Problem 423 - Howe Roof Truss by Method of Sections

 

 

$\Sigma M_A = 0$

$12R_H = 3(2.7) + 6(4.5) + 9(1.8)$

$R_H = 4.275 ~ \text{kN}$
 

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

$6(F_{DF} \sin 30^\circ) + 3(1.8) = 6(4.275)$

$F_{DF} = 6.75 ~ \text{kN compression}$           answer
 

 

$\Sigma M_H = 0$

$6(F_{EF} \sin 30^\circ) = 3(1.8)$

$F_{EF} = 1.8 ~ \text{kN compression}$           answer
 

$y = 3 \tan 30^\circ = \sqrt{3}$
 

$\Sigma M_F = 0$

$y F_{EG} = 3(4.275)$

$\sqrt{3} F_{EG} = 3(4.275)$

$F_{EG} = 7.404 ~ \text{kN tension}$           answer