TurboCAD Forums

The Ultimate Resource for TurboCAD Knowledge

Register
 
When posting a problem be sure to include which version you are using.  Give as much information as possible.  If the problem is with a specific file be sure to attache it to your post.

Area of Inertia to Moment of Inertia
Read 8291 times
* February 16, 2011, 09:25:02 AM
TC14:  Using a torus as an example, metrics gives areas of inertia.  In the attached example I show a simple circle revolved to give a torus.  Using the weight per area, one would think that Area of Inertia * weight/area would give moment of inertia (wt-length^2).  But it doesn't seem to work, or at least it doesn't agree with a lookup formula solution.  Can anyone explain?  Thanks, Roger

Logged


* February 16, 2011, 10:21:57 AM
#1
There's some inconsistency here. The equation for the moment of inertia (m * (R^2 + .75 * r^2)) applies to a solid torus of uniform density. But your computation of the object's mass implies that the torus is a thin membrane. Which shall it be?

Henry H

Logged


* February 16, 2011, 12:17:54 PM
#2
Hi Henry.  Yes you are right, but I could not find a separate equation for a thin membrane.  However, I ran an example of a solid minus a smaller Diameter solid (by 2xthickness) that would represent a membrane of a certain thickness and weight, and the answer came out to be the same, perhaps because of the symetry.  That's why I used it. ... BUT, if you have one specifically for a membrane, please tell me.  Roger

Logged


* February 16, 2011, 07:57:48 PM
#3
Hi Henry.  Yes you are right, but I could not find a separate equation for a thin membrane.  However, I ran an example of a solid minus a smaller Diameter solid (by 2xthickness) that would represent a membrane of a certain thickness and weight, and the answer came out to be the same, perhaps because of the symetry.  That's why I used it. ... BUT, if you have one specifically for a membrane, please tell me.  Roger

If you're going to subtract the I of a small torus from the I of a large torus, you cannot factor out the mass, since it will be different for the two tori.

Not entirely trusting myself to avoid mistakes, I used two different methods to derive the formula for the mass moment of inertia of a thin-membrane torus about its axis of symmetry, and got the same expression from each. (Method #1 was brute-force integration; Method #2 was subtracting a small torus from a large one and taking the limit as t --> 0.) While it's not at all unlikely that I screwed up somewhere, my result is

I = M * (R^2 + 1.5 * r^2)

Henry H
« Last Edit: February 16, 2011, 08:21:30 PM by Henry Hubich »

Logged


* February 17, 2011, 06:37:44 AM
#4
Henry - THANK YOU!  I also tried to calculate it but got really hung up with evaluating sin & cos integrals, nor could I find a derivation on line.  You clearly have excellent analytical skills.  PLEASE - could you email your derivation notes (a scanned copy)?  I would love to see them and learn how it is done.  (roger.emerick@zodiacaerospace.com).  Also, please see attached excel file.  Perhaps a new wrinkle.  I did a simple example for a torus with min/max R at 6&10.  Very interesting!  The TC Moment of Inertia given and divided by the Surface Area yielded the result from (R^2 + .75r^2) rather than your formula, (R^2+1.5r^2), but they were close.  What that tells me is the Area Moment of Inertia in TC can be used for a membrane (wt/area) by multiplying wt/area * area * TC's Area I / area.  Your input greatly appreciated.  Roger

Logged


* February 17, 2011, 07:21:04 PM
#5
Henry - THANK YOU!  I also tried to calculate it but got really hung up with evaluating sin & cos integrals, nor could I find a derivation on line.  You clearly have excellent analytical skills.  PLEASE - could you email your derivation notes (a scanned copy)?  I would love to see them and learn how it is done.  (roger.emerick@zodiacaerospace.com).  Also, please see attached excel file.  Perhaps a new wrinkle.  I did a simple example for a torus with min/max R at 6&10.  Very interesting!  The TC Moment of Inertia given and divided by the Surface Area yielded the result from (R^2 + .75r^2) rather than your formula, (R^2+1.5r^2), but they were close.  What that tells me is the Area Moment of Inertia in TC can be used for a membrane (wt/area) by multiplying wt/area * area * TC's Area I / area.  Your input greatly appreciated.  Roger

On its way.

