Dynamic Mechanics - Frequency Response of the Rotating Pendulum

Thomas_Bécue · April 14

Hello,
I'm trying to understand the frequency response of an articulated rotor.

To simplicity, I modeled a simple pendulum with centrifugal rotation (see image).
I calculated a transient + linear simulation using a Templates script to take in account centrifugal loads in my linear analysis :
I expected to obtain a frequency response of the first mode equal to the centrifuge's rotation frequency (34 Hz). But this frequency is different (22Hz).

Do you know if this is a mechanical or solver issue?

Thanks in advance,
Thomas

Ps : my template's script :

Orestis_YB · April 15

Hello @Thomas_Bécue,

as described in the helicopter example (attached above) coordinate selection will affect frequency result. In your model I added a solver variable using the expression DZ({b_masse.cm.idstring} and I added a template to set this variable as Plant State.

I run the model and in the log file you can see that this variable was chosen as coordinate and the calculated frequency is 33.89Hz.

Hope this helps!

Best regards,
Orestes

penduleV2_upadted.mdl

Uploaded Apr 15, 20252.98 KB

Orestis_YB · April 14

Hello @Thomas_Bécue,

from the shared picture I see that you are using a bushing next to the revolute. Are the stiffness properties there set properly? MotionView default units are Nmm/rad. How did you calculate the 34 Hz? Dis you include the bushing in those calculations?

Best regards,
Orestes

Thomas_Bécue · April 15

Hi @Orestis_YB

Thank you for your answer.
The 34Hz was a random rotation frequency.
I'have simplified the system to have a classic simple pendulum :

A 1kg punctual mass, with no inertia, at 1000 mm from the Ground_body in the x global direction.
The mass is linked to the Ground_body with a revolut joint in the y global direction.
The gravity force is applied : -9810 mm/s² in the z global direction.

In theory, the frequency of the pendulum is : Freq = (1/(2Pi)) *sqrt(g/l) = 0.4985 Hz

But in MotionView, I have 0.2912 Hz.

Have a good day,

Thomas

Orestis_YB · April 15

Hello @Thomas_Bécue,

as described in Linear Simulation MotionSolve will linearize the model to calculate the eigenvalues and eigenmodes.

In the case of the rotating pendulum we are dealing with a highly non-linear model. If you start from an arbitrary position it will affect the calculation. If you set the pendulum in the stable equilibrium we'll see that the eigen frequency is exactly the theoretical one.

Hope this helps!

Best regards,
Orestes

Thomas_Bécue · April 15

Oh sorry @Orestis_YB, you're right, for the simple pendulum, the problem was that i was not in the good initial condition:

The formula Freq = (1/(2Pi)) *sqrt(g/l) = 0.4985 Hz, is correct went small angle are supposed (10 deg). My inertial condition was in perpendicular so it's incorrect. I putted it at 10 deg, and i found a frequency of 0.49 Hz so it's good. Thanks !

If i go back to my rotating pendulum, i still have a problem:
-Gravity disabled
-A rotor body with no mass rotating at 34 Hz (213 rad/s) is linked to the Ground
-A mass of 1kg at 1m from rotation axis, is linked to the rotor body with a revolut joint (pendulum articulation).

In my template for simulation, I first run a Transient simulation to take the system centrifuge states after stabilization, and then, run a Linear simulation to eigen frequency.
But my frequency is equal to 0 Hz
Do you think this method is correct ?

Thanks for your help,

Thomas

Orestis_YB · April 15

Hello @Thomas_Bécue,

Could you please share the model? If you are not using any external file then sharing just the mdl will work just fine. In the mean time here you can see some information regarding PlantState. Reference_PlantState defines a list of user-defined states used in generating a linear representation of a model about an operating point. The linear representation is used for both eigenvalue analysis and state matrix generation. If you scroll down there is the example of a helicopter rotor which is pretty similar to what you are describing. When defining a Reference_PlantState using a solver variable you need to also define a parameter for the Linear solver. If you share the model I could give it a look and share further information.

Best regards,
Orestes

Thomas_Bécue · April 15

Hi,
Here is my model.
Ok, i will check this ! Thanks!

penduleV2.mdl

Uploaded Apr 15, 20252.46 KB

Best regards,

Thomas

Orestis_YB · April 15

Hello @Thomas_Bécue,

as described in the helicopter example (attached above) coordinate selection will affect frequency result. In your model I added a solver variable using the expression DZ({b_masse.cm.idstring} and I added a template to set this variable as Plant State.