Catenaries

[Andy Mel]

Hope someone out there can help me with this one, what I am trying to establish is how to make the calculation for a SWL when using steel wire rope to make catenaties. I am presuming there is a formula / technique!!!! Obviously the loading when correctly terminated is easily established when loading in one direction ie. straight down, but what about two axis???

This is a job I often carry out throughout festival season for creating cable runs and decor supports but out of interest how do I demonstrate the loading potential. At the moment its a case of years of experience of the practice for many of us and while discussing the loading potential none of us could give a hard and fast answer - so heres your chance - what is the answer!!!!

 

Many thanks

 

Andy Melleney

[flypig]

Hi Andy.

 

Here's some calculations that may help you. Before you read on, do consider the fact that you are getting into some fairly serious rigging and engineering territory and I would strongly recommend that you either go and get your rigging ticket or employ a qualified rigger to do the job. That said, I'm a great believer in the fact that knowledge is there to be shared and it's better that you read the following and comprehend the complexities behind what it is you are doing.

 

Also, please be aware that the calculations below do not take into account any form of wind or shear loading which will alter what may be safely required quite considerably.

 

The minimum diameter of Flexible Steel Wire Rope (FSWR) are;

Load Lifting = 5mm

Lifeline = 9mm

Lifting Persons - (power operated) = 13mm - (Hand operated) = 9mm

 

To determine the Safe Working Load (SWL) the following rule of thumb formula should be used;

 

SWL = the diameter mm squared x (times) 8

 

Example; 14mm diameter FSWR. 14 x 14 x 8 = 1568kg

 

To determine the rope size of FSWR the following rule of thumb formula should be used;

 

Rope size = square root of the total load size divided by 8

 

Example; Load = 1568kg. 1568 / (divided) 8 = 196. square root of 196 = 14, therefore 14mm diameter.

 

The formulae above would be used for lifting purposes. As you are using catenaries (or span ropes), there are a whole lot more calculations to take into consideration.

 

The tension in the span rope is critical to size of the FSWR used for the span. Each doubling of the sag halves the tension in the span rope, and conversely each halving of the sag doubles the tension. The minimum sag in a span rope should be no less than 1/20th of the span with no load attached. The sag is measured vertically from the middle of the span to a line joining the opposite end of the span.

 

Formula:

Tension In Span Rope (TSR) = Total head Load x (times) span /(divided by) 4 x (times) sag.

Example:

Total Head Load (THL) = Load + Gear + Load In the Hauling part - Therefore;

THL = 800kg (example load only)

Sapn = 20m (example span only)

Sag = 1m (1/20th of span)

 

TSR = 800 x (times) 20 / (divided by) 4 x (times) 1 (in other words 16000 / (divided by) 4 = 4000kg).

 

The TSR is then used in FSWR size formula (first example) as a total load to calculate the rope size required for the span rope.

 

I must reiterate Andy that you should either become qualified, or engage a qualified rigging company to carry this work out for you.

 

Best regards,

H

[Pete McCrea]

Very useful stuff. I think that's one to add to the reference file for use when checking if something is possible. Thanks Ian.

[kim]

Just remember to take into account the self weight of the cable, on the above example which suggested a 22mm cable it would be a little under 40kg, which if you put that back in to the formula you would come out with a 23m cable and a TRS of 4230kg

 

Having said that I would also suggest you employ professional riggers and engineers when dealing with catenaries.

[Andy Mel]

Hi Ian,

 

that is exactly what I was after!!!

 

Its one of those elusive things that we have all done for years but never had the formula to back it up!!

 

And please don't worry.. I realise lots of people post on this board who really should not be let loose on the real world, in fact I take alot of entertainment from some of the posts during my moments of escape from work, but I am a working profesional (hold back the laughter and insert slacker for those that know me) in the industry - for the last 12 yrs in fact, and have hung pretty much most things in most conditions and continue to do so ever hoping for my phone to stop ringing maybe for a week!!!!!

 

Thanks again Ian - you have helped me regain faith in what the blueroom can produce.

 

Andy

[Pete McCrea]

Likewise, I know what I'm doing, and have practical experience, plus a slightly unrelated Civil Engineering Degree. It's always usefull to have a point of reference that you can point at when telling somebody why it's not safe to do it. Add to that the fact when it's anything more than a simple rig we bring in the experience guys to do it.

[Seano]

That said, I'm a great believer in the fact that knowledge is there to be shared and it's better that you read the following and comprehend the complexities behind what it is you are doing.

