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Whirling speed of a propeller shaft

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schopenhauer

Marine/Ocean
Mar 17, 2018
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Hello everyone,

I have been assigned the task of designing a propeller shaft for a truck, one of the requirements is to avoid the phenomenon of whirling. This is a new topic for me, so I will summarise what I believe I know until now:
- At certain critical speeds, the elastic deflection of the shaft gets very large
- To avoid this problem, the critical speed of the shaft must be above its maximum operating speed
- For some reason, these critical speeds coincide with the natural frequencies of the system

Based on this, I implemented some analytical equations in Excel and built a FEM of the shaft to extract the natural frequencies of the system. The numbers agree very well, but I have a few questions:

- Is there a way to simulate the rotating speed of the shaft in Abaqus to obtain the real displacements at the whirling speed?
- How can I study the influence of torque and twisted geometry on the critical speed of the shaft? I’ve read that the impact can be significant.
- What does it mean in real life for a propeller shaft simply supported and fixed conditions? The boundary conditions have a deep impact on the critical speed of the tube, the only information I’ve read until now is that short bearings act as SS supports while long bearings as fixed supports. Why?

Thanks in advance for any help, please let me know if I can clarify anything.
 
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Does the "truck" have a multi-piece driveshaft, requiring one or more center support bearings? Center support bearings are typically extremely flexible, so the entire system is running super critical, and small diameter driveshafts sized for torque requirements not first bending critical speeds work just fine.

Plenty of understandable theory and closed form solutions here -

"The only information I’ve read until now is that short bearings act as SS supports while long bearings as fixed supports. Why? "
If you look at the design of transmissions and rear axles/differentials the output and input shafts are designed to handle reasonable radial loads, so the boundary condition becomes the stiffness of those component mountings which are in turn relatively soft for reasons of NVH.
Do you have a car? Open the hood and push the engine around a bit to get a feel how flexible engine ( and transmission ) mounts are.

U-joints are very flexible in bending, so often are considered a Simple Support and thus define the "length" of the driveshaft.
 
Hi Tmoose, thanks for your answer.

The old design had a central bearing. The new shaft is going to be made out of carbon fibre, since the material is stiffer, we can go from end to end without a central bearing.

My question regarding the rigidity of the "supports" was oriented from a beam theory point of view, where simply supported are those supports that constraint displacements but not deflection of the beam at the point. I was looking for a similar criteria to apply here to know if I can count on the supports to increase the bending frequency of the shaft (from my calculations, 6000 rpm for SS and around 10000 for fixed supports). This particular shaft is conected by yokes to the spline yoke at one end and to the axle coupling at the other end. From your message, I visualise the possibility to increase the accuracy of my analysis by taking into account the stiffness of the components acting as supports (the spline yoke and the axle coupling?). Do you have any experience translating that idea to a FEM model?

Finally, the disc coupling and rubber coupling are also part of the system, any idea on how these components modify the critical speed of the system? Is a modal analysis (finding the natural frequencies) still the rigt strategy for this?

Again, many thanks for your reply.
 
Hand calculations make all sorts of horrible approximations just so you can get an answer. In reality the termination at each end is elastic and has inertia, since in actual whirl the engine gearbox and axle are all involved. The reason that the whirl frequency is the same as the bending frequency is that it is a positive feedback effect, where the excitation is proportional to the bending in the shaft.

Cheers

Greg Locock


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