Problem 447 - Compound Truss Formed Into Three-Hinged Arch

$2A_H + 8A_V = 6(24) + 2(72)$

$2A_H + 8A_V = 288$

$A_H + 4A_V = 144$   ←   Equation (1)
 

$\Sigma M_A = 0$

$2C_H + 2(24) + 6(72) = 8C_V$

$2C_H + 480 = 8C_V$

$2C_H - 8C_V = -480$

$C_H - 4C_V = -240$   ←   Equation (2)

$4A_H + 2(24) = 4A_V$

$4A_H - 4A_V = -48$

$A_H - A_V = -12$   ←   Equation (3)
 

$\Sigma M_A = 0$

$4B_H = 4B_V + 2(24)$

$4B_H - 4B_V = 48$

$B_H - B_V = 12$   ←   Equation (4)

$6C_H + 2(72) = 4C_V$

$6C_H - 4C_V = -144$

$3C_H - 2C_V = -72$   ←   Equation (5)
 

$\Sigma M_C = 0$

$6B_H + 4B_V = 2(72)$

$6B_H + 4B_V = 144$

$3B_H + 2B_V = 72$   ←   Equation (6)

$A_V = 31.2 ~ \text{kN}$

$C_V = 64.8 ~ \text{kN}$

$B_V = 7.2 ~ \text{kN}$

$19.2 - 19.2 = 0$       (okay!)

$31.2 + 64.8 - 24 - 72 = 0$       (okay!)

$4F_{BD} = 2(24)$

$F_{BD} = 12 ~ \text{kN compression}$

$F_{BE} \sin 45^\circ + 24 = 31.2$

$F_{BE} = 7.2\sqrt{2} ~ \text{kN}$

$F_{BE} = 10.182 ~ \text{kN compression}$

$6F_1 = 2(72)$

$F_1 = 24 ~ \text{kN}$

$\frac{3}{\sqrt{13}}F_2 + 64.8 = 72$

$F_2 = 2.4\sqrt{13} ~ \text{kN}$

$F_2 = 8.653 ~ \text{kN tension}$
 

From the FBD of Joint B
$\Sigma F_V = 0$

$F_{BE} \sin 45^\circ = \frac{3}{\sqrt{13}}F_2$

$7.2\sqrt{2} \sin 45^\circ = \frac{3}{\sqrt{13}}\left( 2.4\sqrt{13} \right)$

$7.2 = 7.2$       (okay!)

$F_{BD} + F_{BE} \cos 45^\circ + \frac{2}{\sqrt{13}}F_2 = F_1$

$12 + 7.2\sqrt{2} \cos 45^\circ + \frac{2}{\sqrt{13}}\left( 2.4\sqrt{13} \right) = 24$

$24 = 24$       (okay!)

$B_V = 7.2 ~ \text{kN}$

$F_{BD} = 12 ~ \text{kN compression}$

$F_{BE} = 10.182 ~ \text{kN compression}$