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Moment of Inertia

  1. Apr 9, 2004 #1
    If I have a Rectangle h high, b wide, I thought I could calculate 2nd Moment of Area, I , about its neutral axis this way: taking the top half of the rectangle, the area is bh/2, and then multiplying this by the distance from the centroid of the top half to the neutral axis, squared ie (h/4)**2. Then because there are two halves of the rectangle, doubling the answer. This all gives I=(bh**3)/16.

    My problem is reconciling this with the answer obtained by integrating
    (y**2) b dy from -h/2 to h/2, which gives I=bh**3)/12.

    I generally thought the concept of second moment of area was an area times the distance of the centroid of the area to an axis it is rotated about, squared. Now I'm not so sure.

    Any assistance appreciated.
     
  2. jcsd
  3. Apr 9, 2004 #2
    [tex] I=\int _{S} r^2 dS [/tex]

    This is the definition, so your second calculation is correct.

    Although this looks much like the definition, and yields the correct units, it differs from it because you multiply instead of integrate. If you insist using multiplication of some sort of 'mean distance' (it's called the radius of gyration) with an area, you should not use h/4. Because the definition involves r^2 instead of r you should use the 'root mean square' instead of the mean distance for your radius of gyration. But calculating the root mean square involves another integration, so I'd just stick to the above integral...
     
  4. Apr 9, 2004 #3
    Thank you da Willem. My concepts were not translating to accurate definition. So thanks for clarifying this for me.
    Imstat
     
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