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Rotated Section - Moment of Inertia

Rotated Section - Moment of Inertia

Rotated Section - Moment of Inertia

(OP)
Hi everybody:

I can see that there are formulas to calculate the moment of inertia of a 2D Area (Second moment of area) here: http://en.wikipedia.org/wiki/Second_moment_of_area

In the same link, there is a formula to calculate the Inertia moments about a ROTATED AXIS (Axis Rotation). But, is there a formula to calculate the Inertia moments of a ROTATED SECTION (rotated area) ??

Thanks in advance.
Eduardo

RE: Rotated Section - Moment of Inertia

What do your Strength of Materials Book equations show?

Mike McCann
MMC Engineering

 

RE: Rotated Section - Moment of Inertia

i don't understand the question ... the difference between rotated axes and rotated section ??  how does it matter if you rotate the axes about a fixed object, or rotate the object about fixed axes ??

RE: Rotated Section - Moment of Inertia

I agree with rb.  It's all relative. Ix of a WF oriented strong is the same as Iy of the same WF oriented weak, as long as the global axes, x and y, are fixed.

RE: Rotated Section - Moment of Inertia

(OP)
Ok, i will explain a little bit:

1. You have a 2D Area in a XY axis. You can calculate the Moments of Inertia Ix, Iy, Pxy.
2. You rotate the 2D Area around the origin, with a tetha angle.
3. So, known Ix, Iy, Pxy and tetha: Is there a formula so i can calculate the new Ix, Iy, Pxy around the same axis?

In wikipedia, you can see that there is a way of calculating the Ix', Iy' and Pxy' around the rotated axis, but that is not what i want, because i am not rotating Axis. I am rotating only the 2D Area and i want calculate Ix, Iy, Pxy in the same axis based on the previously Ix, Iy, Pxgy calculated.

Any ideas?

RE: Rotated Section - Moment of Inertia

are you rotating about the origin or about the centroid ?

rotating about the origin is the same as rotating the axes (only in th eopposite direction).

if rotating about the centroid (ie so a square rotates to become a diamond) i suspect you could account for this by using the parallel axis theorem to move the origin to the centroid, rotate the axes (= rotating the shape), and parallel axis theorem again to move from the centroid to the origin

RE: Rotated Section - Moment of Inertia

Can you please provide a sketch?  

RE: Rotated Section - Moment of Inertia

Quote:

So, known Ix, Iy, Pxy and tetha: Is there a formula so i can calculate the new Ix, Iy, Pxy around the same axis?
I assume Pxy is Ixy and tetha is theta, right?

The answer is yes, there is and you have found it in the Wikipedia article you quoted earlier.  The only thing you must do is change the sign of theta and press on.

BA

RE: Rotated Section - Moment of Inertia

As stated by rb, rotating the section relative to the axes is no different in principle than rotating the axes relative to the section.  The only difference is that the angle changes sign.

BA

RE: Rotated Section - Moment of Inertia

As einstein said, everything is relative and there are no truly fixed points.

In the space time continuum rotating the axis is the same as rotating the section in the opposite direction.


That is unless you are down a black hole and then ..who knows!

RE: Rotated Section - Moment of Inertia

'cept if you're rotating the shape about it's centroid ... does a square have the same Ixx and Iyy as a diamond ?  of course, they both have the same principal I.

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