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5-Regular Member

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

1 ACCEPTED SOLUTION

Accepted Solutions

Mike,
I made some corrections and some clarifications / comparisions.

Different references and mising thick walled and thin walled analysis was getting things messed up. This should be clearer

It looks like what was done in the past was:

1) Assume cyliner was continuoulsy wound with a uniform tenion in the cable, and not reaction or anchoring of the cable

Check stresses for this condition using von Mises.

Check stability of section, the collapse load.

2) Assume the cylinder anchored the cable (for it's rated load) by applying the cable force to the center of the cylinder, the cylinder then spans as a hoirzontal beam from end to end.

Check bending stresses

Check Shear stresses.

For case 2 however, I don't think the check is complete.

The cylinder cannot be loaded at the center, so there must be come torsion, which addes to the shear stress.

The worst case for shear is with the load applied towards on end or the othere.

This design assume no winds of the cable on the cylinder, whcih is probably not reasonable form what I understand. As a upper limit check, these stresses should be superimposed on the stresses from 1 and a von Mises check done.

Additionally, the collapse load is effected by case 2 as well, but much harder to analyize.

So the first thing is to determine what the actual design (loading) criteria is.

Wayne.

42 REPLIES 42

Mike,

See attached.

The principal stresses are a check on the failure criterion, using von Meises, not a different method for obtaining the stresses.

Roarks formulas may look a little different and include calculations for the change in diameter due to pressure, but are the same for determining the stresses.

Both the Lame method and the Roark method (same) are based on continuously wound cylinders, in effect assuming, uniform conditions away from ends.

If you only have one wind, the condition is much better. You could estimate using the above, but use a larger effective width over which the cable force is assumed to act, therby lowering the stresses.

The end caps will go a long way to increasing the buckling, or collapse, load of the cylinder. Again, I believe the buckling equation is based on infinite lenght, or open ends.

A sketch would help, showing the how many windes and how the cable comes in and goes away from the cylinder, because I am still not sure.

Wayne

5-Regular Member
(To:wayne)
 The principal stresses are a check on the failure criterion, using von Meises, not a different method for obtaining the stresses. Roarks formulas may look a little different and include calculations for the change in diameter due to pressure, but are the same for determining the stresses.

Right I get you, Wayne.

 If you only have one wind, the condition is much better. You could estimate using the above, but use a larger effective width over which the cable force is assumed to act, therby lowering the stresses.

As a worst case, I think we would have to look at the barrel fully loaded with wire.

I will obtain a sketch from one of our Draftsman and post sometime today.

Cheers

Mike

Mike,
I made some corrections and some clarifications / comparisions.

Different references and mising thick walled and thin walled analysis was getting things messed up. This should be clearer

It looks like what was done in the past was:

1) Assume cyliner was continuoulsy wound with a uniform tenion in the cable, and not reaction or anchoring of the cable

Check stresses for this condition using von Mises.

Check stability of section, the collapse load.

2) Assume the cylinder anchored the cable (for it's rated load) by applying the cable force to the center of the cylinder, the cylinder then spans as a hoirzontal beam from end to end.

Check bending stresses

Check Shear stresses.

For case 2 however, I don't think the check is complete.

The cylinder cannot be loaded at the center, so there must be come torsion, which addes to the shear stress.

The worst case for shear is with the load applied towards on end or the othere.

This design assume no winds of the cable on the cylinder, whcih is probably not reasonable form what I understand. As a upper limit check, these stresses should be superimposed on the stresses from 1 and a von Mises check done.

Additionally, the collapse load is effected by case 2 as well, but much harder to analyize.

So the first thing is to determine what the actual design (loading) criteria is.

Wayne.

5-Regular Member
(To:wayne)

I thought I was half decent at Structural analysis, but your work has embarrassed me.

Lovely work and descriptive text. Once (if) I finish the calculation I will post my results.

 Additionally, the collapse load is effected by case 2 as well, but much harder to analyize.

We will add lateral stiffeners inside the barrel to prevent collapse. Working out a combined moment of inertia value will be my next task, which I assume could be superimposed into the critical stress calculation.

Cheers

Mike

19-Tanzanite
(To:MikeArmstrong)

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

5-Regular Member
(To:RichardJ)

Why didn't I think of that.