Henry H

Logged


* February 19, 2011, 11:00:58 AM
#6
Henry, Thank you very much.  I see my biggest mistake was I had incorrectly evaluated the integral of
cosine^2.  I didn't realize that Cos^0=theta.  Also, I see instead of evaluating theta integral from 0 to 2pi, you did 2x integral from 0 to pi.  I had done that also, because it didn't make sense that Cos^2 evaluated from 0 to 2pi came out to zero (1^2  - 1^2).  Now that I agree with your answer, please take a look at my spreadsheet file I sent.  Cell e19 has the TC area of inertia divided by the TC surface area of the torus (=67). This 67 = the area of inertia term in cell 11, which has the formula R^2+.75r^2, which is that used for a solid.  Any comment you can share on that would be appreciated.  And thanks again for the derivation.  Roger.

Logged


* February 19, 2011, 06:45:12 PM
#7
  I didn't realize that Cos^0=theta. 

Hm?

Henry H

Logged


* February 20, 2011, 10:31:05 AM
#8
Now that I agree with your answer, please take a look at my spreadsheet file I sent

I haven't received it yet, Roger.

Henry H

Logged


* February 20, 2011, 02:52:12 PM
#9
TC14:  Hi Henry.  Attached is the excel file I made up with a torus example (and which I thought I had originally sent).  Please review my earlier writeup which suggests the value calculated by TC is (R^2 + .75r^2)  by comparing Cells e11 and e19, rather than the (R^2+1.5r^2) you calculated, and which I agree with.  Any comment?  Thanks, Roger

Logged


* February 20, 2011, 07:10:36 PM
#10
TC14:  Hi Henry.  Attached is the excel file I made up with a torus example (and which I thought I had originally sent).  Please review my earlier writeup which suggests the value calculated by TC is (R^2 + .75r^2)  by comparing Cells e11 and e19, rather than the (R^2+1.5r^2) you calculated, and which I agree with.  Any comment?  Thanks, Roger

To make sense of any moment of inertia reported in the Selection Info palette, you must correct for the actual mass by dividing the reported I by the reported volume and then multiplying the result by the actual mass.

Shell the torus to some small thickness (e.g., .001") and TCad will report the correct I, which in this case is very close to that for a thin shell (m * (R^2 + 1.5 * r^2)), after you correct for actual mass as outlined above.

Henry H

Logged


* February 21, 2011, 05:49:24 AM
#11
Thank you Henry.  Your insight on this was very educational.  Also, on the issue of cos integrals in the Area of Inertia derivation ... one of the integrals was Integral of Cos^n = Sin(x)*Cos^n(x)/n + (n-1)/n*integral of Cos^(n-2)x dx.  So for n=2, the second part is the integral of (Cos^(2-2) x) dx; which apparently is just x? So for evaluation over 0 to pi = pi, so (n-1)/2*pi = 1/2*pi.  Am I understanding that correctly?  Thanks, Roger

Logged


* February 21, 2011, 11:29:03 AM
#12
Thank you Henry.  Your insight on this was very educational.  Also, on the issue of cos integrals in the Area of Inertia derivation ... one of the integrals was Integral of Cos^n = Sin(x)*Cos^n(x)/n + (n-1)/n*integral of Cos^(n-2)x dx.  So for n=2, the second part is the integral of (Cos^(2-2) x) dx; which apparently is just x? So for evaluation over 0 to pi = pi, so (n-1)/2*pi = 1/2*pi.  Am I understanding that correctly?  Thanks, Roger

You did it the hard way, but I think you're correct -- except that my Peirce's Integrals shows the n colored red in the above quote to be (n-1).

Henry H

Logged


* February 21, 2011, 12:37:06 PM
#13
Henry:  Yes, the n was supposed to be n-1, in sinx*Cos^(n-1)x.  I wrote it incorrectly. Thanks, Roger.

Logged


* February 23, 2011, 09:09:19 AM
#14
Hi Henry:  Initially I thought I understood how to take the TC's area of inertia and use it based on what you said earlier, and quoted below: 
"To make sense of any moment of inertia reported in the Selection Info palette, you must correct for the actual mass by dividing the reported I by the reported volume and then multiplying the result by the actual mass."

But I cannot make this work, see my examples (2 & 3) in the attached excel file.  Can you give a spreadsheet example for the torus, because that has equations we can count on for checking? I have to do some strange revolved shapes, of which I know the mass & volume, but need to calc the MOI. 

