# Problem 438 - Truss With Redundant Members

**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**

$\Sigma M_H = 0$

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

$R_H = 94 ~ \text{kN}$

**From FBD of Section Through M-M**

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

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

**From FBD of Joint C**

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

**From FBD of Section Through N-N**

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

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

**From FBD of Joint E**

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

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

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

**From FBD of Joint G**

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

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

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