Solucion.
Ecuacion.
\(\sum_{ }^{ }Fy=FBD=110KN\)
Despejamos.
\(MA=MAC\circ-\left(110\ Kn\right)\left(0.5\right)+FBD\ \left(0.6m\right)=0\)
\(FAC+FBD=110\ Kn\ ----\sum_{ }^{ }Fy=0\)
\(-\left(110\ Kn\right)\left(0.5\right)+FBD\left(0.6\ m\right)=0---\sum_{ }^{ }MN=0\)
\(De\ \left(Z\right)\ despejamos\ FBD\)
\(FBD\ \left(0.6\ m\right)=\left(110kn\right)\left(0.5\ m\right)\)
\(FBD=\ \frac{\left(110\ Kn\right)\left(0.5\ m\right)}{\left(0.6\ m\right)}=91.66Kn\)
\(FAC=110\ Kn\ -91.66\ Kn=18.34\ Kn\)
\(SA=-\frac{18.34X10}{\pi\left(0.01\right)}\)\(\frac{N\left(0.4m\right)}{\left(200\ X\ 10\ PA\right)}=0.116mm\)
\(\)=(-41.66 X 103 N) (0.4m) /   π  (0.01m)2 (200 X 104)=0.583 mm