We determine the torsional moments at characteristic intervals

\begin{aligned} &M_{B C}=-20 \mathrm{kNm} \\ &M_{A B}=-20+30=10 \mathrm{kNm} \\ \end{aligned}

We determine the maximum shear stresses

\begin{aligned} &\tau=\frac{M_{S}}{W_{S}} \\ &W_{S}=\frac{\pi \cdot d^{3}}{16} \\ \end{aligned}

To do this, we need to calculate the torsional strength indicators

\begin{aligned} &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} \\ \end{aligned}

Now we can calculate the stresses

\begin{aligned} &\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}

Next, we will calculate the angles of twist in the characteristic sections relative to the supports

\begin{aligned} &\varphi=\frac{M_{S} \cdot l}{G \cdot I} \\ \end{aligned}

For this, we need the moments of inertia for torsion

\begin{aligned} &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} \\ \end{aligned}

For example: the angle of twist of section C relative to section B will be calculated as follows:

\[ \varphi_{B C}=\frac{M_{S_{B C}} \cdot l_{B C}}{G \cdot I_{B C}} \]

However, let’s start from the support at point A, where we know that the angle of rotation is equal to 0

\[ \varphi_{A}=0 \]

Next, the angle of rotation of section B relative to A:

\[ \varphi_{B}=\varphi_{A}+\varphi_{A B} \]

and the total angle of twist, which is the sum of the angle of twist in B and the twist of section C relative to B

\[ \varphi_{C}=\varphi_{B}+\varphi_{B C} \]

Calculations

\begin{aligned} &\varphi_{A B}=\frac{10 \cdot 10^{3} \cdot 3}{80 \cdot 10^{9} \cdot 2,04 \cdot 10^{-5}}=0,0183 \text{ rad} \\ &\varphi_{B C}=\frac{-20 \cdot 10^{3} \cdot 2}{80 \cdot 10^{9} \cdot 1,27 \cdot 10^{-6}}=-0,3937\text{ rad} \\ &\varphi_{B}=0+0,0183=0,0183 \text{ rad} \\ &\varphi_{C}=0,0183-0,3937=-0,3754 \text{ rad} \end{aligned}

After completing all calculations, we can draw the graphs of torsional moments, shear stresses, and the angle of twist

Torsion

Strength of materials - Torsion - Example 2

Example 2