I'm trying to derive eq. (f) in Timoshenko's Theory of Elastic Stability, 2nd Ed. Article 11.1, p. 458. Its for the increase in strain energy during symmetrical buckling in a cylindrical shell. I've got the 1st and 3rd terms, but for the 2nd term, (piA^2Ehl/2a), I have a Poison's ratio squared that doesn't belong. I have the correct expression for the work of compressive forces, eq. (h), so this is the last term I need. Is anyone familiar with this solution?
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buckling of cylindrical shell in compression 3
- Thread starter groveri
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- #3
The last homework I did was 28 years ago. Engineers actually do this kind of stuff. I planned on using this method fro a much more complicated problem. Anyway, I answered my own question so I don't need a response. I'll try to figure out how to end this topic.
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How big of a dent? Actually I'm attempting to predict the buckling of a thin metal liner (can) of a filament wound composite pressure vessel after its autofrettage cycle. This is a rare event, but has been known to occur with very small t/r ratios. During the pressurization, most significant "dents" will be straightened out due to the high internal pressure and tensile yielding of the metal. It is a difficult problem to work by FEA, because the pressure comes from contact with the external shell and the pressure is relieved on the inward buckle waves, and the mode shape is prevented from moving outward by contact with the stiff external shell.
I took the first step by following the Timoshenko method I referenced (after I finally found my mistake). With this energy method, any function that reasonably approximates the actual mode shape will produce a reasonable approximation of the critical load. For the symmetrical buckling of Timoshenko’s problem, he used -sin(m*pi*x/l). I used (1-sin(m*pi*x/l)), which is always greater than 0, for a shell that is constrained from outward deformation by external contact with a rigid structure. The predicted critical load was ~50% higher. (I know it doesn't satisfy the same end conditions, but it's a long cylinder and it won't matter much).
I’m not sure how I’d do this with FEA. Any suggestions (or examples) would be appreciated.
My next step is to extend this method to circumferential buckling with external pressure.
Any discussion or criticism of my approach is appreciated, I can take it.
I took the first step by following the Timoshenko method I referenced (after I finally found my mistake). With this energy method, any function that reasonably approximates the actual mode shape will produce a reasonable approximation of the critical load. For the symmetrical buckling of Timoshenko’s problem, he used -sin(m*pi*x/l). I used (1-sin(m*pi*x/l)), which is always greater than 0, for a shell that is constrained from outward deformation by external contact with a rigid structure. The predicted critical load was ~50% higher. (I know it doesn't satisfy the same end conditions, but it's a long cylinder and it won't matter much).
I’m not sure how I’d do this with FEA. Any suggestions (or examples) would be appreciated.
My next step is to extend this method to circumferential buckling with external pressure.
Any discussion or criticism of my approach is appreciated, I can take it.
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1
- #8
groveri,
Well, as you have found, this is where the classical methods run out. Always worth a try first.
I have used FE for tricky buckling situations a couple of times but not on your type of geometry. I did it by (sometimes) introducing an initial mesh distortion ("dent it first") and then applying a large load with a nonlinear analysis. I used automatic load incrementation and watched for the load fraction to which the load tended asymptotically. The Ricks method is often recommended in this case but I have never found it necessary because I'm not interested in what happens after it buckles.
The mesh distortions were generated by tweaking the node coordinates by projecting them onto proposed dent geometries. These geometries were either long-wavelength or local dents but always small, usually setting the peak of the dent to be the maximum deviation specified on the drawing. Some geometries are self-seeding but not in your case.
In your case you can do the same but you must include contact too. This will cause difficulties at the start of the analysis when the loads are low and the surface contact "chatters". You will probably have to use a few tricks to help it along at the start.
I have checked this method against flat plate formulas and found very good correlation for the simple cases, this would be a good place to start before diving into the contact case.
Regards,
gwolf.
Well, as you have found, this is where the classical methods run out. Always worth a try first.
I have used FE for tricky buckling situations a couple of times but not on your type of geometry. I did it by (sometimes) introducing an initial mesh distortion ("dent it first") and then applying a large load with a nonlinear analysis. I used automatic load incrementation and watched for the load fraction to which the load tended asymptotically. The Ricks method is often recommended in this case but I have never found it necessary because I'm not interested in what happens after it buckles.
