Fatigue Failure Resulting from Variable Loading

6 Fatigue Failure Resulting
from Variable Loading
Chapter Outline
6–1 Introduction to Fatigue in Metals 258
6–2 Approach to Fatigue Failure in Analysis and Design 264
6–3 Fatigue-Life Methods 265
6–4 The Stress-Life Method 265
6–5 The Strain-Life Method 268
6–6 The Linear-Elastic Fracture Mechanics Method 270
6–7 The Endurance Limit 274
6–8 Fatigue Strength 275
6–9 Endurance Limit Modifying Factors 278
6–10 Stress Concentration and Notch Sensitivity 287
6–11 Characterizing Fluctuating Stresses 292
6–12 Fatigue Failure Criteria for Fluctuating Stress 295
6–13 Torsional Fatigue Strength under Fluctuating Stresses 309
6–14 Combinations of Loading Modes 309
6–15 Varying, Fluctuating Stresses; Cumulative Fatigue Damage 313
6–16 Surface Fatigue Strength 319
6–17 Stochastic Analysis 322
6–18 Road Maps and Important Design Equations for the Stress-Life Method 336

257

258 Mechanical Engineering Design

In Chap. 5 we considered the analysis and design of parts subjected to static loading.
The behavior of machine parts is entirely different when they are subjected to time-
varying loading. In this chapter we shall examine how parts fail under variable loading
and how to proportion them to successfully resist such conditions.

6–1 Introduction to Fatigue in Metals

In most testing of those properties of materials that relate to the stress-strain diagram,
the load is applied gradually, to give sufficient time for the strain to fully develop.
Furthermore, the specimen is tested to destruction, and so the stresses are applied only
once. Testing of this kind is applicable, to what are known as static conditions; such
conditions closely approximate the actual conditions to which many structural and
machine members are subjected.

The condition frequently arises, however, in which the stresses vary with time or
they fluctuate between different levels. For example, a particular fiber on the surface of
a rotating shaft subjected to the action of bending loads undergoes both tension and com-
pression for each revolution of the shaft. If the shaft is part of an electric motor rotating
at 1725 rev/min, the fiber is stressed in tension and compression 1725 times each minute.
If, in addition, the shaft is also axially loaded (as it would be, for example, by a helical
or worm gear), an axial component of stress is superposed upon the bending component.
In this case, some stress is always present in any one fiber, but now the level of stress is
fluctuating. These and other kinds of loading occurring in machine members produce
stresses that are called variable, repeated, alternating, or fluctuating stresses.

Often, machine members are found to have failed under the action of repeated or
fluctuating stresses; yet the most careful analysis reveals that the actual maximum
stresses were well below the ultimate strength of the material, and quite frequently even
below the yield strength. The most distinguishing characteristic of these failures is that
the stresses have been repeated a very large number of times. Hence the failure is called
a fatigue failure.

When machine parts fail statically, they usually develop a very large deflection,
because the stress has exceeded the yield strength, and the part is replaced before fracture
actually occurs. Thus many static failures give visible warning in advance. But a fatigue
failure gives no warning! It is sudden and total, and hence dangerous. It is relatively sim-
ple to design against a static failure, because our knowledge is comprehensive. Fatigue is
a much more complicated phenomenon, only partially understood, and the engineer seek-
ing competence must acquire as much knowledge of the subject as possible.

A fatigue failure has an appearance similar to a brittle fracture, as the fracture sur-
faces are flat and perpendicular to the stress axis with the absence of necking. The frac-
ture features of a fatigue failure, however, are quite different from a static brittle fracture
arising from three stages of development. Stage I is the initiation of one or more micro-
cracks due to cyclic plastic deformation followed by crystallographic propagation
extending from two to five grains about the origin. Stage I cracks are not normally dis-
cernible to the naked eye. Stage II progresses from microcracks to macrocracks forming
parallel plateau-like fracture surfaces separated by longitudinal ridges. The plateaus are
generally smooth and normal to the direction of maximum tensile stress. These surfaces
can be wavy dark and light bands referred to as beach marks or clamshell marks, as seen
in Fig. 6–1. During cyclic loading, these cracked surfaces open and close, rubbing
together, and the beach mark appearance depends on the changes in the level or fre-
quency of loading and the corrosive nature of the environment. Stage III occurs during
the final stress cycle when the remaining material cannot support the loads, resulting in

Fatigue Failure Resulting from Variable Loading 259

Figure 6–1

Fatigue failure of a bolt due
to repeated unidirectional
bending. The failure started
at the thread root at A,
propagated across most of
the cross section shown by
the beach marks at B, before
final fast fracture at C. (From
ASM Handbook, Vol. 12:
Fractography, ASM Inter-
national, Materials Park, OH
44073-0002, fig 50, p. 120.
Reprinted by permission of
ASM International ®,
www.asminternational.org.)

a sudden, fast fracture. A stage III fracture can be brittle, ductile, or a combination of
both. Quite often the beach marks, if they exist, and possible patterns in the stage III frac-
ture called chevron lines, point toward the origins of the initial cracks.

There is a good deal to be learned from the fracture patterns of a fatigue failure.1
Figure 6–2 shows representations of failure surfaces of various part geometries under
differing load conditions and levels of stress concentration. Note that, in the case of
rotational bending, even the direction of rotation influences the failure pattern.

Fatigue failure is due to crack formation and propagation. A fatigue crack will typ-
ically initiate at a discontinuity in the material where the cyclic stress is a maximum.
Discontinuities can arise because of:

• Design of rapid changes in cross section, keyways, holes, etc. where stress concen-
trations occur as discussed in Secs. 3–13 and 5–2.

• Elements that roll and/or slide against each other (bearings, gears, cams, etc.) under
high contact pressure, developing concentrated subsurface contact stresses (Sec. 3–19)
that can cause surface pitting or spalling after many cycles of the load.

• Carelessness in locations of stamp marks, tool marks, scratches, and burrs; poor joint
design; improper assembly; and other fabrication faults.

• Composition of the material itself as processed by rolling, forging, casting, extrusion,
drawing, heat treatment, etc. Microscopic and submicroscopic surface and subsurface
discontinuities arise, such as inclusions of foreign material, alloy segregation, voids,
hard precipitated particles, and crystal discontinuities.

Various conditions that can accelerate crack initiation include residual tensile stresses,
elevated temperatures, temperature cycling, a corrosive environment, and high-frequency
cycling.

The rate and direction of fatigue crack propagation is primarily controlled by local-
ized stresses and by the structure of the material at the crack. However, as with crack
formation, other factors may exert a significant influence, such as environment, tem-
perature, and frequency. As stated earlier, cracks will grow along planes normal to the

1See the ASM Handbook, Fractography, ASM International, Metals Park, Ohio, vol. 12, 9th ed., 1987.

260 Mechanical Engineering Design

Figure 6–2

Schematics of fatigue fracture
surfaces produced in smooth
and notched components with
round and rectangular cross
sections under various loading
conditions and nominal stress
levels. (From ASM Handbook,
Vol. 11: Failure Analysis and
Prevention, ASM International,
Materials Park, OH
44073-0002, fig 18, p. 111.
Reprinted by permission
of ASM International ®,
www.asminternational.org.)