Right O.

 

The minimum diameter of Flexible Steel Wire Rope (FSWR) are;

Load Lifting = 5mm

Lifeline = 9mm

Lifting Persons - (power operated) = 13mm  -  (Hand operated) = 9mm

Load lifting - I'm not aware of any lower limit.  Where is this from?  

SWR is readily available below 5mm Rope Assemblies carry several different types down to 3mm, and its perfectly fine to use for lifting things its strong enough to lift.

 

Lifeline - As far as I'm aware, there isn't a specified minimum diameter in the EU as such.  Funnily enough, the only Australian one I could find with a quick Google was actually 10mm (for the Northern Territory construction industry here).

 

Lifting persons - Where did you get that figure from?  Does this mean that "Foy" flying isn't possible in Australia?  Or does Peter Pan fly on a rope you could use to moor a battleship?

 

To determine the Safe Working Load (SWL) the following rule of thumb

formula should be used;

SWL = the diameter mm squared x (times) 8

Example;  14mm diameter FSWR.  14 x 14 x 8 = 1568kg

Seems unduly pessimistic, what kind of rope are you talking about there, and at what safety factor?

A good rule of thumb is one that always errs on the side of caution, but not by much.

 

A fairly standard 1770N/mm^2 fibre cored 6x36 or 6x19 14mm rope is good for 2320Kg at a safety factor of 5:1 - almost half again the load given above. Wasteful.

 

BUT.  The best rule of thumb for this is don't use a rule of thumb.  If you know exactly what kind of steel wire rope it is, ask the manufacturer, or look it up.

 

If you don't know exactly what kind of rope it is, to get the safe working load, multiply the diameter squared by ZERO.  Then put it back in the skip you found it in, because under LOLER (and common sense) you can't hang anything with a rope you can't precisely identify.  

Go and buy a nice new rope, and you'll be told exactly what its SWL in the application you want to use it for.

 

Each doubling of the sag halves the tension in the span rope, and conversely each halving of the sag doubles the tension.

Err... no, not exactly.

 

The horizontal component of the tension in the rope is constant along the length of the rope, and varies inversely with the 'sag'. This isn't quite the same thing as the overall tension in the rope, except where the rope itself is horizontal (right at the middle of the sag). Elsewhere, you also have to allow for the vertical component of the tension - which varies along the length of the rope and is greatest at the anchor, where its equal to half the weight of everything thats hanging on it. (Because there are two anchors, and between them they're holding the whole thing up.)

 

If the rope is within a few degrees of horizontal all the way to the end, then its probably fairly safe to assume (as Flypig seems to have done here) that the vertical component of that force is negligible - but you need to be aware that this is an approximation, and one that gets less accurate as the rope sag increases.

 

The minimum sag in a span rope should be no less than 1/20th of the span with no load attached.

Where does this arbitrary figure come from?  Under some circumstances (strong anchors - light loads), much less than this is acheivable.  (The obvious

'theatrical' example being a tightrope.)  Under other circumstances (weaker anchors - heavier loads) it may be rather too flat.  Perhaps it would be better not to be too dogmatic.

 

Formula:

Tension In Span Rope (TSR) = Total head Load x (times) span /(divided by) 4 x (times) sag.

Example:

Total Head Load (THL) = Load + Gear + Load In the Hauling part - Therefore;

THL = 800kg (example load only)

Sapn = 20m (example span only)

Sag = 1m (1/20th of span)

 

TSR = 800 x (times) 20 / (divided by) 4 x (times) 1 (in other words 16000 / (divided by) 4 = 4000kg).

Just out of interest, care to define the terms "Total Head Load", "Gear" and

"Hauling part" a bit more precisely?

 

I think I see where this formula has come from, but it seems to be out by a factor of 2, which I don't get.

 

Lets call the distance between the anchors (at the same height, btw): L

Weight per unit length of the thing we're hanging up: w

Cable sag: h

Horizontal component of rope tension (constant along the rope): H

 

H = wL^2/8h Ie: Horizontal tension = (weight per unit length*distance between anchors squared) / (8 x sag at centre)

 

This formula is based on the assumption that the rope is loaded evenly in a horizontal plane (ie: its a relatively light cable supporting a drape, or a decorative doodah placed every so many mm across the gap - NOT every so many mm along the length of the cable). A suspension bridge is the obvious example, and the curve the rope makes is a parabola, not a catenary.

