Problem 420 - Howe Truss by Method of Sections

$\Sigma M_G = 0$

$3F_{DF} = 4(210)$

$F_{DF} = 280 ~ \text{kN}$   compression           answer
 

 

$\Sigma F_V = 0$

$\frac{3}{5}F_{DG} + 120 = 210$

$F_{DG} = 150 ~ \text{kN}$   compression           answer
 

$\Sigma M_D = 0$

$3F_{EG} + 4(120) = 8(210)$

$F_{EG} = 400 ~ \text{kN}$   tension           answer