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Euler-Bernoulli bending moment equation, how do you know it's right?

  1. Aug 25, 2012 #1
    [itex]M=EI\frac{∂^2w}{∂x^2}[/itex]

    1. What exactly is this? Is it like a geometric constraint? This equation doesn't seem to depend explicitly on external loading or beam boundary conditions.

    2. Can I derive this starting from the Euler-Bernoulli beam equation?

    I am dealing with a beam with non-uniform cross-section and stiffness. The beam is cantilevered. It also has a point load at the tip. I tried to start with:

    [itex]EI(x)\frac{∂^4w}{∂x^4}=-μ\frac{∂^2w}{∂t^2}+Fδ(x-L)[/itex]

    My goal is to get bending moment as a function of x and t. I thought about integrating with respect to x, because maybe that would get me a "moment" term Fx. But then, integrating the left side seems really ugly, and I am also confused because μ is mass per length? So you get mass per length times acceleration times x, which definitely is not moment? And

    I am hopelessly confused... I just want a way to make sure that I am not misusing the bending moment equation, because it seems weird to me that it would hold for ever single case (for example, shouldn't it be different if the beam had an axial load?).
     
  2. jcsd
  3. Aug 25, 2012 #2
  4. Aug 26, 2012 #3
    OK, so it looks like I can integrate the Euler-Bernoulli equation twice to get moment?

    [tex]EI(x)\frac{∂^4w}{∂x^4}=-μ\frac{∂^2w}{∂t^2}+Fδ(x-L)[/tex]

    Integrate once (using integration by parts, I guess):

    [tex]EI(x)\frac{∂^3w}{∂x^3}-∫\frac{∂^3w}{∂x^3}(\frac{∂}{∂x}EI(x))dx = -μ\frac{∂^2}{∂t^2}(∫wdx) + \left\{\begin{array}{cc}0,&\mbox{ if }
    x≠L\\F, & \mbox{ if } x=L\end{array}\right. [/tex]

    I don't think I integrated the delta function properly (please advise). This becomes:

    [tex]EI(x)\frac{∂^3w}{∂x^3}- \frac{∂^2w}{∂x^2}\frac{∂}{∂x}EI(x)
    + ∫\frac{∂^2w}{∂x^2}\frac{∂^2}{∂x^2}EI(x)dx= -μ\frac{∂^2}{∂t^2}(∫wdx) + \left\{\begin{array}{cc}0,&\mbox{ if }
    x≠L\\F, & \mbox{ if } x=L\end{array}\right. [/tex]

    Am I on the right track? If I integrate once more, will I get the expression for moment? This just doesn't seem right. I can accept the Euler-Bernoulli equation for now, but I still don't understand why second derivative = moment, and whether M= EIw" still applies to cantilevered beams with a transverse point load. I don't think this is leading me anywhere, because even if I isolate the EIw" term, how can I just call everything else "M"? Do you see what I'm getting at?

    OK, I can maybe see how the third derivative is basically a force balance on the beam. So the second derivative equation should be a moment balance? But I can't wrap my head around how you would do a moment balance here.

    (some I theta double dot term)= (reaction moment at the wall) -F*L

    How is this going to get me [tex]M=EI\frac{∂^2w}{∂x^2}[/tex] ? Please forgive me for my pathetically poor understanding of beams...
     
  5. Aug 26, 2012 #4

    AlephZero

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    You seen to have got into a circular argument here. The original derivation (used by Euler and Bermouilli) was based on some experimental results, not just on math.

    The key expermiental observation is that when long thin straight beam bends, every cross section through the beam remains flat and stays perpendicular to the neutral axis of the beam. You can demonstrate that by marking the surface of the beam, bending it, and measuring how the marks move.

    If the the beam bends a small amount, ##d^2w/dx^2## measures the radius of curvature of the beam. The formula for the curvature from calculus also includes the term ##1 + (dw/dx)^2##, but if the beam only bends a small amount ##dw/dx## is small comapred with 1 so the theory ignores it.

    Now, if the beam is curved and plane sections remains plane and perpendicular to the neutral axis, the material on the "outside" of the curve must be stretched and the inside must be compressed. You can apply Hooke's law to find the force distribution across the section that correpond to the amount of curvature. (Note that Hooke's law is also based on experments, not on math!) You then sum the forces to get a bending moment. That's where the equation ##M = EI\,d^2 w/ dx^2## came from.

    Everything else is then derived from that equation, not the other way round.

