Problem 432 - Force in Members of a Truss by Method of Sections

 

$\Sigma M_F = 0$

$16R_A = 12(24) + 8(36)$

$R_A = 36 ~ \text{kN}$
 

 

$\Sigma M_D = 0$

$3F_{BC} + 8(36) = 4(24)$

$F_{BC} = -64 ~ \text{kN}$

$F_{BC} = 64 ~ \text{kN compression}$       answer
 

$\Sigma M_E = 0$

$8(\frac{3}{5}F_{BD}) + 8(24) = 12(36)$

$F_{BD} = 50 ~ \text{kN tension}$       answer
 

$\Sigma M_B = 0$

$3(\frac{4}{5}F_{AD}) + 4(\frac{3}{5}F_{AD}) = 4(36)$

$\frac{24}{5}F_{AD} = 144$

$F_{AD} = 30 ~ \text{kN tension}$       answer
 

 

$\Sigma F_V = 0$

$\frac{9}{\sqrt{97}}F_{AB} + \frac{3}{5}F_{AD} + 36 = 0$

$\frac{9}{\sqrt{97}}F_{AB} + \frac{3}{5}(30) + 36 = 0$

$F_{AB} = -59.09 ~ \text{kN}$

$F_{AB} = 59.09 ~ \text{kN compression}$       answer