Rotordynamics: why does unbalance in rotor only excite forward modes? 1

[trich122]

I’ve seen many anecdotes in papers that suggest that imbalance forces have a component in the forward direction, or unbalance forces can be considered “forward whirl forces” - but I can’t picture it on a FBD. The unbalance force points out radially (to me) and shouldn’t have a component forward or backward.

Anyone have any light to shed on this?

Any help would be greatly appreciated! Thanks!

 

[electricpete]

attach a weight at one location on an otherwise balanced rotor,

the unbalance force points out radially, yes.... but in which direction?

toward the weight. The weight rotates with the rotor. The direction of the unbalanced force rotates (forward) with the rotor.

So the rotation of the unbalance force vector is forward (seems obvious)

Whether it excites only forward modes...I'm not sure about that.


=====================================
(2B)+(2B)' ?

 

[trich122]

My initial interpretation was that any off axis deflection would result in synchronous whirl as the “whirl” would be “driven” by the spin of the shaft that was still on the axis (I.e. at a bearing). However it seems as though whirl and spin can be treated independently (since you can have both forward and backward whirl, and it can be asynchronous) - so if whirl and spin are independent, why would an unbalance force always produce forward whirl? The force does follow the spin of the rotor, but again - are we able to associate spin with whirl?

Just looking for some further clarification here. I’ve read many papers/texts and they all kind of gloss over this fundamental thing. Or maybe I’m overthinking something that is quite simple.

 

[electricpete]

If you are talking about numerical rotodynamic analysis,then the dynamics are captured in the vector equation:

F = KX + MX''
(I'm assuming undamped, no gyroscopic effects to make my life simpler)

The centrifugal force due to non-uniform rotor (no whirl) is captured in the F term and depends on spin. This is separate from the system and considered an external excitation.

The centrifugal force associated with whirling of a uniform rotor (no mass non-uniformity) is captured in the MX''. This is part of the system (M) and system response X''

Since the excitation under this model is purely forward, then no matter what kinds of terms we include on the RHS as coefficients of X or its derivatives, the forced solution will be forward. But the natural solution may be reverse.

That's the formulation that I have seen in textbooks. I haven't given a lot of thought to the aspects you mention. You may be right there may be some significant things glossed over even if the assumptions (linearity) are met.



=====================================
(2B)+(2B)' ?

 

[trich122]

Thank you for the clarification.
I do understand that the forcing function (frequency) will drive the steady state response of the equations of motion.

However I am still stuck on spinning vs. whirling, as the external force in an unbalance analysis is - spinning forward -, not necessarily whirling forward. So I’m still a bit confused as to how backward whirl comes about (for instance, what does the unbalance force vector look like during a backward whirl? - the vector really should be the same)..., yet in some cases there will be a backward whirl induced by unbalance and the “forcing function” will be spinning forward, and the “mass*acceleration” will be whirling backward.... can’t get my head around it.

 

[electricpete]

The above formulation splits the mass into two pieces, a uniform mass, and a mass associated with the unbalance... could physically be something like 1 ounce at 5" from the center.

We assume that 1 ounce at 5" distance from shaft axis is rotating in a 5" radius circle producing a constant centrifugal force, which doesn't change regardless of any rotor whirl/orbit. But in reality that 5" radius circle is pertubed by a few mils as the rotor orbits/whirls. That effect (addition mass acceleration of the unbalance mass due to orbitting/whirling) is clearly neglected in the model I posted. Maybe we can imagine it is negligible since the few mils of pk/pk displacement orbit is much less than the 10" diameter circle path. Then if we start adding bowed rotor things get more complicated as well.

Here's something I can't explain and I think maybe it's the same thing that's bothering you: An overhung disk rotor (with gyroscopic effects) has both forward and reverse modes predicted by modal analysis. And the frequencies of the reverse mode are sometimes observed in the real world during coastdown waterfall plot. It raises a question... what is the exciting those reverse modes during coastdown. By our model I don't think it is unbalance because if we don't have reverse excitation on LHS I don't think we can have reverse response on RHS. Either I am mistaken about my interpretation of the equation, or there is something about the coastdown transient that excites the reverse whirl.


