Calculating Burst Pressure of Tubing 1

[TJK1]

What is the equation for calculating Burst Pressure of a 1/2" I.D. x 5/8" O.D. x 2.40" long piece of LEXAN, polycarbonate tubing? It is open on both ends. The yield strength is 9000 psi and the ultimate tensile strength is 10,000 psi. Room temp water is flowing thru it.

 

[Cockroach]

I suppose you can use the API Burst Equation governing pipe collapse in the oil industry (drilling and downhole). This is typically thin wall pressure vessel with a factor of 7/8 applied to it.
P = 0.875 [2 Sy / (D/t)]
for P=internal pressure, Sy=yield stress, D=pipe OD and t=wall thickness.

I prefer the Von Mises-Hencky Equation for a triaxial state of stress.
S = sqrt(3) P [R^2/(R^2 - 1)] for R=D/d
D=OD and d=ID. You can substitute material yield strength for S and solve for internal pipe pressure given the OD and ID geometry. This model I have found to be extremely accurate and representative of practical, everyday engineering problems.

There are other models, you can use them for cross reference on your calculation. Good luck.

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

 

[TJK1]

I am not sure how to interpret "sqrt(3)P" in your equation. Does that mean take the square root of the quantity 3 times P and then multiply that answer times the quantity in brackets?

 

[Cockroach]

Yeah, you've got it. You can see dimensional consistancy noting "R" is unitless and regarded as a geometrical factor, [P]=psi so that =psi.

The square root of three is simply a scaling factor which drops out of the mathematics as a result of hoop, radial and longitudinal stresses of the thick wall pressure vessel theory.

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

 

[jcfoley]

TJK1,

B31.3 has a section for non-metallic piping. I would suggest that you give this code a look. I do not use nonmetallic piping in my industry but if it applies to your application, it can save you a lot of liability exposure.

Chris Foley
Midland, TX

 

[KenShumway]

Over 20 years ago (before I was finished with engineering school), I did some work with Lexan tubing. I'm not sure and I would talk with your supplier, but it appeared to me that the tubing, an extruded product, might have different strength properties in the longitudinal direction than in the circumferential direction. I also learned that it does not have very good fatigue resistance. A couple of cautions that you might want to consider.
Ken Shumway