Let's say we have a lightly damped 2DOF system as follows:

Ground --- K1 ---- M1 ---- K2----- M2
(damping in parallel with K1 and K2 not shown).

Force is assumed to act on mass M2

For a rotating machine, this could be:
M2 = rotor mass
K2 = rotor and bearing stiffness
M1 = mass of stator that moves with bearing housing
K1 = stiffness of stator supporting bearing housing

For a 2DOF system, we will have two resonant frequencies: call them w1 (lower) and w2 (higher).

Also let's define
L1 = lag angle of position mass M1 with respect to force F
L2 = lag angle of position mass M2 with respect to force F

Now, let's try to piece together the phase response L1 and L2 of the system using intuition:

We know the following items:
Item 1 - w1 represents motion of the two masses in-phase with each other => L1 ~ L2 on both sides of w1
Item 2 - w2 represents motion of the two masses out-of-phase with each other => L1 ~ L2 +/- 180 degrees on both sides of w2
Item 3 - Far below w1, we would expect L1 and L1 ~ 0 since this is the spring-controlled region.
Item 4 - For lightly-damped system, we might expect that as we change frequency, phase doesn't change significantly unless we are close to an important system frequency like a resonant frequency.

So start with L1=L2=0 (item 1) and slowly increase frequency. They would stay constant over most of the interval (item 4), until we get near w1, at which point presumably increase together (L1 and L2 both incrase but remain equal to each other), up until we pass just above w1 (item 2). Now between w1 and w2 we might expect no change because we are not near a resonance (item 4). But by the time we get to the neighborhood of w2 we need to have L1 180 degrees away from L2 (item 3).

Sounds like a contradiction, doesn't it? If it doesn't seem like a contradition, draw a picture of L1 and L2 and see if you can generate L1 and L2 to match items 1-4. The problem is trying to draw the area between w1 and w2 with L1 and L2 equal on the left-hand side, L1 and L2 180-opposite on the right-hand side, and no change in L1 and L2 during the middle of this interval... can't be done.

There is a logical mathematical explanation. It is sort of related to item 4. Would anyone like to take a stab?

This message has been edited. Last edited by: electricpete,

We're you contemplating this while cutting the grass again?

Good one. I hope this thread will get as much response as that crawdad piles one.

But I kind of doubt it. The motivation for this question might not make sense unless you are a math geek.

But I still think the answer and the shape of the phase plots will still be interesting to everyone.

Hint: Attached is a plot of phase angles that would be expected using the following exaample values:
m1 = 1, m2 = 2, k1 = 3, k2 = 5, c1 = 0.1e-1, c2 = 0.1e-1

The magnitudes (not shown) behave as expected. There are peaks at resonant frequencies w1 and w2.

The phase angles behave as expected at resonant frequencies w1 and w2. Specifically, there with 180 phase increase in both L1 and L2 crossing either one of those resonant freuqencies.

But there is an unusual transition in the phase angle L2 at a third frequency labeled w? which is in between w1 and w2. The quiz: What causes that transition?

This message has been edited. Last edited by: electricpete,

TwoMassSpringPhase.ppt (37 Kb, 25 downloads)

If you think of this modally it is simpler. Write the force distribution in terms of modal forces

F=[0;1] force on m2

F= F1+F2 where F1 excites only the first mode and F2 excites only the second mode. We know this can be done from theory, assuming modal damping (approximately so for light damping).

------ VERY LONG ASIDE -------
(For your particular example the forces should be the following [up to my errors].
K=[8 -5
-5 5] written as [8 -5;-5 5] for brevity

M=[1 0;0 2]

With modes e1=0.8779987 (sqrt of eigenvalue) and
e2 = 3.1191534 approximately

Modeshapes,
x1 =[0.4393403; 0.6352087]
x2=[-0.8983207;0.3106605]
The modes are orthogonal (actually orthonormal) with respect to the mass matrix, i.e.
x1’*M*x2=0, etc.

