\(\Sigma Fy=0\)
\(f_A-60\ KN\ +\ F_B=0\) (1)
\(\Sigma M_0=0\)
\(\left(2m\right)\left(-60\ KN\right)\left(6mF_B\right)=\ 0\)\(Despejamos\ FB=\ \frac{60KN\left(2m\right)}{6m}\) FB=20 KN (3)
Sustituimos (3) en (1)
\(FA\ -\ 60\ KN\ +\ 20\ KN\ =\ 0\) \(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ FA=\ 40\ KN\ \ \)
Ahora calculamos los desplazamientos
\(S_A=\ \frac{\left(40\ X10^3\ N\right)\left(2\ m\right)}{2\ \left(10^{-3}\right)\ m^{2\ }\ \left(60\ x10\ Pa\right)}\ =\ 80,000=\ 0.666\ x10^{-3}\)
\(S_B=\frac{\left(20\ X10^{3\ }N\right)\left(4m\right)}{2\left(10^{-3}\right)m^{2\ \left(6x10\right)}}=\ 20,000\ =\ 1.666x10^{-4}\)