non-linear spring as mechanism to limit "resonant" response? 4

[electricpete]

We are installing a Bendix coupling in place of gear coupling in a turbine-driven pump application at or plant.

I am not directly involved, but did have a chance to talk it over with the responsible engineer and it brought up some interesting aspects that I wasn't familiar with.

The turbine and pump have thrust bearings. An axial resonant frequency was calculated under assumption that both shafts are fixed and the coupling mass moves on it's own springiness (diagrpham) between them. If each half has stiffness K, the total stiffness of SDOF model would be K/2 since it is connected to "ground" on both sides.

Quasi-physical model:
Ground ===K===Mass===K===Ground

SDOF Model
Ground ===K/2===Mass

The resonant frequency of the coupling as calculated above falls within the operating speed range. But it is not considered a problem because:
1 - There is not a lot of axial excitation at running speed to be found. (open to comment.. I'm thinking misalignment may create an axial forcing function)
2 - The coupling is axially flexible and for low levels of forces and deflections (far below any endurance limits) it enters the non-linear region of the "spring" (diaphram). Apparently a stiffening spring creates a classic response vs frequency profile shaped like a shark's fin (with the pointy part of the fin pointing toward higher frequencies for stiffening spring and lower frequencis for softening spring). Mechanial Vibration's by Rao discusses this in terms of Duffing's equation. Den Hartog also discusses it in a simpler way.

I can see the shape of this non-linear system response curve is much different than a linear resonant SDOF system and I can see the idea of "resonance" doesn't apply to this non-linear SDOF in exactly the same way as a linear system. I can't particularly judge how effective it is in limiting max vibration magnitude for a system with only structural damping (no fluid or elastomeric damping).

Above is a lot of new concepts for me, I'd be interested in any comments. In particular:
Can anyone provide discussion or link of ability of non-linear spring to limit max vib amplitude for a given forcing function magnitude, variable frequency?
If this is an effective means to limit vibration magnitude, why isn't it used more often or discussed within "vibration control" section of textbooks?













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

 

[electricpete]

The resonant frequency of the coupling as calculated above falls within the operating speed range.

That is resonant frequency calculated using the low-deflection stiffness. It is within the operating speed range and not at the top of the operating speed range.

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

 

[desertfox]

Hi electricpete

A bit of bedtime reading here:-

http://artec-machine.com/_images/documents/Application_of_Flexible_Couplings_for_Turbomachinery.pdf

regards

desertfox

 

[electricpete]

Thanks. The Engineer I talked to had a hardcopy of that article in his hands when we talked about it. Now I’ve got a copy.

The Bendix coupling that we’re using is the “contoured diaphragm” as shown in Figure 24 on page 10.

DETOUR - An interesting side-note, the 1/ri^2 contour shown there on page 24 is also shown in our literature and I was curious about it since I initially wondered whether this feature was intentionally used to create the non-linear stiffness, but it is not…..it is trying to equalize stresses as stated. That raises another question. I was remembering that for a shaft transfmitting torque, the highest resulting stress is at the od (so why would we make it thinnest). The reason is different is that the shaft transmits torque axially (with more of the heavy lifting done by the outer radius elements). The diaphragm transmits the torque / force radially from OD to ID and ID to OD. So a cylinder at each radius must transmit the entire torque.
Torque = F*r = tau*A*r = tau*2*pi*r*L*r = tau*2*pi*L*r^2
solve for shear stress tau:
tau = Torque / 2*pi*L*r^2
So if the profile is 1/r^2, the shear stress tends to be same throughout.
END OF DETOUR


Page 18/19 have relevant discussion:

AXIAL NATURAL FREQUENCY
In the introduction to the section on CRITICAL SPEEDS, it was mentioned that flexible element couplings have an axial natural frequency which gear couplings do not have. This axial natural frequency (ANF) is treated as ifit were a critical speed. The ANF results from the physics of supporting a mass, (the weight of the spacer section of the coupling), between two springs, (the coupling flexible elements) (Figure 56).

