From the FBD of the whole frame

$\Sigma M_E = 0$
$6R_A + 3(30) = 5(24)$

$R_A = 5 ~ \text{kN}$

$\Sigma M_A = 0$
$6E_V + 5(24) = 9(30)$

$E_V = 25 ~ \text{kN}$

$\Sigma F_H = 0$

$E_H = 24 ~ \text{kN}$

From the FBD of member CE

$\Sigma M_D = 0$
$2C_H = 2(24)$

$C_H = 24 ~ \text{kN}$

$\Sigma M_C = 0$
$2D_H = 4(24)$

$D_H = 48 ~ \text{kN}$

From the FBD of member BF

$\Sigma M_B = 0$
$3C_V = 3(24) + 4(24)$

$C_V = 56 ~ \text{kN}$

$\Sigma F_H = 0$
$B_H = 24 + 24$

$B_H = 48 ~ \text{kN}$

$\Sigma F_V = 0$

$B_V = C_V$

$B_V = 56 ~ \text{kN}$

From the FBD of member AG

$\Sigma F_V = 0$
$D_V + 5 = 56 + 30$

$D_V = 81 ~ \text{kN}$

Check at member CE

$\Sigma F_H = 0$
$C_H + E_H = D_H$

$24 + 24 = 48$ (*okay!*)

$\Sigma F_V = 0$

$C_V + E_V = D_V$

$56 + 25 = 81$ (*okay!*)

**Answer Summary**

B_{H} = 48 kN

B_{V} = 56 kN
C_{H} = 24 kN

C_{V} = 56 kN

D_{H} = 48 kN

D_{V} = 81 kN