Castigliano's Second Theorem on Curved Structures

  • Thread starter Triskelion
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  • #1

Homework Statement


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


Homework Equations


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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?
 

Answers and Replies

  • #2
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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.
 
  • #3
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.
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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