Page 459 - Mechanical Behavior of Materials

Section 9.7 Mean Stresses 459

Solution Since the SWT relationship generally works well for aluminum alloys, we will use
this. The form of Eq. 9.24(b) is the most convenient, for which we need
σ min 172
R = = = 0.40
σ max 430

Constants σ = 900 MPa and b =−0.102 are available for this material from Table 9.1. Solving
f
Eq. 9.24(b) for N f and substituting gives
$ % 1/b $ % 1/(−0.102)
! !
1 σ max 1 − R 1 430 1 − 0.40
N f = = = 255,000 cycles Ans.
2 σ 2 2 900 2
f
Example 9.5
Fatigue data for unnotched, axially loaded specimens tested at various mean stresses are given in
Table E9.5 for the same AISI 4340 steel as in Table 9.1. Additional data for zero mean stress are
given in Ex. 9.1. Plot σ ar versus N f for all data for the mean stress equations of (a) Goodman,

Eq. 9.15, and (b) Morrow with σ , Eq. 9.17(b). On each plot, also show the stress–life line from
f
the constants in Table 9.1. Then comment on the success of each equation in correlating the data.
Solution (a) Materials properties σ u = 1172 MPa, σ = 1758 MPa, and b =−0.0977 are

f
needed from Table 9.1. Solving the Goodman relationship of Eq. 9.15 for σ ar gives
σ a
σ ar =
σ m
1 −
σ u
Then calculate σ ar for each (σ a ,σ m ) combination in Table E9.5. For example, for the first test
listed, with N f = 73,780 cycles, the value is
379 MPa
σ ar = = 806.1MPa
621 MPa
1 −
1172 MPa
Table E9.5
σ a ,MPa σ m ,MPa N f , cycles σ a ,MPa σ m ,MPa N f , cycles

379 621 73 780 310 414 445 020
345 621 83 810 552 207 45 490
276 621 567 590 483 207 109 680
517 414 31 280 414 207 510 250
483 414 50 490 586 −207 208 030
414 414 84 420 552 −207 193 220
345 414 437 170 483 −207 901 430
345 414 730 570
Source: Data in [Dowling 73]. Note: All prestrained 10 cycles at ε a = 0.01.