Shock Mechanics Calculations for Rotor Failures 1

[IKBrunel]

Hi,

I am studying the rapid deceleration of a turbo rotor during a catastrophic failure and I need to develop a hand-calc analysis method to help assess the results from tests.

The turbo product is about 150mm dia by 200mm tall. The 1.2Kg rotor spins at 60,000rpm and when it fails it stops in less than half a revolution.

Roughly:
Rotational kinetic energy = 18KJ
Angular momentum = 5.76Kgm2/s
therefore the failure torque for 0.4ms stop = 14.4KNm

The total product mass is 9Kg and is secured by some flange bolts which must be proved to be substantial. There are many flange and product variants to consider and we need a way of proving all products are safe based on the results of a few tests.

Using strain gauges the failure torque and shock duration of one failed product have been found (5.6KNm and 0.4ms) but the the included area under the curve is far from equalling the change in angular velocity.

Furthermore, the indicated failure torque suggests our bolts should have sheared easily but they didn't.

I wondered whether to use a bolt strength hand calc that maybe allows for a torque/acceleration amplitude magnification factor based on the stiffness and coulomb damping of the bolted joint.

Does anyone have any experience with such calculations, or can anyone point me to relevent effort elsewhere?

Many thanks


Matt

 

[desertfox]

Hi IKBrunel

If the bolts didn't fail what did?
If you can put a sketch up of your bolted joint configuration we might be able to help further, all you have told us is that the bolts didn't shear, but how many bolts, bolt material?, bolt preload?, bolt size?

http://www.roymech.co.uk/Useful_Tables/Screws/index_screws.htm

the site above as some good formula for bolted joint strength.

desertfox

 

[gwolf2]

You forgot the plastic strain energy expended in ripping the innards to bits. This is why there is a difference between kinetic energy and reaction torques.

 

[GregLocock]

Once the bolts are partially in tension then you have a whole new thang. Nice analysis thus far, by the way.



Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[GregLocock]

By teh way, what coefficient of friction are you assuming across the flange?

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[IKBrunel]

Thanks for the replies, to answer then:

The strain energy is an interesting element but I am happy that it is a negligible effect here. I calculated the strain energy of a few components and it was quickly apparent that strain energy is a few decades lower than rotational kinetic energy. It would also suggest a larger failure load rather than a smaller one.

To help keep this in perspective, I calculated the strain energy on one rotor part ejects it radially at 30mph but the tangential speed of the (R64mm) rotor tip is 900mph.

Regarding fixings, the bolting configuration is:
4off M6 12.9 (shear strength = 600N/mm2) on a PCD of 178mm torqued to provide a preload of 11KN (although this is only relevant if coulomb work is considered).

This joint design can then be shown to give 6KNm resistance. We also tested with just two bolts that gave a calculated 3.7KNm resistance and found that even they could withstand the full 14KNm torque of the failure.

So I'm missing a trick somewhere, all the above is a static assessment and I wanted to see if I could undertake a dynamic assessment. I think it must be to do with the compliance of the joint where component stiffnesses would limit the peak loads. I am not sure if coulomb damping is releveant here


All comments welcome.

Thanks

 

[IKBrunel]

Thanks, Greg,

I am using 1.4 as a dynamic friction coefficient, it's an aluminium flange on an aluminium interface.

A simple calc shows that if the flange slips round, say, 2mm then only 124J of work has been done. I was surprised by how small this was.

I have found fleeting references to dynamic shock calcs and I'd like to find more but I also wanted to see how anyone else here handled such problems.

Thanks

 

[IKBrunel]

gwolf2,

I misunderstood your point. The loads must still be there though, however the energy is dissipated.

Because we measured the failure torque at the mouting interface for the product we should read the net torque correctly, however the loads are distributed. This is becasue angular momentum must be preserved. I think...

Maybe there are gyroscopic effects translating the axial torque into a different vector....

 

[rb1957]

how'd you get from strain gauges to torque ?

 

[IKBrunel]

Hi rb1957: we calibrated the strain gauge by applying a large torque (dead weight on a lever) to give a KN/V value then zero'd

 

[IRstuff]

It appears that you're applying a static analysis to a transients problem. A transient load has a much lower impact that a static load. You'd go unconcious from prolonged 10-g load transient, but you could stand sufficiently short 2-ms, 60-g impulses for hours, and would only realize the fact when your muscles start complaining.

TTFN

FAQ731-376

 

[IRstuff]

One way to look at it is to consider how much actual work energy is required to torque the flange off, and compare that to the 18kJ of available energy.

TTFN

FAQ731-376

 

[btrueblood]

The torque x time (or energy) required to stop the shaft is (can't be) the same as the torque-time or energy required to spin the shaft up to speed...

Why? Because the rotor is no longer an integral device once it ruptures. The rotational inertia of the broken/damaged stuff that is still hanging onto the shaft is much less than the original, intact rotor. The remaining pieces undergo collisions, fracture, and all kinds of energy-dispersing stuff as they rattle around inside the case.

Regarding angular momentum - momentum is only conserved in elastic (conservative) collisions. A rotor burst is pretty much a non-conservative event.

http://en.wikipedia.org/wiki/Momentum

 

[btrueblood]

Whoops, sorry got it backwards. Momentum is conserved, but kinetic energy is not. The original point about angular momentum being conserved by multiple little bits flying around is correct.

 

[CoryPad]

Any chance you could post a picture. If this is a standard type of assembly configuration, then the bolts are contained within fairly confining holes. Is there any evidence of slip? Any fastener fracture? Hole deformation or tear-out?

