buckling of horizontally mounted column 1

[nickatahi]

I'm driving myself crazy with a seemingly easy problem. I've got a hydraulic cylinder mounted horizontally with no support in the middle (see attached picture) and I'm trying to determine what effect the weight of the cylinder has on the critical buckling load in the gravitational plane.

Ignore the fact that this is a hydraulic cylinder. In general how do I approach this? I'm inclined to find the axial load that would create the same mid-length deflection and subtract that load from the critical buckling load to arrive at the 'adjusted' critical buckling load. I'm just not sure if that is valid to convert the lateral load to an axial one.

 

[rb1957]

look up "beam column analysis" ...

 

[izax1]

I would say it has a tremendous effect. The general Euler column buckling load formulas is assuming a prefect shaped column. Of course that does not exist in real life, and the weight of the cylinder will induce a deflection. Remeber that the buckling load is higly non-linear, and the adjustment you are talking about is a linear substitution. Calculating the critical buckling load is not a trivial matter, and requires a large Factor of Safety, unless you have measured the cylinder geometry and mid-span deflection to a very high accuracy.

 

[rb1957]

depends on P/Pcr ... if small, no big deal; if close to 1, very significant

 

[dvd]

... don't forget to have it built in exactly the same configuration as it was analyzed.

What kind of bearings are supporting the moving load? How perfect is the alignment and rigidity of the bearing rails? I say that the cylinder weight is inconsequential compared with eccentricities and induced moments.

 

[Ron]

You can't neglect that this is a hydraulic cylinder, because you have essentially a restricted hinge upon full extension. While the hinge is restricted in vertical and lateral movement, it nevertheless causes an eccentricity and a vertical/lateral load component to affect the Euler buckling condition.

 

[nickatahi]

Thank you for all the responses! I've never been an online forum user but that might change now that I know how helpful it can be.

Just to add some details to the situation...I work for a cylinder manufacturer so the cylinder under analysis is exactly the one that will be made in the shop downstairs. We have methods to find cylinder buckling loads but only pinned-pinned, vertically-mounted cylinders. I'm developing a program that will find buckling loads for other end conditions and other mounting angles in which gravity becomes important.

It turns out the problem is not as hard as I initially thought… I can just use a variation of the secant column formula adapted to hydraulic cylinders. At first I thought that the moment created by gravitational force would grow non-linearly as deflection increases but now I realize that it remains essentially constant.

It would probably be good to use this variation of the secant column formula instead of our old Euler-derived formulas for other cylinders too. Then we could account for the eccentricity generated by some of the real-world inconsistencies that some of you mentioned. And, as always, a very large safety factor will be used.

 

[bradleyelwood]

nickatahi,
First, you should be aware that the midpoint lateral stiffness of the cylinder will be different in the two directions. Lets say that the cylinder pushes in the z direction and the mounting pin is the x direction, the buckling capacity in the y direction will be a smaller value.

Generally, hydraulic cylinder buckling loads are much greater than the weight of the cylider, and so you dont have to be concerned about the transverse weight.

There is a rather clever way to determine buckling values of non-uniform members...determine lateral stiffness of the member AT THE MIDPOINT, then multiply by length between supports, then divide by 5. Be sure to use consistant units. Lateral stiffness can be found by manual analysis, or computer analysis. Or you can go out in the shop and load the cylinder with a known load at midlength between supports, then measure the deflection with a dial indicator. Be sure to keep the lateral loads below the bending strength of the cylinder. This method works for simple cylinders, telescopic cylinders, cylindrical columns with variable diameter, wood poles, etc. One of the benefits of "going out in the shop" is that you determine E (young's modulus), I (moments of inertia) and things like rod-to-barrell- overlap all at one time.

 

[corus]

I wouldn't treat it as a column as you'll probably get localised buckling of the cylinder wall and not as a whole structural member.

Tata but not yet tara

 

[Ron]

corus...I was trying to say the same thing...you said it better!!

 

[Japo]

The qualitative effect of bending composed with buckling is of course that any compression load will cause additional bending even before the buckling limit is reached, because of the torque caused by the load on the previous deformation. But if you plot buckling displacement versus compression load, the curve will still pass through 0 displacement due to buckling at 0 compression load, and will still tend to infinite displacement as the load approaches the buckling limit. Only it won't go abruptly as in the ideal case, but smoothly. As will always be the case by the way, since in the real world there's always some degree if imperfections: previous displacement, eccentricity or misalignment of the compression load.

As others said, how sooner you'll have unacceptable buckling deformation (even if it doesn't reach instability) depends on the ration between the bending load that's composed with the compression, and the buckling limit for the latter.

This is an easy math problem, only a small modification on Euler's original simple buckling, and fortunately I have the solution right here in a book. This is for bi-articulated beams, for other cases substitute the buckling length.

If you're applying a centered compression force P on a bi-articulated beam, depending on transversal displacement y, you'll have a bending torque

M = P y

and so you have the equation

d²y/dx² = (-1/EI) P y

where x is the axial coordinate, 'E' is the Young's modulus, and 'I' is the area moment of inertia of the cross-section. And the rest is history: you have the solutions y=0 and P=n²·3.1416²·EI/L² where n=1,2,3... (only 1 is interesting), L is the length of the bi-articulated beam (considering small displacements and neglecting second order infinitesimals).

The key here is that if you have previous transversal displacement (bending deformation) you only have to add it to the equation, like this:

d²y/dx² = (-1/EI) P (y+y*)

The book I have relies on y* being sinusoidal on x to provide an analytical solution, since that's a good approximation of deformation caused by real-world manufacturing defects, but helpfully observes that any function can be expressed as a Fourier series addition of sinusoidal terms.

