The GATE Grind

GATE 2025 ME – Question 43

Machine Design · Fatigue strength and S-N diagram · 2 marks · Multiple choice

The endurance limit of a specific grade of steel is same as its yield strength. The ultimate strength of this grade of steel is twice of its yield strength. A component made of this steel is loaded in tension and unloaded periodically. It is required that the component does NOT fail for at least $10^6$ loading cycles, as per the Soderberg law. Considering a factor of safety of 2, the maximum applied tensile principal stress is

  1. one-fourth of the endurance limit
  2. half of the endurance limit
  3. the endurance limit
  4. twice the endurance limit

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: (B) half of the endurance limit

Explanation

The load goes from 0 to a maximum, so the mean stress is $\sigma_m = \frac{\sigma_{max}}{2}$ and the amplitude is $\sigma_a = \frac{\sigma_{max}}{2}$. The Soderberg law with the safety factor is $\frac{\sigma_a}{S_e} + \frac{\sigma_m}{S_y} = \frac{1}{2}$. With $S_e = S_y$, this gives $\frac{\sigma_{max}}{S_e} = \frac{1}{2}$, so the maximum stress is half the endurance limit.