**Problem 409**

Determine the force in members AB, BD, BE, and DE of the Howe roof truss shown in Fig. P-409.

**Solution 409**

$\Sigma M_H = 0$

$12R_A = 9(2.7) + 6(4.5) + 3(1.8)$

$R_A = 4.725 \, \text{kN}$

**At Joint A**

$\Sigma F_V = 0$

$F_{AB} \sin 30^\circ = 4.725$

$F_{AB} = 9.45 \, \text{kN}$ compression *answer*

**At Joint C**

By inspection

$F_{BC} = 2.7 \, \text{kN}$ tension

**At Joint B**

By inspection

$F_{BE} = 2.7 \, \text{kN}$ compression *answer*

$\Sigma F_x = 0$

$F_{BD} + F_{BE} \cos 60^\circ + 2.7 \cos 60^\circ = 9.45$

$F_{BD} + 2.7 \cos 60^\circ + 2.7 \cos 60^\circ = 9.45$

$F_{BD} = 6.75 \, \text{kN}$ compression *answer*

**At Joint D**

By inspection

$F_{DF} = 6.75 \, \text{kN}$ compression

$\Sigma F_V = 0$

$F_{DE} = F_{DF} \sin 30^\circ + 6.75 \sin 30^\circ$

$F_{DE} = 6.75 \sin 30^\circ + 6.75 \sin 30^\circ$

$F_{DE} = 6.75 \, \text{kN}$ tension *answer*