\(\sum_{}^{}\) \(Fy=0\)
\(F_{AC}+F_{AB}=110KN\)
Sustituyendo los valores que tenemos
\(\left(\left(-\left(110KN\right)\left(0.5M\right)+\left(F_{BD}\right)\right)\left(0.6\right)\right)=0\)
Despejando \(F_{BD}\)
\(F_{BD}=\frac{\left(110KN\right)\left(0.5M\right)}{0.6M}=91.66KN\)
Ahora \(F_{AC}\)
\(F_{AC}=110KN-91.66KN=18.34KN\)
El desplazamiento es:
\(\delta_{BD}=\frac{\left(91.66x10^3\right)\left(0.4M\right)}{\pi\left(0.01\right)^2\left(200X10^9Pa\right)}=0.583mm\)
\(\delta_{AC}=\frac{\left(18.34x10^3\right)\left(0.4M\right)}{\pi\left(0.01\right)^2\left(200X10^9Pa\right)}=0.116mm\)