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The Reciprical Theorem - Beams

  1. Apr 6, 2008 #1
    I am trying to learn and figure our the reciprical theorem when applied to beans and the deflection on a beam when a load is applied.

    The beam would be x long, say 10m and would have a weight/load applied at 5m, or dead center. There would be point A (Dead Center) point B (Half way between dead centre and end) or 1/4 beam. Load applied to A and deflection at a and b needed, and then load applied at B and deflection and A and B.

    I need to create a graph with a slope of load cases to varify the theorem.

    I am a little lost on this one, i know what the theorem is, but i am not sure how to apply it to beam deflection and how to create a graph.

    Does anyone know any good links to learn further or some examples.

    This is needed for a test in 48 hours, and i am lost on it completely, the graph part especially

    Thanks for any help
     
  2. jcsd
  3. Apr 6, 2008 #2
    Are you useing the virtual work method to apply a unit load (in the virtual system) and find the deflection at "i" due to a load at "j"? The method I use is, find the virtual bending moments, find the real bending moments, and integrate the product of these 2 and divide by EI to find the flexibility coefficient (fij).

    If you use this method so I can help you in finding the displacements and applying the reciprocal theorem but I haven't had anything to do with graphing a solution.
     
  4. Apr 11, 2008 #3
    Yeah,

    What i am trying to do is find the deflection of 2 when a weight is applied to 1, and the deflection of 2 when weight applied to 2.

    I have the equation for weight applied at 1, and deflection at 1.

    1 is the dead center, 2 is 2/4 or 1/4 of the lengh.

    Any help as i say, is very appreciated.
     
  5. Apr 11, 2008 #4
    I tried to post a link up to the image on another site so you didn't have to wait for the attachments to be approved but i cant do that until I have 15 posts apparently.

    What I have done in the first image is drawn the beam and bending moment diagrams under one unit of load in both the real and virtual system (you might want to check the bmd's, it was a quick sketch).

    In the real system, we are applying a load at L/4, but we want the deflection at L/2 so we apply a load of 1 unit in the virtual system.

    So if l/4 is point B and L/2 is point A, then you will be trying to find fab. (That is, deflection at A due to load at B)

    I have uploaded a table to help you with this. you can do it with integration which is essentially what the tables do, but since you have to break the intergral up into 3 segments you get alot of ugly algerbra to deal with. Its not difficult maths, but it is very easy to make a mistake. You will still need to do the problem in 3 parts with the tables, but it should work out a bit easier/neater.

    Once you get a value for fab, then useing the reciprical theorem you have also found fba. This implies that a the deflection at A due to a unit load at B is the same as the deflection at B due to a unit Load at A. fij=fji

    Since you would have evaluated the flex coeficient in terms of unit loads, all you have to do is multiply your coeficient by P to find the deflection. Ie delta=P*fij. This is true because we assume we are dealing with a linear system (linear analysis).

    I hope I haven't confused you more, it really isn't to hard of a concept to understand after a few problems but as in the example you provided there is often alot of work if you choose to do all the calculations useing integration. Half the problem is generating an appropriate BMD with the correct sign convention in both cases (I dont think it matters what convention you use, as long as you are consistant in both systems).

    Have a go at solving the system again and if you have trouble I will provide a worked solution.
     

    Attached Files:

  6. Apr 12, 2008 #5
    Hey, thanks for that, i think i understand it all, what i will do is have another go and post the results/equations i have used...

    It would be great if you could confirm for me that what i have done is correct when i post it.

    I will wait for the attachments to be approved, and then have a go. I think i understand more from just reading your text.

    Your text also helps me solve the other problem, recipricol theorem.

    Thank you lots.
     
  7. Apr 12, 2008 #6
    No problem - I am currently covering this material for an exam I have on Friday so it will be good practice for me :smile:. Ill wait for your solution and then I will work through it myself and we can compare results, sounds like a good plan to me.

    Elbarto
     
  8. Apr 14, 2008 #7
    Hey...

    I am stuck, really stuck, i have read the stuff you posted, looked over google, and i just cant seem to derive a formula to solve my problems.

    I would be so so greatful if you could post the worked example, i would really appreciate it.

    What i need this for is inclusion in a project i am doing in C Programming, i am a Software engineer, not mechanical and haveing real issues.

