Question

Use the method of sections to determine the forces in members BC, CD, and DE of the truss.

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Answer #1

Take mom ent about point \(A\)

$$ \begin{array}{l} \sum M_{A}=0 \\ -(10 \times 3)-(15 \times 6)+9 E_{y}=0 \\ E_{y}=\frac{120}{9} \\ =13.33 \mathrm{kN} \end{array} $$

Take section vertically between \(B C\) and \(D E\).

Take mom ent about point \(C\).

$$ \begin{array}{l} \sum M_{C}=0 \\ -3 F_{D E}+(13.33 \times 3)=0 \\ F_{D X}=13.33 \mathrm{kN} \end{array} $$

Apply equilibrium equations.

$$ \begin{array}{l} \sum F_{y}=0 \\ -F_{D C} \sin 45^{\circ}-15+13.33=0 \\ F_{D C}=-\frac{1.67}{\sin 45^{\circ}} \\ =-2.36 \mathrm{kN} \\ \sum F_{x}=0 \\ -F_{B C}-F_{D C} \cos 45^{\circ}=0 \end{array} $$

\(F_{B C}=-\left(-2.36 \cos 45^{\circ}\right)\)

$$ =1.67 \mathrm{kN} $$

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