Verification of truss

$M= 2j- r \\ 9 = 2 (6) – 3 \\ 0= 0$

LHS = RHS … Hence, perfect truss.

FBD

To apply method of joints at points A, E, B, F and D. It is essential to find reactions at A and B

$∑M = 0$ (about point A)

$$-6 \times 2 – 9 \times 6 + Rb \times 8 = 0$$

$Rb = 6 kN \\ ∑Fy = 0 \\ -6-9+Ra+Rb = 0 \\ Ra = 9kN $

Solution:

$\text {Angle } α = \tan^{-1}(3/2) = 56.31 \\ \text {Angle } β = \tan^{-1}(3/4) = 36.86$

At joint A

$∑Fy = 0 \\ AC \sin 56.31 + Ra = 0 \\ AC = - 10.82 kN \\ ∑Fx = 0 \\ AC \cos 56.31 + AE = 0 \\ AE = 6kN $

At joint E

By special Case,

$EC = 6kN $ and $EF = AE = 6kN$

At joint B,

$∑Fy = 0 \\ BD \sin 56.31 + Rb = 0 \\ BD = - 7.21 kN \\ ∑Fx = 0 \\ -BD \cos 56.31 -BF = 0 \\ BF = 4kN $

At joint F

$∑Fx = 0 \\ -EF + BF – FC \cos 36.86 =0 \\ FC = -2.5kN \\ ∑Fy = 0 \\ -9 + FD + FC \sin 36.86 = 0 \\ FD = 10.5kN $

At joint D

$∑Fx= 0 \\ -CD + BD \cos 56.31 = 0 \\ CD = -4 kN$