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Axial/Longitudinal Vibrations in beams

  1. Jan 11, 2016 #1
    Hi there,

    I'm studying for an exam in mechanical vibrations, and I'm at the moment looking at longitudinal waves in continuous beams - and I've run into some problems.

    The whole idea is to explain the solutions of the system in terms of eigenvalues and eigenfunctions, which is not that big of a problem, except that I'm having trouble understanding how to solve the problem of a forced response using eigenfunctions.

    The displacement of a plane at x is given by ## \Psi(x,t) ##.
    As I understand it, the general wave equation (1D) for the beam is given by
    [tex]\frac{\partial^2 \Psi}{\partial t^2} - c^2 \frac{\partial^2 \Psi}{\partial x^2} = F(x,t)[/tex]
    Here ## F(x,t) ## is the axial distributed force (if such a thing was physically possible) on all the planes. In the case of free vibrations we set ## F(x,t) = 0 ## and have the homogeneous wave equation.

    Alright. Solving the free-vibrations using separation of variables is simple. And the solutions come out to be
    $$
    \Psi(x,t) = \psi(x)\tau(t)
    \begin{cases}
    \psi(x) & = A\cos(kx) + B\sin(kx)\\
    \tau(t) & = C\cos(\omega t) + D\sin(\omega t)\\
    \end{cases}
    $$
    So far so good. At this point, I make the following comment: The solutions here are the eigenfunctions of the system, and the ##\omega = kc ##, is called the eigenfrequencies. Depending on the boundary conditions we have in place, the eigenfunctions will be limited. For example, if we have a finite beam of length L, with free ends, the eigenfrequencies would be limited to ## \omega_n = \frac{n\pi c}{L} ##.

    The problems start now. Because if I want to solve the inhomogeneous wave-equation for an ## F(x,t) \ne 0 ##, how do I proceed?

    My first idea has been to assume that the solution can be written as a superposition of the eigenfunctions (eg. ## \Psi(x,t) = \sum_n \Psi_n(x,t) = \sum_n \psi_n(x)\tau_n(t) ##.

    However, here I notice that, if I insert this sum into the inhomogeneous wave-equation, I get
    $$
    \frac{\partial^2}{\partial t^2}\sum_n\psi_n(x)\tau_n(t) - c^2 \frac{\partial^2}{\partial x^2}\sum_n\psi_n(x)\tau_n(t) = F(x,t) $$
    $$\sum_n\frac{\partial^2}{\partial t^2}\psi_n(x)\tau_n(t) - c^2 \sum_n\frac{\partial^2}{\partial x^2}\psi_n(x)\tau_n(t) = F(x,t) $$
    $$\sum_n\omega_n^2\psi_n(x)\tau_n(t) - c^2 \sum_n k_n^2\psi_n(x)\tau_n(t) = F(x,t) $$
    $$\sum_n\omega_n^2\psi_n(x)\tau_n(t) - \sum_n c^2 k_n^2\psi_n(x)\tau_n(t) = F(x,t) $$
    $$0 = F(x,t)$$
    I know there are some tricks I need to use involving Fourier's trick (multiplying with an eigensolution and integrating), but I can't get my head around it at the moment o_O

    Could anyone give me some pointers in the right direction? That would be much appreciated!

    Thanks!
     
  2. jcsd
  3. Jan 11, 2016 #2

    Orodruin

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    Your problem is a common one and largely it is caused by the way separation of variables is taught. A priori, you cannot use separation of variables in this fashion - the separation into the x and t directions in this manner requires the PDE to be homogeneous.

    What you can do is to use the same kind of approach to find the eigenfunctions of ##\hat L = -\partial_x^2##. Since ##\hat L## is a Sturm-Liouville operator, you can expand any function of x in terms of the resulting eigenfunctions. In particular, for a fixed t, a function ##f(x,t)## may be written as
    $$
    f(x,t) = \sum_n \tau_n(t) X_n(x),
    $$
    where the expansion coefficients ##\tau_n(t)## are generally time-dependent and ##X_n(x)## are the eigenfunctions of ##\hat L##. See if you can take it from there.
     
