The following video course will help in understanding Example 1.

Strength indicators and moments of inertia of basic flat figures |CLICK|.

Torsional moments and moments of inertia (for calculation simplification, we will check the ratio of moments of inertia to each other).

\begin{aligned}\\ &M_{AB}=-M_A\\ &M_{BC}=-M_A+20\\ \\ &I_{AB}=\frac{\pi \cdot D^4}{32}\\ &I_{BC}=\frac{\pi \cdot (D^4-d^4)}{32}=0.06711D^4\\ &\frac{I_{AB}}{I_{BC}}=1.4639\ \Rightarrow I_{AB}=1.4629I_{BC}\\ \end{aligned}

Geometric condition - the total angle of rotation is equal to 0: \(\varphi_c = 0\)

\begin{aligned}\\ &\varphi=\frac{M_S\cdot l}{G\cdot I}\\ \\ &\varphi_c=\varphi_{AB}+\varphi_{BC}=0\\ &\varphi_c=\frac{-M_A\cdot 2}{G\cdot 1.4629I_{BC}}+\frac{(-M_A+20)\cdot 3}{G\cdot I_{BC}}=0\ |\cdot GI_{BC}\\ &-1.367M_A-3M_A=-60\\ &M_A=13.74\ kNm\\ \end{aligned}

Dimensioning, strength condition: \(\tau_{max}\le k_s\)

\begin{aligned}\\ &M_{AB}=-13.74\ kNm\\ &M_{BC}=-13.74+20=6.26\ kNm\\ \\ &W_{AB}=\frac{\pi D^3}{16}\\ &W_{BC}=\frac{\pi \cdot (D^4-d^4)}{16D}=\frac{175\pi D^3}{4096}\\ \end{aligned}

On the AB interval, there is a greater torsional moment, but the torsional strength indicator is also greater than on the BC interval, so it is safer to check both intervals.

\begin{aligned}\\ &\tau_{AB}=\frac{-13.74\cdot 10^3\cdot}{\frac{\pi D^3}{16}}=-\frac{69977.25}{D^3}\\ &\tau_{BC}=\frac{6.36\cdot 10^3\cdot}{\frac{\pi D^3}{16}}=\frac{47383.72}{D^3}\\ &|\tau_{AB}|\ge |\tau_{BC}|\\ \end{aligned}

Strength condition

\begin{aligned}\\ &\frac{69977.25}{D^3}\ge \ 70\cdot 10^6\\ &d\ge \sqrt[3]{\frac{69977.25}{70\cdot 10^6}}\\ &d\ge 0.0999\ m\\ &d=0.10\ m=10\ cm\\ \end{aligned}

Tensile stresses

\begin{aligned}\\ &\tau =\frac{M_S}{W_S}\\ \\ &W_{S_{AB}}=\frac{\pi \cdot 0.1^3}{16}=1.96\cdot 10^{-4}\ m^3\\ &W_{S_{BC}}=\frac{\pi \cdot (0.1^4-0.075^4)}{16\cdot 0.1}=1.34\cdot 10^{-4}\ m^3\\ \\ &\tau_{AB}=\frac{M_{AB}}{W_{S_{AB}}}=\frac{-13.74\cdot 10^3}{1.96\cdot 10^{-4}}=-70.1\ MPa\\ &\tau_{BC}=\frac{6.26\cdot 10^3}{1.34\cdot 10^{-4}}=46.72\ MPa\\ \end{aligned}

Section B rotation angle

\begin{aligned}\\ &\varphi_B=\varphi_{AB}\\ &\varphi=\frac{M_S\cdot l}{G\cdot I}\\ \\ &I_{AB}=\frac{\pi \cdot 0.1^4}{32}=9.82\cdot 10^{-6}\\ \end{aligned}

Relationship between the Kirchhoff modulus and Young's modulus:

\begin{aligned}\\ &G=\frac{E}{2(1+\nu)}=82.69\ GPa\\ \\ &\varphi_B=\frac{-13.74\cdot 10^3\cdot 2}{82.69\cdot 10^9\cdot 9.82\cdot 10^{-6}}=-0.03384\ rad=-1.939\ [^o]\\ \end{aligned}

Charts

Torsion

Strength of materials - Torsion - Example 5

Example 5