A coupling will vibrate at its ANF only if it is excited by the connected machines. Merely rotating at the same RPM as the ANF will not cause a coupling to vibrate. Although one can make a coupling vibrate at its AN F on a laboratory' vibration machine, it is not easy to excite it enough to get large amplitudes because of some non-linearity exist in most of the flexure stiffness's (some small and some very large) . Many flexible element couplings have been operating for years at speeds at or near to their ANF but no field problems with couplings on their connected machines have been traced to a coupling vibrating at its ANF.
The API Standard 671 requires that couplings be designed
such that the AN F "shall not fall within ten percent of any of the
following:
• Any speed within the range from the minimum allowable
speed to the maximum continuous speed.
• Two times any speed within the above speed range.
• Any other speed or exciting frequency specified by the
purchaser. "
The purpose of forbidding the ANF to two times an operating speed is to prevent vibration occurring on those machines which generate a 2x axial excitation, for example, from a warped thrust runner. There are occasions when a coupling cannot be designed to place the ANF either wholly below or wholly above the forbidden speed ranges. In those cases where the ANF encroaches on part of the forbidden range, the coupling designer may be able to make use of the change in ANF with axial deflection or incorporate some type of damping of the coupling to have the encroachment only occur at a seldom used operating speed such as during a startup transient (Figure 57).

The first bolded sentence implies the non-linearity limits the amplitude.
The second bolded sentence is a different twist, that the non-linearity might be used to alter the “resonant” frequency to a range that is not of concern (not the case for us, the small-deflection resonance falls within our normal speed range and not near the top of it

Interesting that figure 57 of the Kopflex / Mancuso paper is apparently supposed to represent the center curve of Den Hartog’s figure 262 (I have attached the entire relevant portion of Den Hartog, not just for this figure but for general discussion of stiffening spring behavior). Of course that center curve represents idealization for zero excitation which I’m sure is not what is of interest. ok, we’ll cut them some slack in simplifying the figure to omit features not of interest with respect to shifting the frequency.

An important thing the paper is the italicized portion which states that based on experience, even operating at the axial natural resonance doesn't seem to cause problems.

So the bottom line - I really don't have a concern for the design (at least from a distance and not being responsible), but again I'm intrigued by this non-linear explanation of the first bolded sentence. Is it really true and if so why isn't this design principle used elsewhere (using non-linear spring to prevent high resonant-like vibration for machine which may be subject to a range of exciting frequencies)?



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

 

[electricpete]

(I have attached the entire relevant portion of Den Hartog, not just for this figure but for general discussion of stiffening spring behavior)

Forgot attachment - here it is

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

 

[MikeHalloran]

Substituting a non-linear spring for a linear one in a given system might reduce the response at any particular frequency, but I can't see it reducing the peak amplitude significantly, just changing the peak frequency, and not affecting the damping.

I agree that for a diaphragm type coupling, any coupling between torsional/ radial/ axial modes has to be tempered/modulated/magnified by misalignment. ... so remove all possible misalignment. ... which you would do anyway.

<tangent>
The worst example of coupling between torsional and axial modes is found in the Heli-Cal coupling, where torque transmission by means of interleaved helical springs cut from the solid also causes a change in length of the coupling itself.

They don't exactly highlight this behavior in the catalog. You probably wouldn't notice it in many applications of the otherwise fine device. It came to my attention after I designed one into a medical instrument where the thrust bearing for a piston-driving screw was supported by a plastic diaphragm, and the torque was supplied by a stepping motor. Buzzing noises are not easily accepted as normal for medical equipment used in quiet offices.
</tangent>


Mike Halloran
Pembroke Pines, FL, USA

 

[electricpete]

Is it really true and if so why isn't this design principle used elsewhere (using non-linear spring to prevent high resonant-like vibration for machine which may be subject to a range of exciting frequencies)?

I guess the starting point would be understanding what it is about this coupling design that makes it a non-linear spring (it may not be possible to apply that in more traditional rigid structures). I'm not sure what gives it the non-linear characteristic. The diaphram is designed to transmit torque with low axial stiffness (so axial growth etc don't significantly load either thrust bearing).


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

 

[electricpete]

So if the profile is 1/r^2, the shear stress tends to be same throughout.

should have been

So if the profile is L[r) = 1/r^2, the shear stress tends to be same throughout.

