Shape suggestions for known mass & inertia

[mloew]

Hi all,

I am trying to develop a revolved shape that will have a known mass = 130 kg and known I (about the rotation axis) = 3.6e7 kg*mm^2 and must fit inside the cylinder with D_o <= ~1200 mm, D_i >= ~300 mm and Length < 450 mm.

Shape is completely arbitrary and density can be modified as well. I tried many obvious shapes and can not get the inertia high enough with the variables I have. Any suggestions? I am using Pro/ENGINEER's Behavioral Modeling functionality, but the problem solution is not dependant on this tool.

Thanks in advance.

Best regards,

Matthew Ian Loew
&quot;Luck is the residue of design.&quot;
Branch Rickey

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[israelkk]

The best it a cylinder. But I=0.5*M*(Ro^2-Ri^2) meaning that given the mass M=130 kg and given Ro=Outside Radius Ri=inside radius the inertia is 2.194e7 kg*mm^2 and there is nothing you can do about it unless you can increase the mass.

 

[israelkk]

Just want to add that you could keep the mass but change the outside diameter or change both.

 

[IRstuff]

Are you sure about that equation? As Ri increases to Ro, the moment should increase, not decrease.

In the limiting case, I=mass*R^2, for which I get 4.7E7 kg*mm*2, so it should be doable.

The problem, then, is that the density is too low.

Running a Mathcad solve block, I got 542.1 mm and 509.8 mm to match the moment and density of Aluminum.

TTFN

 

[ivymike]

Isrealkk is correct. The smallest OD that will give you the inertia/mass ratio you're looking for is 1488mm (with zero ID). If you can violate the OD constraint, you've got a shot. If you can violate the inertia/mass ratio constraint, you can get your inertia.

 

[israelkk]

The moment of inertia of a solid cylinder is 0.5*M*Ro^2 therefore, the moment of inertia of a hollow cylinder is less as I stated I=0.5*M*(Ro^2-Ri^2). The reason is simple, deduct the moment of inertia of a cylinder which has an outside radius equal the inner radius of hollow cylinder as though it was a solid cylinder from moment of inertia of the solid cylinder with the outside radius of Ro.

I suspect you looked at the Machinery's handbook where there is a typo error, it uses the &quot;+&quot; sign where it should be the &quot;-&quot; sign which is the correct one.

 

[IRstuff]

If you set Ro=Ri the resultant inertia is zero, which is not possible, since that's the condition that should give you the maximum possible inertia

TTFN

 

[minerk]

I don't think the formula from Machinery's handbook is a typo. All of the other sources I have indicate that the radii are squared then added, not subtracted. This would make sense, the mass moment of inertia for a hollow cylinder of diameter D would be greater than that for a solid cylinder of diameter D having the same mass, since the mass is further away from the axis. If this is correct, then I come up with the following:

I = 0.5*M*(Ro^2+Ri^2) = 3.6e7 kg*mm^2
for Ro = 600 mm, solving for Ri gives 440 mm

 

[IRstuff]

Please see:

http://230nsc1.phy-astr.gsu.edu/hbase/ihoop.html

TTFN

 

[israelkk]

minerk:

Sorry for the confusion, I usually calculate the mass from the density multiply by the volume. Now I understand your point you try to keep M constant. You were correct I was mistaken.

I tried to calculate it from the basics using the material density &quot;rou&quot; and came to:

Mass=Pi*rou*(Ro^2-Ri^2)*Length
Inertia=0.5*Pi*rou(Ro^4-Ri^4)*Length

Now you have four unknown, Ro, Ri, Length and the density (rou).

From the Mass equation you can extract Ri^2 and insert it in the Inertia equation. This reduces to only three unknown Ro, the Length and the density (because the Inertia and the Mass are constant).

I assume that you are trying to design a flywheel. Therefore, to reduce stresses and unbalance to the minimum you need the smallest outside radius Ro. Therefore, I would choose the largest Length that you can allow and then draw the Ro against the density and choose a material that will give the minimum outside radius Ro. You also have to calculate the stress and the material has to able to withstand the stresses too.

 

[mloew]

Thanks for all the information and encouragement.

Interesting, my second edition of Fundamentals of Machine Component Design by Robert C. Juvinall & Kurt M. Marshek (

http://www.amazon.com/exec/obidos/t...=sr_1_1/102-2078603-1626553?v=glance&s=books)

has the equation with the “-“ sign. In another page that IRStuff references (

http://230nsc1.phy-astr.gsu.edu/hbase/icyl.html#icyl3)

the opposite equation is found.

In my analysis, however, I am not using the closed form solutions (because I did not want to be restricted to rectangular x-sections) for my general purpose feature. The optimizer could not create the geometry I wanted with the restrictions on max size. I will keep plugging away until I find a solution I like.

