Problem 443 - Analysis of Frame by Method of Members

$2A_H = 2(12)$

$A_H = 12 ~ \text{kN}$       answer
 

$\Sigma F_H = 0$

$E_H = A_H$

$E_H = 12 ~ \text{kN}$       answer
 

$\Sigma F_V = 0$

$E_V + A_V = 12$   †

By Symmetry
$A_V = B_V = \frac{1}{2}(12)$

$A_V = B_V = 6 ~ \text{kN}$       answer
 

Substitute AV = 6 kN to †
$E_V + 6 = 12$

$E_V = 6 ~ \text{kN}$       answer

$1.5F_{BC} = 2.5(6)$

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

$\Sigma F_H = 0$

$\frac{3}{5}F_{BD} = 12$

$F_{BD} = 20 ~ \text{kN compression}$       answer