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Homework Help: 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
    Last edited by a moderator: May 2, 2017
  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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    Last edited by a moderator: Apr 25, 2017
  22. Apr 24, 2011 #21
    I have the exact same problem as robip. "Calculate the equations for the horizontal and vertical deflections taking into account both bending moments and axial loads using castiglianos theorem".

    I have read the website http://school.mech.uwa.edu.au/~dwrig...thin.html#thin. but I am still not 100% certain of my answer.

    My attempt at the solution.
    For vertical deflection due to axial load:
    Paxial = Psin(theta)
    dP(axial)/dP = sin(theta)

    vertical deflection= (1/EA) * int( Paxial * dP(axial)/dP * R )d(theta) with limits 0->Pi
    = (1/EA) * int( Psin(theta)*sin(theta)*R )d(theta) with limits 0->Pi
    =(PR/EA) * int( (sin(theta))^2 )d(theta) with limits 0->Pi
    =(PR/EA) * (Pi/2)
    =(P*PI*R/2EA)

    For horizontal deflection due to axial load:
    Paxial = Pcos(theta)
    dP(axial)/dP = cos(theta)

    horizontal deflection= (1/EA) * int( Paxial * dP(axial)/dP * R )d(theta) with limits 0->Pi
    = (1/EA) * int( Pcos(theta)*cos(theta)*R )d(theta) with limits 0->Pi
    =(PR/EA) * int( (cos(theta))^2 )d(theta) with limits 0->Pi
    =(PR/EA) * (Pi/2)
    =(P*PI*R/2EA)

    I.e. Horizontal deflection and vertical deflection due to axial loading is exactly the same.

    Verification of my solution would be very much appreciated.
     
  23. May 1, 2011 #22

    Pyrrhus

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    I am a little rusty as when I solved this problem I had just graduated with BS in Structural Engineering. However, I went to graduate school in economics (funny, eh?).

    I remember Castigliano Theorem(s) fairly well, and from my solution of vertical displacement, I can tell your vertical displacement is wrong. Your lever arm is wrong! you forgot another R!, also your partial derivative inside the integral is wrong.

    For the second part,

    I believe you don't need to take into "account" the axial load. Take the reference point with respect to the edge of the section, and the apply Castigliano again by putting an horizontal fictious load. The lever arm should be the one I gave in a previous post [itex]R-R \cos \theta [/itex]. The reason is that when you cut a section, you aggregate the stresses into equivalent Force-Couple systems. In this case, you decompose the force into a radial (shear force), and a tangential (axial force), and you also have the couple (or moment). If you take moment with respect to the section, then the lines of action of both shear and axial gives radius vectors of length 0, and thus moments of 0 with respect to it. However, the couple still counts.

    Good luck!
     
  24. May 4, 2011 #23
    I was able to get it in the end. Thankyou for your help!!
     
  25. May 17, 2011 #24
    If i wanted to calculate I in this problem would it simply be as follows:

    I = (pi*P*R^(3)) / (2E) ?

    and thus substituting values in if we take
    E = 69 GPa
    R = 150 mm
    P = 0.1 N

    we get I = 1.28 *10^(-14)m^(4)

    The answer i get in my textbook however reads - 1.63*10^(-28)m^4

    Am i doing something wrong?

    Thank you for your time.
     
  26. May 17, 2011 #25
    vertical deflection = (pi*P*R^3) / (2EI)

    Therefore,
    I=(pi*P*R^3) / (2E*vertical deflection)

    Hope that helps
     
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