broken shaft of 280 KW SQIM 4

[genman196]

We had experienced a broken shaft on our 280 KW,SQIM while it was running under load. We dont know what was the real cause,as it was within its nameplate values and the OLR did not trip. Can anyone please give some hints of possible causes of broken shafts for large motors? thanks a lot.

 

[electricpete]

desertfox - sorry I didn't read your response. I will think about your analysis of the beam problem including torsion stress.

I didn't understand your point about the weight of the pulley creating plane stress. Are you talking about gravity weight or unbalance?

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[desertfox]

Hi electricpete

Yes I was just giving a possible example of plane bending ie a gravity load always acting vertically however in this case it might be very small compared with other loads, but read my post I think that the maximum resultant stresses occur at the fillet.
If you ignore the shear stress due to bending for the minute, whatever bending stress you have at the fillet based on the bearing shaft diameter, it is then multiplied by the stress concentration factor, in addition you also have torsional stress at the fillet, which also as to be multiplied by a stress concentration factor.

desertfox

 

[electricpete]

... the moment (and therefore bending stress) is the same on both sides of the inboard bearing, but the shear is higher on the pulley side.

I was right the moment doens't change from one side to the other. But I was wrong, the stress does change as Tmoose said - the higher area moment of inertia on larger shaft side (inboard) has lower stress for the same moment.

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[desertfox]

Hi electricpete

Yes on the side where the larger diameter is the stress will drop, however upto the fillet radius the value of stress is that calculated for the smaller diameter which is then multiplied by the stress concentration factor to give the actual stress in the fillet. The same applies to the torsional stress, the value of stress is that on the smaller diameter right upto the fillet, which is then multiplied by the stress concentration factor.
Once you have passed the fillet radius and get into the larger diameter the stress drops.
Our interest is where the failure occured which is in the fillet radius due to the stress concentration there and its the resultant of the torsional stress and cyclic tensile stress which causes the failure.

desertfox

 

[electricpete]

I agree - the stress in larger portion of the shaft is not much interest - I was just trying to catch up to understand what Tmoose had said.

Back to the subject of the different types of bending. In the context of rotating shaft discussion there are two types of bending: plane bending and rotating bending. In the way that Sachs uses these terms they are applied only in the reference frame of the rotor.

So rotating bending for the rotor would occur when the plane of bending rotates with respect to the rotor. The most common scenario would be if the direction of force were constant for a stationary observer (such as belt load or gravity). That would lead to the round IZ.

Plane bending occurs when there is a component of bending that remains in a plane that rotates with the rotor. (It looks like plane bending in the rotor reference frame). It is tough for me to imagine how this can occur other than the type of scenario I described...constant direction force in the stationary reference frame (like belt) combined with assymetry in rotor stiffness (like keyway).
1 - When keyway aligns with belt the shaft tilts toward the belt.
2 - Rotate 90 degrees and shaft tilts away from the belt.
3 - Rotate another 90 degrees and shaft again tilts toward the belt.
Comparing 1 and 3, the shaft tilted the same way in our stationary reference frame, but it tilted opposite direction in the rotor reference frame. There is a component of shaft bending back and forth in a plane that always contains the keyway and the point 180 from it. That is the plane bending (to my understanding)


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[electricpete]

Also there is a simple symmetry argument that supports the logic of a round IZ for pure rotating bending... if we start with the ASSUMPTION that the cracks started uniformly around the shaft... then in the presence of symmetric shaft and a load like a belt, there is no reason than anyone one of the cracks will reach a certain radius before the other... by symmetry they all reach a given radius at the same time (circular pattern at any time including final fracture and IZ).

To get anything different (under that starting assumption uniform cracks all the way around), we need some type of assymetry. It's hard to create an assymetry in the applied force if we have something like belt load or gravity, but we can create an assymetry in the shaft (keyway) that leads to assymetric stresses and assymetric crack growth... proceeds at different rates for different portions of the rotor and one portion reaches a given radius before the others and the pattern is not a circle but oval or football.

