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 -150
B +300 -300
C +500 -200
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.

B A and C

Explanation:

Given:

Specimen         σ                      σ

A                       +450                      -150

B                       +300                      -300

C                       +500                      -200

Solution:

Compute the mean stress

σ =  (σ  +  σ)/2

σ =  (450 + (-150)) / 2

       =  (450 - 150) / 2  

       = 300/2

σ = 150 MPa

σ  = (300 + (-300))/2

        = (300 - 300) / 2

        = 0/2  

σ  = 0 MPa

σ  = (500 + (-200))/2

        = (500 - 200) / 2

        = 300/2

σ  = 150 MPa  

Compute stress amplitude:

σ =  (σ  -  σ)/2    

σ =  (450 - (-150)) / 2

       =  (450 + 150) / 2

       = 600/2

σ = 300 MPa

σ =  (300- (-300)) / 2

       =  (300 + 300) / 2

       = 600/2

σ  = 300 MPa

σ  = (500 - (-200))/2

        = (500 + 200) / 2

        = 700 / 2

σ   = 350 MPa

From the above results it is concluded that the longest  fatigue lifetime is of specimen B because it has the minimum mean stress.

Next, the specimen A has the fatigue lifetime which is shorter than B but longer than specimen C.

In the last comes specimen C which has the shortest fatigue lifetime because it has the higher mean stress and highest stress amplitude.

jesus loves you

explanation: