Problem 453 - Analysis of A-frame with Weightless Cylinder

From the FBD of the whole frame

$\dfrac{y}{2}= \dfrac{1.5}{1}$

$y = 3 ~ \text{m}$
 

$\Sigma F_x = 0$

$A_H = 0$

 

 

$\Sigma M_A = 0$

$3R_F = 6(240)$

$R_F = 480 ~ \text{kN}$

 

$\Sigma M_F = 0$

$3A_V = 3(240)$

$A_V = 240 ~ \text{kN}$

 

From the FBD of member BG

$\Sigma M_B = 0$

$2E_V = 4(240)$

$E_V = 480 ~ \text{kN}$

 

 

$\Sigma M_E = 0$

$2B_V = 2(240)$

$B_V = 240 ~ \text{kN}$

 

From the FBD of member CF

$\Sigma F_y = 0$

$C_V + 480 = 480$

$C_V = 0$

 

 

$\Sigma M_E = 0$

$3C_H = 1(240)$

$C_H = 80 ~ \text{kN}$
 

$\Sigma M_C = 0$

$3E_H = (3 - 1)(240)$

$E_H = 160 ~ \text{kN}$

 

From the FBD of member AC
 

 

$\Sigma M_A = 0$

$1.5B_H + 1(240) = 4.5(80)$

$B_H = 80 ~ \text{kN}$

 

Summary
$B_H = 80 ~ \text{kN}$   and   $B_V = 240 ~ \text{kN}$

$C_H = 80 ~ \text{kN}$   and   $C_V = 0$

$E_H = 160 ~ \text{kN}$   and   $E_V = 480 ~ \text{kN}$
 

Thus,
$R_B = \sqrt{80^2 + 240^2} = 252.98 ~ \text{kN}$   ←   answer

$R_C = \sqrt{80^2 + 0^2} = 80 ~ \text{kN}$   ←   answer

$R_E = \sqrt{160^2 + 480^2} = 505.96 ~ \text{kN}$   ←   answer