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[desertfox]

Hi electricpete

Thinking further on this plane bending I can't see on the shaft in question how you could have a plane bending component (which makes my earlier statement about mass of the components incorrect),any variation in shaft stiffness as it rotated would only vary the cyclic stress at that point.
Looking again a the photo of the shaft the IZ is only very slightly elongated which suggests mainly rotating bending fatigue as previously agreed.
Reading Sachs reference again page 66,fig5.7 and 5.8 gives an example of a shaft that initially had fractures due to rotational bending, but after after the fracture got halfway through the shaft, the final failure came from plane bending, the plane bending however appears to come from the blow received from the cutting roll which occured after every revolution.
So in that case the operating forces changed, in regard to the OP's shaft we are not aware of any changes to the operating force which might give rise to plane bending.
I agree with you that the force would have to rotate with the shaft constantly to have any plane bending in this case.

desertfox

 

[desertfox]

hi genman196

Happy new year to you.

I am not sure how familiar you are with bending moment and shear diagrams etc.
Anyway I just sketched out some dimensions and put numbers on them to give a feel for what were talking about.
I haven't included any torsional or bending shear stress in the calcs, but concentrated on demonstrating how the bending stress changes, due to the changes in shaft diameter
and the fillet radius area. It can be seen from my example that the bending stress in the fillet radius is 2.8 times the nominal bending stress in the 2" diameter ie 235.56lb/in^2 increasing to 659.568 lb/in^2 before dropping down to 29.44lb/in^2 in the 4" diameter.
Whilst my example is just a static case if you now imagine the shaft rotating, the bending stress in the fillet radius alternates from tensile stress to compressive stress every 180 degrees of rotation and it is this alternating stress which causes the fatigue. As you have probably read already
this stress does not have to be the yield or tensile stress of the shaft material but can be a value significantly lower.
Also if in my example we assume the 40lb load is correct for this particular set up, but due to an error during maintenance this 40lb is now set at say 80lb we automatically double the stress in the 2" diameter shaft portion, which is then multiplied up by the stress concentration factor of 2.8 in the region of the fillet radius, which could possibly cause a failure.
In your particular case it could be belt over tension or misalignment of the belt that as increased the stresses over the previous motor setup.
Anyway we will wait for you to post some more info as stated in your last post.

desertfox

 

[electricpete]

You neglected shear and included stress concentration factor, and your conclusion was stress was highest on the winding side of the bearing.

Bob included shear stress and excluded stress concentration and concluded the maximum combined (shear and bending) stress occurs on the pulley side of the bearing.

Two different statements, both true.

To find which is more relevant or closer to the true answer, we need to look at magnitudes as you have begun to do. A quick calculation of shear stress: area is pi inch^2, shear force is 40/pi lbf/inch^2 ~ 13 psi... not too important.

Recognizing the relatively small impact of shear stress, it is reasonable to suspect as you said that the highest combined stress will occur at the winding side of the bearing at the shoulder, even if well radiused.

I vote you a lps.

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[desertfox]

Hi electricpete

Well I would say the highest stress,even combined will be on the winding side.
Firstly which is why I asked the OP which side the failure was on, if and only if there was a stress concentration factor on the outer side of the drive bearing, which had a bigger impact than at the fillet radius would it have failed on that side.
Had that been the case though, the chances are that we would have had only one or possibly two fracture sites and clearly we have several or more.
As you rightly pointed out, the shear stress outside the drive bearing due to bending is 13 psi falling to half that value on the winding side, however the only stress concentration factor exists on the winding side so it can only magnify stresses on that side, so your shear stress due to bending as well as the bending stress would get magnified.
More significantly than that consider the torsional shear stress due to the 6" dia pulley and the 40lb force on the 2" shaft, it works out to 76.4lbs/in^2 far more significant than the shear stress due to bending.
Now this torsional shear stress is also subject to a stress concentration factor at the fillet radius which according to my graphs is about 2.
Therefore before we get the resultant combined stress we have to magnify the individual stresses at the fillet radius.
I will look at this later I hadn't said this earlier as I was trying to keep it simple while still making the point.