I agree that for a diaphragm type coupling, any coupling between torsional/ radial/ axial modes has to be tempered/modulated/magnified by misalignment. ... so remove all possible misalignment. ... which you would do anyway.

In this particular case, the pump and turbine are on independent platforms and each carry a high temperature fluid. So thermal growth of machine and support are big factors complicating the alignbment. Machine is intentionally aligned way off at room temperature with intent of growing into alignment at normal operating temperature. At a minimum it operates with misalignment during startup periods and with all the SWAG factors built into thermal growth calcuations, I'm not sure about what it's really like at operating temp.. I do know we have axial prox probes that feeds a thrust position channel on our computer. It would be interesting to look at that signal to see if there is 1x variation on it... I'm not sure if I can get to that data.

Substituting a non-linear spring for a linear one in a given system might reduce the response at any particular frequency, but I can't see it reducing the peak amplitude significantly, just changing the peak frequency, and not affecting the damping.

That is the central question of this thread. Thanks for your opinion. I'm still pondering..




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

 

[electricpete]

I guess the starting point would be understanding what it is about this coupling design that makes it a non-linear spring

That's a silly way to say it since any structure goes non-linear if the forces get high enough. Let me correct that question:

I guess the starting point would be understanding what it is about this coupling design that makes it a non-linear spring at normal operating forces and deflections

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

 

[MikeHalloran]

>>>In this particular case, the pump and turbine are on independent platforms and each carry a high temperature fluid. So thermal growth of machine and support are big factors complicating the alignbment. <<<

Too late now, but if they were each mounted on a hinged platform, and the platforms were linked to each other with hinges whose centerlines intersected the center of the coupling, at least some of that thermal misalignment would go away.

... but that might force you to use flexible joints or long loops in the piping, which you probably want to avoid.

Or, you use a double coupling with short intermediate shaft. ... as I think you are intending.





Mike Halloran
Pembroke Pines, FL, USA

 

[electricpete]

Or, you use a double coupling with short intermediate shaft. ... as I think you are intending.

Yes, there is a short spacer element in the axial center of coupling, supported in part by those flexibile elements (diaphgrams) on each end.

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

 

[GregLocock]

We call them snubbers. Often the force deflection curve can be approximated as a cubic. As such there literally isn't enough force available at any frequency to exceed the desired travel. Mathematically they are a nightmare in the frequency domain, in the time domain we model them at about 1000 times the likely forcing frequency. The problems in analysis come about because a preload will change the effective average stiffness at a given frequency.





Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[Tmoose]

I think I've heard that one of the advantages of the "beehive" springs popular in automotive valvetrains these days is better resistance to spring surge/resonance. Made some sense based on the stuff I'd read previously similar to what e-pete posted.

The first B-hives I saw were on Buick/Olds radical aluminum V8 in the early 60s, but I always figured that was more a continuation of the heroic weight saving effort on that engine.

 

[electricpete]

Thanks Greg and Dan. I had a feeling that if the principle was valid then it must appear in other applications. Do you or anyone have a link to simple discussion of the mechanism by which snubbers, beehive springs or similar devices use stiffening spring behavior to limit max displacement at any frequency?

I did attempt to think through the question using the Duffing's approximate solution to undamped SDOF with cubic spring. I starrted with Rao's textbook approximate solution, rearranged it to suit my preference, and plotted it. Conclusions below, open to comment.

Rao's Mechanical Vibrations 2nd Ed describes the problem as follows:
Frao*cos(w*t) = x'' + w0^2*x + alpha*x^3 [eq 13.44]
where x'' = 2nd derivative of x with respect to time
The approximate solution (implicit relationship between amplitude A and frequency w) is given as:
w^2 = w0 + 3/4*A^2*alpha - Frao/A [eq 13.50]
where
w0 = sqrt(K/M) (sdof)
A = max Amplitude of displacement x(t)


I prefer not to use the symbols alpha and Frao (I don't like Frao because the units are not force; and I don't like alpha because it's not clear exactly what it means).

So to get rid of them, I'll define instead Kc = alpha * M and F = Frao * M
where F is force.

