Question

Determine the force in the member CK of the loaded truss. Ans. CK = 9290 N (Compression)

Please show detailed steps in solving for force.

Answer 1

\(F B D\) of a weight:

Take forces acting along \(y\) -direction \(\Sigma F_{y}=0\)

\(2 T \sin 60^{\circ}-m g=0\)

\(2 T \sin 60^{\circ}=1000 \times 9.81\)

\(T=5663.8 \mathrm{~N}\)

\(F B D\) of truss:

Take point from (0,0) \(C=(0,2.25) \mathrm{m}\)

\(L=(0,4.5) \mathrm{m}\)

\(H=\left(6 \cos 15^{\circ}, 4.5+6 \sin 15^{\circ}\right)\)

\(H=(5.8,6.05) \mathrm{m}\)

\(G=\left(6 \cos 15^{\circ}+1 \cos 75^{\circ}-4.5+6 \sin 75^{\circ}-1 \sin 75^{\circ}\right)\)

\(=(6.05,5.09) \mathrm{m}\)

Ang le \(\theta=\tan ^{-1}\left(\frac{5.09-2.25}{6.05}\right)\)

\(=1.524 \mathrm{~m}\)

Take point at \({ }^{2} E^{\prime}\) \(E=\left(6 \cos 15^{\circ}+1 \cos 75^{\circ}-2\left(1.524 \cos 25.1^{\circ}\right), 4.5+6 \sin 15^{\circ}-1 \sin 75^{\circ}-2\left(1.524 \sin 25.1^{\circ}\right)\right)\)

\(E=(3.30,3.79) \mathrm{m}\)

Take moment at ' \(B^{\prime}\) \(\Sigma M_{B}=0\)

\(-A_{\gamma}(3)-T \sin 60^{\circ}-(3.30)-T \cos 60^{\circ}(3.79)-T \sin 60^{\circ}(6.05)+T \cos 60^{\circ}(5.09)=0\)

\(-3 A_{3}-T 2.85-T(1.895)-5.23 T+2.545 T=0\)

\(-3 A_{y}-7.43 \times 5664=0\)

\(A_{y}=-14070 \mathrm{~N}\)

Take force along \(y\) -direction \(\Sigma F_{2}=0\)

\(-1470+B,-1000 \times 9.81=0\)

\(B_{\gamma}=23900 \mathrm{~N}\)

Cutting at the sections at \(L K, C K,\) and \(C D\) and separate it

$$ \begin{aligned} \beta &=\tan ^{-1}\left(\frac{1.5 \cos 15^{\circ}}{2.25+1.5 \sin 15^{\circ}}\right) \\ &=28.8^{\circ} \end{aligned} $$

Take moment at \(C\) \(\Sigma M_{C}=0\)

\(14070 \times 3-F_{L X}(2.25)\left(\sin 75^{\circ}\right)=0\)

\(F_{L X}=19420 \mathrm{~N}\)

So, \(\Sigma F_{x}=0\)

$$ \begin{aligned} 19420 \cos 15^{\circ}+F_{C X} \sin 28.8^{\circ}+F_{C D} \sin 25.1^{\circ} &=0 \\ 19420 \times 0.96+F_{C X} 0.481+F_{\infty} 0.424 &=0 \end{aligned} $$

Now, \(\Sigma F_{z}=0\)

\(-14070+23960+19420 \sin 15^{\circ}+F_{c x} \cos 28.8^{\circ}+F_{\infty} \sin 25.1^{\circ}=0\)

Solving (1) and (2) equation we get \(F_{C K}=-9290 \mathrm{~N}\)