Problem 004-mj | Method of Joints

$6R_A = 4(100) + 2(120)$

$R_A = 106.67 \, \text{ kN}$
 

At joint F
$F_{FG} = 0$

At joint A
$\Sigma F_V = 0$

$\frac{3}{\sqrt{13}}F_{AG} = R_A$

$\frac{3}{\sqrt{13}}F_{AG} = 106.67$

$F_{AG} = 128.20 \, \text{ kN compression}$
 

$\Sigma F_H = 0$

$F_{AB} = \frac{2}{\sqrt{13}}F_{AG}$

$F_{AB} = \frac{2}{\sqrt{13}}(128.20)$

$F_{AB} = 71.11 \, \text{ kN tension}$
 

At joint B
$\Sigma F_H = 0$

$F_{BC} = F_{AB}$

$F_{BC} = 71.11 \, \text{ kN tension}$
 

$\Sigma F_V = 0$

$F_{BG} = 0$
 

At joint G
$\Sigma F_V = 0$

$\frac{3}{\sqrt{13}}F_{CG} + 100 = \frac{3}{\sqrt{13}}F_{AG}$

$\frac{3}{\sqrt{13}}F_{CG} + 100 = \frac{3}{\sqrt{13}}(128.20)$

$F_{CG} = 8.01 \, \text{ kN tension}$           answer

$6R_D = 2(100) + 4(120)$

$R_D = 113.33 \, \text{ kN}$
 

At joint F
$F_{EF} = 0$

At joint D
$\Sigma F_V = 0$

$\frac{3}{\sqrt{13}}F_{DE} = R_D$

$\frac{3}{\sqrt{13}}F_{DE} = 113.33$

$F_{DE} = 136.21 \, \text{ kN compression}$
 

$\Sigma F_H = 0$

$F_{CD} = \frac{2}{\sqrt{13}}F_{DE}$

$F_{CD} = \frac{2}{\sqrt{13}}(136.21)$

$F_{CD} = 75.56 \, \text{ kN tension}$
 

At joint E
$\Sigma F_V = 0$

$F_{CE} = \frac{3}{\sqrt{13}}F_{DE}$

$F_{CE} = \frac{3}{\sqrt{13}}(136.21)$

$F_{CE} = 113.33 \, \text{ kN tension}$

At joint C
$\Sigma F_V = 0$

$\frac{3}{\sqrt{13}}F_{CG} + F_{CE} = 120$

$\frac{3}{\sqrt{13}}F_{CG} + 113.33 = 120$

$F_{CG} = 8.02 \, \text{ kN tension}$           answer