Problem 438 - Truss With Redundant Members | Engineering Mechanics Review at MATHalino

Problem 438
The center diagonals of the truss in Figure P-438 can support tension only. Compute the force in each center diagonal and the force in BC, DE, and FG.

Solution 438

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$\Sigma M_H = 0$

$10R_H = 4(30) + 3(60) + 6(60) + 14(20)$

$R_H = 94 ~ \text{kN}$

From FBD of Section Through M-M

$\Sigma F_V = 0$

$F_{BE} \sin 45^\circ + 20 = 94$

$F_{BE} = 74\sqrt{2} ~ \text{kN} = 104.65 ~ \text{kN tension}$

From FBD of Joint C

$\Sigma F_V = 0$

$F_{BC} = 94 ~ \text{kN compression}$

From FBD of Section Through N-N

$\Sigma F_V = 0$

$\frac{4}{5}F_{DG} + 20 + 60 = 94$

$F_{DG} = 17.5 ~ \text{kN}$

From FBD of Joint E

$\Sigma F_V = 0$

$F_{DE} + 60 = F_{BE}\sin 45^\circ$

$F_{DE} + 60 = \left( 74\sqrt{2} \right) \sin 45^\circ$

$F_{DE} = 40 ~ \text{kN compression}$

From FBD of Joint G

$\Sigma F_V = 0$

$F_{FG} + \frac{4}{5}F_{DG} = 60$

$F_{FG} + \frac{4}{5}(17.5) = 60$

$F_{FG} = 46 ~ \text{kN tension}$

Problem 438 - Truss With Redundant Members

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