Plotting the S-N curve requires actual fatigue testing of the specimen, and that is why it may not always possible to have an exact S-N diagram for the component of your design interest. Fortunately, there is an work around, an analytical (relatively less accurate) method of calculation for the high cycle fatigue analysis of the components made out of steel.

We will discuss the analytical method with the help of the following fatigue stress design calculation problem:

**Calculate the approximate fatigue strength of a steel component at 10^4 cycles. The following input data is available:**

**The ultimate tensile strength of the steel (Sut)= 85 Kpsi**

**Solution:**

**Step-1-**** ****calculate**** ****the**** ****endurance**** ****strength**** ****(S’e):**** **The endurance strength (S’e) can be calculated from the following equations:

**S’e =**** ****0.5*Sut**** ****for**** ****Sut<=200**** ****Kpsi**** **

=** ****100**** ****Kpsi**** ****for**** ****Sut**** ****>200**** ****Kpsi**

For our example, Sut=85 Kpsi. So, endurance strength for our problem is

S’e=0.5*85 = **42.5**** ****Kpsi**

**Step-2**** – ****Calculate**** ****the**** ****fatigue**** ****strength**** ****coefficient**** ****(σ’f):**** **The fatigue strength coefficient is the true stress corresponding to the fracture of the component in one reversal of the loading cycle. The fatigue strength coefficient can be calculated from the following equation:

**σ’f = Sut+50 kpsi**

For our example, Sut=85 Kpsi. So, the fatigue strength coefficient for our problem is

σ’f = 85+50 =**135**** ****kpsi**

**Step-3**** – ****Calculate**** ****the**** ****fatigue**** ****strength**** ****exponent**** ****(b):** Smaller the fatigue strength exponent of the component, larger the fatigue life of the component. Typically, the values of **b** for the common materials lies in the range of **-0.12**** ****to**** ****-0.05.** The fatigue strength exponent can be calculated from the following equation:

**b=**** ****-[log(σ**** ′****f**** ****)/(S’e)]/[log(2*Ne)]**

**Where,**

**σ’f**** ****=**** **fatigue strength coefficient = **135**** ****Kpsi** in our example

**S’e** = Endurance strength = **42.5**** ****Kpsi** in our example

**Ne**** ****=**** **Number of loading cycles corresponding to the endurance limit = **1000000**

So, by putting all the above values, for our example,

**b = -0.079**

**Step**** – ****4**** – ****Calculate**** ****fatigue**** ****strength**** ****(S’f):**** **The fatigue strength of the component corresponding to the particular number of loading cycles can be calculated from the following equation:

**S**** ′****f**** ****=**** ****(σ’f)***** ****(2n**** ****)^b**

**Where,**

**σ’f**** =**fatigue strength coefficient=**135****Kpsi** in our example

n = The number of loading cycle for which the fatigue strength need to be calculated = 10000 for our example as given in the problem statement

**b ****= **Fatigue strength exponent =** -0.079**

So, by putting all the above values, for our example, the fatigue strength of the component at 10000 cycles is:

**S ′f = 61.33 Kpsi**

The fatigue stress design calculation explained here is useful for a design engineer in case she don’t have a S-N diagram for the material of her design interests. However, the method has the following limitations:

- It should only be used for the component of steel
- It produces reasonable results only for the high cycle fatigue strength analysis

### Shibashis Ghosh

Disclaimer: I work for Altair. mechGuru.com is my personal blog. Although i have tried to put my neutral opinion while writing about different competitor's technologies, still i would like you to read the articles by keeping my background in mind.

#### Latest posts by Shibashis Ghosh (see all)

- INR 0.8 Million per Kg is the Cost of NOT Doing Design Optimization – How? - November 6, 2018
- How to Make Line Follower Robot without Using Microcontroller - September 15, 2018
- Design for Additive Manufacturing Worksheet - July 7, 2018