Answer
$$Area_{AB} = 0.2687 \space in^2$$
Work Step by Step
1) Free Body Diagram of $\overline{BCDE}$
$\space\space\space\bullet$ Moment Equillibrium:
$\sum M_{B} = 0$
$5 * 1.4 + 600* 4.2 *2.1- D_y * 2.8 = 0$
$\Rightarrow D_y = 4.39\space kip \uparrow$
$\space\space\space\bullet$ y-Force Equillibrium:
$\sum F_y = 0$
$AB_{y}-5-600*4.2+4.39 = 0$
$\Rightarrow AB_y=3.13 \space kip \uparrow$
$\space\space\space\bullet$ Total force in AB:
$AB = \frac{AB_y}{sin35} = 5.457\space kip$
2)Allowable axial stress:
$\sigma_{all} = \frac{\sigma_U}{F.S.} = \frac{65}{3.2} = 20.313\space ksi$
3)Design Area of AB:
$\sigma_{all} = \frac{AB_y}{A_{AB}}\space\Rightarrow A_{AB} = \frac{AB_y}{\sigma_{all}} = \frac{5.457}{20.313} = \boxed{0.2687 \space in^2} \space\space\leftarrow\space ANS$
