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Fatigue Failure Resulting from Variable Loading 341

2
2(31.1)(1 + 0.15)
−1 + 1 +
(1 + 0.045) 2 1.26(86.2)(1 + 0.045)
C Sa = − 1 = 0.134
1 + 0.150 2
2(31.1)
−1 + 1 +
1.26(86.2)
Reliability: Since S a = 28.9LN(1, 0.134) kpsi and = 14.5LN(1, 0.121) kpsi,

a
Eq. (5–43), p. 250, gives
⎛ ⎞

2 2
¯ 1 + C 28.9 1 + 0.121
S a
ln σ a ln ⎝ 14.5 1 + 0.134 2 ⎠
2
¯ σ a 1 + C
S a
=−3.83
z =−' ! " =−& 2 2
ln 1 + C 2 1 + C 2 ln[(1 + 0.134 )(1 + 0.121 )]
S a σ a
From Table A–10 the probability of failure is p f = 0.000 065, and the reliability is,
against fatigue,
Answer R = 1 − p f = 1 − 0.000 065 = 0.999 935
The chance of first-cycle yielding is estimated by interfering S y with max . The
quantity max is formed from + . The mean of max is ¯σ +¯σ = 14.5 +

a
a
m
m
11.54 = 26.04 kpsi. The coefficient of variation of the sum is 0.121, since both
COVs are 0.121, thus C σ max = 0.121. We interfere S y = 56LN(1, 0.077) kpsi with
max = 26.04LN (1, 0.121) kpsi. The corresponding z variable is
⎛ ⎞

56 1 + 0.121 2
ln ⎝ ⎠
26.04 1 + 0.077 2
=−5.39
z =−& 2 2
ln[(1 + 0.077 )(1 + 0.121 )]
7
which represents, from Table A–10, a probability of failure of approximately 0.0 358
−8
[which represents 3.58(10 )] of first-cycle yield in the fillet.
The probability of observing a fatigue failure exceeds the probability of a yield
failure, something a deterministic analysis does not foresee and in fact could lead one
to expect a yield failure should a failure occur. Look at the S a interference and the

a
S
max y interference and examine the z expressions. These control the relative proba-

bilities. A deterministic analysis is oblivious to this and can mislead. Check your sta-
tistics text for events that are not mutually exclusive, but are independent, to quantify
the probability of failure:
p f = p(yield) + p(fatigue) − p(yield and fatigue)
= p(yield) + p(fatigue) − p(yield)p(fatigue)
−4
−4
−7
−4
−7
= 0.358(10 ) + 0.65(10 ) − 0.358(10 )0.65(10 ) = 0.650(10 )
−4
R = 1 − 0.650(10 ) = 0.999 935
against either or both modes of failure.