Shaft Support

Not withstanding any "technical" arguments:-
Cost, weight, time.

Ahh, you've added more:

Well, question is...says who?..other than my comment above

 

No AH, I have to disagree. There is a minimum recommended distance between bearings because of misalignment-induced bending loads on the shaft. The shorter the unsupported length, the higher the loads, for the same max deflection. It comes from the elastic-beam theory, as shown through a simple cantilever beam example (which you have rightly criticized as overly simplistic).

When shaft is rotating, this static load becomes rotating bending load, with known results in terms of fatigue strength.

I have witnessed a situation where the guy was trying to insert a shaft of generous diameter into slightly misaligned bearings. He had to give up (fortunately) and modify bearings - such was the bending force required to make the shaft conform to the bushings (or bushings to the shaft, I don't know). For the same amount of misalignment, he might have made it if the bearings were placed further apart- but then he would probably had dynamic problems (vibration). In any case, the sleeves would rapidly get worn out.

By the way, back to the cantilever beam example, it can be used with profit for the static analysis of two basic types of misalignment: parallel and angular. One just has to insert the correct unsupported length and beam displacement in the cantilever beam equation. Check the attached pdf.

Cheers

 

Isn’t that a bit meaningless? If you have a shaft of say 2.0m and a deflection of say 25mm, but now the length is 100mm, a deflection of 25mm on that length is gross..it’s outrageous!! Obviously the stress shall be higher

Well, you can out this another way, again

If we assume the maximum bending and torque = M, then we can work out what the diameter of the shaft is from simple bending theory viz: stress = M/Z

Thus for a circular shaft the stress or sigma = 32.M/(pi.d^3)

Or put another way the diameter, d = 2.15.(M/stress)^0.33

Stress being the material properties of yield stress.

Thus, you can relate that back to the spacing of supports, as this influences the magnitude of the bending moment in the transverse plane.

 

AH, I am not getting the problem here. Isn't it obvious that we are saying the same thing?

Take a round bar, 1/2" diameter. If it is 1 m long, then you will be able to visibly bend it with one finger.
If it is 10 cm long, you won't be able to bend it with all your forces.

1 m vs. 10 cm - that's the unsupported length in the previous posts. Mathematically:

F = K (yEJ / L^3) - force required to displace the bar by amount y;
M = K (yEJ / L^2) - moment at the constrained point;​

where K is a factor depending on the type of supports (boundary conditions).

Clearly, the shorter the length, the bigger the force necessary to bend the bar. That's exactly what I have been writing in all my posts here... I am failing to see the part where you see a contraposition to your examples.

 

I think that part of the point is that at greater distances, greater offset is likely. Ie, put two bearings right next to each other and I can align them pretty well. Put one on a separate surface 4' away and not so much.

 

It is simply a matter of the influence of misalignment of one bearing on the stress in the shaft. It is typical to calculate bearing influence coefficients based on 0.001" misalignment of a bearing. The closer the bearings are the greater the influence coefficients. Bearings experience movement due to temperature, wear, support structure movement, etc. Therefore, no matter how careful your are in initial alignment, there exists the probability that the alignment will change, thus having bearings too close together will lead to problems.

Now we have carbon fiber shafts which changes all the old parameters...

 

In all things you may have to compromise.What is wrong with self aligning beerings?

 

Tom
There's more than two bearings on the shaft.

If the shaft doesn't have enough length between bearings and there are close tolerances then an alternative is a bearing with some elasticity in the mount. As Johneck has pointed out it's axial misalignment that's the problem with short distances. The minimum is 12D. Below that you need to use specific design.

Somebody asked why torque\pwr is not in the bearing spacing formula. The max spacing is to prevent whirling vibration and a simple statics is a good indication for this. Max spacing is based on shifting the first whirling mode away by at least a factor of 2.

 

The theory of the 20 40 Rule does not apply in many cases such as gear boxes and the like and many applications have bearing mounted side by side. Theory just has to be ignored sometimes.

 

In the case of a gear box, both bearings are mounted in the same gear case and aligned in a factory. A much better situation than bearings mounted to a flexible hull. But, you are correct, there are times when compromises must be made. Designers need to be cognizant of the possible effects of making those compromises.

 

By this time I hope Mr Kendall (the origin of the topic) has had his question answered.

 

The questions usually get answered in the first 5-6 replies.
But we just love to beat every problem to death.

 

I think flexible hull (or temp changes) is a good point. It's not just about distances and offsets in a perfectly rigid system.

 

I think pretty much everything has been said about this issue...

 

Not quite all said. Have your shaft drive independent of the hull then you have no worries about alignment at installation or through the life of the boat.A trim able shaft drive.Ridgid drives do not make practical sense, you just create problems.