Looking at the load on the shaft, we can draw a graph of the torsional moments

We calculate the unit change in angle for the individual intervals:

\begin{aligned} &\Delta \varphi_{A B}=\frac{-40 \cdot 10^{3}}{80 \cdot 10^{9} \cdot \frac{\pi D^{4}}{32}}=\frac{-5,09 \cdot 10^{-6}}{D^{4}} \frac{\mathrm{rad}}{\mathrm{m}} \\ &\Delta \varphi_{B C}=\frac{30 \cdot 10^{3}}{80 \cdot 10^{9} \cdot \frac{\pi D^{4}}{32}}=\frac{3,82 \cdot 10^{-6}}{D^{4}} \frac{\mathrm{rad}}{\mathrm{m}} \\ &\Delta \varphi_{C D}=\frac{30 \cdot 10^{3}}{80 \cdot 10^{9} \cdot \frac{\pi\left(\frac{3}{4} D\right)^{4}}{32}}=\frac{1,21 \cdot 10^{-5}}{D^{4}} \frac{\mathrm{rad}}{\mathrm{m}} \end{aligned}

Dimensioning for the permissible angle of twist:

\(\varphi \max \leq \varphi d o p \) \begin{aligned} &\frac{1,21 \cdot 10^{-5}}{D^{4}} \leq 0,2 \cdot \frac{\pi}{180} \\ &D \geq 0,2426 m \\ &D=0,25 m \end{aligned}

Shear stresses:

\begin{aligned} &\tau=\frac{M_{s}}{W_{s}} \\ &W_{A B}=W_{B C}=\frac{\pi 0,25^{3}}{16}=3,07 \cdot 10^{-3} \mathrm{~m}^{3} \\ &W_{C D}=\frac{\pi 0,1875^{3}}{16}=1,29 \cdot 10^{-3} \mathrm{~m}^{3} \\ &\tau_{A B}=\frac{-40 \cdot 10^{3}}{3,07 \cdot 10^{-3}}=-13,03 \mathrm{MPa} \\ &\tau_{B C}=\frac{30 \cdot 10^{3}}{3,07 \cdot 10^{-3}}=9,77 \mathrm{MPa} \\ &\tau_{C D}=\frac{30 \cdot 10^{3}}{1,29 \cdot 10^{-3}}=23,26 \mathrm{MPa} \end{aligned}

Angle of twist of the shaft:

\begin{aligned} &\varphi_{A}=0 \\ &\varphi_{B}=\varphi_{A}+\Delta \varphi_{A B} \cdot 1,5=\frac{-5,09 \cdot 10^{-6}}{0,25^{4}} \cdot 1,5=-1,95 \cdot 10^{-3} \mathrm{rad} \\ &\varphi_{C}=\varphi_{B}+\Delta \varphi_{B C} \cdot 0,5=-1,95 \cdot 10^{-3}+\frac{3,82 \cdot 10^{-6}}{0,25^{4}} \cdot 0,5=-1,46 \cdot 10^{-3} \mathrm{rad} \\ &\varphi_{D}=\varphi_{C}+\Delta \varphi_{C D} \cdot 2=-1,46 \cdot 10^{-3}+\frac{1,21 \cdot 10^{-5}}{0,25^{4}} \cdot 2=4,74 \cdot 10^{-3} \mathrm{rad} \end{aligned}

Graphs:

Torsion

Strength of materials - Torsion - Example 3

Example 3