# Shade tree engineer needs help calculating hoop stress

I have an extruded aluminum tube. I'm trying to come up with an Excel calculator that will help me find out how much force would be exerted on the internal walls if the tube was filled with rubber, and then sent through a vulcanization process; failing that, I'd like to learn how to calculate the hoop stresses that the tube should be able to withstand, with the assumption that the structure of the tube was perfect (weld seams, free of voids, and such).

I know all of the dimensions and mechanical properties of the tube (below); however, absolutely nada! for rubber; much less vulcanized rubber. This is an extruded tube, not rolled and welded. The weld seams I mention above are solid-state welds, fused in the welding chamber of a porthole extrusion die, while under tremendous pressure and at a temperature of about 1,000f. If it makes any difference, the extrusion die has 5-ports so, there are 5 seam-welds approx. 72 degrees apart that travel the length of the tube.

Meet the victim:

• 6061 aluminum alloy tube produced utilizing port-hole extrusion.
• Artificially aged to T6 temper
• Yield Strength = 36.2 ksi
• Ultimate Tensile Strength = 40.0 ksi
• Length = 3.0"
• Outside Diameter = 3.150", with 0.008" ovality
• Wall Thickness = 0.118" with 0.004" wall spread
• I'll also say that I can routinely expand the diameter by 15%+ via drift-tests

If you can help me, I'll add you to the liner-notes on my upcoming record release; it's been in the works for ten years so, we've got time.

• So, could you treat the tube as a pressure vessel, find the thermal expansion of the rubber and the aluminum tube, then find the delta expansion, and compute stresses based on that? – GisMofx Dec 22 '20 at 20:49

You just need to calculate the hoop stress of a cylinder. The formula is really easy:

$$\sigma =\frac{p r}{t}$$

where:

• $$\sigma$$: is the hoop stress
• $$p$$: the pressure
• $$r= d/2$$ : the radius of the tube (I would ignore the ovality if its small)
• $$t$$: the thickness of the tube.
• Thanks, I did read that very page this morning, but I don't know the thermal expansion of the rubber.. – jimmymustang06 Dec 22 '20 at 21:27
• I did just find this.. The coeficient of thermal expansion of rubber is approximately 4.8 × 10 –4/K, similar to a hydrocarbon liquid. Addition of ﬁllers reduces the value slightly. Comparing this expansivity to steel (3.5 × 10–5/K), a tenfold diﬀerence, one begins to understand that built-in interfacial strains may occur in a bonded rubber-metal or composite structure – jimmymustang06 Dec 22 '20 at 21:37
• @jimmymustang06: You also need to see how the rubber behaves during the vulcanization process (I haven't the slightest idea). Vulcanization significantly changes the mechanical properties of rubber, so it may be that the rubber expands or contracts significantly during the process as it shifts from one "phase" to another, in a way not accurately described by its simple thermal expansion coefficient. – Wasabi Dec 22 '20 at 22:40
• That's the problem I have no way of finding that information.That process is not ours. The issue is that tubes are subjected to this vulcanizing process. Randomly, the tubes will split. There is pressure in that process, when I receive samples back, the rubber has squeezed though the split. We are not completely innocent, as we may not have perfect seam-welds, but out of millions of feet, it's a only a couple feet that ever has an issue. The problem jumps around different parts (OD's/Wall's). We may have one part that suffers a few splits, and then not have an issue until years later. – jimmymustang06 Dec 22 '20 at 23:12
• I found the first page an article which reports on the expansion of vulcanised rubber. – NMech Dec 23 '20 at 18:09

If this nails your problem, I want to play bass on your release.

To split an extruded tube indicates two things: 1) severe overpressure, in excess of the yield point hoop stress quoted above, OR 2) defective extrusions. Let's look at each.

1. Rubber is far, far more compliant than aluminum. This suggests to me that the "bulk" overpressure during the vulcanization process would have to be of order ~hundreds of PSI (!!!) in order to develop the ~tens of thousands of PSI stress in the Al tube required to split an unflawed sample of it. (Note also that the stored strain energy "pent up" in the chunk of (presumably compressed) rubber inside the tube will be small even for large values of deflection because of how soft the rubber is; it is this strain energy that provides the work needed to open a fissure in the aluminum tube.)

2. closed-die extrusions (in which there is no seam at all) will be as strong as the parent material only in the absence of inclusions in the parent material, which are going to get pulled into linear flaws while going through the die- and those flaws can ruin the mechanical strength of the extrusion, causing the pipes to burst as you describe. Eddy-current tests can find those flaws for you, and it is common to employ semi-automatic continuous eddy-current testing on seamless extruded tube to find such flaws- and then excise them from the extrusion run. Are you doing this?

Now if the vulcanization process were consistently overstressing your extrusions then I would expect a higher failure rate. But the spotty nature of the fail suggests instead the occasional presence of a flaw in the pipe which compromises its pressure-holding capacity.

Here is what to do:

1. perform microscopic examination of the fracture zone in the split pipes, to see if the crack started at a visible inclusion or followed a linear flaw

2. build you a test die set which can be placed in a hydraulic press, where the die faces compress only the rubber slug inside the aluminum tube, and then measure the die forces required to squeeze the rubber hard enough to split a GOOD piece of tubing.

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The modulus of elasticity of the vulcanized rubber is in the range of 218psi that compared to 9990ksi of 6061 aluminum alloy negates any concerns about the difference in thermal expansion.

But the ratio of the radius to the wall thickness of approx 25, classifies the pipe as a thick wall and we use thick wall cylinder formulas.

For the Excel spreadsheet, these the formulas for thick pipe stress from Roark's Roark’s Formulas for Stress and Strain, 7th ed. pp123 can be used. I recommend reading chapter seven and checking the solved problems.

$$\sigma Y_{external}=P\frac{r_i^2(r_0^2+r_0^2) }{r_0^2(r_o^2-r_i^2)}$$

$$\sigma Y{internal=P \frac{r_0^2+r_i^2}{r_0^2-r_i^2}}$$

$$r_i=inner\ radius\\r_0=outer\ radius\\P=internal\ pressure-external\ pressure\\ \sigma_Y=\text{radial or hoop stress}$$