Solution to Problem 113 Normal Stress

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