Substituting these definitions into 13.44 leads to more familiar forms:
F*cos(w*t)/M = x'' + w0^2*x + (Kc/M)*x^3
multiply by M
F*cos(w*t) = M*x'' + K*x + Kc*x^3
We can now see that the spring would satisfy
Fspring = K*x + Kc*x^3
and the meaning of Kc as a cubic stiffness term is now clear

Now let's rewrite the entire thing in dimensionless form.
Select "base" quantities: M, w0, F0 and express all other variables as Unitless ratios of these three base quantities (subscript u for Unitless):
wu = w/w0
Fu = F/F0
Xu(t) = X(t)*M*w0^2 / F0
Au = A*M*w0^2 / F0
Kcu = Kc * F0^2 / (M^3*w0^6)

Then the differential equation becomes:
Fu*cos(wu*t) = Xu'' + Xu + Kcu*Xu^3
And the corresponding approximate solution becomes:
wu^2 = 1 + 3/4*Au^2*Kcu - Fu/Au

This form doesn't look much different than Rao's original, but it has one less parameter (w0 is gone…buried in the definitions of the remaining unitless quantities). Also this form has the advantage that all possible approximate solutions can be represented with one family of curves in the Au vs wu plane, paramterized by Kcu and Fu. Side note - we also could get rid of yet another variable (Fu) by declaring that the base (normalizing) force F0 will always be set equal to applied force F (so Fu=1 always), but we choose not to do this - see asterisk (*) below.

Slide 1 shows a family of three curves for Kcu = 0.03, 0.1, 0.3
(Kcu>0 represents stiffening sytem rather than softening system)
Also added on the same plot for comparision is the more familiar solution of dimensionless damped linear SDOF
|Au| = Fu/sqrt([wu^2-1]^2 + [2*zeta*wu]^2)
where zeta is damping factor

For low frequencies, the behavior is similar to a linear system. For high frequencies, the lower-magnitude branch of the cubic system is similar to linear system. In between (near "resonance"), it gives the general impression that we take the linear undamped system and drag the peak toward the right. Instead of approaching a vertical asymptote for undamped system, there is now a sloped asymptote for the cubic system. Higher Kcu is a smaller slope in the Au vs wu plane.

* In theory, we could also deduce effect of changing force for a given system from slide 1 by altering F0. Fu would stay as 1, but in looking at a given system with fixed Kc we would have to change our Kcu value when we change F0. And also when interpretting results the translation of Au back into A would vary if we changed F0. The results is well defined from slide 1, but cumbersome to envision.

Therefore, in slide 2 I took the more straightforward approach and kept F0 the same (preserving base quantities defining Kcu and Au) and varied Fu = F/F0.

On the linear system, we see expected results, factor of 3 change in Fu causes factor of 3 change in response at all frequencies. For the cubic system with this particular Kcu (Kcu = 0.1), factor of 3 change in Fu causes much less than factor of 3 change in amplitude in the "resonant" region.

What is practical meaning?
Looking back to slide 1, if Kcu is high and/or we don't go far above resonant frequency, the magnitude doesn't grow very high. That may cover our coupling situation since operating range doesn't go too far above coupling resonant frequency. Even if excitation were to increase far above resonant frequency, apparently if some some tiny non-zero amount of damping is present, the system will at high enough frequency eventually transition from the upper value to the lower value for that frequency.

I'm also still curious about features of this coupling that lead to this non-linear/stiffening axial spring constant within the normal operating range. It's built out of steel.


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

 

[Tmoose]

Simply parrot repeated benefits of bee-hive springs

http://www.lunatipower.com/ProductGroup.aspx?id=192&cid=23

http://www.enginelabs.com/engine-tech/beehive-springs-sound-great-but-will-they-work-for-you/

================

David Vizard is a most prolific publisher of engine tech articles and books. Much of what he publishes appears to be based on his own testing. Here he says that in one test replacing standard cylindrical wound springs with slightly softer beehive springs controlled valve motion a most significant 1000 rpm higher. Normally stiffer springs are the means to higher rpm control

http://www.crankshaftcoalition.com/wiki/Valve_spring_tech

Whether simply due to weight reduction, or magic damping is unclear.

