Anchor point for fall arrest device 3

[haggis]

I wonder if anyone can help with my problem.

We are installing a fall arrest device for maintenance personnel within the plant.

The specs on the device are as follows:

Drop forged carabiner - tensile strength 5000lbs.
Drop forged harness snap hook - tensile strength 5000lbs.
Capacity - One person - 310lbs.

My question is this. Why would the literature say that the supporting structure to which the device is attached, be capable of supporting a load of 5000lbs.

The wearer of the harness attached to the device would free fall approx. 3 ft. when the fall is arrested.

Does this fall under the same premise that given the rated capacity of a hoisting device, the support structure must be capable of sustaining that load. I can see this in the case of a hoist, but a fall arrest device?

The device is going to be hung mid span from a clevis welded to a member spanning 9'-0", clamped both ends to building joists. Now in choosing a member, and keeping the bending stress at around 14.5 ksi, I come up with say a C8 x 11.5 channel.

Just seems awfully big for what it has to do. Any help greatly appreciated.

Thanks in advance.

 

[Focht3]

Philosophically, you don't want the beam (and/or connecting point) to be the weak point in the safety system. And while it does seem to be heavily overdesigned, what standard would you expect if you were the 310 pound guy taking that 3 foot plunge? And what if the plunge were say, 6 feet instead of 3?

On the other hand, you are talking about a member that could conceivably be a "use once and replace" unit rather than designed for repeated loads. (Assuming, of course, that the plant manager will accept the cost implications and the clamping system can hold the beam when the beam yields i.e. the beam deflects but does not suffer a catastrophic failure.) This would allow the use of a lighter member -

If the beam will be a permanent fixture and is not temporary or must sustain repeated loads without being replaced, then I would use the larger beam and forget it.



Please see FAQ731-376 for great suggestions on how to make the best use of Eng-Tips Fora.

 

[SlideRuleEra]

Since tensile strength is the force needed for the device to fail why not use a bending stress of 36 ksi (for A36 steel) instead of 14.5 ksi to compute the size of the steel member? Then you are making a direct comparison between the failure strength of the device and the failure (yield) strength of the steel member. Of course the true allowable load on each will be lower.

 

[ERV]

haggis,

Fall arrest devices are under the jurisdiction of OSHA, check here for requirements:

http://www.osha.gov/pls/oshaweb/owa...able=STANDARDS&p_id=9730&p_text_version=FALSE

Useful information, hope this helps.

 

[haggis]

Thanks to all for the replies.

However, a question for SlideRuleEra. Shouldn't the stresses for bending in A36 be limited to 0.6*36000 ?

ERV,

You provided a great link. I found something that pretty well determines the member selection.

The anchorage should be rigid, and should not have a deflection greater than .04 inches (1 mm) when a force of 2,250 pounds (10 kN) is applied.

Thanks again.

 

[ERV]

haggis,

Glad I was of help. This is what this forum is all about.

ERV

 

[SlideRuleEra]

haggis - The yield strength (36 ksi for A36 steel) is the value where the steel would be permanently deformed by an applied force. The tensile strength, where the steel actually fails is quite higher. Using 60% (or 67% depending on the member shape) of the yield strength is normally the limit for routine applications.

However your product's information gives the tensile strength for the items (5000 lb). This is the value where they could be expected to fail (break). Since you do not have the manufacturer's safe loading on these parts to compare with the steel member's safe loading (60% of 36ksi), I am suggesting that you compare the failure load of harness parts (5000 lb) to one measure of the failure load of the steel member (the load at 36 ksi yield strength).

This would be in addition to the excellent information that ERV has funished.

 

[haggis]

SlideRuleEra

Point taken. I arrived at a much smaller member size (C4 x 5.4) using the 36ksi figure (or close to it) using readily available shapes or shapes we have on hand.

Under the circumstances though, the loads that would result in failure of the member would never be experienced and talking with the millwright who would be using the device, I find that he would have no hesitation hooking on to a C4 channel. This however is beside the point as the OHSA regulations are clear.

It does raise an interesting point though. Does the OHSA differentiate between Engineered anchor points and convenient anchor points in the existing building structure. In this particular case, the member will be designed and installed for the operation at hand as set out in the regulations and it will be blessed by a Licensed Structural Engineer. In other cases, where the millwright has to climb on conveyor support steel at various locations for different reasons, and apart from having the common sense not to tie off to an electrical conduit or sprinkler pipe, who is to say what he ties off to and do we end up in violation of the OHSA if he ties off to some 2 x 2 angle joist bridging?

Interesting.......

 

[structuresguy]

Not that I disagree with the above discussion, but I just wanted to put out my experience with fall protection. I spent a few years working out at Cape Canaveral Air Force Station doing among other things, installing and certifying existing fall protection anchor points on the rocket launch pads and assembly facilites. Some areas I worked in were 250 ft clear height drop inside rocket assembly buildings. Now worker safety is paramount out there, rightly so. The first 6 months I worked out there, 3 workers were killed in accidents, two relating to falling. In my time out there, I probably installed about 75 points, certified maybe 100 more, and installed several perimeter cable systems.