The maximum I value I can obtain is ~2.5e7 kg mm² with
D_o <= 914 mm and Length < 450 mm. (note: I was off with the earlier figure on the D_o)

I can meet both goals with D_o allowed to go to a value of 1083 mm.

Again, the solver in Pro/ENGINEER works great for me, but I can’t find a better geometry to fulfill the goal. I am still interested in additional commentary on the discrepancies on the equations.

Thanks all!


Best regards,

Matthew Ian Loew
&quot;Luck is the residue of design.&quot;
Branch Rickey

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[IRstuff]

The geometry is pretty much a given. The ideal case for maximum moment of inertia is an infinitely thin hoop at the maximum radius.

However, your optimizer seems to be giving out goofy answers, but that may be because you've fixed Di at 300mm. That page I referenced has a moment calculator and the answers it gave matched my Mathcad calc. With all the mass at a diameter of 1200mm, you should get 4.7E7 kg*mm^2. That's the maximum mathematically possible for that diameter.


Is there some reason that Di can't go past 300mm? If there's some sort of limit, then the solution would be to have a hollow cylinder with a small number of flat annuli to get the required inner diameter, but put all remaining mass at the maximum radius. I guess the worst-case configuration would require a cylinder at the inner radius, a cylinder some maximum radius and a bunch of annuli to hold the two cylinders together.





TTFN

 

[mloew]

IRstuff,

Thanks for your continued interest. The optimizer seems spot on to me, the D_o <= 914 mm and Length < 450 mm works out with the closed form equations, but I am short on the I. I am not forcing the D_i to a value, the 300 mm figure was the minimum value that the value could take. In reality, I have the model as a thin revolved protrusion, so here are the actual variables:

D_o is d335
Length is d330
Thickness is d323 so D_i is derived from D_o - 2 * Thickness

Here is the screenshot of the optimizer:

Again, my real limit on D_i is 914 mm. It simply will not work with densities of earthly materials.





Best regards,

Matthew Ian Loew
&quot;Luck is the residue of design.&quot;
Branch Rickey

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[IRstuff]

Sorry, you're right, I hadn't realized that you had reduced the radius.

The minimum mathematically possible radius is 526 mm.

Alternately, the mass must increase beyond 172 kg, or some combination thereof

TTFN

 

[borjame]

Maybe my math is wrong here but what density are you using? I calculate that you have 4.77E8 mm^3 available and if divide that by the 130 kilos I get a density of 2.7E-7 kg/mm^3. Steel's density is ~7.84E-6 kg/mm^3 (I usually work in English so my density might be wrong here) If I use that density and multiply by the volume available I get a mass of 3743kg, a magnitude larger than the 130kg limit. Again, check me b/c I never work in SI but it might be that the 130kg is a bogus number and you can get a suitable answer. Yes, no?

 

[mloew]

borjame,

Here is where my model stands now:

D_o = 1051.6 mm
Thickness = 0.82 mm
Length = 476.73 mm
Density = 0.0001 kg / mm³

Mass = 130 kg
I = 3.6 e7 kg * mm²



Best regards,

Matthew Ian Loew
&quot;Luck is the residue of design.&quot;
Branch Rickey

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[IRstuff]

The only problem I see is that your density is about 5 times higher than any material around.

Osmium is 22.6 gm/cm^3, which is more like 0.00002257 kg/mm^3 in your units.

With osmium, you can get Do = 1056.2, thickness = 3.4mm, and same length as you have

TTFN

 

[IRstuff]

I just noticed that you had a slight round-up from 3.5885, which suggests that if you play that card, you can get by with Do = 1048.8, thickness = 3.68 mm, resulting in 3.55e7 kg*mm^2.

Sloppy job on subtracting in the previous post, thickness = 3.65 mm

TTFN

 

[mloew]

IRstuff,

I am just posting the current numbers from the model solution. The actual numbers in the optimization go off to about 15 significant figures. The density is not a problem, as I am trying to represent with simple geometry the properties of a far more complex system. The size restriction I was attempting to put on the analysis was to keep the solid geometry “hidden” within a certain cosmetic envelope. I use Pro/ENGINEER as an engineering tool, not just a modeling tool. So I am allowed certain freedoms in how certain engineering attributes get represented in the engineering models.


Best regards,

Matthew Ian Loew
&quot;Luck is the residue of design.&quot;
Branch Rickey

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[IRstuff]

I understand what you are saying, but I'm not seeing how you reconcile the effective density of the envelope being 5 times higher than any natural material.

If whatever you have does not occupy the full volume, then you need to have the envelope density be lower than the density of the material being used.

TTFN