Rotor Analysis -- Big deformation problem

AI Thread Summary
The rotor analysis using modal analysis in Ansys shows excessive deformation values, which may be attributed to mass-normalized values rather than actual physical displacements. The discussion clarifies that modal analysis provides eigenvectors, where absolute magnitudes are not meaningful without specified excitation conditions. The analysis of an ideal rotor does not yield a definitive vibration amplitude, as it remains indeterminate without initial or boundary conditions. The conversation emphasizes that real systems include damping, which affects the transient solutions. Understanding these principles is crucial for interpreting modal analysis results accurately.
Stef
Messages
3
Reaction score
0
Hi everyone,
i am currently doing a rotor analysis with "modal analysis" on Ansys and even though my rotor's specs are L=145mm D=10mm and the rotating velocity is 10000rpm i get a deformation of 100mm at most of the modes. I have read that this might be cause of a mass-normalized value but i am not sure if that's the case. Does anyone know why is this happening and if so, can i change it so i can see the real value?
Thanks in advance.
 

Attachments

  • Capture.PNG
    Capture.PNG
    60.2 KB · Views: 468
Engineering news on Phys.org
Modal analysis only gives you the eigenvectors. The absolute magnitudes mean nothing at all, only the relative magnitudes are significant. I suspect that this is what you are seeing; I've seen it myself with FEA modal analysis.
 
Yes it looks like it. Maybe I have to use some other way. Thank you anyway.
 
You speak of wanting to see the "real value." The difficulty with that is that there is no "real value" until you specify the excitation. As long as you analyze an ideal, perfectly balanced rotor, there is not way to evaluate the vibration amplitude; it is indeterminate.

Consider a simple example to clarify this situation. Consider a mass M on a spring with constant K, with position described by x. The equation of motion is
M*DDx + K*x = 0
where D = d/dt.
As you know, the solution is x = A*cos(omega_n*t) + B*sin(omega_n*t)
where
omega_n^2 = K/M
But, and this is the point, A and B cannot be determined without initial or boundary conditions. The solution just described is called the transient (homogeneous) solution. You may wonder why it is called the transient solution when it clearly persists forever.

In the discussion above, damping was omitted, but in all real systems, damping is present. The inclusion of positive damping of any type (viscous, V^2, dry friction, hysteretic, etc) will cause this solution to disappear as the time becomes large.

Now, if we put an excitation term on the right side of the equation, the steady state (particular) solution will have a definite amplitude, but the homogeneous solution is still of undefined amplitude. In many cases, we simply say that enough time has elapsed to cause the transient to disappear.
 
That was very helpful thank you very much for your explanation. Have a nice day.
 
Thread 'Turbocharging carbureted petrol 2 stroke engines'
Hi everyone, online I ve seen some images about 2 stroke carbureted turbo (motorcycle derivation engine). Now.. In the past in this forum some members spoke about turbocharging 2 stroke but not in sufficient detail. The intake and the exhaust are open at the same time and there are no valves like a 4 stroke. But if you search online you can find carbureted 2stroke turbo sled or the Am6 turbo. The question is: Is really possible turbocharge a 2 stroke carburated(NOT EFI)petrol engine and...
I need some assistance with calculating hp requirements for moving a load. - The 4000lb load is resting on ball bearing rails so friction is effectively zero and will be covered by my added power contingencies. Load: 4000lbs Distance to travel: 10 meters. Time to Travel: 7.5 seconds Need to accelerate the load from a stop to a nominal speed then decelerate coming to a stop. My power delivery method will be a gearmotor driving a gear rack. - I suspect the pinion gear to be about 3-4in in...
Thread 'Calculate minimum RPM to self-balance a CMG on two legs'
Here is a photo of a rough drawing of my apparatus that I have built many times and works. I would like to have a formula to give me the RPM necessary for the gyroscope to balance itself on the two legs (screws). I asked Claude to give me a formula and it gave me the following: Let me calculate the required RPM foreffective stabilization. I'll use the principles of gyroscopicprecession and the moment of inertia. First, let's calculate the keyparameters: 1. Moment of inertia of...

Similar threads

Back
Top