General Mechanical

General Mechanical

clarification on Modal analysis and Transient analysis

    • saravanakumar

      Hi all,

      I have a shaft with 3 hydrodynamic bearings and one thrust collar.

      I need to find the shaft natural frequency in

           1)      Lateral direction

           2)      Axial direction

      I know the bearing stiffness in y & z direction and axial stiffness (x direction) at thrust collar.


      1)      Do I need to do two independent modal analysis one with lateral bearing stiffness alone and another with axial stiffness


      One single analysis with both the axial and lateral stiffness.



    • peteroznewman

      One single Modal analysis. Request multiple modes. Look at the mode shapes of each mode to categorize it as Lateral or Axial.  The first mode will have a near-zero frequency, which is the rotational DOF.

      • saravanakumar

        Hello peter,

        Thanks for the reply.

        Below sketch shows the actual problem and actually I have to do shock analysis with 10g at 27ms as a full sine curve at radial bearing 1 near to thrust bearing in Y direction.



        To understand the procedure to do transient structural analysis. I have made certain assumptions as all the radial bearing as rigid support by inserting displacement boundary conditions.

        At all radial bearings 1, 2 and 3, displacement in Y & Z direction is set to 0 mm and axially set to free. And at thrust bearing, displacement in X direction is restricted by giving 0 mm and free in Y & Z direction. At avoid rigid mode, a remote displacement BC is given at the end of shaft and rotation about X alone is set to 0, rest is free.



        With this BCs I have done modal analysis and ensured the ratio of effective mass to total mass above 0.9 in Y direction since my shock load is in Y direction.


        Under transient structural analysis:

        Pl have a look at the Analysis setting in below image.

        1.      I have included the standand earth gravity in –Y direction (is this necessary to include or solver takes from modal analysis, as we have linked modal analysis under initial condition)

        2.      Should I include the residual vector.

        Inserted acceleration as Base excitation and scoped to BC bearing 1. Here I don’t know how to give acceleration as full sine curve instead for study purpose I have given as linear as of now.

        Output screen:1


        Output screen:2


        Output screen:3




        I have seen directional acceleration in Y direction for all body here I got 128.8 m/s2 as peak acceleration. But my max input acceleration is only 98.1 m/s2. Is there any concept behind this?

        I have also seen the total deformation, I got max 11.76 mm at coupling end and it stays at that 11.76 mm until the end of analysis. What does it mean? (Correct me if I am wrong, my thinking is it should come to its original position after certain time)

        Finally I have seen the directional deformation (Y axis) at bearing 1 where the acceleration is given as an input. I got 9 mm maximum deformation. How could this be possible? Because under BC bearing-1 in modal analysis it is given that displacement in Y direction is 0 mm? please clarify.

        please ask for any other details required for you to clarify my doubts.

        Thanks & Regards


      • saravanakumar

        how to upload my achived analysis file in this forum

    • peteroznewman

      Is it really a full sine curve?  A half-sine curve is more typical for a shock load.

      If the 10G shock is a half-sine with a 27 ms duration, it would look like this:

      You can create the data for this curve in Excel. The time increment for this data should be no larger than 0.002 s.

      A 27 ms half-sine means that a full period is 54 ms, or 18.5 Hz.  Compare this with the first natural frequency of the shaft of 580 Hz.  The shock load is much, much lower than the first natural frequency of the shaft. That means it cannot do much to excite vibrations in the shaft.

      There is no ability to upload files on this new site. You need to upload to an external file share site and put the link in your post. You can use sites such as Google Drive, Dropbox or Jumpshare.

      • saravanakumar

        Dear peter,

        yes, it is a full sine curve and it looks like below image.

        the details given in previous message were created just to understand the procedures and output of the analysis results.

        now i have increased the shaft length and the point mass to bring down the natural frequency values.

        i have attached the link of my analysis file.

        can you kindly please go through the analysis procedure and clarify my doubts in previous message.




    • peteroznewman

      What version of ANSYS are you using?

