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\begin{document}

\title{Problemas sobre Vigas.}



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


\vspace{-1em}



  
  \date{May 07, 2024}


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La viga r\selectlanguage{ngerman}ígida soporta la carga de 60 kn, determina el desplazamiento en
B. tomar E = 60 GPa y ~\(A_{BC\ }=2\left(10^{-3}\right)m^2\)

Datos

Carga= 60 kn

E= 60 GPa

\(A_{BC\ }=2\left(10-^3\right)m^2\)

Diagrama de Cuerpo Libre~

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Las fuerzas de comprension~ ~que act\selectlanguage{ngerman}úan en la parte superior de cada
poste se determinan a partir del equilibrio del miembro~ A, B.

Por lo cual esas fuerzas se puede decir que son iguales a las fuerzas
internas en cada uno de los postes.

\par\null

Necesitamos conocer las fuerzas.

\(\Sigma Fy=0\)

\(F_{AD+}F_{BC}-60kn=0\ \ \ \ \ \left(1\right)\)

\(-60kn\left(2m\right)+F_{BC}\left(6m\right)=0\ \ \ \ \ \left(2\right)\)

Despejar~\(F_{BC}\) en la ecuación 2

\(F_{BC}\left(6m\right)=120kn\ \cdot\ m\)

\(F_{BC}=\frac{120kn\ m}{6m}\)

\(F_{BC}=20kn\)

Sustituiremos en la ecuación 3 en (1)

\(F_{AD}+F_{BC}-60kn=0\)

\(F_{AD}+20kn-60kn=0\)

\(F_{AD}-40kn=0\)

\(F_{AD}=40kn\)

\(S_{BC}=\frac{PL}{AE}=\frac{2\left(10-^3\right)N\left(3m\right)}{2\left(10-^3\right)m^2\left(60x10\right)n}=-5\ x\ 10-^4\ m^{\ \ \ \ \ }\)

El desplazamiento en B es de ~\(-5\ x\ 10-\ ^4m\).

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