moment of inertia

[grunt58]

Need help calculating moment of inertia. I need to flip my part and frame 90deg. Im hoping someone would be kind enough to reply with the formula needed to calculate it.

I think the uploaded drawing should be enough.

PS I know my CAD software (SolidWorks) can calculate this but SolidWorks can't this time because some of the parts have an assigned mass prop. and not a material.

Thanks

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[rb1957]

you've got an x-and y- axes system on the section ? just change the axis ...

Ix'x' = Iyy
Iy'y' = Ixx

no?

 

[hydtools]

If you have mass properties assigned, that should be sufficient without assigning a material. Assigning a material assigns mass properties for that material and you say you have mass properties assigned.

Do you know the rotation axis? Mass centers of gravity?

Ted

 

[grunt58]

It's going on a gantry robot and the wrist will flip up and down 90deg. I'm checking to make sure the part and framing don't exceed the moment rating of the quick adaptor. The quick adaptor mounts between the flat black plate and the gantry robots mounting interface. The robot will come in wrist flipped down and flip up to place the part.

Certified SolidWorks Associate
SW2009 X64 SP 1.0
Dell Precision T5400
Nvidia Quadro FX 5600
Xeon 2.5GHz Quad Core, 4GB RAM
XP Pro X64 SP2.0

 

[rb1957]

yeah, you have the arm in the horizontal position, with the x-axis running along it, and the y-axis normal.

if the arm is rotated 90deg, the horizontal x-axis inertia Ix'x' is the original Iyy ... no ?

 

[grunt58]

If I'm following you correctly then yes, X and Y would be the same since its flipping 90deg.

I don't know the formula to calculate the inertia.

Certified SolidWorks Associate
SW2009 X64 SP 1.0
Dell Precision T5400
Nvidia Quadro FX 5600
Xeon 2.5GHz Quad Core, 4GB RAM
XP Pro X64 SP2.0

 

[rb1957]

sorry, i thought you had the properties for the original position ...

Ixx = int(y^2dA) = sum(Io+y^2*A)

the 1st is mathematically correct, but not stunningly usefull.
the 2nd is for summing rectangles (or other convenient pieces of the section) ... Io is the self moment of the piece, A it's area, y it's distance for the NA.

 

[grunt58]

Can you break it down some more? What info am I missing on the attached drawing above. What gets plugged in where?

Certified SolidWorks Associate
SW2009 X64 SP 1.0
Dell Precision T5400
Nvidia Quadro FX 5600
Xeon 2.5GHz Quad Core, 4GB RAM
XP Pro X64 SP2.0

 

[rb1957]

looking at your pic, you're rotating the bucket, this is different than calc'ing I at 0deg and at 90deg. i remember something about "rotating frames of reference" but not much more than that !? i'd start with wiki (shiver)

 

[tygerdawg]

The easy way is to let SolidWorks calculate an estimate for you for the assembly. You'd just have to add an axis to your assembly from which to base the total inertia.

The rigorous way is to use the Parallel Axis Theorem and sum the individual component inertias. I don't have my references handy, but I found this on the net that will do well enough:

http://www.efunda.com/math/areas/ParallelAxisTheorem.cfm

In your situation,

M[small]i[/small] = mass of individual component

d[small]i[/small] = distance from individual component's centroid axis to the parallel axis

I[small]i[/small] = mass moment of inertia of individual components

I[small]total[/small] = [Σ]( I[small]i[/small] + M[small]i[/small]*d[small]i[/small]^2)

Of course you could estimate the extrusions as simple blocks...the difference won't be that much, and will give you a conservative estimate. It simply turns into a tedious arithmetic excercise.

TygerDawg
Blue Technik LLC
Virtuoso Robotics Engineering

http://www.bluetechnik.com