Axial Collapse of thin walled cylinders
Axial Collapse of thin walled cylinders
(OP)
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.





RE: Axial Collapse of thin walled cylinders
Fb = (π2 E I)/l2
and
I = (π/4)(ro4 - ri4)
where
π = 3.141 592 654
E = elastic modulus
I = second area moment
ro = tube outer radius
ri = tube inner radius
l = tube height
Local buckling occurs when the axial stress σ exceeds a critical value σc:
σ = (0.6 α E t )/r
where
α = knockdown factor = 0.5
t = tube wall thickness
r = tube outer radius
Regards,
Cory
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RE: Axial Collapse of thin walled cylinders
Steve Braune
Tank Industry Consultants
www.tankindustry.com
RE: Axial Collapse of thin walled cylinders
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
RE: Axial Collapse of thin walled cylinders
RE: Axial Collapse of thin walled cylinders
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.
RE: Axial Collapse of thin walled cylinders
http://euler9.tripod.com/analysis/asm.html
RE: Axial Collapse of thin walled cylinders
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
RE: Axial Collapse of thin walled cylinders
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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