Page 350 - Shigley's Mechanical Engineering Design

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Fatigue Failure Resulting from Variable Loading 325
Figure 6–34 shows that the material has a life N 1 = 8520 cycles at 60 kpsi, and conse-
quently, after the application of σ 1 for 3000 cycles, there are N 1 − n 1 = 5520 cycles of
life remaining at σ 1 . This locates the finite-life strength S f,1 of the damaged material, as
shown in Fig. 6–34. To get a second point, we ask the question: With n 1 and N 1 given,
how many cycles of stress σ 2 = S can be applied before the damaged material fails?
e,0
This corresponds to n 2 cycles of stress reversal, and hence, from Eq. (6–58), we have
n 1 n 2
+ = 1 (a)
N 1 N 2
Solving for n 2 gives
N 2
n 2 = (N 1 − n 1 ) (b)
N 1
Then
10 6
! 3 3 " 6
n 2 = 8.52 10 − 3 10 = 0.648 10 cycles
8.52 10 3
This corresponds to the finite-life strength S f,2 in Fig. 6–34. A line through S f,1 and S f,2
is the log S–log N diagram of the damaged material according to the Miner rule. Two

b

points, (N 1 − n 1, σ 1 ) and (n 2, σ 2 ), determine the new equation for the line, S f = a N .

b
b

Thus, σ 1 = a (N 1 − n 1 ) , and σ 2 = a n . Dividing the two equations, taking the loga-
2

rithm of the results, and solving for b gives
log (σ 1 /σ 2 )

b =
N 1 − n 1
log
n 2
Substituting n 2 from Eq. (b) and simplifying gives
log (σ 1 /σ 2 )

b =
log (N 1 /N 2 )
For the undamaged material, N 1 = (σ 1 /a) 1/b and N 2 = (σ 2 /a) 1/b , then
log (σ 1 /σ 2 ) log (σ 1 /σ 2 )

b = ! 1/b 1/b " = = b
log (σ 1 /a) /(σ 2 /a) (1/b) log (σ 1 /σ 2 )
This means that the damaged material line has the same slope as the virgin material line,
b
and the two lines are parallel. The value of a is then found from a = S f /N .

3 −0.085 091
For the case we are illustrating, a = 60/[5.52(10) ] = 124.898 kpsi, and

b
6 −0.085 091
thus the new endurance limit is S = a N = 124.898[(10) ] = 38.6 kpsi.

e,1 e
Though the Miner rule is quite generally used, it fails in two ways to agree with
experiment. First, note that this theory states that the static strength S ut is damaged, that
3
is, decreased, because of the application of σ 1 ; see Fig. 6–34 at N = 10 cycles.
Experiments fail to verify this prediction.
The Miner rule, as given by Eq. (6–58), does not account for the order in which the

stresses are applied, and hence ignores any stresses less than S . But it can be seen in
e,0
Fig. 6–34 that a stress σ 3 in the range S <σ 3 < S would cause damage if applied
e,1 e,0
after the endurance limit had been damaged by the application of σ 1 .
25
Manson’s approach overcomes both of the deficiencies noted for the Palmgren-
Miner method; historically it is a much more recent approach, and it is just as easy to
25
S. S. Manson, A. J. Nachtigall, C. R. Ensign, and J. C. Fresche, “Further Investigation of a Relation for
Cumulative Fatigue Damage in Bending,” Trans. ASME, J. Eng. Ind., ser. B, vol. 87, No. 1, pp. 25–35,
February 1965.