Mike

1-Newbie
(To:RichardJ)

In my first engineering class, FEES 101, they tought us

1) make things as convoluted and difficult as possible,

2) mispell a lot, that way they will think you are a genius from another contry

3) I am not sure what goes here?

4) then they will pay you a lot of money.

I go the first 2 down, could someone provide some insite on 3)

5-Regular Member
(To:wayne)
 I go the first 2 down, could someone provide some insite on 3)

No, sorry.

 4) then they will pay you a lot of money.

I wish someone would help me with this one.

Mike

19-Tanzanite
(To:wayne)
 I go the first 2 down, could someone provide some insite on 3)

Change career from being an engineer or scientist, otherwise you will never get to 4)

1-Newbie
(To:RichardJ)

which leeds me to the old saying:

How do you make a small fortune in the Engineering business?

5-Regular Member
(To:RichardJ)
 Change career from being an engineer or scientist, otherwise you will never get to 4)

Bit late now

Mike

4-Participant
(To:MikeArmstrong)

Mike,

I don't think you need to worry about overall elastic instability. A drum will not buckle under a wound-on rope any more than a concrete column would buckle under a prestressing cable. The tension in the rope and the compression in the cylinder shell are mutually bound, the one to the other, for buckling to occur the load must be independent of the structure so that it still has effect after a putative buckle has occurred. A few minutes playing with some string and a kitchen roll tube shows that even if you grossly distort a tightly wound tube there is no tendency to buckle; you can induce some localized creasing under the string where it first contacts the tube which, I think, is the main design problem with the shell. Especially if the item being wound in snags on something so that the cable force rises to the clutch-slip load, or whatever other cut-off device limits the system, (hopefully, before the rope breaks!).

Bill.

Lester,

Thoughtful respose.

I am not an expert in this area, this is more of a mechanical engineering field, but a few comments; (many of wich you allude to)

1) There is global buckling, local buckling and stability (which may be elastic or inealstic). An internally generated stress, like the prestressed column, will not buckle in a global sense from the internal prestress, as you point out, but it can buckle from local instability and it can be unstable.

2) In a perfect world, a column won't buckle either , just uniformily compress.

3) A beam- olumn, under transverse load, does not really buckle because it starts out displaced, but can become unstable and can have local instabilities.

4) The pretensioned column can be unsymetrically overstressed from the pretensioning and become unstable and fail, all by itslef

5) The barrel will not be perfectly round, will have significat internal stresses from the bending process (note that it already has plastic deformation) and will have at least one weld seam (which is why it may be inelastically unstable). Additionally the barrel must span from end to end to the bearing supports, so there are additional stresses. It is noted that the elastic stress check (nom Mises) is not really valid because of this, but it is still has merrit because it is a measure of the average stress levels, noting that stesses begin to even out as higher stressed areas become inelastic.

6) In the Richard Test (sorry Richard, but you invented the toilet roll test, so it's named after you), If you kink the role and then tension the string, It will have a load under which it will collapse. If the tension dissapeared as the tube begins to collapse, then I agree that it would not contnue to fail from the initial tension in the string, but if externally applied, as would need to be assumed in this case, it would. (I note that you also allude to this)

7) So what to do: Short of doing a bunch of tests, or a finite element analyisis, you could apply some general width /thickness criteria or local instability criteria as is done in the structural engineering field, and I assume is done in the mechanical engineering field as well. In this case, the engineer that did the calculations for which Mike is adapting, has a collapse equation, which on the face of it appears to have the necessary variables. I assume that this equation came from a handbook or text that addresses similar condiitons. It would be a good idea to find the origon if that equation, which is probabley in a mechanical engineering text, machine design, or something like that.

😎 You can also add stiffners, for example a donut shaped cylinder midway between the ends, which are already stiffened, along with some thickness criteria to prevent the local instability. (The end stiffners may already accomplish that, but I don't know).

Anyway, some more to think about. But in the end, there is some form of widht/thickness criteria that must be applied. (at least I definatly probably think that could possibily be the most likely possibility )

Wayne

4-Participant
(To:wayne)

Wayne,

Yes, I agree with nearly all of that.

In a perfect world, a column won't buckle either , just uniformly compress. I don't know about that, I thought the Euler load was the buckling load of a 'perfect' strut, not often, if ever, seen in the real world I grant you, but in a perfect world I would expect to see it. The whole point of the critical load is that it is the load at which a perfect column will buckle, even if there is an infinitely high material yield stress.