    All i seem to do is the maximum deflection.

    My case's are however.

    Point 1 = 0.5m
    point 2 = 0.75m

    Beam = 1m

    Simple support

    case's

    1) Load at 1, what is deflection at 1 and 2?

    2) Load at 2, what is the deflection at 1 and 2

    3) Load at 1 and 2, what is the deflection at 1 and 2

    Load = 3kg or 3 * 9.81 = 29.43newtons?

    I have the real life data, so if you post the worked example, i can give you the "real" figures and compare them to the worked example?

    I have to show this to my tutor on Wednesday, so i will obtain information on if the formulas and answers are corrent and report back.

    I feel bad that i have nothing to post, however, i really do need and appreciate the help and really am stuck.

    Thanks again.
     
  9. Apr 14, 2008 #8
    I derived both these values from the equation sheet I posted, I haven't got time at the moment to post a full solution but here you are. If you apply a load (any point load, positive or negative in value applied at point A or point B), you can find the deflection for that particular case using these 3 equations (well 4 equations really, since f_ab=f_ba).



    [tex]
    \delta_{aa}=P \cdot f_{aa}=\frac{PL^3}{48EI} [/tex]
    [tex]
    \delta_{ab}=P \cdot f_{ab}=\frac{PL^3}{64EI}
    [/tex]
    from the reciprocal therom
    [tex]
    \delta_{ba}=P \cdot f_{ba}=\frac{PL^3}{64EI}
    [/tex]
    also to solve the deflection at b when load is applied at b we need 1 more coeficient
    [tex]
    \delta_{bb}=P \cdot f_{bb}=\frac{3PL^3}{256EI}
    [/tex]
    If you want the delection due to a load P at A, use the top equation. Then, if there is also a load at B, use the second equation and add to 2 values together to show you the total deflection at point A (the centre of the beam).

    note here that we derived the flexibility coeficient [tex]f_{ab}[/tex] for a load applied at L/4, this value will also be true for a load applied at 3L/4 (symmetrical loading properties).

    If you know I and E, you have already found your load P and you have your legnth, its just a matter of putting the numbers together for each case.

    so for the first part of case 3 (point 1 = center = point a in my workings)
    [tex]
    \delta_{a}=\frac{PL^3}{48EI}+\frac{PL^3}{64EI} [/tex]
    for case 2
    [tex]
    \delta_{b}=\frac{3PL^3}{256EI}+\frac{PL^3}{64EI} [/tex]

    Since you need to know the basic process for you assignment, here is how I solved for f_bb

    draw the BMD's under a unit load(both virtual and Real)
    >you will see they are identical
    >each gBMD has to be broken up into 2 triangles
    >use the sheet supplied to integrate the area of triangle #1 in the real with triangle #1 in the virtual
    >do the same for the second triangle and sum the 2 values.

    the sheet say the formuala for 2 triangles is [tex]\frac{LAC}{3}[/tex]
    so, by applying this to both parts of the BMD (real and virtual)
    we get
    [tex]f_{bb}=\frac{1}{EI} \cdot (\frac{1}{3} \cdot \frac{L}{4} \cdot \frac{3L}{16} \cdot \frac{3L}{16}+\frac{1}{3} \cdot \frac{3L}{4} \cdot \frac{3L}{16} \cdot \frac{3L}{16})=\frac{3L^3}{256EI}
    [/tex]

    Note: L in the equation is the segment of the beam you are considering, not the length of the beam itself. Also note how I have 2 parts in this equation, this is because I had to break the problem into 2 segments.

    I hope that helps a little bit, I might be able to do a full worked solution for one flex coeficient tonight if your still haveing trouble but im a bit pressed for time at the moment (I will warn you that there is quite a bit of mucking around with [tex]f_{ab}[/tex], its not hard its just there is alot of working as you have to solve it in 3 segments so if you are working on that one keep that in mind).

    Good Luck
    Elbarto
     
  10. Apr 14, 2008 #9
    EDIT:

    When I say that L is the length of the segment of the beam you are considering, this is only for the term "L" given in the tables, where you have to substitute L/2 in for L if you break the BMD up into equal sections etc.

    The "L" in the final equations like I have derived will correspond to the full length of the beam. Sorry for the confusion.
     
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