Step 1: Determination of stability.

Here, m = 9, j = 6, r = 3

$\therefore$It satisfies the relation m = 2j - r. This implies that the given truss is a perfect truss.

Step 2: Determination of Support Reactions.

• Applying Conditions of Equilibrium to the entire truss,

$i) \space \sum F_x = 0$

$\mathbf{\therefore H_A = 0\space kN}$

$ii) \sum F_y = 0$

$6 \times 2 + 9 \times 6 - V_B \times 8 = 0$

$\mathbf{\therefore V_B = 8.25 k N \space \cdots (\uparrow)}$

$iii) \sum F_y = 0$

$\therefore V_A -6 - 9 +V_B= 0$

$V_A -15 +8.25 = 0 $

$\mathbf{\therefore V_A = 6.75 k N \space \cdots (\uparrow)}$

Step 3: Isolation of Joints.

Isolating Joint A →

Apply Conditions of Equilibrium.

$i) \sum F_y = 0$

$F_{AC} \space sin\space 56.31^\circ +6.75 =0$

$\mathbf {\therefore F_{AC} =-8.11 kN\space \cdots (C)}$

$ii) \sum F_x = 0$

$F_{AC} \space cos \space 56.31^\circ +F_{AE} = 0$

$\therefore(-8.11) \space cos \space 56.31^\circ +F_{AE} = 0$

$\mathbf{\therefore F_{AE} = 4.5 kN \space \cdots (T)}$

Isolating Joint E →

Apply Conditions of Equilibrium.

$i) \sum Fy = 0$

$F_{EC} -6 =0$

$\mathbf {\therefore F_{EC} = 6 kN \space \cdots (T)}$

$ii) \sum F_x =0$

$F_{EF} - F_{AE} =0$

$\therefore F_{EF} - (-4.5)=0$

$\mathbf {\therefore F_{EF} = - 4.5 kN \space \cdots (C)}$

Isolating Joint C →

Apply Conditions of Equilibrium

$i) \sum F_y = 0$

$- F_{AC} \space sin \space 33.69^\circ -F_{EC}-F_{CF}\space sin\space36.87^\circ=0$

$\therefore -(-8.11) \space sin \space 33.69^\circ -(-6)-F_{CF}\space sin\space36.87^\circ=0$

$\mathbf {\therefore F_{CF} = 17.5 kN \space \cdots (T)}$

$ii) \sum F_x=0$

$- F_{AC} \space cos \space 33.69^\circ +F_{CF}\space cos\space36.87^\circ + F_{CD}=0$

$\therefore - (-8.11) \space cos \space 33.69^\circ +(17.5)\space cos\space36.87^\circ + F_{CD}=0$

$\mathbf {\therefore F_{CD} = -20.74 kN \space \cdots (C)}$

Isolating Joint B →