Critical speed - vertical or horizontal positin

[JUL74000]

Hi all,

I have to calculate the critical speed of a fan shaft for an oven.
Belt driven on a cantilever beam.
what is the best formula to calculate the first critical speed for an overhung fan ?
I need to define the minimal distance between the bearings.
The overhang is large.
The centrifugal fan can be on horizontal or vertical position.
Is the critical speed of the shaft line the same in the two cases?.
the deformation of the shaft is not the same (weight of the wheel).

Many thanks in advance

Julian

 

[electricpete]

In most cases the effects of gravity are small and not included in the model.
Yes, vertical position may cause some change in axial shaft thrust (which has a stiffening effect).
Also there may be some change in behavior of the bearings as they are somewhat non-linear... meaning stiffness is not a constant but depends on the loading.

what is the best formula to calculate the first critical speed for an overhung fan ?

Best in what way? Most accurate would surely involve numerical analysis. Simplest would involve many simplifications. Tough to know what simplifications might be appropriate without more info. You might want to tell us more about the machine especially the type bearings and their support, perhaps rough rotor dimensions including the fan. Is machine speed fixed? What speed?

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[JUL74000]

Hi Electricpete
Thank you very much for your answer.

« In most cases the effects of gravity are small and not included in the model »
In horizontal position, the deflection of the end of the shaft can reach 0.015 inches.
It seemed to be incident for speed limit (deflection >> than vertical position).

What is the best formula to calculate the first critical speed for an overhung fan ?
For a cantilever shaft (weight of the wheel at the end).

Please see the PDF file.

Many thanks in advance

Julian

 

[electricpete]

If you have a particular limit for machine deflection, that will be a consideration not related to resonant frequency calc.

For the resonant frequency calc, we are considering (by definition) a linear or linearized system. The effects of gravity on calculated critical speed are irrelevant except to the extent they may change the linearized coefficients (like bearing stiffness changing with load).

As a side note, static deflection of 0.015" would correspond to w = 8hz for a SDOF system. It will certainly be different for your system but it's a first very preliminary guess at what the lowest natural frequency might be.

I still don't think there is enough to go on to define critical speed in terms of understanding your geometry.

We still don't know about the support stiffness. One way I like to get around that is simply model the rotor (*) and map the behavior with varying support/bearing stiffness (critical speed map). But we don't really know enough about the rotor yet (is gyroscopic effect important? beats me because I have no idea of diameter and speed range).

* That first approach of assumes there is no significant mass movement in the support which may or may not be the case,.. don't know.

More info is better to get better answer about critical speed calcualtion. If you want to continue on that aspect with more info, probably better to move it out of the pump forum (it's a fan) and into this forum:
forum384
where you may get great input from reg Locock and lots of others.


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(2B)+(2B)' ?

 

[electricpete]

As a side note, static deflection of 0.015" would correspond to w = 8hz for a SDOF system. It will certainly be different for your system but it's a first very preliminary guess at what the lowest natural frequency might be

Whooops I goofed that up. Again SDOF is not directly applicable, but if you wanted to calculate it would be
w = sqrt(g/static deflection) = sqrt(0.015 / 386.4) = 160 rad / sec
f = w / (2*pi) = 25 hz


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[JUL74000]

Hi Electricpete.
Thank you for these important Informations
- The effects of gravity on calculated critical speed are irrelevant (except bearing stiffness / load – shaft thrust)
- Little effect of the loading in first approach
- Formula for the first critical speed for an overhung fan is : w = sqrt(g/static deflection) f = w / (2*pi

I realized that I had to define the static deflection considering only the weight of the wheel at the end of the shaft.
Please see simplified representation.

To improve the stiffness and limit the static deflection of the shaft end, i think to reduce distance between the 2 bearings (spherical roller or pillow block RHP). The bearings are oversized.

Thank you Electripete, MikeHalloran and Tmoose

Nice day

Julian

 

[electricpete]

Is the machine built yet or just working on paper?

What does this 0.015 inch “static deflection” represent?
Is this a change in fan position when the belt load is applied?
Was the deflection seen in up direction or down direction?

Attached I did a quick analysis assuming very stiff bearings and support structure: K2 = K3 = 1E10 N/m.
The resulting static deflections are 4E-5 meter ~ 0.00015 inch ~ 10 times lower than what you mentioned.

So you may have some pretty flexible supports. Will be very relevant to any critical speed calculation (rigid bearing support assumption would be waaay off

I could try lower stiffnesses to try to match your 0.015", which would give a stiffness estimate as input to critical speed calc, but that would be a shot in the dark without knowing what the static deflection represents (questions above). Also do you have an idea about whether the stiffness of support for both bearings would be similar?

Is the machine in the other thread the same machine?
thread821-342851
Is fan weight 25 pounds or 80 pounds?
Is the fan wheel diameter 16" or 18"

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[electricpete]

By the way, the previous attachment is from MuPad... may be tough to decipher, but the results are in the four graphs at the end.

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[JUL74000]

Hi Electricpete

The example given is indicative of our current machines.

