Autodesk Inventor

[#1]

Hi,

I am trying to determine the moment of inertia of an L-section using IV 2009. By selecting Tools>Region Properties of a closed loop profile in the sketch environment the moment of inertia is given, however the principal axes through the centroid are at 45 degrees to the toe of the angle (see attached image). I want Ixx and Iyy where the X and Y axes are parallel to the horizontal and vertical toe respectively.

Coincidentally, AutoCAD orientates the principal axes the same when viewing the Region Mass Properties of a similar Region.

Any help on this would be greatly appreciated.

Thanks

Rich

[#2]

Hi! The moment of inertia is based on the sketch coordinate system of the particular sketch. If you want the coordinate to be evaluated at a different place, you can edit the coordinate system (finish sketch->right-click on the sketch in the browser->Edit Coordinate System). Then you can redefine the origin of the sketch coordinate system and also realign X or Y axis.

The triad in the middle is where the centroid is located and the arrows are pointing to the principal directions.

Thanks!

[#3]

I have tried rotating the sketch co-ordinates, and also rotating the geometry but leaving the sketch co-ordinates (effectively rotating the principal axis of the geometric shape), yet the principal axes in the Region Properties are always given 45 degrees from the global or sketch co-ordinates "Roatation Angle from projected Sketch Origin to Principal Axes (degrees): About z axis = -45

I have also tried this for a symetrical shape (rectangle), and the principal axes are not rotated from the origin. It would seem that this particular geometric shape causes problems.

Thanks

Rich

[#4]

Hi! Based on my limited understanding of how principal axes are computed, I think the result is correct. Basically, one of the principal axes has to be on the symmetric axis. Is my understanding incorrect?

Thanks!

[#5]

Hi,

I believe the principal axes intersect at the centroid of the shape, but do not have to correspond with any axis of symmetry, that depends on where you take moments from to determine the moment of inertia.

I cant find the formula from the Engineers Handbook of how IV calculates I, so below is a link to a web page which shows how Engineers calculate I about an axis parrallel to the toe of the angle. This is what I am trying to determine.

http://www.engineersedge.com/material_science/moment-inertia-gyration-7.htm

Thanks

Rich

[#6 · Accepted solution]

Hi Rich,

Just to demonstrate that you can have Inventor's region property results correspond litterally with the results in your engineering handbook, here is what you could do (granted it requires some manual work)

The moment of inertia values as shown in the handbook shot are calculated with the COG positioned in the origin.

If I move the entire Sketch so that the COG is coinciding with the origin I get following value from the Region properties for

Ixx= Iyy = 9112722  mm^4   (because of the square 150x150  shape Ixx = Iyy).

This corresponds nicely with what your theoretical mathematical formula gives me:

I =Ixx=Iyy= 1/3 *(t*y^3 + a*(a -y)^3 - (a-t)*(a-y-t)^3)      with      a = 150 mm

                                                                                                      t = 15 mm

                                                                                                      y = 150 -43.026 mm = 106.974 mm

I =   9112722 mm^4  (compare this to the value in the screenshot)

Cheers

Bob

[#7]

Hi Bob,

Thank you for you reply this is exactly what I was lloking for.

Regards

Rich

[#8]

HI AI,

I HAD MODELED FULL PUMP ASSEMBLY. HERE IAM IN NEED TO HYDRALIC FLOW VOLUME (WATER  ). FOR THAT I DERIVED THE WHOLE ASSEMBLY AND REVOLED A SKETCH. THEN BY COMBINE OPTION , I HAD DONE THE BOOLEAN OPERATION . PLEASE LET ME TO KNOE WHY IT TAKES SO MUCH TIME .

[#9]

Hi! The issue you are raising here does not seem to be related to the original thread here. Do you mind starting a new thread with an example exhibiting the behavior?

Thanks!

[#10]

Alternately, you could have used the parallel-axis theorem: I = I_overbar +A*d^2