(i) Reactions at D and C

(ii) Zero force members

(iii) Forces in members FE and DC by method of sections and

(iv) Forces in other members by method of joints.

Hinge support is D and Roller support is C

Now $\sin30 = \dfrac xL$

Applying static conditions of equilibrium,

$∴ ΣM_D = 0 ( +ve) \\ ∴ -20 ×\dfrac L2+ R_C× L-50×1.5=0 \\ ∴R_C=75+10=85 kN (↑) \\ ∴Σ F_x (→ +ve)=0 \\ ∴H_D=0 \\ ∴Σ F_y (↑ +ve)=0 \\ ∴V_D=20+50-85= -15kN \\ ∴V_D=15 kN (↓) \\ ∴ R_D= \sqrt{H_D^2+V_D^2 }=15 kN (↓) \\ θ=90° $

By method of inspection,

$F_{FC}= F_{FB}=0 kN $

Applying static conditions of equilibrium,

$∴Σ F_y (↑ +ve)=0 \\ F_{DE} \sin30-15=0 \\ ∴F_{DE}=\dfrac {15}{\sin30}=\dfrac {15}{0.5}=30 kN (↑) \\ ∴Σ F_x (→ +ve)=0 \\ ∴F_{DE} \cos30-F_{Dc}=0 \\ ∴F_{Dc}= -30 \cos30= -25.98 kN $

Applying static conditions of equilibrium,

$∴Σ F_y (↑ +ve)=0 \\ ∴ -20-F_{FE} \sin30-15-F_{EC} \sin30=0 \\ ∴F_{FE} \sin30+ F_{EC} \sin30= -35 \\ ∴ F_{FE}+ F_{EC}= -70……………….(I) \\ ∴Σ F_x (→ +ve)=0\\ ∴-F_{FE} \cos30+ F_{EC} \cos30 + F_{DC}= 0 \\ ∴-F_{FE} \cos30+ F_{EC} \cos30= -F_{DC}= -(-30 \cos30) \\ ∴ -F_{FE}+ F_{EC}= +30……………………(II) $

Solving (I) and (II) simultaneously ,

$F_{FE}= -50kN , F_{EC}= -20kN \\ ∴F_{FE}= 50kN(↑) , F_{EC}= 20kN (C) $

Method of joints :

Isolating joint A :

Applying static conditions of equilibrium,

$∴Σ F_x (→ +ve)=0 \\ ∴-F_{AF} \sin60-F_{AB} \sin30 = 0 \\ ∴F_{AF} \sin60+F_{AB} \sin30 = 0……………..(III) \\ ∴Σ F_y (↑ +ve)=0 \\ ∴-F_{AB} \cos30-F_{AF} \cos60-50 = 0\\ ∴F_{AB} \cos30+F_{AF} \cos60=-50 \space \space or \space \space F_{AF} \cos60+F_{AB} \cos30=-50…………….(IV) $

Solving equations (III) and (IV) simultaneously,

$∴F_{AF}=50kN (↑)F_{AB}=-86.6kN \\ ∴F_{AB}=86.6kN (c) $

Isolating joint B :

Applying static conditions of equilibrium,

$∴Σ F_y (↑ +ve)=0 \\ ∴F_{AB} \sin60-F_{BC} \cos30 = 0 \\ ∴F_{AB}=F_{BC}………………..{ Since, \sin60 = \cos30} \\ ∴F_{BC}=86.6 kN (c) \\ Ans: F_{AB}=86.6kN (c),F_{BC}=86.6 kN (c),F_{AF}=50kN (↑), F_{FE}= 50kN(↑) , \\ F_{ED}=30 kN (↑),F_{Dc}=-25.98 kN ,F_{EC}= 20kN (C),F_{FC}= F_{FB}=0 kN$