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Castigliano's Theorem on Curved Beams

  1. May 10, 2007 #1
    1. The problem statement, all variables and given/known data

    You have a semi-circular cantilever with a point radial load at its free end.
    i.e load acting down from top of semi circle, with bottom attached to something.

    Calculate the equations for the horizontal and vertical deflections taking into account both bending moments and axial loads using castiglianos theorem

    2. Relevant equations

    -

    3. The attempt at a solution

    Ok I gave it a go and compared it to some experimental data and found I was a bit off.

    I got dV=LR^3/2EI
    with L = load
    R = Radius
    E = Youngs Modulus
    I = moment of inertia bh^3/12
     
  2. jcsd
  3. May 10, 2007 #2

    Pyrrhus

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    You could show your attempt so we can determine where you went wrong.
     
  4. May 10, 2007 #3
    Well here is how I got it:

    M(theta)=L*R*Sin(theta)
    dM/dL=RSin(theta)

    dV=dU/dL=d(int(M^2/(2*E*I)*r*dTheta,Pi->0)/dP
    =1/EI * int(M*dm*R*dTheta,Pi->0)

    Substituing values for M(theta) and dM

    PR^3/EI * int(sin^2(theta)dTheta,Pi->0)

    dV=PR^3/2EI
     
  5. May 11, 2007 #4

    Pyrrhus

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    Hey robip, a few questions when you mean the load is radially wouldn't the load be horizontal (because of the geometry properties of a semi circle)?

    Also would you care to post a diagram with the geometry of the problem?, i'm not seeing what are the variables on your work, besides the obvious L.
     
  6. May 11, 2007 #5
    [​IMG]
    Here is a pic of it
     
  7. May 11, 2007 #6

    Pyrrhus

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    Vertical Displacement case:

    So for a thin curved-beam, taking as our ind. variable theta and L will be the force by Castigliano, then we have:

    [tex] M(\theta) = (-R \sin \theta)(L) [/tex]

    [tex] M(\theta) = (-R \sin \theta)(L) [/tex]

    [tex] \frac{\partial M}{\partial L} = (-R \sin \theta) [/tex]

    Ok, then L = P,

    [tex] M(\theta) = (-R \sin \theta)(P) [/tex]

    [tex] \frac{\partial M}{\partial L} = (-R \sin \theta) [/tex]

    and

    [tex] \Delta = \int^{\theta}_{0} M (\frac{\partial M}{\partial L}) \frac{R d \theta}{EI} [/tex]

    [tex] \Delta = \int^{\pi}_{0} ((-R \sin \theta)(P)) (-R \sin \theta}) \frac{R d \theta}{EI} [/tex]

    [tex] \Delta_{free-end} = \frac{P \pi R^{3}}{2EI} [/tex]

    Anyway, it looks like your work is accurate. This is what i got.
     
    Last edited: May 11, 2007
  8. May 11, 2007 #7

    Pyrrhus

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    You forgot the pi in your answer.
     
  9. May 11, 2007 #8
    i can't believe i didn't notice all my values were out by about 3... :rofl:

    so for the horizontal i would just use Cos instead right?
     
  10. May 11, 2007 #9

    Pyrrhus

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    More like [itex] R - R \cos \theta [/itex].
     
  11. May 11, 2007 #10
    oh yea, i see, thanks so much for the help!

    really appreciated
     
  12. May 11, 2007 #11

    Pyrrhus

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    No problem :biggrin: , and welcome to PF!.
     
  13. May 12, 2007 #12
    hmmm i just realised it is also asking for the axial load to be included in the equation, how would i calculate the axial load at any point with respect to angle theta?
     
  14. May 12, 2007 #13

    Pyrrhus

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    Use your geometry knowledge to decompose the axial component of the stresses in x and y, and then do sums of forces in y and x.

    Remember the axial component is tangential to the circle, and the radius is perpendicular to tangents.
     
  15. May 17, 2007 #14
    hi robip,

    I would be interested in seeing the rest of your workings and results for this particular problem as i am doing something similar atm.

    Much appreciated

    Cheers
     
  16. May 17, 2008 #15
    Hi robip,

    Are you a student at curtin university of technology in Perth, Australia, by any chance? :)

    .Ksum006
     
  17. May 25, 2008 #16
    OK, so for horizontal this is what i got:

    (3*pi*P*R^3)/(2EI)

    I did this using the same process as the vertical, but using M(theta)=(R-Rcos@)*P.

    Is the above answer correct? im not very good at calulus and this is where i may have come stuck:

    dh=(PR/EI)*[Integration between Pi and 0 of (R-Rcos@)^2.d@]
     
  18. Jun 17, 2009 #17
    I have a similar question but with a 3/4 cantilever. The answers are on the attachment but can't get them using the formula supplied - Any ideas?
     

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  19. May 19, 2010 #18
    i seem to be doing the same question and i am able to do the first part of the vertical displacement, as stated above, but i cant get an equation for the axial load in terms of the applied load P. and having it as a function of theta.
     
  20. Jul 20, 2010 #19
    anybody can help me how to derive an equation of deflection of circular bar by using castigliano theorem ???...the final answer is like this:

    δ_v=∂U/(∂P_v )=(P_v R^3)/EI (π/4-2/π) )
     
  21. Jul 25, 2010 #20

    Pyrrhus

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