The mesh distortions were generated by tweaking the node coordinates by projecting them onto proposed dent geometries. These geometries were either long-wavelength or local dents but always small, usually setting the peak of the dent to be the maximum deviation specified on the drawing. Some geometries are self-seeding but not in your case.
In your case you can do the same but you must include contact too. This will cause difficulties at the start of the analysis when the loads are low and the surface contact "chatters". You will probably have to use a few tricks to help it along at the start.
I have checked this method against flat plate formulas and found very good correlation for the simple cases, this would be a good place to start before diving into the contact case.
Regards,
gwolf.
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1
- #9
groveri,
a point you possibly didn't think to up to now, is that, in the axial case, a significant radial reaction should be generated at the point of contact of each outward wave with the casing. This would have an effect on the critical load (increasing it), as the vertical reaction generated by friction would have to travel with the deformation progression, and so would suck energy. Also, if the sum of the friction reactions would prove to be higher than the critical load (and likely would be so for a very long cylinder with many waves), then buckling of the whole cylinder would become impossible, but could remain possible with only the first part becoming unstable.
This way of reasoning wouldn't apply to circumferential buckling, as there is no relative sliding there.
However here too I think you'll have no luck.
In fact, taking, to simplify things, a long cylinder, the buckling may be calculated as for a circular arch radially loaded, taking a clamped arch with an opening angle equal to 2[π] divided by the number of waves (that will be determined to give the minimum critical load).
But if you look at the deformed shape calculated for the buckled arch, it always includes half of the arch going inwardly and the other half outwardly. This means that the only possible instability of a constrained pressurized circular tube is with pure membrane stress and no bending. This is something similar to the snap through instability that you have with two compressed struts at low inclination.
However, due to the curved shape of the arch, I guess that the instability, if ever possible, would be in all cases well beyound the plastic strength of tube material.
prex
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a point you possibly didn't think to up to now, is that, in the axial case, a significant radial reaction should be generated at the point of contact of each outward wave with the casing. This would have an effect on the critical load (increasing it), as the vertical reaction generated by friction would have to travel with the deformation progression, and so would suck energy. Also, if the sum of the friction reactions would prove to be higher than the critical load (and likely would be so for a very long cylinder with many waves), then buckling of the whole cylinder would become impossible, but could remain possible with only the first part becoming unstable.
This way of reasoning wouldn't apply to circumferential buckling, as there is no relative sliding there.
However here too I think you'll have no luck.
In fact, taking, to simplify things, a long cylinder, the buckling may be calculated as for a circular arch radially loaded, taking a clamped arch with an opening angle equal to 2[π] divided by the number of waves (that will be determined to give the minimum critical load).
But if you look at the deformed shape calculated for the buckled arch, it always includes half of the arch going inwardly and the other half outwardly. This means that the only possible instability of a constrained pressurized circular tube is with pure membrane stress and no bending. This is something similar to the snap through instability that you have with two compressed struts at low inclination.
However, due to the curved shape of the arch, I guess that the instability, if ever possible, would be in all cases well beyound the plastic strength of tube material.
prex
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Prex,
I did think about the radial reactions, but decided they don't affect the energy solution (see Timoshenko in my initial question) I used because they do no work and thus do not enter into the energy balance. As for the friction, remember that my cylinder is a pressure vessel liner, and the compression is developing because the liner has yielded on pressurization to the autofrettage pressure and as the external shell contracts during decompression, it compresses the now larger liner. As my FEA shows, there is very little relative movement between the two. Anyway, I'd probably go the conservative route and neglect friction because I've found it's very hard to convince some people that it exists.
Your idea about the arch is an approach I looked at yesterday. I skipped the 2 half wave configuration and looked at the shallow arch with a snap through buckle. For my configuration, there are low pressure mode shapes where the half wave is about 2 deg of the circumference. The chord height is very shallow. Of course the problem with this is that as soon as the arch starts to straighten, it loses contact with the external wall and the pressure goes away. Also, there is positive internal pressure until depressurization is complete. I did learn, though, that the snap through pressure is larger than the contact pressure between the internal and external shells. So, it was a valuable exercise.
