Calculating natural frequency of a shaft

[gearman1234]

Is there any book or internet source available where I can find a formula for finding natural frequency of vibration of a shaft?
I found ageneral formula f = sq root of(Stiffness/mass) . The question is "How to calculate the stiffness of a member?"

 

[electricpete]

1 - On one extreme is rotor very rigid compared to the bearings. In that case sqrt(k/m) would work.

In that case k would be twice stiffness of a single bearing (assuming you have between bearings machine with identical bearings on each end).

Very rough values for bearing stiffness per API684 are
Sleeve bearigns - 500,000 lbf/inch
Tilting pad bearings - 700,000 lbf/inch
rolling bearings - 5,000,000 lbf/inch

I emphasize rough (approximate) and there is variation among bearings and preload. If support is not very rigid compared to bearing, add that in too.

2 - On the other extreme would be bearings much stiffer than the rotor. In that case we can model the rotor as a pinned/pinned beam and use analytical solutions which are fairly simple.

2A - if all the mass is uniformly spread on the shaft, the pinned/pinned beam solution is:
F1 = (1/[2*Pi]) * 9.87*sqrt(E*I/[mu*L^4])
F2 = (1/[2*Pi]) * 39.5*sqrt(E*I/[mu*L^4])
where mu is mass per lenghth.

2B - if all the mass were concentrated at the center of the shaft, use
f1 = (1/[2*Pi])*sqrt(48*E*I/[M*L^3])
There may be a few more simple cases in this category.

3 - If the shaft is not at either extreme (not a lot stiffer or more flexible than the bearings), than the solution is not as straightforward and the approach is typically numerical. For example the transfer matrix method. I have programmed that in sci-lab. If you give the shaft information I can run it for you (when I have the time) and post my program and results.

My transfer matrix method is still pretty darned simple. There are a lot of variations on the numerical methods.

I will look around for links. Tom Irvine has some great info at his site

http://www.vibrationdata.com

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[electricpete]

"My transfer matrix method is still pretty darned simple."

That's not exactly what I meant. It takes awhile to program these and if I showed you my program you would have to study it... many pages long.

What I meant to say is the program I have is a lot less sophisticated and accurate than a lot of the other programs out there.

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[GregLocock]

Have you ever measured the pedestal stiffness at the frequency of rotation? I find it very hard to believe that 5e6 lb/inch is attainable.

Admittedly on smaller structures, I very rarely see anything over 100 kN/mm. Which is 100 000/4*25.4=

6e5 lbf/inch

to get that would need good cast iron brackets mounted to the slab.

Cheers

Greg Locock

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

 

[electricpete]

I have never measured it. I’m not that sure how to estimate it either . I believe it would require FEA for all but the most simple geometries.

Note I did say “If support is not very rigid compared to bearing, add that in too.”... which really should have been “add in that effecttoo”.

To add springs in series we all know it’s the product of spring constants over the sum.

Another thing to note in situation 2B (mass concentrated at the center of massless shaft), by inspection of the formula, the spring constant is effectively 48*E*I/L^3.

You could take this spring constant (stiffness) and combine in series with a bearing stiffness and support stiffness, again using product over sum of stifnesses. This would be valid for the first critical speed.

Greg mentioned pedestal stiffness at frequency of rotation. I am talking static stiffnesses under the assumption there is not significant mass moving with the bearing support. I think this would be a reasonable assumption in some cases such as large horizontal motor... the bearing is supported by a hollow frame mounted to the foundation. The stator/core is mounted to the bottom of the frame which may be assumed stationary. The bearing housing is mounted a few feet up on the frame which is a point which can move without moving much stator mass other than the hollow frame itself.

If there is significant mass moving with the bearing support, than it cannot be modeled simply as a stiffness.

I did not at all mean to imply that estimating critical speed is a simple excercize .. was just trying to provide some simple examples. The key point will be to understand the assumptions of any approach you use and how appropriate they are to your situation. Sometimes there will be a tradeoff between complexity of the calculation and the accuracy you require.

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[GregLocock]

Our most critical shafts are mounted on flexible supports, one way or another, so I'm probably more used to thinking about this as an issue. For instance the tailshaft is mounted to a diff which is on say three 1000 N/mm bushes, at one end, a 100 N/mm centre bearing, and a gearbox mount that is probably about the same or a little more, vertically. The actual bearings are pretty stiff, but the rubber bushes are not.

Dynamically of course things get horrid very quickly. An impact (bump) test is a great way of estimating your pedestal stiffness.

Cheers

Greg Locock

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

 

[rob768]

I agree a bump test is a good way to measure pedestal stiffness. However, in that stage it is often difficult and costly to alter anything, if the test indicate problems are to be expected. FEA is in our experience the best option to predict stiffness and vibration behaviour.

 

[gearman1234]

Thanks

Gearman1234

 

[electricpete]

A more general formulation to handle the case 3 above (neither rotor nor bearings can be considered rigid) seems to be suggested by a number of references (applicable for computing the FIRST critical speed only)
f1 = (1/[2*pi] * w1
w1 = sqrt(K_equivalent/M_equivalent)
Mequivalent = M + 0.5*m where M is concentrated mass at axial center of shaft and m is shaft mass.
K_equivalent = series combination of shaft and two parallel-bearing-stiffnesses
K_equivalent = Kshaft*2*Kbrg/(Kshaft+2*Kbrg)
Kshaft = 48*E*I/L^3
I = pi * R^4/ 4 =pi*D^4/64 as before.

The basis for developing Kshaft and K_equivalent from the massless shaft system with lumped-mass in the middle seems clear since it is directly analogous to a discrete SDOF mass spring system for purposes of computing first cirtical speed.

But the basis for developing M_equivalent is a little muddy to me.

I can derive it as follows but I have some reservations:
Assume rigid bearings for simplicity.
Assume Mequivalent = M + X*m (X is unknown to be solved)
w1 = sqrt([48EI/L^3]/[M+X*m])

In the limit M->0 we want w1 to resemble the distributed-mass pinned beam solution (Remember distributed-mass pinned beam solution was w1=sqrt([Pi^4*E*I/L^3]/[m]))

Equate the two frequencies and solve for X
sqrt([48EI/L^3]/[X*m]) = sqrt([Pi^4*E*I/L^3]/[m])
48/X = Pi^4
X = 48/Pi^4 = 0.493 ~ 0.5

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[electricpete]

1 – Does anyone seen any proof or justification for incorporating shaft mass using M_equivalent = M + .5*m (M is lumped mass in center and m is shaft mass) and applying this with the lumped-center-mass stiffness to give w1=sqrt(48*E*I/[M_equivalent*L^3]))?

2 – What assumptions / approximations are made to arrive at this formulation (beyond Euler/Bernoulli beam assumptions)?

I see this method suggested in Harris’ Shock and Vib Handbook, Mark’s Mechanical Engineering Handbook, Rao’s Mechanical Vibration, and Ehrlich’s Rotordynamics Handbook, but none of them provides any proof.

I am a little uncomfortable with my proof above, especially since it seems that the deflection pattern of the shaft as a function of axial distance would change somewhat depending on relative weights of the center mass and the distributed mass.


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[electricpete]

I am going to post my question about M+0.5m in a new thread.

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