It was the 'collapse equation' which caught my eye, it is for the 'elastic critical stress in a short tube with ends held circular but not otherwise restrained'. It is in Roark, 4th ed. p 354. It was this which led me to respond and it is the point of principle that a wire wound round a cylinder might induce overall instability such as might arise in a cofferdam, for example; it will not.

I was reminded of a paper in the ICE Proceedings many years ago in which it was proposed that a prestressed turbo-alternator block could be 're-tuned' to avoid resonance with the machine by altering the prestress. Just like tuning a violin. Many professors of engineering replied to the discussion to say that this was rubbish, the wires might have been re-tuned but the block just got a mite shorter. I know this is vibration, not stability, but they are related effects.

regards, Bill

Bill,

Thanks for finding the equation. I will look at later FMI. I assume the load is to the side of the tube, applied thru the center, so I am not sure it applies, but on the other hand, it has some relation to the proportions of the barrel.

A perfect column must have a displacement, however small, to initate buckling. You find the buckling load by applying a small or virtual displacement.

Thanks again

Wayne

4-Participant
(To:wayne)

Wayne,

I don't want to harp on about this but, I looked in my copy of The stability of Frames, Horne & Merchant, where in para. 1.2 I read...

The behaviour of a member subjected to axial load only (they had previously been discussing a beam-column with end moments) may conveniently be obtained by allowing M to approach zero. The curves shown then approach the limit represented by the axes indicating that no deformation occurs until P=Pe at which load deflexions of indefinite magnitude occur...so although the analysis starts by assuming a small deflexion of the strut it then considers what happens when this vanishes.

Then again, in Coates, Coutie & Kong, when describing the Euler strut they say that:

a) the ends are in perfectly smooth pinned supports

b) the strut is perfectly straight

c) it carries only a perfectly axial compressive load

d) the strut material is elastic and does not yield

Sounds to me like a perfect strut in a perfect world, (you did say 'in a perfect world'!).

I agree practical struts will always have imperfections and that we use small perturbations to calculate critical loads; but the Euler strut, a wholly theoretical concept, is perfect in every way and yet it will buckle but it won't squash (by that I mean shorten without buckling).

again, regards,

Bill

19-Tanzanite

I will throw in my 2 cents worth as a physicist. A perfect column (i.e. a perfectly uniform structure with at least orthotropic symmetry that is aligned with the axis of the column) subjected to only a purely axial load cannot buckle. Buckling could only occur if there is some sideways force (even if it is extremely small) that causes the column to preferentially deflect in one direction. A purely axial load, by definition, cannot provide such a force. As the column compresses some of the axial force will be deflected outwards (otherwise the column could not squash either!), but for a perfect column the outward force will have perfect radial symmetry, and so also cannot cause buckling. Even if the column undergoes catastrophic failure, that would also have perfect radial symmetry! Of course, in practice no such column exists, and even the slightest imperfection could result in bucking or failure with no radial symmetry (at this point we get into chaos theory!)

19-Tanzanite
(To:RichardJ)

Actually, a slight edit is needed. The outward forces will of course only have perfect radial symmetry if the column has perfect radial symmetry. I am fairly sure I am right that the column will not buckle even if it has orthotropic symmetry though, as long as the column has a physical symmetry that is aligned with the orthotropic symmetry.

5-Regular Member

This has turned into a very interesting discussion.

Are we all in agreement that the Lame's equation are the right way to go?

Mike

Lester,

I don't think what I said is contrary to what you quoted. At the buckling load, the column is not stable, any virtural displacement will result in collapse, which is why one way to find the critical load is by assuming a virtual dispacement (or load), or deflected shape, and see what happens as the displacement (or load)apporaches 0. I also agree (as usual) with Richard, after all there is a test named after him.

Mike,

If your client in not opposed to stiffners, wether internal donut shaped or transverse recantgular stiffners at intervals around the perimeter, then I suggest that you use Lema' and von Mises as a check on stresses, if they are high, then thicken the plate. This is a first check, not a sufficient one.