The liner does exceed the compressive yield stress, but Timoshenko accounts for this with the use of a reduced modulus of elasticity. In my case, the post yield curve continues to rise and the reduced modulus is always sufficiently high.
Intuitively, buckling does not seem possible because, as you observed, outward buckle half waves are impossible and membrane shortening must occur to permit the inward half waves. I think this would cause a simultaneous increase in membrane and bending strain energy without any external work (I think). I would like to express this in quantitative terms. I think there is a lower energy state with inward buckles (high bending energy, low membrane energy). I will try to write the correct energy balance equations using various assumed inward only modes shape equations and show that the transition from the initial shape to the final shape can only be reached by passing through higher energy states when the inward mode amplitude is low. Like pushing a weight up a small hill before it falls off a cliff.
I think I will still have to follow gwolf2’s advice and revert to FEA in the end. I can even imagine a type of failure similar to the sliding of the continental plates over and under each other. Since yielding is occurring I can see one liner section sliding over the other on the 45 deg max shear plane. This would require a very detailed local nonlinear (material and geometric) analysis, probably setup to cause this behavior. Also, my assumption that the external shell is rigid is false. Althoug it is relatively stiff, it can flex locally and may need to be acounted for.
Thanks to prex and gwolf2 – good responses. This is all very involved. I wish I had a group of grad student slaves to do it. I think this would make a good thesis subject.
Also, thanks to ivymike! When I was in school, aluminum cans were pretty new and we weren’t that bright (about the dent). Got a lot of headaches! Before aluminum cans, I don’t think the dent did much good.
I did think about the radial reactions, but decided they don't affect the energy solution (see Timoshenko in my initial question) I used because they do no work and thus do not enter into the energy balance. As for the friction, remember that my cylinder is a pressure vessel liner, and the compression is developing because the liner has yielded on pressurization to the autofrettage pressure and as the external shell contracts during decompression, it compresses the now larger liner. As my FEA shows, there is very little relative movement between the two. Anyway, I'd probably go the conservative route and neglect friction because I've found it's very hard to convince some people that it exists.
Your idea about the arch is an approach I looked at yesterday. I skipped the 2 half wave configuration and looked at the shallow arch with a snap through buckle. For my configuration, there are low pressure mode shapes where the half wave is about 2 deg of the circumference. The chord height is very shallow. Of course the problem with this is that as soon as the arch starts to straighten, it loses contact with the external wall and the pressure goes away. Also, there is positive internal pressure until depressurization is complete. I did learn, though, that the snap through pressure is larger than the contact pressure between the internal and external shells. So, it was a valuable exercise.
The liner does exceed the compressive yield stress, but Timoshenko accounts for this with the use of a reduced modulus of elasticity. In my case, the post yield curve continues to rise and the reduced modulus is always sufficiently high.
Intuitively, buckling does not seem possible because, as you observed, outward buckle half waves are impossible and membrane shortening must occur to permit the inward half waves. I think this would cause a simultaneous increase in membrane and bending strain energy without any external work (I think). I would like to express this in quantitative terms. I think there is a lower energy state with inward buckles (high bending energy, low membrane energy). I will try to write the correct energy balance equations using various assumed inward only modes shape equations and show that the transition from the initial shape to the final shape can only be reached by passing through higher energy states when the inward mode amplitude is low. Like pushing a weight up a small hill before it falls off a cliff.
I think I will still have to follow gwolf2’s advice and revert to FEA in the end. I can even imagine a type of failure similar to the sliding of the continental plates over and under each other. Since yielding is occurring I can see one liner section sliding over the other on the 45 deg max shear plane. This would require a very detailed local nonlinear (material and geometric) analysis, probably setup to cause this behavior. Also, my assumption that the external shell is rigid is false. Althoug it is relatively stiff, it can flex locally and may need to be acounted for.
Thanks to prex and gwolf2 – good responses. This is all very involved. I wish I had a group of grad student slaves to do it. I think this would make a good thesis subject.
Also, thanks to ivymike! When I was in school, aluminum cans were pretty new and we weren’t that bright (about the dent). Got a lot of headaches! Before aluminum cans, I don’t think the dent did much good.
