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 _(n)=60ksi Goodman (sigma _(a))/(S_(e))+(sigma _(m))/(sigma _(uts))=1 sigma _(m)=(F_(m))/(A)=(10)/(0.5)=20 sigma _(a)=S_(e)-(S_(e))/(sigma _(u))(sigma _(m)) =32-(32)/(60)20=21.3ksithusF_(A)=21.3ksi

To solve this problem, we need to use the Goodman criterion for fatigue stress amplitude.<br /><br />Given information:<br />- Cross-sectional area of the rod: $A = 0.5in^{2}$<br />- Static mean tensile load: $F_m = 10kips$<br />- Yield strength: $\sigma_y = 32ksi$<br />- Ultimate tensile strength: $\sigma_u = 60ksi$<br />- Number of cycles: $N = 10^6$<br /><br />Step 1: Calculate the mean stress, $\sigma_m$, using the formula:<br />$\sigma_m = \frac{F_m}{A} = \frac{10}{0.5} = 20ksi$<br /><br />Step 2: Use the Goodman criterion to calculate the fatigue stress amplitude, $\sigma_a$:<br />$\frac{\sigma_a}{\sigma_s} + \frac{\sigma_m}{\sigma_{uts}} = 1$<br />where $\sigma_s$ is the yield strength and $\sigma_{uts}$ is the ultimate tensile strength.<br /><br />Substituting the values:<br />$\frac{\sigma_a}{32} + \frac{20}{60} = 1$<br />$\sigma_a = 32 - \frac{32}{60} \times 20 = 21.3ksi$<br /><br />Therefore, the fatigue stress amplitude, $\sigma_a$, that will produce failure after $10^6$ cycles is 21.3ksi.