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
BH = 48 kN
BV = 56 kN
CH = 24 kN
CV = 56 kN
DH = 48 kN
DV = 81 kN
