Datos :
\(E\ =\ 60GPa\)
\(A_{BC\ =\ 2\left(10^{-3}\right)m^2}\)
\(\Sigma FY=0\)
\(\Sigma M=0\)
\(\Sigma Fy\)
\(F_{AD}\ +F_{BC\ }-60\ KN=0\) (1)
Para el momento :
\(M_{AD\ }=0\)
\(M=-\left(2m\right)\left(60KN\right)\)
\(M_{BC\ }=\left(6m\right)F_{BC}\)
\(-\left(2m\right)\left(60KN\right)+\left(6m\right)F_{BC\ }=0\) (2)
De la ecuacion (2) se despeja \(F_{BC}\)
\(-\left(2m\right)\left(60KN\right)+\left(6m\right)F_{BC}=0\)
\(6m\ F_{BC}=120KNm\)
\(6m\ F_{BC}=120KNm\)
\(F_{BC=\frac{120KN}{6\ }}\)
\(F_{BC}=20KN\)
\(S=\frac{PL}{AE}\)
\(S=\frac{\left(-20KN\right)\left(3m\right)}{2\left(10^{-3}\right)m^2\ \left(60\left(10^9\right)\right)}\ =\ -5\left(10^{-4}\right)KN\)