAL START ISOTROPIC STEEL E 2.05e+008 POISSON 0.3 DENSITY 76.8195 ALPHA 1.2e-005 DAMP 0.03 ISOTROPIC CONCRETE E 2.17185e+007 POISSON 0.17 DENSITY 23.5616 ALPHA 1e-005 DAMP 0.05 END DEFINE MATERIAL MEMBER PROPERTY BRITISH 1 TO 25 TABLE ST PIPE OD 0.6 ID 0.4 CONSTANTS MATERIAL CONCRETE ALL SUPPORTS 26 FIXED 1 14 TO 25 FIXED BUT FY MX MY MZ 2 FIXED BUT FY MX MY MZ KFX 38857.5 KFZ 38857.5 3 4 FIXED BUT FY MX MY MZ KFX 36738 KFZ 36738 5 FIXED BUT FY MX MY MZ KFX 29673 KFZ 29673 6 FIXED BUT FY MX MY MZ KFX 22608 KFZ 22608 7 FIXED BUT FY MX MY MZ KFX 17662.5 KFZ 17662.5 8 FIXED BUT FY MX MY MZ KFX 14836.5 KFZ 14836.5 9 FIXED BUT FY MX MY MZ KFX 12010.5 KFZ 12010.5 10 FIXED BUT FY MX MY MZ KFX 8831.25 KFZ 8831.25 11 FIXED BUT FY MX MY MZ KFX 3532.5 KFZ 3532.5 12 FIXED BUT FY MX MY MZ KFX 2472.8 KFZ 2472.8 13 FIXED BUT FY MX MY MZ KFX 1413 KFZ 1413 LOAD 1 LOADTYPE None TITLE LOAD CASE 1 JOINT LOAD 1 FY -12480 PERFORM BUCKLING ANALYSIS FINISH

Biswatosh · Accepted Answer

The load is acting ideally along the column axis. There is no eccentricity (lever arm) from the support to the point of application of load and hence no flexure is developed along the column line.

Raymond Curtis · Answer

Use the basic solver with a perturbation load. 
 Add a command 
 SET STAR 0 
 before the JOINT COORDINATES 
 Change 
 PERFORM BUCKLING ANALYSIS to 
 PERFORM BUCKLING ANALYSIS MAXSTEP 25 
 PRINT MEMBER FORCE 
 Change loads to 
 JOINT LOAD 
 1 FY -12480 
 7 FX 100.0 
 -------------------------- 
 The result will be a load case with buckling factor, displacements, member forces as a static load case with a applied load of buckling factor times the User applied loads. 
 Joint 7 was chosen because it is in the middle of the nodes that can displace. The force is less than 1% of the actual load so it should be just a perturbation. 
 The results have the same buckling factor but have reasonable displacements, shears and member moments and spring reactions.

Modestas · Answer

Yes, you need to use a PERFORM BUCKLING ANALYSIS command. 
 Probably you know that there are two types of buckling analysis in STAAD.Pro: Advanced Buckling Analysis and Basic Buckling Analysis. You can read more about both in the Technical Reference manual chapter 1.18.2.2 Buckling Analysis. 
 If you are using the Advanced Analysis Engine, then few buckling factors with buckling shapes will be reported in the postprocessing mode -> Buckling page. 
 If you are not using the advanced analysis engine - the basic buckling analysis will be used and you will get only the factor of the first buckling mode which can be found in the Output file. Note that in this case you need to use a command PERFORM BUCKLING ANALYSIS MAXSTEP 15 instead of PERFORM BUCKLING ANALYSIS in order to get a converged buckling factor. 
 Regards, 
 Modestas

DSANJU · Answer

Basic buckling solves for the buckling factor by repeated static analysis (using K+Kg) . The final results are the buckling factor & displacements/forces/reactions commensurate with the loads scaled up by the buckling factor. The displacement will not be a classic buckling shape

Sye · Answer

Yes you sure can. The message means that you need to reverse the direction of applied loads for load case 10, for the structure to buckle. One way of doing it is to create a REPEAT LOAD case ( say LOAD 100 ) in which you may use a load factor of -1 for the load 10. The software would then report a buckling factor for the load 100.

Sye · Answer

I mentioned about load case 10 as the warning included that load case number. So if this load is something that in reality, cannot act in opposite direction, you don’t need to worry about it from the buckling standpoint. 
 
Regarding your other question, the load can only reach the support at the bottom if the intermediate supports do NOT offer vertical resistance. The applied load at the top of the member will be transmitted to the bottom through the member which is a continuous column. The intermediate supports are just lateral restraints. On the other hand if you restrain the FY at these supports, then the entire vertical load will be picked up by the top most of these supports which is not the intent.

Sye · Answer

When you use SET STAR 0 you are telling the software to use the basic solver. The basic solver calculates buckling factor for one mode whereas the advanced solver solves for 4 modes and the higher modes obviously would have higher buckling factors. Moreover the basic solver actually analyzes the structure for the load multiplied by the buckling load factor to give you the results for the buckling load case whereas advanced solver does not do that. For an advanced solver, if you want to look at the results at or near the buckling loads, you have to create a REPEAT LOAD case with the load case multiplied by a load factor, which is just smaller than the buckling load factor. You can then analyze the structure using a first order or a second order analysis to get the results at near buckling load.

Sye · Answer

A wiki article on why the buckling analysis results may differ between two solvers is now available at 
 
 communities.bentley.com/.../25250.buckling-analysis-results-differ-between-basic-and-advanced-solver

Sye · Answer

I am not familiar with any specific requirements that EC may have on this but on a general note, buckling analysis can give you an idea of how close your loading are to initiatestructural buckling. When the structure is close to buckling you may see some larger displacements and if these are large enough to affect the structural stiffness, then nonlinear analysis can be carried out.