 

[zekeman]

"Using strain gauges the failure torque and shock duration of one failed product have been found (5.6KNm and 0.4ms) but the the included area under the curve is far from equalling the change in angular velocity."

I believe you meant angular momentum .

I came late to the party and I have a few questions.

Have you done this test with systems that did not fail?

You mention that the stopping angle is less than 180 degrees which means that that for a uniform deceleration the time would be less than 1msec by my calcs. So. when you give a value to the strain gauge torque, is it the peak or the average?

Summarizing, the problem is, assuming all the momentum is in the impeller, the area under the torque- time curve should be approximately 5.76Kgm^2/sec

Now if the .4ms is correct, the average torque by your calcs should be 14.4KNm. BTW, do you have the time history of the strain gauge torque?

But, how much faith do you have in this measurement, or for that matter, the time duration. I believe the problem is in these measurements, including the dynamic performance of the strain gauge. You might get a better handle on this if you could accurately measure the angular stopping distance.
Together with the energy, you could calculate the average torque with much more precision.

T*@=energy in impeller
T= average torque
@= stopping angle

 

[IKBrunel]

Hi,

Thanks for the responses.

I did mean angular momentum rather than velocity (which is what you get from the acceleration/time graph), sorry for the confusion.

Corypad,
I haven't got useful pics at hand right now but I'll look for them. Simply imagine a cyclindrical product with four tapped holes in the base. Sometimes the four bolts will exhibit sideways strain, being slightly dog-legged.

We don't stipulate what through-hole dia is used to mount the product but we have seen slip and slight fastener deformation. Hole tear-out has not occured ... yet.

Zekeman, to answer:
We have generally good success in passing this test. The mounting bolts are not thus far the failure mode but I do need to prove each product's bolt strength. When we've overcooked the design or the test it is the product that disintegrates in a wholly catastrophic manner - most envelope parts are blown to bits. During a contained failure, the rotor disintegrates anything inside the enclosure and the bols have the above-mentiond deformation.

The value I gave was peak torque (5.6KNm). I have posted a picture of the strain-gauge response. A second strain gauge was used and this shows a lower torque value.

so you can see that we have chosen the initial peak as being indicative of the failure torque. I have seen that in MIL-STD 810 there is quite a different treatment of shock-reponse plots and I do not yet have a good understanding of that.

Other Thoughts

Strain Rate Sensitivity
I am now aware of the effect of strain rate sensitivity and how the strength of the bolt material will increase under v high strain rates such as these. It seems there is not much data available on this - can anyone help with data for 12.9 or A2-70 bolts? This may explain some test results we have had where the bolting configuration should have failed with ~25% of the failure load but the bolts survived with little deformation.

Gyroscopic Effect
I am trying to determine if the torque could be translated into an out-of-plane torque simply by the moments applied during crash. It's hard to imagine as the rotor always appears to have spun around it's axis with very little pitching, but then I appreciate that doesn't count for much. This is quite a departure from my assessment so far but it would explain why we are not picking up the full torque expected from measurements made around the rotor axis. so we would need to measure torque in three planes to identify how the full angular momentum is transferred into mounting strain. And I would then also need to factor the resultant load into the mounting bolt calcs. I confess to be being confused about this right now.

Thanks again for the help, your comments have been invaluable.


Matt

 

[zekeman]

I have a problem with your assertion that the area under the strain gauge curve is the peak* 0.4ms.
Unless I am mistaken, your pulse duration is at least 4-5 msec wide as seen by the first trace, so the area under the curve would be considerably more than your stated value.

 

[zekeman]

"I am studying the rapid deceleration of a turbo rotor during a catastrophic failure and I need to develop a hand-calc analysis method to help assess the results from tests.

The turbo product is about 150mm dia by 200mm tall. The 1.2Kg rotor spins at 60,000rpm and when it fails it stops in less than half a revolution."


Exactly how do you induce this failure, or does this happen spontaneously? Please explain.

You also say the rotor disintegrates. Does it happen by increasing the speed? Or spontaneously.

What inyour opinion is the cause of the deceleration?

I'm not too clear on this. This should yield to a hand calculation for sure, but something is amiss here.

 

[IKBrunel]

Zekeman, Thanks for your thoughts.

From your questions I can see how much it would help to explain this:

Inducing Failure
The rotor hub is drilled axially to weaken it. Experience tells us how much to drill. We spin the rotor to 60krpm, run it some hours and if it doesn't fail we take it off and drill it a bit more. And repeat. This works well as we often have failure at or near full speed and hence can burst with the full rotation kinetic energy. We are trying to simulate a fatigue failure from poor quality hub material.

Failure Mechanism
The turbo runs with sub-millimetre tolerances between the rotor and stator stages and a burst is commonly arrested within < half a revolution. The turbo blades are stripped off around half of the impeller and clearly show how far the rotor has turned between failure and stop. We can corroborate the duration read on the strain-gauge traces with observation.

So the response after ~0.4ms is seen as an elastic vibration.

So I hope you agree that with such a useful visual record we can try and regard the response as simplistic.

I do wonder if strain gauges are best. I know our ICP accelerometers are not ideal for the job but I did read that MEMS accelerometers may be best.

However, I am still left with two curious and coherent facts:

1) The measured failure torque is much lower than the expected 14.4KNm average.
2) Mounting bolts that should fail do not, they survive this.

So perhaps I need to consider out-of-plane measuremenmts?

Thanks


Matt