Fatigue Failure
Fatigue refers to failure due to time-varying loads. This is a very complicated subject,
and in this lesson we will attempt to develop an overview. Please note that Shigley devotes more
than 80 pages to this subject. Multiple-semester courses are taught on this subject.
Section 6 – 1, Introduction to Fatigue in Metals
This section contains a great deal of detailed material. I asked you to skim it. However, I
want to emphasize one very important concept.
When a machine part made of a ductile material fails statically, the stress in the material
has usually exceeded the yield strength, and the material has developed a large deflection. Static
failure of ductile material gives a warning. For example, suppose a shaft with a large central
rotor was improperly designed such that the bending stresses in the shaft exceeded the yield
strength of the material when the shaft was not rotating. The shaft would sag, and eventually the
rotor would rub against the housing when the machine was turned on. This would lead to the
discovery of the ductile failure, and the defective shaft would be replaced.
As an aside, this is a real phenomenon with respect to large steam turbines. If the turbine
shaft were stationary, it would develop a permanent sag due to gravity. The solution is to slowly
“roll” the turbine at a fractional rpm, using a very small electric motor.
However, suppose the shaft did not yield plastically, and everything appeared to be fine,
until the shaft failed catastrophically from fatigue. This could be a much more dangerous
situation. Fatigue failures tend to be sudden and total, with no warning. For this reason, it is
very important to consider fatigue failure during a design process.
Unfortunately, fatigue is a much-more-complicated subject than static loads.
Section 6 – 4, The Stress-Life Method
The strength of a material under fatigue loads is determined by subjecting a specimen of
the material to a large number of cycles of a known load, and counting the number of cycles until
failure. Unfortunately, a large number of known loads must be tested, and the resulting test
program is time-consuming and costly.
MAE3501 - 19 - 1
A rotating specimen, such as the one shown in the following figure, is subjected to forces
as shown. These forces develop a constant bending moment at the center of the specimen
resulting in compression along the top of the specimen, and tension along the bottom of the
specimen. As the specimen rotates, the stress in any longitudinal fiber oscillates back and forth
between tension and compression.
After a large number of tests have been conducted, the logs of the magnitudes of the
reversed stress are plotted vs. the log of the number of cycles to failure, as shown in the
following figure.
MAE3501 - 19 - 2
This is referred to as an S – N diagram. There are a number of important aspects that
need to be emphasized.
As would be expected, for a reversing stress approaching the ultimate strength of the
material, only a few cycles are required for failure. As the reversing stress decreases in
magnitude, more cycles are required for failure.
There is a “knee” in the curve, usually somewhere around 10 3 cycles for steels. This
knee separates “low-cycle fatigue” from “high-cycle fatigue.”
For steels, there is a second “knee” in the curve, usually between 10 6 and 10 7 cycles. To
the right of this second “knee” the stress levels form a horizontal asymptote. The material
strength corresponding to this stress level is referred to as the endurance limit of the material
and is denoted by S E . For reversing stresses below the endurance limit, fatigue will not occur
regardless of the number of cycles. Please note that, in general, non-ferrous metals do not
possess an endurance limit.
The above figure does not specifically label the region between the “knees” of the curve.
I will refer to this region of the figure as the “central region.”
N, the number of cycles in an S – N diagram, refers to the number of completely reversed
cycles, such as would occur with the rotating specimen discussed earlier. However, if a stress is
applied then removed, then applied then removed, each cycle counts as one-half a reversed cycle.
MAE3501 - 19 - 3
Section 6 – 7, The Endurance Limit
Consider the following figure, in which the endurance limit is plotted against the ultimate
strength for a large number of carbon and alloy steels.
It is reasonable to conclude that the endurance limit is about one-half the ultimate tensile
strength, for ultimate tensile strengths below about 200 ksi. For ultimate tensile strengths above
200 ksi, the endurance limit is about 100 ksi.
This is an extremely important approximation for mechanical engineers. It is relatively
straightforward to find a source listing ultimate tensile strengths of steels, permitting the
endurance limit to be estimated with reasonable accuracy.
Shigley contains material properties for a large number of steels in Tables A-20 through
A-27.
MAE3501 - 19 - 4
Section 6 – 8, Fatigue Strength
This section presents a relatively simple, extremely powerful method of estimating
points on an S – N diagram, for high-cycle fatigue of steels, without conducting fatigue tests.
In the central region, between the “knees” in the curve, the S – N curve is close to a
straight line. The S value at the left end of the central region is fairly close to S UT . The S value
at the right end of the central region is equal to S E . This leads to the following relationships for
this central region.
S  a Nb
(19.1)
a and b in equation (19.1) are determined from the following relationships.
a
(f S UT ) 2
SE
(19.2)
fS 
1
b     log10  UT 
 3
 SE 
(19.3)
f is determined from the following figure, Figure 6-18 in Shigley. f is the value of S, the
fatigue strength at the left knee of the S – N diagram, divided by S UT .
MAE3501 - 19 - 5
Example
For 1050 HR steel, estimate the following.
(a)
(b)
(c)
The endurance limit.
The fatigue strength corresponding to 10 4 completely reversed cycles.
The number of completely reversed cycles to failure under a stress of 55 ksi.
(a)
From Table A – 20,
(b)
S UT  90 ksi
(E.1)
S E  0.5 S UT  45 ksi 
(E.2)
Use the figure on the previous page.
f  0.86
(E.3)
Substitute equations (E.1), (E.2), and (E.3) into equations (19.2) and (19.3).
(0.86  90) 2
a
 133 ksi
45
(E.4)
1
 0.86  90 
b     log10 
   0.0785
 3
 45 
(E.5)
Use equation (19.1).
S  133 (10 4 )  0.0785  64.5 ksi 
(c)
(E.6)
Use equation (19.1).
55  133 ( N)  0.0785
 55 
N

