Answer
$$ \sigma_{AB} = 84.88 \space\space MPa\space\space (Tension)$$
$$ \sigma_{BC} = -96.77 \space\space MPa\space\space (Compression)$$
Work Step by Step
Solution:
1)Set up:
Area AB: $A_{AB} = \frac{1}{4}\pi d_1^2 = 7.0686 * 10^{-4} \space \space m^2$
Area BC: $A_{BC} = \frac{1}{4}\pi d_1^2 = 1.9635 * 10^{-3} \space\space m^2$
2) Internal Forces:
$\bullet AB:$
Taking a cut between points A and B and applying the equilibrium equation, we get:
$\sum{F_x} = 0; \space\space -60 \space kN + N_{AB} = 0$
$N_{AB} = 60 \space kN \space (Tension)$ $\space\space\space\space\leftarrow N_{AB}$
$\bullet BC:$
Taking a cut between points B and C and applying the equilibrium equation, we get:
$\sum{F_x} = 0; \space\space -60 \space kN + 2*125 \space kN + N_{BC} = 0$
$N_{BC} = -190 \space kN \space (Compression)$ $\space\space\space\space\leftarrow N_{BC}$
3)Average Normal Stress
Average Normal stress is equal to :
$\sigma_{avg} = \frac{N}{A}$
$\bullet AB:$
$\sigma_{AB} = \frac{N_{AB}}{A_{AB}} = \frac {60*10^3 \space N}{7.0686*10^{-4} \space m^2} = 84.88*10^6 \space Pa = \boxed{84.88 \space MPa\space (Tension)}$ $\space\space\space\space\leftarrow ANS1$
$\bullet BC:$
$\sigma_{BC} = \frac{N_{BC}}{A_{BC}} = \frac {-190*10^3 \space N}{1.9635*10^{-3} \space m^2} = -96.77*10^6 \space Pa = \boxed{-96.77 \space MPa\space (Compression)}$ $\leftarrow ANS2$
