Question

8.26 Three identical fatigue specimens (denoted A, B, and C) are fabricated from a nonferrous alloy. Each is subjected to one of the maximum-minimum stress cycles listed below; the frequency is the same for all three tests.Specimenmax (MPa) min (MPa) A +450 –350 B +400 –300 C +340 –340 (a) Rank the fatigue lifetimes of these three specimens from the longest to the shortest. (b) Now justify this ranking using a schematic S–N plot.

Answer 1

Answer:

Fatigue lifetimes will be ranked as B>A>C

Explanation:

Start by calculating the mean stress for all samples

σa= (σamax + σamin)/2 = (450 - 350)/2 = 50MPa

σb= (σbmax + σbmin)/2 = (400 - 300)/2 = 50MPa

σa= (σcmax + σcmin)/2 = (340 - 340)/2 = 0MPa

Now calculate the stress amplitudes of all three samples

σa= (σamax - σamin)/2 = (450 + 350)/2 = 400MPa

σb= (σbmax - σbmin)/2 = (400 + 300)/2 = 350MPa

σc= (σcmax - σcmin)/2 = (340 + 340)/2 = 340MPa

The mean stress of samples A and B is the same which is higher than sample C. Hence samples A and B will have a higher fatigue life than sample C. However, the higher stress amplitude means lower fatigue life, hence, sample B will have a higher fatigue life than sample A. So the order will be B> A> C.

Attached picture shows the justification using and S- N plot.