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How to estimate the force of impact of a whirling shaft on the ID of a containment 1

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WYSIWYG67

WYSIWYG67

Mechanical
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One of my colleagues at work approached me about a shaft in a test rig that has a crack and he was wondering if the rig case would be able to contain the shaft if it boke at the location of the crack. I've done analysis of various ballistic impacts from flying debris, but even though I have designed shaft coupling guards I have never approached it from this point of view. Is there a standard for containment of failed drive shafts? I haven't found one. Most of the literature seems to be centered around not allowing an operator to touch a moving shaft, not impact containment.

Anyone out there have experience with this. I can't share the specifics, but I'm looking for some guidance on methods and containment energy for what I would assume would be a whirling/whipping shaft as the rig spools down. There are vibe sensors and torque measurement plus they would hear it through the cell walls so the operator would shut down when a failure occurs.

And before anyone asks why they want to use a cracked shaft, its mainly schedule driven and no personnel are exposed to any danger.
 
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dvd - Thanks for the reply. That looks like one of the below ground spin pits we use to use for containment evaluation. I am familiar with ballistic containment analysis from a disk burst into three parts. Defnitely some interesting papers on that search. It doesn't really address what I was asking though.

The question I was posing in this thread is more about a shaft impact rather than a burst disk and really about the shaft letting go at one side near a span between two bearings - perhaps still anchor at the other bearing and whipping or even break and being flung into the case wall. Think cantilever with some imbalance that is spinning at some RPM.
 
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If the clearance between the shaft and guard is small, there is almost no impact, just sliding contact.
 
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yes, restricting the ends from turning into propellers is very important. As such you really want at least 3 if not 4 restraints if you don't have a full cage and aren't sure where it'll break

has a picture of the sort of thing i was familiar with.





Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376
 
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Curious what is attached to the shaft and at what rate it is spinning. My view on this is that all of the kinetic energy of the shaft and its attachments should be anticipated to go somewhere that it is not wanted - so whether it is a rupturing flywheel, or a failed shaft, there could well be a large energy release.
 
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Agreed. Definitely a potentially huge energy release. I am familiar with these types of problems, but I've only approached them from the stand point of fragments - the 1/3 disk type of problem like a ruptured flywheel or a turbine/fan blade type of release. I've never seen a treatment of a failed shaft addressed.

I'm thinking the most conservative approach is to assume the shaft fails at one end and assume the a small portion of the "free" end strikes the interior of the case. The real problem is how much of the energy remains rotational versus part of the strike. The answer might be somehow related to a pendulum swing.

The other alternative is assume both ends of the shaft fail and the entire shaft is flung. The rotational vs straight line velocity energy seems to be an issue though. How do you decide how much kinetic energy goes into the impact compared to friction as the shaft slows down rolling around the inside.

I am not a liberty to give the technical details.
 
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WYSIWYG67
Can we consider the rotation energy converted to linear kinetic energy at the impact and solve for the force? I am simply assuming that part of shaft/total shaft fracturing and impacting casing with same tangential velocity v =rω. Maybe we can consider same for bent cantilever shaft and impacting on the casing. You can increase that by adjusting either mass or velocity or by applying some sort of safety factor.
 
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