regards

desertfox

 

[desertfox]

hi electricpete

Thanks for the lps


desertfox

 

[Tmoose]

e-pete posted "Plane bending occurs when there is a component of bending that remains in a plane that rotates with the rotor. "

I believe an example of that would be "force" unbalance, and could only cause bending stress in the shaft between the bearings. Since the magnitude of that force would likely be less than 20% of the rotor weight (lest the vibration analysts storm the castle with pitch forks and torches), the effect would be superimposed on the constant vertical gravity load ( assuming the popular horizontal shaft configuration ). So although the unbalance force is ~ constant value and ~ constant direction relative to the shaft, the vertical load would vary from 80% rotor weight > 120% rotor weight once per rev. It seems to me if that was a significant load the ODE of motor shafts would be failing all the time. I base this on ODE bearing typically being smaller than De bearings. If the bearing IDs are the same then my assumption is flawed.

Here is a link to A Baldor premium efficiency motor spec.

http://www.baldor.com/pdf/TEFCSpec.pdf

"All shafts shall be precision machined from high-strength carbon steel suitable for belt and pulley drives (except as limited by 3600 RPM motors)."

Section 3.4.18 seems to be based on belt drive loading being the extreme case that effectively defines the motor shaft dimensions and material. The effect of unavoidable stress concentrations etc vs allowable endurance ( fatigue resisisting) stress for the material would all be part of the shaft design. I'd bet a dollar that the shaft design uses endurance limits for "infinite" theoretical life, so stress concentrations are harmless, UNLESS design loads are exceeded.

 

[desertfox]

Hi T.moose

According to my Ball & Roller Bearings Theory,Design and Application (Wiley Publishers) the Drive End and Non Drive end bearings for motrs are normally the same size for economical reasons, associated with the high precision machining of the housings.
The plane bending failure came from the Sachs links above which discusses how the beach marks are observed on the fracture surface as compared with the skewed beach marks with a rotational bending failure.
The plane bending failure shows beach marks progressing straight down till it encounters the IZ zone and involves one way loading similiar to a leaf spring or beam under a static load.
I don't think in this case plane bending is involved with the failure but if it were involved the load would of had to rotate with the shaft to produce the beach marks as described by Sachs and I agree with electricpete on that.
I also agree with you, that the motor shaft would have been designed for a life span and its likely that either the belt tension or misalignment of the belt as resulted in the failure, however we cannot be 100% sure.

desertfox

 

[electricpete]

Tmoose - unbalance load would cause the shaft to deflect and then stay that way (not continue to flex) since the bow rotates unchanged with the shaft. Imo, a rotating assymetry such as a keyway (combined with load such as belt) provides a good credible mechanism for the plane bending. In this case the bow is sensitive to shaft position, but reverses direction twice per revolution creating what I think Sachs calls plane bending.

desertfox
It is common practice to treat a ball bearing as a simple support. However it can provide some moment resistance such as described here (click on download)

http://www.nsk.com/services/basicknowledge/technicalreport/05distribution.html

On page 5.8.2 is a discussion of ball bearing moment response as a function of angle. I haven't studied it much. I expect it should be a function of not only angle but load because I have read in another reference that a ball bearing accomodates misalignment through it's clearances... there is a free angle of misalignment where no moment resistance is offerered, under no radial load. You can't bend it that far without moment when radial load is present because the radial load also wants to use up the clearance ina certain way. I am a little bit confused about the way they have treated radial loading on this page. In the figure 4 there is a plot of moment load vs angle for 6208 40mm bore, just a little smaller than desertfox' example 2" shaft (about 50mm). In theory with some work, the bending moment diagram might be integrated to give an angle to see what kind of moment is present.... if it is small then we know the simply supported approximation is good here. If it is large, then I think some iterative solution is necessary to adjust the diagram to account for the moment, which in turn depends on the diagram (slope). In any case I'm still confused why radial load is not explicitly accounted in Figure 4.

