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Minimum Energy Curve Derivation Help

  1. Nov 10, 2013 #1
    I've been trying to find the equation of the line of no strain of a bent linearly elastic beam of fixed length.
    (in a plane, given just its endpoints, or both the end points and the slopes at the endpoints)
    I imagine this problem has been solved many times, but I couldn't find a solution online that I could understand.

    Using the property of linear elasticity it follows that [itex]dE/ds \propto (dθ/ds)^2[/itex]

    (Where E is the bending energy, s is the arclength, and θ, the angle of the tangent line)

    What it comes down to is finding the curve which minimizes the integral of the square of the curvature across it.

    \begin{equation} \int_0^l \frac{(x'y'' - y'x'')^2}{(x'^2 + y'^2)^3} + λ(x'^2 + y'^2)^{1/2}\,dt \end{equation}

    λ is a Lagrange multiplier used to meet the length constraint.
    (' here denotes a derivative with respect to t)

    I take the first variation with respect to x,
    So for all functions v:
    \begin{equation} \int_0^l \frac{-2y'(x'y'' - y'x'')}{(x'^2 + y'^2)^3}v'' + \bigg(\frac{2y''(x'y'' - y'x'')}{(x'^2 + y'^2)^3} + \frac{-6x'(x'y'' - y'x'')^2}{(x'^2 + y'^2)^4} + \frac{λx'}{(x'^2 + y'^2)^{1/2}}\bigg)v'\,dt = 0 \end{equation}
    Which I guess means that:
    \begin{equation} \frac{2y''(x'y'' - y'x'')}{(x'^2 + y'^2)^3} + \frac{-6x'(x'y'' - y'x'')^2}{(x'^2 + y'^2)^4} -\bigg(\frac{-2y'(x'y'' - y'x'')}{(x'^2 + y'^2)^3}\bigg)' + \frac{λx'}{(x'^2 + y'^2)^{1/2}} = c_1 \end{equation}
    I figure I am free to choose x and y to be parametrized by arclength
    so I set:
    \begin{equation}x'^2 + y'^2 = 1
    \end{equation}
    Taking the derivative:

    [itex]2x'x'' + 2y'y'' = 0[/itex] so [itex]y'y'' + x'x'' = 0[/itex]

    [itex]y''y'' + y'y''' + x''x'' + x'x''' = 0[/itex]

    Also:

    [itex] x'y'' - y'x'' = (x'^2 + y'^2)y''/x' = y''/x' = -x''/y' [/itex]

    Substituting this in:

    [itex] - 4y''^2/x' - 2x''' + λx' = c_1 [/itex]

    Analogously:

    [itex] - 4x''^2/y' - 2y''' + λy' = c_2 [/itex]

    Multiplying the first equation by x', the second by y' and combining gives:

    [itex] λ = c_1x' + c_2y' + 2x''^2 + 2y''^2 [/itex]

    Dividing the second equation by x':

    [itex] λy'/x' = c_2/x' + 2x''^2/y'x' + (2y'''x' - 2x''y'')/x'^2 [/itex]

    Substituting in the value of λ:

    [itex] c_1y' + 2x''^2/y'x' = c_2(1-y'^2)/x' + 2x''^2/y'x' + 2(y''/x')' [/itex]

    [itex] c_1y' - c_2x' = 2(y''/x')' [/itex]

    Which can be integrated to give:

    [itex] c_1y - c_2x + c = 2y''/x' [/itex]

    Which means that the curvature along this curve is proportional to the distance from a straight line. This sounds somewhat intuitvely pleasing.

    However, taking the derivative of [itex] λ = c_1x' + c_2y' + 2(x''/y')^2 [/itex]

    [itex] 0 = c_1x'' + c_2y'' + 2(y''/x')*2(y''/x')' [/itex]

    [itex] c_1x'' + c_2y'' + 2(y''/x')*(c_1y' - c_2x') = 0[/itex]

    [itex] c_1x'' + c_2y'' - 2c_1x'' - 2c_2y'' = 0[/itex]

    [itex] c_1x'' + c_2y'' = 0[/itex]

    So:

    [itex] (y''/x')(y''/x')' = 0[/itex]

    Which means that either the curvature is zero (A straight line) or the change in the curvature is zero (A circle)

    Based on physical experience, this doesn't seem to be the general solution to this problem.

    Assuming my algebraic manipulations are correct (I've checked them over), can anyone point out the flaws in my derivation? I'm new to variational calculus to be honest.
     
    Last edited: Nov 10, 2013
  2. jcsd
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