Buckling of rings and arches

In 2007 I wrote a paper on buckling of confined steel tunnel liners for the ANZ Geomechanics Conference in Brisbane, and in the course of that I noticed that the calculated buckling load of an unconstrained pipe subject to uniform radial load was significantly different for different methods of analysis.  The most widely quoted theoretical solution (for instance, Roark’s Formulas for Stress and Strain, Edition 7, Table 15.1.8) gives a buckling load (radial load per unit length) of 3EI/R³, and this was supported by an incremental non-linear analysis, including geometric non-linear effects.  On the other hand a linear buckling analysis, based on an eigenvalue analysis, gave a buckling load very close to 4EI/R³.  I did not find the reason for this difference at the time, but I was reminded of the problem by a recent thread at the Eng-Tips forum, and decided to investigate it further using the buckling analysis spreadsheet presented here earlier this year.

I have modified the spreadsheet to allow the input of a quarter circular arch (which with appropriate end conditions is equivalent to symmetrical effects in a full ring), or for a smaller portion of an arc.  The spreadsheet also allows different options for the change in direction of the applied load after deformation of the ring (for reasons that will become apparent), and the ring may be either specified with a constant cross section or a series of stepped sections.  The modified spreadsheet, including full open source code, is available for free download.

For the purposes of this investigation the end conditions were set as follows:

  • Base: restrained against rotation and vertical translation
  • Top: restrained against rotation and horizontal translation

These end conditions make the quarter circle equivalent to a full circle for actions that are symmetrical about the X and Y axes.

The options for the application of the load after displacement of the point of application are:

  1. The load remains normal to the point of application, appropriate for loading by a fluid pressure.
  2. The load direction remains unchanged.
  3. The load remains directed towards the original centre point.

Option 1 may be modelled with either uniform loads applied to each element of the ring, or with point loads applied to the mid point of each element.  Options 2 and 3 require the use of point loads.

The model may be subdivided into any number of straight elements, but there is little increase in accuracy with more than about 40 elements, and the analysis becomes very slow if 100 or more elements are used.

The procedure for the analysis is:

  • Calculate node coordinates for the specified radius and number of elements
  • Apply linearly increasing deflections to each node.
  • Apply radial unit load/unit length, either to each element as a distributed load, or to the mid-point of each element.
  • Analyse the structure for this loading, to find the X and Y deflections at each node.
  • Scale the deflections so that the maximum resultant deflection is 1 unit, and add these normalised deflections to the original circular arc node positions.
  • If the loads are being applied as node loads, adjust the load direction according to the chosen option (distributed beam loads are automatically applied as perpendicular to the member direction).
  • Repeat until the change in maximum resultant deflection is less than the specified tolerance.
  • The buckling load is then 1/maximum resultant deflection

Input and results for the example given in the Eng-tips discussion are shown in the screen shot below.  This is a steel ring of 25 mm square cross section and 1000 mm radius to the centroidal axis.  The output shown is for the case where the applied load remains radial.

Ring Buckling Input and Output

The spreadsheet gives the following results for the three load direction cases (k = PR³/EI):

  1. Load remains perpendicular to elements; k = 3.01
  2. Load direction remains unchanged; k = 4.01
  3. Load direction remains radial; k = 4.51

These values compare well with the theoretical values quoted in Reference (1), which gives:

  1. k= 3.0
  2. k = 4.0, most common value (but see below)
  3. k = 4.5

It should be emphasised that these values are for symmetrical buckling shapes.  If the ring is free to deform asymmetrically the buckling load may be greatly reduced, and Reference (1) quotes a k value of only 0.701 for a ring subject to constant direction loading and free to buckle in an asymmetric manner.

References: (1) “Critical Constant-“Directional Pressure on Circular Rings and Hingeless Arches”; Robert Schmidt; Journal of Applied Mathematics and Physics (ZAMP), Vol. 31, 1980

This entry was posted in Arch structures, Beam Bending, Excel, Newton, VBA and tagged , , , . Bookmark the permalink.

15 Responses to Buckling of rings and arches

  1. dik coates says:

    You may want to check with the University of Sheffield… They have done a lot of research on historic masonry rings and arches.

    Dik

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  2. Rod Pinna says:

    Hi,

    There is a good discussion of this in “Buckling of Bars, Plates and Shells”, Brush and Almroth 1975. Unfortunately out of print. They take a variational approach to it, but end up with the same results as above.

