2.9. THE PROBABILISTIC FATIGUE DAMAGE THEORY (THE K-D MODEL) 107
e results from the P-S-N curve and the K-D model are almost identical. e relative difference
is 0.036%.
ese three examples cover typical cyclic loadings. Results from these three examples have
shown that the reliability obtained from the P-S-N curves and the K-D models for the same
issue are almost the same with a maximum relative error 0.036%. ese have approved and
verified that the probabilistic fatigue damage model (the K-D model) can be used to describe
the fatigue test data of the same material specimen under different fatigue test stress levels.
e K-D model can be used to solve the reliability of a component under any type of
cyclic loadings. e following Example 2.35 could not be directly solved by the P-S-N curve
approach and but can be solved by the K-D model.
Example 2.35
e aluminum 6061-T6 10 Gauge sheet-type fatigue specimen as shown in Figure 2.8 is sub-
jected to cyclic axial stress with a stress amplitude
a
D 25 (ksi) and a mean stress
m
D 15 (ksi).
e number of cycles of this cyclic axial stress is 50,000 (cycles). e ultimate tensile strength
of this material is S
u
D 51:2 (ksi). e P-S-N curves of this material is listed in Table 2.13. e
K-D model is listed in Table 2.35. Use the P-S-N curve and the K-D probabilistic model to
calculate its reliability.
Solution:
(1) e P-S-N curve approach.
e P-S-N curves of this material listed in Table 2.13 do not have a probabilistic distri-
bution function for this specific cyclic axial stress which is a stress amplitude
a
D 25 (ksi) and a
mean stress
m
D 15 (ksi). So, the P-S-N curve approach could not be directly used to calculate
the reliability of this problem.
(2) e K-D model.
e K-D probabilistic fatigue damage model can be used to solve this problem.
e equivalent stress amplitude of the cyclic stress with a stress amplitude
a
D 25 (ksi)
and a mean stress
m
D 15 (ksi) per Equation (2.21) is:
eq
D
a
S
u
.
S
u
m
/
D
25 51:2
.51:2 15/
D 35:359.ksi/: (a)
e fatigue damage index D due to the specified cyclic axial stress per Equation (2.85) will be:
D D n
L
m
eq
D 50;000
.
35:35912
/
3:8812
D 5:11698 10
10
: (b)
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