Fatigue Failure Resulting from Variable Loading 261

maximum tensile stresses. The crack growth process can be explained by fracture
mechanics (see Sec. 6–6).

A major reference source in the study of fatigue failure is the 21-volume
ASM Metals Handbook. Figures 6–1 to 6–8, reproduced with permission from ASM
International, are but a minuscule sample of examples of fatigue failures for a great
variety of conditions included in the handbook. Comparing Fig. 6–3 with Fig. 6–2, we
see that failure occurred by rotating bending stresses, with the direction of rotation
being clockwise with respect to the view and with a mild stress concentration and low
nominal stress.

Figure 6–3

Fatigue fracture of an AISI
4320 drive shaft. The fatigue
failure initiated at the end of
the keyway at points B and
progressed to final rupture at
C. The final rupture zone is
small, indicating that loads
were low. (From ASM
Handbook, Vol. 11: Failure
Analysis and Prevention, ASM
International, Materials Park,
OH 44073-0002, fig 18,
p. 111. Reprinted by
permission of ASM
International ®,
www.asminternational.org.)

Figure 6–4

Fatigue fracture surface of an
AISI 8640 pin. Sharp corners
of the mismatched grease
holes provided stress
concentrations that initiated
two fatigue cracks indicated
by the arrows. (From ASM
Handbook, Vol. 12:
Fractography, ASM
International, Materials Park,
OH 44073-0002, fig 520,
p. 331. Reprinted by
permission of ASM
International ®,
www.asminternational.org.)

262 Mechanical Engineering Design

Figure 6–5

Fatigue fracture surface of a
forged connecting rod of AISI
8640 steel. The fatigue crack
origin is at the left edge, at the
flash line of the forging, but no
unusual roughness of the flash
trim was indicated. The
fatigue crack progressed
halfway around the oil hole
at the left, indicated by the
beach marks, before final fast
fracture occurred. Note the
pronounced shear lip in the
final fracture at the right edge.
(From ASM Handbook,
Vol. 12: Fractography, ASM
International, Materials Park,
OH 44073-0002, fig 523,
p. 332. Reprinted by
permission of ASM
International ®,
www.asminternational.org.)

Figure 6–6

Fatigue fracture surface of a 200-mm (8-in) diameter piston rod of an alloy
steel steam hammer used for forging. This is an example of a fatigue fracture
caused by pure tension where surface stress concentrations are absent and
a crack may initiate anywhere in the cross section. In this instance, the initial
crack formed at a forging flake slightly below center, grew outward
symmetrically, and ultimately produced a brittle fracture without warning.
(From ASM Handbook, Vol. 12: Fractography, ASM International, Materials
Park, OH 44073-0002, fig 570, p. 342. Reprinted by permission of ASM
International ®, www.asminternational.org.)

Fatigue Failure Resulting from Variable Loading 263

Medium-carbon steel
(ASTM A186)

30 dia
Web

Fracture Flange

Fracture Tread (1 of 2)

(a) Coke-oven-car wheel

Figure 6–7

Fatigue failure of an ASTM A186 steel double-flange trailer wheel caused by stamp marks. (a) Coke-oven car wheel showing position of
stamp marks and fractures in the rib and web. (b) Stamp mark showing heavy impression and fracture extending along the base of the lower
row of numbers. (c) Notches, indicated by arrows, created from the heavily indented stamp marks from which cracks initiated along the top
at the fracture surface. (From ASM Handbook, Vol. 11: Failure Analysis and Prevention, ASM International, Materials Park, OH 44073-
0002, fig 51, p. 130. Reprinted by permission of ASM International ®, www.asminternational.org.)

Figure 6–8 Aluminum alloy 7075-T73
4.94 Rockwell B 85.5
Aluminum alloy 7075-T73
landing-gear torque-arm 25.5
assembly redesign to eliminate
fatigue fracture at a lubrication 10.200
hole. (a) Arm configuration,
original and improved design Fracture Lug
(dimensions given in inches). (1 of 2)
(b) Fracture surface where
arrows indicate multiple crack A
origins. (From ASM
Handbook, Vol. 11: Failure Lubrication hole Primary-fracture
Analysis and Prevention, ASM surface
International, Materials Park,
OH 44073-0002, fig 23, 1.750-in.-dia Lubrication hole
p. 114. Reprinted bushing,
by permission of ASM 0.090-in. wall
International ®,
www.asminternational.org.) 1 in

3.62 dia Secondary
fracture

Original design Improved design

Detail A
(a)

264 Mechanical Engineering Design

6–2 Approach to Fatigue Failure in Analysis and Design

As noted in the previous section, there are a great many factors to be considered, even
for very simple load cases. The methods of fatigue failure analysis represent a combi-
nation of engineering and science. Often science fails to provide the complete answers
that are needed. But the airplane must still be made to fly—safely. And the automobile
must be manufactured with a reliability that will ensure a long and troublefree life and
at the same time produce profits for the stockholders of the industry. Thus, while sci-
ence has not yet completely explained the complete mechanism of fatigue, the engineer
must still design things that will not fail. In a sense this is a classic example of the true
meaning of engineering as contrasted with science. Engineers use science to solve their
problems if the science is available. But available or not, the problem must be solved,
and whatever form the solution takes under these conditions is called engineering.

In this chapter, we will take a structured approach in the design against fatigue
failure. As with static failure, we will attempt to relate to test results performed on sim-
ply loaded specimens. However, because of the complex nature of fatigue, there is
much more to account for. From this point, we will proceed methodically, and in stages.
In an attempt to provide some insight as to what follows in this chapter, a brief descrip-
tion of the remaining sections will be given here.

Fatigue-Life Methods (Secs. 6–3 to 6–6)

Three major approaches used in design and analysis to predict when, if ever, a cyclically
loaded machine component will fail in fatigue over a period of time are presented. The
premises of each approach are quite different but each adds to our understanding of the
mechanisms associated with fatigue. The application, advantages, and disadvantages of
each method are indicated. Beyond Sec. 6–6, only one of the methods, the stress-life
method, will be pursued for further design applications.

Fatigue Strength and the Endurance Limit (Secs. 6–7 and 6–8)

The strength-life (S-N) diagram provides the fatigue strength Sf versus cycle life N of a
material. The results are generated from tests using a simple loading of standard laboratory-

controlled specimens. The loading often is that of sinusoidally reversing pure bending.

The laboratory-controlled specimens are polished without geometric stress concentra-

tion at the region of minimum area.
For steel and iron, the S-N diagram becomes horizontal at some point. The strength

at this point is called the endurance limit Se and occurs somewhere between 106 and 107
cycles. The prime mark on Se refers to the endurance limit of the controlled laboratory
specimen. For nonferrous materials that do not exhibit an endurance limit, a fatigue
strength at a specific number of cycles, Sf , may be given, where again, the prime denotes
the fatigue strength of the laboratory-controlled specimen.

The strength data are based on many controlled conditions that will not be the same

as that for an actual machine part. What follows are practices used to account for the

differences between the loading and physical conditions of the specimen and the actual

machine part.

