$\Sigma M_G = 0$

$12R_C + 6(60) = 6(100) + 9(80) + 18(40)$

$R_C = 140 ~ \text{kN}$

**At section through M-M**

$\Sigma F_V = 0$
$\frac{2}{\sqrt{5}}F_{BC} = 40$

$F_{BC} = 44.721 ~ \text{kN}$ compression *answer*

**At section through N-N**

$\Sigma M_D = 0$
$6F_{CE} + 9(40) = 3(140)$

$F_{CE} = 10 ~ \text{kN}$ tension *answer*

$\Sigma M_E = 0$
$6F_{DF} + 3(80) + 12(40) = 6(140) + 6(60)$

$F_{DF} = 80 ~ \text{kN}$ compression *answer*

## Comments

## can you please elobarate ,

can you please elobarate , how to get 1 and 2 , i know the process but i dont know how you get 1 and 2