Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations SDETERS on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Strength of Safety Lanyard Clip-In Ring

Status
Not open for further replies.

jondon

Mechanical
Jul 10, 2014
12
I am looking for some guidance on an apparent discrepancy between my own calculated strength values for a u-shaped clip-in ring (similar to a shackle), and the rated strengths of commercially available shackles. Can anyone provide a reference for some equations to use in shackle design? All of the references I've been able to find simply assume that the rated load of the shackle is already known, but don't provide any guidance on how to actually calculate the stresses in a shackle for a given load.

Some background: I am designing a structure which will have an overhead attachment point for a maintenance worker's fall-arrest safety lanyard. The attachment point will consist of a u-shaped half ring made of bent 316 stainless steel round bar, with the upper ends welded to the steel supporting structure. The shape of the clip-in ring will be very similar to a D shackle

The ring will have an inner diameter of about 7/8", and needs to support a fall-arrest load of ~1250lb. Due to space and aesthetic issues, I need to limit the maximum diameter of the round bar to 1/2".

I am running into some contradictory information when solving this problem:

- First, a quick Google search indicates that the safe working load for a 1/2" 316 SS shackle of similar dimensions has a safe working load of 2,500 lb. Assuming a typical yield safety factor of 2, this would indicate that the shackle would yield at a load of 5,000lb, which would seem to be more than adequate for my purposes.

- However, as a sanity check, I performed a quick hand calc using a curved beam bending stress equation (Mechanical Engineering Design, 7th Ed. Shigley et. al. Equation 4-66), which indicates that the yield load for my overhead ring is only 950 lb. I calculated the bending moment assuming a point force at the center of a simply supported beam with a length equal to the width of my ring. This is in line with the method suggested by BigInch in another thread here, but it seems that either the moment calc or the stress calc is far too conservative in this case.
 
Replies continue below

Recommended for you

It may be that the shackle strength is confirmed or even determined by testing.
It may be that the shackle strength is enhanced by cold working or by forging or heat treating or by tweaking the alloy used. The manufacturer wouldn't necessarily be limited to the minimum yield strength of the annealed alloy.

Note that stainless steels typically have low yield strengths and relatively high tensile strength, so there is a possibility of a considerable amount of improvement from cold working.

I would think the shackle you're finding actually has a FS of 4 or 5, not 2, for what that's worth.
 
And it is even better than that....

When the person is falling, he/she is arrested and that impulse load is, of course, transmitted to the shackle. And that shackle and harness could be at any number of "weird" and twisted and unusual positions when it gets hit by the impact load. (Obviously, a "theoretical person" of "average weight" wearing the harness theoretically correct in the right manner would not have fallen in the first place.) it is the goofballs and dare devils who will fall in real life.

But a considerable amount of energy is "used up" deforming the shackle and links as they stretch and straighten as the person decelerates. In fact, a lot of the "safety" involved is in the deceleration time - which greatly reduces the internal damaging stresses on organs and muscles.

So, you really should not use the "theoretical" yield or break strength of the "pure metal" rod, nor even of the bent metal shackle, but rather use simple testing to find out the yield point (how much it stretches) under impact.

THEN - you go back and in the instructions - require the user to replace a shackle once it has been loaded after a fall.
 
Is the 'fall-arrest load" according to OSHA? For some reason I thought the fall- arrest for one person was higher than that. If 2 people attach to that ring it is not enough.

Richard A. Cornelius, P.E.
 
In Canada, the fall arrest load is 5000 lbs. Based on what I have read here on E-Tips, that is the standard in North America. In my jurisdiction, the maximum arrest force can be reduced if a shock-absorber is used (with some appropriate factor of safety applied, don't recall exact number right now) but I've never personally had to rely on that and would require very specific circumstances to allow it. I especially would not allow it for a permanent anchor over which I would not have personal control over the use.

1/2" may very well be enough...but if you're unsure don't hesitate to bump it up. Aesthetics don't trump safety. You may want to look into the requirements for your inner diameter, there are minimums and 7/8" sounds much too small to me.

Crosby lists a max proof load of 2 times the working limit and an ultimate the load of 5-6 times the working limit. The way the requirements for fall arrest anchors are written in Canada, you require a load of 2500 lbs without permanent deformation and a load of 5000 lbs without failure.
 
Thanks for the responses, everyone.

Just for some additional context, the fall-arrest ring in question will be used for maintenance at the top of a mast/flagpole on a large luxury yacht, so aesthetics trumps quite a lot in this case.

You are all correct that the fall arrest load I am using is quite a bit lower than what is normally used. The trouble here is that we are trying to essentially re-design a flagpole (that was originally intended only to take wind loads) to now support a fall-arrest load from a deckhand performing equipment maintenance. We had originally intended to design the mast and safety equipment to EN795.2012, which specifies a 12kN (~2700lb) fall arrest load, but this was simply too onerous for the structure as a whole, so after some internal discussion, we agreed to reduce the rated capacity of the system. Note that in this case, the deckhand will be suspended from a pulley system at the top of the flagpole, so the only way the fall-arrest lanyard would come into play would be if the pulley system fails completely.

