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

 

[SnTMan]

gtg477f, to get a feel for these problems forget summing the moments for now, just work the triangle. A relatively small vertical force can yield a huge horizontal component. The tighter (more horizontal) the cable, the worse...

Regards,

Mike

 

[BAretired]

For a sag of only 2", the horizontal component of the cable, neglecting cable weight, is 30,000#. The horizontal component of the cable force H is given by M/s where M is simple span moment and s is sag. Since M = PL/4, H = PL/4s = 500*40*12/(4*2) = 30,000#.

BA

 

[msquared48]

Think of it this way. If you relax your belly muscles, your belly sags. If you tighten your muscles, the sag is reduced. Same principle. However, I have never seen a cable with a six-pack. Lots of bellies consume six packs though. Moving on...

Mike McCann
MMC Engineering

http://mmcengineering.tripod.com

 

[Denial]

gtg477f,

I have a spreadsheet that calculates the forces and deflections in a cable under the actions the cable's self weight plus a point load anywhere along the cable. It allows for the extensibility of the cable, which might be important in your case because your cable seems quite taut. You might well find it useful, given the way you state you intend to solve the problem.

Recent changes in my employment circumstances mean that I have recently decided to make this spreadsheet generally available. You can download it from the "downloads" page under

http://rmniall.com

Your particular problem is a dynamic one, but perhaps not really an impact one. The factor of 2 that you propose using is, strictly speaking, applicable only to problems of a "suddenly applied" load. It might not be appropriate here.

My initial thoughts on your problem are that it is an energy balancing exercise. Your man falls a known distance, thereby acquiring a known amount of kinetic energy. His fall is then "gradually" arrested by the cable. In this process his kinetic energy is absorbed by the cable and manifests itself as strain energy in the cable. Because of its initial tension, the cable already has some strain energy before the man hits it. How much more can it absorb before it yields? Before it breaks?

If the roof beams can move laterally as the tension in the cable increases, that will help you by being another absorber of the man's kinetic energy. And if the harness that the man is wearing (??) has an ability to stretch under load, it will be another potential absorber.

 

[IFRs]

If this is a safety cable being used for fall arrest in the USA, you need to look at OSHA, which requires fall arrest anchorages to sustain a 5,000 pound force without breaking. If this is a fall prevention, it is only 1000 pounds. ANSI also has similar guidelines. If you are unsure or are making your own assumptions, please contact an engineer who is familiar with safety devices, rules and rigging. When human life is at risk there is no room to guessing.

SnTMan said it well - draw the cable to scale. The lengths of the sides of the triangle are proportional to the forces. So if the vertical distance is 2 units and the diagonal distance is 20 units then the force in the cable is 10x the vertical load. Look at the extremes - if the cable were vertical it would have 1x the load in it, if it were horizontal it would have infinite load. Sag is necessary to prevent over-stressing the cable and the anchorage. In your case the horizontal is 240 inches and the vertical is 2 inches. Our Greek friend Pythagorea tells us that the diagonal is 240.008 inches. The ratio is therefore 2 to 240. If the load is 200 pounds then the cable force is 24000 pounds. For a center load, that is.

 

[rb1957]

same as many above ... force triangle is all you need ... draw a free body of the load point ... you have two tension forces reacting the side load (and they react each other along the length of the cable).

2":240" = 500/2 lbs:30,000 lbs

 

[gtg477f]

I am not so worried about the anchorage, cable, etc. itself as it is being spec'd by a well known safety harness supplier. What I am worried about is the beam buckling laterally. I am trying to get the load smaller so I can manage the LTB/torsion of the roof beam. I understand the geometry and how the answer is derived from it, I just cant grasp in my head that such a small load amplifies 60X when the cable isnt even tensioned due to 2" of sag. Obviously, you cant deny geometry and statics. Thanks all for the responses.

 

[Archie264]

gtg477f,

Don't miss what IFRs pointed out. If your project is in the US, and if it's intended to stop a falling person, per OSHA it has to risist 5,000#, not 500#.

