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How to evaluate the resistance of a wire mesh by calcs ? 1

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gasma1975

Structural
Sep 19, 2006
53
Hi,

I have a client who asked me to evaluate how much pressure his steel cage can handle. It consists of HSS beams/columns with wire mesh. He asked me if it can "WITHSTAND A MAXIMUM IMPACT OF 295 psi"

My question is regarding the wire mesh, is it possible to calculate?

Its a cage where industrial tires will be inflated...

thx

Gasma1975
 
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I have attached an image of the cage. How do you evaluate the pressure on the structural elements. I mean that when the pressure is released, some will pass through and some will be on the structure, but how much ?
 
I would say you need to re-define the problem.

What he means is "I want this to restrain a tire/rim combo that blows apart at 295 psi". The impact force of a split rim on the wire mesh could be a lot higher than that. The gas pressure would be a lot lower than that. The 295 psi comes from the requirement in 29 CFR 1910.177 about 1.5 times the maximum pressure, which is shown on the drawing as 195 psi or so.

I don't know that you can get a fool-proof approach to this without a lot of testing. You can come up with some estimate of the maximum speed of a wheel or a split rim or a wheel-tire combo if it blows apart, and design to absorb that much energy from the impact. If the cage fits close around the tire, you could possibly assume that the tire blew out against the cage on both sides and design for that static force as an additional load case. If a tire blows out (as opposed to blowing off the rim), the whole assembly could jump straight up or to either side. I would think that the customer would need to either pay for a bunch of testing, or accept some or all of the liability for the design of the cage. It would be prudent to look at commercially available cages, too. The successful use of the cage involves using it properly, and is not just a structural issue. (For example, have any part of your body inside that cage at the time, and it's liable to be missing, even if the cage "works".) If the rims or wheels are in current production, it might be worthwhile to try to see if those manufacturers have any recommendations on cage construction or use as well.

Note that tires of the same general physical size can vary considerably in weight based on construction and load rating. Note that wheel/rim parts can vary in weight as well (budd vs spoke, 1 pc/2pc rims, etc.)

I would consider grating or plate rather than expanded metal in the construction. If it's supposed to be portable, that could get tricky.

Long ago, I worked in a dump-truck company and did a lot of flat-fixing, and all the wheels were split rims. It is possible for these things to blow apart, and is a serious concern, but it's not something that happens twice a week either, so that makes it harder to have a good feel for the effects.
 
By way of trivia, I submit the attached picture. As I went along in my flat-fixing career, I started saving all the metal crap that I pulled out of tires. This is it. Each piece represents about 40 minutes of work. About half the flats were caused by rocks or glass pieces working through a tire, and I didn't save those, just the metal.

Normally, the action of an inflated tire keeps the split rim in place, so you can't pop the rim off an inflated tire. And normally, the first thing we did when we brought a tire in was to remove the valve core. But one time, I neglected to do that and pried the split rim off a tire that had just a very little amount of pressure still in it. When it came off, the rim went about 5' in the air. I still have a scar about a 1/2" long on one finger from that. That wasn't enough pressure to pop the tube.
 
 http://files.engineering.com/getfile.aspx?folder=2cce34ca-38de-4843-8e7e-49abdb42384b&file=Flats.jpg
JS...nice collection!

I agree with your comments. They come from practical experience and engineering...usually the better combination.

Split rims are inherently dangerous. As I'm sure you know, people have been killed by them during re-inflation of truck/equipment tires.

All in all, the frame seems light compared to the homemade ones I've seen that were severely dented from mishaps. As you noted the forces can be tremendous. If you take the 295 psi and apply it to the contact area of a split rim (which might be around 100 square inches) you get a reasonably high force!!
 
Thank you all for the tips, I guess you cannot design it and verify it on paper you need real life testing.

As suggested by Ron, is it ok to assume if the tire is inflated at 295 psi and the total contact surface is 100 sq.in you get a force of 29.5 kips, Is it ok to apply this force on the HSS tubing of the cage ?

 
SB75...that might be a place to start, but I would look at different tire sizes, different rim configurations and restraint conditions. Most of the cages I've seen have been made from HSS sections with expanded metal, not wire mesh. Perhaps some of them were overdesigned, but they were dented in ways that showed they worked but that the forces were quite large.

I have no real feel for the forces....I gave you a guess as to an approach. JS might have a better feel for it.

My experience comes from getting tires repaired, and standing there while they were completed. I've seen a split rim retaining ring pop over the rim. It lifted the cage off the floor. That was on a 22-inch diameter rim on a typical semi-trailer.
 
This doesn’t seem to me to be a problem which is easily amenable to any fairly simple set of calcs. or engineering assumptions. As Jstephen suggests it probably has to finally be approached by some testing method, and even then you won’t know what loads to apply or how to apply them in your testing methods. There are systems like this on the market, and in use, and it would be nice to know more about them and how they have been proven, proof tested. The first question is how will the tire explode at 195 or 295psi. You could get a 10"x24" piece of tread hitting the cage over some area, a distributed load and force on the cage; or if the split ring unzipped, from the rest of the rim, you might get the mass of that split ring hitting the cage at the ring’s open end and trying to stop the entire mass of the split ring. This concentrated loading on one or two wires would undoubtedly shear right through those few wires, in the process of slowing the split ring down. And, all you can do is guess at the magnitudes of the forces distributed over some area or concentrated on a few wires. Your extension of Ron’s thinking to arrive at 29.5k is not likely correct, but does show that you probably do not have a good handle on this problem.

You can make some reasonable engineering assumptions and with some experience and judgement start to hone in one the capacity of the cage. You must know all of the details of the cage construction; mechanical properties of the various materials used, the welding details, etc. Then I would start by thinking about and calculating the capacity of a single wire at its yield or ultimate strength and a tolerable deformation; then two or three wires. Then, under the worst conditions how do these wire forces affect the tube structural members and the wire welds to the tube structure. You would like everything (all structural elements) to start to yield, or collapse around the failing tire at about the same time to absorb the max. energy of this dynamic action (problem). And, still the big engineering difficulty is determining these actual dynamic forces and displacements, and setting some limit for an acceptable outcome. I would think it might be reasonable and acceptable to assume that the cage was essentially destroyed after a tire or rim failure, and having saved a life. And, I would expect some pretty big dents and deformations from lesser failures. For a seemingly simple apparatus, this is really quite a complex engineering problem the way you have asked it.
 
The only contact area that the 295 psi can apply to is the interior surface area of the tire. You could assume a 10x10 chunk of tire hitting the mesh, being accelerated by the 295 psi. You'd then need to assume a certain deceleration time caused by the mesh, which would then give you the deceleration force applied to the tire chunk. That force would then need to be less than the strength of the mesh.

TTFN

FAQ731-376
Chinese prisoner wins Nobel Peace Prize
 
How do you determine the acceleration of a 295 psi ?
 
A brute-force, worst-case, approach would be to assume that the pressure does not lessen during the entire travel time from rupture to impact. 295psi*area of chunk/mass of chuck is the acceleration.

A more rigorous approach would take into account the velocity of the chunk and the dissipation of the pressure wave as it expands.

TTFN

FAQ731-376
Chinese prisoner wins Nobel Peace Prize
 
IRstuff....the distance from the tire to the cage is so small that dissipation (other than the initial loss of energy) would be nil. I would certainly opt for the "brute force" approach; however, proof testing is more appropriate for this device.
 
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