Step 1

We are given the horizontal force at point D is ${F_1} = 3\;{\rm{kN}}$, the vertical force at point B is ${F_2} = 4\;{\rm{kN}}$, and the angle $\theta $ is $\theta = 30^\circ $.

We are asked the force in each member of the truss and state if the members are in the tension or compression.

 
Step 2

The following is the free body diagram of the system:

We have the length of member BD is $BD = 1.5\;{\rm{m}}$.

We have the length of member AB is $AB = 2\;{\rm{m}}$.

We have the length of member BC is $BC = 2\;{\rm{m}}$.

 
Step 3

The magnitude of angle $\theta $ is,

\[\tan \phi = \frac{{BD}}{{AB}}\]
 
Step 4

Substitute the values in the above expression.

\[\begin{array}{c} \tan \phi = \frac{{1.5\;{\rm{m}}}}{{2\;{\rm{m}}}}\\ \phi = 36.87^\circ \end{array}\]
 
Step 5

The reaction at point C by moment of all the forces about point A is,

\[\begin{array}{c} \Sigma {M_A} = 0\\ - {F_2}\left( {AB} \right) - {F_1}\left( {BD} \right) + {R_C}\cos \theta \left( {AB + BC} \right) = 0 \end{array}\] Step 6

Substitute the values in the above expression.

\[\begin{array}{c} \left[ \begin{array}{c} - \left( {4\;{\rm{kN}}} \right)\left( {2\;{\rm{m}}} \right) - \left( {3\;{\rm{kN}}} \right)\left( {1.5\;{\rm{m}}} \right)\\ + {R_C}\cos 30^\circ \left( {2\;{\rm{m}} + 2\;{\rm{m}}} \right) \end{array} \right] = 0\\ {R_C} = \frac{{12.5\;\;{\rm{kN}} \cdot {\rm{m}}}}{{4\cos 30^\circ \;{\rm{m}}}}\\ {R_C} = 3.61\;{\rm{kN}} \end{array}\]
 
Step 7

The horizontal force at joint A of the truss by equilibrium condition in horizontal direction is,

\[\begin{array}{c} \Sigma {F_x} = 0\\ {A_x} + {F_1} - {R_C}\sin \theta = 0 \end{array}\]
 
Step 8

Substitute the values in the above expression.

\[\begin{array}{c} {A_x} + \left( {3\;{\rm{kN}}} \right) - \left( {3.61\;{\rm{kN}}} \right)\sin 30^\circ = 0\\ {A_x} = - 1.20\;{\rm{kN}} \end{array}\]
 
Step 9

The vertical force at joint A of the truss by equilibrium condition in vertical direction is,

\[\begin{array}{c} \Sigma {F_y} = 0\\ {A_y} - {F_2} + {R_C}\cos \theta = 0 \end{array}\] Step 10

Substitute the values in the above expression.

\[\begin{array}{c} {A_y} - \left( {4\;{\rm{kN}}} \right) + \left( {3.61\;{\rm{kN}}} \right)\cos 30^\circ = 0\\ {A_y} = - 0.87\;{\rm{kN}} \end{array}\]
 
Step 11

The free body diagram of forces acting at point A is:

 
Step 12

The force on member AD by equilibrium condition is,

\[\begin{array}{c} \Sigma {F_y} = 0\\ {A_y} + {F_{AD}}\sin \phi = 0 \end{array}\]
 
Step 13

Substitute the values in the above expression.

\[\begin{array}{c} - 0.87\;{\rm{kN}} + {F_{AD}}\sin 36.87^\circ = 0\\ {F_{AD}} = 1.45\;{\rm{kN}}\left( {{\rm{Compression}}} \right) \end{array}\]
 
Step 14

The force on member AB by equilibrium condition is,

\[\begin{array}{c} \Sigma {F_x} = 0\\ {A_x} + {F_{AD}}\cos \phi + {F_{AB}} = 0 \end{array}\]
 
Step 15

Substitute the values in the above expression.

\[\begin{array}{c} - 1.20\;{\rm{kN}} - \left( {1.45\;{\rm{kN}}} \right)\cos 36.87^\circ + {F_{AB}} = 0\\ {F_{AB}} = 2.36\;{\rm{kN}}\left( {{\rm{Compression}}} \right) \end{array}\]
 
Step 16

The free body diagram of forces acting at point B is:

 
Step 17

The force on member BD by equilibrium condition is,

\[ - {F_2} + {F_{BD}} = 0\]
 
Step 18

Substitute the values in the above expression.

\[\begin{array}{c} - 4\;{\rm{kN}} + {F_{BD}} = 0\\ {F_{BD}} = 4\;{\rm{kN}}\left( {{\rm{Compression}}} \right) \end{array}\]
 
Step 19

The force on member BC by equilibrium condition is,

\[ - {F_{AB}} + {F_{BC}} = 0\]
 
Step 20

Substitute the values in the above expression.

\[\begin{array}{c} - 2.36\;{\rm{kN}} + {F_{BC}} = 0\\ {F_{BC}} = 2.36\;{\rm{kN}}\left( {{\rm{Compression}}} \right) \end{array}\]
 
Step 21

The free body diagram of forces acting at point C is:

 
Step 22

The force on member CD by equilibrium condition is,

\[{R_C}\cos \theta + {F_{CD}}\sin \phi = 0\]
 
Step 23

Substitute the values in the above expression.

\[\begin{array}{c} \left( {3.61\;{\rm{kN}}} \right)\cos 30^\circ + {F_{CD}}\sin 36.87^\circ = 0\\ {F_{CD}} = 5.21\;{\rm{kN}}\left( {{\rm{Compression}}} \right) \end{array}\]