 

[hacksaw]

Hydraulic cylinders have a number of common modes of failure, buckling (as a nonlinear effect) as a beam is not one of them. Given the design of the piston with multiple seals, and the hydraulic seal/bushing at the end, as the piston is run out,it can be modeled as a pinned joint with a spring. It can be unstable at some load conditions, and visually look like buckling may have occurred. Traditional buckling would occur at much higher loads.

 

[nickatahi]

Thanks for the extra considerations guys...

Yeah Japo it looks like you basically derived the secant column formula and that's exactly how I approached half of my problem.

There are a mind-boggling amount of failure modes in a long, horizontally-mounted, unsupported hydraulic cylinder such as: failure of mounting welds, mount-breakage, rod "buckling" (more correctly termed "bending-failure"), bending failure in barrel caused by moment between head gland and piston, compressive failure in head gland/piston causing fluid bypass/leakage, etc... On top of this is the fact "buckling" calculations become a whole different animal because half of the "column" (the barrel) is not experiencing a compressive load while the other half (the rod) is experiencing a compressive load. So, in a nutshell, my approach is to use the overall mount-to-mount length to find steady-state gravity induced deflection, then use my variation of secant formula to find the axial load at which compressive failure will occur in the rod (taking into account the moment created by the gravitational force at the cylinder COG as well as any frictional/misalignment effects I can quantify)and then kind of over-design the other components so that the rod would be the first to fail. In special cases I could carry out a full analysis which would likely take some serious time.

I'm sure as I get more into this I'll refine a better approach but this is purely analytical at this point anyway.

And I'm still open to any suggestions or criticisms to my approach! It has all been much appreciated so far!

 

[Canadieng]

I haven't been closely following this thread, but maybe this will help.

Have a look into:

ISO/TS 13725:2001 Hydraulic fluid power -- Cylinders -- Method for determining the buckling load

http://www.iso.org/iso/iso_catalogu...s.htm?ics1=23&ics2=100&ics3=20&csnumber=33432

(it contains some pretty intense analysis)

 

[bradleyelwood]

Nicatahi,
You seemed to have a good knowledge of cylinder failure modes until your comment that "half of the column is not experiencing compressive load". Don't think of it that way. I once had a guy tell me that a hollow telescopic cylinder was not subject to buckling because the oil carried all the compressive load, strange thinking.

Long cylinders do indeed "buckle". I have firsthand knowledge that they do so. Think of a cylinder the same way you would think of a column made of two concentric materials. In your case, one material is probably steel having a Youngs modulus and the seccond material is oil which makes no contribution whatsoever to bending stiffness. Whether the coulumn goes into Euler buckling or secant buckling depends on its slenderness ratio.

Next comment, be sure to look at buckling in both "x" and "y" directions as I explained before. Then apply the appropriate factor of safety. Using my "rule of thumb method" above to four stage, 380 inch stroke telescopic cylinders, the calculated value (k)(L)/5 agreed quite closely to the buckling determined from Cosmos. However, when we went into the shop and actually measured "k", we found that "k" was considerably lower than the Cosmos or calculated value, probably due to loacalized effects of packing, minimal overlap, etc. We did the shop lateral test with dead weights and dial indicators. If you don't have dial indicators, use dial calipers, both tools are way cheaper than a good reference book.

Incidently, the reason we did all the investigation is because we had a machine with buckled cylinders.

 

[nickatahi]

I see what you're saying bradleyelwood and i definitely agree that the statement "hollow telescopic cylinders cannot buckle" is strange thinking. I know it would also be ideal to get the lateral stiffness measurements of the actual cylinder as you have done. Maybe in the future I'll be able to take lateral stiffness measurements of various cylinder configurations and use them to improve my analysis. But, at this point, my analysis is purely theoretical and I'd like to make it general enough to apply to a large range of cylinder configurations with different barrel/rod/stage sizes, overlaps, and seal configurations.

Just to clear up my comment that "half the column is not in compression"... i was referring to how my buckling analysis MAY need to be altered to account for this. But I haven't looked deeply into the matter yet. What I know is that the barrel of a hydraulic cylinder (and all stages except for the plunger in a telescopic cylinder) are not experiencing the axial load that is involved in standard buckling equations. So the bending moment that non-linearly increases causing catastrophic buckling (P*y----where P=axial load and y=midspan deflection) is missing its "P" when there is no axial load. But the rod is experiencing a P*y bending moment (I'm just still trying to determine what 'y' is) and this moment is directly transferred to the barrel by the reaction forces at the head gland and piston. So although the barrel doesn't directly see the non-linearly increasing bending moment I guess it's getting transferred from the rod to the barrel anyway.

Maybe in the end I will find out that I can just treat hydraulic cylinders as bi-articulated beams in compression or pinned-pinned beams with a spring or whatever but I have to prove that to myself! I'm definitely going to get out the dial indicator too and start getting some 'real world' data. Thanks for that tip!!

 

[Canadieng]

I suggest not brushing off the ISO 13725 Tech spec. The analysis has been done already:

This Technical Specification specifies a method for the determination of buckling load which

- takes into account the complete geometry of the fluid power cylinder, meaning it does not treat the fluid power cylinders as an equivalent bar,
- can be extended to be used for all types of cylinder mounting and rod end connection,
- includes a factor of safety, k, to be determined by the person performing the calculations and reported with the results of the calculations,
- takes into account a possible off-axis loading,
- takes into account the weight of the fluid power cylinder, meaning it does not neglect all transverse loads applied on the fluid power cylinder,
- can take into account a misalignment, but only if it is situated in the plane of cylinder selfweight, and is easy to transcribe under the form of a simple computer program.