 

For a true catenary, where the rope is evenly loaded along its length, for example when flying a cable - this formula is a little bit wrong. The shape the rope hangs in is actually called a "catenary" and its not a parabola, its a hyperbolic cosine curve. There's a small error introduced by this assumption, probably nothing to worry about, but you should know its there.

 

Ok, so, looking at Flypig's example.

We have a 20m span, and we're going to hang 800kg evenly distributed across the space - thats a 40kg per meter. (The mother of all drapes.) For a 1m sag, that gives us

H = 800kg x 20m / (8 x 1m) = 2000kg

 

(Strictly, I should be using Newtons for force - including weight - rather than Kilograms but lets not worry about that.)

 

Ok, thats the horizontal component of the tension in the rope. But remember that at each end there's a vertical component too, half the weight of the whole thing. Neglecting the weight of the cable, that would be 400kg. To get the actual tension, you can combine the horizontal and vertical components using Pythagoras:

 

Tension = Square root of (horizontal tension squared + vertical tension squared).

 

So, the tension at the end of the cable is SQRT (2000^2 + 400^2) = 2040kg

 

Remember that this assumes an evenly distributed load (and that there are other assumptions involved too). If you want a real worst-case scenario to base a rough calculation on (always a good idea) - you could assume that the load is actually a point load right in the middle of the rope. That turns the maths into the much simpler calcs for a (very flat) even bridle. In the case of 800kg hanging 1m below the centre of a 20m span, you're looking at 400kg x sqrt(10m squared + 1m squared) / 1m = 4020kg

 

Here are a couple of handy links about catenaries:

http://www.du.edu/~jcalvert/math/catenary.htm

http://whistleralley.com/hanging/hanging.htm

 

and here's a handy calculator thingy:

http://www.spaceagecontrol.com/calccabm.htm

its designed to calculate the error caused by cable sag in a positional encoder, but will do just as well for getting a feel for how the forces stack up in a true catenary. (Like a cable run - it doesn't assume that the curve is a parabola the way the sums above do.)

 

or engage a qualified rigging company to carry this work out for you

There is no such thing as a "qualified rigging company". A 'reputable' rigging company is the best you can hope for, which should be employing qualified people, for various definitions of "qualified".

 

Just remember to take into account the self weight of the cable

The derivation of the formula I've used above (and presumably also the thingy that Flypig posted) is based on the assumption that the self-weight of the rope is neglible (compared to the load, which is evenly distributed across the gap). You're right to suggest that you need to take into account the self-weight of the rope, but if its significant you also need to come up with a more accurate formula (which is a bit of a nightmare of hyperbolic functions), or make sure you're erring on the side of caution by looking at a worst-case scenario.

 

And please don't worry..  I realise lots of people post on this board who

really should not be let loose on the real world

The posters who I find a bit worrying are the ones who post authoritative, convincing but wrong advice. The really scary ones are those who accept said advice without really questioning it for themselves. I get a mental image of the look on the coroner's face as they explain that some random person on the internet told them how to do it.

 

Incidentally, theres a chapter in Harry Donovan's book which deals with cable sag in bridle legs and is sort of relevant to all this malarky. Worth a look.

 

Sean

x

[flypig]

Hi Seano.

 

Thank you for your observations, questions and information. Good to see you sharing the knowledge.

 

I certainly don't want to get into an argument over methods and formulae of any sort here, but just to set the record straight. All rigging companies here in Australia are qualified under state and territory laws. We pay massive insurance premiums and face very heavy fines for any breach of conduct and/or accident. It is illegal to employ a person to conduct any form of rigging in Australia if they do not posses a current licence to perform the task they are asked to do. Rigging licences are generally provided by courses run by The Department of Education.

 

Unfortunately, some of the regulations and/or requirements are not standard across the whole of Australia, something we are currently working on.

 

In endeavouring to answer Andy Mel's question as simply as possible, I used the general

?rule(s) of thumb as indicated in the course requirements here in New South Wales. I appreciate that they may not be identical in the EU community.

He asked for a formula, I gave him one to work with.

 

"?Foy"? flying is most certainly done in Australia and as I am sure you are aware, this is normally done on Stainless Steel Wire Rope ranging anywhere between 2mm to 10mm and beyond, which require a whole different set of calculations, so I don't think we will confuse Andy with that one. I might just casually mention the last couple of Superman movies, the Olympic games and the Commonwealth games. Enough said?

