Page 362 - Shigley's Mechanical Engineering Design

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Fatigue Failure Resulting from Variable Loading 337
From Eq. (6–73),

k c = 1.23S ¯ −0.0778 LN(1, 0.125) = 1.23(87.6) −0.0778 LN(1, 0.125)
ut
= 0.869LN(1, 0.125)
k d = k f = (1, 0)

The endurance strength, from Eq. (6–71), is
S e = k a k b k c k d k f S e
S e = 0.816LN(1, 0.058)(1)0.869LN(1, 0.125)(1)(1)44.3LN(1, 0.138)
The parameters of S e are
S e = 0.816(0.869)44.3 = 31.4 kpsi
¯
2
2 1/2
2
C Se = (0.058 + 0.125 + 0.138 ) = 0.195
so S e = 31.4LN(1, 0.195) kpsi.
In computing the stress, the section at the hole governs. Using the terminology
.
of Table A–15–1 we find d/w = 0.50, therefore K t = 2.18. From Table 6–15,
√
a = 5/S ut = 5/87.6 = 0.0571 and C kf = 0.10. From Eqs. (6–78) and (6–79) with
r = 0.375 in,
2.18
K t
K f = √ LN 1, C K f = LN(1, 0.10)
2(K t − 1) a 2(2.18 − 1) 0.0571
1 + √ 1 + √
K t r 2.18 0.375
= 1.98LN(1, 0.10)
The stress at the hole is
F 1000LN(1, 0.12)
= K f = 1.98LN(1, 0.10)
A 0.25(0.75)
1000 −3
¯ σ = 1.98 10 = 10.56 kpsi
0.25(0.75)
2
2 1/2
C σ = (0.10 + 0.12 ) = 0.156
so stress can be expressed as = 10.56LN(1, 0.156) kpsi. 34
The endurance limit is considerably greater than the load-induced stress, indicat-
ing that finite life is not a problem. For interfering lognormal-lognormal distributions,
Eq. (5–43), p. 250, gives
⎛ ⎞

2
¯ 2 31.4 1 + 0.156
S e 1 + C σ
ln ln ⎝ 2 ⎠
¯ σ 1 + C 2 10.56 1 + 0.195
S e
=−4.37
z =−' =−&
2
2
! 2 2 " ln[(1 + 0.195 )(1 + 0.156 )]
ln 1 + C 1 + C
S e σ
From Table A–10 the probability of failure p f = (−4.37) = .000 006 35, and the
reliability is
Answer R = 1 − 0.000 006 35 = 0.999 993 65
34 Note that there is a simplification here. The area is not a deterministic quantity. It will have a statistical
distribution also. However no information was given here, and so it was treated as being deterministic.