Step 1

We are given that the load at point B is ${P_1} = 10\;{\rm{kN}}$ and the load at point D is ${P_2} = 8\;{\rm{kN}}$.

We are asked to calculate the force in each member of truss.

We have the distance between point A and B is ${d_1} = 1\;{\rm{m}}$.

We have the distance between point B and C is ${d_2} = 2\;{\rm{m}}$.

We have the distance between point C and D is ${d_3} = 1\;{\rm{m}}$.

We have the perpendicular distance between point A and G is $h = 2\;{\rm{m}}$.

 
Step 2

The free body diagram of the system is shown as:

 
Step 3

According to Pythagoras theorem,

\[DE = \sqrt {d_3^2 + {h^2}} \]
 
Step 4

Substitute the known value in the equation:

\[\begin{array}{c} DE = \sqrt {{{\left( {1\;{\rm{m}}} \right)}^2} + {{\left( {2\;{\rm{m}}} \right)}^2}} \\ = \sqrt {1\;{{\rm{m}}^2} + 4\;{{\rm{m}}^2}} \\ = \sqrt 5 \;{\rm{m}} \end{array}\]
 
Step 5

Therefore, the value of first angle is given as,

\[\sin {\theta _1} = \frac{2}{{\sqrt 5 }},\cos {\theta _1} = \frac{1}{{\sqrt 5 }}\]
 
Step 6

Applying the equilibrium of force along y-axis at point D:

\[\begin{array}{c} \sum {F_y} = 0\\ {F_{DE}}\sin {\theta _1} - {P_2} = 0 \end{array}\]
 
Step 7

Substitute the known values in the equation:

\[\begin{array}{c} {F_{DE}}\left( {\frac{2}{{\sqrt 5 }}} \right) - 8\;{\rm{kN}} = 0\\ {F_{DE}} = \left( {8\;{\rm{kN}}} \right)\left( {\frac{{2.236}}{2}} \right)\\ = 8.94\;{\rm{kN}} \end{array}\]

Therefore, the force is tension in member DE.

 
Step 8

Applying the force equilibrium along x-axis at point D:

\[\begin{array}{c} \sum {F_x} = 0\\ {F_{DC}} - {F_{DE}}\cos {\theta _1} = 0 \end{array}\]
 
Step 9

Substitute the known values in the equation:

\[\begin{array}{c} {F_{DC}} - \left( {8.94\;{\rm{kN}}} \right)\left( {\frac{1}{{\sqrt 5 }}} \right) = 0\\ {F_{DC}} = \left( {8.94\;{\rm{kN}}} \right)\left( {\frac{1}{{2.236}}} \right)\\ = 4\;{\rm{kN}} \end{array}\]

Therefore, the force is compression in member DC.

 
Step 10

Applying the force equilibrium along x-axis at point C:

\[\begin{array}{c} \sum {F_x} = 0\\ {F_{CB}} - {F_{DC}} = 0\\ {F_{CB}} = {F_{DC}} \end{array}\]
 
Step 11

Substitute the known value in the equation:

\[{F_{CB}} = 4\;{\rm{kN}}\]

Therefore, the force is compression in member CB.

 
Step 12

Applying the force equilibrium along y-axis at point C:

\[\begin{array}{c} \sum {F_y} = 0\\ {F_{CE}} = 0 \end{array}\]
 
Step 13

According to Pythagoras theorem,

\[BE = \sqrt {d_2^2 + {h^2}} \]
 
Step 14

Substitute the known values in the equation:

\[\begin{array}{c} BE = \sqrt {{{\left( {2\;{\rm{m}}} \right)}^2} + {{\left( {2\;{\rm{m}}} \right)}^2}} \\ = \sqrt {4\;{{\rm{m}}^2} + 4\;{{\rm{m}}^2}} \\ = 2\sqrt 2 \;{\rm{m}} \end{array}\]
 
Step 15

Therefore, the second angle of the system is given as,

\[\sin {\theta _2} = \frac{2}{{2\sqrt 2 }},\cos {\theta _2} = \frac{2}{{2\sqrt 2 }}\]
 
Step 16

Applying the force equilibrium along y-axis at point E:

\[\begin{array}{c} \sum {F_y} = 0\\ {F_{EB}}\sin {\theta _2} - {F_{DE}}\sin {\theta _1} = 0 \end{array}\]
 
