If I have a pipe with external radius R and internal radius r. Maximum compressive force will be given by the same formula Fmax=σS as for cylinder or different? where $S = πR^{2}-πr^{2}$, and σ is compressive strength of material. In other words, if I have some shape with the same mass per meter length as cylinder, maximum compressive force will be the same or different? For example, if I have solid cylinder from steel with 10 cm radius, area will be 314 square cm, and if I have pipe from the same steel with external radius 20 cm and internal radius 17.32 cm, area and mass per meter length will be the same as cylinder 10 cm radius. What can handle more compressive force pipe or cylinder or it will be the same? If it is different, maybe someone can give me useful links and formulas?

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    $\begingroup$ It matters how tall the column is. For a very short column, it's pretty much just governed by area. For most columns of useful height, buckling will be the consideration and you hollow pipe will do better than the cylinder of equal area. $\endgroup$
    – Ethan48
    Aug 23, 2016 at 20:51
  • $\begingroup$ Thanks for your comment. But I need numbers or formula? How many time hollow pipe will be better than cylinder? May be there is well know formula for this? Or you mean it will be better from point of view of stability from wind and etc, but pure compressive strength will be the same? $\endgroup$
    – Zlelik
    Aug 25, 2016 at 6:29

1 Answer 1


In my opinion the comment of Ethan48 should give you enough information to answer your question. Hereby a bit more elaboration.

With a column under axial compression the stress is uniformly distributed*. Thus the stress is indeed given by $\sigma = F/S $

When designing a column under axial compression, you do not only check whether the stress is not too high, but also check if the column will not buckle. The formula for euler buckling is $F_{buckling} = {(\pi^2* E*I)} / {(K*L)^2}$ Have a look at the Wikipedia page for details on this formula and how to use it.

The force at which a column buckles depends on the area moment of inertia $I$. For a circular hollow section with the same area as a circular solid section, $I$ is larger. Thus, with all else being equal, the solid column will buckle under a smaller load.

Have a look at the Wikipedia page on beam theory for more background and details.

*Depending on how the column is loaded, it takes some distance for the force to distribute itself uniformly. So the maximum stress could well be higher than the standard formula gives. Never forget that theory is a simplification of reality.

  • $\begingroup$ Thanks for the answer. Buckling is interesting, but for now I am interesting only about pure compressive stress, like ideal conditions, something like indoor column, no wind, no earthquake and etc. I still do not understand, if max compressive force will be the same for pipe and cylinder of the same area. Especially I am interesting about very tall tower. $\endgroup$
    – Zlelik
    Aug 26, 2016 at 9:45
  • $\begingroup$ You are aware that buckling is an effect that can occur under pure compressive stress and that it needs no wind, earthquake etc.? And that buckling is often the defining load case for tall and slender columns? $\endgroup$
    – jos
    Aug 26, 2016 at 10:13
  • $\begingroup$ I was not aware about buckling. Thanks. But anyway, if I consider ideal case and only compressive strength, no buckling or something else. If steel material as compressive strength 300MPa, it will be the same for any shape, like cylinder or pipe? $\endgroup$
    – Zlelik
    Aug 26, 2016 at 13:28
  • $\begingroup$ Having said what I have said, I leave that to you to find out. $\endgroup$
    – jos
    Aug 26, 2016 at 16:09
  • $\begingroup$ Thanks a lot for many valuable comments. Especially about buckling. I read wikipedia link and now it is more clear. Looks like pure compression force will be the same for solid cylinder, than for pipe of the same area, but because of buckling, pipe will work better. Anyway looks like I need to simulate it to understand better in some professional software like SolidWorks, because there is no simple answer. $\endgroup$
    – Zlelik
    Aug 27, 2016 at 21:25

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