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$6–31$. Determine the force in members $CD$, $CJ$, $KJ$, and  $DJ$ of the truss which serves to support the deck of a bridge.  State if these members are in tension or compression.

$6–31$. Determine the force in members $CD$, $CJ$, $KJ$, and $DJ$ of the truss which serves to support the deck of a bridge. State if these members are in tension or compression.

$6–31$. Determine the force in members $CD$, $CJ$, $KJ$, and $DJ$ of the truss which serves to support the deck of a bridge. State if these members are in tension or compression.

​$ \text{Step by Step Solution​} $

$ \text{Step 1: Given Data} $

Truss with vertical loads:
\begin{array}{c}
W_B = 4000\,{\rm{lb}} \\
W_C = 8000\,{\rm{lb}} \\
W_F = 5000\,{\rm{lb}}
\end{array}

Key dimensions:
\begin{array}{c}
AB = BC = FG = 9\,{\rm{ft}} \\
BL = CK = 12\,{\rm{ft}} \\
CG = 36\,{\rm{ft}} \\
BG = 45\,{\rm{ft}} \\
AG = 54\,{\rm{ft}}
\end{array}

$ \text{Step 2: Support Reactions} $

Moment equilibrium about G:
\begin{array}{c}
\sum M_G = 0 \\
5000(9) + 8000(36) + 4000(45) – A_y(54) = 0 \\
45,\!000 + 288,\!000 + 180,\!000 = 54A_y \\
\hline
A_y = 9500\,{\rm{lb}}\,(\uparrow)
\end{array}

$ \text{Step 3: Section ACKL Analysis} $

Free body diagram of the truss part ACKL

Angle calculation for member $CJ$:
\begin{array}{c}
\tan \theta = \frac{CK}{BC} = \frac{12}{9} \\
\theta = \tan^{-1}\left(\frac{12}{9}\right) \\
\hline
\theta \approx 53.1^\circ
\end{array}

$ \text{Step 4: Force in KJ} $

Moment equilibrium about C:
\begin{array}{c}
\sum M_C = 0 \\
4000(9) + F_{KJ}(12) – 9500(18) = 0 \\
36,\!000 + 12F_{KJ} = 171,\!000 \\
\hline
F_{KJ} = 11,\!250\,{\rm{lb}}\,(\rm{T})
\end{array}

$ \text{Step 5: Force in CJ} $

Vertical equilibrium at C:
\begin{array}{c}
\sum F_y = 0 \\
9500 – F_{CJ}\sin(53.1^\circ) – 4000 – 8000 = 0 \\
-2500 = 0.8F_{CJ} \\
\hline
F_{CJ} = -3125\,{\rm{lb}}\,(\rm{C})
\end{array}

$ \text{Step 6: Force in CD} $

Horizontal equilibrium at C:
\begin{array}{c}
\sum F_x = 0 \\
F_{CD} + 11,\!250 + (-3125\cos 53.1^\circ) = 0 \\
F_{CD} = -11,\!250 + 1875 \\
\hline
F_{CD} = -9375\,{\rm{lb}}\,(\rm{C})
\end{array}

$ \text{Step 7: Force in DJ} $

Vertical equilibrium at D:
\begin{array}{c}
\sum F_y = 0 \\
F_{DJ} = 0\,{\rm{lb}}\,(\text{Inactive})
\end{array}

$ \text{Step 8: Results Summary} $

\begin{array}{c}
\text{KJ: } 11,\!250\,{\rm{lb}}\,(\rm{T}) \\
\text{CJ: } 3,\!125\,{\rm{lb}}\,(\rm{C}) \\
\text{CD: } 9,\!375\,{\rm{lb}}\,(\rm{C}) \\
\text{DJ: } 0\,{\rm{lb}}\,(\text{Inactive})
\end{array}

$ \text{Step 9: Verification} $

\begin{array}{c}
\text{Moment Check: } \sum M_G = 0 \quad \checkmark \\
\text{Force Balance: } \sum F_x = \sum F_y = 0 \quad \checkmark \\
\hline
\text{All equilibrium conditions are satisfied}
\end{array}

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