Then I suggest the following lower bound approach (Lower bound - implies that the desing is at least this strong and most likely stronger)

1) take a uint stirp of plate and load with the appropiate winding pressure. Determine plate thickness and distance between stiffners by assuming the strip spans as a continuous beam loaded with the pressure (I would just use qL^2 / 12. Then check the ability of the stiffeners to carry the reaction from the continuous span of the barrel plate, For the plate, let the bending stresses reach Fy, for the stiffners, limit to .66Fy (or something like that).

For rectangular stiffners, you can design the plate thickness and required welding as if a beam spanning between the ends.

For donut shaped, design teh donu as a thick shell and check the collapse equation and stresses using Lema`.

The width/thickness of the stiffners, limit to 0.2.45*(E/Fy)^1/2, which is the width/thickness limiting ratio to achieve plastic deformation that is applied to a beam web, which is a similar condition.

2) Check for the maximum load that might develpe in the cable by calculating the pressure, much the same as for the winding, but use the pressure calculated for only one wind and assume an effective width of plate equal to, say, the diameter of the cable pluse 2 times the plate tickness, with rest unloaded. Then Check the plate and stiffeners above for the resulting concentrated load on the effective width of plate spanning between the stiffeners.

In effect, I am suggesting that you design the barrel wall not as a barrel, but as strips of plate spanning along the barrel for donut shaped stiffners,

or as straight segments of plate spanning to stiffners which are spaced around the circumference. If such an apporach works, then then actual strength of the barrel is at least this strong and probably much stronger.

If you needed to minimize the material, then you could try to refine the above a little; But to truly minimize, you would probably need to do a finite element analysis. It will take a long time to do an appropriate analysis including second order effects and possibly even accounting for fabrication stesses.

Maby Lester could make some suggestions also.

Wayne

19-Tanzanite
(To:wayne)
 6) In the Richard Test (sorry Richard, but you invented the toilet roll test, so it's named after you),

Cool! I've never had a test named after me before

Sure, I would have preferred something with a more impressive sounding name than "the toilet roll test", but it's better than nothing! I'm going to update my resume and my LinkedIn profile immediately

5-Regular Member
 I don't think you need to worry about overall elastic instability. A drum will not buckle under a wound-on rope any more than a concrete column would buckle under a prestressing cable. The tension in the rope and the compression in the cylinder shell are mutually bound, the one to the other, for buckling to occur the load must be independent of the structure so that it still has effect after a putative buckle has occurred.

Very interesting, but I was under the impressed that an external pressure (Hoop stress) is generated by the tension in the wire, thus generating a compressive stress within the barrel.

You are correct when you said the critical collapse stress has been extracted from Roarks and the worksheet is attached.

Do you think this is the right route to go down, becuase I still feel Wayne has hit the nail on the head.

19-Tanzanite
(To:MikeArmstrong)

Unless I misunderstand something (which is quite possible!) it is not the wound cable that will cause the cylinder to buckle, but rqather the force applied to the cylinder by the cable that is being wound in. Right?

5-Regular Member
(To:RichardJ)
 Unless I misunderstand something (which is quite possible!) it is not the wound cable that will cause the cylinder to buckle, but rqather the force applied to the cylinder by the cable that is being wound in. Right?

Yes, as the wire is wound in and paid out a tensile force is generated in the wire, thus creating a compressive force in the barrel.

Mike

19-Tanzanite
(To:MikeArmstrong)
 Yes, as the wire is wound in and paid out a tensile force is generated in the wire, thus creating a compressive force in the barrel.

Right, but that compressive force will not cause the barrel to buckle. It will just squash it. It's the tensile force that will cause the buckling. Or does the compressive force somehow factor into this as well?

5-Regular Member
(To:RichardJ)
 Right, but that compressive force will not cause the barrel to buckle. It will just squash it. It's the tensile force that will cause the buckling. Or does the compressive force somehow factor into this as well?

I'm sure a radial load is generated due to the external pressure, which in-turn can cause the barrel to buckle. This is a very vague subject and I doubt this has been checked within my company before.

They have probably just added internal stiffening to the barrel to guarantee it won't fail.

Mike

19-Tanzanite
(To:MikeArmstrong)

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

5-Regular Member
(To:RichardJ)

That would work .

One problem = \$\$\$\$\$\$\$\$\$\$

Can't see my boss being pleased with a solution like that when we're planning on making several different sized barrels.

Mike

Mike,

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

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