**Problem 419**

Use the method of sections to determine the force in members BD, CD, and CE of the Warren truss described in Problem 408.

**Solution 419**

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$\Sigma M_A = 0$

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

$R_E = 4500 ~ \text{kN}$

**At section through M-M**

CDE is equilateral triangle, thus, CD = 10 ft.

$y = 10 \sin 60^\circ = 5\sqrt{3} ~ \text{ft}$

$\Sigma M_C = 0$

$yF_{BD} + 5(3000) = 10(4500)$

$5\sqrt{3}F_{BD} + 5(3000) = 10(4500)$

$F_{BD} = 3464.10 ~ \text{lb}$ compression *answer*

$\Sigma F_V = 0$

$F_{CD} \sin 60^\circ + 3000 = 4500$

$F_{CD} = 1732.05 ~ \text{lb}$ tension *answer*

$\Sigma M_D = 0$

$yF_{CE} = 5(4500)$

$5\sqrt{3}F_{CE} = 5(4500)$

$F_{CE} = 2598.08 ~ \text{lb}$ tension *answer*