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\begin{document}

\title{Problemas sobre Columnas.}



\author[1]{Josue Israel Esquivel Chavez}%
\affil[1]{Affiliation not available}%


\vspace{-1em}



  
  \date{May 10, 2024}


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la formula para la carga el\selectlanguage{ngerman}éctrica de una columna fue derivada en 1757
por Leonhard Euler, el gran matemático suizo. El análisis de Euler se
baso en la ecuación diferencial de la curva elástica:

\par\null

~\(\frac{d\ ^2v^{\ ^{ }}}{dx^2}+\frac{P}{EI}=0\)

Encuentre la soluciona a esta ecuación y aplique las siguientes
condiciones para obtener los valores para las constantes de integración:

\par\null

\(u\ _{x=0}=0\)

\(u_{x=L}=0\)

Finalmente explique como obtener el siguiente resultado:

\par\null

el primer paso es derivar

\(u=C_1\sin Lx+C_2\cos Lx\)

\(v^1=\frac{dv}{dx}=C_1L\cos-C_2L\sin\ Lx\)

\(u^n=\frac{d^2v}{dx^2}=C_1L^2\sin\ Lx-C_2L^2\cos Lx\)

el segundo paso es factorizar

\(-C_1L^2\sin\ Lx-\ C_2\ Lx\ \cos Lx+\)

\(\left(\frac{P}{EI}\right)\left(C_1\sin\ Lx+\ C_2\ \cos Lx\right)\ =0\)

\(C_1L^2\ \sin\ Lx-C_2L^2\cos\ Lx+C_1\ \left(\frac{P}{EI}\right)\sin\ Lx+C\ _2\ \left(\frac{P}{EI}\right)\ \cos\ Lx=0\)

\(C_1\sin Lx\left(\frac{P}{EI}-x^2\right)+C_2\ \cos\ Lx\left(\frac{P}{EI}-x^2\right)=0\)

para resolver una ecuación diferencial debemos poner una solución que la
satisfaga:

\(\frac{d^2u}{dx^2}+\left(\frac{P}{EI}\right)u=0\)

se deriva.

\(u=C_1\sin\gamma\ x\ +C_2\cos\gamma x\)

\(u^.=\frac{dv}{dx}=C_1\gamma\cos-C_2\gamma\sin\gamma x\)

\(u^n=\frac{d^2\ u}{dx^2}=C_1\gamma^2\ \sin\ \gamma x-C_2\gamma\cos\ \gamma x\)

se factoriza

\(-C_1\gamma^2\sin\gamma x-C_2\gamma^2\cos\gamma x+\left(\frac{P}{EI}\right)\left(C_1\sin\gamma x+C_2\cos\gamma x\right)=0\)

\(C_1\gamma^2\sin\gamma x-C_2\gamma^2\cos\gamma x\ +C_1\left(\frac{P}{EI}\right)\sin\gamma x+C_2\left(\frac{P}{EI}\right)\cos\gamma x=0\)

\(C_1\sin\gamma x\left(\frac{P}{EI}-x^2\right)+C_2\cos\gamma x\left(\frac{P}{EI}-x^2\right)=0\)

\(\frac{P}{EI}=\gamma^2\ \ \ \ \gamma=\sqrt{\frac{P}{EI}}\)

\(u=C_1\sin\sqrt{\frac{P}{EI}}x+C_2\cos\sqrt{\frac{P}{EI}\ }\ x\)

\(u=O\ I\ \ \ x=0\)

\(u=O\ \ \ I\ \ \ \ \ x=L\)

\par\null

\(C_1\sin\sqrt{\frac{P}{EI}}\left(0\right)+C_2\cos\sqrt{\frac{P}{EI}}\left(0\right)=0\)

Para:~\(u=0\ \ \ \ x=L\)

\(u\left(x=L\right)=C_1\sin\sqrt{\frac{P}{EI}}=0u\left(x=L\right)=C_1\sin\sqrt{\frac{P}{EI}\ \ \ \ }\ \ \ L=0\)

\(\sin\left(\sqrt{\frac{P}{EI}}L\right)=0\)

\(\sqrt{\frac{P}{EI}}=L=n\pi\)

se despeja P

\(\frac{P}{EI}L^2=n^{2\ }\pi^2\)

\(p=\frac{n^2\pi^2EI\ \ \ }{L^2}\)

Para calcular la critica:

\(n=1\)

\(P_{cr}=\frac{n^2\ EI\ }{L^2}\)

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