Another question on shaft critical speed 3

[Oakley]

I find in technical books an literature that computation of critical speed deals with a shaft with two supports and one or more rotating masses attached to the shaft.

In dealing with design of a line shaft where one only has torsional loads (plus the weight of the shaft for shear and bending), and given, say, that the shaft is 16 feet long between supports and couplings....it is pretty straight forward to calculate the critical speed.

However, how is the critical speed impacted if one puts a support bearing at mid span of the shaft? Nothing else changes. The shaft diameter, total length, and torsional load remain the same, only the support bearing is added at mid-span.

First, do you know of a technical reference that deals with this type of loading?

Second, how is the critical speed compare of the shaft with and without the mid-span bearing support.

Thanks,

oakley408

 

[GregLocock]

It will normally be much higher.

I don't know of a technical reference in particular, the answer is pretty obvious if you have a dynamics background. Big clue: what is the same frequency as the first critical speed?



Cheers

Greg Locock

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

 

[alexit]

If I understand correct, you make shaft with two supports, 16' apart. This has 1st critical frequency X (Half-sine mode shape) and second critical is 2X (Full-sine mode shape.)

Now you add third support at exact center, so 8' (minus support width) from front to middle to rear. Theory is 1st critical is now 2X and second critical is ~4X, but you still have sub-critical first mode shape at X, this can be measured as whirl or whip usually.

Better results occur when added support is not in center, like 5' or 6' from one end...you have lower 1st but the sub critical is not so active.

 

[GregLocock]

Good answer, except the frequency of the second mode is not double that of the first one. That only works for certain systems, not, generally, shafts in bending.

Cheers

Greg Locock

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

 

[Oakley]

Thanks, Greg and Alexit for two very helpful answers. Being a retired engineer my studies in dynamics and vibrations were more years in the past than I care to admit. So, forgive if I ask what to you is the obvious.

Alex, I believe your clue has to do with the natural frequency of the shaft, right? And I thought about that. some time back I downloaded a structural software package that computes natural frequency with the uniform weight per foot, moment of inertia, and input concerning the span. I tried simulating my post above by inputing both the single span and then the span equally divided with a center support. Both answers came up at 10.64 hz. Which, with my study in the distant past, seemed irrational. I would have thought the center support raised the natural frequency by virtue of shortening the unsupported spans.

Where am I going wrong on that issue. Of course, I understand the dynamic component of the rotating mass makes it a different ball game.

Thanks again for your help.

oakley

 

[GregLocock]

You must have done something wrong in the model (or the program was faulty). It should have worked as Alexit described.

Cheers

Greg Locock

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

 

[sms]

Unless of course you were calculating the torsional natural frequency rather than the lateral. That center support bearing would likly have no effect on the torsional critical.

 

[Oakley]

Greg, I think you are correct. With regard to the input the program gives a data base for shape selection (and it is a structural program, not mechanical for shafts), when one selects the shape it inputs the moment of intertia, Young's modulus, and provides the uniform weight per foot. Then I set the 16 foot span and run the program (single span) and get the natural frequency. Then keeping the same overall length and clicking on a button for "two or more spans) it provides a diagram for a two span beam (the program doesn't seem to have a means of inputting a triple equal span). Click the run button and up comes the same hz as the single span. I have emailed the software provider to ask why this is the case.

sms, thanks for your input as well. And I understand your point, the torsional moment is transferred right through the support bearing. My difficulty, however, is: why wouldn't the critical speed (where destructive vibration would be manifested) be increased significantly as the mid support bearing cuts the span in half, thus increasing the stiffness coeficient (k) by a factor that is inverse to the length of the beam (L) to the third power (L^3). Although the torsional critical speed would not be impacted the resistance to the resulting destructive vibration, it would appear, should increase significantly. I know I'm probably missing something. Any enlightenment is appreciated.

oakley

 

[GregLocock]

Don't confuse whirl with torsionals (althought to be fair in automotive crankshafts they occur at roughly the same frequency). Whirl is a BENDING phenomenon.

"Then keeping the same overall length and clicking on a button for "two or more spans) it provides a diagram for a two span beam (the program doesn't seem to have a means of inputting a triple equal span). "

I wonder if your beam is now 48 feet long? Even then the maths is wrong.

Here's the equation for a pin jointed single span, uniformly loaded beam. A=9.87

F=A/2/pi*sqrt(E*I/mu/L^4)

mu=mass per unit length

For a centre bearing, A=39.5, and for a 3 span, A=88.9





Cheers

Greg Locock

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

 

[Oakley]

Greg, I sent the company an email and they acknowledged that the input window was unclear (and to me it was misleading). I did get proper information for running the program for multiple spans.

And thanks, also, for the equation. I was confusing the "whirl" which Alexit referred to for torsionals...my mistake. You "pros" have been a tremendous help.

oakley