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Statical moment of area, Q, for Semi-Circle

  1. Jul 22, 2009 #1
    Question Exactly from Text
    Determine the location e of the shear center, pint O, for the thin-walled member having the cross section shown.

    http://img268.imageshack.us/img268/8732/scannedimage058.jpg [Broken]

    Work Done So Far
    I'd like to find the statical moment of area, Q, for a semi-circle in general; using this I should be able to generate Q = Q(R_outer) - Q(R_inner); I've been provided with such a Q as a hint but have failed to produce this value on my own ( 1/3 * (R_outer^3 - R_inner^3)) sin (theta) ).

    I begin by determining the area A' by noting that its:
    (area of circle)/4 - (area of circle in proportion to theta) - (triangle underneath shaded area)
    http://img268.imageshack.us/img268/34/scannedimage059.jpg [Broken]
    I took its derivative and use C for theta:
    http://img268.imageshack.us/img268/2466/scannedimage060.jpg [Broken]

    I used a calculator for integration.

    What am I doing wrong?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 23, 2009 #2

    nvn

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    lizzyb: Your integral looks correct. But I didn't understand yet why you are integrating from zero to theta. Maybe I'm missing something, but shouldn't that be integrated from theta to 0.5*pi?

    I am currently getting Q(theta) = 0.3333(r_outer^3 - r_inner^3)*cos(theta)^3. I currently have no idea how they obtained that hint value for Q; so far, it does not look correct. Are you relatively certain the hint value is correct?
     
  4. Jul 23, 2009 #3
    Its correct in that I dutifully copied the hint value but perhaps someone somewhere messed up; I don't see how they got it.

    The hints assign r_i = r - t/2 and r_o as r + t/2.
     
  5. Jul 25, 2009 #4

    nvn

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    lizzyb: Did you uncover any other information regarding why they claim Q is that hint value you listed in post 1?
     
  6. Jul 25, 2009 #5
    This is from the solutions:
    http://img195.imageshack.us/img195/9210/scannedimage061.jpg [Broken]
    http://img188.imageshack.us/img188/1713/scannedimage062.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
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