|
|
George Mason UniversityConor Philip NelsonSource: Home > Classes > Math 447 > Project 1 > Step 4Math 447: Numerical Analysis
4. Add a sinusoidal pile to the beam. This means adding a function of form \(s(x)=-pgsin(\frac{\pi}{L}x)\) to the force term \(f(x)\). Prove that the solution: $$y(x)=\frac{f}{24EI}x^2(x^2-4Lx+6L^2)-\frac{pgL}{EI\pi}(\frac{L^3}{\pi^3}sin(\frac{\pi}{L}x)-\frac{x^3}{6}+\frac{L}{2}x^2-\frac{L^2}{\pi^2}x)$$ satisfies the Euler-Bernoulli beam equation and the clamped-free boundary conditions. We wish to find \(y''''(x)\). We begin by rearranging \(y(x)\) $$y(x)=\frac{f}{24EI}x^4-\frac{f}{6EI}Lx^3+\frac{f}{4EI}L^2x^2-\frac{pgL^4}{EI\pi^4}sin(\frac{\pi}{L}x)+\frac{pgLx^3}{EI6\pi}-\frac{pgL^2x^2}{EI2\pi}+\frac{pgL^3x}{EI\pi^3}$$ Here we can check if \(y(0)=0\): $$(1)...y(0)=\frac{f}{24EI}0^4-\frac{f}{6EI}L0^3+\frac{f}{4EI}L^20^2-\frac{pgL^4}{EI\pi^4}sin(\frac{\pi}{L}0)+\frac{pgL0^3}{EI6\pi}-\frac{pgL^20^2}{EI2\pi}+\frac{pgL^30}{EI\pi^3}$$ $$=0$$ Derive \(y(x)\) once: $$y'(x)=\frac{f}{6EI}x^3-\frac{f}{2EI}Lx^2+\frac{f}{2EI}x-\frac{pgL^3}{EI\pi^3}cos(\frac{\pi}{L}x)+\frac{pgLx^2}{EI2\pi}-\frac{pgL^2x}{EI\pi}+\frac{pgL^3}{EI\pi^3}$$ Here we shall check if \(y'(0)=0\) $$(2)...y'(0)=\frac{f}{6EI}0^3-\frac{f}{2EI}L0^2+\frac{f}{2EI}0-\frac{pgL^3}{EI\pi^3}cos(\frac{\pi}{L}0)+\frac{pgL0^2}{EI2\pi}-\frac{pgL^20}{EI\pi}+\frac{pgL^3}{EI\pi^3}$$ $$=-\frac{pgL^3}{EI\pi^3}+\frac{pgL^3}{EI\pi^3}=0$$ Deriving \(y(x)\) twice: $$y''(x)=\frac{f}{2EI}x^2-\frac{f}{EI}Lx+\frac{f}{2EI}+\frac{pgL^2}{EI\pi^2}sin(\frac{\pi}{L}x)+\frac{pgLx}{EI\pi}-\frac{pgL^2}{EI\pi}$$ Here we shall check if \(y''(L)=0\): $$(3)...y''(L)=\frac{f}{2EI}L^2-\frac{f}{EI}LL+\frac{f}{2EI}+\frac{pgL^2}{EI\pi^2}sin(\frac{\pi}{L}L)+\frac{pgLL}{EI\pi}-\frac{pgL^2}{EI\pi}$$ $$=0+0=0$$ Deriving \(y(x)\) thrice: $$y'''(x)=\frac{f}{EI}x-\frac{f}{EI}L+\frac{pgL}{EI\pi}cos(\frac{\pi}{L}x)+\frac{pgL}{EI\pi}$$ Here we can check if \(y'''(L)=0\): $$(4)...y'''(L)=\frac{f}{EI}L-\frac{f}{EI}L+\frac{pgL}{EI\pi}cos(\frac{\pi}{L}L)+\frac{pgL}{EI\pi}$$ $$=0-\frac{pgL}{EI\pi}+\frac{pgL}{EI\pi}=0$$ Finally, deriving the fourth time, we arrive at: $$y''''(x)=\frac{f}{EI}-\frac{pg}{EI}sin(\frac{\pi}{L}x)$$ Rearranging yields: $$y''''(x)EI=f-pgsin(\frac{\pi}{L}x)$$ Note: \(s(x)=-pgsin(\frac{\pi}{L}x)\). Therefore: $$(5)...y''''(x)EI= f+s(x)$$ Consider (1), (2), (3), (4). By these, we have: $$(6)...y(0)=y'(0)=y''(L)=y'''(L)=0$$ This proves that (5) function satisfies the Euler-Bernoulli equations. This also proves that (6) satisfies the clamped-free boundary conditions.
|