**Problem 433**

Compute the forces in bars AB, AC, DF, and DE of the scissors truss shown in Fig. P433.

**Solution 433**

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

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*