Critical Shaft Speed, Overhung Load 1

[iainuts]

I'm doing an analysis for the critical speed of a shaft with overhung load. The overhung load is a small gear, about 8 lb. Load on the gear is 300 lb which is what dictates shaft deflection and stress. Shaft rotation about 13,000 RPM.

Question: Should the load used in the critical speed equation be the weight of the gear or the applied load?

 

[Heckler]

Have you calculated the static deflection with regards to the load and gear weights? Critical Speed calculations require you to sum of all static components which would mean the gear and load

Best Regards,

Heckler
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[Tmoose]

Not sure of the equation you are using, but I believe weight or mass (depending on the constants) is the answer.

Critical speed of single degree of freedom system is related to (k/m)^1/2.
Note there is no force term in there.
The exception would be if an applied load somehow stiffened things up (or down) as in axial force increasing restoring forces in a guitar string or inducing buckling (zero critical speed) in a column or shaft.

The importance and usefulness of static deflection is through its relationship to stiffness (k). If static deflection was "the problem" then space craft at zero gravity would be free of resonant frequencies .

 

[iainuts]

Per "Pump Handbook",
The critical speed of a drive shaft is determined by the deflection, or "sag," of the shaft in a horizontal position under its own weight. The less the sag, the higher the critical speed. ... Knowing the natural deflection of the shaft, it is possible to calculate the first critical speed from the equation:
Ncrit = 187.7 * ( 1/ deflection)^.5

It isn't clear if this 'sag' should take into account loads imposed on the shaft other than the weight. Looking through other literature left this question open. Machinery handbook for example, discusses load and seems to imply weight only, but is equally unclear.

Knowing the vibration is set up because of the mass revolving around the centerline of the shaft, I might think it is only the mass itself that is important, but deflection of the shaft is also a function of imposed loads due to gearing, belt tension, or other loads perpendicular to the shaft axis. Both weight and imposed loads are of course forces and equally capable of deflecting the shaft, so it is unclear to me what loads should be considered for critical speed.

 

[BigInch]

Your equations are only valid for a mass rotating about its own center of gravity. If the mass has an additional offset to its center of rotation, caused by some lateral load, the offset caused by that load must be considered.

BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com

 

[iainuts]

Thanks for the responces. Because of the seeming disagreement, I tracked down an expert in our company.

The answer I got was that the load used for the calculation is only the weight of the spinning mass as Tmoose said. Reason is that this mass creates a force which, if we consider it a vector, is rotating 360 degrees. This force will cause the shaft to wobble or gyrate around the shaft axis, and if it wobbles at it's natural frequency, it will induce stress above and beyond what it normally would induce due to a static load.

The applied load from gears or belts is a vector in a constant direction, unchanging. This load will not force the shaft to wobble, so it need not be considered for the critical frequency.

The total bend in the shaft can be seen as a static bend caused by external loads along with another rotating bend superimposed on that one caused by the rotating mass.

Does that sound about right?

 

[BigInch]

I was just checking back to clarify, but looks likes its already resolving itself. The offset masses don't change the natural frequency, just the amplitudes of the deflections.

BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com

 

[GregLocock]

Well I sort of agree with your expert, but would add that your static preload is caused by interaction with another system, whose masses and stiffnesses may well interact with your simple system, so rendering any estimate of the critical seed somewhat suspect.



Cheers

Greg Locock

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

 

[zekeman]

In the dynamic state you have the centerline of the gear deflection from the bearing centerline, say r and the CM distance of the gear to its center =e, the formula I found for r is (x=rcoswt;y=rsinwt)
r=e*w^2/[1-(w/wn)^2]=Ze
the classical 1 degree of freedom response.
where k is the spring constant at the shaft end.
m(x+ecoswt)"+kx=0 in x direction
m(y+esinwt)"+kr=0 in y direction
mx"+kx=mw^2coswt
my"+ky=mw^2sinwt
wn=(k/m)^.5
Steady state solutions are
x=eZcoswt
y=eZsinwt
At resonnance w=wn , causind r to blow up.
adding these two vectorially you get
the result above for r

 

[Heckler]

I also agree with your expert but like I stated earlier you should also look at the static deflection and how that will affect the system in the dynamic state.

Let me ask you this are you dynamicaly balancing the shaft assembly? I'm working on a miniature system that spins at 90K RPM. After a lot of testing I decided that getting the entire shaft assembly minus bearings balanced was the best thing. I also had a problem with shaft deflection and bearing preloads but have since solved those problems.