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Area moment of inertia-circular cross section

  1. Jul 7, 2004 #1
    Area moment of inertia--circular cross section

    From the bending beam calculation, the moment of inertia of the cross section with regard to a coplanor axis of rotation is used. If we have a circular "beam", the area moment of inertia of a circular disk of radius a about a diameter is [tex] I_d = \frac{\pi a^4}{4}[/tex] according to two separate references. I believe the integral involved can be generally stated as [tex]I_d = \int y^2 dA [/tex] if y is the distance to the diameter d perpendicular to y which diameter (as all diameters of a uniformly mass distributed disk) passes through the centroid of the disk.

    Now my stab at actually evaluating this is to do a double integral in polar coordinates and long story short the only way I can come up with the agreed upon answer is [tex] \int_0^\pi\int_0^a r^2\; r\; \mathrm{dr}\;\mathrm{d\theta}[/tex] where [tex] r\;\mathrm{dr}\;\mathrm{d\theta}=dA[/tex] and [tex]r^2[/tex] is the distance to the origin (centroid)

    If I were in a creative writing glass I might get a passing grade for this fudge but I really would like to understand what I am doing better than backing into an answer like this. Any help would truly be appreciated.
     
  2. jcsd
  3. Jul 7, 2004 #2

    arildno

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    Your formula is non-sense, and seems to have been contrived in order to get the agreed upon answer.

    Here's a correct derivation:
    We have:
    [tex]y=r\sin\theta[/tex]
    Hence, you have:
    [tex]I=\int_{A}y^{2}dA=\int_{0}^{2\pi}\int_{0}^{a}r^{3}\sin^{2}\theta{drd\theta}=[/tex]
    [tex]\frac{\pi{a}^4}{4}[/tex]
    as required.
     
    Last edited: Jul 7, 2004
  4. Jul 7, 2004 #3
    Thank you arildno,
    I did that integral sometime in the past 3 years and was racking my memory to come up with it; I found another nonsense solution that agreed with the known answer but as in the one I posted (after 2 days and maybe 6 hours surfing the web searching for answers), the limits of integration didn't make sense.
    Even so , the polar coordinate translations are basic and I am chagrined. It was not a waste however as this is the first time I have heard the term area moment of inertia even though I have done a lot of mass moments and radii of gyration calculations. The Euler-Bernoulli Beam Equation was under review here and it is fascinating:

    [tex]\frac{d^2}{dx^2}\left[E I \frac{d^2w}{dx^2}\right]=\rho[/tex] where E is Young's Modulus, I is the area moment of inertia, w is the out of plane displacement and [tex]\rho[/tex] is force acting downward on a very short segment and has units of Force per unit length (distributed loading). The x-axis is the lengthwise polar axis passing through the center of the beam. If E and I do not vary with x, then [tex] E I \frac{d^4w}{dx^4} = \rho [/tex] This is the first ODE I have come across that utilizes the fourth derivative and since the boundary conditions, depending on how the beam is supported include up to the third derivative, I wanted to make sure that I understood what the equation was saying and be able to relate to the area moment integral was critical. Thank you again.
     
  5. Jul 8, 2004 #4

    arildno

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    It's been quite a while since I did bending problems; however, if I remember correctly, 4th derivatives are rather common there.
    I think, for example, that the fourth order biharmonic equation occurs naturally (that is "the Laplacian of the Laplacian")
     
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