Problem 421 - Cantilever Truss by Method of Sections

$\Sigma M_E = 0$

$\frac{3}{\sqrt{13}}F_{DF}(\frac{40}{3}) = 100(20) + 200(10)$

$F_{DF} = 360.5551 ~ \text{kN}$   tension           answer
 

 

$\Sigma M_A = 0$

$\frac{2}{\sqrt{5}}F_{EF}(20) = 200(20) + 200(10)$

$F_{EF} = 335.4102 ~ \text{kN}$   tension           answer
 

$\Sigma M_F = 0$

$20F_{EG} = 100(30) + 200(20) + 200(10)$

$F_{EG} = 450 ~ \text{kN}$   compression           answer