Page 424 - Mechanical Behavior of Materials

Section 9.2 Definitions and Concepts 425

Table E9.1

σ a ,MPa N f , cycles
948 222
834 992
703 6 004
631 14 130
579 43 860
524 132 150

Source: Data in [Dowling 73].
Solution (a) The plotted data are shown in Fig. E9.1. They do seem to fall along a straight
line, and the first and last points represent the line well. Denote these points as (σ 1 , N 1 ) and
B
(σ 2 , N 2 ), and apply the equation σ a = AN to both:
f
B B
σ 1 = AN , σ 2 = AN 2
1
Then divide the second equation into the first, and take logarithms of both sides:
B

σ 1 N 1 σ 1 N 1
= , log = B log
σ 2 N 2 σ 2 N 2
Solving for B gives
log σ 1 − log σ 2 log (948 MPa) − log (524 MPa)
B = = =−0.0928 Ans.
log N 1 − log N 2 log 222 − log 132,150

3000
σ´ = 1695 MPa
f
2000 AISI 4340 steel
σ a , Stress Amplitude, MPa 1000 A = 1587 MPa h v h
σ = 1172 MPa
u

500
v D
B = = – 0.0945
300
200 b = B h D v D v
D
h
100
0.5 1 10 10 2 10 3 10 4 10 5 10 6
N , Cycles to Failure
f
Figure E9.1