  4. Jan 11, 2016 #3

    Krylov

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    While in principle I very much agree with @Orodruin's suggestion above, I would like to point out that, in order to be able to manipulate (in particular: differentiate) the expansion for ##f(\cdot,t)## without any problems, you cannot take "any function": some smoothness as well as uniformity of convergence to ##f(\cdot,t)## and its derivatives is required. For this it is necessary (but clearly not sufficient) that ##f(\cdot,t)## satisfies the boundary conditions.

    The most general (but weakest) statement can be made when ##f(\cdot,x)## is merely square integrable. In that case the series converges in the ##L^2## ("mean square") sense. Otherwise, the series expansion is meaningless.

    Note that in the above it was assumed that the spatial domain is a bounded interval.

    Also, when considering ##\hat{L}## it is important to keep in mind that the operator is not specified merely by its action (##-\partial_x^2##) but also by the aforementioned boundary conditions. Without these, there is not enough information to determine the eigenfunctions.
     
    Last edited: Jan 11, 2016
  5. Jan 11, 2016 #4
    Thanks guys!

    So, I'm giving it a go here, and I just wanna see if what I'm about to do is the right thing...

    As far as I understand, I still assume that the solution is separable, in the way that we solution I'm looking for ##\Psi(x,t)=\sum\tau_n(t)\psi_n(x)##.

    And here, the spatial parts are eigenfunctions of the ##\partial^2/\partial x## operator as follows:
    $$\frac{\partial^2\psi(x)}{\partial x^2} = -k^2 \psi(x)$$
    In the above equation, the eigenvalues are all assume to be less than 1. Otherwise the solutions would not be physical. And the solutions could be described in a number of ways, I've always preferred ##A\cos(kx+\phi_a)##, or sometimes ##A\cos(kx)+B\sin(kx)##.
    Using the idea that the solution at any given time t is a sum of these eigenfunctions, I can insert it into the differential equation and I get.
    $$\frac{\partial\Psi(x,t)}{\partial t^2} - c^2 \frac{\partial\Psi(x,t)}{\partial x^2} = F(x,t)$$
    $$\sum\psi_n(x)\ddot \tau_n(t) - c^2 \sum\tau_n(t)(-k_n^2)\psi_n(x) = F(x,t)$$
    What we can do next is to use Fourier's trick and arrive at
    $$\ddot \tau_m(t)\int_0^L\psi_m^2(x) + \omega_m^2 \tau_m(t)\int_0^L\psi_m^2(x) = \int_0^L\psi_m(x)F(x,t)$$

    Is this sort of the right idea? Or am I sinking fast? :sorry:
    I think perhaps the problem is that I'm stumbling a bit in the dark, and a bit unsure of what the end-goal is.

    Is it correct that, the idea is to get to an equation that gives you the coefficients ##\tau_n(t)## from basically knowing the function ##F(x,t)## and the eigenfunctions to the homogeneous spatial equation?

    Thanks again!
     
  6. Jan 11, 2016 #5

    Krylov

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    Is the beam fixed at both ends? If so, and if for simplicity the length of the beam is 1, then you have to impose the boundary conditions ##\psi(0) = 0 = \psi(1)## to obtain the eigenfunctions. This yields ##A = 0## and ##k_n = n\pi## with ##n = 1,2,\ldots##, so the eigenvalues are ##\lambda_n = n^2\pi^2##. (I don't see why you say they cannot be ##\ge 1##. In fact, all of them are.) The eigenfunctions are ##\psi_n(x) = \sin{n \pi x}## for ##n = 1,2,\ldots##.

    Any reasonably well-behaved (I hate this phrasing, also see post #3) function on ##[0,1]## that satisfies the boundary conditions can then be expanded in terms of the ##\psi_n##, like you did for ##\Psi(\cdot,t)##. (Because ##\Psi## also depends on time, the coefficients are time-dependent.)

