Solution
\begin{aligned}
&M_{B C}=-20 \mathrm{kNm} \\
&M_{A B}=-20+30=10 \mathrm{kNm} \\
&\tau=\frac{M_{S}}{W_{S}} \\
&W_{S}=\frac{\pi \cdot d^{3}}{16} \\
&W_{S_{B C}}=\frac{\pi \cdot 0,06^{3}}{16}=4,24 \cdot 10^{-5} \mathrm{~m}^{3} \\
&W_{S_{A B}}=\frac{\pi \cdot 0,12^{3}}{16}=3,39 \cdot 10^{-4} \mathrm{~m}^{3} \\
&\tau=\frac{M_{B C}}{W_{S_{B C}}}=\frac{-20 \cdot 10^{3}}{4,24 \cdot 10^{-5}}=-471,57 \mathrm{MPa} \\
&\tau=\frac{M_{A B}}{W_{S_{A B}}}=\frac{10 \cdot 10^{3}}{3,39 \cdot 10^{-4}}=29,5 \mathrm{MPa}
\end{aligned}
\begin{aligned}
&\varphi=\frac{M_{S} \cdot l}{G \cdot I} \\
&I=\frac{\pi \cdot d^{4}}{32} \\
&I_{B C}=\frac{\pi \cdot 0,06^{4}}{32}=1,27 \cdot 10^{-6} \mathrm{~m}^{4} \\
&I_{A B}=\frac{\pi \cdot 0,12^{4}}{32}=2,04 \cdot 10^{-5} \mathrm{~m}^{4} \\
&\varphi_{B C}=\frac{M_{S_{B C}} \cdot l_{B C}}{G \cdot I_{B C}} \\
&\varphi_{A}=0 \\
&\varphi_{B}=\varphi_{A}+\varphi_{A B} \\
&\varphi_{C}=\varphi_{B}+\varphi_{B C} \\
&\varphi_{A B}=\frac{10 \cdot 10^{3} \cdot 3}{80 \cdot 10^{9} \cdot 2,04 \cdot 10^{-5}}=0,0183 \mathrm{rad} \\
&\varphi_{B C}=\frac{-20 \cdot 10^{3} \cdot 2}{80 \cdot 10^{9} \cdot 1,27 \cdot 10^{-6}}=-0,3937 \mathrm{rad} \\
&\varphi_{B}=0+0,0183=0,0183 \\
&\varphi_{C}=0,0183-0,3937=-0,3754
\end{aligned}
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