Buckling of a Beam of Circular Cross Section

Hi AP1,

Are you talking about compressive force as shown here? (although this is not a "to-scale" drawing, I tried to make the circular cross-section dia. appear the same length as the square cross-section diagonal).

If so, there are others that can address this better than I can (and I'm sure they will be chiming in).

If not please explain further.

Well yes,

it is exactly the way you've perceived it.

And I think I may have found my answer in the formula for the Critical Buckling load. But I hope someone can throw some more light on this.

If you are in the elastic buckling the section inertia gives the load limit. If you are in the elasto-plastic loading range then the section area gives the limit.

In the 1st case the circular section has a higher J thus accepts higher loads for same slenderness

In this, you don't say if the wall thickness or the total area is the same.

I hate to disagree with Nickname (he's usually right) but...

In the elastic range, the strength is inversely proportional to the slenderness ratio which is kL/r where k is a factor governed by the end fixity, L is the length and r is the radius of gyration. With k and L being the same, r is the determinant. The square has a greater r so the square is stronger.

Huh? The cross-section of the specified cylinder lies entirely equal or outside the cross-section of the square tube.

The equation for the critical LOAD in case of elastic buckling is :

P*= π^2*E*J/(k*L)^2 where E-modulus of Young;J-Inertia of the section considered as constant; k - form factor depending on the boundary conditions at beam end and L- beam length.

For a square section a x a there are a minimal Jmin = a^4/12 and with respect to a reference system at 45° Jmax= 5/24*a^4.

For the circular section J= π/64* d^4 if d= a*√2 (as mentioned in the text) J= π/16*a^4

For same conditions the only variable parameter is J we have now to compare

the MINIMAL J for the square section with the one for the circular:(π/16)/(1/12) >1

Thus the circular section will carry a higher load.

The error was made by neglecting the fact that d=a*√2 = diagonal of square!

If d=a then of course the square has a higher critical load but this was NOT considered in this case.

By the way thanks for the compliment !

(Do I deserve a GA from you too ?)

If you're talking about a column where an applied axial load is compressive, then Passington is absolutely correct in his statements and I couldn't agree more with him!

BTW, in your case it's not a beam, but rather a column, although a beam can undergo both bending and an axial loading...

If it is actually a beam, it will at least have it's own weight distributed along it's entire length, and therefore, be bending (ie: in flexure),if ever so slightly, in addition to the axial load you mentioned.

Beams buckle where stress is concentrated enough to exceed the beams' material yield strength.

maximum stress would be along the extreme edge of length of the beam, which would be at a greater distance from the neutral axis of the circular cross-section than the square cross section.

So, assuming both round and square beams are of exactly equally strong material, then, yes, the circular cross-section beam would fail at a lower stress than would the square cross-section beam, under identical loading conditions.

In buckling it is demonstrated that transverse loads do NOT affect the critical axial load.

It is not a problem if it is a column (vertical beam) or an horizontal beam the stability depends mainly of the axial force applied upon.

nick name: Your comments in post 9 are incorrect. Transverse loading can significantly reduce the buckling strength of a member.

passingtongreen: There is no wall thickness, as stated in post 1. The relative area is given in post 1. Your answer in post 4 is incorrect. The member of circular cross section has a higher buckling strength.

CaptMoosie: You said in post 5 that the statements by passingtongreen in post 4 are "absolutely correct." This is incorrect. The member of circular cross section has a higher buckling strength, not the square member.

MR. Guest: The statement in your fourth paragraph in post 8 is incorrect. The member of circular cross section would fail at a higher axial stress (1.50 times higher) than would the square member (if this is a long column), just the opposite of what you said. Nonetheless, the correct answer to the given question is, the member of circular cross section has a higher buckling strength (meaning it will carry a higher axial compressive load) than the square member. Therefore, the square member will buckle before the circular member does.

I suggest you review your manuals and other SERIOUS sources on the buckling problem.

After you done it you may come back and recognize you wrong comment.

I expect you to do it. By the way the elastic buckling (called also Euler buckling) is considered in this case.

If the transverse load is sooo big that it generates a plastic "hinge" then of course with such a big deformed beam the stability is compromised.

I notice that you only mention "incorrect" answers why do you not mention that in my comment I already demonstrated that the circular section as defined by the OP will carry a higher load ? If one considers the circular section with d= square side a then the load is higher for the square section since its J is bigger. Radius of giration √J/A is only valid for the slenderness computation, the critical load in the elastic buckling domain is as I wrote above proportional to J/leq^2 and the square section has a bigger J since it has more stuff far from the center line.

So all you write is not 100% correct or am I wrong ?

nick name: My apologies for not mentioning that your seventh paragraph in post 7 is a correct answer. However, it is immediately preceded by incorrect information in your fifth and sixth paragraphs, where you again said the buckling strength is determined only by the moment of inertia, even after passingtongreen explained that that is incorrect. Radius of gyration, r, is the determinant parameter here. For the given question, the cross section having a greater minimum radius of gyration has higher buckling strength. Nonetheless, it is easy for any of us to make mistakes when trying to post to a forum that does not allow sufficient time to correct typos and mistakes.

I am not stubborn and accept correct correction, I gave several times the proof for it, but I have to insist and ask you to have a second look at the different documents available about buckling in the elastic domain.

The gyration radius (sorry for the orthographic error ) is important to determine if the beam/column is or not in the elastic domain. It is the factor which shows the validity limit of the equation. If the ratio l/rg is too small then the load is not any more to be computed by the Euler formula. Which as I wrote above is function as far as the limits are respected of the ratio J/l^2.

As for the transverse loads may I make a remark, buckling is the loss of stability of a beam/column which is "straight". A transverse load will have an effect only if its bending moment leads to high deformation values and those increase the bending from the axial load this is not any more buckling but combined loading which can have as effect the generation of a "plastic hinge" at the highest bending moment place.

In the plastic range which is valid for short beam/column the stress level IS function of slenderness but NOT in the elastic range.

Please check and if you find a text where this is infirmed then please give me the coordinates and I shall correct my point of view.