calculating shaft critical speed by hand 1

[hector002]

I am trying to calculate(estimate) the critical shaft speed of a rotating shaft.

When i did this back in school we solved for the eiganvalues of the 4th order diff equation:

d^4/dx^4 - B^4*y=0 where B^4=rho*A/(E*I)*omega

however this assumes A (shaft area) to be constant. The shaft i an interested in steps from 1" to 2" and back down to 1" between the bearings. Does anyone have some tips or references to solve eigenvalue problems for a shaft with non uniform diameter. thanks

 

[hector002]

the sad thing is neither am I, ive been struggling through all this bearing stuff, hopefully ill get it right and actually get to taking some measurements.

Your correct-- whirl is loss of load capacity when an external load oscillates at .47 times the rotating frequency. Whip is fluid film bearing instabilty at starting ~ 2x crital speed. Thanks

Also that shaft mode shape program is great! Did you make that or is it a commerically avail program?

I plan on calculating a pinned uniform shaft both 50mm and 4" and see where the critical speeds of those fall just out of curiousity.

 

[electricpete]

Returning to the subject of bearing stiffness - I used 3,000,000 lbf / inch.

API 684 Tutorial on rotodynamics doesn't say much on the subject. They have a table 1-3 of "Typical Stiffness and Damping Properties of Common Bearings (Comparison Only)" which lists 3,000,000 lbf / inch for the broad category of antifriction bearings.

I don't have any better references for selecting bearing stiffness. Does anyone know of good references on this subject?

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[ivymike]

However is that sound logic that if a 50mm uniform shaft is OK that my 50,4",50 shaft will be OK too, i would guess that adding mass in the middle would lower the critical speed, but maybe not since the 4" section runs almost the majority of the length.
good point - you could throw in some extra mass in your calculation if you're worried about it, but then you might be "fudging" a bit too much for comfort.

 

[electricpete]

Correction - the API document lists 5 million lbf/inch. (I used 3 million).

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[electricpete]

Being a stiff rotor, the first critical speed is heavily influenced by bearing stiffness. Increasing stiffness up to 5 million lbf/inch caused first critical to increase to 82,000rpm.

You can also tell that by looking at the mode shape... there is a lot of displacement at the bearing compared to the bending of the shaft. If the bearings were stiff compared to the rotor, the 1st critical mode shape trace would go through shaft centerline near the bearings... but in the attachment the whole mode shape trace is offset from the centerline indicating substantial flexing of the bearings.

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[electricpete]

Another consequence of the fact this is a rigid rotor and critical speed heavily dependent on bearing... if you use the "quick check" and the the underlying analytical solution does not account for bearing stiffness you can't get any reasonable lower bound for critical speed (You will get an upper bound using analytical solutions that assume infinite bearing stiffness).

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[electricpete]

For a very stiff rotor, you can approximate the critical speed as (1/(2PI)) * sqrt(2*k/M) where k is bearing stiffness per bearing and M is rotor mass.

Unfortunately, to the extent that the rotor flexes, this formulation also overestimates the critical speed and does not provide a lower bound.

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[hector002]

Pete, i found a copy of Feildings Books- is that were you got the methods for your mode/critical speed program?

 

[electricpete]

The program is critspd by Rotdyn inc

I didn't write that program.

Based on the inputs and outputs produced by the program I have deduced that it uses the same algorithm shown in Chapter 1 of Fielding's book.

I am planning on programming it within excel in the near future but I have some other pressing projects in my spare computer time first.

I am very happy with Fielding's book. My review:
TOC:
Lateral Critical Speed Calc
Torsional Crit Speed Calc
Rotor Bearing Stability Program
Blad Vibration (Flexural, Torsional)
Disk Weight and Inertia Properities Progra
Disk Stresses
Blade Stresses
Stationary Shrink Fit calcs
Rotating Shrink Fit calcs
Fluid dynamics stuff that I haven't looked at.

Each of the chapters has a mathematical development of the algorithm and then some examples of output expected for a given input so you can double-check your program.

The exact programs are written in TI-59 programming language which is meaningless to me. There is enough info to figure it out from the math descritpion without looking at the code but the discussion is a little terse (brief) and I have had to work at it. It's a good book to use along with other texts and the price is definitely right.

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[electricpete]

Going back to the (1/(2Pi) sqrt(2*k/m)

For the case in the word file above k = 3E6 lbf/ft and
M = 37.4*lbm
With extra factor of 60 to convert to rpm it gives:

f = (60/2*Pi)*sqrt(2*3E6*lbf/inch/(37.4*lbm)*(32.2*12*inch/sec^2*lbm/lbf))
= 74,000 rpm.

As expected, higher than the transfer matrix model which included shaft flexing. But only about 10%. Again not good for finding a lower bound but I just wanted to mention it.

