From the FBD of the truss:
$\Sigma M_A = 0$
$24R_F = 16(30)$
$R_F = 20^k$
At joint F:
$\Sigma F_V = 0$
$\frac{3}{5} DF = 20$
$DF = 33\frac{1}{3}^k \, \text{(Compression)}$
At joint D:
By symmetry
$BD = DF = 33\frac{1}{3}^k \, \text{(Compression)}$
$\Sigma F_V = 0$
$DE = \frac{3}{5} BD + \frac{3}{5} DF$
$DE = \frac{3}{5}(33\frac{1}{3}) + \frac{3}{5}(33\frac{1}{3})$
$DE = 40^k \, \text{(Tension)}$
At joint E:
$\Sigma F_V = 0$
$\frac{3}{5} CE + 30 = 40$
$CE = 16\frac{2}{3}^k \, \text{(Tension)}$
Stresses: (Stress = Force/Area)
$\sigma_{CE} = \dfrac{16\frac{2}{3}}{1.8} = 9.26 \, \text{ksi (Tension)}$ answer
$\sigma_{DE} = \dfrac{40}{1.8} = 22.22 \, \text{ksi (Tension)}$ answer
$\sigma_{DF} = \dfrac{33\frac{1}{3}}{1.8} = 18.52 \, \text{ksi (Compression)}$ answer
