Answer
$$\sigma_{BE}=25.55\space ksi$$
Work Step by Step
1)Support Reactions
Moment Equilibrium:
$\sum M_{A} = 0$
$H_y * 36 - 80*9 - 80*18-80*27 = 0$
$\rightarrow H_y = 120 \space kip$
By symmetry of loading and gemoetry,
$A_y = H_y= 120 \space kip$
2)By Inspection, elements AC and CE are zero force members.
3)Equilibrium of forces at joint A
$\space\space\space\bullet$ Y-forces Equilibrium:
$\sum F_y = 0.8 F_{AB} + 120 = 0$
$\rightarrow F_{AB} = -150\space kip $ (Compression)
$\space\space\space\bullet$ X-forces Equilibrium:
$\sum F_x = A_x-150*0.6=0$
$\rightarrow A_x=90 \space kip$ (Horizontal Rxn at A)
4)Make a cut through elements BD, BE, CE.
$\space\space\space\bullet$ Y-forces Equilibrium:
$\sum F_y=0$
$ 120 - F_{BE} * 0.8 = 0 $
$\Rightarrow F_{BE} = 150\space kip$ (Tension)
5)Axial Stress in element BE
$\sigma_{BE} = \frac{F_{BE}}{A_{BE}} = \frac{150\space kip}{5.87\space in^2} = \boxed{25.55 \space ksi}\space\space \leftarrow ANS$
