Page 338 - Shigley's Mechanical Engineering Design

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Fatigue Failure Resulting from Variable Loading 313
For δ = 2in, σ a = 11 kpsi, σ m = 10.99(1 + 2) = 33 kpsi
For δ = 5in, σ a = 11 kpsi, σ m = 10.99(1 + 5) = 65.9 kpsi
(a) A plot of the Gerber and Langer criteria is shown in Fig. 6–30b. The three preload

deflections of 0, 2, and 5 in are shown as points A, A , and A . Note that since σ a is
constant at 11 kpsi, the load line is horizontal and does not contain the origin. The
intersection between the Gerber line and the load line is found from solving Eq. (6–42)
for S m and substituting 11 kpsi for S a :

S a 11
S m = S ut 1 − = 150 1 − = 116.9 kpsi
S e 28
The intersection of the Langer line and the load line is found from solving Eq. (6–44)
for S m and substituting 11 kpsi for S a :
S m = S y − S a = 127 − 11 = 116 kpsi
The threats from fatigue and first-cycle yielding are approximately equal.
(b) For δ = 2 in,

Answer n f = S m = 116.9 = 3.54 n y = 116 = 3.52
σ m 33 33
and for δ = 5 in,
116.9 116
Answer n f = = 1.77 n y = = 1.76
65.9 65.9

EXAMPLE 6–12 A steel bar undergoes cyclic loading such that σ max = 60 kpsi and σ min =−20 kpsi. For
the material, S ut = 80 kpsi, S y = 65 kpsi, a fully corrected endurance limit of S e =
40 kpsi, and f = 0.9. Estimate the number of cycles to a fatigue failure using:
(a) Modified Goodman criterion.
(b) Gerber criterion.

Solution From the given stresses,

60 − (−20) 60 + (−20)
σ a = = 40 kpsi σ m = = 20 kpsi
2 2
(a) For the modified Goodman criterion, Eq. (6–46), the fatigue factor of safety based
on infinite life is
1 1
n f = = = 0.8
σ a σ m 40 20
+ +
S e S ut 40 80
This indicates a finite life is predicted. The S-N diagram is only applicable for completely
reversed stresses. To estimate the finite life for a fluctuating stress, we will obtain an
equivalent completely reversed stress that is expected to be as damaging as the fluctuat-
ing stress. A commonly used approach is to assume that since the modified Goodman
6
line represents all stress situations with a constant life of 10 cycles, other constant-life
lines can be generated by passing a line through (S ut , 0) and a fluctuating stress point
(σ m , σ a ). The point where this line intersects the σ a axis represents a completely reversed
stress (since at this point σ m = 0), which predicts the same life as the fluctuating stress.