Hollow Shaft Critical Speed

  • Thread starter sallama
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  • #1
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I want to figure out critical speed of hollow shaft. When I calculate from equation, critical speed of this shaft is approximately 10000 RPM. But when I start to increase speed of shaft, at ~2500 RPM I see that the critical speed occured. What can I do to increase this critical speed to double, (to add mass under hollow shaft?). And what is the problem that I can not calculate real critical values. Assy Configuration, Fixed Shaft, Grooves, Length??

System can be seen at attachment. On the center axis of the shaft, a non turn shaft placed. On this shaft there are two 6308 Ball bearings to allow turn the hollow shaft. On the hollow shaft there are grooves.

Hollow Shaft info:
Material: Steel (I dont know exactly which type, but producer said that ST37/ASTM A283)
Do= 160mm
Di=129mm
Length (Center From Bearing)=1350mm
Total Length=1500mm
Distribution Force (approximate) =1500N
Fixed Shaft D=45mm
Bearing Type: Ball Bearing 6308Motor: 15 KW

Hollow_Shaft-FBD.jpg
 

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Answers and Replies

  • #2
AlephZero
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From a hand calc, I get the first vibration mode of the central shaft = about 50 Hz = 3000 RPM.

I guess that is where your resonance at 2500 RPM is coming from.
 
  • #3
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Thanks AlephZero,

What is your advice to increase this critical speed to 90-95 Hz without new production. Because i want to run the hollow shaft with half of the critical speed, and this will be 2500x2=5000 RPM. But above 4000 RPM will be enough. Where and which sizes can l add mass to the inner side of the hollow shaft. The inner diameter of hollow shaft is 129mm, so that my adding cylinder' outer diameter will be 129mm. And what should be the length and inner diameter of these cylinders?

I use software and hand formula to get critical speed. Which formula did you use with which datas?

Thank a lot for your interest and favour.
Serkan
 
  • #4
AlephZero
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I assumed the center shaft was a beam pinned at the bearing positions. That is not quite right because presumably it is also supported at the ends and the mass of the outer shaft will have a small effect. That would make the actual frequency lower than my 50 Hz which and that is what you are seeing.

Changing the hollow shaft won't make much difference. You need to change the frequency of the center shaft. If its only purpose is to support the bearings at the ends, you could probably make a different design with no center shaft at all. It's not very clear from your drawings if it has some other purpose as well.

Maybe you could also fix the problem by making the center shaft more flexible to reduce the 2500RPM frequency. Accelerate through it and run the machine above it.
 
  • #5
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Yes, you are right.
Center shaft doesn't rotate.Center Shaft only for fixing and bearings mounted on it. This was done because I have limited hollow shaft material/tube (Length & Diameter is limited per my design so that I should use inner shaft for unbalance and fix). I can not change design that any new hollow/outer shaft. But I have a chance to make different inner/center shaft if this doesn't take too much time. Or any adding some mass between outer & inner shaft without any unbalance.

I must not run above its critical speed because of machine/fatigue/bearing life. This is special machine to deburring&grinding hard material. So, I should increase critical speed.

Will "Change inner shaft with high strength steel" be a solution for increase critical speed? or any other opinions.
Thanks for your reply...
Serkan
Hollow_Shaft-FBD_Rev.jpg
 

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  • #6
AlephZero
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Changing the material of the inner shaft won't make much difference. For the same shape of object, the material property that controls the vibration frequency is the ratio of [itex]E/\rho[/itex], and that is very similar for any structural metal.

You need to make the inner shaft stiffer, and/or lighter. For a beam the frequencies are proportional to [itex]\sqrt{EI/m}[/itex] where m is the mass per unit length. For a circular beam I is proportional to [itex]d^4[/itex] and m is proportional to [itex]d^2[/itex] so the frequency is proportional to the diameter.

You could make the center shaft lighter by fabricating it in three parts, with most of the length as another hollow tube. Or you could machine the solid shaft into a cross-shape to reduce the mass but keep the bending stiffness large.
 
  • #7
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From your answers, I should apply some solutions...

-inner shaft will be lighter with decreasing diameter between bearings. Or machine shaft to cross shape between bearings? will this be effect worst fatigue life in the future??
-Is there a good choice to apply heat treatment to get harder steel?
-If inner shaft produce in three parts, I think I will have trouble with rigidity when assy of these part will be done.

I have only one choice to apply one of the solution, because I dont have any time and any extra material. Which solution must I select?

Thank yo very much AlephZero...
For your fast response...
Serkan
 
  • #8
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And is there any solution for rotating hollow shaft. Not for the non rotate center shaft
 
  • #9
AlephZero
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I have only one choice to apply one of the solution, because I dont have any time and any extra material. Which solution must I select?
Nobody here can answer that question from the limited amount of information we have here.

I've told you what I think the problem is, and if that really is the problem some ideas about how to fix it. If it was my job to sort this out, I would want to spend a lot more time working on it than I'm going to do here, for free.
 
  • #10
GT1
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From a hand calc, I get the first vibration mode of the central shaft = about 50 Hz = 3000 RPM.

I guess that is where your resonance at 2500 RPM is coming from.

How did you calculate it? I get frequency of ~176Hz.
As far as i know-for steel shafts f=2.01*10^6*(d/L^2)
 
  • #13
AlephZero
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What units is your formula in? The constant must be units dependent.

FWIW if the lengths are in mm, I get 2.01*10^6* 45 / 1350^2 = 49.6 Hz

I used d = 0.045m L = 1.35m E = 200 GPa density = 7800 Kg/m^3

I = 2.01* 10^-7 m^4
m = 12.4 kg/m

9.87 / (2 pi * 1.35^2) sqrt(200 x 10^9 x 2.01 * 10^-7 / 12.4)
= 0.862 x sqrt(3242) = 49.1 Hz
 
  • #14
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I got it, you used d=45mm for the shaft diameter. I used the hollow shaft diameter.
So such a system have 2 frequencies - one of the inner non rotating shaft and one for the hollow rotating shaft?
 
  • #15
AlephZero
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I got it, you used d=45mm for the shaft diameter. I used the hollow shaft diameter.
So such a system have 2 frequencies - one of the inner non rotating shaft and one for the hollow rotating shaft?

Actually it has two sets of frequencies (sometimes called families of frequencies), one for the outer shaft and one for the inner.

They two sets will be slightly coupled together by the bearings of course, but it's hard to take that into account with a simple hand calculation.

You may have found another problem here though. You get 176 Hz = 10,560 RPM for a solid outer shaft, but the OP calculated the first whirl mode for the hollow shaft at "about 10,000 RPM". Those two numbers look suspiciously close to each other!
 
  • #16
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Yes, for the hollow shaft it should be:
f=2.01*10^6*(160/1350^2)*(160/129)=219Hz
 

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