Endurance Limit Modifying Factors (Sec. 6–9)

Modifying factors are defined and used to account for differences between the speci-
men and the actual machine part with regard to surface conditions, size, loading, tem-
perature, reliability, and miscellaneous factors. Loading is still considered to be simple
and reversing.

Fatigue Failure Resulting from Variable Loading 265

Stress Concentration and Notch Sensitivity (Sec. 6–10)
The actual part may have a geometric stress concentration by which the fatigue behav-
ior depends on the static stress concentration factor and the component material’s sensi-
tivity to fatigue damage.

Fluctuating Stresses (Secs. 6–11 to 6–13)
These sections account for simple stress states from fluctuating load conditions that are
not purely sinusoidally reversing axial, bending, or torsional stresses.

Combinations of Loading Modes (Sec. 6–14)
Here a procedure based on the distortion-energy theory is presented for analyzing com-
bined fluctuating stress states, such as combined bending and torsion. Here it is
assumed that the levels of the fluctuating stresses are in phase and not time varying.

Varying, Fluctuating Stresses; Cumulative
Fatigue Damage (Sec. 6–15)
The fluctuating stress levels on a machine part may be time varying. Methods are pro-
vided to assess the fatigue damage on a cumulative basis.

Remaining Sections
The remaining three sections of the chapter pertain to the special topics of surface
fatigue strength, stochastic analysis, and roadmaps with important equations.

6–3 Fatigue-Life Methods

The three major fatigue life methods used in design and analysis are the stress-life
method, the strain-life method, and the linear-elastic fracture mechanics method. These
methods attempt to predict the life in number of cycles to failure, N, for a specific level
of loading. Life of 1 ≤ N ≤ 103 cycles is generally classified as low-cycle fatigue,
whereas high-cycle fatigue is considered to be N > 103 cycles. The stress-life method,
based on stress levels only, is the least accurate approach, especially for low-cycle
applications. However, it is the most traditional method, since it is the easiest to imple-
ment for a wide range of design applications, has ample supporting data, and represents
high-cycle applications adequately.

The strain-life method involves more detailed analysis of the plastic deformation at
localized regions where the stresses and strains are considered for life estimates. This
method is especially good for low-cycle fatigue applications. In applying this method,
several idealizations must be compounded, and so some uncertainties will exist in the
results. For this reason, it will be discussed only because of its value in adding to the
understanding of the nature of fatigue.

The fracture mechanics method assumes a crack is already present and detected. It
is then employed to predict crack growth with respect to stress intensity. It is most prac-
tical when applied to large structures in conjunction with computer codes and a peri-
odic inspection program.

6–4 The Stress-Life Method

To determine the strength of materials under the action of fatigue loads, specimens are
subjected to repeated or varying forces of specified magnitudes while the cycles or
stress reversals are counted to destruction. The most widely used fatigue-testing device

266 Mechanical Engineering Design

is the R. R. Moore high-speed rotating-beam machine. This machine subjects the specimen
to pure bending (no transverse shear) by means of weights. The specimen, shown in
Fig. 6–9, is very carefully machined and polished, with a final polishing in an axial
direction to avoid circumferential scratches. Other fatigue-testing machines are avail-
able for applying fluctuating or reversed axial stresses, torsional stresses, or combined
stresses to the test specimens.

To establish the fatigue strength of a material, quite a number of tests are necessary
because of the statistical nature of fatigue. For the rotating-beam test, a constant bend-
ing load is applied, and the number of revolutions (stress reversals) of the beam required
for failure is recorded. The first test is made at a stress that is somewhat under the ulti-
mate strength of the material. The second test is made at a stress that is less than that
used in the first. This process is continued, and the results are plotted as an S-N diagram
(Fig. 6–10). This chart may be plotted on semilog paper or on log-log paper. In the case
of ferrous metals and alloys, the graph becomes horizontal after the material has been
stressed for a certain number of cycles. Plotting on log paper emphasizes the bend in
the curve, which might not be apparent if the results were plotted by using Cartesian
coordinates.

3 7 in
16

0.30 in

9 7 in R.
8

Figure 6–9

Test-specimen geometry for the R. R. Moore rotating-
beam machine. The bending moment is uniform over the
curved at the highest-stressed portion, a valid test of
material, whereas a fracture elsewhere (not at the highest-
stress level) is grounds for suspicion of material flaw.

Figure 6–10 Low cycle High cycle

An S-N diagram plotted from Finite life Infinite
the results of completely life
reversed axial fatigue tests. Sut
Material: UNS G41300 100 Se
steel, normalized;
Sut = 116 kpsi; maximum Fatigue strength Sf , kpsi
Sut = 125 kpsi. (Data from
NACA Tech. Note 3866,
December 1966.)

50

100 101 102 103 104 105 106 107 108
Number of stress cycles, N

Fatigue Failure Resulting from Variable Loading 267

Figure 6–11 Peak alternating bending stress S, kpsi (log) 80 Wrought
70
S-N bands for representative 60 Permanent mold cast
aluminum alloys, excluding 50 Sand cast
wrought alloys with
Sut < 38 kpsi. (From R. C. 40
Juvinall, Engineering 35
Considerations of Stress, 30
Strain and Strength. Copyright 25
© 1967 by The McGraw-Hill
Companies, Inc. Reprinted by 20
permission.) 18
16
14
12

10

8
7
6

5
103 104 105 106 107 108 109

Life N, cycles (log)

The ordinate of the S-N diagram is called the fatigue strength Sf ; a statement of

this strength value must always be accompanied by a statement of the number of cycles

N to which it corresponds.

Soon we shall learn that S-N diagrams can be determined either for a test specimen

or for an actual mechanical element. Even when the material of the test specimen and

that of the mechanical element are identical, there will be significant differences

between the diagrams for the two.

In the case of the steels, a knee occurs in the graph, and beyond this knee failure

will not occur, no matter how great the number of cycles. The strength corresponding

to the knee is called the endurance limit Se, or the fatigue limit. The graph of Fig. 6–10

never does become horizontal for nonferrous metals and alloys, and hence these mate-

rials do not have an endurance limit. Figure 6–11 shows scatter bands indicating the S-N

curves for most common aluminum alloys excluding wrought alloys having a tensile

strength below 38 kpsi. Since aluminum does not have an endurance limit, normally the

fatigue strength Sf is reported at a specific number of cycles, normally N = 5(108)

cycles of reversed stress (see Table A–24).

We note that a stress cycle (N = 1) constitutes a single application and removal of

a load and then another application and removal of the load in the opposite direction.

Thus N = 1 means the load is applied once and then removed, which is the case with
2
the simple tension test.

The body of knowledge available on fatigue failure from N = 1 to N = 1000

cycles is generally classified as low-cycle fatigue, as indicated in Fig. 6–10. High-cycle

fatigue, then, is concerned with failure corresponding to stress cycles greater than 103

cycles.

We also distinguish a finite-life region and an infinite-life region in Fig. 6–10. The

boundary between these regions cannot be clearly defined except for a specific material;

but it lies somewhere between 106 and 107 cycles for steels, as shown in Fig. 6–10.

As noted previously, it is always good engineering practice to conduct a testing

program on the materials to be employed in design and manufacture. This, in fact, is a

requirement, not an option, in guarding against the possibility of a fatigue failure.

