FB and BE using method of section, rest by using method of joints

Applying static conditions of equilibrium,

$∴ ΣM_A = 0 ( +ve ) \\ ∴ -40 ×2-20 ×3+ R_D ×6=0 \\ R_D ×6×6=80+60 \\ R_D=\dfrac {140}6=\dfrac {70}3=23.33 kN ( ↑ ) \\ ∴Σ F_y (↑ +ve)=0 \\ ∴ V_A-40+23.33=0 \\ ∴ V_A=40-23.33=16.67 kN( ↑ ) \\ ∴Σ F_x (→ +ve)=0 \\ ∴ H_A+20=0 \\ ∴ H_A= -20kN \space\space or \space\space H_A=20kN (←) \\ R_A= \sqrt{H_A^2+V_A^2}= \sqrt{20^2+16.67^2 }=26.03 kN \\ θ= \tan^{-1}(\dfrac {V_A}{H_A} )= \tan^{-1}(\dfrac {16.67}{20})= 39.81° $

Method of joints:-

Isolating joint A :-

$∴Σ F_y (↑ +ve)=0 \\ F_{AB} \sin45+16.67=0 \\ ∴ F_{AB}= \dfrac {-16.67}{\sin45} = -23.57 kN \\ ∴ F_{AB}=23.57 kN (c) \\ ∴Σ F_x (→ +ve)=0 \\ ∴ F_{AB} \cos45-20+ F_{AC}=0 \\ ∴-23.57 \cos45-20+ F_{AC}=0 \\ ∴ F_{AC}=20+16.67=36.67 kN (↑) $

Isolating joint D:-

$∴Σ F_y (↑ +ve)=0 \\ F_{DC} \sin56.31+23.33=0 \\ ∴ F_{DC}= \dfrac {-23.33}{\sin56.31 }= -28.04kN \\ ∴ F_{DC}=28.04 kN (c) \\ ∴Σ F_x (→ +ve)=0 \\ ∴ - F_{DE}- F_{DC} \cos56.31=0 \\ ∴ F_{DE}= - F_{DC} \cos56.31= -( -28.04)\cos56.31 \\ ∴ F_{DE}=15.55 kN (↑) $

Isolating joint C:-

$∴Σ F_x (→ +ve)=0 \\ ∴ - F_{CB} \cos45+ F_{DC} \sin33.69+20=0 \\ ∴ F_{CB} \cos45= F_{CD} \sin33.69+20= -28.04 \sin33.69+20 \\ ∴ F_{CB}=6.288 kN (↑) \\ ∴Σ F_y (↑ +ve)=0 \\ - F_{DC} \cos33.69-F_{CE}- F_{CB} \cos45 =0 \\ ∴ F_{CE}= +28.04 \cos33.69-6.288 \cos⁡45 \\ ∴ F_{CE}=18.88 kN (↑) $

Method of sections :

$∴Σ F_x (→ +ve)=0 \\ ∴ - H_A- F_{AB} \cos45 - F_{FE}=0 \\ ∴ -20-23.57 \cos⁡45- F_{FE}=0 \\ ∴ F_{FE}= -36.67 kN=36.67 kN (c) \\ ∴Σ F_y (↑ +ve)=0 \\ ∴ F_{AB} \sin45+ V_A-40- F_{FB}=0 \\ ∴ F_{FB}=-23.57 \sin45-40+16.67 \\ ∴ F_{FB}= -40 kN=40 kN (c)$

Applying static conditions of equilibrium,

$∴Σ F_x (→ +ve)=0 \\ ∴F_{BC} \cos26.57+ F_{BE} \cos45+ F_{FE}-20=0 \\ ∴6.288 \cos26.57+ F_{BE} \cos45+36.67-20=0 \\ ∴ F_{BE}= \dfrac { -22.29}{\cos45}= -31.53 kN \\ ∴ F_{BE}=31.53 kN (c)$