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Fatigue Failure Resulting from Variable Loading 281
9
If this should happen, then Reemsnyder suggests the use of numerical integration
employing the algorithm
m
δa j = C( K I ) (δN) j
j
a j+1 = a j + δa j
(6–7)
N j+1 = N j + δN j

N f = δN j

Here δa j and δN j are increments of the crack length and the number of cycles. The pro-
cedure is to select a value of δN j , using a i determine β and compute K I , determine
δa j , and then find the next value of a. Repeat the procedure until a = a f .
The following example is highly simplified with β constant in order to give some
understanding of the procedure. Normally, one uses fatigue crack growth computer pro-
grams such as NASA/FLAGRO 2.0 with more comprehensive theoretical models to
solve these problems.

EXAMPLE 6–1 The bar shown in Fig. 6–16 is subjected to a repeated moment 0 ≤ M ≤ 1200 lbf · in.
√
The bar is AISI 4430 steel with S ut = 185 kpsi, S y = 170 kpsi, and K Ic = 73 kpsi in.
Material tests on various specimens of this material with identical heat treatment
√
indicate worst-case constants of C = 3.8(10 −11 )(in/cycle)/(kpsi in) m and m = 3.0.
As shown, a nick of size 0.004 in has been discovered on the bottom of the bar. Estimate
the number of cycles of life remaining.

Solution The stress range σ is always computed by using the nominal (uncracked) area. Thus
I bh 2 0.25(0.5) 2 3
= = = 0.010 42 in
c 6 6
Therefore, before the crack initiates, the stress range is
M 1200 3
σ = = = 115.2(10 ) psi = 115.2 kpsi
I/c 0.010 42
which is below the yield strength. As the crack grows, it will eventually become long
enough such that the bar will completely yield or undergo a brittle fracture. For the ratio
of S y /S ut it is highly unlikely that the bar will reach complete yield. For brittle fracture,
designate the crack length as a f . If β = 1, then from Eq. (5–37) with K I = K Ic , we
approximate a f as
2 2
1 K Ic . 1 73

a f = = = 0.1278 in
π βσ max π 115.2
Figure 6–16 1 4 in

M M 1 in
2
Nick

9 Op. cit.