A few more remarks on your problem.
1)For the axial case Timoshenko uses an assumption that I think is no more valid in your conditions. The assumption is that the axial compressive force in the buckled shell is constant and equal to the value before buckling.
This condition may clearly be approximated only for a buckling deformation that has equal inward and outward waves, as the outward waves will have less compression and the inward ones more, but the average stays yhe same.
This may be no more valid for inward waves only, so I think that, if you wanted to go on with a theoretical approach, those equations would need to be made more general.
2)I'm not sure to fully understand your description of the interaction between the liner and the can, but guess you have something similar to a buckling due to differential expansion between the two: in this case I agree that friction is unrelevant.
3)Concerning the circumferential buckling, the calculation can be performed also theoretically, though this is quite cumbersome, but not really complex.
However here too it is necessary to use the equations for circular arches that fully include the contribution of membrane and shear deformations (Timoshenko's approach is with an inextensional bending deformation superimposed to a uniform membrane strain)
You can see a solution of this kind here at Xcalcs and obtain the numerical coefficients for the radial deformation corresponding to any opening angle of the arch (an analytical approach is also possible, as I said above).
Now clearly the buckling condition would correspond to a radial deflection in the middle equal to the rise of the arch.
prex
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1)For the axial case Timoshenko uses an assumption that I think is no more valid in your conditions. The assumption is that the axial compressive force in the buckled shell is constant and equal to the value before buckling.
This condition may clearly be approximated only for a buckling deformation that has equal inward and outward waves, as the outward waves will have less compression and the inward ones more, but the average stays yhe same.
This may be no more valid for inward waves only, so I think that, if you wanted to go on with a theoretical approach, those equations would need to be made more general.
2)I'm not sure to fully understand your description of the interaction between the liner and the can, but guess you have something similar to a buckling due to differential expansion between the two: in this case I agree that friction is unrelevant.
3)Concerning the circumferential buckling, the calculation can be performed also theoretically, though this is quite cumbersome, but not really complex.
However here too it is necessary to use the equations for circular arches that fully include the contribution of membrane and shear deformations (Timoshenko's approach is with an inextensional bending deformation superimposed to a uniform membrane strain)
You can see a solution of this kind here at Xcalcs and obtain the numerical coefficients for the radial deformation corresponding to any opening angle of the arch (an analytical approach is also possible, as I said above).
Now clearly the buckling condition would correspond to a radial deflection in the middle equal to the rise of the arch.
prex
: Online engineering calculations
: Magnetic brakes and launchers for fun rides
: Air bearing pads
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- #12
Haven't had time to think about prex's remarks, I've been on extthended travel. I think I might disagree with remark one. The energy state before and after buckling is for small deflections and the amplitude of the bucklle waves can be arbitrarily small. The axial compression is just the vector component tangent to the buckle wave and the tangent angle is arbitrarily small. I don't see that it makes any difference whether the waves are in and out or just in.
Will have to think about it some more and items 2 and 3 also.
Will have to think about it some more and items 2 and 3 also.
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Hansmeister
Aerospace
When I did some consulting at Structural Composite Industries, we were working on very high pressure vessels having a very thin liner. The liner would inward buckle after autofrettage. You need a very good adhesive to the composite, and there will be a minimum thickness that will work.
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- #14
Hansmeister - Would it be possible to get some details on the SCI liner buckling - vessel diameter & length, liner material & thickness, composite material and thickness, autofrettage pressure, description of buckle (picture if possible), etc. I also have done work for SCI. Can you give me a contact name?
Hansmeister
Aerospace
Dear Groveri,
The work being done back then was for the original Star Wars programs, which was about 20 years ago. The principals doing the work are all gone, one way or the other.
The buckling solution you should be considering is the cylindrical shell with a radial load, i.e., hydrostatic pressure.
Try Batdorf, NACA TN 1341.
The work being done back then was for the original Star Wars programs, which was about 20 years ago. The principals doing the work are all gone, one way or the other.
The buckling solution you should be considering is the cylindrical shell with a radial load, i.e., hydrostatic pressure.
Try Batdorf, NACA TN 1341.
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