It is the value of the deflection of the wheel (center) under the effect of the single loading of the wheel and the shaft. In down direction. 0.015 inch was not correct. Sorry. Correct value is 0.0008 inch.

I don’t find any information for stiffness of pillow block RHP (self-aligning).
For a first calculation, stiffness of the support is considered very rigid.

Thank you Electripete.

Julian

 

[electricpete]

Attached is numerical analysis of your rotor. The big assumption here is that any motion of support mass neglected (that’s an assumption, don’t really know anything about what is below this fan).

It uses numerical transfer matrix method (Myklestad method). Assumptions listed at the bottom of the instructions tab. I used to have a companion website for this spreadsheet until I changed ISP’s. It included test cases that have been used to validate the spreadsheet. I can put that website back up online if anyone knows of free website.

Tab Main is the basic options for the calculations. Note that I used Gryoscopic option 1 which includes gyroscopic effects (more later).

RotorSection and RotorGeometryGraph show the model of the rotor used (tried to recreate your drawing). Units have been converted to SI (much easier calculations that way imo).

Outsheet is a summary of the results. Table at the top has columns corresponding to calculated 1st, 2nd, 3rd etc natural frequencies. The rows of the table corresond to a “bearing stiffness multiplier” which is the factor by which the bearing stiffness in “main” (1E7 N/m) is multiplied for the calculation (to see effect of a range of bearing stifnesses).

Below the table in outsheet is a “critical speed map”. It shows how each natural frequency changes as a function of bearing stiffness.

I chose further examination of bearing stiffness multipliers 0.1 and 1000.

For bearing stiffness multiplier 0.1, we see from the critical speed map that the frequencies of the first and 2nd modes are increasing as sqrt(Kb) (slope of ½ on log log plot), indicating the behavior of the bearing spring dominates in this stiffness range for these modes (rigid rotor modes, flexible supports).

For bearing stiffness multiplier 1000, we see from the critical speed map that the frequencies of the first three modes are constant, indicating the behavior of the bearing springs are irrelevant in this stiffness range (flexible rotor modes, rigid supports).

The tabs labeled 0.1 and 1000 show the first three calculated modes which confirm earlier observations about rigid and flexible modes. You’ll also notice as the calculated natural frequency goes up, the left end of modeshape tends to approach displacement and slope of zero, indicating the gyroscopic effect is dominating to create that type of boundary condition.

You’re probably interested in rigid support (mult = 1000), first mode. It has low enough frequency that the gryoscopic effect is not dominating as discussed above, so we can’t replace it with simple slope=0, displacement = 0 boundary condition like you could for the higher modes. How important is the gyroscopic effect for this rigid support (mult = 1000), first mode? Change the calculation mode from 1 (critical speed including gryo) to 3 (no disk effects) on the main tab, re-run the calcuation and you’d see the calculated rigid-support first critical decreases from 49 (with gyroscopic effects) to 44 (without gyroscopic effects). So more than 10% change in natural frequency means gyroscopic effect is still pretty important even though you wouldn’t have guessed it from the modeshape.

The next step if you wanted to improve the model would be to model the support mass (not an easy thing with transfer matrix approach, but can be done with other numerical approaches).

I’m suspecting as natural you’d rather start with simpler models to build and understanding / feel for things. I am short on ideas of simple model to capture essential elements of model so far.

Do you feel comfortable to neglect the mass of the shaft?
Then you can use the mechanics of materials approach described in my earlier post to give you the 2x2 stiffness coefficients of a 2DOF system (mass on each end of the rotor.. one for the fan and one for the sheave).
K*X = -M*X’’ for 2*2 matrix.
K*X = w^2*X solve for w using eigenvalue problem

If you want to add gyropscopic effects, that’s trickier. Can still be done with 2*2 matrices. Something like
K*X + M*X’’ + w*G* X’ = 0
K*X – w^2* M*X + w^2*G* X = 0
Again 2x 2 matrices
where G is a skew symmetric matrix.... only off-diagonal terms are zero and they are opposite signs. Magnitude is the polar moment of inertia for thin disks. Figuring the signs always requires some thought or looking in a book.

I’m sure there is analytical treatment in textbooks somewhere if that’s what you’re interested in. In fact I do remember reading something like that but I’m not going to go digging for it. Again if you post in the vib forum you might get more input.


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[electricpete]

If you want to add gyropscopic effects, that’s trickier. Can still be done with 2*2 matrices. Something like
K*X + M*X’’ + w*G* X’ = 0
K*X – w^2* M*X + w^2*G* X = 0
Again 2x 2 matrices
where G is a skew symmetric matrix.... only off-diagonal terms are zero and they are opposite signs. Magnitude is the polar moment of inertia for thin disks. Figuring the signs always requires some thought or looking in a book.

I'm sorry, disregard this piece. It was an oversimplification on my part. There are a number of ways to present it, but matrix solution of gryoscopic typically is shown as four coordinates per position: (X, Y, ThetaX, ThetaY). The X and Y because the gryoscopic couples from one radial coordinate to another. The Theta because d/dt(theta) is what results in a moment. There are two non-zero terms in the 4x4 G matrix. See for example Adams' Rotating Machinery Vibration, equation 2.3

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