=====================================
(2B)+(2B)' ?

 

[electricpete]

You got me curious, so I went back to the books. "Rotordynamics Prediction in Engineering" by Lalanne, Figure 7.47 on page 177 shows unbalance response predicted for a particular "propfan" at three different speeds. At two of the three speeds, the predicted steady state unbalance response is forward, at a third speed it is backward.

That contradicts what I surmised from inspection of the matrix equations of motion. I dont' have a good explanation... would have to dig into the math (I'm not THAT curious at the moment).


=====================================
(2B)+(2B)' ?

 

[electricpete]

ok, I just thought of a simple way that unbalance (forward rotating excitation) can result in reverse rotating orbit (reverse rotating response).
Since I can't figure out how to indent for organization, I'm going to set it off in a quote box as follows:

Let's say we have a machine which rotates CCW in the x-y plane

Let's say it is a simple enough machine to model as an undamped SDOF in each direciton consisting of rotor mass M and stiffnesses Kx for x direction and Ky for y direction.

There is a resonant frequency for each direction
wx = sqrt(Ky/M), wy = sqrt(Kx/M)

Let's say Kx << Ky, and accordingly wx << wy

Let's say the machine rotational frequency is w and wx << w << wy

So... the x direction SDOF system is excited far above it's resonant frequency ("mass controlled") and the displacement in the x direction (Dx) is approximately 180 degrees opposite the force in the x direction (Fx).

And...the y direction SDOF system is operating far below it's resonant frequency ("spring controlled"), and the displacement Dy in the y direction is in phase with the force Fy in the y direction.

Our unbalance is forward sense (CCW in the xy plane).
So Fy lags Fx by 90 degrees.

Let's use lag phase angle convention and arbitrarily assign the phase of Fx as 0 degrees.
The phase of Fy is then 90 degrees (lagging).

What are the responses Dx and Dy to these forces Fx and Fy?
In the x direction, Dx is 180 opposite from Fx (mass controlled), so the phase of Dx is 180 (180 lag from Fx)

In the y direction, Dy is in phase with Fy (spring controlled), so the phase of Dy is 90 (90 lag from Fx)

Comparing the phase of Dx and Dy, Dx has the greater lag angle. So Dx is lagging 90 degrees behind Dy. That corresponds to a CW orbit (reverse rotating).

The flaw of my previous logic concerning the matrix equations of motion was that I imagined we could just use a phasor approach, multiplying the force by something like exp(i*(theta-wt)) and exp(i*(theta+wt)) to capture the forward and reverse aspects. But I was really mixing things up.... that's not the way it works. The state vector X includes both x and y components, and the phase relation between these components gives us forwards or backwards rotation sense. I'm sorry that it took me awhile to get there.

=========
By the way, above I am referring to the single rotation sense that we would assign to the single frequency orbit tracing it's path in the x / y plane. Distinct from this, there is a transformation of a given single-frequency orbit or spectrum into forward and reverse components (search for "full spectrum"). That's a completely different take on forward and backward than I'm using.


=====================================
(2B)+(2B)' ?

 

[trich122]

Thank you for that explanation, it is quite clear that the phase shift associated with passing the first critical speed in 1 direction (for asymmetric bearing stiffness) will cause synchronous reverse whirl leading up to the second critical speed (in the opposing direction).

I think the way i can rationalize it in my head is as follows:

In absence of bearing support stiffness asymmetry, and nonlinear or cross-coupling forces, the rotor will whirl synchronously in the forward direction, because at the bearings where the shaft is effectively "pinned" it is spinning in the forward direction, which drives the remainder of the shaft in the same direction when travelling off-center.