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

 

[gwolf2]

1) It is the plastic deformation of the internal components which is reducing the peak torque which you are reading externally.

As an example calculation I just straightened out the problem into a straight line impact between an 18000J object and a 100mm square aluminium block. With a purely elastic material the block compresses 2.3mm with a peak load of 15.6 kN before coming to rest, if I account for plasticity in the material I get 5.5mm compression and 3.6 kN peak load.

I think something similar is happening internally to your rotor as it deforms metal plastically whilst coming to rest.

gwolf

 

[IKBrunel]

Thanks gwolf2,

In your example the plastic deformation has elongated the work distance and so the peak force should be lower. I have a known stopping angle and a known angular momentum which leads to the purile result that I should have a known average torque. As you'll know, your average forces, between the elastic and plastic cases, should differ by the ratio 5.3/2.3. The nub of my problem is that I'm not accounting for all the angular momentum by either the failure strength of the fasteners or the shock response

But I'm very interested to know, how did you calculate that?

I am leaning towards surmising that the angular momentum vector must change notably upon a rotor crash and that our measurements need to also be out of that initial vector plane.

 

[zekeman]

Could you upload a sketch of the assembly, stator, rotor and indicate the location of strain gauges, etc.

My confusion starts with you stopping the rotor within 180 degrees; so you are mounting the strain gauges in the stator.
And the secondary "shock" due to vibration is also confusing, since the traces should show reversal.

Please correct me if I am wrong.

I think that there is a much simpler explanation than out- of- plane motion which I believe is second order

 

[gwolf2]

I assumed a compression impact on a 100mm cube made of a medium strength aluminium alloy. I compared 18000J with EE=1/2.K.X2 where K=A.E/L for the elastic case (kitetic energy=spring energy). For the plastic case I used ramberg osgood to get an elastic-plastic stress strain curve and then did the same as for elastic but as a piecewise integration.

It becomes obvious that plasticity soaks up lots of energy internally. The strain gauges which you have on the outside are just measuring the elastic response of the casing to the elastic plastic loads developed internally. They will never add up to the kinetic energy. As has previously been stated, the conservation of momentum is applicable only for elastic systems.

Your numbers for the bolt shear strength don't add up. I would check them, remembering to use the ultimate shear strength and not the yield shear strength. Even given that I am still surprised that in a test you got M6 bolts to react a 14KNm torque. Worth double checking the calcs and test results.

I don't think it is practical to predict the failure torque in this scenario because of the complexities of the internal collisions. I have seen it done on big engines and they use a combination of large DYNA3D models and test. It doesn't sound like you have the resources to do this.

I recommend that you use some sort of worst case analysis and stress the case and flanges to this load.

gwolf.