F1= [0.5581456;0.8069801] force and phase relation for m1 and m2 respectively
F2=[-0.5581456;0.1930199]

F1 should be orthogonal to the second mode and F2 should be orthogonal to the 1st mode.


…. Long winded, but modal forces can be found)
-----------------------

This is then a modal superposition problem. F1 draws a modal (polar plot) circle starting in the modal direction of F1 (x1). F2 draws a modal circle in the modal direction corresponding to it, i.e. x2.

The polar plots are modeshape dependent and modally damping dependent (how much response at resonance).

If we are talking about constant forces (not speed squared), then the first circle heads to 180 degrees, but doesn’t get there because we didn’t let the forcing frequency get to infinity. The second modal force wants to modally go towards 0 degrees or draw the second mass to 0 degrees and the first mass towards 180 degrees.

Thus the polar plot for the second mass does a loop, two modal circles in phase. The first mass does the figure 8.

If one does this for a tuned vibration absorber (usually done with force on mass one of your configuration, but the principle is similar.) the polar circle for the mass with the force goes through the origin; in control terms it has a zero.

I should have known that would be an easy one for you Bill.

I solved it differently than you did. (It would be a good excercize for me to try it as an eigenvalue problem like you did).

I used the two equations of motion, took the fourier transform of those equations, and ended up with two equations in two unknowns X1(w) and X2(w) which can be solved algebraically. Then I calculated the transfer function H1(w)=X1(w)/F(w) and H2=X2(w)/F(w). Then plotted the magnitudes and phases of the transfer function (actually I took the negative of the phase of the transfer function before I plotted it, in order to turn it into a lag angle).

The results are linked in an attached pdf (sorry - it's 2MB). If the math is not interesting for you folks, the plots of magnitudes and lag angles are at the end.

I found w1 = 0.88 and w2=3.1. Bill your results are different on w1, I'm not sure why (you mentioned an error but I'm not sure what it is).

Examination of the numerator of the transfer function X2(w) shows it has a zero at wz=2.8.

That corresponds to w=sqrt(k1+k2/m1)

To physically interpret it, I agree with your analogy to dynamic absorber. In that case (dynamic absorber) we apply force on m1 and at the tuned frequency create a zero of movement at m1 at a frequency which corresponds roughly (I think) to resonant frequency of m2 with mass 1 held still (it must be still at a zero), which would be sqrt(k2/m2).

In this case (2DOF stator/rotor), we apply the force on m2 and create a zero of movement at m2 at a frequency which corresponds to the resonant frequency of m1 with m2 held still. With m2 held still, then m1 is supported from rigid base below by k1 and from stationary mass above by k2. Parallel combination of these two springs is k1+k2. The resonant frequency of m1 on these springs would be wz=sqrt((k1+k2)/m1).

Now here is a difference between those two cases. In the 2DOF stator/rotor case, the resonant behavior of m1 at the zero of m2 does NOT show up of excessive amplitude of m1. That makes some sense because the force is applied to m2 and m2 isn't moving...so the force can't get through to m1. But in the case of dynamic absorber, the resonance of mass m2 DOES show up as movement of m2 at that resonant frequency.... even though the force is applied in that case at m1 which is stationary... so I'm not sure how the force gets thru to m2 and why these two systems seem different in that respect.

** The really interesting part for me is that you would never guess there was a zero (of the 2DOF rotor/stator syste) looking at the amplitudes (ok...X2 does go to 0, but it was close to 0 anyway), but yet that hidden zero has a big effect on the phase. I am used to thinking about resonances when I look at phase plots but not used to thinking about zero's. If the system is acting like a discrete multi-dof system, then the zero's could have a dramatic effect on the phases. I'm not sure whether any real machines act like that.

TwoMassTwoSpring.pdf (2,266 Kb, 16 downloads)

Mine seem to satisfy the equation
-omegai^2 M Xi + K Xi = 0 which is the equation for the eigenvalues (also Fourier transform of equation).

e1=0.8779987 which is approximately 0.88.?

