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Second moment area of symmetrical shape

  1. May 18, 2010 #1
    1. The problem statement, all variables and given/known data

    -a beam with square cross section, side lengths bXb is rotated 45 degree.
    -can we assume the second moment area (I) is the same if this is done?

    2. Relevant equations

    I=b^4/12 (cube)
    I=integral(r^2dr)

    3. The attempt at a solution
     
  2. jcsd
  3. May 18, 2010 #2
    If you know how the formula I=b^4/12 (cube) has been derived from an integral, then you should be able to do the same for the diamond shape. Or, if you accept the formula for a triangle and apply the parallel axis theorem, you can answer the question yourself. I don't see the relevance of I=integral(r^2dr).
     
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