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Buckling of a thin-walled cylinder 2
- Thread starter davhar
- Start date
Skogsgurra
Electrical
Roark's standard text "Formulas for Stress and Strain" gives a formula for the buckling of a ring:
p' = 3EI/(r^3)
where p' is the buckling load per unit length of cricumference. In his Edition 5, it is Table 34, Case 8.
This is readily extendable to your case of a LONG thin-walled cylinder. Don't forget the (1-n^2) correction, where n is Poisson's ratio.
NOTE however that most texts warn that formulae such as this one are critically (no pun intended) dependent upon the shape having perfect geometry. Slight imperfections lead to large reduction in load-carrying capacity.
HTH.
p' = 3EI/(r^3)
where p' is the buckling load per unit length of cricumference. In his Edition 5, it is Table 34, Case 8.
This is readily extendable to your case of a LONG thin-walled cylinder. Don't forget the (1-n^2) correction, where n is Poisson's ratio.
NOTE however that most texts warn that formulae such as this one are critically (no pun intended) dependent upon the shape having perfect geometry. Slight imperfections lead to large reduction in load-carrying capacity.
HTH.
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1
- #4
GregLocock
Automotive
No, that is the wrong formula, that is for out of plane buckling of a ring. I think the OP is referring to local buckling of the skin (oil canning).
table 35 case 19 is probably the one you need.
Cheers
Greg Locock
table 35 case 19 is probably the one you need.
Cheers
Greg Locock
Greg,
Are you sure Table 34 / Case 8 of Edition 5, "Uniform circular ring under uniform radial pressure", is for out-of-plane buckling of the ring? I have always taken it as applying to in-plane buckling.
And now that you draw my attention to Case 19 of Table 35, "Thin tube under uniform lateral external pressure", for which my thanks, I believe that the two formulae are the same (mutatis mutandis).
As an aside to a fellow user of Table 34, there is an error in Case 4, "Uniform straight bar under end load P ... elastically supported by lateral pressure p proportional to deflection". The pi^2 in the denominator of the formula for P' should actually be pi^4. I passed this error on to the Australian office of the publisher (McGraw-Hill) in 1993, but am not sure whether the error has been corrected in any subsequent edition. Does anyone out there know?
Are you sure Table 34 / Case 8 of Edition 5, "Uniform circular ring under uniform radial pressure", is for out-of-plane buckling of the ring? I have always taken it as applying to in-plane buckling.
And now that you draw my attention to Case 19 of Table 35, "Thin tube under uniform lateral external pressure", for which my thanks, I believe that the two formulae are the same (mutatis mutandis).
As an aside to a fellow user of Table 34, there is an error in Case 4, "Uniform straight bar under end load P ... elastically supported by lateral pressure p proportional to deflection". The pi^2 in the denominator of the formula for P' should actually be pi^4. I passed this error on to the Australian office of the publisher (McGraw-Hill) in 1993, but am not sure whether the error has been corrected in any subsequent edition. Does anyone out there know?
GregLocock
Automotive
Interesting gotcha there, he's using p for linear loading in one, and q for area loading in the other. Getting rid of that idiotic confusioan I get:
34/8 is pcrit=3*E*I/r^3/b, ie 1/4*E*t^3/r^3
35/9 is pcrit=E*t^3/(4*(1-nu^2)*r^3)
That's about a 10% difference.
However, since they are so similar I think you are right, the 34/8 is for an in-plane buckle.
Cheers
Greg Locock
34/8 is pcrit=3*E*I/r^3/b, ie 1/4*E*t^3/r^3
35/9 is pcrit=E*t^3/(4*(1-nu^2)*r^3)
That's about a 10% difference.
However, since they are so similar I think you are right, the 34/8 is for an in-plane buckle.
Cheers
Greg Locock
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1
- #7
The full formula for the critical pressure of a thin cylinder of length L is given by a complex formula that is used in the german and italian codes for pressure vessels. In the formula n([≥]2) is the number of waves of the buckled shape and must be determined in order to minimize the critical pressure. The factor of safety adopted with respect to this theoretical value is normally 3, but shape imperfections should also stay within code limits.
[tt] 2Et 1 t2 2n2-1-[ν]
pcr= ----{--------------------- + ---------- [n2-1 + -----------]}
Do (n2-1)[1+(2nL/[π]Do)2]2 3(1-[ν]2)Do2 (2nL/[π]Do)2-1
[/tt]
prex
Online tools for structural design
[tt] 2Et 1 t2 2n2-1-[ν]
pcr= ----{--------------------- + ---------- [n2-1 + -----------]}
Do (n2-1)[1+(2nL/[π]Do)2]2 3(1-[ν]2)Do2 (2nL/[π]Do)2-1
[/tt]
prex
Online tools for structural design
This small difference is to be expected, and the formulae are totally consistent with each other. The ~10% comes entirely from the (1-nu^2) term. The absence of this term in the ring case reflects the fact that the ring's cross-section is free to expand and contract in the direction normal to the plane of the ring. The expansion / contraction in this direction is purely a Poisson effect resulting from the circumferential stresses directly induced by any in-plane bending in the ring.
The tube is constrained against deforming in this manner, by virtue of its large axial length. It is therefore stiffer in response to any such bending. This additional effective stiffness leads to the slightly larger resistance to buckling.
I alluded to this effect in both my above posts, perhaps too cryptically.
The tube is constrained against deforming in this manner, by virtue of its large axial length. It is therefore stiffer in response to any such bending. This additional effective stiffness leads to the slightly larger resistance to buckling.
I alluded to this effect in both my above posts, perhaps too cryptically.
GregLocock
Automotive
At a physical level I find this rather confusing. Collapsing the thin wall tube, in, typically, a diamond shaped pattern, seems to me to bear little resemblance to that required to create a crease the entire length of the pipe.
Or am I concentrating too much on the idea of pushing my thumb into the side of Pepsi (tm) can?
Cheers
Greg Locock
Or am I concentrating too much on the idea of pushing my thumb into the side of Pepsi (tm) can?
Cheers
Greg Locock
Burst and Collapse of Circular Cross Sections can be developed from API 5C3 Bulletin. Pressure governing the three transition points are:
Burst: Pb = (7/8)(2 Yp [t/D])
Yield Strength Collapse: Pyc = 2Yp([(D/t) - 1]/(D/t)^2)
Elastic Collapse: Pe = [2E/(1-v^2)](1/[(D/t)(D/t - 1)^2]
for Yp=yield strength (2% offset), D=Outside Diameter, t=wall thickness, E=Young's Modulus, v=Poisson Ratio.
I prefer a wrap fist around a Budweiser can as an analogy myself, I've never personally taken the Pepsi Challenge.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
Burst: Pb = (7/8)(2 Yp [t/D])
Yield Strength Collapse: Pyc = 2Yp([(D/t) - 1]/(D/t)^2)
Elastic Collapse: Pe = [2E/(1-v^2)](1/[(D/t)(D/t - 1)^2]
for Yp=yield strength (2% offset), D=Outside Diameter, t=wall thickness, E=Young's Modulus, v=Poisson Ratio.
I prefer a wrap fist around a Budweiser can as an analogy myself, I've never personally taken the Pepsi Challenge.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
The original poster didn't mention whether the cylinder is stiffened with rings or longitudinal stringers. Either of these would make a big difference. Just in case it is so, Davhar should look up:
Admittedly, it's for internal pressure, but it's a starting point.
Steven Fahey, CET
"Simplicate, and add more lightness" - Bill Stout
Admittedly, it's for internal pressure, but it's a starting point.
Steven Fahey, CET
"Simplicate, and add more lightness" - Bill Stout
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