Geometry: We assume, $BE \perp AD, \ CD \perp AD,$

$AE = ED = 3m, \ CD= 5m, \ BE= 2m$

$AD = 3+3 = 6 m$

In Δ AEB, by Pythagoras theorem,

$AB = \sqrt{AE^2 + BE^2} = \sqrt{3^2 + 2^2} = \sqrt{13}$

$\theta = \tan ^{-1} \left ( \dfrac {2}{3} \right ) = 33.69^\circ$

Similarly, $BD=\sqrt{13}$

In $\Delta BCD, \ \angle BDC=90^\circ - 33.69^\circ = 56.31^\circ$

By cosine rule,

$BC^2 = BD^2 + CD^2 - 2 \times BD \times CD \times \cos \angle BDC$

$= 13 + 5^2 - 2 \times \sqrt{13} \times 5 \times \cos 56.31$

$= 18$

$\therefore BC= \sqrt{18}= 3 \sqrt{2} = 4.2426$

$\therefore \cos \beta = \dfrac {BC^2 + CD^2 - BD^2}{2 \times BC \times CD} = \dfrac {18+25-13}{2 \times \sqrt{18}\times 5} = \dfrac {1}{\sqrt{2}} $

$\therefore \beta = \cos ^{-1} \left ( \dfrac {1}{\sqrt{2}} \right ) = 45^\circ$

Since the truss is in equilibrium, $\sum M_A = 0$

$\therefore -100 \times AE - 200 \times AE - 200 \times AD - 150 \times 3D + R_B \times AD =0$

$\therefore R_B \times 6 = 100 \times 3 + 200 \times 3 + 200 \times 6 + 150 \times 5$

$\therefore R_B = 475 \ N$

Also $\sum F_Y = 0$

$\therefore R_A \sin \alpha + R_B - 100 -200 - 200 = 0$

$\therefore R_A \sin \alpha + 475 - 500 = 0$

$\therefore R_A \sin \alpha = 25 N \ \to (1)$

And, $\sum F_X = 0$

$\therefore R_A \cos \alpha + 150 = 0$

$\therefore R_A \cos \alpha = - 150 \to 2$

Squaring and adding (1) and (2)

$R_A^2 \sin^2\alpha + R_A^2 \cos^2 \alpha = 25 ^2 + (-150)^2$

$\therefore R_A^2 = 23125$

$\therefore R_A = 152.0691 N$

Dividing (1) by (2)

$\dfrac {R_A \sin \alpha}{R_A \cos \alpha} = \dfrac {25}{-150} $

$\therefore \tan \alpha = - \dfrac {1}{6}$

$\therefore \alpha \tan^{-1} \left ( \dfrac {-1}{6} \right )= 170.5377^\circ$

We use method of sections:

Applying conditions of equilibrium to the followin section, $\sum M_D = 0$

$\therefore F_{CB} \sin \beta \times CD - 150 \times CD = 0$

$\therefore F_{CB} \sin 45 \times CD = 150 \times CD$

$\therefore F_{CB} = \dfrac {150}{\sin 45} = 212.1320 \ N$

$Also, \ \sum F_Y = 0$

$\therefore R_B + F_{DB} \sin \theta - F_{CB} \cos \beta - 200 = 0$

$\therefore 475 + F_{DB} \sin 33.69 - 212.1320 \cos 45-200 = 0$

$\therefore F_{DB} = \dfrac {212.132 \cos 45+200-475}{\sin 33.69} = -225.3474 \ N$

And, $\sum F_X =0$

$\therefore - F_{DE} - F_{DB} \cos \theta - F_{CB} \sin \beta + 150 =0$

$\therefore 225.3474\cos 33.69 - 212.1320 \sin 45+150 = F_{DE}$

$\therefore F_{DE} = 187.5 \ N$

We use method of Joints:

Joint A:

$\sum F_{Y} = 0$

$\therefore F_{AB} \sin \theta + R_A \sin \alpha = 0$

$\therefore F_{AB} \sin 33.69 + 2 5 = 0 \ (From \ 1)$

$\therefore F_{AB} = \dfrac {-25}{\sin 33.69} = -45.0695 \ N$

Also, $\sum F_X=0$

$\therefore F_{AE} + R_A \cos \alpha + F_{AB} \cos \theta =0$

$\therefore F_{AE} - 150 - 45.0695 \cos 33.69 =0 \ (from \ 2)$

$\therefore F_{AE} = 187.5 \ N$

Joint E:

$\sum F_Y = 0$

$\therefore F_{EB} - 200 = 0$

$\therefore F_{EB} = 200 \ N$

Hence,

Hence, Support Reaction at B=475 N ↑,

Support Reaction at A=152.0691 N ∠ 9.4323°

| Sr. No. | Member | Magnitude of Force (in N) | Nature of Forse | | --- | --- | --- | --- | | 1) | BC | 212.1320 | Tension | | 2) | BD | 225.3474 | Compression | | 3) | DE | 187.5 | Tension | | 4) | AB | 45.0694 | Compression | | 5) | AE | 187.5 | Tension | | 6) | BE | 200 | Tension |