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Finding second moment of inertia about x axis

  1. Dec 3, 2015 #1
    1. The problem statement, all variables and given/known data
    The cubic is divided into 4 parts , A, B and C, D , each with thickness of 1mm , i am sked to find thesecond moment of inertia about x -axis

    2. Relevant equations


    3. The attempt at a solution
    i' m using the formula Ixx = Ix +A(d^2)

    for part CD, Ixx = 50(1^3) / 12 + 50 ((0.5-0.5)^2)
    or Ixx = 50(1^3) / 12 + 50 ((25-0.5)^2)
    is correct ?
     

    Attached Files:

  2. jcsd
  3. Dec 3, 2015 #2
    Can anyone tried to ans? I need it urgently.
     
  4. Dec 3, 2015 #3

    SteamKing

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    If you're trying to calculate the second moment of area about the x-axis, no, your calculations are not correct.

    Hint: Since everything is symmetrical about the x- and y-axes and no material is located inside the 1 mm thick plates, wouldn't it be easier to dispense with the parallel axis theorem and just subtract the MOI of the interior from the MOI of the exterior?
     
  5. Dec 3, 2015 #4
    Ixx = 50(1^3) / 12 + 50 ((25-25)^2) this is the correct one??
     
  6. Dec 3, 2015 #5

    SteamKing

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    Like your original calculations, that expression doesn't make sense

    What is (25 - 25) ?
     
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