Cylinder under external load

[MikeArmstrong]

I am currently having trouble designing a reel winder in work, picture included below to help with the explanation.

If you can imagine the wire coming off the bottom of the barrel and attached to a pipeline thus, putting the wire in tension. The tension in the wire creates a compressive stress in the barrel (hoop stress).

From here I have used to Hoop stress (also known as the circumferential stress) to find the external pressure using Lame’s equation, which is shown in the attached *.pdf.

From the solve block the external pressure seems acceptable, but the resultant tangential stress seems excessive.

Can anyone confirm if my Maths or logic is correct? I have been told that this could be calculated using the principle stress theory or a method out of Roarks, any ideas?

Mike

[MikeArmstrong]

Mike,

How are we going to make a lot of money with Richard around.

Maybe we could combine both of Richards idea's.

<Can't you just build a model out of a toilet roll tube and some string, and then scale everything up?

<Make the barrel out of a solid cylinder of steel. I bet it won't buckle then

Mike

[wayne]

By the way,
The cylinder will not have a unifrom compression, it may tend to average out, but at times could be very non-uniform, and for small amounts of time, could have a sever tangental load if the goiing gets tough. The checks I suggested may prove overily conservative, but they do not realy on uniform compression and can, at lest in some measure, deal with the tangential load condition.

[MikeArmstrong]

By the way,

The cylinder will not have a unifrom compression, it may tend to average out, but at times could be very non-uniform, and for small amounts of time, could have a sever tangental load if the goiing gets tough. The checks I suggested may prove overily conservative, but they do not realy on uniform compression and can, at lest in some measure, deal with the tangential load condition.

Doesn't Lame's theorem deal with tangential loads?

A conservative analysis isn't an issue because there could end up being additional loads while in operation.

Mike

[wayne]

Mike,

No it doesn't. only normal pressure, no shear on surface.
That's why it is only one check, and only for the uniforminly loaded condition, which, after all this discussion, is acutally a minor case.

See may last response to Lester about a lower bound solution, if you havn't already seen. If you wish, I cold sketch out and send tomorrow.

(again, just a recommendation, I think that's how I would proceed if I had limited tools and/or time.)

Wayne

[MikeArmstrong]

See may last response to Lester about a lower bound solution, if you havn't already seen. If you wish, I cold sketch out and send tomorrow.

I did see your response and was planning on adopting it tomorrow. I have been given other work and therefore haven't had chance to progress with the design. If you don't mind spending a bit of time a sketch would be much appreciated.

(again, just a recommendation, I think that's how I would proceed if I had limited tools and/or time.)

The only limited tool is me

Mike

[BillWadsworth]

You are in the garden, making your grandson a bow and arrow out of a garden cane, you get a cane, buckle it and tie a string between the ends, OK? You have a buckled strut, buckled because the load still acts directly between the ends of the strut. Now tie some loops of garden twine around the (straight) cane at intervals so that a loose string between the ends is bound to the cane. Now, however hard you pull on the string, the cane will not buckle. This is because the force in the string is bound to the cane and does not act between the ends in a straight line. My point is that a wire wound around a cylinder is just the same, you will get a direct compression in the shall and local effects where the wire first impinges and overall bending and possible local buckling effects but what you won't get is any elastic instability of the cylinder due to the inward compression from the wires.

Euler's mathematics proves that an absolutely perfect strut cannot carry more than pi^2EI/L^2 and stay straight. Granted, no real strut could do this but the imaginary Euler strut is perfect and does buckle without any encouragement, that's the whole point of his analysis. Its a brave person who contradicts Euler.

Someone asked for my suggestions for the case in point. Mike's original photo shows a relatively short barrel between large end plates. I would think in terms of a tube axle with radial plates at about 45 deg. intervals between the axle and the shell so that the barrel can span between the end plates as a tubular beam when the major force, the wire load at clutch-slip acts. I suppose the plate thickness could be assessed from the number of wires in the fully wound drum and the average wire force this will produce a balancing circumferential compression in the shell (but no buckling effect) to be combined with the longitudinal bending stresses using von Mises. Not exactly my line of country, but them's my thoughts.

Bill.

[wayne]

Bill,

Thanks for your thoughts,

In regards to stability, I believe we agree that the condition in question does not exist, so I am going to leave it at that.

We are pretty much in agreement on the drum and generally on a way foward for Mike.

FYI, the picture, at least for me, is a little missleading. Everything is relative, but

The rope is 77mm, or 3" in diameter. The UDL, is 75 tonne, or 165kips

The Drum's radius is 1.232m or 48.5" in diameter and the plate thickness is 20mm or 0.79"

The Drum's span is 1.3m or 51.81"

So realtively speaking, the plate is thin.

Thanks,

Wayne

[BillWadsworth]

Just been out walking on the hills in the sun, its great to be retired!

Seen your extra comments and thought perhaps we could agree that:

1) a perfect strut loaded less than the critical Euler load will remain straight

2) the same strut loaded with more than the Euler load will have buckled

I accept that a very, very, minute disturbance might be needed to induce the buckle but I insist that however carefully it is applied the strut can never carry any more than the critical load and remain straight, (which is what I thought your original comment was implying). You could say the disturbing force is vanishingly small so that I think it is not there at all but others, obviously, think it is there, just a bit.

I note the sizes involved you mention, it is big beast.

Bill

[RichardJ]

Just been out walking on the hills in the sun, its great to be retired!

I've just been out shoveling several inches of wet snow. I think you got the better deal!

You could say the disturbing force is vanishingly small so that I think it is not there at all but others, obviously, think it is there, just a bit.

Exactly. a very very small bit, but still a bit.

[RichardJ]

An object (or part of an object) cannot move in a given direction unless there is a force acting upon it in that direction.

A perfectly axial force acting down the axis of a perfect strut or column cannot produce a net external force on part of the strut or column in a direction other than straight down the load axis. The symmetry is perfect so there can be no preference for one direction over another. Therefore there can be no net movement of any part of the strut or column perpendicular to the load direction.

It's an even braver person that contradicts Newton.

[RichardJ]

Where are you going to get a solid steel toilet roll "tube"?

[MikeArmstrong]

Where are you going to get a solid steel toilet roll "tube"?

Amazon?????

Mike

[f.kohlhepp]

Richard Jackson wrote:

Where are you going to get a solid steel toilet roll "tube"?

"seamlesss drawn steel tubing"

http://www.karaymetals.com/steel.html

(for one of thousands)