    All the above is only approximate. The basic theory only applies to thin straight beams (say length / depth > 20) with small deflections and curvature, and no axial load. It can be extended to deal with thick short beams, curved beams, axial loads, large deflections, the fact that the plane cross sections do not remain perfectly flat, etc, but that would lead to Timoshenko beam theory (which was formulated about 150 years after Euler) and beyond.
     
  6. Aug 26, 2012 #5
    Sorry, I am still confused. So which equation is the actual Euler-Bernoulli beam equation?

    So this:[tex]EI(x)\frac{∂^4w}{∂x^4}=-μ\frac{∂^2w}{∂t^2}+Fδ(x-L)[/tex] is derived from this:
    [tex]M=EI\frac{∂^2w}{∂x^2}[/tex]

    Did I understand correctly?

    Or is [tex]EI(x)\frac{∂^4w}{∂x^4}=-μ\frac{∂^2w}{∂t^2}+Fδ(x-L)[/tex] the Euler-Bernoulli equation and [tex]M=EI\frac{∂^2w}{∂x^2}[/tex] is just some separate empirical observation?

    For some reason I thought I did a homework problem once for a beam with an eccentric axial load, and I couldn't just use [tex]M=EI\frac{∂^2w}{∂x^2}[/tex] but instead had to get the proper shear equation and then integrate to get moment. This is why I thought I needed to start with a higher-order ODE and integrate to get bending moment.

    Regardless, the main problem I'm struggling with is: is there any way to determine without experiment the proper expression for bending moment of a non-uniform beam with transverse point loads?

    Thanks so much for all your responses.
     
  7. Aug 26, 2012 #6

    AlephZero

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    If ##M(x)## is the bending moment at any section of the beam, ##dM/dx## is the shear force ##Q## on that section and ##dQ/dx## is the distributed load ##q##. (That's what Stdiot's links said).

    That's all you need to get the static beam equation from ##M = EI\,d^2 w /dx^2##. For dynamics you also have inertia forces, which give you the ##\partial^2w/\partial t^2## term.

    Of course you can justify Euler's "empitcal obervation" by finding the 3-dimensional stress field in the beam using continuum mechanics, and investigating how closely it approximates to Euler beam theory, but that's not what Euler did back in 1750 amd it's way outside the scope of a first course on the static and dynamic analysis of beams.
     
  8. Aug 26, 2012 #7
    I see, thank you for clarifying. Your explanation helps me see where these equations come from. So if I understand correctly, the moment equation, shear equation, and [tex]EI(x)\frac{∂^4w}{∂x^4}=-μ\frac{∂^2w}{∂t^2}+...[/tex] are all basically just different derivatives of the same equation?

    Indeed, I have seen that the moment equation holds for uniform beams, but should it hold for non-uniform beams as well? I have been finding that it definitely does not.
    [tex]M(x) ≠ EI(x)\,d^2 w(x) /dx^2[/tex]
    I need to figure out what the correct moment equation is for a non-uniform beam.

    This is how I got into this whole mess, trying to see if I could somehow derive the moment equation. I thought, "If my moment equation isn't working for non-uniform beams, then there must be something wrong with my equation. It must only work for uniform beams."

    Do you know what I mean? How can I compute moment for a non-uniform beam?
     
  9. Aug 26, 2012 #8

    AlephZero

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    $$M(x) = EI(x) \frac{d^2 w(x)}{dx^2}$$ is the correct equation, unless your "beam" is so nonuniform that it doesn't behave like a beam any more.

    Remember that when you differentiate it, you get
    $$\frac{d}{dx}M(x) = \frac{d}{dx}\left(EI(x) \frac{d^2 w(x)}{dx^2}\right)$$
    and you have differnetiate everything that varies with x on the right hand side, including the ##I(x)## term. ##\mu## will probably vary with x as well.
     
  10. Aug 26, 2012 #9
    Depends what is non uniform about your beam.

    Aleph Zero is quite correct in observing that I may be a function of x so has to be included as such in the integration.

    This also applies to beams that vary in composition along their length. In this case you have a function E(x).

    Another type of beam might be a reinforced concrete beam with varying reinforcement.

    You may note that the linked set of differential equations in my reference may be divided into two sets of three, linked by a common equation. I do not have time tonight to do this for you but hope to answer your question tomorrow about this.

    As regards your attempted use of the impulse function try googling 'macaulay brackets' Macaulay introduced this method and notation, which is useful for manipulatig the differential equations.
     
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