Example 1

A two-stage shaft with dimensions d = 6 cm was permanently attached with the left end and loaded with moments of 20 and 30 kNm as shown in the drawing. Assume G = 80 GPa. Calculate and draw diagrams of torsional moments, tangential stresses and torsion angle: \ (M (x), \ tau (x), \ varphi (x). \)

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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}