268 Mechanical Engineering Design

Because of this necessity for testing, it would really be unnecessary for us to proceed
any further in the study of fatigue failure except for one important reason: the desire to
know why fatigue failures occur so that the most effective method or methods can be
used to improve fatigue strength. Thus our primary purpose in studying fatigue is to
understand why failures occur so that we can guard against them in an optimum man-
ner. For this reason, the analytical design approaches presented in this book, or in any
other book, for that matter, do not yield absolutely precise results. The results should be
taken as a guide, as something that indicates what is important and what is not impor-
tant in designing against fatigue failure.

As stated earlier, the stress-life method is the least accurate approach especially
for low-cycle applications. However, it is the most traditional method, with much
published data available. It is the easiest to implement for a wide range of design
applications and represents high-cycle applications adequately. For these reasons the
stress-life method will be emphasized in subsequent sections of this chapter.
However, care should be exercised when applying the method for low-cycle applications,
as the method does not account for the true stress-strain behavior when localized
yielding occurs.

6–5 The Strain-Life Method

The best approach yet advanced to explain the nature of fatigue failure is called by some
the strain-life method. The approach can be used to estimate fatigue strengths, but when
it is so used it is necessary to compound several idealizations, and so some uncertain-
ties will exist in the results. For this reason, the method is presented here only because
of its value in explaining the nature of fatigue.

A fatigue failure almost always begins at a local discontinuity such as a notch,
crack, or other area of stress concentration. When the stress at the discontinuity exceeds
the elastic limit, plastic strain occurs. If a fatigue fracture is to occur, there must exist
cyclic plastic strains. Thus we shall need to investigate the behavior of materials sub-
ject to cyclic deformation.

In 1910, Bairstow verified by experiment Bauschinger’s theory that the elastic lim-
its of iron and steel can be changed, either up or down, by the cyclic variations of stress.2
In general, the elastic limits of annealed steels are likely to increase when subjected to
cycles of stress reversals, while cold-drawn steels exhibit a decreasing elastic limit.

R. W. Landgraf has investigated the low-cycle fatigue behavior of a large number
of very high-strength steels, and during his research he made many cyclic stress-strain
plots.3 Figure 6–12 has been constructed to show the general appearance of these plots
for the first few cycles of controlled cyclic strain. In this case the strength decreases
with stress repetitions, as evidenced by the fact that the reversals occur at ever-smaller
stress levels. As previously noted, other materials may be strengthened, instead, by
cyclic stress reversals.

The SAE Fatigue Design and Evaluation Steering Committee released a report in
1975 in which the life in reversals to failure is related to the strain amplitude ε/2.4

2L. Bairstow, “The Elastic Limits of Iron and Steel under Cyclic Variations of Stress,” Philosophical
Transactions, Series A, vol. 210, Royal Society of London, 1910, pp. 35–55.
3R. W. Landgraf, Cyclic Deformation and Fatigue Behavior of Hardened Steels, Report no. 320, Department
of Theoretical and Applied Mechanics, University of Illinois, Urbana, 1968, pp. 84–90.
4Technical Report on Fatigue Properties, SAE J1099, 1975.

Figure 6–12 Δ␴ Fatigue Failure Resulting from Variable Loading 269
1st reversal
True stress–true strain hysteresis 4th ␴ A 3d
loops showing the first five 2d
stress reversals of a cyclic- 5th
softening material. The graph
is slightly exaggerated for ␧
clarity. Note that the slope of
the line AB is the modulus of B
elasticity E. The stress range is Δ␧p Δ␧e

σ , εp is the plastic-strain Δ␧
range, and εe is the
elastic strain range. The
total-strain range is

ε = εp + εe.

Figure 6–13 100 ␧'F

A log-log plot showing how Strain amplitude, Δ␧/2 10–1
the fatigue life is related to c
the true-strain amplitude for
hot-rolled SAE 1020 steel. 1.0
(Reprinted with permission
from SAE J1099_200208 10–2 ␴ F'
© 2002 SAE International.) E

Plastic strain Total strain
b

1.0

10–3
Elastic strain

10– 4
100 101 102 103 104 105 106
Reversals to failure, 2N

The report contains a plot of this relationship for SAE 1020 hot-rolled steel; the graph
has been reproduced as Fig. 6–13. To explain the graph, we first define the following
terms:

• Fatigue ductility coefficient εF is the true strain corresponding to fracture in one re-
versal (point A in Fig. 6–12). The plastic-strain line begins at this point in Fig. 6–13.

• Fatigue strength coefficient σF is the true stress corresponding to fracture in one
reversal (point A in Fig. 6–12). Note in Fig. 6–13 that the elastic-strain line begins at
σF/E.

• Fatigue ductility exponent c is the slope of the plastic-strain line in Fig. 6–13 and is
the power to which the life 2N must be raised to be proportional to the true plastic-
strain amplitude. If the number of stress reversals is 2N, then N is the number of
cycles.

270 Mechanical Engineering Design

• Fatigue strength exponent b is the slope of the elastic-strain line, and is the power to
which the life 2N must be raised to be proportional to the true-stress amplitude.

Now, from Fig. 6–12, we see that the total strain is the sum of the elastic and plastic
components. Therefore the total strain amplitude is half the total strain range

ε = εe + εp (a)
22 2

The equation of the plastic-strain line in Fig. 6–13 is

εp = εF (2N )c (6–1)
2

The equation of the elastic strain line is

εe = σF (2N )b (6–2)
2E

Therefore, from Eq. (a), we have for the total-strain amplitude

ε = σF (2N )b + εF (2N )c (6–3)
2 E

which is the Manson-Coffin relationship between fatigue life and total strain.5 Some
values of the coefficients and exponents are listed in Table A–23. Many more are
included in the SAE J1099 report.6

Though Eq. (6–3) is a perfectly legitimate equation for obtaining the fatigue life of
a part when the strain and other cyclic characteristics are given, it appears to be of lit-
tle use to the designer. The question of how to determine the total strain at the bottom
of a notch or discontinuity has not been answered. There are no tables or charts of strain
concentration factors in the literature. It is possible that strain concentration factors will
become available in research literature very soon because of the increase in the use of
finite-element analysis. Moreover, finite element analysis can of itself approximate the
strains that will occur at all points in the subject structure.7

6–6 The Linear-Elastic Fracture Mechanics Method

The first phase of fatigue cracking is designated as stage I fatigue. Crystal slip that
extends through several contiguous grains, inclusions, and surface imperfections is pre-
sumed to play a role. Since most of this is invisible to the observer, we just say that stage
I involves several grains. The second phase, that of crack extension, is called stage II
fatigue. The advance of the crack (that is, new crack area is created) does produce evi-
dence that can be observed on micrographs from an electron microscope. The growth of

5J. F. Tavernelli and L. F. Coffin, Jr., “Experimental Support for Generalized Equation Predicting Low Cycle
Fatigue,’’ and S. S. Manson, discussion, Trans. ASME, J. Basic Eng., vol. 84, no. 4, pp. 533–537.
6See also, Landgraf, Ibid.
7For further discussion of the strain-life method see N. E. Dowling, Mechanical Behavior of Materials,
2nd ed., Prentice-Hall, Englewood Cliffs, N.J., 1999, Chap. 14.

