Horizontal Lifeline Anchorage - Cable Analysis 9

[RFreund]

Unfortunately I didn't pay much attention during the cable/catenary analysis portion of my structures class and now that has come back to bite me. Attached is a schematic layout of a horizontal lifeline. It is anchored at the end and has two midpoints which the rope/cable just passes through.

I'm trying to determine the worst case reactions due to two persons using the horizontal lifeline assuming the lanyard provides a maximum force of 2*900lbs = 1800lbs to the lifeline. My question is twofold I suppose.

1. What information would you need in order to "accurately" analyze this situation (i.e. E of the "rope/cable" i suppose). How would you do it? Does it seem reasonable to do by hand?

2. Is there a conservative force to apply to the anchorage points which skips the "difficult" (IMO) analysis (see attached).

Somewhat unrelated, it appears the manufacture suggests to design the support for a minimum of 5,000lbs. So I suppose I could design the support for 5,000lbs tension and 5,000lbs shear applied at seperate times and this would also cover the anchor design.

EIT
www.HowToEngineer.com

 

[CANPRO]

Hello RFreund, I've done quite a few of these designs recently. I can offer the following info to help you out:

- First off, I suggest you put all of your calc's into a workable spreadsheet. You will want to play around with different arrangements. The calc's for this aren't overly difficult, but they can take time to run through multiple scenarios.
- Google search "effective modulus for wire rope". there will be lots of hits on google. From what I have read, it is approximately 1/2 of E for steel. This takes into account the relaxation of the wire rope.
- You really should buy/acquire your local/national codes that govern this. In Canada, applicable codes are CSA Z259.13-04, CSA Z259.16-04, and they go on to reference a couple other codes for the anchor design. The previously mentioned codes will give you modification factors for the applied loads in different scenarios (see below).
- Your design will be a balance between limiting the tension in the cable to a reasonable value that can be accommodated by the anchors, and limiting the sag of the cable in the event of a fall. These factors change depending on your pretension - Higher pretension reduces your sag, but increases the tension in the anchors. If you are limited in your allowable fall distance, you need to apply more tension in the cable. The expected sag in the cable during a fall is something that must be clearly communicated to the end user of the system. Codes reduced the maximum applied force for the sag calculations (0.8 in the Canadian Codes). Codes will also modify the applied loads for multiple falls.
- Canadian codes require you take into account the change in temperature. Doesn't affect the design much for short cables, but in the range of 100'+ it can become a factor. Positive temperature swing from time of installation is worst case for sag and negative swing worst case for tension in cable.
- My spreadsheet takes the following steps for the design (its setup to calc out your exact arrangement) - specify the pretension in the cable and use that to determine the sag in the spans (I assume equal horizontal tension across the entire length of cable). This can be used to determine the total length of cable. During a fall in any span, I assume that in the event of a fall the cable turns to straight-line segments (perfectly horizontal between anchors and then a triangle in the span where the fall occurs). The geometry of the cable can be used to determine the tension. Use the tension to determine strain and the new length cable. Iterate this process until you're satisfied that you've reached an accurate enough result. Once that is complete you will have the max tension/sag for the cable. You need to run two scenarios with the same pre-tension to determine the max tension and max sag (load factors change and so does temperature swing).
- I have attached cables to existing anchors where their ultimate strength was 5000lbs. You just need to limit your tension to 2500lbs.

Hopefully this helps. There is a fair amount to go into this, I definitely forgot to mention some stuff. Can't write much more right now. I'll check later to see if there is any follow up questions that I can help with.

 

[RFreund]

CANPro - First thanks for the response. Great information.

My first question - have you come across the use of non-steel type systems? In my current situation I beleive they want to use a kernmantle rope by Guardian here. I suppose I could ask Guardian for the 'effective modulus'. Also from my understanding of the product there is a certain max pretension that the system is set to.

I haven't got into the calcs yet, but if you have a higher pretension you would have a higher force on your anchors for the same fall force, right? Or am I getting myself confused here. At first I wanted to say no, as I was thinking it would be similar to applying tension to pre-tensioned anchors.

Is there a reference or example you could point me towards as far as how to determine the deflection when applying the fall force in any span?

Regarding codes - I will have to look into that, there is OSHA but there may be additional codes that I'm not aware of. This is in the US.

Thanks!


EIT

http://www.HowToEngineer.com

 

[Archie264]

>>>Is there a reference or example you could point me towards as far as how to determine the deflection when applying the fall force in any span?<<<

I use an iterative approach:

1. Calculate the energy of the falling object, E=.5mv^2
2. Set energy = work = Fd, with d being an assumed initial deflection.
3. Solve for F.
4. Use F (i.e. "P") to calculate deflection, i.e. deflection = PL^3/48EI if it hits midspan.
5. Use the new deflection to repeat the process.
6. Lather, rinse, repeat to the degree of accuracy required.

Some mechanical engineers have told me that this is not quite theoretically accurate but it seems to work insofar as I can tell. But watch your units. Mass in imperial units is expressed as slugs, which is a pound-force divided by the acceleration of gravity, I believe.

 

[theonlynamenottaken]

if you have a higher pretension you would have a higher force on your anchors for the same fall force, right?

Yes. Sag and tension are directly related. Higher pretension = lower sag = higher live load tension

If you are new to this you should go ahead and buy Nigel Ellis', "Introduction To Fall Protection". He works several examples and provides several insights that are not instinctual, such as how to allow for more than one person on one or several spans of a single horizontal lifeline cable.

You can work these problems by hand, but it begins to get complicated when you include shock absorbing devices... especially ones that will extend and add to the total length of cable as originally installed.

Also, you can end up with cable end termination reactions that exceed 5000 lb. 5000 lb is the standard personnel support anchor design force, so make sure to determine whether or not you'll need end anchors that can withstand more than the 5000 lb standard.

