calcular las ecuaciones  de equilibrio
   \(\sum_{}^{}\) \(Fy\)\(FAC+FBD=110KN\)
\(\sum_{}^{}\)\(MA=MAC=0-\left(110KN\right)\left(0.5m\right)+FBD\left(0.6m\right)=0\)
\(FAC+FBD=110KN\ \ \ \ \ \)
\(-\left(110kN\right)\left(0.5\right)+FBD\left(0.6\right)=0\)  
                       \(De\ \left(2\right)\ Despejar\ FBD\)
\(FBD\left(0.6m\right)=\left(110kN\right)\left(0.5m\right)\)
\(FBD=\frac{\left(110kN9\right)\left(0.5m\right)}{\left(0.6m\right)}=91.66KN\)
  
                    \(Sustituir\ en\ 1\)
\(FAC=110kN-91.66KN=18.34KN\)
\(SA=\frac{\left(PAC\right)\left(LAC\right)}{\left(AAC\right)\left(EAC\right)}=\frac{\left(-18.34x10\wedge\ 3N\right)\left(0.4m\right)}{\pi\left(0.01m\right)\wedge2\left(\left(200x10\wedge9pa\right)\right)}=-1.16x10\wedge-4m=0.116mm\)
\(SB=\frac{\left(-91.66x10\wedge3N\right)\left(0.4m\right)}{\pi\left(0.01m\right)\wedge2\left(200x10\wedge9Pa\right)}=-5.83x10\wedge-9=0.583mm\)
Problema #2
\(FAC+FBD=110KN\)La viga mostrada en la figura soporta una  carga 60KN determine el desplazamiento en B considerando \(E=60GPa\)  y  \(ABC=2x10\wedge-5m\)