Answer
$$\sigma_{AB} = 17.93 \space ksi \space (Tension)$$
$$\sigma_{BC} = 22.63 \space ksi \space (Tension)$$
Work Step by Step
1) Setup:
$A_{AB} = \frac{1}{4} \pi d_{AB}^2 =1.227 \space in^2$
$A_{BC} = \frac{1}{4} \pi d_{BC}^2 =0.4418 \space in^2$
2) Internal Forces:
$\bullet AB:$
Taking a cut between points A and B and applying the equilibrium equations:
$\sum{F_y} = 0; \space\space N_{AB} -10 -12 = 0 \Rightarrow N_{AB} = 22 kip \space\space(Tension)$
$\bullet BC:$
Taking a cut between points B and C and applying the equilibrium equations:
$\sum{F_y} = 0; \space\space N_{BC} -10 = 0 \Rightarrow N_{BC} = 10 kip \space\space(Tension)$
3) Average Normal Stress:
From the formula for average normal Stress:
$\sigma_{avg} = \frac{N_{int}}{A}$
$\bullet AB:$
$\sigma_{AB} = \frac{N_{AB}}{A_{AB}} = \frac{22 \space kip}{1.227 \space in^2} = \boxed{17.93 \space ksi\space(Tension)}$ $\space\space\leftarrow ANS1$
$\bullet BC:$
$\sigma_{BC} = \frac{N_{BC}}{A_{BC}} = \frac{10 \space kip}{0.4418\space in^2} = \boxed{22.63 \space ksi\space(Tension)}$ $\space\space\leftarrow ANS1$