Also see:
Fall protection system from elevated platform - cable analysis
Horizontal (with sag) lifeline
Fall Protection
fall protection system
Concentrated Load on a Prestressed Cable

 

[Denial]

You are right to ask about the cable's axial stiffness.[&nbsp;] This is an important parameter, and ideally an actual value should be obtained from the manufacturer, or determined from tests on a sample of the particular cable.

However for the purposes of preliminary design you might be able to use some rough rules:

» The effective cross sectional area of actual material in a twisted wire rope will usually be between 0.48 and 0.76 times the gross area, where the gross area is calculated on the basis of the nominal diameter.[&nbsp;] Values low in this range tend to apply to hollow-cored cables, while values high in the range tend to apply to strand-cored cables.

» The effective Young's modulus will usually be between 0.41 and 0.72 times the material's actual value.[&nbsp;] This factor allows for the fact that the individual strands of material are twisted into a spiral pattern rather than running directly from end to end.[&nbsp;] The actual value depends upon the winding pattern that is used.[&nbsp;] If the cable has already been tensioned to a high value at least once in its life prior to installation the value will probably have increased slightly because the strands will tend to have "nestled into" their preferred arrangement.

 

[thaidavid40]

RFreund,
Also, you don't need to consider two persons falling at the same exact instant. Even if they fall somewhat at the same time, in the same incident, the load won't statistically add up to the maximum lanyard load doubled. Figure on one pull of the maximum 900 pounds, with the other lanyard loading a static 300 pounds (the at-rest load for a single person). That would make your FACTORED load equal to 2*(900#+300#), or 2,400#.

Thaidavid

 

[RFreund]

Wow, great posts, thanks for the info. I'll check out the book, thanks.

As far as the actual analysis of the cable goes, in order to determine the deflection, is there a good reference or example for this (again, I must have missed this day of class).



EIT

http://www.HowToEngineer.com

 

[thaidavid40]

Are you using a "hard" cable (firmly anchored at each end), or a "cushioned" cable (one with a spring compensator at the end(s))? That would make a great deal of difference in your loads on the anchor arms. If it is a cushioned type, the particular major manufacturers will have their own calculation programs to help you. Some of these can be had just for the (polite!) asking of a kind company representative.
Dave

Thaidavid

 

[Denial]

You ask "is there a good reference or example for this".[&nbsp;] Assuming you are prepared to ignore gravitational sag in the cable, and also to ignore dynamic effects, the solution is not all that hard to derive.[&nbsp;] See attached, wherein you end up with a pair of nonlinear simultaneous equations.[&nbsp;] If you are further prepared to assume that the deflection is small, the problem reduces to a single cubic equation in T (the cable's tension).

 

[RFreund]

Thanks Denial! However a bit of an oxymoron here: "the solution is not all that hard to derive... you end up with a pair of nonlinear simultaneous equations"

Curious to others - Is this the approach you take or is this the approach taken in Nigel Ellis book?

EIT

http://www.HowToEngineer.com

 

[Denial]

(Would my comments have been slightly less oxymoronic if I had bolded the "not all that hard" bit?)

I had a bit of spare time yesterday, so I drank my own Kool-Aid and implemented my approach in a spreadsheet.[&nbsp;] See attached.[&nbsp;] The spreadsheet starts with a direct solution to the "small deflection" approach, then it uses an iterative approach to extend that solution to the large deflection situation.[&nbsp;] Note that in the spreadsheet as attached all cells are locked against change except for those cells into which the user would normally enter data.[&nbsp;] However this protection is NOT enforced by a password, so it can easily be deactivated.[&nbsp;] (Should I bold that "easily"?)

 

[Ingenuity]

Nice work, Denial. A star for you!

 

[RFreund]

Wow, great work Denial!




EIT

http://www.HowToEngineer.com

 

[Denial]

I have made some changes to the spreadsheet I uploaded on 31st August.[&nbsp;] The new version can accommodate slack cables, and is generally more robust.[&nbsp;] Note that the original version still works fine, within its limitations.

Rather than upload the new version to Eng-Tips I have added it as an additional worksheet in the "Loaded Cable" spreadsheet that I make available from my website (rmniall.com).[&nbsp;] It can be downloaded from there.

 

[KootK]

If you're in the market for an excellent primer on cable design and behaviour fundamentals, consider this reference which I can vouch for: Link. $5 tops. And it's just a cool retro add to your library.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[bridgebuster]

You can download the classic USS Wirerope Engineering Handbook (1968)from SlideRuleEra's site:

http://www.slideruleera.net/miscellaneous.html

 

[miningman]

Methinks theres a bit of wheel reinventing going on here. As has been posted elsewhere previously, Canadian codes require a lifeline to be capable of withstanding a shock force of 5000 lb. Previous posters have confirmed that USA regulations require the same.

 

[RFreund]

Miningman - I'd be careful with that line of thinking. Actually that was my first thought - design it for 5 kips and your good to go. Nope. I bought that book by Ellis (a bit pricey for sure). The first thing he says about horizontal lifelines "The non-engineered 5,000 lb anchorage strength requirement is often implemented in association with the ends of horizontal lifelines. This is wrong and potentially dangerous, as end forces on a lifeline system often far exceed 5,000lb."

If you think about it, I suppose it makes sense. If you have a long line, you need quite a bit of pretention which will add to the anchorage force.

Denial - Thanks again, I had hoped to of had more time this weekend as I wanted to convert that spreadsheet into an Smath calculation. After going though some of your work I realized you end up with a pair of non-linear equations (as you said). Before I was thinking it was a pair of differential equations, so I guess not as bad as I was originally thinking.

Kootk - nice, thank you.

EIT

http://www.HowToEngineer.com