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Shaft Critical Speed Considerations

  1. May 7, 2013 #1
    Hi all I'm a bit confused on shaft critical speed theories and I need some clarification. I'll just say what I know and point out if I am wrong in reasoning or something. I really appreciate the help.

    As I understand it, we are considered with the critical speed of the shaft because it is the point where an unbalance in the shaft will cause the shaft to "whirl" which damages bearings and the shaft, starts vibrations etc. because the deflection "tends to infinity".

    There are two methods of analysis for combined loadings Dunkerly and Rayleigh with one underestimating and the other overestimating, respectively, the first critical speed. This is because we don't know how the shaft's deflection behaves dynamically since our beam stiffness equations are based on static/quasi-static conditions.

    The problem I'm having trouble understanding is I never see any analysis of the components of the shaft considering the forces they transmit. For example a problem I see commonly is determining the proper diameter of a shaft for a certain critical speed safety with say a gear on the shaft. However these analyses only consider the weights of the gears and never the forces they transmit to the shaft. Obviously the gear transmits a radial force to the shaft but that is never shown to be considered in these analyses. I can't imagine that the effects can be negligible. Or simply by specifying that we stay a certain percentage below the first calculated critical speed the way we deal with that? Can someone explain? I'd appreciate it! Thank you!
  2. jcsd
  3. May 7, 2013 #2


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    The force that excites the vibration comes form the fact that a real shaft is never perfectly balanced. This creates a radial force at the shart rotation speed.

    This force causes the shaft to bend. At the critical speed, as the bending increases the amount of unbalance increases, which means the force increases more, which causes more bending .... until something faiils.

    Any forces acting on the shaft from gears etc are irrelevant to this.

    You can calculate the forces in the shaft caused by the unbalance load when it is NOT at critical speed (this is just steady state forced response situation, except you need ot include gyroscopic effects). But AT the critical speed, the forces will increase without limit until something fails, so there is not much point in trying to calculate them. The important thing is not to run the shaft at its critical speed - or only do that for a very short time, as the rotor accelerates or decelerates through the critical speed, and there isn't time for the resonance to build up and cause a falure.
    Last edited: May 7, 2013
  4. May 7, 2013 #3
    Right I understand why the critical speed is to be avoided. But I don't understand why forces acting on the shaft besides weight are not considered. The radial force on the shaft from the gear's radial load would contribute a force to deflection. I know tangential doesn't because that is seen by the shaft as torque.
  5. May 8, 2013 #4


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    It's not clear what you mean by 'radial force'.
  6. May 8, 2013 #5


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    With the usual meaning of "weight" (i.e. the force on the shaft caused by gravity, acting in a constant dowwards direction), the weight is also irrelevant to shaft whirling.

    The relevant forces are those where the direction of the force rotates at the same speed as the shaft.
  7. May 9, 2013 #6
    When a gear transmits power it does so through a force which is NOT completely tangential to the gear circle. The portion which is tangential of this force is the transmitted load. The other portion of the force acts vertically downward and is equivalent to the transmitted load multiplied by the gear's pressure angle (at least for spur gears). This force is the radial force because it acts in the radial direction of the gear. This is the force I am wondering about why it isn't included in the analysis.

    Can you explain this? This doesn't make any sense to me at all especially considering that all textbooks I am using as well as any online resources all do analysis of the shaft based on the weight of the attached mechanism and the weight of the shaft. Your response is contradictory to everything I have been taught.
  8. May 13, 2013 #7
    I'll take a stab at this, though I'm still pretty early on in my rotordynamics analysis career, so I wouldn't call anything I say gospel without fact-checking.

    I think you're getting some stuff mixed up. The natural frequency of the rotor is a property inherent to the design of that rotor, which, as you have already noted, is based on static/quasi-static calculations. Therefore, since we're static, dynamic forces such as the ones imparted by gears, pulleys, impellers, etc. aren't considered here. We consider their mass effect on the system (remember, the natural frequency is the square root of the rotor stiffness divided by the rotor mass), but for the purpose of calculating the natural frequency of the system, we do not look at the dynamic forces. Yet.

