Problem 425 - Fink Truss by Method of Sections

 

$\Sigma M_A = 0$

$12R_G = 6(80)$

$R_G = 40 ~ \text{kN}$
 

From the section to the right of M-N
$\Sigma M_D = 0$

$3F_{CE} + 3(20) + 6(10) = 6(40)$

$F_{CE} = 40 ~ \text{kN tension}$           answer
 

 

$\Sigma M_G = 0$

$4F_{DEv} = 3(20)$

$4F_{DE}(\frac{3}{\sqrt{13}}) = 60$

$F_{DE} = 5\sqrt{13} ~ \text{ kN} = 18.03 ~ \text{kN tension}$           answer
 

$\Sigma M_E = 0$

$4F_{DFv} + (3 - 2)(20) + 4(10) = 4(40)$

$4F_{DF}(\frac{1}{\sqrt{5}}) = 100$

$F_{DF} = 25\sqrt{5} ~ \text{ kN} = 55.90 ~ \text{kN compression}$           answer