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Buckling Load of Simply Supported Beam + Midpoint Transverse Spring

  1. Dec 7, 2013 #1
    Alright, I took a while setting this problem up for you guys, because I am absolutely stumped and have hit a brick wall and have nowhere else to turn for assistance. Please excuse the shoddy diagram. I made it from scratch to try to illustrate the problem better.

    1. The problem statement, all variables and given/known data
    Find the buckling load exactly.


    2. Relevant equations
    Here is the diagram.
    yirpNxP.png
    The beam has a length L, spring is at midpoint. Left support is pinned (no moment). Right support has rollers. The beam is simply supported. The stiffness of the beam is EI. The spring has a stiffness k.
    I am supposed to calculate the maximum load P that can occur before the system becomes unstable (buckles).


    3. The attempt at a solution

    I split the problem into two parts, yielding two differential equations for the vertical displacement.

    These are the variables I used for simplification:
    ##\lambda =\sqrt{\frac{P}{\text{E} \text{I}}}##
    ##b=v_1({\frac{L}{2}})##
    1st: Left of the spring.
    Differential equation:
    ##\frac{P v(x)}{\text{E} \text{I}}+v''(x)=\frac{b k x}{2 \text{E}
    \text{I}}##
    Solution:
    ##v_1(x)=A \cos (\lambda x)+\frac{b k x}{2 P}+B \sin (\lambda x)##

    2nd: Right of the spring.
    Differential Equation:
    ##\frac{P v(x)}{\text{E} \text{I}}+v''(x)=\frac{b k (L-x)}{2
    \text{E} \text{I}}##
    Solution:
    ##v_2(x)=\frac{b k (L-x)}{2 P}+C \cos (\lambda x)+D \sin (\lambda x)##

    Now, I apply boundary conditions!
    I have 4 unknowns to solve for, so I need 4 boundary conditions.

    BC 1: Displacement at left is 0, so A = 0.
    BC 2: Displacement at right is 0, so
    ##C \cos (\lambda L)+D \sin (\lambda L) = 0##
    BC 3: Displacements match at L/2, so
    ##B \sin \left(\frac{\lambda L}{2}\right)=C \cos
    \left(\frac{\lambda L}{2}\right)+D \sin \left(\frac{\lambda
    L}{2}\right)##
    BC 4: Slopes match at L/2, so
    ##\frac{b k}{2 P}+B \lambda \cos \left(\frac{\lambda
    L}{2}\right)=-\frac{b k}{2 P}-C \lambda \sin
    \left(\frac{\lambda L}{2}\right)+D \lambda \cos
    \left(\frac{\lambda L}{2}\right)##

    Put all terms of B, C, D on left hand side, put all the remaining terms on the right hand side, and end up with an equation...

    ##\left(
    \begin{array}{ccc}
    0 & \cos (L \lambda ) & \sin (L \lambda ) \\
    \sin \left(\frac{L \lambda }{2}\right) & -\cos \left(\frac{L
    \lambda }{2}\right) & -\sin \left(\frac{L \lambda }{2}\right)
    \\
    \lambda \cos \left(\frac{L \lambda }{2}\right) & \lambda \sin
    \left(\frac{L \lambda }{2}\right) & -\lambda \cos
    \left(\frac{L \lambda }{2}\right) \\
    \end{array}
    \right)\left(
    \begin{array}{c}
    B \\
    C \\
    D \\
    \end{array}
    \right)=\left(
    \begin{array}{c}
    0 \\
    0 \\
    -\frac{k b}{P} \\
    \end{array}
    \right)##

    This is the point where I am stuck. I am not sure where to continue from here. When I attempt to solve for B, C, D, I do not get any possible solution, not even 0. I know I am setting up the problem wrong, somehow. Someone pointed out to me that the matrix on the right hand side should be all 0. That kind of makes sense to me, but I do not see where in the algebra I messed up that would result in this occurring.
     
    Last edited: Dec 7, 2013
  2. jcsd
  3. Dec 7, 2013 #2
    I realized that there is a substitution that I can do, substituting v1(L/2) for b and recombining terms. This is my new linear system that I need to solve for[STRIKE], and, given the complexity of this problem and the propensity for people to get this involved with something, I might just pull this linear system out and post it into the linear algebra section.[/STRIKE]

    ##\left(
    \begin{array}{ccc}
    0 & \cos (L \lambda ) & \sin (L \lambda ) \\
    \sin \left(\frac{L \lambda }{2}\right) & -\cos \left(\frac{L
    \lambda }{2}\right) & -\sin \left(\frac{L \lambda }{2}\right)
    \\
    \lambda \cos \left(\frac{L \lambda }{2}\right)+\frac{4 \sin
    \left(\frac{L \lambda }{2}\right)}{L} & \lambda \sin
    \left(\frac{L \lambda }{2}\right) & -\lambda \cos
    \left(\frac{L \lambda }{2}\right) \\
    \end{array}
    \right)##

    Solving for P, using the above definitions, results in...

    ##P=\frac{(4.0575^2) \text{E} \text{I}}{L^2}##

    The only problem now stems from the lack of a "k" in my buckling load. Looking over the problem, I think of the problem as behaving like this:

    1: k is very small. Thus, the beam will buckle in the first mode, no problem.

    2: k is very big. The beam will be compelled to buckle in the second mode.

    Isn't this right? If so, then how can I say that the critical buckling load, P, lacks a k in its term? I am thinking I will need a relationship between the beam's stiffness and the spring's stiffness, but something more subtle is going on. Because, for different stiffness ratios, different buckling modes will be induced.
     
    Last edited: Dec 7, 2013
  4. Dec 7, 2013 #3
    Oddly, upon further analysis of the beam itself, it seems that this might be the proper answer for the critical load. The matrix yields multiple solutions for lambda, since the characteristic equation can be expressed as a growing sinusoidal function. The zeroes of the characteristic "function" are the solutions to the characteristic equation. The values for P which correspond to possible antisymmetric buckling modes are unaffected by the spring. The values which correspond to symmetric buckling modes are affected. Even better, the symmetric modes are "pushed" to a higher P value to lesser degrees with subsequent symmetric buckling modes. This makes sense, because, with higher modes, there is less midpoint displacement, so, the spring has a smaller effect.
     
  5. Dec 8, 2013 #4
    Can you check your results by letting k be very small, and very big, as two extreme cases to which you may know the 'standard' answer?
     
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