Problem 419 - Warren Truss by Method of Sections

 

 

$\Sigma M_A = 0$

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

$R_E = 4500 ~ \text{kN}$
 

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