Moment of Inertia Clarification

  • #1

Main Question or Discussion Point

Hi all. It's been a while since I've taken mechanics of solids, I was hoping I could get some confusion ironed out here.

I have a long thin plate with C channels welded along its length, below is a cross sectional view.
[PLAIN]http://img375.imageshack.us/img375/6672/channelexcel.jpg [Broken]
As you can see, I have placed my axis about the neutral axis of the plate as well as its left edge. In my scenario I have an upwards pressure load on the bottom face of the plate.

I treated this entire setup as multiple rectangular bodies, computed the MOI about their centroids then used the parallel axis theorem to relate them to my chosen axis and combined them for the overall Ix and Iy about that origin.

My problem arises when I don't comprehend what I have found. Is Ix the resistance of this entire cross section bending in and out of the page about the x axis? So for example, a vertical strip would become a 3 dimensional U shape with the "prongs" facing the viewer?

If so, this is not what I am looking for. I need the resistance of the above shape transforming into a "rainbow" shape; it's resistance to buckling. Is this Ixy?
 
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Answers and Replies

  • #2
nvn
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laforzadiment: No, you need Ix about the cross section neutral axis, not about your xy coordinate system axes. Did you do this correctly, to already obtain Ix about the cross section neutral axis? If so, that is what you need.

You do not need product of inertia, Ixy.

No, Ix will not cause a vertical strip to become a U shape with the prongs facing the viewer. It will instead cause a vertical strip to rotate about the horizontal neutral axis. We cannot really say what shape your structure will transform into, because we do not know your boundary conditions. But in general, Ix will allow your cross section to deflect vertically upward.
 
  • #3
laforzadiment: No, you need Ix about the cross section neutral axis, not about your xy coordinate system axes. Did you do this correctly, to already obtain Ix about the cross section neutral axis? If so, that is what you need.

You do not need Ixy. Product of inertia is zero here.

No, Ix will not cause a vertical strip to become a U shape with the prongs facing the viewer. It will instead cause a vertical strip to rotate about the horizontal neutral axis. We cannot really say what shape your structure will transform into, because we do not know your boundary conditions. But in general, Ix will allow your cross section to deflect vertically upward.
Thanks for your reply. Ok, so you're saying that I cannot arbitrarilly place the axis and that it must go at the centroid of the mass as a whole? Alright, I'll update my spreadsheet.

Consider the left and right end to be fixed in place so any upward uniformly distributed load will exclusively cause first order bowing until failure.

In the text you wrote it seems like you were saying Ix resists this motion:
[PLAIN]http://img806.imageshack.us/img806/9324/85217762.jpg [Broken]
If this is the case, wouldn't I need Iy?
 
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  • #4
nvn
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laforzadiment: No, what you drew in post 3 is bending rotation about the z axis. Ix resists rotation of the vertical strip about the x axis; therefore, the top of the strip would move toward the viewer, and the bottom of the strip would move away from the viewer.

Because the left and right edges of your structure are fixed, your structure will deflect into somewhat of a rainbow shape, in the view you showed in post 1. But it will do this mainly because of Iz, instead of Ix. Ix would be more important if your structure were constrained along its front and back edge. So you need Iz of the yz cross section about its z neutral axis.
 
  • #5
Ah I gotcha now. Awesome, thanks for the help!
 
  • #6
86
1
I treated this entire setup as multiple rectangular bodies, computed the MOI about their centroids then used the parallel axis theorem to relate them to my chosen axis and combined them for the overall Ix and Iy about that origin.
Parallel axis theorem WON'T work in this case. The theorem only works when the sectional property is uniform over the span of the bending member. You have non uniformity, as each segment has different properties from its adjacent neighbors.
 
  • #7
Parallel axis theorem WON'T work in this case. The theorem only works when the sectional property is uniform over the span of the bending member. You have non uniformity, as each segment has different properties from its adjacent neighbors.
I don't follow. The area is uniform in and out of the page, I follow the same method used for a T or I-Beam just with more components.
 
  • #8
86
1
Oops. I mistakenly thought you were trying to determine the stiffness from left to right of the picture. Evidently from your comment your interest is perpendicular. In that case, I agree the theorem does apply.
 
  • #9
Oops. I mistakenly thought you were trying to determine the stiffness from left to right of the picture. Evidently from your comment your interest is perpendicular. In that case, I agree the theorem does apply.
No prob :) it can be difficult to convey the original problem setup only through text.
 

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