## Approximate expansion diameter of latex tubing

I previously wrote about how to estimate the expansion diameter of latex tubing, but I made no measurements to verify that the formula I suggested, $8.5 D_i + 2 t$ where $D_i$ is the unexpanded inner diameter of the tube and $t$ is the unexpanded tube wall thickness, was accurate.

Back in 2010, I made some measurements for tests that were aborted as the tubes continually burst. I was testing higher pressure tubes. These tubes have a tendency to burst and they are extremely loud when they burst if you fill them with air. I do not suggest testing latex tubes with air for this reason.

So, I have three data points and a reasonable understanding of the geometry of these tubes. I’ll make a better estimate for expansion diameter.

I’ll start with a few assumptions. When the tube expands, it is assumed to stop expanding when its expanded inner diameter equals a certain multiple of the unexpanded diameter. This is justifiable based on the observed behavior of the rubber. The Gent hyperelastic model makes the same assumption. My second assumption is that when the tube is fully expanded the cross sectional area of the expanded tube is a multiple of the cross sectional area of the unexpanded tube. I originally anticipated these areas would be equal as I know for small deformations, rubber is approximately incompressible (i.e. the volume is preserved). But that’s only applicable for volume, not area; the tube is expanding in length too. Also, it is only applicable for small deformations. I do not know how the rubber will act for large deformations. $D_o$ is the outer diameter of the tube. $D_i$ is the inner diameter of the tube. $t$ is the tube wall thickness. The superscript $e$ refers to the expanded state. A variable or constant without $e$ is in the unexpanded state or it does not refer to any state (like the constants). $A = \tfrac{\pi}{4} (D_o^2-D_i^2)$ is the equation for the cross sectional area of the unexpanded tube. $A^e = \tfrac{\pi}{4} [(D_o^e)^2-(D_i^e)^2]$ is the equation for the cross sectional area of the expanded tube. $D_i^e = C_1 D_i$ is my assumption about when the tubes stop expanding. $A^e = C_2 A$, where $C_2$ is my assumption about the cross sectional areas of the tubes when they stop expanding.

Plugging all these equations together and solving for $D_o^e$, I find that $D_o^e = \sqrt{C_1^2 D_i^2 + 4 t C_2 (D_i + t_i)}$.

The data points I have available:

• Unexpanded ID: 3/8″, wall thickness: 3/16″, expanded OD: a bit more than 3″ (from memory)
• Unexpanded ID: 1/8″, wall thickness: 3/16″, expanded OD: 1.25″
• Unexpanded ID: 1/8″, wall thickness: 7/32″, expanded OD: 1.375″

A linear regression leads to $C_1$ = 7.35 and $C_2$ = 3.33 ( $R^2$ = 0.9999, which would definitely be lower if there were more data points). These constants seem reasonable given my understanding of the phenomena, so it is reasonable to accept that the tubes’ inner diameters expand to about 7.35 times their original inner diameter and the cross sectional area increases by 3.33 times.

The equation above with these constants can be used to find the expanded outer diameter. The equation with $C_1$ can find the expanded inner diameter. The definition $D_o^e = D_i^e + 2 t^e$ can find the expanded wall thickness.

Further tests are necessary, but this formula is the best I can do with the data I have on hand.

An equation for the inner diameter of a tube if the outer diameter is restricted (i.e. does not fully expand) follows. This is based on the assumption that the area ratio scales linearly with the inner diameter ratio. $D_i^e = (C_1 D_i)^{-1} [\sqrt{4(t C_2)^2 (D_i + t)^2 + (C_1 D_i)^2 (D_o^e)^2} - 2 t C_2 (D_i + t)]$