Step 17

Substitute the known values in the equation:

\[\begin{array}{c} {F_{EB}}\left( {\frac{2}{{2\sqrt 2 }}} \right) - \left( {8.94\;{\rm{kN}}} \right)\left( {\frac{2}{{\sqrt 5 }}} \right) = 0\\ {F_{EB}}\left( {\frac{1}{{\sqrt 2 }}} \right) = \left( {8.94\;{\rm{kN}}} \right)\left( {\frac{2}{{\sqrt 5 }}} \right)\\ {F_{EB}} = 11.3\;{\rm{kN}} \end{array}\]

Therefore, the force is compression in member EB.

 
Step 18

Applying the force equilibrium along x-axis at point E:

\[\begin{array}{c} \sum {F_x} = 0\\ {F_{EB}}\cos {\theta _2} + {F_{DE}}\cos {\theta _1} - {F_{EF}} = 0 \end{array}\]
 
Step 19

Substitute the known values in the equation:

\[\begin{array}{c} \left( {11.3\;{\rm{kN}}} \right)\left( {\frac{1}{{\sqrt 2 }}} \right) + \left( {8.94\;{\rm{kN}}} \right)\left( {\frac{1}{{\sqrt 5 }}} \right) - {F_{EF}} = 0\\ {F_{EF}} = 8\;{\rm{kN}} + 4\;{\rm{kN}}\\ = 12\;{\rm{kN}} \end{array}\]

Therefore, the force is tension in member DC.

 
Step 20

Applying the force equilibrium along x-axis at point B:

\[\begin{array}{c} \sum {F_x} = 0\\ {F_{BA}} - {F_{CB}} - {F_{EB}}\cos {\theta _2} = 0 \end{array}\]
 
Step 21

Substitute the known values in the equation:

\[\begin{array}{c} {F_{BA}} - 4\;{\rm{kN}} - \left( {11.3\;{\rm{kN}}} \right)\left( {\frac{2}{{2\sqrt 2 }}} \right) = 0\\ {F_{BA}} = 8\;{\rm{kN}} + 4\;{\rm{kN}}\\ = 12\;{\rm{kN}} \end{array}\]

Therefore, the force is compression in member BA.

 
Step 22

Applying the force equilibrium along y-axis at point B:

\[\begin{array}{c} \sum {F_y} = 0\\ {F_{BF}} - {P_1} - {F_{EB}}\sin {\theta _2} = 0 \end{array}\]
 
Step 23

Substitute the known values in the equation:

\[\begin{array}{c} {F_{BF}} - 10\;{\rm{kN}} - \left( {11.3\;{\rm{kN}}} \right)\left( {\frac{2}{{2\sqrt 2 }}} \right) = 0\\ {F_{BF}} = 8\;{\rm{kN}} + 10\;{\rm{kN}}\\ = 18\;{\rm{kN}} \end{array}\]

Therefore, the force is tension in member BF.

 
Step 24

Applying the force equilibrium along y-axis at point F:

\[\begin{array}{c} \sum {F_y} = 0\\ {F_{FA}}\sin {\theta _1} - {F_{BF}} = 0 \end{array}\]
 
Step 25

Substitute the known values in the equation:

\[\begin{array}{c} {F_{FA}}\left( {\frac{2}{{\sqrt 5 }}} \right) - 18\;{\rm{kN}} = 0\\ {F_{FA}} = \left( {18\;{\rm{kN}}} \right)\left( {\frac{{\sqrt 5 }}{2}} \right)\\ = 20.1\;{\rm{kN}} \end{array}\]

Therefore, the force is compression in member FA.

 
Step 26

Applying the force equilibrium along x-axis at point B:

\[\begin{array}{c} \sum {F_x} = 0\\ {F_{EF}} + {F_{FA}}\sin {\theta _2} - {F_{FG}} = 0 \end{array}\]
 
Step 27

Substitute the known values in the equation:

\[\begin{array}{c} 12\;{\rm{kN}} + \left( {20.1\;{\rm{kN}}} \right)\left( {\frac{1}{{\sqrt 5 }}} \right) - {F_{FG}} = 0\\ {F_{FG}} = 12\;{\rm{kN}} + 9\;{\rm{kN}}\\ = 21\;{\rm{kN}} \end{array}\]

Therefore, the force is tension in member FG.