Dirac delta function (DE problem) solved

[ESPCerpinTaxt]

NOTE: I actually found the correct answer while I was typing this 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 $$w_{0}$$ at $$x=\frac{1}2{L}$$. 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 $$EI\frac{d^4y}{dx^4}=w_{0}\delta(x-\frac{L}2)$$ where $$y(0)=0, y'(0)=0, y''(L)=0, y'''(L)=0$$.




Here is what I did: let $$y''(0)=c_{1}, y'''(0)=c_{2}$$

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

$$s^4Y(s)-sc_1-c_2=\frac{w_0}{EI}e^\frac{-LS}2$$

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

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

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

$$y''(x)=\frac{w_0}{EI}(x-\frac{L}2)U(x-\frac{L}2)+c_1+c_2x$$

$$y'''(x)=\frac{w_0}{EI}U(x-\frac{L}2)+c_2$$



$$y''(L)=\frac{Lw_0}{2EI}+c_1+c_2L=0$$

$$y'''(L)=\frac{w_0}{EI}+c_2=0$$

$$c_2=\frac{-w_0}{EI}$$

$$c_1=\frac{Lw_0}{2EI}$$

$$y(x)=\frac{w_0}{6EI}(x-\frac{L}2)^3U(x-\frac{L}2)+\frac{Lw_0x^2}{4EI}-\frac{w_0x^3}{6EI}$$

 

[phinds]

Glad to hear you got it solved. We love the easy ones