Step 1

We are given that force applied at point D is ${F_D} = 5\;{\rm{kN}}$, the force applied at point C is ${F_C} = 10\;{\rm{kN}}$, the force applied at point B is ${F_B} = 10\;{\rm{kN}}$ and the force applied at point E is ${F_E} = 5\;{\rm{kN}}$.

We are asked to calculate the force in members BF, BG and AB of the truss.

 
Step 2

The free body diagram of the point F is shown as:

Here, ${F_{EF}}$ is the force in member EF, ${F_{BF}}$ is the force in member BF and ${F_{FG}}$ is the force in member FG.

 
Step 3

Applying the equilibrium force of equation along x-axis at point F:

\[\begin{array}{c} \sum {F_x} = 0\\ {F_{BF}} = 0 \end{array}\]

Therefore, the force in member BF is neither compression nor tension.

 
Step 4

The free body diagram of the system is shown as:

Here, ${F_{AB}}$ is the force in member AB, ${F_{BG}}$ is the force in member BG and ${F_{FG}}$ is the force in member FG.

We have the angle made by force BG with the horizontal which is $\theta = 45^\circ $.

We have the vertical distance between points D and C which is ${h_1} = 4\;{\rm{m}}$.

We have the vertical distance between points C and B which is ${h_2} = 4\;{\rm{m}}$.

We have the vertical distance between points B and A which is ${h_3} = 4\;{\rm{m}}$.

We have the horizontal distance between points A and G which is $d = 4\;{\rm{m}}$.

 
Step 5

Applying the equilibrium force of equation along x-axis:

\[\begin{array}{c} \sum {F_x} = 0\\ {F_D} + {F_C} + {F_B} - {F_{BG}}\cos \theta = 0 \end{array}\]
 
Step 6

Substitute the known values in the equation:

\[\begin{array}{c} 5\;{\rm{kN}} + 10\;{\rm{kN}} + 10\;{\rm{kN}} - {F_{BG}}\cos 45^\circ = 0\\ - {F_{BG}}\left( {0.707} \right) = - 25\;{\rm{kN}}\\ {F_{BG}} = 35.4\;{\rm{kN}} \end{array}\]

Therefore, the force is compression in member BG.

 
Step 7

Applying the moment of force equation about point G:

\[{F_{AB}}d - {F_B}{h_3} - {F_C}\left( {{h_2} + {h_3}} \right) - {F_D}\left( {{h_1} + {h_2} + {h_3}} \right) = 0\]
 
Step 8

Substitute the known values in the equation:

\[\begin{array}{c} \left\{ \begin{array}{l} {F_{AB}}\left( {4\;{\rm{m}}} \right) - \left( {10\;{\rm{kN}}} \right)\left( {4\;{\rm{m}}} \right) - \\ \left( {10\;{\rm{kN}}} \right)\left( {4\;{\rm{m}} + 4\;{\rm{m}}} \right) - \\ \left( {5\;{\rm{kN}}} \right)\left( {4\;{\rm{m}} + 4\;{\rm{m}} + 4\;{\rm{m}}} \right) \end{array} \right\} = 0\\ {F_{AB}}\left( {4\;{\rm{m}}} \right) = 40\;{\rm{kN}} \cdot {\rm{m}} + 80\;{\rm{kN}} \cdot {\rm{m}} + 60\;{\rm{kN}} \cdot {\rm{m}}\\ {F_{AB}} = \frac{{\left( {180\;{\rm{kN}} \cdot {\rm{m}}} \right)}}{{\left( {4\;{\rm{m}}} \right)}}\\ = 45\;{\rm{kN}} \end{array}\]

Therefore, the force is tension in member AB.