This concept if very important when balancing. Proper interpretation of the polar plot will help placing a balance weight on the correct end. If you take a symmetric rotor (or somewhat symmetric) operating between the first and second modes like many petrochemical machines, the end with the high vibration may not be the end to place a balance weight if there is primarily imbalance at one end. The zero (dynamic absorber analogy) makes the end with the force have the lower vibration.

Power generation turbine generators act this way too. One has to determine if it is best to try to balance two modes with a single ended shot or if more weight is needed to balance a particular mode (usually the second, because the first requires too much weight to be effective in the end planes.). If you balance often this senario will arise.

This message has been edited. Last edited by: William_C._Foiles,

If you get rid of damping for a second, it is easy to find the transfer function.

K=[8 -5
-5 5] written as [8 -5;-5 5] for brevity

M=[1 0;0 2]

Let L=-w^2 – frequency squared

The steady state equation of motion at frequency w (Fourier transform if you like) is.

L M +K =F – with forces

A=LM+K=
[ 8+L -5
-5 5+2L]

The transfer function is the inverse of A above. To get the inverse we use Crammer’s rule

Inv(A)= 1/det(A) [5+2L 5
5 8+L]

The det(A) gives the poles solving (8+L)(5+2L)-25 as in previous post using eigenvalue routine.

The zero with a force at 1 appears to be L=2.5 or w=sqrt(-2.5)=1.58i

The zero with a force at 2 is L=8 or w=sqrt(-8)=2.828i.

Not difficult enough to need symbolic calculation.
--------

I think I mis-spelled Cramer if anyone looks it up, but I didn't check.

This message has been edited. Last edited by: William_C._Foiles,

I was mistaken before, there is no difference between our results on the poles.

Also no difference in the zero's: One zero for the rotor/stator system with force on m2, the zero occurs at w=2.8 = sqrt((k1+k2)/m1). The other zero for force applied at m1, you found at w=1.58 ... I didn't calculate it, but it works out to sqrt(k2/m2) as I thought.

I went back and solved the poles with an eigenvalue problem. That part was pretty easy. I didn't follow how you solved the zero's using a matrix or eigenvalue problem, but it's easy enough for me to do it with a transfer function.

In your final message, you came up with an imaginary frequency. But w should be real of course (s=j*w would be imaginary). Maybe there was a minus sign error somewhere between L and w^2?

As far as symbolic vs numeric solution - I would suggest that's a matter of preference. The symbolic program told me that the zero occurs at sqrt((k1+k2)/m1). While that may be somewhat logical in retrospect, I don't think I ever would have figured that out solving it numerically. Besides, it's kind of like having a favorite tool that's a hammer... it doesn't matter what needs to be fixed, you just try fixing it with a hammer first ;-)

I gave the actual pole which lies on the imaginary axis if undamped. The real assumption in the FT is iw; I stuck in the i (i^2=-1).

When you add damping the zero has both imaginary and real components, which means that the forced response never really gets to zero - only close. The undamped solution has a zero ont he imaginary axis, and thus has an actual zero that is achievable - although an undamped system isn't achievable.

Epete,

You have H21 with a zero (with damping yes, but light damping it is at a high frequency). I got H22 to have the zero, which is what your plots state - force on mass 2 creating zero response on mass 2.

--------------

This occurs in rotor systems regularly (ok, generally flexible rotor systems and not every time, but if it happens one can get lucky. Don't we all like to get lucky? I ran a balance lab last year that I set up with 2 modes and imbalance on one plane. I let the students put a weight on the wrong end, because this end had the high vibration. Well ... they bent my rotor when it went through the first critical, ouch. This was not exactly the lesson I had hoped for; we now had to balance a bent rotor (first mode) with another imbalance.).

Almost every time, when people see this type of balance issue, they attack the wrong end. A little training might prevent this on the real thing.

There is only one force in my system (on mass 2). H21 is response of mass 2 to that force (at mass 2). The 1 stands for "system1" which was the example system values I plugged in. (That was probably not the best choice of terminology). So it would correspond to what you're calling H22.

I can imagine the behavior you mentioned could make balancing very tricky.

This message has been edited. Last edited by: electricpete,