It seems that my problem (see attached), is I had a drawing in mm.  It gave me #'s in mm^4.  I converted to m^4 by dividing by 1000^4.  Makes sense to me.  However you get the wrong answer when multiplying *mass(kg)/vol(m^3).  You get the correct answer if you convert mm^4 to m^4 by dividing by 1000^5!  In other words its a 1000 too big.  Can you explain this?

Thanks, Roger

Logged


* February 23, 2011, 07:19:52 PM
#15
Hi Henry:  Initially I thought I understood how to take the TC's area of inertia and use it based on what you said earlier, and quoted below: 
"To make sense of any moment of inertia reported in the Selection Info palette, you must correct for the actual mass by dividing the reported I by the reported volume and then multiplying the result by the actual mass."

But I cannot make this work, see my examples (2 & 3) in the attached excel file.  Can you give a spreadsheet example for the torus, because that has equations we can count on for checking? I have to do some strange revolved shapes, of which I know the mass & volume, but need to calc the MOI. 

It seems that my problem (see attached), is I had a drawing in mm.  It gave me #'s in mm^4.  I converted to m^4 by dividing by 1000^4.  Makes sense to me.  However you get the wrong answer when multiplying *mass(kg)/vol(m^3).  You get the correct answer if you convert mm^4 to m^4 by dividing by 1000^5!  In other words its a 1000 too big.  Can you explain this?

Thanks, Roger

Roger...

The units of mass moment of inertia are kg-m^2, or gm-mm^2, or slug-ft^2. No fourth power is involved. Mass moment of inertia is a property of three-dimensional objects, and it defines an object's resistance to angular acceleration according to the formula t = I * alpha, where t is torque about a specific axis, I is the mass moment of inertia about that same axis, and alpha is the resulting angular acceleration about that axis. This is the rotational equivalent of the well-known f = ma. We have been discussing the mass moment of inertia of a specific 3D body; viz., a torus.

The units of area moment of inertia are length^4. No units of mass are involved. Area moment of inertia is a property of a plane figure, usually the cross section of a beam, in which case it defines the beam's stiffness when it is subjected to bending.

Henry H

Logged


* February 24, 2011, 05:39:22 AM
#16
Henry:  Thank you for the insight.  Regarding the case in point, a torus (a circle revolved).  I wanted to get a handle on what the TC Area of Inertia was and how to use it for two types of torus, a solid and a membrane, in each case I know the mass, the surface area, and the volume.  Then progress into irregular revolved 'solid' shapes of which I also would know the mass, SA, Vol.  Perhaps the TC area of inertia, (as you indicated, is the integral of x^2*dA), is meant only for a surface, not a volume.  As seen in my spreadsheet example it did provide the correct answer when used for a surface (membrane configuration), and correlated exactly with the answer provided by your equation:  mass*(R^2+1.5r^2).  I think I'll just have to play around with different examples to feel more comfortable with it, and understanding of it and its use for solids.  For an irregular shape, perhaps the key lies in obtaining a 'radius of gyration' from the 2D shape.  Appreciate your erudite contributions to my learning process.  Really!  Thanks again.  Roger

Logged


* March 08, 2011, 12:46:07 PM
#17
Henry (et al):  I believe the area moments of inertia in TC are mislabled.  I do not believe they are units in 10^4 (e.g. Integral of dArea * X^2 = X*X*X^2=X^4).  I believe what is really calculated is a volume of inertia (e.g. Integral dVolume*X^2 = x*x*x*X^2).  As indicted in prior writeups, to convert units properly (like between mm to m, you have to divide by 1000^5 not 1000^4.  Also, the only way to get the values for a membrane are to shell the solid and use that inertia term * mass.  Also, to get Moments of Inertia from TCs 'Area' of Intertia requires Mass * inertia/volume.  In other words Mass/vol * TC's Inertia = density * TC's Inertia.  So that means that TC's Inertia must be a Vol*X^2 # and have units of X^5.  That way when you multiply density * TC's inertia (mass/vol * Vol*X^2) the volumes cancel and you get mass*X^2.  Hope this makes sense and is correct.  If so, the TC program should be listing units in X^5 not X^4.  Then it makes sense and gives results consistent with lookup formulas.  Perhaps they should list both, Area of Inertia and Volume of Inertia depending if it is an area or a volume.  Roger

Logged