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 106 loading cycles, as per the Soderberg law. Considering a factor of safety of 2, the maximum applied tensile principal stress is

The correct option is half of the endurance limit.

Step 1: Identify the given data
Let:
- Yield strength of the steel grade = ${\sigma }_y$
- Endurance limit of the steel grade = ${\sigma }_e$
- Ultimate strength of the steel grade = ${\sigma }_u$
- Factor of safety = FOS = 2

From the problem description:
1. The endurance limit is equal to the yield strength:
${\sigma }_e={\sigma }_y$
2. The ultimate strength is twice the yield strength:
${\sigma }_u=2{\sigma }_y$

Step 2: Understand the loading condition
The component is loaded in tension and unloaded periodically (zero-to-tension fluctuation). This means:
- Minimum tensile stress, ${\sigma }_{\text{min}}=0$
- Maximum tensile stress, ${\sigma }_{\text{max}}=\sigma$ (the maximum applied tensile principal stress)

We calculate the mean stress (${\sigma }_m$) and stress amplitude (${\sigma }_a$) as follows:
${\sigma }_m=\frac{{\sigma }_{\text{max}}+{\sigma }_{\text{min}}}{2}=\frac{\sigma }{2}$
${\sigma }_a=\frac{{\sigma }_{\text{max}}-{\sigma }_{\text{min}}}{2}=\frac{\sigma }{2}$

Step 3: Apply Soderberg's relation
According to Soderberg's law for fatigue design:
$\frac{{\sigma }_a}{{\sigma }_e}+\frac{{\sigma }_m}{{\sigma }_y}=\frac{1}{\text{FOS}}$

Step 4: Substitution and calculation
Substitute ${\sigma }_m=\frac{\sigma }{2}$, ${\sigma }_a=\frac{\sigma }{2}$, and $\text{FOS}=2$ into Soderberg's equation:
$\frac{\sigma /2}{{\sigma }_e}+\frac{\sigma /2}{{\sigma }_y}=\frac{1}{2}$
Since ${\sigma }_e={\sigma }_y$, we can replace ${\sigma }_y$ with ${\sigma }_e$ in the equation:
$\frac{\sigma }{2{\sigma }_e}+\frac{\sigma }{2{\sigma }_e}=\frac{1}{2}$
Combine the terms on the left-hand side:
$\frac{\sigma }{{\sigma }_e}=\frac{1}{2}$
Solving for $\sigma$:
$\sigma =\frac{{\sigma }_e}{2}$
Thus, the maximum applied tensile principal stress is half of the endurance limit.