Winch drum and cable wrap question

[dpsmith85]

I'm in the process of writing a report on a winch drum failure. Our winches are wrapped around the drum and then terminated with a u-bolt on the flange of the drum. The customer is stating that the cable slipped out from under the u-bolt. This may have happened but I want to know how I could estimate the force required to do this. The question really comes down to the number of wraps (not layers) on the drum itself. The more wraps the less force is actually exerted on the u-bolt. I'm estimating that there were 3 total wraps on a 5" diameter winch using 3/8" cable. I know this is a loaded question where many outside factors could come into play, however, I'm just trying to understand the basic math behind the idea. Any thoughts?

-Thanks

 

[BrianE22]

The ratio of tension on the two ends of the wrapped cable would be:

T2/T1 = e ^ (mu*B)

where mu = coefficient of friction
B = arc of contact in radians

 

[dpsmith85]

Brian

I'm confused on what this equation is really giving me. I've attached a picture for more clarification of what I'm talking about. What I am gathering (which could be wrong) is that if I make a single wrap around a winch drum (cylinder) and input a force T2, then T1 is the force on the opposing side of the winch drum. I looked up a steel on steel coefficient of friction and found it to be 0.8. The arc radius if I assume 1 wrap is 2pi. I used the equation posted above to calculate the "output tension" of the cable and got 0.05 lbs? The strange part is that based on this equation the output tension would increase with wraps not decrease. I know I could be twisting this up and thinking about it all wrong so hopefully you can provide some insight.

Thanks

 

[BrianE22]

If the load on the winch was hanging on the end called T2 (from your sketch), and the torque applied by the winch motor to the drum was clockwise, then T1 = 3000/152 = 19.7#.

Your value of .8 for the coefficient of friction is not very conservative. If you want to know the worst case load on the clamp (attached to the end called T1 in your sketch and connected to the rotating drum) then you should assume a low value for the c.o.f. (.1 is fairly common)

 

[Compositepro]

Your clamp needs to be able to hold the full tension of the cable. How do you know how many wraps were on the drum when slippage in the clamp (not necessarily failure) occurred? The friction equations apply when things are allowed to slip. But with slippage the cable will eventually come out of the clamp. Theoretically a piece of Scotch tape will hold a cable with 10 wraps around the drum. I wouldn't suggest doing that though.

 

[btrueblood]

A u-bolt type cable clamp is probably the worst in terms of long term holding capability, especially if there are cyclic loads applied. It depends on development of crushing force on the cable; the problem is that applying tension on the cable tends to make it contract in the cross section dimension, which then reduces the applied force from the clamp bolts. Crimped end terminals are better. Better yet are spelted fittings.

 

[dpsmith85]

Brian,

Just out of curiosity do you know where you got that equation? Since I am writing a report I would like to back it up with some type of source just in case anyone questions the validity of my findings. Thanks again for the help.

 

[BrianE22]

Out of my college dynamics book. I'm sure it's on the web somewhere, like here:

http://www.dtic.mil/cgi-bin/GetTRDoc?Location=U2&doc=GetTRDoc.pdf&AD=AD0297309

 

[Windward]

This topic is covered nicely in Mark's Standard Handbook for Mechanical Engineers, I have the Fifth and the Eighth Editions, and the information is at the end of Chapter 3 in both.