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Castigliano's Second Theorem on Curved Structures

  1. Jan 27, 2015 #1
    1. The problem statement, all variables and given/known data
    A semi-circular ring of stiffness EI and radius R is supported on an anchored hinge and on a roller hinge. A vertical load F is applied at the center. Determine the horizontal displacement of the roller support. 62QT1EH.jpg

    2. Relevant equations
    vVmJPC5.jpg

    3. The attempt at a solution
    So I apply a horizontal force H (to the right) at the roller. Sectioning the first quadrant gives [tex]M_1=HR sinθ_1[/tex] (moment is taken as positive clockwise). So [tex]\frac{∂M_1}{∂H}=R sinθ_1[/tex]. Similarly, [tex]M_2=HR cosθ_2-FRsinθ_2[/tex] and [tex]\frac{∂M_2}{∂H}=R cosθ_2[/tex]. Because H is an imaginary load and setting it to 0, [tex]M_1=0[/tex], [tex]\frac{∂M_1}{∂H}=R sinθ_1[/tex], [tex]M_2=-FRsinθ_2[/tex] and [tex]\frac{∂M_2}{∂H}=R cosθ_2[/tex]. So the integral is [tex]\int_0^ \frac{π}{2} \frac{M_1}{EI} \frac{∂M_1}{∂H}R dθ_1 + \int_0^ \frac{π}{2} \frac{M_2}{EI} \frac{∂M_2}{∂H}R dθ_2[/tex]. The first integral is zero. The second gives [tex]\int_0^ \frac{π}{2} \frac{-FR^3}{2EI} sin2θ_2 dθ_2=\frac{-FR^3}{2EI}[/tex]. I checked my workings, and I have no idea why there is a negative sign there. If my understanding is correct, the negative sign implies roller movement to the left, and this is really counter intuitive. Can anyone shed some light on this?
     
  2. jcsd
  3. Jan 28, 2015 #2
    It might be clearer if you include a free body diagram for the derivation of your equations but my first comment is that M2 doesn't seem right.
     
  4. Jan 29, 2015 #3
    Hey, thank you for the hint. It was like an epiphany. Turns out that I've forgotten the reaction forces all along. Funny how the human mind works, one can keep checking the thing for days without figuring what's wrong.
     
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