Page 458 - Mechanical Behavior of Materials

458 Chapter 9 Fatigue of Materials: Introduction and Stress-Based Approach

Solving the life equation for N f and substituting σ ar produces the desired result:

$ % 1/b 1/(−0.0977)
1 σ ar 1 507.8
N f = = = 166,000 cycles Ans.
2 σ 2 1758
f
Noting that Eqs. 9.7 and 9.21 were combined to obtain Eq. 9.23, we get the same result in fewer
steps by solving the latter for N f :

$ % 1/b 1/(−0.0977)
1 σ a 1 450
N f = = = 166,000 cycles Ans.
2 σ − σ m 2 1758 − 200

f
(b) The σ a versus N f curve for σ m = 200 MPa may be obtained from Eq. 9.23, with σ a left
as a variable.

b

σ a = (σ − σ m )(2N f ) = (1758 − 200)(2N f ) −0.0977 = 1558(2N f ) −0.0977 MPa Ans.
f
Second Solution (a) An alternative is to employ Eq. 9.24 with σ max = σ m + σ a = 650 MPa.
First, solve Eq. 9.24 for N f , and then substitute as appropriate:

$ √ % 1/b $ √ % 1/(−0.0977)
1 σ max σ a 1 650 × 450
N f = = = 86,900 cycles Ans.
2 σ 2 1758
f
(b) The σ a versus N f curve for σ m = 200 MPa may be obtained from the foregoing by
retaining σ a in variable form:

√ 1/(−0.0977)
1 (200 + σ a )σ a
N f = (σ a in MPa) Ans.
2 1758

Discussion For (a), the N f value is seen to differ between the two solutions, which were
based on the Morrow and SWT mean stress equations, respectively. In general, these approaches
can be expected to agree only roughly. The equations for σ a versus N f for part (b) also differ,
with the one for the second solution being a gradual curve, rather than a straight line, on log–log
coordinates.

Example 9.4
The aluminum alloy 2024-T4 is subjected to cyclic loading between σ min = 172 and σ max =
430 MPa. What life is expected?