STAAD Natural Frequency vs Classical Methods 1

[nicoga3000]

I am trying to determine why STAAD provides a different frequency for a simple beam with concentrated mass model versus proven classical methods for the same model. The results are different enough to merit a discussion.

If any of you have STAAD, I have attached my input file. You can perform a Rayleigh or Modal analysis as the results are the same for this structure (just uncomment the request).

The method I used is a cantilever beam of mass Mb with a concentrated mass of mass M at the end. The Blevins book on Natural Frequencies lists the frequency as:

Here's my quick and dirty Mathcad showing the results:

As you can see, the difference is not insignificant. A nearly 1s difference in the period is tremendous for our structures.

Any input on the difference would be awesome. Thank you!

 

[Denial]

I have never used either STAAD or MathCad.[ ] However two lines in your attached STAAD file
[ ][ ][ ]JOINT LOAD
[ ][ ][ ]2 FY -10 FZ 10
make me suspect you are not making a proper distinction between force and mass.

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[GregLocock]

One obvious check is the static end deflection.

Cheers

Greg Locock


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[nicoga3000]

@Denial I'll double check that. It could be exactly that, which means I'll need to look at how to define a large mass at the end of the cantilever.

@GregLocock I'll check my output in the morning once I'm back at the office. I believe we were looking at this and noticing that they didn't check out either. Which, yes, would suggest that the way we are looking to define the end mass is incorrect.

E: However, upon looking at the Bentley documentation, I think I am doing it right?

https://docs.bentley.com/LiveConten...UID-90E51C9A-E909-442F-983C-2240FF9A95AA.html

 

[Denial]

Applied forces are irrelevant to natural frequency analyses.[ ] Such analyses require knowledge only of the stiffness "distribution" thru the structure and the mass "distribution".[ ] Most (all!) structural analysis systems I have worked with treat mass as part of the definition of the structure, not part of the definitions of the loadings to which the structure will subjected.[ ] Hence my querying of those two lines above.[ ] I would expect STAAD to have a way to specify "lumped masses" located at nodes.

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[nicoga3000]

I'll definitely peak at it tomorrow.

Really, I need to figure out how to model in FEA the Blevins example. That's where I'm at with it. I'm happy to admit my model is wrong. But if that's the case, I really want to know what the correct answer is.

 

[GregLocock]

Assuming the inputs to your Mathcad model are OK then a 4535 kg mass on the end of a steel beam with your dimensions agrees with 0.272 Hz. Tip deflection should be 1.69m due to the end mass and 2.57 m due to the self weight of the beam. (roughly, I looked up the values rather than converting yours from Imperial to sensible).

Cheers

Greg Locock


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[nicoga3000]

So problem has been mostly solved!

The answer was not at all what I expected.

My model was correct, my math was correct, but the ratio of end mass to beam mass was not.

The Blevins calculation is for a "slender beam with a concentrated end mass". What it doesn't really explain is that there's some inherent assumption regarding how heavy that end mass is in relation to the beam. When I changed my end force to 1000 kips and left the beam as 41-ish kips, the two periods were within 0.5%.

So I THINK this only proves to me that the FEA results are correct and verifiable IF I can find the correct classical method to replicate the modeled situation.

 

[Celt83]

Is it possible that STAAD is considering the shear deformation of the beam which would slightly increase the system flexibility. The "slender" in the Blevin's formulation may indicate Euler-Bernoulli beam assumptions.

 

[nicoga3000]

Great question!

I modified my input file to include the command SET SHEAR, which forces STAAD to ignore Shear Deformation. That solved one of our other unrelated questions.

 

[ThomasH]

ake me suspect you are not making a proper distinction between force and mass.

It was a long time since I used STAAD. But one thing that irritated me was that STAAD did not make a proper distinction between mass and force.

Mass is defined as a force. And that means that mass actually has a direction in STAAD. If I remember correct

 

[GregLocock]

You can check Blevins by assuming no tip mass or a massless beam. I very much doubt your explanation is correct.

Cheers

Greg Locock


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[centondollar]

There should be no input of any forces in a natural frequency (eigenvalue & eigenmode) calculation. The ratio of length to thickness (almost 100) also indicates that classical beam theory formulas should provide sufficiently accurate results. The elastic modulus you use seems to be for steel (at least that is what the google unit converter told me), which would further indicate that shear deformations should be insignificant in both static analysis and for the stiffness in a natural frequency analysis.

You should find out how to model a point mass in STAAD - it cannot be replaced by a point force (a part of an external load vector). It's a basic feature, really, since many FE softwares use mass lumping for dynamic analysis due to simplicity and a shorter calculation time when compared to calculation time obtained by applying consistent mass models.

 

[Celt83]

Didn't see anyone else ask did you model the beam as a single element for the full length or did you subdivide it into multiple elements? The accuracy of the finite element formulation and mass lumping will depend on how many elements you used to model the beam.

 

[Denial]

Celt.[ ] OP used a single beam model, which I managed to deduce from his attachment of his STAAD "input file".[ ] You are right.[ ] If STAAD simply threw half the beam mass to the fixed end and half to the free end the results would be significantly affected.

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[Denial]

Taking Celt's highly relevant warning a bit further, and quantifying my "significantly affected" comment.

Ignore any added lumped mass at the free end of the cantilever beam.[ ] A quick back-of-envelope doodle tells me that if you model the system with a single beam element using a software that treats the beam's self-mass by lumping half at one end and half at the other, then the resulting analysis will underestimate the beam's natural frequency by about 30%.[ ] Significant indeed!

Any actual added lumped mass will reduce this underestimation.

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[ThomasH]

I did a quick analysis of the beam in Nastran after converting everything to SI units. Just to feel more comfortable with the size of the numbers .

I get the frequency 0.272 Hz, if I remove the beams mass I get 0.384 Hz.

Like I said before, I think STAAD can be confusing regarding how mass and force is used. But I think your setup is correct.
But I noticed something else in your input file, (as mentioned by others) you have 1 element and 2 joints. I have not used STAAD in years but I think there was a minimum of 5 masses for this type of analysis. Perhaps they have "improved" the analysis but I would check without the beams mass. There is probably some crude mass lumping in your analysis.

 

[IDS]

I analysed the beam as defined in the OP using Strand7 with the following options:

Model as a single beam or subdivided into 12 elements.
Include or exclude beam mass.
Lumped or consistent beam mass.
Shear stiffness included or excluded.

Results were:

Results with 12 beams and/or consistent beam mass were the same as ThomasH results.
Results with lumped mass and a single beam were close to the STAAD result in the OP.
Adding shear stiffness to the beam made negligible difference.

So subdividing the beams in the STAAD analysis, or specifying a consistent matrix if that is an option, should make the STAAD results consistent with the classical solution.

Strand7 users note that the lumped mass matrix is the default option for some reason.

Doug Jenkins
Interactive Design Services

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[ThomasH]

I can add that I used 10 elements and a lumped mass formulation. I think that the reason for lumped mass is that there is a small impact on accuracy but the calculations are faster. There are situations when accuracy may suffer but not for this type of problem.

 

[IDS]

I just checked the Roark formula for this case. It is effectively as in the OP, except the factor on the distributed mass is 0.236, rather than 0.24.

Table 16.1.3c in the 7th Edition.


Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/