 

[fegenbush]

IFRs,

OSHA 29 CFR Appendix C (c)(10) allows for a factor of safety of 2, if designed, installed, and used under the supervision of a qualified person. I think that gtg477f is trying to do that, but will need to apply the factor of safety to the maximum load including impact. [note: impact load is defined in paragraph (d)(1)(iv)]. I believe it was specifically intended for cases such as this where the 5,000 lb criteria would cause problems.

 

[Archie264]

PS: Lest you think that's excessive, that's what's required to stop a 200# man who's fallen 6' and stop him within 6" once the arresting gear is engaged, all with a factor of safety of approximately 2. My understanding of where this comes from is the following:

Using those equations that gave me so much trouble in calculus, physics, diff. eq., and dynamics...

For an object falling from rest...

d^2x/dt^2 = g

dx/dt = gt

x = .5gt^2

...and...

kenetic energy = .5mv^2

Work = Force x distance

...so, using imperial units...

An object falling 6' falls for

t = sqrt(2x/g) = sqrt [(2)(6)/32.2] = .610468 seconds

...which yeilds a velocity of...

v = dx/dt = gt = (32.2)(.610468) = 19.657 ft/sec = 13.40 m/hr

...a 200# man that has fallen this distance has generated...

E = .5mv^2 = (.5)(200/32.2)(19.657^2) = 1,200 ft-#

...setting energy equal to the work done on the system by stoping this in 6" yields a force of...

F = E/d = 1,200/.5 = 2,400#

Using a factor of safety of 2 gets you to 4,800#, which appears to have been rounded to 5,000# for the regulation.

These calculations are probably easier to do in metric, and our mechanical brethern can correct whatever mistakes are made above, but that's my understanding of where OSHA's requirement came from. Regardless, the point remains: the anchorage required for fall arrest in the US is legally required to resist 5,000#. So my suggestion is to proceed with that in mind.

 

[Archie264]

fegenbush,

Whoops, it looks like our posts crossed in the mail. It sounds like you have a handle on OSHA specific requirments. I was listing what I understood them to be. I don't have their requirements in front of me and I could be wrong.

 

[rb1957]

"such a small load amplifies 60X when the cable isnt even tensioned" ...

it's tied to the requirement to keep the 40' for all intents straight, so the component of cable tension reacting the applied load is very small.

and you've got to deal with 10x the load !

 

[hvacpiper]

I just cant grasp in my head that such a small load amplifies 60X when the cable isnt even tensioned due to 2" of sag.

I think what the OP doesn't understand is ,that a 40' 1/2" cable with 2" of sag is ALREADY a highly tensioned cable...

 

[MiketheEngineer]

While this looks like it has been beaten to death and mostly all good answers - remember this

1. Tighten the winch on your jeep to a tree - you can then use your little finger to move the truck
2. When your mom told you to quite fooling around with the clothes line becasue your weight would pull the poles inward.

Both good examples of what everyone is trying to tell you!!

 

[rb1957]

"i'm not dead yet, ... i think i'll go for a walk"

 

[BAretired]

It seems to me that a cable with 2" sag is not a practical answer to the problem. Even a larger sag will produce significant lateral force on the two existing beams. A better solution might be to provide a flexural member spanning between the two beams with cable and harness attached.

BA

 

[grantstructure]

The cable won't be deflected 2" in the final position, so the loads will be closer to those in the 24" position, as you noted in your original post.

I believe you can use the principles of cable barrier design to get what you're after:

http://pti.mountaininternet.com/reportFiles/28-Technote14.pdf

My final suggestion would be to not make your beam the final anchor point. Carry the cable past the beam and perhaps downward, and anchor to something else that can take the tensile force in the cable, and your beam sees only a vertical load.

 

[dhengr]

Grant:
What do you think takes the horiz. reaction from the cable at the beam, where the cable now makes a 90° turn to an anchor point below and adds a new vert. load to the beam?

 

[grantstructure]

Well, if he takes it straight down, that's true. And even if he continues the cable in the same plane, there's some horizontal component applied to the support as the cable deflects, but it's going to be much more manageable. When I said "downward" I was thinking laterally and downward, such that the angles in the final position are balanced such that there is no computed horizontal component.