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Dirac delta function (DE problem) solved

  1. May 23, 2005 #1
    NOTE: I actually found the correct answer while I was typing this :rolleyes: and since I already had it typed, I figured i would post anyway. mods you can do with it as you please or leave it for reference. thanks

    Here's the problem:



    A uniform beam of length L carries a concentrated load [tex]w_{0}[/tex] at [tex]x=\frac{1}2{L}[/tex]. The beam is embedded at its left end and is free at its right end. Use the Laplace transform to determine the deflection y(x) from [tex]EI\frac{d^4y}{dx^4}=w_{0}\delta(x-\frac{L}2)[/tex] where [tex]y(0)=0, y'(0)=0, y''(L)=0, y'''(L)=0[/tex].




    Here is what I did: let [tex]y''(0)=c_{1}, y'''(0)=c_{2}[/tex]

    [tex]EI(s^4Y(s)-s^3y(0)-s^2y'(0)-sy''(0)-y'''(0))=w_0e^{\frac{-LS}2[/tex]

    [tex]s^4Y(s)-sc_1-c_2=\frac{w_0}{EI}e^\frac{-LS}2[/tex]

    [tex]Y(s)=\frac{w_0}{EIs^4}e^{\frac{-LS}2}+\frac{c_1}{s^3}+\frac{c_2}{s^4}[/tex]

    [tex]y(x)=\frac{w_0}{6EI}(x-\frac{L}2)^3U(x-\frac{L}2)+\frac{c_1x^2}2+\frac{c_2x^3}6[/tex]

    [tex]y'(x)=\frac{w_0}{2EI}(x-\frac{L}2)^2U(x-\frac{L}2)+c_1x+\frac{c_2x^2}2[/tex]

    [tex]y''(x)=\frac{w_0}{EI}(x-\frac{L}2)U(x-\frac{L}2)+c_1+c_2x[/tex]

    [tex]y'''(x)=\frac{w_0}{EI}U(x-\frac{L}2)+c_2[/tex]



    [tex]y''(L)=\frac{Lw_0}{2EI}+c_1+c_2L=0[/tex]

    [tex]y'''(L)=\frac{w_0}{EI}+c_2=0[/tex]

    [tex]c_2=\frac{-w_0}{EI}[/tex]

    [tex]c_1=\frac{Lw_0}{2EI}[/tex]

    [tex]y(x)=\frac{w_0}{6EI}(x-\frac{L}2)^3U(x-\frac{L}2)+\frac{Lw_0x^2}{4EI}-\frac{w_0x^3}{6EI}[/tex]
     
    Last edited: May 23, 2005
  2. jcsd
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