Page 303 - Materials Science and Engineering An Introduction

8.8 The S–N Curve • 275

For example, at a stress of 200 MPa (30,000 psi), we would expect 1% of the specimens
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to fail at about 10 cycles, 50% to fail at about 2 10 cycles, and so on. Remember
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that S–N curves represented in the literature are normally average values, unless noted
otherwise.
The fatigue behaviors represented in Figures 8.19a and 8.19b may be classified into
two domains. One is associated with relatively high loads that produce not only elastic
strain but also some plastic strain during each cycle. Consequently, fatigue lives are
relatively short; this domain is termed low-cycle fatigue and occurs at less than about
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5
10 to 10 cycles. For lower stress levels wherein deformations are totally elastic, longer
lives result. This is called high-cycle fatigue because relatively large numbers of cycles
are required to produce fatigue failure. High-cycle fatigue is associated with fatigue lives
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4
greater than about 10 to 10 cycles.

EXAMPLE PROBLEM 8.2
Maximum Load Computation to Avert Fatigue for Rotating-Bending Tests

A cylindrical bar of 1045 steel having the S–N behavior shown in Figure 8.20 is subjected to
rotating-bending tests with reversed-stress cycles (per Figure 8.18). If the bar diameter is 15.0
mm, determine the maximum cyclic load that may be applied to ensure that fatigue failure will
not occur. Assume a factor of safety of 2.0 and that the distance between loadbearing points is
60.0 mm (0.0600 m).

Solution
From Figure 8.20, the 1045 steel has a fatigue limit (maximum stress) of magnitude 310 MPa.
For a cylindrical bar of diameter d 0 (Figure 8.18b), maximum stress for rotating–bending tests
may be determined using the following expression:
16FL
s = (8.18)
3
pd 0
Here, L is equal to the distance between the two loadbearing points (Figure 8.18b), s is the
maximum stress (in our case the fatigue limit), and F is the maximum applied load. When s is
divided by the factor of safety (N), Equation 8.18 takes the form

s 16FL
= (8.19)
N pd 0 3
and solving for F leads to
3
F = spd 0 (8.20)
16NL
Incorporating values for d 0 , L, and N provided in the problem statement as well as the fatigue
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2
limit taken from Figure 8.20 (310 MPa, or 310 10 N/m ) yields the following:
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2
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(310 * 10 N/m )(p)(15 * 10 m) 3
F =
(16)(2)(0.0600 m)
= 1712 N
Therefore, for cyclic reversed and rotating–bending, a maximum load of 1712 N may be applied
without causing the 1045 steel bar to fail by fatigue.