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Moment of inertia of a beam

  1. Dec 6, 2013 #1
    1. The problem statement, all variables and given/known data
    I'm supposed to solve for the maximum moment assuming the beam bends about the y axis (not the z axis as shown in the image. Same image for different questions). I don't understand how to find the moment of inertia in this case. The solution gives the moment of inertia for the 80 x 16 mm part of the bar to be
    [tex] I =\frac{1}{12}(80)(16^3) + 80(16)(16^3)[/tex]
    What I don't understand is where the 80(16)(163) comes from. Can some one explain?
     

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  3. Dec 6, 2013 #2

    SteamKing

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    I don't understand either. Sometimes, the authors of these solutions use obscure methods to find the answers. If in doubt about the I calculation, use a method which you understand better.

    You could calculate I for an 80x48 rectangle and then subtract out the I for each of the 24x16 insets:

    Iy = (1/12)*80*48^3 - 2*(1/12)*24*16^3 mm^4

    This is a simple method because the parallel axis theorem is not required (everything has the same centroidal location w.r.t. the y-axis.
     
  4. Dec 6, 2013 #3
    It's the stupid parallel axis theorem. I've only used that a few times before so I had completely forgotten about it. Your post reminded me of it. The second area moment of inertia using the parallel axis theorem is
    [tex] I = I_x+Ar^2[/tex]
    here A is the area of the region. Anyways thanks for the help
     
  5. Dec 6, 2013 #4

    SteamKing

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    The only problem is, it doesn't match the calculation in the OP.

    If you wanted to split the Iy calculation into three segments, then you would have the following:

    center segment: A = 16 * 32 mm^2 Iy = (1/12)*32*16^3 mm^4; no PAT required

    upper, lower segments: A = 16*80 mm^2 Iy = (1/12)*80*16^3 mm^4
    PAT = 16*80*24^2 mm^4,

    So, the total Iy =

    (1/12)*32*16^3 + 2*[(1/12)*80*16^3 + 16*80*16^2] = 720,896 mm^4

    Alternately, by subtracting the two 16 x 24 mm cutouts from the 80 x 48 mm rectangle:

    Iy = (1/12)*80*48^3 - 2*[(1/12)*24*16^3] = 720,896 mm^4
     
  6. Dec 7, 2013 #5
    I realize I made a mistake in the OP. It's supposed to be 80(16)(16^2). However I like your second method better than the PAT.
     
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