When doing stress life fatigue analysis of spectral loading, interpolating S-N data becomes necessary. MEboost has three methods to do this, depending on the data available.

### Background on Cumulative Fatigue Damage

Before proceeding, let's briefly explain why we need to interpolate. With spectral loading, the stress cycles are varying with time. The first step is to count the stress cycles and their amplitudes using a counting algorithm such as the rain flow method. The cycle counts are binned according to stress range from peak to valley. Finally, we use Miner's rule to calculate the amount of stress life (cycles) consumed for each stress range bin.

At each stress level, we need to find the corresponding fatigue life cycles. Therefore, we need to interpolate the S-N data which is shown graphically below.

### Interpolating S-N Data

Interpolation of S-N data can be done in one of three ways in MEboost.

- Log-log
- Semi-log
- Linear

In the case of limited S-N data points, the method used can have drastically different results as we'll see. Shigley & Mischke [1] show that when limited data is available, the S-N curve can be approximated by a power law where the data is a straight line when plotted on log-log axes.

S = aN^b where a and b are constants

Note that slope is generally different for low cycle (N < 1000) and high cycle (N > 1000) fatigue.

Thus, if there is limited data, we could assume the power law interpretation of the data and use log-log interpolation. Log-log interpolation is determined by:

log(N) = log(Na) + (log(Nb)-log(Na))[(log(S)-log(Sa))/(log(Sb)-log(Sa))]

The subscripts a and b refer to the data points that we are interpolating between.

Semi-log interpolation uses linear interpolation of S, and log interpolation of N. It's used when the plot of S vs. log(N) is a straight line. Semi-log interpolation is determined by:

log(N) = log(Na) + (log(Nb)-log(Na))[(S-Sa)/(Sb-Sa)]

Linear interpolation is used when there are many data points. If this method is used with few data points, it is far less conservative than the other two methods. Linear interpolation is determined by:

N = Na + (Nb-Na)[(S-Sa)/(Sb-Sa)]

### Example

To illustrate differences in the methods, we'll use the data from the S-N curve shown earlier.

S | N |

63000 | 1 |

58727 | 100 |

56700 | 1000 |

52502 | 2500 |

49534 | 5000 |

47876 | 7500 |

46734 | 10000 |

43274 | 25000 |

40829 | 50000 |

39462 | 75000 |

38521 | 100000 |

35669 | 250000 |

33653 | 500000 |

32527 | 750000 |

31751 | 1000000 |

31751 | 10000000 |

For our example, let's say S is 40,000 psi. We need to interpolate and find N. First, we'll find N using all available data. S = 40,000 falls between [S=40,829, N = 50,000] and [S=39,462, N=75,000]

Log-log: N = 63,832

Semi-log: N = 63,938

Linear: N = 65,161

The differences between the three methods are relatively small. Now let's take it to the extreme and say we only have two data points: [S=56,700, N = 1000] and [S=31,751, N = 1,000,000]. We'll interpolate again at S = 40,000 using the three methods.

Log-log: N = 63,838

Semi-log: N = 101,879

Linear: N = 669,692

In this case, there is a significant different between log-log and semi-log. There is a drastic difference between log-log and linear.

### Visualizing the Difference

The data from the example was intentionally generated using the power law. Therefore, the log-log interpolation agrees with the power law and gives the same values if the power law was used to calculate N. We'll use this information to visualize the differences between the methods.

First, we'll replot the S-N curve from N = 1000 to N = 1,000,000. This time the axes are semi-log so the S-N curve is not a straight line. In the case where we used only two data points, we will draw a straight line between the two points (the red line).

We can see the source of the error between log-log and semi-log. If the S-N curve is a straight line when plotted on semi-log axes, then semi-log would be the appropriate method of interpolation.

Again, we'll replot the S-N curve from N = 1000 to N = 1,000,000. This time the axes are not log-log so the S-N curve is not a straight line. In the case where we used only two data points, we will draw a straight line between the two points (the red line).

Because we used an extreme example of two points very far apart on the x axis, the difference between the actual S-N curve and the straight line is stark. Linear interpolation is only appropriate with a large number of data points and when the S-N curve isn't a straight line when plotted on log-log axes.

### Conclusion

We've covered the differences when interpolating S-N data and when each method is appropriate. If in doubt about which method to use, log-log is the conservative choice.

#### Reference

[1] Shigley, J., Mischke, C., *Mechanical Engineering Design*, 5th Edition, 1989, McGraw-Hill, pp. 278-280.