Tube stiffness perpendicular to it's axis

[PCS74]

I need to design a die that will allow me to crush a 3.0in. diameter 0.096in. wall 6.0ft. long DOM tube into an oval ~1.75in. x 3.XXin.

Can someone point me towards something that will teach me how to calculate a close estimate of the stiffness of a cylindrical tube perpendicular to it's axis. In addition links to any websites that would give me any additional info would be greatly appreciated

I don't have access to any FEA software, or testing equipment.

Thanks for any help you can give,

Chris Skarzenski

 

[israelkk]

Roark's book has the formulations for ring loading per unit length or the ring and load per unit length. A tube is just a long ring. However, in your case you have to go beyond the elastic zone into the plastic zone. Therefore, use the maximum load that yields the tube material as the minimum needed. As fas as I recal Bruhn's book has the formulations too.

 

[JStephen]

I don't have an analysis, but it seems as the tube ovals out, the load may reduce- meaning that if you apply Roark's formuas, find the load corresponding to the yield stress in the tube, that may be the maximum load as well.

This response and the one above are both assuming that you apply pressure along the whole length of the tube. If you are applying a point load, and just crushing the tube in one place, that would be different.

Also, when you compress the tube, there's no guarantee that it will go into a nice oval shape- at some point, it would deform into an 8-shaped cross section, sort of.

 

[MikeHalloran]

Have you tried just beating it with a urethane mallet?

If you fill it with sand or salt and weld the ends closed, you can probably crush it in a big press brake with a flat punch and a flat die. Better and probably safer if you have a flat urethane die. Restrain it so it doesn't pop out.

You might be able to crush it to a defined shape with multiple passes through a two-roll forming machine. The required custom rolls would be a bit cheaper than a press brake punch and die.





Mike Halloran
Pembroke Pines, FL, USA

 

[GregLocock]

If this is for large quantities then it might be cheaper to get it extruded in the new shape.

However, the simplest mechanism that can give plastic collaps is a 4 bar link, so you can work out the Force required to fail the metal at 4 plastic hinges by using a work equation and the fully-plastic moment of the wall.

In practice I suspect this is an upper bound estimate.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[vonlueke]

PCS74: I haven't checked Roark, but I think you can estimate the maximum line load per unit length, w, required to deform the ring by setting the bending stress at +/-77 deg (and +/-103 deg), divided by plastic shape factor, equal to tensile yield strength, and solving for load w. This works out to w = 1.47(Sy)(t^2)/r, where r = cylinder mean radius, t = wall thickness, and Sy = tensile yield strength. And the initial ring stiffness per unit cylinder length works out to be k = 0.1126(E)(t/r)^3, where E = modulus of elasticity.

 

[dik]

I think that you have to use a hydraulic press for control and so the load falls off quickly. I would probably use a test piece compressing it to 1-1/2" or whatever, allowing it to recover to 1-3/4"...

Dik

 

[PCS74]

Thank you everyone for your help. I have ordered Roark's book and it will be here Thursday.

I will only be making the tube in 12ft section two to four times a year so anything excessively expensive or overly complicated is not desirable. The two roll forming machine is a good idea and I will look into doing rolls instead of dies.

Greg,

I am interested in the method you mention but I am not familiar with it. Can you recommend some good reference materials?

Once again, I greatly appreciate everyones help.

Chris Skarzenski

 

[dik]

I think you have to do this by trial and error. I don't think you can analyze this because of internal stresses within the tube and the areas of contained plastic flow. I think it's too variable to model by FEM or formulae.

dik

 

[GregLocock]

Method of plastic moments was taught to me as part of my structures course. Section 8-6 in "Mechanics of Materials" by Higdon, Ohlson,Stiles,weese,riley.

The fully plastic moment of a hinge in tube of length L is Mp=yield stress*t/2*2*t/4*L, I think

Then set up a mechanism that will collapse, work out the angular deflection of each hinge, for a given deflection at the loading point delta, and its length. Use Mp as above.

Then you know F*delta=sum(Mp*theta) for all hinges

Choosing the appropriate mechanism is the trick - you can get very clever and differentiate a parametrical shape to find a minimum for F, but your 3d geometry better be really good.

Alternatively 20 minutes in the workshop with a hydraulic press would probably pay dividends. Grin.









Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[RPstress]

Use the formula in Roark (the most appropriate is probably in the curved beam section, where it covers rings) to predict onset of yield. Then multiply this by 1.5 for the load to fully plastically deform the wall all the way through its thickness. If you use this with an average material property divided by about 0.8 for an upper bound on compressive yield strength, then it should give you something like an upper bound for the force.

 

[GregLocock]

Sure, for a plate that is unconstrained. The point about this problem is that ONE plastic hinge is not enough to give collapse, you must have 4 or more.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[GregLocock]

Having said that you can certainly get permanent plastic deformation with yielding just at one point, obviously.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.