With regards to the lifting ring stresses - I am still trying to figure out how a shackle designer would design new shackles. There must be some calculation involved that is not as conservative as the one I have performed. It can't be solely based on testing - and even if it was originally, there must be some empirical formulas out there that could give a more accurate idea of stresses in a shackle based on load and bend radius. Are any of you aware of any such formulas?
 
jondon, I would suspect that new shackle designs would be handled through FEA and testing, but I do not know. As I noted earlier, Crosby does state the proof load is twice the working load and the ultimate load is 5 times or more the working load...so it seems obvious that the their final published capacities are based on testing.

Did you consider a point load in your analysis? In reality it will be far from a point load, especially with your inner diameter of 7/8". The hook from the lanyard will almost be fully bearing on the inside of your D-ring. That may bring your calculated capacity up a bit.

jondon said:
Just for some additional context, the fall-arrest ring in question will be used for maintenance at the top of a mast/flagpole on a large luxury yacht, so aesthetics trumps quite a lot in this case.

But it doesn't trump safety. That is your number one priority. You can't jeopardize the safety of the worker so some rich guy doesn't have to see a 3" ring 25 feet above his head.

The Canadian code requires you to design for at least 2x the maximum arrest force (MAF). Regardless of the code requirements, I think it would unreasonable to go below this. The typical shock absorbing lanyards I've seen limit the arrest force to 900 lbs, meaning your 1250 lbs is a little short (based on my experience). If you can find a lanyard that reduces the MAF to 625 lbs, I will be surprised...and I will be even more surprised if that style lanyard is consistently used with this fall arrest system.

I also think your inner diameter is too small. The worker will have a difficult time getting his lanyard attached to that small of an opening. And if I understand correctly, he will have to ascend the pole and put himself in a dangerous position before he can safely tie off...I think the last thing you want to make him do is fumble and struggle with his lanyard to get safely secured. Saying that the pulley would have to fail for the anchor to be engaged is not justification for reducing the rated load on the anchor...that is the purpose of the anchor. No one plans to fall. In this emergency situation you have to have confidence the anchor is going to perform.

I know I went on a bit of a tangent with this, and I don't mean to make it sound like you're not considering the safety of the worker...but I see too often that fall arrest is left as a minor item and anchors are located and designed without consideration for the people they're meant to protect.
 
Canpro - naturally aesthetics doesn't trump safety, it's just one more thing I need to consider in the design in this case. I'm quite concerned with safety here, which is exactly why I've put the question to this forum. It's often frustrating in this industry to have to constantly balance aesthetics with strength. I'd love to have a big, 3/4" dia ring with a large inner bend radius to make clipping in as easy and safe as possible, but we're somewhat limited by the shape of our anchor points and considerations to make it as unobtrusive as possible.

In my hand calculations, I am considering the fall-arrest load as a point force at the center of the ring. I am also calculating the moment assuming that the ring is essentially simply-supported, so the moment is F*w/4, where F is the force, and w is the width of the ring. I know that this is a conservative assumption, since the ring should perform somewhere in between a simply supported beam and a fixed-end beam.

One other interesting thing I have found is that, at least in my hand calculations, increasing the inner bend radius actually seems to make the stress worse. This is because, although the increased bend radius decreases the stress concentration at the inner fiber of the ring, the increased width results in a larger bending moment at the ring center.

This is why I was hoping to find some sort of empirical formula to at least serve as a sanity check on my stress levels. For such a small detail as this, FEA seems a bit like overkill, but in the meantime, I will probably throw together a simple FE model, just to satisfy my own curiosity.
 
Years back the firm I was with tested a frayed binding strap since an OSHA inspector rejected it. Our test showed it had all the manufacturer's rated strength, so OSHA accepted that.

In this case, I would construct the same model hook or ring that you propose and secure it to a similar flag pole support. then I'd take to a testing firm and have them run a load test on it. You can try several ways of loading to duplicate possibly field situations.

You can run all the calcs you want, but a true test of what goes on is to test the device in a qualified testing lab for an identical attachment method and get a PE seal the report.

 
Jondon:
In light of the lack of sketches, details with dimensions and piece sizes, etc., which you have not bothered to provided, you are pretty much on your own, because you are the only one who can see it. For the most part, this kind of hardware is highly dependant upon testing for setting the working loads and breaking strengths. But, it does seem to me that I have seen solutions for the stresses in rings, hooks and curved beams, and the like in various Strength of Materials and Theory of Elasticity text books, look around. This is exactly the kind of problem where FEA should shine, when modeled properly. The closed form solutions to these typed of problems are on the simplest geometries and the simplest boundary conditions, and you probably have neither. I would also assume that a company which makes this type of hardware as a regular item has some rules-of-thumb and first-cut analysis methods which they’ve developed over time. And, they’ve done enough of these so they have some sense of the trade-offs btwn. bigger ring dia. vs. longer apparent span length, that you’ve mentioned, etc. But, they still finally test a sampling of each run of their product. Then you have to worry about any cracking/tearing/necking from the ring forming process. Is it forged or just cold bent from bar stock? Then, there are going to be stress concentration issues every place you make connections or change shape, etc. Given the flag pole/mast nature of your problem, I’d be as worried about the strength of the mast with 250lbs. hanging eccentrically off the mast, at the same time there are other lateral loads on the mast.
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor