Topic: Machine Design
A 4 meter long solid circular cylindrical shaft transmits 1600 kW at 500 rpm. The endurance limit of the material is 210 MPa and a factor of safety of 3 is to be used with respect to fatigue failure. Use G = 77.5 GPa. The design diameter 'D' of the shaft in meter is
- A. \sqrt[3]{\frac{69}{4375\pi^{2}}}
- B. \sqrt[3]{\frac{69}{4735\pi^{2}}}
- C. \sqrt[3]{\frac{4375}{96\pi^{2}}}
- D. \sqrt[3]{\frac{96}{4375\pi^{2}}}
Answer: D. \sqrt[3]{\frac{96}{4375\pi^{2}}}
Explanation: First, find the torque transmitted: T = (60 * P) / (2 * pi * N) = (60 * 1600 * 10^3) / (2 * pi * 500) = 96000/pi Nm. Using the Soderberg equation for fatigue failure in shear (as the load is purely torsional, mean stress is 0 and alternating stress is tau), the allowable shear stress tau_v = Endurance limit / FOS = 210 MPa / 3 = 70 MPa. Using the torsion equation tau = 16T / (pi * d³), we get 70 * 10^6 = (16 * (96000/pi)) / (pi * d³). Solving for d³ yields d³ = (16 * 96000) / (70 * 10^6 * pi²) = 96 / (4375 * pi²). Taking the cube root gives D = \sqrt[3]{\frac{96}{4375\pi^{2}}}.