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[electricpete]

I should have mentioned the reason I mention all this: to the extent the bearing offers a moment reaction to the shaft slope, it tends to reduce the bendings stress on winding side compared to pulley side.

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[Tmoose]

Moment is moment. To get stress or deflection/slope the shaft geometry and material E must be entered. It would take a heap of shaft bending or slope just to use up the clearance in a single ball bearing. I believe the bearing would run hot, and full of HFBD type vibration, and maybe even noisy the instant it was asked to run while handling any real moment.

 

[Tmoose]

Hi Desertfox,

FAG used to give that Wiley book away for free off their website around 1998. That's where I got mine. Nice survey type text.

EPete and others undoubtabley have way more current motor experience than I do, and Genman could report what size bearings his motor has, but I think it is not uncommon for motors in the ~ under 200 HP range to have different DE and ODE bearings. I believe some are even set up so a roller bearing can be directly substituted for the standard ball bearing when serious radial loads (belt drive) are anticipated.

I could not find an online motor catalog with the level of detail necessary to correct my misconception.

 

[genman196]

Thanks desertfox for helping me out about the discussion and thanks all for sharing the information, I try to learn as much as i can from this thread and its a whole lot!
Sorry for the delay in posting more details,
shaft diameter where pulley is mounted : 100mm
shaft diameter where bearing is mounted : 110mm
shaft lenght extending from bearing : 250mm
shaft lenght where pulley is mounted : 210mm
pulley diameter : 470mm
pulley width : 360mm
flywheel diameter :1200mm
motor rpm : 980 rpm
Pulley is coupled by 11 belts to the flywheel.I have no details about the shaft material.
Also sent a pic which shows keyway position relative to the break.im compiling some pics of the installation which could be of help and will post again soon. thanks.

 

[electricpete]

Tmoose - I suspect you are right since I checked several references and they all treated a ball bearing as a simple support. I will try solving it lataer tonight to satisfy my curiosity
y' = Integral {M(x)}dx + C1
y = Integral {y'(x)}dx + C2
M(x) as solved already
C1 and C2 solved satisfy y=0 at both bearings.
convert y' to moment using the chart and compare to moment diagram

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[desertfox]

Hi electricpete
Thanks for the interesting link.
Right, have looked at the link its quite heavy reading, in fact some of it goes right over my head, I would agree that the bearing may take off some of the bending stress at the winding side of shaft, but I doubt it would take it all off, if it did then the shaft from the drive bearing outwards could be treated as a cantilever beam and the shear force and bending moment would stop at the drive bearing. In addition whatever stress is in the shaft at the winding side, it is magnified by the stress concentration factor, this also should include the rotor mass which has been omitted from my simple calculation.
Now in truth the majority of motor failures are due to the drive end bearing failing for one reason or another,I saw a pi chart for motor failures and I think 51% was drive bearing failure and about 2% shaft/coupling failures.
The design methods we use however are a simplification and truthfully the shaft is supported by bearings which are neither simple supports nor "built in" so its not easy to be 100% sure exactly where the maximum stress is, we can only approximate and I think the simple supported model is the best as you have indicated.
Just going back to the rigidity of the bearing I would also go along with T.moose's statement regarding excessive bending of the shaft to take up all the bearing clearance without knowing about it fairly soon, because just looking at figure2 on page13, the inner to outer race angular misalignment (dependant on load) can be about 1/6 to 1/2 a degree which when taken to the end of the motor shaft would be quite a deflection from the horizontal. Also looking at fig4 on page 14, for a small angular misalignment in the range above,it would generate a large moment in the bearing which I feel wouldn't go unseen.

this link posted earlier shows critical areas for shaft failure and its anywhere from the drive bearing to the shaft end.

http://vibration.pknu.ac.kr/vibration_pds/motor_diagnosis/Shaft_failure.pdf

Hi T.moose I am jealous you got the book for nothing!
I had to pay for mine at a discarded library book sale, but it was only a £1 and at least its hardback. I looked for information on motor bearing arrangements too, but drew a blank like you, I'll carry on looking though.

Regards

desertfox