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Second moment of area - Need walkthrough

  1. Mar 12, 2009 #1
    Hi all,

    I'm at uni starting off engineering, and I'm looking for a walk through in this algebra involved in the second moment of area. While I know it's probably pretty basic I'm undertaking a bridging course to try and keep up with my maths.

    http://home.exetel.com.au/peleus/smoa.jpg

    Here's a picture of the most relevant lecture slide discussing the problem.

    I'll type out the steps they undertook to get the final formula for the second moment of area for a rectangle.

    On the next page we take the integral of this, which I can do fine.

    This gives

    1. [tex]I = \frac{b}{3}[y^3][/tex] with limits +h/2 and -h/2

    Taking it further we end up with

    2. [tex]I = \frac{b}{3}[\frac{h^3}{8}-(-\frac{h^3}{8})][/tex]

    and finally we take it to the step

    [tex]I = \frac{bh^3}{12}[/tex]

    Ok, I can understand a bit about this but here are my questions.

    - Why are the limits h/2 and -h/2, isn't this simply the middle of the rectangle?
    - Why is [tex]dI = y^2 dA[/tex], where does the [tex]y^2[/tex] come from?

    Any help is greatly appreciated.
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Mar 12, 2009 #2

    minger

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    Because its the second moment of area. The first moment of area is defined as:
    [tex]M_x = \int_A y\,dA[/tex]
    and is commonly used to find the centroid of an object.

    It's y² simply by definition.

    edit: I don't quite understand your first question. I'll take a stab though. The limits are h/2 because you're finding the moment around the centroid of the object. If you were finding it about an axis on the base of the part, the limits would be 0->h. For practice, try finding the second moment of area about an axis other than the the centroidial axis.

    hint: the area moment of inertia is 4 times greater about the base rather than the centroid (having y squared helps!).
     
  4. Mar 12, 2009 #3

    nvn

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    peleus: Answers to your two questions:
    - The limits are +/-0.5*h because you are computing second moment of area about the horizontal centerline of the rectangle, in this case.
    - Moment (M) is summation of moment arm y times force on each infinitesimal area. But force on each infinitesimal area is stress times dA. However, linear elastic stress at any point on a cross section is a linear function of y times the extreme fiber stress. Now substitute all of this together and notice within this expression for M you have summation of y*y*dA, which is called second moment of area.
     
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