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Design for a static Trapeze

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simonmags

Computer
Oct 27, 2010
7
I have need to work out how to design a permanent static trapeze with specifications as follows.

Height 6000mm
Width 4500mm
Load less than 190kg (2 people)
Load may swing to a maximum of 45degrees
I'm guessing guy wires will be necessary?
Soil is medium reactive clay

Having only done a Computer IT degree, I am familiar with some design tools, just looking for a place to start in order to work out material strengths needed to build said trapeze.

I have found one company that makes self standing frames ( ) But have not seen any specifications on how to design a permanent rig.

Thanks in advance, sorry if i've missed any critical information, but please don't hesitate to ask.

Simon
 
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OK...first of all, a trapeze will not be static. It has to be designed for dynamic effects. Second...you're dealing with people and potential personal injury...get someone involved who knows this stuff...you are putting yourself in a liability position you don't want.
 
Thanks Ron, who would you recommend to design something of this nature. Liability will only be an issue if I let someone else use it, which i don't intend to.

Whilst the load won't be static, the name of the trapeze is a static trapeze as it is not designed to swing, more to have someone hang from and move about on, I imagine this will cause some movement and as such specified that it would not go past 45 degrees as the maximum. Likley it would not even swing 30 degrees, but I prefer the idea of building in a bigger tolerance.

So if I was to continue with the idea of working out a design would you be able to point me in the direction of ho to specify the material for the uprights and horizontal beam?

 
Look around for a structural engineer with dynamic design experience or a mechanical engineer with similar. Either should have product analysis and design experience.

Those involved with amusement rides and similar will have the necessary tools to handle something like this. You'll have to look around a bit, because there are not many with such experience.

I have the experience with both dynamic product analysis/design and amusement rides/devices; however, I will not solicit work here and typically do not do this type of work anymore. The best dynamic product designers are captive to the permanent amusement parks (Disney, Universal, Busch, Six Flags); but they often moonlight. The suppliers of such are very knowledgeable, but less likely to moonlight.
 
So no tips on what formula's to use to work out the requirements for the steel posts or horizontals then.
 
Simonmags,

Would you really risk your life on something you designed/built from some stranger's suggested formula on an internet forum? I understand you probably don't want to spend money on something so "simple," but what you are asking is more complicated than you probably realize.

I don't know much about statistics, but I do know that if something has a 50-50 chance of going wrong, 9 times out of 10 it will.
 
Thanks Johnny, I was interested in the challenge of actually working it out.

Anyone happen to know if there are Australian Standadrs for this sort of thing?

I guess I was thinking it would just be broken down into a series of formula's, i.e. you would use something like this to calculate the steel requirements for the columns ( ) and so on for each component. But maybe not
 
Simon,

You're structure will need to comply with numerous Australian Standards. I'm assuming this equipment is outside? You will need to design it for wind loads combined with gravity loads, you should also be familiar with the safety requirements of such a structure.

I'm guessing AS4685.2-2004 would be a good place to start but get someone who knows what they're doing!
 
simonmags,

Structural codes are not 'how to' guides, you need to understand the fundamentals behind it in order to safely use any of the codes.

 
c'mon guys ... he doing this for himself.

start with wiki "strength of structures", "strength of materials", "beam stress analysis", "truss analysis", "free body diagram".

i don't really see what the apparatus is going to do. but it'll be loaded by the inertia of the people it's supporting, and it sounds like they're bouncing over the place, in different directions. this'll make for lateral loads, so yes, you'll need guy wires.
 
A ray of hope :) Thanks rb1957.

My better half used to be an acrobat and I'd like to at least work out how hard it would be to get one of these built. I'll start poking about on wikipedia and look at those terms. Thanks for the tips.

Static Trapeze is really just a giant fancy monkey bar :)
 
Simonmags,

There is also this article which may be of interest to you -


Read the first paragraph in the `Elastic Energy` section, which gives an indication of the bar deflection required in order to achieve the desired bar `springiness`. The bar is usually 28.5mm (1-1/8") diameter (needs to be small enough to grip properly) and with your 2-people and span of 4500mm, I think you will struggle to achieve this!! Your bar needs to be much shorter, nearer 2000mm and 1-person only, I would suggest.

Some calculations are in order I think.

Regards
 
Hi Neilmo,

That was some interesting information in regards to the Gymnastics parlell and uneven bars.

The Trapeze bar that I want to hang from the Frame does not require deflection in the bar. It should be stable and have a minimum of deflection. But thanks for those links they were interesting to read.

This idea is all about doing some calculations, I'm still looking at the information on wikipedia to determine the best structure :)

Cheers

 
oh ... that'll be much easier to design ... being limited to a hand hold (as above) was really limiting what you could do.

so you what a bar 6m long that is b'jesus rigid and you're going to suspend a trapeze bar from this.

let's say you've got 2 people at (say) 1/3 pts on our beam(2m apart). let's say they're "whipping" around at 2g. so the beam has reactions at both ends of 100kgs*2 (= 1000N*2) in pretty much any direction ... sounds like a job for square or round tube. the maximum moment in the beam is (2000)*2m = 4000Nm. now this beam is going to be designed by stiffness (rather than strength). Conservatively (and 'cause it simplifies the sums enormously) assume the load is applied in the middle of the beam. max deflection = W*L^3/(48*E*I) = (2*2000)*6^3/(48*E*I) assume steel, E = 200E9Nm^2) ... so d = 4000*216/(48*200E9*I)

so a d = 0.001m requires I = 1e-4m^4 = 1e8mm^4

a 100mm sq tube, 10mm thick wall has I = 5E6mm^4 ... I square tube = (100^4-80^4)/12 ... so 1mm is probably too stiff. also consider the weight of this tube 6000mm*3600mm^2 = 2.2E6mm^3 = 1800N = 180 kgs ... this'll increase the deflections some.

think about this, see where you want to go ...
 
so you can see the bracing required for the up-rights.

you could replace the 3 bracing wires with 2, at 45deg to the cross beam, yes?
 
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