Problem 348 | Equilibrium of Non-Concurrent Force System

Length of DF
${L_{DF}}^2 = 4^2 + 3^2$

${L_{DF}}^2 = 25$

$L_{DF} = 5 \, \text{ m}$
 

Weights of members
$W_{AB} = 6(5) = 30 \, \text{ kN}$

$W_{CE} = 6(5) = 30 \, \text{ kN}$

$W_{DF} = 5(5) = 25 \, \text{ kN}$
 

 

$\Sigma M_B = 0$

$6A_H = 3W_{CE} + 2W_{DF} + 6(200)$

$6A_H = 3(30) + 2(25) + 6(200)$

$A_H = 223.33 \, \text{ kN}$           answer
 

$\Sigma F_H = 0$

$B_H = A_H$

$B_H = 223.33 \, \text{ kN}$           answer
 

$\Sigma F_V = 0$

$B_V = W_{AB} + W_{CE} + W_{DF} + 200$

$B_V = 30 + 30 + 25 + 200$

$B_V = 285 \, \text{ kN}$           answer