Bending and torsion of a cylinder

[Willers]

A stage in one of our manufacturing processes involves the rolling of a cylinder. Inevitably, the cylinders are not fed into the roller perfectly square to the backstop.
Attached is a pitiful attempt at explaining the resulting shape of the cylinder (exaggerated).

The next process in production is welding of the cylinder seam, and stops on the weld machine force the cylinder back into shape. Although tests have been carried out on the force required for this, and some FEA, i'm attempting to validate them numerically.

While I believe the cylinder can be modelled as large helix/spring with one revolution, I cannot find any formulas relating this to a rectangular x-sectional helix.

Is modelling this as a combined bending and torsion problem, in order to determine vertical deflection/force possible, or is it too complex?

Thanks in advance

 

[CoryPad]

I don't think your attempt was pitiful - I found its simplicity quite descriptive.

Can you give us an approximation of the dimensions? Is the thickness much smaller than height and circumference?

 

[hydtools]

Here is a machine design reference for rectangular wire helical springs. Might help.

http://books.google.com/books?id=08...&hl=en&sa=X&oi=book_result&resnum=2&ct=result

Ted

 

[Willers]

The wall/sheet thickness is 4mm, outer diameter 300mm, and height approximately 400mm. I hope this helps you help me!

 

[GregLocock]

Is it not strongly related to the shear force required to shear a flat panel, 400*900, by whatever the joggle distance is?



Cheers

Greg Locock

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

 

[rb1957]

i agree with corypad, the simple pic explained the question very well.

basically you've got a rectangle that's rolled about an axis that's not parallel to a side. if you try and force the corners together you won't end up with circular end-faces (if that's an issue) ... they'll be elliptical, and they may not be flat (again may not matter too much).

i'm guessing you're using steel.

to answer the question, it looks like you'd be trying to enforce a displacement (of 1/2 the difference) on a panel 400x470(=pi*150). there should be a theoretical solution and FE would be simple as well.

 

[swearingen]

Greg, it will be far less than the plane shape shear force because you are also twisting the sheet to accomplish the alignment, which requires much less force.

"Although tests have been carried out on the force required for this, and some FEA, i'm attempting to validate them numerically." - does anybody else have an issue with this statement? How can numerical approximations ever validate anything? Isn't it supposed to be the other way around?

You already have tests showing what the force needs to be - are you really being required to somehow back this up with numbers?

Try taking the data from testing and comparing it to helix-type formulas to see if the data fits. For small deflections, it may be close...


If you "heard" it on the internet, it's guilty until proven innocent. - DCS

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[Willers]

"You already have tests showing what the force needs to be - are you really being required to somehow back this up with numbers?"

I am only required to prove our manufacturing process to a customer, and i'm trying to do it as well as possible. You are right, if FEA matches with test data, more than likely it will be correct.

However, we are trying to validate our manufacturing procedure to an engineering buff, and a numerical answer will mean a lot.

As stated in the subject, i believe this is a combined bending and torsion problem. While as a flat plate, the deflection is simple to find, once it is subjected to torsion it is more complicated. I guess i must accept the empirical data and the FEA.

 

[chicopee]

About welding the rolled plate the way it comes out and then trim off the ends, you'll then have a "circular" cylinder.

 

[GregLocock]

... of indeterminate diameter!

Cheers

Greg Locock

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

 

[prex]

I'm away from my machine handbook at the moment, so will attempt a solution from scratch (no guarantee).
With obvious meanings for symbols, assuming one side of the cut supported, the other one loaded with P, the torsional moment at any location angle [θ] (from 0 to 2[π]) is M[sub]t[/sub]=PR(1-cos[θ]), the corresponding angle of torsion for an elementary length of cylinder ds=Rd[θ] is d[Θ]=(PR[sup]2[/sup]/GK)(1-cos[θ])d[θ].
This angle produces a displacement at the loaded end that's given by dh=d[Θ](1-cos[θ]) and this leaves us with dh=(PR[sup]3[/sup]/GK)(1-cos[θ])[sup]2[/sup]d[θ] and after integration
[Δ]h=3[π]PR[sup]3[/sup]/GK
Now, for a very elongated rectangle, K=ab[sup]3[/sup]/3 (a is the long side) and this gives the result.
There's however another phenomenon.
In a very thin rectangle under torsion the relationship between the moment and the unit angle of torsion (d[Θ]/ds) is no more linear, it has the form (taking now [θ]=d[Θ]/ds):
M[sub]t[/sub]=Gab[sup]3[/sup][θ]/3+Ea[sup]5[/sup]b[θ][sup]3[/sup]/360
Can't verify at the moment whether this influences appreciably the result.



prex

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[Willers]

Thanks very much Prex, thats the kind of thing im looking for, ill put some numbers in and see if i get a similar result to the FEA.

"About welding the rolled plate the way it comes out and then trim off the ends, you'll then have a "circular" cylinder."

This will only add another process to manufacture and increase cost unfortunately.

 

[Willers]

Thankyou all so much for your help

Using the calculation for the deflection of a rectangular x-section helix produced a deflection/force within a few newtons of the FEA and testing, and im sure I can justify the difference by the fact that 1 rotation of a helix isnt technically 1 link in a spring, due to it also acting as the ends (i think)

Problem solved!

 

[Willers]

Just one more thing!

The book hydtools linked give the formula for deflection of a helical rectangular x-section spring (given below)

http://books.google.com/books?id=08...hl=en&sa=X&oi=book_result&resnum=2&;ct=result

For the deflection, everything is self explanatory apart from the 'i' after the 2.83constant. I was assuming that this is the number of coils in the system, as apart from that, there is no term for it, however I have only ever seen the letter n used for this.

Could anyone verify that this is correct?

 

[rb1957]

personally, i'd follow prex's calc ... does your situation look like a helical spring ?

 

[Willers]

It doesnt look like one persay, although as a system (forces etc) I believe it should behave in the same way.

 

[chicopee]

I believe that no type of formulae will correct your problem of straightening the seam by applying external forces. You may get undalations or other abnormalities.
I would suggest the extra step that I mentioned above or try readjusting the rollers so that the gap between is not quiet uniform to get the effect that you want.

 

[hydtools]

Willers, I had to go all the way back to chapter 2 in the referenced book to find the author's definition of i as the number of active coils.
Chapter 2, para 2.10.

Ted

 

[prex]

Willers, helical springs are calculated assuming the force is along the centerline of the helix. In my formula above this would cause the factor 3 being replace by a 2 (50% difference!). Don't want to cool down your enthusiasm, but your result could be a matter of coincidence.

prex

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[Willers]

"In my formula above this would cause the factor 3 being replace by a 2 (50% difference!)"

Testing, FEA, and your calculation Prex (and also the helical spring, although i appreciate that the use of this is tenuous at best) show a corrective force in the region of 10s of newtons, while the machine correcting the step is capable of potentially 1000s. 50% different is fairly irrelevent, as would a 1000% difference for the purpose of this, but then I should have probably made that more clear at the start.

Any attempt to find a force numerically is only an effort to show that this is the case, whereby the corrective force needed is many times less than the potential corrective force.

Thanks again!