Fatigue Failure Resulting from Variable Loading 271

the crack is orderly. Final fracture occurs during stage III fatigue, although fatigue is not
involved. When the crack is sufficiently long that KI = KIc for the stress amplitude
involved, then KIc is the critical stress intensity for the undamaged metal, and there is
sudden, catastrophic failure of the remaining cross section in tensile overload (see
Sec. 5–12). Stage III fatigue is associated with rapid acceleration of crack growth then
fracture.

Crack Growth

Fatigue cracks nucleate and grow when stresses vary and there is some tension in

each stress cycle. Consider the stress to be fluctuating between the limits of σmin and

σmax, where the stress range is defi√ned as σ = σmax − σmin. From Eq. (5–37) the
stress intensity is given by KI = βσ πa. Thus, for σ, the stress intensity range per

cycle is

√√ (6–4)
KI = β(σmax − σmin) π a = β σ π a

To develop fatigue strength data, a number of specimens of the same material are tested
at various levels of σ. Cracks nucleate at or very near a free surface or large discon-
tinuity. Assuming an initial crack length of ai , crack growth as a function of the num-
ber of stress cycles N will depend on σ, that is, KI. For KI below some threshold
value ( KI)th a crack will not grow. Figure 6–14 represents the crack length a as a
function of N for three stress levels ( σ )3 > ( σ )2 > ( σ )1, where ( KI)3 >
( KI)2 > ( KI)1. Notice the effect of the higher stress range in Fig. 6–14 in the pro-
duction of longer cracks at a particular cycle count.

When the rate of crack growth per cycle, da/d N in Fig. 6–14, is plotted as shown

in Fig. 6–15, the data from all three stress range levels superpose to give a sigmoidal

curve. The three stages of crack development are observable, and the stage II data are

linear on log-log coordinates, within the domain of linear elastic fracture mechanics

(LEFM) validity. A group of similar curves can be generated by changing the stress
ratio R = σmin/σmax of the experiment.

Here we present a simplified procedure for estimating the remaining life of a cycli-

cally stressed part after discovery of a crack. This requires the assumption that plane strain

Figure 6–14 (ΔKI)3 (ΔKI)2 (ΔKI)1
da
The increase in crack length a Crack length a a
from an initial length of ai as dN
a function of cycle count for
three stress ranges, ( σ ) 3 >
( σ)2 > ( σ)1.

ai

Log N

Stress cycles N

272 Mechanical Engineering Design

Figure 6–15 Log da
dN
When da/dN is measured in
Fig. 6–14 and plotted on Region I Region II Region III
loglog coordinates, the data
for different stress ranges Crack Crack Crack
superpose, giving rise to a initiation propagation unstable
sigmoid curve as shown.
( K I) th is the threshold value Increasing
of K I, below which a crack stress ratio
does not grow. From threshold
to rupture an aluminum alloy R
will spend 85--90 percent of
life in region I, 5--8 percent in
region II, and 1--2 percent in
region III.

(ΔK )th Kc
Log ΔK

Table 6–1 C, m/cycle C, in/cycle
MPa√m m √m
Conservative Values of Material kpsi in m
Factor C and Exponent
m in Eq. (6–5) for Ferritic-pearlitic steels 6.89(10−12 ) 3.60(10−10 ) 3.00
Various Forms of Steel Martensitic steels 1.36(10−10 ) 6.60(10−9 ) 2.25
(R =. 0) Austenitic stainless steels 5.61(10−12 ) 3.00(10−10 ) 3.25

From J.M. Barsom and S.T. Rolfe, Fatigue and Fracture Control in Structures, 2nd ed., Prentice Hall, Upper Saddle River, NJ, 1987,
pp. 288–291, Copyright ASTM International. Reprinted with permission.

conditions prevail.8 Assuming a crack is discovered early in stage II, the crack growth in
region II of Fig. 6–15 can be approximated by the Paris equation, which is of the form

da = C( KI)m (6–5)
dN

where C and m are empirical material constants and KI is given by Eq. (6–4).
Representative, but conservative, values of C and m for various classes of steels are
listed in Table 6–1. Substituting Eq. (6–4) and integrating gives

Nf 1 af da

d N = N f = C ai (β σ √ a )m (6–6)
π
0

Here ai is the initial crack length, af is the final crack length corresponding to failure,
and Nf is the estimated number of cycles to produce a failure after the initial crack is
formed. Note that β may vary in the integration variable (e.g., see Figs. 5–25 to 5–30).

8Recommended references are: Dowling, op. cit.; J. A. Collins, Failure of Materials in Mechanical Design,
John Wiley & Sons, New York, 1981; H. O. Fuchs and R. I. Stephens, Metal Fatigue in Engineering, John
Wiley & Sons, New York, 1980; and Harold S. Reemsnyder, “Constant Amplitude Fatigue Life Assessment
Models,” SAE Trans. 820688, vol. 91, Nov. 1983.

Fatigue Failure Resulting from Variable Loading 273

If this should happen, then Reemsnyder9 suggests the use of numerical integration
employing the algorithm

δaj = C( K I )mj (δ N )j

aj+1 = aj + δaj

Nj+1 = Nj + δ Nj (6–7)

Nf = δNj

Here δaj and δ Nj are increments of the crack length and the number of cycles. The pro-
cedure is to select a value of δ Nj , using ai determine β and compute KI, determine
δaj , and then find the next value of a. Repeat the procedure until a = af .

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.

9Op. cit.

EXAMPLE 6–1 The bar shown in Fig. 6–16 is subjected to a repeated moment 0 ≤ M ≤ 1200 lbf√· in.
Solution The bar is AISI 4430 steel with Sut = 185 kpsi, Sy = 170 kpsi, and KIc = 73 kpsi in.

Material tests on various specimens of this material with id√entical heat treatment
indicate worst-case constants of C = 3.8(10−11 )(in/cycle)ր(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.

The stress range σ is always computed by using the nominal (uncracked) area. Thus

I = bh2 = 0.25(0.5)2 = 0.010 42 in3
c6 6

Therefore, before the crack initiates, the stress range is

σ= M = 1200 = 115.2(103) 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 Sy/Sut it is highly unlikely that the bar will reach complete yield. For brittle fracture,

designate the crack length as af . If β = 1, then from Eq. (5–37) with KI = KIc, we

approximate af as

af = 1 KIc 2 =. 1 73 2
π
β σmax π 115.2 = 0.1278 in

Figure 6–16 1 in
4

M M 1 in
2

Nick

274 Mechanical Engineering Design

From Fig. 5–27, we compute the ratio af / h as
af = 0.1278 = 0.256
h 0.5

Thus af / h varies from near zero to approximately 0.256. From Fig. 5–27, for this range
β is nearly constant at approximately 1.07. We will assume it to be so, and re-evaluate

af as

af = 1 73 2
π 1.07(115.2)
= 0.112 in

Thus, from Eq. (6–6), the estimated remaining life is

1 af da 1 0.112 da
Nf = C ai (β σ √π a)m = 3.8(10−11) 0.004 [1.07(115.2)√π a]3

5.047(103 ) 0.112
√
= − a 0.004 = 64.7 (103) cycles

6–7 The Endurance Limit

The determination of endurance limits by fatigue testing is now routine, though a lengthy

procedure. Generally, stress testing is preferred to strain testing for endurance limits.

