Analysis for frequency based transfer functions

tcarsontcarson
Is there anything in Mecway like the harmonic analysis in Ansys? I would like to compare FRF's of physical testing to simulations.

A couple of links describing what I am after

http://www.mece.ualberta.ca/tutorials/ansys/IT/Harmonic/Print.pdf

http://www.deskeng.com/de/practical-frequency-response-analysis/

Thanks

Comments

  • scottr99scottr99
    Isn't just

    *STEADY STATE DYNAMICS, HARMONIC = YES
    Low f, Upr f, num data pts, bias

    Google "steady state dynamics calculix".
  • SergioSergio
    edited June 2016
    I did some similar (if not the same) on another solver, but guess that CCX has implemented the same. What I use to make is first run a modal analysis to find the Eigen frequencies, and then run a second analysis called Modal Superposition, using the Eigen values computed before and appling a displacement in one part of the model. Then I can plot the displacemente on the movil side of the model against the frecuencie and see the peacks at resonance frecuencies and how is amplified this impossed displacement.

    We compare the measured curves on a electrodinamic actuator (shaker) against this computed curves, and remember that was very easy and accurate to correlate with lab. measurements.

    I will look in the CCX documentation to see if is possible.
  • scottr99scottr99
    In Ansys the natural frequencies are calculated automaticly in/for a Harmonic Response analysis so there's no need to perform both. I just did a search to see if Calculix does it the same way but I can't find anything either way. So yeah you may have to do the modal first (but I doubt it). I'd try what I have given you above and see what it gives you, easy enough for a cantilever beam especially.
  • SergioSergio
    I have checked and found that is possible, you must use first a step with *FREQUENCY card to find and store the Eigen frecuencies, and then a second step with a *MODAL DYNAMIC card to apply the impposed displacement.

    By the way, you can do it also with the Mecway's built in solver, is called Modal Vibration and Dynamic Response.

    Will try to run some example to compare against other solvers during the wekeend.




  • SergioSergio
    Mecway documentation:
  • scottr99scottr99
    edited June 2016
    Haha Sergio I just found that myself. Just do the following prior to the above card..

    *FREQUENCY, STORAGE = YES
    num modes

    :)
  • tcarsontcarson
    Thanks for the help

    I made this model but I am having trouble getting it to work. I can solve the model with calculix but I am having trouble adding the custom step. I add the step but it does not seem to show up in the .inp file. any ideas where I am going wrong?

  • SergioSergio
    I don't understant, when you choose Dynamic step, Mecway doesn't include the custom step data. I have changed to Static 3d and it include the custom data as usual, so maybe is a mistake in Mecway.

    A walkaround could it be define the test as Static 3d, then add several "don't generate kewords" to remove the static step and add then the complete definition for the two steps as custom data. In any case, you should remove the Mecway dynamic keyword to include the storage option to save the Eigen frecuencies.
  • tcarsontcarson
    I used your suggestion to switch over to Static 3D and I can now run the model. I am still having issues getting results but I am getting closer.

    Thanks for the help.
  • SergioSergio
    I have checked the Mecway and CCX documentation and examples, and in fact I didn't find nothing similar (or understood) to what I have done in the past (modal superposition), even if the Mecway interface for Dynamic Response there is a Mode Superposition check box.

    What I would like to make is first compute the eigen frecuencies, then make a dynamic sweep covering this frecuencies and applying a fixed displacement, to see how this displacement is amplified at the resonance frecuencies. This is what is done physically on a dynamic test with a shaker.
  • VictorVictor
    Mecway doesn't yet have any features for this, either with the internal solver or CCX. You'd have to define it by hand using *FREQUENCY and *STEADY STATE DYNAMICS steps. There could also be problems reading the solution data in because it's a function of frequency instead of time, and that might get ignored or interpreted as time.

    The mode superposition checkbox does the same transient dynamic response as without it, but uses the mode superposition method instead of direct integration. Both methods largely solve the same problems as each other. The main differences are in efficiency and the kinds of damping allowed.
  • SergioSergio
    Víctor , you are right, the STEADY STATE DYNAMIC card looks like the kind of steps that I have in my old solver. Will give a try tomorrow. Thanks!
  • SergioSergio
    Hi Victor, sorry for posting this CCX exclusive question here, but maybe (well... sureley) your understanding is far better than mine.

