Problem 422 - Right-triangular Truss by Method of Sections

$\Sigma M_F = 0$

$15R_A = 12(400 + 200) + 6(800 + 200)$

$R_A = 880 ~ \text{kN}$
 

From section through MN
$\Sigma M_C = 0$

$6(\frac{2}{\sqrt{5}}F_{BD}) = 3(880)$

$F_{BD} = 491.93 ~ \text{kN}$   compression           answer
 

$\Sigma M_F = 0$

$12(\frac{1}{\sqrt{5}}F_{CD}) + 12(400 + 200) = 15(880)$

$F_{CD} = 1118.03 ~ \text{kN}$   compression           answer
 

$\Sigma M_D = 0$

$3F_{CE} + 6(400 + 200) = 9(880)$

$F_{CE} = 1440 ~ \text{kN}$   tension           answer