**Problem 408**

Compute the force in each member of the Warren truss shown in Fig. P-408.

**Solution 408**

$20R_A = 1000(15) + 4000(10) + 3000(5)$

$R_A = 3500 \, \text{lb}$

$M_E = 0$

$20R_E = 1000(5) + 4000(10) + 3000(15)$

$R_E = 4500 \, \text{lb}$

**At Joint A**

$\Sigma F_V = 0$

$F_{AB} \sin 60^\circ = 3500$

$F_{AB} = 4041.45 \, \text{lb}$ compression

$\Sigma F_H = 0$

$F_{AC} = F_{AB} \cos 60^\circ$

$F_{AC} = 4041.45 \cos 60^\circ$

$F_{AC} = 2020.72 \, \text{lb}$ tension

**At Joint B**

$\Sigma F_V = 0$

$F_{BC} \sin 60^\circ + 1000 = 4041.45 \sin 60^\circ$

$F_{BC} = 2886.75 \, \text{lb}$ tension

$\Sigma F_H = 0$

$F_{BD} = 4041.45 \cos 60^\circ + F_{BC} \cos 60^\circ$

$F_{BD} = 4041.45 \cos 60^\circ + 2886.75 \cos 60^\circ$

$F_{BD} = 3464.10 \, \text{lb}$ compression

**At Joint C**

$\Sigma F_V = 0$

$F_{CD} \sin 60^\circ + 2886.75 \sin 60^\circ = 4000$

$F_{CD} = 1732.05 \, \text{lb}$ tension

$\Sigma F_H = 0$

$F_{CE} + F_{CD} \cos 60^\circ = 2020.72 + 2886.75 \cos 60^\circ$

$F_{CE} + 1732.05 \cos 60^\circ = 2020.72 + 2886.75 \cos 60^\circ$

$F_{CE} = 2598.07 \, \text{lb}$ tension

**At Joint D**

$\Sigma F_V = 0$

$F_{DE} \sin 60^\circ = 1732.05 \sin 60^\circ + 3000$

$F_{DE} = 5196.15 \, \text{lb}$ compression

$\Sigma F_H = 0$

$F_{DE} \cos 60^\circ + 1732.05 \cos 60^\circ = 3464.10$

$5196.15 \cos 60^\circ + 1732.05 \cos 60^\circ = 3464.10$

$3464.10 = 3464.10$ *Check!*

**At Joint E**

$\Sigma F_V = 0$

$5196.15 \sin 60^\circ = 4500$

$4500 = 4500$ *Check!*

$\Sigma F_H = 0$

$5196.15 \cos 60^\circ = 2598.07$

$2598.07 = 2598.07$ *Check!*

**Summary**

AB = 4041.45 lb compression

AC = 2020.72 lb tension

BC = 2886.75 lb tension

BD = 3464.10 lb compression

CD = 1732.05 lb tension

CE = 2598.07 lb tension

DE = 5196.15 lb compression