Double Bullwheel Shaft Loading

[1athoti]

Hi,

I'm working on a bullwheel capstan system for underground cable pulling.
Can someone give input on whether I'm approaching the loads on the shafts correctly?

From capstan calcs (mainly US Navy wire rope handbook) I'm aiming for 100,000 lbf on final wrap and should be able to achieve it with the 2 sheaves driven by hydraulic motors, geared down etc.
Here I'm considering each individual shaft (both being similar in loading).
The gear shown is driven by another pinion gear with a torque which would create aprox. Wt = 36000 lbf and Wr = 13000 lbf on each gear.
Can I approximate Fpull at center of shaft as 2 x wire rope tension = 200,000 lbf to calculate total loads on the shaft?

Thank you!

 

[3DDave]

Look at the capstan equation to see how the loads will be distributed from wrap to wrap.

See https://en.wikipedia.org/wiki/Capstan_equation for more information.

 

[mfgenggear]

OP
Look at other capstan designs. Your design looks weak. Not robust.
No gears on top of the capstan. Notice the design of the capstan on other are flared.
Both ends.

 

[1athoti]

@mfgenggear Yes, that isn't the full assembly. Here is a better snapshot. However, can you elaborate on "Notice the design of the capstan on other are flared. Both ends." ?



Yes thank you, I went through the capstan equation.

I have 6 wire rope winds on each sheave. Considering Steel on Steel and greasy conditions, navy handbook suggests a coeficient of friction of 0.07. With 1000 holding force I would achieve full pulling force within about 3-4 wraps. But, I understand having more wraps is beneficial for wire rope life and I could also decrease the holding force (tension achieved by the storage drum rotation).

Since the maximum torque of drive system develops a wire rope tension of 100,000 lbf, I should be OK to approximate Fpull as 200,000 lbf rougly at the center of the shaft?

 

[3DDave]

I would have worked backwards from the 100,000 lbf output, subtracting the capstan amount; also I think excess wraps = more fatigue cycles in this application. If they were onto a single drum where the bending happens only once and remains bent around the drum that would be different, but this is bend-unbend-bend-unbend... the wire rope is getting an extra workout.

Anyway, you can add up the cumulative forces to see what the load is on each drum.

 

[1athoti]

Thank-you @3DDave good observation! I'm afraid I made a mistake on the calcs I had previously shared above. I think I should not have added the angle (in the e^f*angle) to each subsequent wire rope wind as I did. It should look more like this, in which case with a higher coefficient of friction and value of holding tension, I do need all the wraps. Not to mention I may need more holding tension if the wire rope happens to be lubricated.

But, to answer my first question, if I add up all the tensions for one of the sheaves I indeed get the 200,000 lbf value so I would expect it should be safe to approximating it at center of the shaft or distribute the load over the width of the sheave.