Problem 434 - Scissors Truss by Method of Sections | Engineering Mechanics Review at MATHalino

Problem 434
Compute the force in bars GI, GH, EH, and HI for the scissors truss shown in Fig. P-433.

Solution 434

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

By ratio and proportion

$\dfrac{a}{10} = \dfrac{24}{15}$

$a = 16 ~ \text{m}$

$\dfrac{b}{15} = \dfrac{a}{20}$

$\dfrac{b}{15} = \dfrac{16}{20}$

$\dfrac{c}{5} = \dfrac{24}{15}$

$c = 8 ~ \text{m}$

From FBD Through Section M-M

$\Sigma M_H = 0$

$5\left( \frac{8}{\sqrt{89}}F_{GI} \right) + 10(30) = 5(12)$

$F_{GI} = -6\sqrt{89} ~ \text{kN}$

$F_{GI} = 56.60 ~ \text{kN compression}$ answer

$\Sigma M_J = 0$

$10F_{GH} = 5(12)$

$F_{GH} = 6 ~ \text{kN tension}$ answer

$\Sigma M_G = 0$

$(16 - 8)\left( \frac{5}{\sqrt{41}}F_{EH} \right) + 5(12) = 10(30)$

$F_{EH} = 6\sqrt{41} ~ \text{kN} = 38.42 ~ \text{kN tension}$ answer

From FBD Through Section N-N

$\Sigma M_J = 0$

$8F_{HI} + 5(12) = 0$

$F_{HI} = -7.5 ~ \text{kN}$

$F_{HI} = 7.5 ~ \text{kN compression}$ answer

Problem 434 - Scissors Truss by Method of Sections