HELP!!! Forces in cable for safety rigging

[gtg477f]

I have been asked by a client to analyze installing a 1/2" steel cable that spans 40 feet to existing roof beams. Before I analyze the roof beams, I need to know the lateral force caused by the cable. The client asked if the cable could sag 2". I take a 250# person falling at midpoint of cable (multiply by 2 for impact for total of 500#) and neglect weight of cable. I sum moments at support to find downward reaction (naturally half of total) and then sum moments at center of cable to determine lateral reaction at support and I get 29.4 kips? So your telling me if a 500# load is hung from a cable that is only sagging 2" the lateral force at the ends is 29.4 kips??? If I let cable sag 24" the load becomes more manageable. Why is this? Does anyone know? I have read up a little on the catenary curve but I still cant grasp how 500# turns to over 29000#.

Calcs:
Cable span = 40 feet, sag is 2" (.17'). Supports are "A" and "B" and midpoint of cable is "C"
Sum moments at "B" = Ax(0 FT)-Ay(40 FT) + 0.5 kips(20 FT) => Ay = 0.25K
Sum moments at "C" = -Ax(.17 FT) - Ay(20 FT) + 0.5 kips(0 FT) => 0.17Ax = -Ay (20 FT)
Substitute 0.25 kips for Ay, Ax = -0.25 kips(20 FT) / 0.17 FT = 29.4 kips

If cable sage 2 FT, force is only 2.5 kips

 

[zcp]

If this is for fall protection, please be sure to have it reviewed by someone experienced in that area. The 5000# requirement is for vertical anchors. Horizontal lifelines (HLL) require a factor of safety of 2 to 1. And yes, that is on the extremely large end reaction of the horizontal cable which is often MUCH greater than 5000#. There are options such as energy absorbers that can be placed in the HLL to help lower the end reactions.

Suggestion, take this to some of the fall protection vendors (DBI / Miller / etc.) and see if you can get a standard HLL system w/ absorber to lower the end reactions to an acceptable level. Also see Nigel Ellis's textbook on fall protection which includes discussions about multiple spans and multiple people (don't forget the rescuer!). At the very least, find a qualified professional engineer (P.E.) experienced in fall protection to help you with this.

ZCP

http://www.phoenix-engineer.com

 

[MiketheEngineer]

I think the basic simple equation is P = wl^2/8 d --- d= sag. Watch units. You will be shocked by the numbers!!

 

[BAretired]

I think the basic simple equation is P = wl^2/8 d --- d= sag. Watch units. You will be shocked by the numbers!!

That would be approximately correct for the horizontal component of cable force due to a uniform load 'w' if the sag is small. It assumes a parabola instead of a catenary.

It is not applicable in the above situation as the primary load is concentrated, the weight of cable being neglected.

BA

 

[MiketheEngineer]

When I look up catenary cable - that's the equation I find.... Do I know if it is correct - a couple of sources say it is close enough!! And W is a plf load - so translate it into a point load...

 

[BAretired]

For a point load P at midspan, the simple span moment is PL/4. Neglecting the weight of cable, the horizontal component of the cable is PL/4d where d is the sag.



BA