Example.A rod of 0.5in^2 is subjected to static mean tensile load of 10 kips. What fatigue stress amplitude sigma _(a) will produce failure after 10^6 cycles? Assume sigma _(n)=32ksi,sigma _(u)=60ksi.

To solve this problem, we need to use the fatigue strength reduction equation, which is commonly used in engineering to estimate the fatigue strength of a material under cyclic loading.<br /><br />Given information:<br />- Cross-sectional area of the rod: $A = 0.5 in^2$<br />- Static mean tensile load: $F = 10 kips = 10,000 lbs$<br />- Number of cycles to failure: $N = 10^6$<br /><br />The fatigue stress amplitude $\sigma_a$ can be calculated using the following equation:<br /><br />$\sigma_a = \frac{\sigma_n}{1 + \frac{S_n}{2N}}$<br /><br />Where:<br />- $\sigma_n$ is the yield strength of the material (given as 32 ksi)<br />- $S_n$ is the mean stress level (given as 60 ksi)<br />- $N$ is the number of cycles to failure (given as $10^6$)<br /><br />Substituting the given values, we get:<br /><br />$\sigma_a = \frac{32}{1 + \frac{60}{2 \times 10^6}}$<br /><br />Simplifying the equation, we get:<br /><br />$\sigma_a = \frac{32}{1 + \frac{60}{2,000,000}}$<br /><br />$\sigma_a = \frac{32}{1 + 0.00003}$<br /><br />$\sigma_a = \frac{32}{1.00003}$<br /><br />$\sigma_a \approx 31.99997 \text{ ksi}$<br /><br />Therefore, the fatigue stress amplitude $\sigma_a$ that will produce failure after $10^6$ cycles is approximately 31.99997 ksi.