A shaft with a diameter of 50 mm and a length of 1300 mm with a safe shear yield strength of 20 N / mm2 is subjected to 565 Nm of torque. What is the torsion angle and applicability? (G = 80GPa)

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 Question

A shaft with a diameter of 50 mm and a length of 1300 mm with a safe shear yield strength of 20 N / mm2 is subjected to 565 Nm of torque. What is the torsion angle and applicability? (G = 80GPa)

Answer

Step 1

$$\begin{gathered} Given: \\ d=50 mm \\ l=130 mm \\ {S}_{ys}=20 MPa \\ T=565 Nm \\ G=80 GPa \end{gathered}$$

Step 2

$$\begin{gathered} J=\frac{\pi }{32}{d}^4 \\ J=\frac{\pi }{32}*{50}^4 \\ J=613592.31 m{m}^4 \\ Torsion angle,\theta =\frac{TL}{GJ} \\ \theta =\frac{565*1000*130}{80*1000*613592.31} \\ \theta =0.001496 rad =0.0857° \\ Safe shear stress,\zeta =S_{ys}=20 MPa \\ \zeta =\frac{16T}{\pi d^3} \\ 20=\frac{16T}{\pi {\left(50\right)}^3} \\ T=490873 N.mm \\ T=490.873 N.m \left(Applicability\right) \end{gathered}$$