Thin-walled cylinder under pressure

[Alatza]

Hello everyone!

I am designing a glass vacuum chamber to work with 0.001 Pa for a sputter coater.

The method I found is Von Mises equation for thin-walled cylinder buckling. More precisely, I am following this paper:
"Prediction of vacuum-induced buckling pressures of thin-walled cylinders".

My parameters are:
E = 7.00E+10 Young's modulus for glass (Pa or N/m)
u = 0.2 Poisson's ratio for glass
r = 0.15 Inner radius (m)
t = 1.00E-03 Thickness (m)
l = 0.5 Length (m)
n = 6
Outside pressure= 1.01E+05 Pa (1 atm)
Inside pressure = 1.00E-03 Pa
Dif. pressure = 1.01E+05 Pa

The solution I found is: Critical pressure = 1.24E+07 Pa
This does not make sense since a glass only 1mm thick should not be able to handle all this pressure. Especially a piece so large.
I need to find the correct glass thickness.

Does anyone have a clue? What am I doing wrong?

Any help will be highly appreciated!

 

[TGS4]

For starters, you've posted in a piping forum, when there is a perfectly functioning pressure vessel forum.

Next, where in the world are you? There may be rules/regulations/laws pertaining to your situation.

Finally, there are perfectly good Codes/Standards that could be followed for your situation. The appropriate one depends on where in the world you are.

 

[LittleInch]

Input error?

Mathcad error?

Did you try the same inputs as in the paper supplied and came up with the same results as a check?

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[mbt22]

The introduction to the paper you linked to explains why you should not be designing using Von Mises equation. Next, look at pressure vessel design, where full vacuum ratings are common.

Matt

 

[Alatza]

LittleInch,

Thank you, man! I tried your advice on using the paper's data as input and found my mistake (Excel should not be used for this).
Now I got 1.2 mm as the minimum thickness for 10-3 Pa. So I intend to use 5 or 6 mm. What makes sense now.

mbt22,

Sorry, I didn't get why Von Mises equation should not be used in this case. The paper's introduction shows discrepancies on other studies but the paper's analytical part itself is based on the Von Mises formula. The results they found using it are not too far the experimental one, considering the precision I need for my vacuum chamber.
My cylinder is short and thin enough to fit on the Von Mises formula, but I am not an expert on this.
The main difference I found is the material, steel x glass. But I am assuming this is not a reason for not using the same method. Am I wrong?
Anyway, thank you for your comments!

 

[BigInch]

Von Mises does not apply especially well to brittle materials.
More to think about along that thread here,

http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3123583/

 

[Alatza]

BigInch, thank you! You are right. So I have to use another method. Von Mises is not applicable to glass.

Most books I found so far shows the stress in thin-walled cylinder as:

s = p * r / t
s = stress
p = pressure
r = cylinder radius
t = wall thickness
They say the formula is the same for external or internal pressure.
But I assume the strength for internal pressure is tensile and for external pressure is compressive.

Glass compressive strength is 10^9 Pa.

Using my data...
10^9 = 10^5 * 0.15 / t
t = 1.5 x 10^-5 m

This doesn't make any sense. What is wrong here?

Does anyone has any reference on calculating a "brittle material thin-walled cylinder subject to external pressure", please?

Thank you!

 

[BigInch]

That formula would only account for radial stress caused by internal and external pressure. Are redial stresses the only stresses here? How is this tube being supported? Are there any bending stresses being applied? Does the tube have closed ends? How is the external pressure applied and how is any internal pressure contained within the tube. Are there any stress concentrations? Is there a possibility of defects? Is this tube and contents at uniform temperatures?

 

[Alatza]

BigInch,

Thanks again for your help.

That formula would only account for radial stress caused by internal and external pressure. Are redial stresses the only stresses here?
The axial stress formula is p*r/ 2t, so the hoop stress is bigger.

How is this tube being supported? Are there any bending stresses being applied?
No, the only extra stress would be the weight of the cylinder itself. This is much lower than the hoop stress.

Does the tube have closed ends?
Top and bottom are closed. The situation is similar to most books cases I read so far.

How is the external pressure applied and how is any internal pressure contained within the tube.
External pressure and internal pressure is equally distributed all over the surfaces.

Are there any stress concentrations?
No.

Is there a possibility of defects?
Of course no material is perfect, but I am assuming no defects to simplify the problem.

Is this tube and contents at uniform temperatures?
Yes.

A simple method for calculation would be perfect. I do not need a super precise or ultra professional model for now.
Just something to have the minimum thickness with an error of around 20%.

Thanks!

 

[TGS4]

Stress is NOT the important consideration for external pressure. The failure mode is buckling. The important consideration for brittle material is the minimum design margin against buckling failure.

Note that out-of-roundess tolerances are critical for buckling assessments.

As I asked before:

Next, where in the world are you? There may be rules/regulations/laws pertaining to your situation.

Finally, there are perfectly good Codes/Standards that could be followed for your situation. The appropriate one depends on where in the world you are.

 

[Alatza]

TGS4, thanks! This explains a lot, indeed.

Guys,

I found new references. Both uses the same method (NOT Von Mises since glass is a brittle material and NOT stress since the collapse is due to buckling)

- Process Equipment Design - Lloyd E. Brownell, Edwin H. Young. John Wiley & Sons, Inc. (Chapter 8)
This paper makes a lot of sense too, since the project is using a brittle cylinder under external pressure.
- New Ceramic Pressure Hull Design for Deep Water Applications - Shinichi Takagawa, University of Tokyo - IEEE 2010.

Right track now?

Thanks for the comments and help! All of them were very useful to me.

 

[BigInch]

Top and bottom closed, or in other words a cylinder with closed ends develops longitudinal stress in additional to radial stress.

 

[Alatza]

Top and bottom closed, or in other words a cylinder with closed ends develops longitudinal stress in additional to radial stress.

Good point! But by what I read so far the longitudinal stress is half the hoop stress. So thickness calculated for hoop will cover longitudinal as well.

 

[TGS4]

Not necessarily so for buckling.

What if you were to design this thing using the rules in ASME Section VIII, Division 1?

 

[Alatza]

TGS4,

This one? [URL unfurl="true"]https://www.asme.org/products/codes-standards/bpvcviii1-2015-bpvc-section-viiirules[/url]
To be honest, the problem is the price. I don't have $660 to invest on this part of my project since I still have to pay for the glass, vacuum pump, aluminum parts, etc...

But probably it would be an excellent reference.

Thanks for the suggestion anyway!

 

[TGS4]

Where in the world are you located? Depending on that, manufacturing such a device may be required to meet the ASME Code, by law. Where I live, that is a requirement! What about your local laws and regulations?

 

[BigInch]

Stresses are not separate. Principal stresses are combined according to the various stress theories to get the maximum shear stress.

 

[Alatza]

TGS4,
I'm in Brazil. I will check the regulations here on this subject.

BigInch,
I will check this question further. After reading your comment I also remember that there will be some load supported vertically by the cylinder. There will be a thick aluminum disk and this weight certainly will influence the stresses supported by the glass.

So, more readings, formulas and calculations to have a better view on this problem. But this is the coolest part!

Again, thank you guys for helping me so much! I will post here the next steps I take (maybe this thread help other people in the future too).

Best!