    The approximate solution that gives you a coefficient of 4 matches the simplifications Donnell made in 1933 to the equations for a cylindrical shell so that he could arrive at a solution. As you note above, it corresponds to treating the load as a “dead load”.

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  3. dougaj4 says:

    Rod – I’ll see if I can find the Brush and Almroth (when I have time!)

    The Donnell approach was mentioned in the Eng-tips discussion, but is it a “dead-load”? My understanding was a radial load that stays at a constant angle at each load point.

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  4. Rod Pinna says:

    Hi Doug,
    Great blog by the way.

    Essentially yes. The Donnell approximation ignores the tangential displacement component of the ring – so loading remains at a constant angle. If the ring buckles into more than an n=2 mode (such as a ring on an elastic foundation) then the Donnell and fluid pressure load solutions become close.

    Cheers

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  5. John Clark says:

    Tried to download your files but never got to anything that would open in Excel 2007. Is there a trick?

    I am looking for recent solutions for buckling of a ring (shell) on elastic foundation eg an FRP cylinder buried vertically w/ hydostatic pressure and possibly active submerged soil pressure. AWWA M 45 1st and 2nd ed’s give solutions for horizontal pipe, not vertical pipe.

    One solution in Brush and Almroth pg 133. Ref to P Seide (4.5) – The Stability under Axial Compression and Lateral Pressure of Circular Cylindrical Shells w/ a soft elastic core, J Aerospace Sci, Vol 29 pp 851-862, July 1962. I am ordering this paper. Hope it is helpful. Looking for newer publications on this problem

    Thanks

    John

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  6. dougaj4 says:

    John – thanks for letting me know about the problem. I have just downloaded the file myself and I find that although it is saved on the web as an xlsb file (Excel 2007 binary format) it downloads with the extension .zip. Seems like a case of Microsoft being too clever for its own good again. If you re-name the file to an xlsb extension it should open in Excel 2007 or 2010. I will put the file into a “proper” zip file, and that should avoid the problem.

    Regarding the question of buckling of a vertical buried pipe; the solutions for a horizontal pipe assume uniform pressure around the pipe, and neglect the self weight of the pipe, so wouldn’t they also be applicable to a vertical pipe?

    I did a paper on buckling of buried pipes a few years ago. It doesn’t look at the vertical vs horizontal issue, but you might find it interesting:

    Click to access Buckling07.pdf

    http://interactiveds.com.au/Publications/Buckling07.ppt

    Doug

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  7. dougaj4 says:

    John – just re-read your comment, and yes if the axial load is significant then that would certainly interact with the lateral pressure. Interesting problem, but I haven’t seen it examined anywhere.

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  8. asma says:

    hi I need the document of buckling of buried flexible pipe help me please
    thank you

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  9. katy says:

    please help me i need any document about buckling of buried flexible culverts

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    • dougaj4 says:

      The link for the article refrred to in the post is given in the previous comment. That has references, some of which are available on the internet, and others will be available from university or technical libraries.

      Like

      • katy says:

        Elastic buckling of buried flexible tubes ― A review of theory and experiment any one have this article
        if anyone have this article please he gave me

        Like

  10. Pingback: Daily Download 1: Buckling of columns, rings and arches | Newton Excel Bach, not (just) an Excel Blog

  11. Jeff Donville says:

    Doug,
    I’m looking at this problem to design circular walers.
    I downloaded your spreadsheet, and find that if I change the just the length unit from meters to mm (that is, maintaining a consistent set of input units and keeping the equivalent of the original input values), that the calculated k value changes (0.19?) and the buckling load is not scaled by the expected value i.e, a simple factor of 1000.

    So the inputs become A = 625mm2, I = ~32552mm4, E = 1.93×10^2 kN/mm2, R = 1000mm.

    Since it has been some time since I’ve worked in SI, would you be able to 1) verify that I’ve converted the input values correctly and 2) verify that the results change? If true, it would seem that the consistency of length units is not maintained in the analysis someplace.
    Jeff

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    • dougaj4 says:

      Jeff – As you say, the buckling factor should stay constant, and the buckling force/unit length should be in proportion to the length unit.

      The problem is that the method used to find the buckling load is based on deformations due to flexure only, but the code used includes axial deformation, which sometimes corrupts the analysis. If you convert I, E and R to mm and factor up the true area in mm2 by 1000 it will find a K of 3.0. Factoring up the area by too much can also cause numerical problems, so any results from the spreadsheet do need to be cross-checked before being used in practice.

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