Area Moment of Thin-Walled Cylinder, I

[pvwander]

Hello All,

I was asked by my immediate superior to calculate the area moment of Inertia for the thin-walled cylinder to be used in the calculation of Wind axial stress, S = Mc / I.

I am not sure if my calculated I is correct, I = 2 pi R^3 t, using the general eq. I = y^2 dA.

Can anyone confirm if is correct or not?

Please pardon my very basic question, I am very new in pressure vessel design. I am still learning all the basics of my job.

Thanks for all your replies.

pvwander

 

[jamesl]

The formula is I=pi*R^3*t. There is no factor 2.
You can find this formula in many handbooks.

 

[prex]

And don't forget, as you mention thin wall, that the allowable axial compressive stress need be checked too.

prex

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[pvwander]

@ Jamesl & Prex;

Thank you very much for your replies.

I now figured-out which (I) I computed. It is the Polar moment of Inertia, Ip = 2*pi*R^3*t (approx. only) for thin walled cylinder (t < 0.1R).

The (I) which I am looking for is the inertia along X or Y axis. Since Ip = Ix + Iy and Ix is equal to Iy for circular section, I can calculate the value of Ix or Iy.

Ix or Iy = Ip/2 = pi*R^3*t. So, I got it now.

Again, Thanks.

-pvwander.

 

[TERIO]

As an aside, the formula for any cylinder is

I = pi/64*(do^4-di^4)

if you substitute do=2*rm+t and di=2*rm-t (rm is the mean radius) and do some algebra you can get to

I = pi*rm^3*t*(1+0.25(t/rm)^2)

so when (t/rm) is small I ~ pi*rm^3*t.