 

Anchor points are another ball game entirely. I (we) always look to an engineer for the recommendations and approvals on this one, even if we are providing them.

 

I found the penultimate paragraph personal and offencive by the way, and reminded of the saying about "?people who live in glass houses".

 

Chris Higgs' book(s) are also worth a read.

 

H

[Seano]

Thank you for your observations, questions and information.

I see you've chosen to ignore most of the questions, but thats cool.

 

Good to see you sharing the knowledge.

Actually this isn't an area where I really had any knowledge. I just had a gut feeling that the calc you posted wasn't quite right somehow, and spent a little time looking some stuff up. Like the OP, whenever I've been involved with catenaries before its either been (lets just say) an empirical process, or the calcs have already been taken care of by a proper structural engineer.

 

I found the penultimate paragraph personal and offencive by the way, and reminded of the saying about "?people who live in glass houses".

I thought you might, though that wasn't my intention. I meant it quite earnestly, if you think that makes me a hypocrite well, thats your opinion. I doubt I'll be losing any sleep over it.

 

Chris Higgs' book(s) are also worth a read.

Yes indeed, well worth a read. But there's nothing in there specifically to do with catenaries and cable sag.

The reason I mentioned the Harry Donovan book was the (unique as far as I know) chapter dealing with exactly that, mainly in relation to very long bridle legs.

 

Sean

x

[andy_s]

The difference between catenery and parabolic curve is explained in "Structural Design for the Stage" by Alys E. Holden and Bronislaw J Sammler, published by Focal Press. The method of working out the stresses involved is also gone into in some detail on page 258 and following, "analyzing cable systems".

 

A very interesting book, even if the maths is a bit dense for an arts & lit chap like me.

[Seano]

A very interesting book, even if the maths is a bit dense for an arts & lit chap like me.

Great, thanks for posting that, I'll see if I can get hold of a copy. (I may even resort to buying it!)

 

Edit:

Just checked it out on Amazon, it looks fantastic, so what the hell - I've just ordered a copy. My maths is horribly rusty, but used to be pretty good - I was always hopeless at arts and lit, but I got a half decent engineering degree some (where did the time go?) years ago.

Hopefully I can brush up a bit, it'd be really good to get a better understanding of all that stuff.

Thanks again for the tip.

[Chris Higgs]

Seano has it right, I think.

 

Harry is qualified, btw, and although he is sharing his hard-won knowledge, in the 'wrong hands' some of his formulae could be a dangerous thing. His view is (and we have often compared notes) that if you carry out rigging work, you should know you can do it safely before you start. Study, practice, research, design, etc.

Just knowing some relevant formulae doesn't make rigging safe.

As is so often said - if you need to ask, you probably aren't competent.

(That's not a personal slight, just that if you really know, you get on and do it; you probably run or work for a reputable rigging company)

 

I avoided catenary wires because the circumstances are critical to the calculation. It's too risky to offer rules of thumb and definitely something your insurers would assume had been calculated by a competent person before you carried it out.

For anything involving serious consequences as the result of failure (I.e not a bit of twin and earth being strung across a car park) you should take advice.

 

I would also recommend using a load cell to measure forces when next rigging catenary wires.

Get a firm loading figure for the anchors first.

Rig the load cell between the Klein tool and the BeBee, then it can come off the finished line easily.

Keep your eye on the numbers with regard to the rigging as you take up the slack, you may be surprised!

[Andy Mel]

Thanks for the responses everyone!!

 

its always nice to see a can of worms opened up from what for many of us is a common everyday practice - at least as far as tents and decor / trs runs go!!

 

As Chris mentioned the forces generated as the line is tightened are large and it was the way of calculating this that prompted my original post. I've rigged with various people doing these sorts of jobs and we've always wondered quite how the calculation would be done but never needed to apply it - in the real world of work ( if decor qualifies!!) we know what can be done with a span and equally we know when to draw the line and look at alternatives!!! Indeed catenaries for any more demanding work is something I don't want to have to get involved with!! (as was decor..)

 

Thanks again guys, and I'll try and catch up with you (chris) at plasa for a quick catch up

 

Andy Mel

[dbuckley]

Y'know, what I think I like most about the Staging and Rigging forum is it constantly reminds me how much easier us folks who play with electrons have it then you folks who try really quite hard not to drop heavy stuff on people. Having a show sound or look bad is absolutely nothing compared to having a truss on your head.