In the late 50s Chevy introduced valve springs with friction dampers to buy some reliability (broken valves!) that had been lost to spring surge or perhaps valve float on a passenger car engine.

 

[MikeHalloran]

I believe the beehive springs resist surge because each different diameter turn has a different natural frequency, so it's just flat not possible to excite the entire spring into axial resonance at any single frequency.

That's not quite the same as having damping effective over the operating frequency range, but it has to give a similar response, and without giving up so much heat, and without worrying about the dynamics of the damper, or having to deal with the wear products of the damper/spring interaction.







Mike Halloran
Pembroke Pines, FL, USA

 

[GregLocock]

That can't be the explanation, there will be some axial resonant frequency of the spring, even if it is hard to work out exactly what it is.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[electricpete]

I’m going to argue the other side – you guys tell me where I’m wrong.

Resonance is well-known and defined for linear systems (and non-linear systems operating in a linear range). It occurs for SDOF because there ealways exists a frequency at which force to accelerate a mass through a given sinusoidal motion is equal/opposite to the spring force associated with that sinudoidal motion.

What about non-linear systems? We know their behavior is not always as predictable and intuitive. I’m sure we have seen some non-linear systems that exhibit resonance-like symptoms, but that doesn’t mean all non-linear systems will. The whole idea of “out of phase” gets more complicated with non-linear system because even if we apply a sinusoidal force, the response will be non-sinusoidal. How do we define phase for non-sinusoidal signal? Most likely (but not guaranteed) the response will be periodic with the same period as the original sinusoid, but will have harmonics at 1x, 3x, 5x etc.

x(t) = A1*sin(w1*t+theta1) + A3*sin(3*w1+theta3) + A5*sin(3*w1+theta5)
Fspring = k*x(t)

x’’(t) = -w1^2*A1*sin(w1*t+theta1) + -9*w1^2*A3*sin(3*w1+theta3) + -25*w1^2*A5*sin(5*w1+theta5)
Fmass_accel = m*x’’(t)

Using above expressions, let's solve for what frequency causes Fspring = -Fmass_accel?
We have to cancel each of the frequencies
Let's start with canceling the fundamental.
k* A1*sin(w*t+theta1) = M*w1^2*A*sin(w*t+theta1)
w = sqrt(k/m). ok, we knew that.

Now see if this frequency cancels the harmonics.
K* A3*sin(3*w1+theta3) ?=? M *9*(K/M)*A3*sin(3*w1+theta3)
1 ?=? 9 Nope, doesn’t work.

K* A5*sin(5*w1+theta5) ?=? M 25*(K/M)*A5*sin(5*w1+theta3)
1 ? =? 25. Nope, doesn’t work. If we select the frequency that canceled the fundamental, it won't cancel the harmonics.

There is NO frequency that will cause Fspring = - Fmass

The magical frequency where x(t) ~ -x’’(t) does not exist for non-sinusoidal periodic signals. Mass and spring forces won’t cancel if the motion is non-sinusoidal

Maybe I have oversimplified in setting up the problem? It would not be the first time - feel free to point it out.

Also, the Duffing result seems well accepted (even though approximate). That curve really doesn’t show a resonant frequency. It shows that as we increase frequency above wu = 1 (w=w0, the magnitude of response will continue to increase monotonically, until we reach some “jump” frequency where the system will transition to the lower-magnitude path (also depending on magnitude of excitation).


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

 

[electricpete]

Maybe I have oversimplified in setting up the problem?

Yes, I did miss something. The spring force will not be K*x(t).
It might be something like K*x(t) + Kc*x(t)^3
That makes it harder to analyse.

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

 

[GregLocock]

I've just looked up resonant frequency of beehive springs and can confidently say that the vast majority of posters on the subject are quoting each other and don't have the faintest idea of the truth. However saying that I don't particularly know why beehive springs are so good for spring surge, it may be the the change in length (and hence rate and hence frequency) during one cycle of operation prevents the resonance from building up.

I'll have a go at working out the resonant frequency of a 2 stage spring today.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?