Problem 433 - Scissors Truss by Method of Sections

By symmetry
$R_A = R_J = \frac{1}{2}(12 \times 5) = 30 ~ \text{kN}$
 

 

By ratio and proportion

$\dfrac{a}{10} = \dfrac{24}{15}$ $a = 16 ~ \text{m}$

$\dfrac{b}{15} = \dfrac{a}{20}$ $\dfrac{b}{15} = \dfrac{16}{20}$ $b = 12 ~ \text{m}$

 

From FBD of Joint A
$\Sigma F_H = 0$

$\frac{5}{\sqrt{89}}F_{AB} + \frac{5}{\sqrt{41}}F_{AC} = 0$   ←   Equation (1)
 

$\Sigma F_V = 0$

$\frac{8}{\sqrt{89}}F_{AB} + \frac{4}{\sqrt{89}}F_{AC} + 30 = 0$

$\frac{8}{\sqrt{89}}F_{AB} + \frac{4}{\sqrt{89}}F_{AC} = -30$   ←   Equation (2)
 

From Equations (1) and (2)
$F_{AB} = -70.75 ~ \text{kN} = 70.75 ~ \text{kN compression}$       answer

$F_{AC} = 48.02 ~ \text{kN tension}$       answer
 

From FBD of Section Through M-M
$\Sigma M_E = 0$

$(24 - 12)\left( \frac{5}{\sqrt{89}}F_{DF} \right) + 15(30) = 5(12) + 10(12)$

$F_{DF} = - \frac{9}{2}\sqrt{89} ~ \text{kN}$

$F_{DF} = 42.45 ~ \text{kN compression}$       answer
 

 

$\Sigma M_A = 0$

$30\left( \frac{4}{\sqrt{41}}F_{DE} \right) + 10(12) + 5(12) = 0$

$F_{DE} = -\frac{3}{2}\sqrt{41} ~ \text{kN}$

$F_{DE} = 9.60 ~ \text{kN compression}$       answer