If you want to see what is the lowest that the critical speed can be, you need to estimate how low can bearing stiffness be (and put it into transfer matrix)

A side note - this algorithm models bearing as simple spring stiffness with spring perpendicular to shaft. There was some discussion about resistance of bearing to angular bending perpendicular to original shaft. I would think this assumption would be pretty good for simple conrad deep groove bearing or for face-to-face duplex angle contact, and certainly for self-aligning bearings like spherical roller bearing. But for typical duplex angle contact bearings mounted back to back, that configuration is very intolerant to misalignment (produces reaction moment) and this assumption of the model is least accurate in this bearing configuraiton.

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[electricpete]

Typographical correction to last message... should have been 3E6 lbf/inch

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[electricpete]

The error from modeling the bearing as simple linear spring and not considering the reaction moment to "misalignment" for back-to-back angle bearings (discused above) would presumably cause calculated critical speed lower than actual... a conservative error if you are trying to esimate a bounding minimum critical speed.

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[GregLocock]

Bearing radial stiffness is heavily dependent on the preload, and is fairly non linear. Your values seem rather high to me.

In particular, the stiffnss of the bearing will be bounded by the stiffness of its mounting system. I rarely see a structural part with a stiffness in excess of 100 000 N/mm, and half that would be more typical.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[electricpete]

Thanks Greg. I think you are right. I have seen examples worked with a range of stiffness far lower than what I used. I thought my value was high until I saw the API number. Maybe that number is supposed to be combined with a support stiffness.

Non-linearity of bearing does not compute (literally... no room for that in the model).

100 000 N / mm * (1lbf/4.482N) * (25.4mm/inch) = 570,000 lbf/inch.

If I plug 500,000 lbf/inch into the program (all other inputs same as the word file), the first critical comes down to

As I mentioned 5E6 lbf/inch came out of the API document. Maybe this value was intended to be combined with another value for support stiffness.

If you run the program again with K = 500,000 lbf/inch, then first critical speed becomes 30,131 rpm. Mode shape looks even more like rigid rotor.

The rigid rotor approximation which had 10% error at K=3E6 gets better and better as we reduce K: For K=5E5
f = evalf(60/2/Pi)*sqrt(2*5E5*lbf/inch/(37.4*lbm)*(32.2*12*inch/sec^2*lbm/lbf)) = 30,694 rpm.
Now only a 2% error compared to the program.

If you want to consider K an unknown and plot the values for various of first critical speed for values of K, use the rigid body approximation above. We know the accuracy (compared to program) is 10% for K=3E6, 2% for K=5E5 and even better as we reduce K further.

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[electricpete]

Disregard my whole previous message which was garbled (repeated things). Here it is again.

Thanks Greg. I think you are right. I have seen examples worked with a range of stiffness far lower than what I used. I thought my value was high until I saw the API number. Maybe that number is supposed to be combined with a support stiffness.

Non-linearity of bearing does not compute (literally... no room for that in the model).

100 000 N / mm * (1lbf/4.482N) * (25.4mm/inch) = 570,000 lbf/inch.

If you run the program again with K = 500,000 lbf/inch, then first critical speed becomes 30,131 rpm. Mode shape looks even more like rigid rotor.

The rigid rotor approximation which had 10% error at K=3E6 gets better and better as we reduce K: For K=5E5
f = evalf(60/2/Pi)*sqrt(2*5E5*lbf/inch/(37.4*lbm)*(32.2*12*inch/sec^2*lbm/lbf)) = 30,694 rpm.
Now only a 2% error compared to the program.

If you want to consider K an unknown and plot the values for various of first critical speed for values of K, use the rigid body approximation above. We know the accuracy (compared to program) is 10% for K=3E6, 2% for K=5E5 and even better as we reduce K further.

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[GregLocock]

I'm a bit worried by the 4 inch shaft. There's nothing wrong with a 4 inch diameter, but it seems very odd proportions for a shaft.

Cheers

Greg Locock

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[electricpete]

"this shaft is going to be used for a fluid bearing test rig "

I presume the large shaft diameter is to simulate the journal of a larger machine.

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[GregLocock]

I'd forgotten that. OK, in that case I think you are chasing the most likely mode - rigid shaft on flexible pedestals.

Can the OP decribe how the bearings are mounted? All the way to the ground.



Cheers

Greg Locock

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[electricpete]

I agree 100% with Greg. And once again I'm glad you interjected some reality into my numbers.

I will be interested to see how the support stiffness is calculated. The only form I can come up with is for a uniformly loaded rectangular block of steel... the vertical stiffness would be E * A / h. That suggests starting with the numerical value of E English units 3E7, and then multiply it by the ratio A/h where A is horizontal area and h is height. Unfortunately not only is the geometry simplistic, but the lower stiffness in horizontal direction associated with bending of the support will be more relevant and that will quickly get a lot more complicated.

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[electricpete]

I would be interested from curiosity to know more about the machine.

How will it be driven... belt?

If you have two antifriction bearings on the end and a sleeve bearing tested in the middle, how will you control the load on the sleeve bearing?

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