    No, you are on the right path. Follow @Orodruin's hint and expand ##F(\cdot,t)## as well. Then use the orthogonality properties of the eigenfunctions (what you call "Fourier's trick"), which gives you an infinite system of second order ODEs, one for each ##\tau_n##. The initial conditions for these ODEs come from the coefficients in the expansions of the initial conditions for the PDE (the wave equation), which are functions on ##[0,1]##.
     
  7. Jan 11, 2016 #6
    Thanks Krylov!

    Well, I'm supposed to delve into 1) Forced Response as the superposition of modes (and I'm guessing that is what I am doing right now), and then I need to 2) Look at the response to a harmonic point force. Now, I was planning on going through an example of a beam with one "free end" (where the point force is applied), and one fixed end (making the boundary condition pretty simple). And I'm having the sensation that if I just get over this kink, then it suddenly all makes sense :-)

    Any thoughts on the two points?

    EDIT: And I guess one problem further down the road is that the harmonic point force gives rise of non-homogeneous boundary conditions?
     
  8. Jan 11, 2016 #7

    Krylov

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    Yes, indeed. In this case, can you confirm that you have the homogeneous BCs ##\Psi(0,t) = 0 = \Psi(1,t)##? If so, then for this question you can proceed exactly as discussed so far.
    For this case you have ##F \equiv 0## but you have inhomogeneous BCs, of the type
    $$
    \Psi(0,t) = 0, \qquad \Psi(1,t) = \beta(t) \qquad \forall\,t \ge 0
    $$
    where ##\beta## is your point force. One then typically introduces a new unknown for which the BCs are again homogeneous. This is always possible. (Hint: look at ##\Phi(x,t) = \Psi(x,t) - x\beta(t)##.) The new PDE will now have a non-zero forcing function, so you can proceed as in the first part. (Caveat: The forcing function will typically no longer satisfy the homogeneous BCs, so you can no longer expect pointwise convergence of the Fourier series at the boundary of ##[0,1]##.)
     
  9. Jan 11, 2016 #8
    Thanks a lot mate! :-)

    The first choice was actually to look at the boundary conditions ## \Psi_x(0,t) = 0 = \Psi_x(L,t) ##. This corresponds to a beam with free ends since, as far as I understand, since the stress in the ends will be equal to 0, and the stress proportional to the spatial derivative of the displacement. Is this kind of boundary conditions still considered homogeneous?
     
  10. Jan 11, 2016 #9

    Krylov

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    Yes, but your eigenfunctions will now be cosines (because ##\cos' = -\sin##, so this lets you satisfy the BCs on the derivative now.) The ideas don't change, but the actual formulas of course become different.
     
  11. Jan 11, 2016 #10
    Great! The fog is beginning to rise :-)

    Another question emerges, however. If I am interested in the "steady state solution", for example in the case of the point harmonic force. That is, in case the initial conditions are not known and/or not of interest, how would the procedure then be?
     
  12. Jan 11, 2016 #11

    Krylov

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    Can you explain what you mean exactly by "steady state solution" in this context?
     
  13. Jan 11, 2016 #12
    Sure!

    It would be how the solution looked if an infinite amount of time had passed.

    The idea is that, if we excite one end of the beam with a harmonic point for, for example we might want to know the amplitude of displacement at the other end of the beam. So in other words - finding the Frequency Response of the system.
     
  14. Jan 11, 2016 #13
  15. Jan 11, 2016 #14
    This is an example from the horrible, horrible, horrendous book that I've have been handed!

    Here they have added a damping by making the Youngs Modulus a complex number, but the basic idea is that there will be resonant frequencies in the structure.

    My question were regarding how one would go from this eigenfunction-based solution to the wave-equation, and get to a description of these resonant frequencies.
     
  16. Jan 11, 2016 #15

    Krylov

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    Thanks for the information.

