Discussion
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
Answer: B
The maximum stress is half of the endurance limit.
Loading in tension and unloading gives a repeated cycle from 0 to \(\sigma\), so \(\sigma_{a} = \sigma_{m} = \sigma/2\). Soderberg's line with factor of safety N reads \(\sigma_{a}/S_{e} + \sigma_{m}/S_{y} = 1/N\). With \(S_{e} = S_{y}\):
\[ \begin{aligned} \frac{\sigma/2}{S_{e}} + \frac{\sigma/2}{S_{e}} &= \frac{1}{2} \\ \sigma &= S_{e}/2 \end{aligned} \]
The ultimate strength plays no part in Soderberg's criterion, which uses yield. One-fourth applies the factor of safety twice, the endurance limit itself ignores it, and twice the endurance limit exceeds the yield strength.
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