Problem 006-fr | Analysis of Simple Frame

$\Sigma M_A = 0$

$4R_E = 6(120)$

$R_E = 180 \, \text{ kN}$           answer
 

 

$\Sigma F_H = 0$

$A_H = R_E$

$A_H = 180 \, \text{ kN}$           answer
 

$\Sigma F_V = 0$

$A_V = 120 \, \text{ kN}$           answer
 

 

$\Sigma M_A = 0$

$3(\frac{2}{\sqrt{13}}F_{BD}) = 6(120)$

$F_{BD} = 432.67 \, \text{ kN}$           answer