      • saravanakumar

        I m using ANSYS 2021R2 version

    • peteroznewman

      You ask how is it possible to input 10g of acceleration at one point and see more than that somewhere else on the structure. You actually input more than 10g because you are also suddenly turning on 1g at t=0 by including Standard Earth Gravity.  I more refined analysis would allow the shaft to come to static equilibrium with the gravity load, then apply the transient shock.  A simpler approach is to turn off gravity.

      With the sinusoidal input, the peak Y acceleration is 115 m/s^2.
      Without gravity, the peak Y acceleration is                 108 m/s^2

      Another reason is that the mass and stiffness of the structure create amplification of input acceleration to create more or less acceleration at various points along the length. Every dynamics textbook starts with a simple mass, spring, damper system. When the input frequency is close to the first natural frequency of the system, the ratio of the acceleration of the mass to the acceleration of the base is greater than 1.0

      You noticed that the point you applied the full sine curve of acceleration ended up 11 mm higher than it started. Well you didn’t need to run a dynamic simulation to know that. Simply integrate acceleration twice with respect to time to get displacement and you will see exactly the curve for displacement that you plot in the simulation.

      If you want the point to end at the same point it started from, you need a different waveform.  You could use a wavelet that has an amplitude of 10g in one direction, but it first goes in the opposite direction at about -6g before it turns around and hits 10g and has another -6g at the end.

      If you perform an FFT on this acceleration waveform, you will see the peak frequency content is at 37 Hz, which is the same as the full sine curve that was 0.027s long.

      Your final quesiton noted that a displacement BC sets Y=0  in the modal but that point moved in the transient. That is because it was a modal superposition analysis that applied a base acceleration BC on that same point. Base acceleration shakes the boundary condition you told it to, so it moved.

    • peteroznewman
      Here is the wavelet data in units of m/s^2
      0 0
      0.0005 -0.4407
      0.001 -1.7496
      0.0015 -3.8875
      0.002 -6.7899
      0.0025 -10.369
      0.003 -14.515
      0.0035 -19.102
      0.004 -23.985
      0.0045 -29.011
      0.005 -34.017
      0.0055 -38.837
      0.006 -43.307
      0.0065 -47.265
      0.007 -50.561
      0.0075 -53.057
      0.008 -54.630
      0.0085 -55.181
      0.009 -54.629
      0.0095 -52.923
      0.01 -50.034
      0.0105 -45.964
      0.011 -40.742
      0.0115 -34.426
      0.012 -27.100
      0.0125 -18.873
      0.013 -9.8777
      0.0135 -0.2667
      0.014 9.7897
      0.0145 20.108
      0.015 30.497
      0.0155 40.758
      0.016 50.693
      0.0165 60.108
      0.017 68.817
      0.0175 76.648
      0.018 83.442
      0.0185 89.064
      0.019 93.399
      0.0195 96.359
      0.02 97.884
      0.0205 97.944
      0.021 96.537
      0.0215 93.691
      0.022 89.464
      0.0225 83.943
      0.023 77.239
      0.0235 69.487
      0.024 60.843
      0.0245 51.478
      0.025 41.578
      0.0255 31.336
      0.026 20.950
      0.0265 10.619
      0.027 0.5344
      0.0275 -9.119
      0.028 -18.170
      0.0285 -26.465
      0.029 -33.869
      0.0295 -40.270
      0.03 -45.583
      0.0305 -49.748
      0.031 -52.732
      0.0315 -54.534
      0.032 -55.178
      0.0325 -54.714
      0.033 -53.220
      0.0335 -50.795
      0.034 -47.559
      0.0345 -43.649
      0.035 -39.215
      0.0355 -34.417
      0.036 -29.420
      0.0365 -24.390
      0.037 -19.489
      0.0375 -14.873
      0.038 -10.685
      0.0385 -7.0563
      0.039 -4.0954
      0.0395 -1.8928
      0.04 -0.5148
      0.0405 -0.0029
      0.041 0
      0.0415 0
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