How do i determine the wall thickness of a Cylinder if it gets expanded.
February 27, 2014 11:41 AM   Subscribe

I would like to use a mathematical formula in Excel to try and determine how much the wall thickness of a Cylinder or tube changes if it is expanded.

Example of what I would like to be able to calculate if possible.

If I have the following:

Outside Diameter (OD): 0.101

Inside Diameter (ID): 0.077

Wall Thickness (WALL): 0.012

How could I calculate the ID and/or the Wall Thickness if I expand the (OD) to 0.203
posted by ndillon to Education (11 answers total)
Wall Thickness = (Outside Diameter - Inside Diameter)/2
posted by rancidchickn at 11:45 AM on February 27, 2014

rancidchicken, that only works if you know two of the variables. The OP only knows one.

Assuming the density of the material doesn't change, you'll want to calculate the total volume of the wall in the initial state, and use that as the second constant to calculate the new inner diameter and wall thickness based on the new outer diameter. At least I think that's what the OP is asking for.

No, I can't write an Excel formula for that, unfortunately.
posted by alms at 11:49 AM on February 27, 2014 [1 favorite]

What mechanical process are you using to expand the cylinder? Is the material itself expanding or are you just stretching it?

If the latter, I would just assume that the cross-sectional area stays the same, so this equation will hold:

pi*(OD1/2)^2 - pi*(ID1/2)^2 = pi*(OD2/2)^2 - pi*(ID2/2)^2

You have OD1, ID1, and OD2. Solve for ID2 algebraically and plug that formula into Excel.
posted by supercres at 11:52 AM on February 27, 2014 [2 favorites]

That is correct, rancidchickn, but the problem is that I do not have the ID of the expanded part. I am trying to determine what it the wall thickness would be if I expanded it out equally.

alms, Do you have an equation for that. I could probably use that to create a formula in excel.
posted by ndillon at 11:52 AM on February 27, 2014

(Note that this is the same as a volume-based calculation, except assuming that length is constant, so it cancels from both sides.)
posted by supercres at 11:53 AM on February 27, 2014

It simplifies to:

OD2^2 - OD1^2 + ID1^2 = ID2^2
posted by muddgirl at 12:13 PM on February 27, 2014

Supercres is right, so give that the best answer. The cross sectional area of a cylinder is (pi/4)*(OD^2 - ID^2). That's how I typically write it and you'll notice that it's identical to what supercres writes. The area will not change with an expansion, so if it's expanded over the full length of the cylinders, you know ID1, ID2, and OD2, so equate the CSAs and solves for ID2. Sorry for repeating what you said, supercres; I just wanted to provide confirmation. /engineer

Also, if you're planning on actually expanding something instead of just doing a math problem, you might want to talk about that as well or head on over to and ask what the best method would be to expand your cylinder.
posted by KinoAndHermes at 12:16 PM on February 27, 2014

alms, Do you have an equation for that. I could probably use that to create a formula in excel.

No, but luckily I don't have to because supercres and muddgirl did it, simplifying by taking advantage of the fact that the length is constant.
posted by alms at 12:26 PM on February 27, 2014

KinoAndHermes, This is for the expansion of PTFE tubing. I am trying to calculate the weight per foot change in the material before and after expansion. I have an equation that does this for me, but it requires the Expanded ID size, which is after the fact.

So what if the length is not constant?

0.078 is the Material Density

I calculate the Weight Per Foot as


I then calculate the Length change by

(Weight Per Foot)/(0.078*((PI()*(OD2/2)^2)-(PI()*(ID2/2)^2)))

This tells me that for every 1 Foot of material that I feed in, I get out 7.752" of Expanded material out.

I want to be able to calculate that Expanded ID that I need to do the second equation that I listed.
posted by ndillon at 12:35 PM on February 27, 2014

It sounds like you have two unknowns-- L2 (the length after expansion) and ID2. One of those is necessary, unless you have some other information like the wall thickness after expansion.
posted by supercres at 1:57 PM on February 27, 2014 [1 favorite]

What you are missing here is the Poisson's ratio for PTFE which is 0.46. This basically means that the dL/L = 0.46 x dT/T where L is the original length and T is the original thickness.

IF the outer diameter be D1 and then D2

L1(D1^2 - (D1-T1)^2) = L2(D2^2 - (D2-T2)^2)


L2 - L1 = L1*0.46*(T1-T2)

Some signs may be confused but you have 2 eqns and two unknowns (T2 and L2) that you should be able to solve for.
posted by hariya at 10:14 AM on February 28, 2014

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