For member BD: (See FBD 01)
$\Sigma M_C = 0$
$3(\frac{4}{5} BD) = 3(60)$
$BD = 75 \, \text{kN}$ Tension
$BD = \sigma_{BD} A$
$75 (1000) = \sigma_{BD} (1600)$
$\sigma_{BD} = 46.875 \, \text{MPa (Tension)}$ answer
For member CF: (See FBD 01)
$\Sigma M_D = 0$
$4(\frac{1}{\sqrt{2}} CF) = 4(90) + 7(60)$
$CF = 275.77 \, \text{kN}$ Compression
$CF = \sigma_{CF} \, A$
$275.77 (1000) = \sigma_{CF} (1600)$
$\sigma_{CF} = 172.357 \, \text{MPa (Compression)}$ answer
For member BC: (See FBD 02)
$\Sigma M_D = 0$
$4BC = 7(60)$
$BC = 105 \, \text{kN}$ Compression
$BC = \sigma_{BC} A$
$105 (1000) = \sigma_{BC} (1600)$
$\sigma_{BC} = 65.625 \, \text{MPa (Compression)}$ answer