The requirements we used were based on OSHA and some other CFR requirements. Basically, if the fall protection anchor point was on a structural component of the facility, then the component must be within allowable design stresses under the imposed falling load, in addition to service loads expected to be on the structure. The imposed load for the design was 5000 lb per person. This is even though the gear either had break away stitching or inertia reels which physically limit the impact force to around 500 lb or less. But the code requires the 5000 lbs, regardless. This is in case some jambony hooks up his belt to a 6 ft piece of wire rope, without an inertia reel or other approved device.

There were many cases where a beam brace was intended, but because it was unbraced for a long length, the allowable stresses would be exceed by a fall. It was not uncommon to be looking at the 50 lb/ft range as a minimum size for any beam span over 20 ft, if unbraced.

 

[jetmaker]

Not being familiar with fall protection devices, I have a question.

If the capability of the harness and carabiner is 5000#, is that the ultimate tensile capability of the equipment, or a working limit? If it is ultimate, should the structure you are tieing into be analyzed on an ultimate and not limit basis?

I realize that it all depends on how throwaway you want the structure to be. What's important should be that your structure is not the weak link.

Looking forward to the discussion.

jetmaker

 

[haggis]

jetmaker

Yes, the 5000lbs is the ultimate strength of the fall arrester components and I agree that the support structure should not be the weak link.

Now, you may have noticed that I may be a little conservative on the bolt torquing issue, but I'm not like that on everything.

Although I don't want the supporting member to be throw away, by designing it on an ultimate strenght basis, I could have used a much smaller member than the C8 x 11.5 channel.

However, thanks to the link that ERV provided, I found the following as mentioned in a previous post:

The anchorage should be rigid, and should not have a deflection greater than .04 inches (1 mm) when a force of 2,250 pounds (10 kN) is applied.

This is an OHSA regulation and I won't argue with this.

In some cases, in the course of my work, I will take into account probability and likelyhood.

First of all, the fall arrest gear will never see a force of 5000lbs, secondly, everytime the guy goes up to change a gearbox, he isn't going to fall so the repetition of stress on the member is negligent.

Therefore, theoretically, I could have made the member much smaller without fear of catastrophic failure or having to replace it in the event of a fall situation.

Two things govern the decision...the OHSA ruling and the cost. Considering we are talking about 9 feet of channel, cost is not a factor so we go with what is mandated and use the bigger member. Although, no matter how many we were installing, in this case, cost would have to take a back seat.

To be honest, if it wasn't for the OHSA mandate, I would have used a smaller member. Not that I'm being cheap or like pushing the envelope with other people's welfare, but I'm not a "work out the size and double it" or "bigger is better" kind of guy.

As a point of interest, using the C8 and applying a load of 5000lbs mid span (more than double the OHSA requirement), results in a bending stress of approx 16.6 ksi, well within the allowable stress of approx 22 ksi and the ultimate of 36 ksi. for A36 steel.

Quite a safety margin.

 

[FFIELD]

The use of the OSHA requirments is mandatory, but to fully understand this problem, go back to physics and strength of materials.

The distance a body falls in free gravity is
Dist= (axtxt)/2 a= gravity accelertion (Known)
Dist= 3 feet
Solve for time.

The velocity reached is
V= Dist X time Dist and time are known

Kinetic energy is
KE= 1/2 x m x V (squared)
m= 310 lbs

Now set this value equal to the area underneath the area of the force deflection curve of a given beam size and span and limiting it to a stress level. This is the energy absorbed when material deflects. If you want it to remain elastic this woud be yield stress over a safty factor. I have used anywhere from 5 to 10 for such a case. Lifting equipment is usually 5.

This is a coservative approach and does not include dynamic effects on the beam which usually helps.

Any comments?

 

[FFIELD]

Correction to my thread. mass= 310 Lbs/ a

 

[krus1972]

Haggis,

Without getting two involved is CODES and heavy structural theory a much more objective view into this discussion will help you out more then anything else.

The loads in the specifications that are provided to you from the fall arrest provider are most likely static (constant & non-moving) loads. They were calculated from their own internal engineering & testing department during the development of their product.

Giving engineers and other buyer’s specifications in the form of static loads is the easiest way for the majority of the people to understand the product requirements for safe operation.

You are assuming the 310 lbs is a simple static load applied to the supporting structure. This is not the case. When a person (or any object falls) even 3 FEET there is a sudden impact load that is applied to the structure. The effect of impact load on the structure when something falls is not so obvious. People who have kids can best understand this.

