A uniform solid round shaft transmits 25 kW at 540 rpm. Allowable shear stress is 35 MPa and maximum torque is 20% above mean torque. Neglect bending. Find minimum diameter in mm, to two decimals.
Mean torque is $T_m=P/\omega=25000(60)/(2\pi540)=442.097$ N·m. Design torque is $T_{max}=1.2T_m=530.516$ N·m. For a solid circular shaft,$\tau_{max}=16T_{max}/(\pi d^3)$. Thus $d=[16(530.516)/(\pi35\times10^6)]^{1/3}=.0425795$ m, or **42.58 mm**.
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