    As per your recomendation I look in the CCX documentation and set of examples, I took the beamdy10.inp (attached), a simple cantilever beam under steady state dynamic loading, and make some small editing to have as output results, the displacement at one node where the movement is impossed (node set N2, node 95), and also the displacement at one free node (node set N1, node 100).


    When I see the results in the dat file (attached), I found that for every time, there are two lines of data. I have ploted in the left side only the first lines, as I see and understand that for the node 95 it has a 1. displacement (that is the impossed displacement). But the plot is not what I suposs it should be, if you look the blue line (impossed) is constant along the frecuencies, but the red one (a node at the free end of the beam) shows a sinus shape. In my understanding should show a peack at the eigen value!

    I look again in the .dat and think.... well, could it be that CCX is given me the peack and valley results for every frecuency? Then a make a third set of data making the diference between the first line and the second (again, for the excitation side the resuls is 1.0 as the second line is 0.), and make a new plot on the right. This new plot shows what I guess that must be the results, is the clasical peack at the Eigen and then the displacement fall bellow the excitation value.

    a) Do you think that this procedure is ok or just a coincidence?
    b) Why CCX shows the result as time if is a frecuency?
    c) Do you know if there are a way to ask for the acelerations also?

    Best Regards!

  • SergioSergio
    The beamdy10.inp example must be run from the command line, don't know if Mecway will import and solve.
  • Tim_GowingTim_Gowing
    Hi Sergio
    I've used *STEADY STATE DYNAMICS recently and I'm convinced that the results are in the frequency domain. I believe that the two sets of results are real and imaginary, those displacements that are in phase with the excitation and those that are 90 degrees lagging in phase.

    The manual entry, below, defines the way in which the results will be produced across the frequency range:-

    "For harmonic loading the steady state response is calculated for the frequency
    range specified by the user. The number of data points within this
    range n can also be defined by the user, default is 20, minimum is 2 (if the user
    specifies n to be less than 2, the default is taken)".

    "Example:
    *STEADY STATE DYNAMICS
    12000.,14000.,5,4.
    defines a steady state dynamics procedure in the frequency interval [12000., 14000.]
    with 5 data points and a bias of 4".

    I hope that I haven't misinterpreted the aspect that you're discussing.

    Tim
  • SergioSergio
    edited June 2016
    Hi Tim, thinking again, guess that you are right. Even if the documentation don't talk about real and imaginary components, in somwhere I remember have been reading that.
    Then, for having the real (measured displacement with an accelerometer), should we combine this two values? The third graph is now with the combined values.


    Oooops, reading a little bit deep (page 212, the theory for steady state dynamics), is confirmated, by default are the components, real and complex of the displacement.

  • SergioSergio
    Another thing, about the damping... how must be taken in count this value? Will depend on the material or the structure/bc? I work normally with steel and aluminium parts, would these parts have different damping????

    Regards!
  • VictorVictor
    I don't understand this much at all. What you say sounds reasonable about the frequency domain and that you'd need to do a bit of math to transform to real valued displacements which would be in the time domain.

    I think it only allows Rayleigh damping which is quite artificial and you need to choose parameters to match both the material and structure close to the frequencies of interest.
  • Tim_GowingTim_Gowing
    I would also really like to understand the relationship between the 2% damping that is often quoted for steel structures, and the alpha and beta coefficients used in *MODAL DAMPING.
    Tim
  • VictorVictor
    There's a couple of formulas to convert damping ratio (ie 2%) to alpha and beta. The tricky part is that it only works correctly at two frequencies so you have to choose frequencies that are significant and tolerate increasing error further from them.

    They're listed on p70 of the Mecway manual under Rayleigh damping.

    Here's a document showing them being applied to steel and concrete structures:
    http://www.iitk.ac.in/nicee/wcee/article/0529.pdf

  • Tim_GowingTim_Gowing
    Victor,
    Thank you for the guidance, I'll read further.
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