    It depends on the precise boundary conditions (in particular: the harmonic point force at one end), but I would start by carefully writing out (for a particular choice of BC at ##x = 0## and point forcing at the other end) what the system of ODEs exactly looks like. Most likely you get an infinite system of harmonically driven (harmonic) oscillators, each of which has its own separate resonance frequency. The infinite sequence that you thus obtain then corresponds to the sequence of peaks that you see in your graph. In other words: depending on the frequency with which you drive the system at one end, you will excite one of the available modes.

    Of course, the above is just a qualitative sketch. To see what really happens to the ##n##th mode as you drive the system, you need to write down the corresponding ODE.

    Just out of curiosity, what book are you using?
     
  17. Jan 11, 2016 #16

    Orodruin

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    Note that in order for the steady state to be the solution after a long time regardless of the initial conditions, it is necessary that your wave equation is damped. If not, your general solution will be a sum of the steady state solution and the oscillations in the eigenmodes.
     
  18. Jan 12, 2016 #17
    Hey guys! Thanks so much the help and interest!

    @Krylov - I'm using a book that has been translated from danish called "Structure-Borne Sound and Vibration". I don't really like it though, the math feels a bit rushed. But perhaps it's just me.
    I just discovered another book, however, which seems more thorough: "Vibrations of Continuous Systems" by Arthur W. Leissa and Mohamad S. Qatu.

    @Orodruin - I was actually thinking along the lines of lumped parameter systems, where it is possible to define closed form solutions of an undamped system (with singularities at the resonances) - but I guess for continuous systems, since there are infinite resonant frequencies, it wont work that way... However in the new book I found, they treat damped oscillations and forced oscillations using eigenfunction development as well.
     
  19. Jan 12, 2016 #18

    Krylov

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    My pleasure, I like this topic quite a bit.
    Thank you, I'm always curious about book titles and I'm also rather particular about my preferences, so I understand. Hopefully the above hints have put you on the right track. Feel free to discuss more with us if the need arises.
     
  20. Jan 12, 2016 #19
    Okay, I've gone and done some work now, and I think I'm starting to get an overview of this thing.

    There's a few kinks however - But here goes. The system I'm using as an example is the following:

    Snapshot.jpg

    So the boundary conditions are: ##\Psi(0,t) = 0## and the first derivative ##\Psi^{(1)}(L,t) = -\frac{1}{EA}F(t)##.
    The condition at L comes from the relationship between force, stress and strain.
    $$\frac{\partial^2\Psi}{\partial t^2} - c^2\frac{\partial^2\Psi}{\partial x^2} = 0 $$
    Since we are dealing with in-homogeneous boundary conditions we look for a solution of the type:
    $$\Psi(x,t) = \Psi_i(x,t) + \Psi_h(x,t) $$
    Here the subscript "i" expresses the fact that the function satisfies the inhomogeneous boundary conditions, and "h" satisfies the homogeneous boundary conditions.
    $$\Psi_i(x,t) = - \frac{L}{EA\pi}F(t)sin(x\pi/L) $$
    This function satisfies the boundary conditions. When we insert this into the differential equation we get:
    $$\frac{\partial^2\Psi_h}{\partial t^2}-c^2\frac{\partial^2\Psi_h}{\partial x^2} = c^2\frac{\partial^2\Psi_i}{\partial x^2}-\frac{\partial^2\Psi_i}{\partial t^2}$$
    The function on the RHS is a known function (because we solved the boundaries), so we can say that it is a function ##q(x,t)##.

    Next we solve this PDE by using the eigenfunctions/modes of the system. The boundary conditions for this system are homogeneous and ##\Psi_h(0,t) = 0## and the first derivative ##\Psi_h^{(1)}(L,t) = 0##.

    These boundary conditions lead to eigenfunctions of the form: ##\psi_n(x) = \sin(\frac{2n-1}{L}\frac{\pi}{2}x)##, and we call the spatial wave numbers ##k_n##.

