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‏A 50 mm diameter shaft is made from carbon steel having ultimate tensile strength of 630...

‏A 50 mm diameter shaft is made from carbon steel having ultimate tensile strength of 630 MPa. It is subjected to a torque which fluctuates between 2000 N-m to - 800 N-m. Using Soderberg me thod, calcukate the factor of safety. Assume suitable values for any other data needed.
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Answer #1

1/FOS = σm / σy + σa / σe  

Mmax = (π/32) σmax d3

2000 = (π/32) σmax 0.053

σmax = 162974661.7261 N/m2 or 162.9746 N/mm2 (MPa)

Mmin = (π/32) σmin d3

800 = (π/32) σmin 0.053

σmin = 65189864.69044 N/m2 or 65.1898 N/mm2 (MPa)

σm = σmax + σmin / 2 = 162.9746 + 65.1898 / 2 = 114.0822 N/mm2 (MPa)

σa = σmax - σmin / 2 = 162.9746 - 65.1898 / 2 = 48.8924 N/mm2 (MPa)

σe = 0.504 σy = 0.504 * 630 = 317.52 N/mm2 (MPa)

By Soderberg equation,

1/FOS = σm / σy + σa / σe

1/FOS = 114.0822 / 630 + 48.8924 / 317.52

FOS = 2.9844

Please rate my answer. Thank You.

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