 133 


1


  0.0785 
(E.7)
76,800 cycles 
MAE3501 - 19 - 6
(E.8)
It would appear that the method presented in this section resulted from complicated
analyses using a large number of S – N curves for different steels. However, this is not the case.
The figure at the bottom of page 5, used to determine f, did result from a large number of
S – N curves. For each curve, S, the fatigue strength, at the left knee of the curve was divided by
S UT and the resulting fraction correlated against S UT . The figure presents the correlation.
From that point, the method presented is almost trivial. For the previous example,
determine the values of material fatigue strength corresponding to 10 3 and 10 6 cycles.
S(10 3 )  133 (10 3 )  0.0785  77.3 ksi  f S UT
(19.4)
S(10 6 )  133 (10 6 )  0.0785  45.0 ksi  S E
(19.5)
The method consists of no more than mathematically plotting the solid line in the
following figure. Note that, instead of the complicated formulas, the solid line could actually be
plotted, but that would require log-log graph paper. The equations appear simpler.
For steels, the left knee generally occurs between 10 3 and 10 4 cycles, and the right knee
generally occurs between 10 6 and 10 7 cycles. If these ranges are assumed to be “correct”, the
parallelogram in the above figure represents a boundary that contains all central-region fatigue
lines for steels.
For any given value of N, the number of cycles to failure, the solid line represents the
lowest possible value for S, the fatigue strength corresponding to that number of cycles. For any
given value of S, the fatigue strength, the solid line represents the lowest possible value for N,
the number of cycles to failure corresponding to that value of fatigue strength.
The method is therefore conservative for either calculation.
MAE3501 - 19 - 7
Section 6-15, Cumulative Fatigue Damage
The core subject of this section is the following. Suppose a material is subjected to a
number of cycles, n 1 , at a reversing stress level corresponding to N 1 cycles to failure, then
subjected to a number of cycles, n 2 , at a reversing stress level corresponding to N 2 cycles to
failure, etc. Determine how to predict fatigue failure from the material S – N diagram.
Miner’s rule states that fatigue failure will occur when the following condition is
satisfied.
n1 n 2
n

  I  1
N1 N 2
NI
(19.6)
Miner’s rule was the method I was taught in the early 1960s. However, Shigley
recommends a modification of Miner’s rule due to Manson that was published in 1965. This
modification is significantly more complicated, and there is no reason to present it in this course,
but I want you to be aware that a “better” method exists.
Shigley also states that Miner’s rule “is quite generally used.”
Homework
Problems 6-3, 6-5.
MAE3501 - 19 - 8