    So the function ##\Psi_h(x,t)## can be written as an eigenfunction-expansion of the form
    $$ \Psi_h(x,t) = \sum\limits_n \tau_n(t)\psi_n(x) $$
    Here the functions ##\tau_n(t)## are the time varying expansion coefficients. The eigenfunctions can be arbitrarily scaled to have a norm (over the domain) of 1. This fact is used when the next step is taken - insert the expansion sum into the differential equation, multiply from the left by a given eigenfunction ##\psi_m(x)## and integrate over the domain. Because of the orthogonality between eigenfunctions and the norm of 1 - we end up with the following:
    $$\ddot\tau_m(t) - c^2k_m^2\tau_m(t) = \int\limits_0^L\psi_m(x)q(x,t)dx$$.
    The term on the RHS is a function only of t, and is equal to the Fourier coefficient of the function q(x,t) in this domain. Also, the whole system is a second order ODE, which can be solved with relevant techniques (for example Laplace). But it of course depends on the function q(x,t). But in our case, it is assumed that we know it (because we know F(t) ).

    Great. I can do one more step: If we assume that the force F(t) is harmonic and given by the function ##F(t) = F_0 sin(\omega t)##, we can insert this into the relevant places. First of all, we know have a specific knowledge of the function q(x,t), because now we have:
    $$\Psi_i(x,t) = -\frac{LF_0}{EA\pi}\sin(\omega t)\sin(\frac{x\pi}{L})$$
    $$q(x,t) = c^2\frac{\partial^2\Psi_i}{\partial x^2}-\frac{\partial^2\Psi_i}{\partial t^2} = (-c^2\frac{\pi^2}{L^2}+\omega^2)\Psi_i(x,t)$$
    So we can calculate the integral that gives us the RHS in the ODE for ##\tau(t)##.
    $$\int\limits_0^L\psi_m(x)(\omega^2-c^2\frac{\pi^2}{L^2})\frac{LF_0}{EA\pi}\sin(\omega t)\sin(\frac{x\pi}{L})dx$$
    $$ = (\omega^2-c^2\frac{\pi^2}{L^2})\frac{LF_0}{EA\pi}\sin(\omega t)\int\limits_0^L\sin(\frac{2n-1}{L}\frac{\pi}{2}x)\sin(\frac{x\pi}{L})dx $$

    And let me stop it here. The idea is then to solve the integral(s) (one for each mode) - then solve the ODE for tau, which gives all the expansion coefficients for the homogeneous equation ##\Psi_h(x,t)##. Finally - the particular and complementary solutions are added together to give the full solution.

    Phew. Did I make any grievous mistakes? I had hoped the integral looked a bit nicer...
     
  21. Jan 15, 2016 #20

    Krylov

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    Up to a minus sign at ##x = L## since ##D_x\Psi_i(L,t) = \frac{F(t)}{EA}##.
    Correct, and it is good to keep in mind that ##n = 1,2,\ldots## with ##k_n := \frac{(2n-1)\pi}{2L}## and eigenvalues ##\lambda_n = k_n^2##, where the eigenvalue problem read
    $$
    \psi''(x) + \lambda\psi(x) = 0, \qquad \psi(0) = 0,\, \psi'(L) = 0 \qquad (1)
    $$
    The idea is now, very roughly, that every function on ##[0,L]## that is sufficiently well-behaved and satisfies the homogeneous boundary conditions can be written as an expansion in terms of the ##\psi_n## and differentiated term-by-term.
    Almost. I don't agree with the minus sign on the left, it should be a "+" I believe. To check this, it is easiest to use (1). It also makes sense, because we expect to end up with an ODE that describes a harmonic oscillator, so the term with ##\tau_m## should appear as positive on the left.
    I think in both lines there is a multiplicative factor ##-1## missing. In the second line the ##n## in the integrand should be ##m##.
    Exactly.
    No, as far as I can see you did well! Using Maple, I managed to evaluate the last integral a bit further to get
    $$
    \int\limits_0^L\sin(\frac{(2m-1)\pi x}{2L})\sin(\frac{x\pi}{L})dx =\frac{(-1)^m4 L}{(4m(m-1) -3)\pi}
    $$
    which looks somewhat nicer.

    Sorry for taking some time to respond.
     
    Last edited: Jan 15, 2016
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