Cable griper axial load calculation
Cable griper axial load calculation
(OP)
Hi Guys,
We need to design a small electrically operated unit to clamp and unclamp onto 12 or 14mm diameter 7/19 IWRC stainless steel cable and act as an emergency stop. The unit will need to be clamped and repositionned at the same point on the cable many times a day so we must not fatigue the cable. I have been unable to locate any calculations for the clamping force required on a cable to achieve a certain axial SWL. Does any body have any experience in how much clamping stress can be applied to the cable with out long term damage and a frormula for the axial load capability for this force as I understand in practice it is not just force x coeficient of friction. Our cable supplier has been unable to help.
Thanks
Kevin
We need to design a small electrically operated unit to clamp and unclamp onto 12 or 14mm diameter 7/19 IWRC stainless steel cable and act as an emergency stop. The unit will need to be clamped and repositionned at the same point on the cable many times a day so we must not fatigue the cable. I have been unable to locate any calculations for the clamping force required on a cable to achieve a certain axial SWL. Does any body have any experience in how much clamping stress can be applied to the cable with out long term damage and a frormula for the axial load capability for this force as I understand in practice it is not just force x coeficient of friction. Our cable supplier has been unable to help.
Thanks
Kevin





RE: Cable griper axial load calculation
This clamping device may act hundreds of times a day, but never act in an emergency situation, so it shouldn’t damage the cable. It must apply some pressure to hold it in place during each cycle. I don’t have any cookbook formulas off the top of my head for this situation. In an emergency stop you may not care about damaged cable. But, I would consider the following: do continue to involve your cable supplier, they know their product better than we do, you’re asking the wrong questions, talk to their engineering people not a salesman; the real gripping part has teeth on it, but they are covered by some material (cable protecting covering, maybe just a couple layers of plastic electricians tape, not very elegant but makes my point) that they will bite right through when needed; any emergency loading is applied in a way, through a cam and lever arm, that causes the teeth to activate, pinch and dig in; maybe it even grips the cable more because the whole device rotates under load and puts the cable in a tight S shape btwn. the teeth. The device might be spring loaded to apply normal working (holding) pressure on the cable, but then the cam action would drive the teeth into the cable in an emergency application. You need to allow some time and distance for this to occur, and the cable will stretch and tend to twist under this sudden loading.
RE: Cable griper axial load calculation
the pictures in these articles are confusing to me, mainly because the view isn't from where I would normal see them:
http://www.skilifts.org/old/tech_tb41maintenance.h...
http://www.skilifts.org/old/tech_omega.htm
http://www.polxwest.com/Articles/MEmagazine/Gettin...
In any case, the manufacturers of those ski lifts might provide some information.
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RE: Cable griper axial load calculation
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RE: Cable griper axial load calculation
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RE: Cable griper axial load calculation
Kevin
RE: Cable griper axial load calculation
Thanks
Kevin
RE: Cable griper axial load calculation
http://www.apsltd.com/c-308-harkencamcleatsaccesso...
RE: Cable griper axial load calculation
RE: Cable griper axial load calculation
The wedge and taper driven down by the trolley momentum sounds as good or better than what I was thinking. It should be round so it grips the max. area (circumference) of cable, and it should be long enough, with gradual enough taper, so that it grips a good length of cable. The thought being that too short a grip might sever the cable, or individual wires, rather than load many wires and making the cable stretch. You want to grab all of the wires in the outer strands, so that takes some length to accomplish. You do not want to be cutting individual wires before many are brought into play. You might do this in two steps, with two mechanisms, a rubber shoe under some pressure, electrically actuated, which holds things in place dozens of times a day. Then, take two mating cone shapes, both truncated, cut square to the central axis, of some length, with very slight taper; the inner tapered cone has an inner straight bore less than the cable dia.; cut this inner cone lengthwise into three 120° segments, and cut teeth on the inside bore. Tooth depth should be a fraction of outer strand wire dia. This inner cone stands higher than the outer cone in your whole motor driven mechanism This inner cone never touches the cable unless it is driven into the outer cone by the momentum of your trolley. I may have this upside-down, I have to think on that a bit. If your rubber shoe holds the inner cone longitudinally on the cable, and just above the shoe, but not contacting the cable, then the outer cone is driven down on the inner cone. This may be a better solution and orientation. At first the shoe holds the inner cone, but as soon as the teeth start engaging the cable the shoe is no longer needed, so it doesn’t need excessive holding power.
RE: Cable griper axial load calculation
Regards,
Cockroach