Area Moment of Thin-Walled Cylinder, I
Area Moment of Thin-Walled Cylinder, I
(OP)
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
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





RE: Area Moment of Thin-Walled Cylinder, I
You can find this formula in many handbooks.
RE: Area Moment of Thin-Walled Cylinder, I
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RE: Area Moment of Thin-Walled Cylinder, I
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.
RE: Area Moment of Thin-Walled Cylinder, I
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.