$ \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 = FH = 12\,{\rm{ft}} \\
AC = 18\,{\rm{ft}} \\
AF = 45\,{\rm{ft}} \\
AG = 54\,{\rm{ft}}
\end{array}

$ \text{Step 2: Support Reactions} $

Moment equilibrium about A:
\begin{array}{c}
\sum M_A = 0 \\
G_y(54) – 5000(45) – 8000(18) – 4000(9) = 0 \\
54G_y = 225,\!000 + 144,\!000 + 36,\!000 \\
\hline
G_y = 7500\,{\rm{lb}}\,(\uparrow)
\end{array}

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

Angle calculation:
\begin{array}{c}
\tan \theta = \frac{FH}{FG} = \frac{12}{9} \\
\theta = \tan^{-1}\left(\frac{12}{9}\right) \\
\hline
\theta \approx 53.1^\circ
\end{array}

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

Vertical equilibrium at F:
\begin{array}{c}
\sum F_y = 0 \\
F_{EI} + 7500 – 5000 = 0 \\
\hline
F_{EI} = -2500\,{\rm{lb}}\,(\rm{C})
\end{array}

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

Moment equilibrium about G:
\begin{array}{c}
\sum M_G = 0 \\
5000(9) – (-2500)(18) – F_{JI}(12) = 0 \\
45,\!000 + 45,\!000 = 12F_{JI} \\
\hline
F_{JI} = 7500\,{\rm{lb}}\,(\rm{T})
\end{array}

$ \text{Step 6: Results Summary} $

\begin{array}{c}
\text{EI: } 2500\,{\rm{lb}}\,(\rm{C}) \\
\text{JI: } 7500\,{\rm{lb}}\,(\rm{T})
\end{array}

$ \text{Step 7: Verification} $

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