Problem 002-mj | Method of Joints | Engineering Mechanics Review at MATHalino

Problem 002-mj
The structure in Fig. T-02 is a truss which is pinned to the floor at point A, and supported by a roller at point D. Determine the force to all members of the truss.

Solution 002-mj

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$\Sigma M_D = 0$

$6R_A = 5(12) + 3(20)$

$R_A = 20 \, \text{ kN}$

$\Sigma M_A = 0$

$6R_D = 1(12) + 3(20)$

$R_D = 12 \, \text{ kN}$

At joint A

$\Sigma F_V = 0$

$\frac{\sqrt{21}}{5}F_{AG} = R_A$

$\frac{\sqrt{21}}{5}F_{AG} = 20$

$F_{AG} = 21.82 \, \text{ kN compression}$

$\Sigma F_H = 0$

$F_{AB} = \frac{2}{5}F_{AG}$

$F_{AB} = \frac{2}{5}(21.82)$

$F_{AB} = 8.73 \, \text{ kN tension}$

At joint G

$\Sigma F_V = 0$

$\frac{\sqrt{21}}{5}F_{BG} + 12 = \frac{\sqrt{21}}{5}F_{AG}$

$\frac{\sqrt{21}}{5}F_{BG} + 1 2 = \frac{\sqrt{21}}{5}(21.82)$

$F_{BG} = 8.73 \, \text{ kN tension}$

$\Sigma F_H = 0$

$F_{FG} = \frac{2}{5}F_{AG} + \frac{2}{5}F_{BG}$

$F_{FG} = \frac{2}{5}(21.82) + \frac{2}{5}(8.73)$

$F_{FG} = 12.22 \, \text{ kN compression}$

At joint B

$\Sigma F_V = 0$

$\frac{\sqrt{21}}{5}F_{BF} = \frac{\sqrt{21}}{5}F_{BG}$

$F_{BF} = F_{BG}$

$F_{BF} = 8.73 \, \text{ kN compression}$

$\Sigma F_H = 0$

$F_{BC} = F_{AB} + \frac{2}{5}F_{BG} + \frac{2}{5}F_{BF}$

$F_{BC} = 8.73 + \frac{2}{5}(8.73) + \frac{2}{5}(8.73)$

$F_{BC} = 15.71 \, \text{ kN tension}$

At joint F

$\Sigma F_V = 0$

$\frac{\sqrt{21}}{5}F_{CF} + \frac{\sqrt{21}}{5}F_{BF} = 20$

$\frac{\sqrt{21}}{5}F_{CF} + \frac{\sqrt{21}}{5}(8.73) = 20$

$F_{CF} = 13.09 \, \text{ kN compression}$

$\Sigma F_H = 0$

$F_{EF} + \frac{2}{5}F_{CF} = \frac{2}{5}F_{BF} + F_{FG}$

$F_{EF} + \frac{2}{5}(13.09) = \frac{2}{5}(8.73) + 12.22$

$F_{EF} = 10.48 \, \text{ kN compression}$

At joint C

$\Sigma F_V = 0$

$\frac{\sqrt{21}}{5}F_{CE} = \frac{\sqrt{21}}{5}F_{CF}$

$F_{CE} = F_{CF}$

$F_{CE} = 13.09 \, \text{ kN tension}$

$\Sigma F_H = 0$

$F_{CD} + \frac{2}{5}F_{CE} + \frac{2}{5}F_{CF} = F_{BC}$

$F_{CD} + \frac{2}{5}(13.09) + \frac{2}{5}(13.09) = 15.71$

$F_{CD} = 5.24 \, \text{ kN tension}$

At joint E

$\Sigma F_V = 0$

$\frac{\sqrt{21}}{5}F_{DE} = \frac{\sqrt{21}}{5}F_{CE}$

$F_{DE} = F_{CE}$

$F_{DE} = 13.09 \, \text{ kN compression}$

$\Sigma F_H = 0$

$F_{EF} = \frac{2}{5}F_{CE} + \frac{2}{5}F_{DE}$

$10.48 = \frac{2}{5}(13.09) + \frac{2}{5}(13.09)$

$10.5 = 10.5$ check

At joint D

$\Sigma F_V = 0$

$R_D = \frac{\sqrt{21}}{5}F_{DE}$

$12 = \frac{\sqrt{21}}{5}(13.09)$

$12 = 12$ check

$\Sigma F_{H} = 0$

$F_{CD} = \frac{2}{5}F_{DE}$

$5.24 = \frac{2}{5}(13.09)$

$5.24 = 5.24$ check

Summary

F_AB = 8.73 kN tension
F_AG = 21.82 kN compression
F_BC = 15.71 kN tension
F_BF = 8.73 kN compression
F_BG = 8.73 kN tension
F_CD = 5.24 kN tension
F_CE = 13.09 kN tension
F_CF = 13.09 kN compression
F_DE = 13.09 kN compression
F_EF = 10.48 kN compression
F_FG = 12.22 kN compression

Problem 002-mj | Method of Joints

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