Page 348 - Shigley's Mechanical Engineering Design

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Fatigue Failure Resulting from Variable Loading 323
Using the deterministic formulation as a linear damage rule we write

n i

D = (6–58)
N i
where D is the accumulated damage. When D = c = 1, failure ensues.

EXAMPLE 6–15 Given a part with S ut = 151 kpsi and at the critical location of the part, S e = 67.5 kpsi.
For the loading of Fig. 6–33, estimate the number of repetitions of the stress-time block
in Fig. 6–33 that can be made before failure.

Solution From Fig. 6–18, p. 285, for S ut = 151 kpsi, f = 0.795. From Eq. (6–14),
( fS ut ) 2 [0.795(151)] 2
a = = = 213.5 kpsi
S e 67.5
From Eq. (6–15),
1 fS ut 1 0.795(151)
b =− log =− log =−0.0833
3 S e 3 67.5
So,
−1/0.0833

S f = 213.5N −0.0833 N = S f (1), (2)
213.5
We prepare to add two columns to the previous table. Using the Gerber fatigue criterion,
Eq. (6–47), p. 306, with S e = S f , and n = 1, we can write
σ a

σ m > 0
S f = 1 − (σ m /S ut ) 2 (3)
S e σ m ≤ 0
where S f is the fatigue strength associated with a completely reversed stress, σ rev ,
equivalent to the fluctuating stresses [see Ex. 6–12, part (b)].
Cycle 1: r = σ a /σ m = 70/10 = 7, and the strength amplitude from Table 6–7, p. 307, is
⎧ ⎫

2
7 151 2 ⎨ 2(67.5) 2⎬
S a = −1 + 1 + = 67.2 kpsi
7(151)
2(67.5) ⎩ ⎭
Since σ a > S a , that is, 70 > 67.2, life is reduced. From Eq. (3),
70
S f = 2 = 70.3 kpsi
1 − (10/151)
and from Eq. (2)
−1/0.0833
70.3 3

N = = 619(10 ) cycles
213.5
Cycle 2: r = 10/50 = 0.2, and the strength amplitude is
⎧ ⎫

2
0.2 151 2 ⎨ 2(67.5) 2⎬
S a = −1 + 1 + = 24.2 kpsi
0.2(151)
2(67.5) ⎩ ⎭
Since σ a < S a , that is 10 < 24.2, then S f = S e and indefinite life follows. Thus,
N ← ∞ .