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Calculating deflection

  1. Feb 3, 2014 #1
    Hi,

    I'd very much like to know how to calculate deflection of a beam by integrating when flexural rigidity EI is not constant. I tried finding an example on the internet but couldn't find any. Can anyone provide me with one? The problem is driving me nuts.
     
  2. jcsd
  3. Feb 3, 2014 #2
    Hi Samppa. Welcome to Physics Forums!!

    I assume you know EI as a function of distance x along the beam, correct? And, you know the bending moment as a function of x. You divide M by EI, and you have a new function of x. That's the function you integrate.
     
  4. Feb 4, 2014 #3
    Yes, but the problem is how do I make EI a fucntion of x when it changes half-way in the beam. I know the equation of moment obviously. I mean I don't know what to do with the changing EI.
     
  5. Feb 4, 2014 #4
    It doesn't have to be continuous. Just integrate it across the discontinuous change.

    Chet
     
  6. Feb 4, 2014 #5

    SteamKing

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    With stepped beams or other non-prismatic members, you may have to write more than one integral.

    After all, [itex]\int^{b}_{a}[/itex] = [itex]\int^{c}_{a}[/itex] + [itex]\int^{b}_{c}[/itex]
     
  7. Feb 5, 2014 #6
    I sort of did something like that but I don't get it right. That's why I was hoping someone could provide and example so I could see where I'm going wrong.
     
  8. Feb 5, 2014 #7

    SteamKing

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    It would be easier if you show what you have been doing. That would give you a quicker diagnosis to your actual problem.
     
  9. Feb 5, 2014 #8
    Fine.
    I attached a picture of the structural model.
    F = 23 kN
    L1 = 2.1 m
    E = 210 000 MPa
    IA-B = 7*10^-4 m^4
    IB-C = 3.5*10^-4 m^4
    EIA-B = 147000 kNm^2
    EIB-C = 73500 kNm^2

    Equation of moment: M''(X) = -23X + 96.6
    M(X) = (-23/6)X^3 + 48.3X^2

    Then I tried: M(2.52) = {[(-23/6)/147000]*2.52^3 + (48.3/147000)*2,52^2} + {[(-23/6)/73500]*2.52^3 + (48.3/73500)*2,52^2} = ....

    This is just one of the ways I tried to calculate.
     

    Attached Files:

  10. Feb 5, 2014 #9
    Your equation for the moment us incorrect. What you call M''(x) is actually the bending moment M(x). What boundary conditions are you using at x = 0 for the displacement?
     
  11. Feb 5, 2014 #10
    It is not incorrect it is exactly what it is supposed to be and as you say it is a bending moment.
    Displacement at X = 0 is 0. This is why there are no C1X and such in case that is what you are wondering. The support is fixed.
     
  12. Feb 5, 2014 #11
    I marked it M''(x) just to make it simple so I don't have to put those integration marks.
     
  13. Feb 5, 2014 #12

    SteamKing

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    You have to include the constants of integration at each step. Since this is a cantilever, you will have two constants of integration in your expression for the deflection. To determine the value of these constants, you will have to apply the boundary conditions for this problem: one BC for the slope, and one BC for the deflection.
     
  14. Feb 5, 2014 #13
    I don't quite know what these boundary conditions would be.
     
  15. Feb 5, 2014 #14
    SteamKing: Do you agree with his determination of the bending moment? I get 23(4.2-x) for the bending moment. In this case, I would have EIy" =23(4.2-x).

    Samppa: Again, what are the two boundary conditions on the displacement y for this cantilever beam at x = 0?

    Chet
     
  16. Feb 5, 2014 #15
    Dude come on...
    23(4.2-x) = -23X + 96.6

    I said: Displacement at X = 0 is 0. Since the support is fixed this means M'(0) = 0.

    The only problem is what to do once the integration has been done. There is no reason to waist time with "wrong" markings, this isn't preliminary school. At least I haven't been there for quite some time.
     
    Last edited by a moderator: Feb 6, 2014
  17. Feb 5, 2014 #16
    Actually, Dude, you said M''(x) is equal to -23x +96.6, not M(x). I assumed you meant that M''(x) is the second derivative of M(x), where M is the moment.

    Regarding the boundary conditions at x = 0, I felt that you failed to recognize a very important one that I was attempting to get you to identify. That boundary condition is that dy/dx = 0 at a "built-in" cantilever boundary. Is that what you meant by M'(0)? I assumed you were saying that the derivative of the moment was equal to zero at x = 0. Again, I though you were using the symbol M for moment.

    If you want me to do the whole problem for you, I can do that. But, one of the objectives here is to help the person figure out how to do it themselves. Here is my next hint:

    [itex]y'(x)=\int_0^x{\frac{M(x)}{147000}dx}[/itex] for x≤2.1

    [itex]y'(x)=\int_0^{2.1}{\frac{M(x)}{147000}dx}+\int_{2.1}^x{\frac{M(x)}{73500}dx}[/itex] for 2.1<x≤4.2

    where M(x) is the moment.

    Hope this helps.
     
    Last edited by a moderator: Feb 6, 2014
  18. Feb 6, 2014 #17
    I can't believe this, again with the markings.
    I told you that I marked it M''(X) so I don't have to use integration marks, meaning integration S if this is more understandable. In my calculations I also state that M''(X) is the moment. Please read the posts before you answer. This shouldn't even need to be said.

    Regarding the boundary conditions at X = 0, those are obvious and that is why I did not tell them the first time you asked. If you looked at my calculations you would see that they have been taken into account. I also said: "Displacement at X = 0 is 0. This is why there are no C1X and such in case that is what you are wondering. The support is fixed".

    Chet please, tell me where do I exactly ask you to solve "the whole problem"? You can see my calculations, you know where I've gotten and all along I've been asking what to do at the end. You've finally understandably answered, thank you for that.

    I tried to be civil, I really tried but you've been harassing me Chet.
     
    Last edited by a moderator: Feb 6, 2014
  19. Feb 6, 2014 #18
    Dear Samppa,

    Please excuse me if you felt that I was harassing you. That was not my intention.

    The source of my confusion had to do with your use of the symbol M. In a beam bending problem, when someone uses the symbol M, it is natural to assume that they are talking about the bending moment. Yes, to be fair, you did identify M'' as the bending moment, but I just naturally thought that this was an oversight. (I still haven't figured out what the symbol M represents physically in your terminology, other than the bending moment integrated twice.)

    The people who help on PF are expected to be patient with the people they are trying to help, even if the people they are helping use confusing terminology. By the same token, the people who are being helped need to be patient with the helpers until everyone is on the same page. I'll try to be more patient in the future. I hope you will agree to do the same.

    So anyhow, was my hint helpful to you? Were you able to get to a correct final answer?

    Chet
     
  20. Feb 6, 2014 #19
    As you say M(x) is usually the bending moment but I didn't want to bother with the integration marks so I decided to name it M''(x). That's all there is to it. Why do you need M to mean something so badly. It shouldn't make a difference if I had named the bending moment SANTA or REINDEER.

    I had given you the explanation what M''(X) was and you can see the equation and it clearly is the bending moment.

    Well my calculations get stuck at the point when I've calculated y'(x) (as chest named it). Just integrating y'(x) didn't give the right answer.
     
    Last edited by a moderator: Feb 6, 2014
  21. Feb 6, 2014 #20
    I should have mentioned that y' is the derivative of the displacement y. After you got y', did you try to integrate again to get y? If not, please try and see how this works.

    Chet
     
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