Question

PROBLEM \( 3.93 \) A 30-mm diameter solid rod is made of an elastoplastic material with \( \tau_{Y}=3.5 \mathrm{MPa} \). Knowing that the elastic core of the rod is \( 25 \mathrm{~mm} \) in diameter, determine the magnitude of the applied torque \( \mathbf{T} \). SOLUTION \[ \begin{aligned} \tau_{Y} & =3.5 \cdot 10^{6} \mathrm{~Pa}, \quad c=\frac{1}{2}(30 \mathrm{~mm})=15 \mathrm{~mm}=0.015 \mathrm{~m} \\ \rho_{Y} & =\frac{1}{2}(25 \mathrm{~mm})=12.5 \mathrm{~mm}=0.0125 \mathrm{~m} \\ T_{Y} & =\frac{J}{c} \tau_{Y}=\frac{\pi}{2} c^{3} \tau_{Y}=\frac{\pi}{2}(0.015)^{3}\left(3.5 \cdot 10^{6}\right)=18.555 \mathrm{~N} \cdot \mathrm{m} \\ T & =\frac{4}{3} T_{Y}\left(1-\frac{\rho_{Y}^{3}}{c^{3}}\right)=\frac{4}{3}(18.555)\left[1-\frac{(0.0125)^{3}}{(0.015)^{3}}\right] \\ & =21.2 \mathrm{~N} \cdot \mathrm{m} \quad T=21.2 \mathrm{~N} \cdot \mathrm{m} \end{aligned} \]

          PROBLEM \( 3.93 \)
A 30-mm diameter solid rod is made of an elastoplastic material with \( \tau_{Y}=3.5 \mathrm{MPa} \). Knowing that the elastic core of the rod is \( 25 \mathrm{~mm} \) in diameter, determine the magnitude of the applied torque \( \mathbf{T} \).
SOLUTION
\[
\begin{aligned}
\tau_{Y} & =3.5 \cdot 10^{6} \mathrm{~Pa}, \quad c=\frac{1}{2}(30 \mathrm{~mm})=15 \mathrm{~mm}=0.015 \mathrm{~m} \\
\rho_{Y} & =\frac{1}{2}(25 \mathrm{~mm})=12.5 \mathrm{~mm}=0.0125 \mathrm{~m} \\
T_{Y} & =\frac{J}{c} \tau_{Y}=\frac{\pi}{2} c^{3} \tau_{Y}=\frac{\pi}{2}(0.015)^{3}\left(3.5 \cdot 10^{6}\right)=18.555 \mathrm{~N} \cdot \mathrm{m} \\
T & =\frac{4}{3} T_{Y}\left(1-\frac{\rho_{Y}^{3}}{c^{3}}\right)=\frac{4}{3}(18.555)\left[1-\frac{(0.0125)^{3}}{(0.015)^{3}}\right] \\
& =21.2 \mathrm{~N} \cdot \mathrm{m} \quad T=21.2 \mathrm{~N} \cdot \mathrm{m}
\end{aligned}
\]

PROBLEM 3.93
A 30-mm diameter solid rod is made of an elastoplastic material with τY=3.5 MPa. Knowing that the elastic core of the rod is 25 mm in diameter, determine the magnitude of the applied torque 𝐓.
SOLUTION

τY =3.5 · 10^6 Pa, c=(1)/(2)(30 mm)=15 mm=0.015 m
ρY =(1)/(2)(25 mm)=12.5 mm=0.0125 m

TY =(J)/(c)τY=(π)/(2) c^3τY=(π)/(2)(0.015)^3(3.5 · 10^6)=18.555 N·m

T =(4)/(3) TY(1-(ρY^3)/(c^3))=(4)/(3)(18.555)[1-((0.0125)^3)/((0.015)^3)]
=21.2 N·m T=21.2 N·m

Added by Ibrahim A.

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