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Thread: 2nd Moment of Area

  1. #1
    Super Member
    Dec 2008

    2nd Moment of Area

    Hello. I'm having a bit of a problem calculating the 2nd moment of area of 2D shapes etc.

    The Question:

    Calculate  I_x for the section sketched in Figure 1 to within about 2%.

    (I have attached the figure as a JPEG - all measurements are in millimetres).

    Useful equations

    In my notes, there are 3 moments of area defined:

     I_z = \int_A y^2 \,dA

     I_y = \int_A z^2 \,dA

     J = I_x = \int_A r^2 \,dA = \int_A (y^2+z^2)dA = I_z+I_y

    Where J is the POLAR 2nd moment of area.

    My Instructor's Solution

    I_{zz}  = \bigg[ \frac{1}{12} \times 300 \times 25^3 + 300 \times 25 \times \bigg(\frac{500-25}{2}\bigg)^2\bigg] \times 2 + \frac{1}{12} \times 12 \times 450^3

    = 9.38 \times 10^8 \, mm^4

    My Confusion

    1) In my instructors solution he calculated  I_{zz} which is a notation he has never used in his notes? Does this mean the same as  I_x , which is what the question asked for? Or has he calculated what he calls I_z in his notes?

    2) In his solution he seems to be pulling figures straight out of his arse, and there is no integration involved at all, despite the fact that the definitions of 2nd moments of intertia are all integrals.

    Can anyone offer an explanation?
    Attached Thumbnails Attached Thumbnails 2nd Moment of Area-untitled.bmp  
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