Step 1

We are given the loads ${W_B} = 1500\;{\rm{lb}}$, ${W_C} = 1500\;{\rm{lb}}$, ${W_D} = 1500\;{\rm{lb}}$, ${W_E} = 1500\;{\rm{lb}}$, ${W_F} = 1500\;{\rm{lb}}$, the length $EF$ as $EF = 3\;{\rm{ft}}$, the length $GE$ as $GE = 4\;{\rm{ft}}$, the length $DE$ as $DE = 3\;{\rm{ft}}$, and the length $HD$ as $HD = 4\;{\rm{ft}}$.

We are asked to determine the force in member $CD$, $HI$ and $CH$, and the state of force.

 
Step 2

We will draw a free body diagram of the truss part $EFG$.

Here, ${F_{HI}}$ is the force in the member $HI$, ${F_{HC}}$ is the force in the member $HC$, and ${F_{CD}}$ is the force in the member $CD$.

 
Step 3

We will find the angle $\theta $ from the figure.

\[\tan \theta = \frac{{GE}}{{EF}}\]

Substitute the given value in the above equation.

\[\begin{array}{c} \tan \theta = \frac{{{\rm{4}}\;{\rm{ft}}}}{{{\rm{3}}\;{\rm{ft}}}}\\ \theta = {\tan ^{ - 1}}\left( {\frac{{{\rm{4}}\;{\rm{ft}}}}{{{\rm{3}}\;{\rm{ft}}}}} \right)\\ \approx 53.1^\circ \end{array}\]
 
Step 4

We will take the moment about point $H$ to find the force in the member $HG$.

\[\begin{array}{c} \sum {{M_H}} = 0\\ - {F_{CD}} \times HD - {W_E} \times DE - {W_F} \times \left( {DE + EF} \right) = 0\\ {F_{CD}} \times HD = - {W_E} \times DE - {W_F} \times \left( {DE + EF} \right) \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} {F_{CD}} \times \left( {4\;{\rm{ft}}} \right) = - \left( {1500\;{\rm{lb}}} \right) \times \left( {3\;{\rm{ft}}} \right) - \left( {1500\;{\rm{lb}}} \right) \times \left( {3\;{\rm{ft}} + 3\;{\rm{ft}}} \right)\\ {F_{CD}} = \frac{{ - \left( {1500\;{\rm{lb}}} \right) \times \left( {3\;{\rm{ft}}} \right) - \left( {1500\;{\rm{lb}}} \right) \times \left( {3\;{\rm{ft}} + 3\;{\rm{ft}}} \right)}}{{4\;{\rm{ft}}}}\\ = - 3375\;{\rm{lb}} \end{array}\]

The force in the member $CD$ is compressive.

 
Step 5

We will resolve the force in vertical direction to find the force in the member $CH$.

\[\begin{array}{c} \sum {{F_y}} = 0\\ - {F_{CH}}\sin \theta - {W_D} - {W_F} - {W_F} = 0\\ {F_{CH}}\sin \theta = - {W_D} - {W_E} - {W_F} \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} {F_{CH}}\left( {\sin 53.1^\circ } \right) = - 1500\;{\rm{lb}} - 1500\;{\rm{lb}} - 1500\;{\rm{lb}}\\ {F_{CH}} = \frac{{ - 4500\;{\rm{lb}}}}{{\left( {\sin 53.1^\circ } \right)}}\\ \approx - 5627.2\;{\rm{lb}} \end{array}\]

The force in the member $CH$ is compressive.

 
Step 6

We will resolve the force in horizontal direction to find the force in the member $HI$.

\[\begin{array}{c} \sum {{F_x}} = 0\\ - {F_{HI}} - {F_{CD}} - {F_{CH}}\cos \theta = 0 \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} - {F_{HI}} - \left( { - 3375\;{\rm{lb}}} \right) - \left( { - 5627.2\;{\rm{lb}}} \right)\left( {\cos 53.1^\circ } \right) = 0\\ - {F_{HI}} + 3375\;{\rm{lb}} + \left( {5627.2\;{\rm{lb}}} \right)\left( {\cos 53.1^\circ } \right) = 0\\ {F_{HI}} = \left( {3375\;{\rm{lb}}} \right) + \left( {5627.2\;{\rm{lb}}} \right)\left( {\cos 53.1^\circ } \right)\\ \approx 6753.7\;{\rm{lb}} \end{array}\]

The force in the member $HI$ is tensile.