For preliminary and prototype design and for some failure analysis as well, a quick

method of estimating endurance limits is needed. There are great quantities of data in

the literature on the results of rotating-beam tests and simple tension tests of specimens

taken from the same bar or ingot. By plotting these as in Fig. 6–17, it is possible to see

whether there is any correlation between the two sets of results. The graph appears to

suggest that the endurance limit ranges from about 40 to 60 percent of the tensile

strength for steels up to about 210 kpsi (1450 MPa). Beginning at about Sut = 210 kpsi

(1450 MPa), the scatter appears to increase, but the trend seems to level off, as sug-

gested by the dashed horizontal line at Se = 105 kpsi.
We wish now to present a method for estimating endurance limits. Note that esti-

mates obtained from quantities of data obtained from many sources probably have a

large spread and might deviate significantly from the results of actual laboratory tests of

the mechanical properties of specimens obtained through strict purchase-order specifi-

cations. Since the area of uncertainty is greater, compensation must be made by employ-

ing larger design factors than would be used for static design.

For steels, simplifying our observation of Fig. 6–17, we will estimate the endurance

limit as

⎧ 0.5Sut Sut ≤ 200 kpsi (1400 MPa)
⎨

Se = ⎩ 100 kpsi Sut > 200 kpsi (6–8)
700 MPa Sut > 1400 MPa

where Sut is the minimum tensile strength. The prime mark on Se in this equation refers
to the rotating-beam specimen itself. We wish to reserve the unprimed symbol Se for the
endurance limit of any particular machine element subjected to any kind of loading.

Soon we shall learn that the two strengths may be quite different.

Fatigue Failure Resulting from Variable Loading 275

140 S 'e = 0.6 0.5
Su 0.4
120 Carbon steels
Endurance limit S 'e , kpsi Alloy steels 105 kpsi
Wrought irons

100

80

60

40

20

0
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300
Tensile strength Sut, kpsi

Figure 6–17

Graph of endurance limits versus tensile strengths from actual test results for a large number of wrought
irons and steels. Ratios of Se/Sut of 0.60, 0.50, and 0.40 are shown by the solid and dashed lines.
Note also the horizontal dashed line for Se = 105 kpsi. Points shown having a tensile strength greater
than 210 kpsi have a mean endurance limit of Se = 105 kpsi and a standard deviation of 13.5 kpsi.
(Collated from data compiled by H. J. Grover, S. A. Gordon, and L. R. Jackson in Fatigue of Metals
and Structures, Bureau of Naval Weapons Document NAVWEPS 00-25-534, 1960; and from Fatigue
Design Handbook, SAE, 1968, p. 42.)

Steels treated to give different microstructures have different Se/Sut ratios. It
appears that the more ductile microstructures have a higher ratio. Martensite has a very
brittle nature and is highly susceptible to fatigue-induced cracking; thus the ratio is low.
When designs include detailed heat-treating specifications to obtain specific
microstructures, it is possible to use an estimate of the endurance limit based on test
data for the particular microstructure; such estimates are much more reliable and indeed
should be used.

The endurance limits for various classes of cast irons, polished or machined, are
given in Table A–24. Aluminum alloys do not have an endurance limit. The fatigue
strengths of some aluminum alloys at 5(108) cycles of reversed stress are given in
Table A–24.

6–8 Fatigue Strength

As shown in Fig. 6–10, a region of low-cycle fatigue extends from N = 1 to about
103 cycles. In this region the fatigue strength Sf is only slightly smaller than the ten-
sile strength Sut . An analytical approach has been given by Mischke10 for both

10J. E. Shigley, C. R. Mischke, and T. H. Brown, Jr., Standard Handbook of Machine Design, 3rd ed.,
McGraw-Hill, New York, 2004, pp. 29.25–29.27.

276 Mechanical Engineering Design

high-cycle and low-cycle regions, requiring the parameters of the Manson-Coffin

equation plus the strain-strengthening exponent m. Engineers often have to work

with less information.
Figure 6–10 indicates that the high-cycle fatigue domain extends from 103 cycles

for steels to the endurance limit life Ne, which is about 106 to 107 cycles. The purpose
of this section is to develop methods of approximation of the S-N diagram in the high-

cycle region, when information may be as sparse as the results of a simple tension test.

Experience has shown high-cycle fatigue data are rectified by a logarithmic transform

to both stress and cycles-to-failure. Equation (6–2) can be used to determine the fatigue
strength at 103 cycles. Defining the specimen fatigue strength at a specific number of
cycles as (Sf )N = E εe/2, write Eq. (6–2) as

(Sf )N = σF (2N )b (6–9)

At 103 cycles,

(Sf )103 = σF (2.103)b = f Sut

where f is the fraction of Sut represented by (S f )103 cycles. Solving for f gives

f = σF (2 · 103)b (6–10)
Sut

Now, from Eq. (2–11), σF = σ0εm , with ε = εF . If this true-stress–true-strain equation
is not known, the SAE approximation11 for steels with HB ≤ 500 may be used:

σF = Sut + 50 kpsi or σF = Sut + 345 MPa (6–11)

To find b, substitute the endurance strength and corresponding cycles, Se and Ne,
respectively into Eq. (6–9) and solving for b

log σF /Se (6–12)
b=−
log (2N e)

Thus, the equation Sf = σF (2N )b is known. For example, if Sut = 105 kpsi and
Se = 52.5 kpsi at failure,

Eq. (6–11) σF = 105 + 50 = 155 kpsi

Eq. (6–12) b = − log(155/52.5) = −0.0746
log 2 · 106

Eq. (6–10) f = 155 2 · 103 −0.0746 = 0.837
105

and for Eq. (6–9), with Sf = (Sf )N ,

Sf = 155(2N )−0.0746 = 147 N −0.0746 (a)

11Fatigue Design Handbook, vol. 4, Society of Automotive Engineers, New York, 1958, p. 27.

Figure 6–18 Fatigue Failure Resulting from Variable Loading 277

Fatigue strength fraction, f, of f 0.9
Sut at 103 cycles for 0.88
Se = Se = 0.5Sut . 0.86
0.84
0.82
0.8
0.78
0.76
70 80 90 100 110 120 130 140 150 160 170 180 190 200
Sut, kpsi

The process given for finding f can be repeated for various ultimate strengths.

Figure 6–18 is a plot of f for 70 ≤ Sut ≤ 200 kpsi. To be conservative, for Sut < 70 kpsi,
let f ϭ 0.9.

