Question

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.

Answer 1

B A and C

Step-by-step explanation:

Given:

Specimen σ
`_(max)` σ
`_(min)`

A +450 -150

B +300 -300

C +500 -200

Solution:

Compute the mean stress

σ
`_(m)` = (σ
`_(max)` + σ
`_(min)`)/2

σ
`_(mA)` = (450 + (-150)) / 2

= (450 - 150) / 2

= 300/2

σ
`_(mA)` = 150 MPa

σ
`_(mB)` = (300 + (-300))/2

= (300 - 300) / 2

= 0/2

σ
`_(mB)` = 0 MPa

σ
`_(mC)` = (500 + (-200))/2

= (500 - 200) / 2

= 300/2

σ
`_(mC)` = 150 MPa

Compute stress amplitude:

σ
`_(a)` = (σ
`_(max)` - σ
`_(min)`)/2

σ
`_(aA)` = (450 - (-150)) / 2

= (450 + 150) / 2

= 600/2

σ
`_(aA)` = 300 MPa

σ
`_(aB)` = (300- (-300)) / 2

= (300 + 300) / 2

= 600/2

σ
`_(aB)` = 300 MPa

σ
`_(aC)` = (500 - (-200))/2

= (500 + 200) / 2

= 700 / 2

σ
`_(aC)` = 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.