    Every rotor has multiple "modes" for that natural frequency. Depending on the application, you may only be concerned with one or two, or you may need to consider more (the highest I've dealt with so far was 6, though jet engines do quite a few more). A critical speed is when some excitation frequency on the rotor equals the frequency of one of the modes of the natural frequency of the rotor. Here is where we start to consider the dynamic forces. For instance, every single rotor will have an excitation frequency that is equal to 1 times the running frequency, this due to imbalance causing one complete oscillation/vibration of the rotor for every revolution (for simplicity's sake, I will use 1X for 1 times, 2X for 2 times the running frequency, and so on). When the 1X frequency coincides with a mode of the natural frequency, that is a critical speed. If I have a 5 vane impeller, I will have an excitation frequency of 5X the running speed. So where that 5X frequency equals a natural frequency mode, I have another critical speed.

    The gear system you're concerned with is another excitation. As the gears work together, as you've noted, forces are imparted into the rotor and this naturally does cause deflection, which results in vibration. The term we use for this particular excitation is the gear meshing frequency, and it's determined by the gear ratios. So, with what I've already said, a rotor with a gear system installed with experience a critical speed where the gear meshing frequency coincides with one of the modes of the natural frequency of the rotor.

    So as you can see, we do consider those forces. But only in as much as they excite the rotor. Hopefully I got all that right and it helps.

  9. May 13, 2013 #8
    Perhaps I need to reexamine this question after I take vibrations because I do keep seeing the idea of a natural frequency pop up which I am vaguely familiar with but it seems to be a different concept than the one I am dealing with currently. The ideas do seem to coincide somewhat though. Or perhaps they're one in the same and I'm completely missing it. I'd like to thank everyone for their responses so far.

    I do vaguely remember from a mechanical dynamics class the idea of the natural frequency of a system but like I mentioned above I thought that they were two different ideas but you may be right that I am mixing things up. What I've found in my mechanical analysis books (various editions of Shigley's Mechanical Design, handouts from class, Roark's) is that the critical speed is being defined similarly, but is: [itex]ω_{i}=\sqrt{\frac{g}{y_{i}}}[/itex], where g is gravitational acceleration and y is the deflection at that point along the shaft. This is for each element (eg gear, pulley, etc.), then the entire shaft's critical speed is then [itex]ω_{n}=\sqrt{(Ʃω_{i}^{-2})^{-1}}[/itex] via the Dunkerly method. I don't really understand why the radial force from the gear is not considered in this equation for the gear since the load the gear applies to the shaft downwards is the radial force plus the weight of the gear. I wouldn't really consider the gear's radial force a dynamic one anyways because once a steady state operation mode has been established which is common for shafts, that force becomes a constantly applied and static load. Hence why I am confused on it's exclusion in shaft deflection calculation. Those deflection formulas are derived under the assumption of quasistatic loadings which means they're valid if jerk is assumed zero. Once steady state is achieved like I said, that assumption should be valid. And if we aren't including that radial force, the critical speed is actually lower than calculated. That seems to be an issue to me.

    Yea I've also seen this in Shigley's chapter for shafts, essentially it says that we only are looking for the first critical speed since once it reaches the first speed catastrophic failure will occur anyways and therefore it's pointless to look for the other critical speeds (frequencies?? ugh).

    See this idea seems to address my primary concern. Is there a reading source or book from which you've found this information? I'd greatly appreciate the reading materials.

    Who knows, maybe I am infinitely over thinking this problem. Thank you so much for the response.
  10. May 13, 2013 #9

    jim hardy

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    See if you can find "Practical Vibration Primer" published by Bentley Nevada corp (now part of GE) . It is just what the title describes.
    Bentley was a pioneer in monitoring equipment for rotating machinery.
    Look for a used copy. I gave mine to son when he graduated BSME and am having difficulty locating another.

    It's a big yellow book maybe 3/4 inch thick. It's rich in "how to figure this out" examples, so it makes a good complement to more academic works . It may help you in your class, my experience is modern textbooks are gravitating away from the practical.

    Torsional mode vibration is fascinating to me - I only recently discovered that flexure in automobile crankshaft pulley dampers.
  11. May 13, 2013 #10


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    I don't think you are "overthinking" it, but you don't seem to have got hold of a basic differences between the vibration of the system when it is rotatiing or not rotating.

    When it is not rotating, you can calculate the frequencies and mode shapes the same way as for any other system (e.g. a beam pinned at the bearing positions), but except for special designs (e.g. a Jeffcott rotor) the dynamics is completely different when it is rotating, because of gyroscopic effects etc.

    That's not true. There are some good reasons to run rotors at "supercritical" speeds above the first critical speed. The important thing is not to run the rotor at a constant speed close to the the critical speed. "Supercritical" rotor designs usually have the first critical speed as low as possible, so the rotor accelerates quickly through the critical speed when the machine is started and stopped, but it operates well above the critical speed.
  12. May 13, 2013 #11
    There's a lot of stuff to deal with here...

    Let me define what I mean by "dynamic." What I mean to say is that it is a force that only occurs when the system is in operation, not static or motionless.

    Let's think on this force some more (by the way, I got confused in my last post, I was talking about the gear meshing frequency, which is different than the force you're referring to. My bad). As I said, we are calculating the natural frequency when of the rotor when the rotor is sitting still. No external forces, just the mass and stiffness of the rotor. That's all. So how do we account for this force that's deflecting the rotor from the gear?

    Let's pretend I have a shaft with one gear on it, with bearings on both ends of the shaft. In a static position, the shaft has a deflection, and from this we can calculate the natural frequency of the rotor. Let's stop and think for a second here: even if that gear is perfectly balanced (it's not, but let's just pretend it is), the shaft is deflected. Therefore, if we take a point on the shaft, just beneath the surface of the shaft, and consider the stress at that point, it will experience a tension and compression force in one full revolution (think of it like a sine curve). Therefore, we say the vibration has a frequency of 1X rotational speed. If the shaft turns 10 times per minute, you have 10 complete oscillations per minute.

    Now then, let's mesh a driving gear to our system. Now you have that gear force (called a gear separation force, by the way) to contend with. But think about what this force is doing- it is only deflecting the shaft more. The natural frequency of the rotor has not changed- again, we calculate that based SOLELY on the rotor's inherent physical properties due to its design, in a static position. The increased deflection, then, is only increasing the amplitude of our 1X vibration.

    I see you haven't taken a vibrations and dynamics course yet. That would account for your confusion on the natural frequency. Hopefully your professors will be able to explain it better. But the gear separation force is not a factor in the natural frequency of the rotor.

    Ok, let me preface what I'm about to say: I like Shigley's book. I really do, it's a good book, great for the fundamentals of mechanical design. That said, there are a couple of topics that he covers in a general sense, but not very detailed, and the problem is that this can lead to oversimplifications like the one above.

    Recall what I said in my first post. A critical speed for a rotating shaft system is any speed where an excitation frequency coincides with a natural frequency or its modes. So it is possible to have multiple critical speeds, in fact most real systems often do. Now, there are several things above that aren't quite right. It is true that we are definitely concerned with the first critical speed, because this is where the vibration amplitudes are going to (most likely) be strongest. It's also often the one that is easiest to run into. However, it's not a magical landmine that will destroy your system the second your operating speed reaches it. Furthermore, catastrophic failure can occur due to operation at other critical speeds besides the first one (though usually the higher the order, the less likely this is to happen).

    This is where vibration and rotordynamic theory start to rocket towards the deep end of the pool, yet at the same time tie in with the basics of mechanical design in a beautiful way.

    Believe it or not, it is possible to operate at a critical speed without failure. It's true! You might be asking how. The thing that Shigley ignores (because he didn't want to devote another 100 pages to it), is that rotordynamic systems also include damping. Furthermore, with proper bearing location and shaft design, the stresses induced into the shaft can be reduced to the point where immediate catastrophic failure will not occur, and you will most likely have a fatigue failure. The design theory is not easy, and I don't think I can explain it all here. In fact, I probably shouldn't have put this paragraph here, because it's liable to confuse you more, but hopefully it will help some.

    Jim already pointed out one of Bently's books, I also have his Fundamentals of Rotating Machinery Diagnostics. The book I used in college was by Dr. Dara Childs and I think it was Engineering Dynamics Applications or something like that. I also refer to J. Vance's Rotordynamics of Turbomachinery quite a bit (it's a bit more practical). And there are a vast multitude of really good papers out there. Texas A&M University Turbomachinery Laboratory hosts a Turbomachinery Conference every year that has had some fantastic papers on rotordynamics fundamentals (Corbo and Malanoski especially).

    Hope this helps.
  13. May 16, 2015 #12
    Hello every one I am new to this forum
    I would like to know if anyone has info. on paper machine rolls that encounter there critical at the first harmonic of running speed or 2x running speed. I have run
    into rolls that will vibrate excessively at the 2x running speed or at exactly half of there first critical not only in the vertical plane but also the horizontal plane.
    is there anyone who can verify this. Thanks Steve T
  14. May 16, 2015 #13


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    Hi, Steve:

    You're new here, so I'll give you a friendly heads up.

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