Fall Arrest Safety Cables

[iv63]

I am evaluating existing fall arrest safety cables running along walkway and attached to W4x13 posts spaced at 20ft. The worst case would be if a force is applied on cable at middle between cable posts. Correct? How would you calculate tension force in the cable? Does force of 5,000 lbs need to be used? Any advice would be appreciated.
iv

 

[rb1957]

i'd assume that the cable becomes two straight lines to the anchor points, and the component of the tension reacts the applied load

Quando Omni Flunkus Moritati

 

[wadavis]

5,000 lbs is the reaction at supports that the horizontal life line suppliers I've dealt with design their systems for. They've wrote their max span and cable slack tolerances so that once the slack in the line and the cable stretch is accounted for, the relatively small arresting force causes a 5000 lb load at the anchor locations.

Of course this is how the HLL suppliers I've worked with designed their system, contact the system's original manufacturer for technical support.

wadavis
E.I.T.

 

[rb1957]

assume a 5000lb cable tension.

assume a 500lb load, each half of the cable reacts 250lbs

assume a span of 20', so each half cable span is 10'

the cable makes an angle to the straight line between the anchors of asin(250/5000) = 0.05rad = 3deg
cable length = 10'/cos(0.05) = 10.0125', slack = 0.15" !!

with 2" of slack, then half the cable is 121" long
and the deflected angle is acos(120/121) = 0.13rad = 7deg
and a 5000 lbs cable tension can support a load of 2*5000*sin(0.13) = 1283 lbs


it's a slow day !

Quando Omni Flunkus Moritati

 

[CCox]

I have done fall protection engineering. There are several issues going on with tension cables to be aware of. For clearance issues, sometimes the angle between the deflected state (when someone falls) and horizontal plane at rest is very small. This will lead to an enormous amount of tension being induced into the cables. This has to do with basic geometry. You are basically limiting the fall distance by developing the force through tension into the cables. The lower the fall distance, the higher the tension in the cables. If, however you are allowed a larger angle, say 30 deg., then the tension in the cables is much more manageable. The anchorage forces are simply the cable tension resolved into components. I have found that in many cases, going to a rigid type of system is required where fall distance is the limiting factor.

The 5,000# force does apply if you do not use energy dissipating devices or devices that limit the fall distance and kinetic energy being built up. Energy dissipating devices are shock absorbing lanyards. Self-retracting lanyards lock the worker in before they can fall a great distance. Both will limit what is called a max. arresting force. I have seen this force as low as 900#. Using the OSHA S.F. of 2, that would bring the load to 1800#.

Do an internet search for USS Wire Rope Engineering Handbook. It has calculations in it for exactly what you are looking for. Someone on here might have it and posted it. You should review the handbook to make sure you understand the theory of cable behavior. One last thing: if the cable is large in diameter and spanning a long distance, you should investigate the sag of the cable from self-weight and if outdoors, from ice on it.

CJC

 

[EngineeringEric]

Look at miller fall protection, they will provide this information. One needs to account for the pre-tension, the height of the fall, the weight of the person, how many people are on the line, how far apart the anchor are, ... and more
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I only mention Miller since i use them when i design these applications

 

[IFRs]

You are in a very dangerous and litigious area of engineering. Extreme caution is advised. You are dealing with human life and death. Please be very careful, conservative and get your design and calcs reviewed. You absolutely can not use a 500 pound load. The arresting force is much larger. A small sag in the wire equals a huge force in the cable and the anchor. If the cable had no sag, the force would be infinite. The OSHA 5000 pounds is the breaking strength of the anchor or a straight cable going from the anchor to the worker. This can be reduced if a qualified engineer designs a different system. If the cable has little sag, the anchor forces can be much larger. With 0.15 inches of slack the force in the cable will be huge, way more than 5000 pounds for a 250 pound load. Simple geometry suggests it would be 120 / 0.15 = 800 x 250 = 200,000 pounds. When you are all done, decide if you would trust the system with your wife or child.

 

[CCox]

IFRs is right. This is a very dangerous part of engineering. I only stumbled into this form of engineering because I work in the manufacturing industry. But you should be an experienced PE or SE if you are designing/evaluating systems.

Bear in mind the max. arresting force I mentioned is the vertical "fall" load applied to the cable, not the tension in the cable. They are very different. IFRs is correct in that the max. arresting force can be reduced only if the system is designed by a qualified PE.

CJC

 

[XR250]

Seems like the force on these is somewhat self-limiting. If the force gets too large the posts will simply bend to reduce the force accordingly.

 

[IFRs]

In general, I consider that bending is not a good thing for a safety anchor because it is yielding and that leads to failure. Yet, I believe that at least some commercial varieties are designed to do just that - bend over, but not break, thence increasing the sag in the cable, which lowers the force in the cable and the anchor to that which does not break the anchor or cable.

 

[iv63]

Does safety cable anchorage have to be designed for anchor force of 5,000 lbs or cable breaking strength? What force perpendicular to post should be used (please see attached)? BTW Shock absorbing lanyards are used.
iv

 

[IFRs]

I interpret OSHA to say that anchor points have to have a breaking strength of 5000 pounds per worker unless an engineered system is used. However, if your fall arrest system generates more than 5000 pounds per worker because of unfavorable geometry, cable sag, etc, then you must account for that. Cables are 1-force members, capable of only tension. Unless the worker is tied off between the anchor post and the next post, the direction of the force is self evident assuming there is no deflection of the either post or system. If the worker is tied off after the second post, the force should be perpendicular to the anchor post assuming no deflection of any post or other part of the system. Since human life is in hanging the balance, be sure of your geometry, the forces involved and the nature of people to do the unexpected.