Take a child who weighs say only 60 lbs and tell him/her to stand up-right on in his/her bed or a trampoline. Take a close look underneath. Take note of the deflection of the springs and how far down the springs and the mat are deflecting. THEN tell him/her to jump up and down to about a point where they are free falling about 3 feet down on the and pay attention to the maximum deflection again. Big difference ah? Would you say the fall makes the deflects the trampoline 5 times, 10 times, 20 times then if the child was just simply standing there not moving? Then imagine how this effect would increase using a person of 300 lbs (5 times the child's weight). Without getting deep involved in momentum theory the springs dissipate the additional impact by elongating much further (5 times?) then when the person is at rest. This will account for SOME of the increase of required static support structure load beyond 300 lbs in your question. I would suspect the springs would elongate a maximum of 5 times when jumping occurs. If we take 5 times 300 we are now at 1500lbs of equivalent static load.

Now in the supporting structure for many fall arrest products the support structure does not have the luxury of deflecting as much as a spring during an IMPACT load. The support structure is much more firm and rigid then the trampoline. The structure must be able to receive and dissipate this impact throughout the internal molecules of the steel, bolts, connections, etc... so it does not deflect and bounce like a spring. Because of this and the effect that a free fall (impact load) will never load the beam in it's perfect center (creating a twist on the beam) you need to add to the 1500lb number above. Without getting to deep into vibration, dynamic, and twisting, you can safely assume that all of this will be a factor of 2 bringing the number to 3000lb static load requirement for the structure.

Lastly people who make these types of products would have to abide by many SAFTEY guidelines such as OSHA in addition to what’s we required by the steel design guides for IMPACT design. I would not doubt it if governing codes for this type of product would require manufactures to apply additional safety factors over and above what I have listed above.

My example listed above is not exact and is intended to give you a perspective and objective feeling of how a free fall impact is much much more complicated then a static load. As a result the required impact load is converted to an equivalent static load requirement having a higher value then many people would suspect.

If you have more interest in this call the manufacture and ask them if the design info for this product is available. If you study it you will find their design methology are somewhere along the lines I have explained above.

I hope this all helps.

Jeffrey A. Krus P.E.

 

[jetmaker]

haggis,

Thanks for the reply. I know the codes dictate how things are suppose to be designed, but often I wonder where these numbers have come from. I have found in my industry that much of the knowledge on how some of the numbers/methods of analysis have been lost as the people who developed them left. As a result, I see too many people just applying the numbers, not really knowing where or how they originated. I tend to drive my bosses crazy by always questioning a method/number.

The paramount issue as you have stated is one of safety, not cost. And in general, it does not hurt to be a little heavier. But again, in my industry, weight is a hugh cost. But then again, we also work to much smaller margins than most industries.

Thanks again, looking forward to more informative discussions.

jetmaker

 

[haggis]

Hi to all again,

Thanks for your repeat replies and opinions on a subject that a lot of people would just say "do what the book says and leave it."

krus1972, I did realise that the loads encountered in this situation were not of a static nature but was merely saying that the OHSA requirements seemed to be high, even from the impact loads.

FFIELD, Your post was very helpful and right on. On calculating the loads, and depending on whether the lanyard extends 3 feet or the full 8 feet in the event of a fall, the OHSA mandate gives this beam a safety factor of 10 or slightly more. Not that I'm arguing it could or should be less as preventing serious injury or even death are of paramount importance.

Jetmaker, I agree with what you said in your post about questioning or not questioning numbers. Again, It's always interesting to delve into the reasoning for the "established numbers". I think we all learn a little more by doing so. Many moons ago, I was told never just accept something, ask why.

By the way, as of today, the beam is up. Let's hope it never serves it's purpose.

Haggis

 

[FFIELD]

Corrextion to my thread
V= accel. x time

 

[haggis]

THanks FFIELD, I already picked up on it.

Haggis

 

[dutchie]

Something else that I believe should factor into this discussion is this.

In a lot of cases with fall arrest equipement, the wearer will hook onto a cable, which is attached to permanent anchor points on the structure. I realize this isn't the case in your situation, but I believe the 5000lb number is intended to cover this situation as well. Depending on the initial sag of the cable and the load, the axial force in the cable can be quite high, even for a 300 lb guy.

I have designed numerous anchor points for fall arrest equipment, and when hooking up directly to structure, I use a safety factor of 5 or so, and design the anchor point accordingly. Anytime I've used a horizontal cable, the manufacuturer has provided me with the 5000 lb number.

As it can be easily shown, if the initial tension in the cable is high, meaning there is little sag, connection points can have much higher loads than 5000lbs, which is why i am not sure on my assumption.

any comments are welcome, as this is something i sometimes struggle with

dutchie

 

[dicksewerrat]

If you read enough OSHA manuals, you might find the section that says the safety factor for personnel lifting is 10. If 5000 is what the manual says then that's what you build. I never dropped people down a shafthole with a cable that was rated less than 10 times the load. never will.