For an actual mechanical component, Se is reduced to Se (see Sec. 6–9) which is
less than 0.5 Sut . However, unless actual data is available, we recommend using the
value of f found from Fig. 6–18. Equation (a), for the actual mechanical component, can

be written in the form

Sf = a Nb (6–13)

where N is cycles to failure and the constants a and b are defined by the points
103, Sf 103 and 106, Se with Sf 103 = f Sut . Substituting these two points in Eq.
(6–13) gives

a = ( f Sut )2 (6–14)
Se

b = − 1 log f Sut (6–15)
3 Se

If a completely reversed stress σa is given, setting Sf = σa in Eq. (6–13), the number
of cycles-to-failure can be expressed as

N = σa 1/b (6–16)
a

Low-cycle fatigue is often defined (see Fig. 6–10) as failure that occurs in a range
of 1 ≤ N ≤ 103 cycles. On a loglog plot such as Fig. 6–10 the failure locus in this range
is nearly linear below 103 cycles. A straight line between 103, f Sut and 1, Sut (trans-

formed) is conservative, and it is given by

Sf ≥ Sut N (log f )/3 1 ≤ N ≤ 103 (6–17)

278 Mechanical Engineering Design

EXAMPLE 6–2 Given a 1050 HR steel, estimate
(a) the rotating-beam endurance limit at 106 cycles.
Solution (b) the endurance strength of a polished rotating-beam specimen corresponding to 104
Answer
cycles to failure
(c) the expected life of a polished rotating-beam specimen under a completely reversed

stress of 55 kpsi.

(a) From Table A–20, Sut = 90 kpsi. From Eq. (6–8),

Se = 0.5(90) = 45 kpsi
(b) From Fig. 6–18, for Sut = 90 kpsi, f =. 0.86. From Eq. (6–14),

a = [0.86(90)2] = 133.1 kpsi
45

From Eq. (6–15),

b = − 1 log 0.86(90) = −0.0785
3 45

Thus, Eq. (6–13) is
Sf = 133.1 N −0.0785

Answer For 104 cycles to failure, Sf = 133.1(104)−0.0785 = 64.6 kpsi
Answer (c) From Eq. (6–16), with σa = 55 kpsi,

N= 55 1/−0.0785
133.1
= 77 500 = 7.75(104)cycles

Keep in mind that these are only estimates. So expressing the answers using three-place
accuracy is a little misleading.

6–9 Endurance Limit Modifying Factors

We have seen that the rotating-beam specimen used in the laboratory to determine
endurance limits is prepared very carefully and tested under closely controlled condi-
tions. It is unrealistic to expect the endurance limit of a mechanical or structural mem-
ber to match the values obtained in the laboratory. Some differences include

• Material: composition, basis of failure, variability

• Manufacturing: method, heat treatment, fretting corrosion, surface condition, stress
concentration

• Environment: corrosion, temperature, stress state, relaxation times

• Design: size, shape, life, stress state, stress concentration, speed, fretting, galling

Fatigue Failure Resulting from Variable Loading 279

Marin12 identified factors that quantified the effects of surface condition, size, loading,
temperature, and miscellaneous items. The question of whether to adjust the endurance
limit by subtractive corrections or multiplicative corrections was resolved by an exten-
sive statistical analysis of a 4340 (electric furnace, aircraft quality) steel, in which a
correlation coefficient of 0.85 was found for the multiplicative form and 0.40 for the
additive form. A Marin equation is therefore written as

Se = kakbkckd kek f Se (6–18)

where ka = surface condition modification factor
kb = size modification factor
kc = load modification factor
kd = temperature modification factor
ke = reliability factor13
kf = miscellaneous-effects modification factor
Se = rotary-beam test specimen endurance limit
Se = endurance limit at the critical location of a machine part in the geom-

etry and condition of use

When endurance tests of parts are not available, estimations are made by applying
Marin factors to the endurance limit.

Surface Factor ka

The surface of a rotating-beam specimen is highly polished, with a final polishing in the
axial direction to smooth out any circumferential scratches. The surface modification
factor depends on the quality of the finish of the actual part surface and on the tensile
strength of the part material. To find quantitative expressions for common finishes of
machine parts (ground, machined, or cold-drawn, hot-rolled, and as-forged), the coordi-
nates of data points were recaptured from a plot of endurance limit versus ultimate
tensile strength of data gathered by Lipson and Noll and reproduced by Horger.14 The
data can be represented by

ka = a Subt (6–19)

where Sut is the minimum tensile strength and a and b are to be found in Table 6–2.

12Joseph Marin, Mechanical Behavior of Engineering Materials, Prentice-Hall, Englewood Cliffs, N.J.,
1962, p. 224.

13Complete stochastic analysis is presented in Sec. 6–17. Until that point the presentation here is one of a
deterministic nature. However, we must take care of the known scatter in the fatigue data. This means that
we will not carry out a true reliability analysis at this time but will attempt to answer the question: What is
the probability that a known (assumed) stress will exceed the strength of a randomly selected component
made from this material population?

14C. J. Noll and C. Lipson, “Allowable Working Stresses,” Society for Experimental Stress Analysis, vol. 3,
no. 2, 1946, p. 29. Reproduced by O. J. Horger (ed.), Metals Engineering Design ASME Handbook,
McGraw-Hill, New York, 1953, p. 102.

280 Mechanical Engineering Design

Table 6–2 Surface Factor a Exponent
Finish b
Parameters for Marin Sut, kpsi Sut, MPa
Surface Modification Ground −0.085
Factor, Eq. (6–19) Machined or cold-drawn 1.34 1.58 −0.265
Hot-rolled 2.70 4.51 −0.718
As-forged 14.4 57.7 −0.995
39.9 272.

From C.J. Noll and C. Lipson, “Allowable Working Stresses,” Society for Experimental Stress Analysis, vol. 3,
no. 2, 1946 p. 29. Reproduced by O.J. Horger (ed.) Metals Engineering Design ASME Handbook, McGraw-Hill,
New York. Copyright © 1953 by The McGraw-Hill Companies, Inc. Reprinted by permission.

EXAMPLE 6–3 A steel has a minimum ultimate strength of 520 MPa and a machined surface.
Solution Estimate ka.

From Table 6–2, a = 4.51 and b = −0.265. Then, from Eq. (6–19)

Answer ka = 4.51(520)−0.265 = 0.860

Again, it is important to note that this is an approximation as the data is typically
quite scattered. Furthermore, this is not a correction to take lightly. For example, if in
the previous example the steel was forged, the correction factor would be 0.540, a sig-
nificant reduction of strength.

Size Factor kb
The size factor has been evaluated using 133 sets of data points.15 The results for bend-
ing and torsion may be expressed as

⎧ (d /0.3)−0.107 = 0.879d −0.107 0.11 ≤ d ≤ 2 in
⎨⎪⎪⎪ 0.91d −0.157 2 < d ≤ 10 in
2.79 ≤ d ≤ 51 mm
kb = ⎪⎪⎩⎪ (d/7.62)−0.107 = 1.24d−0.107 51 < d ≤ 254 mm ( 6–20)

1.51d −0.157

For axial loading there is no size effect, so

kb = 1 (6–21)

but see kc.
One of the problems that arises in using Eq. (6–20) is what to do when a round bar

in bending is not rotating, or when a noncircular cross section is used. For example,
what is the size factor for a bar 6 mm thick and 40 mm wide? The approach to be used

15Charles R. Mischke, “Prediction of Stochastic Endurance Strength,” Trans. of ASME, Journal of Vibration,
Acoustics, Stress, and Reliability in Design, vol. 109, no. 1, January 1987, Table 3.

