(Solved): A solid shaft with a round cross section is subjected to a torque of \( T=168 \mathrm{Nm} \). The ...

A solid shaft with a round cross section is subjected to a torque of \( T=168 \mathrm{Nm} \). The torsion limit stress of the material is \( \tau_{l i m}=111 \frac{\mathrm{N}}{\mathrm{mm}^{2}} \). Calculate the allowable torsion shear stress for a safety factor of \( 1.8 \) : \[ \tau_{a l l}=\quad(\pm 0.5)\left[\frac{N}{m m^{2}}\right] \] Calculate the necessary diameter of the shaft for an allowable torsion shear stress of \( \tau_{a l l}=54 \frac{N}{\mathrm{~mm}^{2}} \) : \[ d_{n e c}=\quad(\pm 0.5)[m m] \] Calculate the related twisting angle in degrees per meter shaft length for a shaft diameter of \( d=63 m m \) and a shear modulus of \( G=80366 \frac{N}{m m^{2}} \) : \( \varphi^{\prime}=\quad(\pm 0.01)\left[\frac{*}{m}\right] \)