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Math Help - Catenary differential equation (suspension bridge)

  1. #1
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    Catenary differential equation (suspension bridge)

    So I am getting stuck trying to solve this. pgA and Tx are constants
    d^2h/dx^2 = (pgA/Tx)*sqrt(1+(dh/dx)^2)

    I substitute dh/dx = sinh(u), d^2h/dx^2 = coshu(du/dx)

    then you get (pgA/Tx)*cosh(u) = cosh(u)(du/dx)

    ... and I'm unsure where to go after this. I tried a couple more things but I'm not sure if it's the right track:

    (pgA/Tx) * cosh(u) = cosh(u)(du/dx) ... integrate and solve for u = (pgAx/Tx)+C
    and then substitute back in for u??? Sorry this isn't in Latex but I couldn't figure it out.
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  2. #2
    Grand Panjandrum
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    Re: Catenary differential equation (suspension bridge)

    Quote Originally Posted by cbjohn1 View Post
    So I am getting stuck trying to solve this. pgA and Tx are constants
    d^2h/dx^2 = (pgA/Tx)*sqrt(1+(dh/dx)^2)

    I substitute dh/dx = sinh(u), d^2h/dx^2 = coshu(du/dx)

    then you get (pgA/Tx)*cosh(u) = cosh(u)(du/dx)

    ... and I'm unsure where to go after this. I tried a couple more things but I'm not sure if it's the right track:

    (pgA/Tx) * cosh(u) = cosh(u)(du/dx) ... integrate and solve for u = (pgAx/Tx)+C
    and then substitute back in for u??? Sorry this isn't in Latex but I couldn't figure it out.
    1. Lump the constants, so then you have:

    \frac{d^2h}{dx^2}=K\left(1+\left(\frac{dh}{dx} \right) ^2\right)^{1/2}

    where K=pgA/Tx.

    Now consider reducing the order by putting u=dh/dx

    CB
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