Axial Collapse of thin walled cylinders 7

[klibnog]

Can anyone give me a pointer as to how to calculate the theorectical axial load needed to begin the collapse of a thin walled (~0.1-0.2mm) cylinder produced from steel or aluminium? I want to predict the effect of cylinder height and diameter on the load needed. Many thanks in advance for any help.

 

[CoryPad]

General buckling with free end rotation occurs when the applied force F exceeds the critical buckling force F_b:

F_b = ([π]² E I)/l²

and

I = ([π]/4)(r_o⁴ - r_i⁴)

where

[π] = 3.141 592 654
E = elastic modulus
I = second area moment
r_o = tube outer radius
r_i = tube inner radius
l = tube height

Local buckling occurs when the axial stress [σ] exceeds a critical value [σ]_c:

[σ] = (0.6 [α] E t )/r

where

[α] = knockdown factor = 0.5
t = tube wall thickness
r = tube outer radius

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[JoeTank]

The thickness range you mention will be very susceptble to minor geometric imperfections due to fabrication. I suggest that you refer to NASA Space Bulletin SP-8007 for the best info that may be aavailable to you free on the web.

Steve Braune
Tank Industry Consultants

http://www.tankindustry.com

 

[Cockroach]

Actually you can use the classical equations from Thick Wall Pressure Vessel and apply the tri-axial stress analysis using Von Mises or Hencky, i.e. twice the stess squared equals the gradient stress vector. You would need to add the wall loading effect to the LONGITUDINAL STRESS since it acts parallel to that axis.

So assuming D=outer diameter, d=inner diameter of the pressure vessel and noting longitudinal stress is simply the reaction of the end caps of the vessel:

HOOP STRESS: Sh = P [(D^2 + d^2) / (D^2 - d^2)]
RADIAL STRESS: Sr = -P
LONGITUDINAL STRESS: Sl = P [d^2 / (D^2 - d^2)]

The internal pressure P acts to stretch the vessel, call this direction in the positive. Since you specifically mention COLLAPSE, your load acts in the opposite direction on the cross sectional area of the wall. Therefore:

LONGITUDINAL STRESS: Sl' = Sl - F / A

Given wall area is simply A = (Pi/4)(D^2 - d^2), you can easily determine the stress vector interms of principle axis or S = Sh i + Sr j + Sl' k, the vector basis being <i,j,k>.

Recalling vector algebra, the Von Mises-Hencky Equation is simply: 2 S^2 = S X S so use the cyclic permutations to resolve the gradient, i.e. S X S cross product. This is considerable mathematics, but without the influence of your F/A factor, that is the original hoop, radial and longitudinal stress terms, you would get:

S = sqrt(3) P [ R^2 / (R^2 - 1)] given R=D/d

This is the classical result found in most advanced books in solid mechanics. Obviously the equation will not clean up as nicely for your F/A term, but you do have the closed form solution set.

I've used this method extensively in the last twenty odd years. It works absolutely without question, most of my strain gauge results have been well within scientific error. Note that for Thin Walled Pressure Vessels, your's may be this case, you could make allowances for the outer/inner diametrical terms to simply the mathematics. This would come with loss of generality, this is typically why there are limitations like D/t < 10 for t=wall thickness.

Hopefully you can follow this mathematical logic. The limitations in space do not lend themselves well to this type of discussion.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[Drej]

Remember also that the classical Euler equations give you unconservative magnitudes regarding the buckling load. If you want to be more accurate with your predictions, taking into account imperfections (material, geometry, connections...) in the structure, you'll need to go non-linear.

 

[Denial]

I believe that CoryPad's second formula is for the sort of local wall buckling where the wall behaves like a concertina. At one location the circular cross-section will have moved radially inwards, whilst at a close-by location the circular cross-section will have moved radially outwards. An axial line drawn along the cylinder will distort into an approximately sinusoidal shape in the radial plane. The structure's defelected shape retains full radial symmetry (so every cross section remains circular).

If this assumption is correct, then the factor of 0.6 in CoryPad's formula is actually 2/(12*sqrt(1-n^2)) where n is Poisson's ratio.

However, also if this assumption is correct, there is another "local" buckling mode to consider. This is where radial symmetry is breached, and the cross-section does not remain circular. At one location the cross-section will take up a slightly square shape, whilst at a nearby location it will take up a similar squarish shape but with the square rotated by 45 degrees. The surface of the cylinder will be "lozenged" by a set of helical ridges.

I do not know the formula for the critical stress than corresponds to this potential buckling mode. I do not even know whether a formula exists. But the mode certainly exists, and its critical stress is often less than that of the radially symmetric local buckling mode.

 

[unclesyd]

You might like to checkout this site. Reas the instructions on how to download.

http://euler9.tripod.com/analysis/asm.html

 

[Cockroach]

I was interested with the outcome with compressive axial load parallel to the longitudinal axis on this pressure vessel. If 'P' is internal pressure, 'F' is compressive load for a pressure vessel of geometry 'D' as outside diameter, 'd' inside diameter then the Von Mises-Hencky Equation would lead to:

S = sqrt(3 Pi^2 P^2 D^4 + 16F^2)/[Pi (D^2 - d^2)

for S=element stress and Pi=3.14156....

Clearly if F=0, then the Von Mises-Hencky relation falls out of the solution set,

S = sqrt(3) P D^2 / (D^2 - d^2) = sqrt(3) P [R^2/(R^2-1)]

for R = D/d as stated in the earlier discussion.

As expected, force acts to greatly increase stress in the wall of the pressure vessel. Dimensional consistency is also preserved, stress would be in units of force divided by area, clearly stress by definition.

So this would be your triaxial state of stress for the wall element of a pressure vessel of internal pressure P subject to a compressive axial load of F.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[mloew]

You may want to consider the following SAE Technical Papers:
880901 Crush Characteristics of Thin-Walled Cylindrical Tubing
880894 Crash Analysis of Thin Walled Beam-Type Structures
811302 Design of Thin Walled Columns for Crash Energy Management--Their Strength and Mode of Collapse
840727 Axial Collapse of Thin Wall Cylindrical Column
860820 Analytical Technique for Simulating Crash Response of Vehicle Structures Composed of Beam Elements

There are many more as well. You can also try ASME International, as much research on this topic may be published by them.


Best regards,

Matthew Ian Loew
"I don't grow up. In me is the small child of my early days" -- M.C. Escher

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