Fatigue Failure Resulting from Variable Loading 281

here employs an effective dimension de obtained by equating the volume of material

stressed at and above 95 percent of the maximum stress to the same volume in the
rotating-beam specimen.16 It turns out that when these two volumes are equated,

the lengths cancel, and so we need only consider the areas. For a rotating round section,

the 95 percent stress area is the area in a ring having an outside diameter d and an inside

diameter of 0.95d. So, designating the 95 percent stress area A0.95σ , we have

A0.95σ = π [d2 − (0.95d)2] = 0.0766d 2 (6–22)
4

This equation is also valid for a rotating hollow round. For nonrotating solid or hollow
rounds, the 95 percent stress area is twice the area outside of two parallel chords hav-
ing a spacing of 0.95d, where d is the diameter. Using an exact computation, this is

A0.95σ = 0.01046d2 (6–23)

with de in Eq. (6–22), setting Eqs. (6–22) and (6–23) equal to each other enables us to
solve for the effective diameter. This gives

de = 0.370d (6–24)

as the effective size of a round corresponding to a nonrotating solid or hollow round.
A rectangular section of dimensions h × b has A0.95σ = 0.05hb. Using the same

approach as before,

de = 0.808(hb)1/2 (6–25)

Table 6–3 provides A0.95σ areas of common structural shapes undergoing non-
rotating bending.

16See R. Kuguel, “A Relation between Theoretical Stress Concentration Factor and Fatigue Notch Factor
Deduced from the Concept of Highly Stressed Volume,” Proc. ASTM, vol. 61, 1961, pp. 732–748.

EXAMPLE 6–4 A steel shaft loaded in bending is 32 mm in diameter, abutting a filleted shoulder 38 mm
Solution in diameter. The shaft material has a mean ultimate tensile strength of 690 MPa.
Answer Estimate the Marin size factor kb if the shaft is used in
(a) A rotating mode.
Answer (b) A nonrotating mode.

(a) From Eq. (6–20)

d −0.107 32 −0.107
kb = 7.62 = 7.62 = 0.858

(b) From Table 6–3,

de = 0.37d = 0.37(32) = 11.84 mm

From Eq. (6–20),

11.84 −0.107
kb = 7.62 = 0.954

282 Mechanical Engineering Design d A0.95σ = 0.01046d 2
de = 0.370d
Table 6–3
A0.95σ = 0.05hb√
A0.95σ Areas of de = 0.808 hb
Common Nonrotating
Structural Shapes

b
2

h
11

2

a
1

A0.95σ = 0.10at f t f > 0.025a axis 1-1
0.05ba axis 2-2
b2 2

tf

1 x2 0.05ab axis 1-1
A0.95σ = 0.052xa + 0.1t f (b − x) axis 2-2
a
1

2
b tf

1

Loading Factor kc

When fatigue tests are carried out with rotating bending, axial (push-pull), and torsion-
al loading, the endurance limits differ with Sut. This is discussed further in Sec. 6–17.
Here, we will specify average values of the load factor as

1 bending (6–26)
kc = 0.85
axial
0.59 torsion17

Temperature Factor kd

When operating temperatures are below room temperature, brittle fracture is a strong
possibility and should be investigated first. When the operating temperatures are high-
er than room temperature, yielding should be investigated first because the yield
strength drops off so rapidly with temperature; see Fig. 2–9. Any stress will induce
creep in a material operating at high temperatures; so this factor must be considered too.

17Use this only for pure torsional fatigue loading. When torsion is combined with other stresses, such
as bending, kc = 1 and the combined loading is managed by using the effective von Mises stress as in
Sec. 5–5. Note: For pure torsion, the distortion energy predicts that (kc)torsion = 0.577.

Fatigue Failure Resulting from Variable Loading 283

Table 6–4 Temperature, °C ST/SRT Temperature, °F ST/SRT

Effect of Operating 20 1.000 70 1.000
Temperature on the 50 1.010 100 1.008
Tensile Strength of 100 1.020 200 1.020
Steel.* (ST = tensile 150 1.025 300 1.024
strength at operating 200 1.020 400 1.018
temperature; 250 1.000 500 0.995
SRT = tensile strength 300 0.975 600 0.963
at room temperature; 350 0.943 700 0.927
0.099 ≤ σˆ ≤ 0.110) 400 0.900 800 0.872
450 0.843 900 0.797
500 0.768 1000 0.698
550 0.672 1100 0.567
600 0.549

*Data source: Fig. 2–9.

Finally, it may be true that there is no fatigue limit for materials operating at high tem-
peratures. Because of the reduced fatigue resistance, the failure process is, to some
extent, dependent on time.

The limited amount of data available show that the endurance limit for steels
increases slightly as the temperature rises and then begins to fall off in the 400 to 700°F
range, not unlike the behavior of the tensile strength shown in Fig. 2–9. For this reason
it is probably true that the endurance limit is related to tensile strength at elevated tem-
peratures in the same manner as at room temperature.18 It seems quite logical, therefore,
to employ the same relations to predict endurance limit at elevated temperatures as are
used at room temperature, at least until more comprehensive data become available. At
the very least, this practice will provide a useful standard against which the perfor-
mance of various materials can be compared.

Table 6–4 has been obtained from Fig. 2–9 by using only the tensile-strength data.
Note that the table represents 145 tests of 21 different carbon and alloy steels. A fourth-
order polynomial curve fit to the data underlying Fig. 2–9 gives

kd = 0.975 + 0.432(10−3)TF − 0.115(10−5)TF2 ( 6–27)
+ 0.104(10−8)TF3 − 0.595(10−12)TF4

where 70 ≤ TF ≤ 1000◦F.
Two types of problems arise when temperature is a consideration. If the rotating-

beam endurance limit is known at room temperature, then use

kd = ST (6–28)
SRT

18For more, see Table 2 of ANSI/ASME B106. 1M-1985 shaft standard, and E. A. Brandes (ed.), Smithell’s
Metals Reference Book, 6th ed., Butterworth, London, 1983, pp. 22–134 to 22–136, where endurance limits
from 100 to 650°C are tabulated.

284 Mechanical Engineering Design

from Table 6–4 or Eq. (6–27) and proceed as usual. If the rotating-beam endurance limit
is not given, then compute it using Eq. (6–8) and the temperature-corrected tensile
strength obtained by using the factor from Table 6–4. Then use kd = 1.

EXAMPLE 6–5 A 1035 steel has a tensile strength of 70 kpsi and is to be used for a part that sees 450°F
Solution
Answer in service. Estimate the Marin temperature modification factor and (Se)450◦ if
(a) The room-temperature endurance limit by test is (Se)70◦ = 39.0 kpsi.
Answer (b) Only the tensile strength at room temperature is known.

(a) First, from Eq. (6–27),
kd = 0.975 + 0.432(10−3)(450) − 0.115(10−5)(4502)
+ 0.104(10−8)(4503) − 0.595(10−12)(4504) = 1.007

Thus,

(Se)450◦ = kd (Se)70◦ = 1.007(39.0) = 39.3 kpsi
(b) Interpolating from Table 6–4 gives

(ST /SRT )450◦ = 1.018 + (0.995 − 1.018) 450 − 400 = 1.007
500 − 400

Thus, the tensile strength at 450°F is estimated as

(Sut )450◦ = (ST /SRT )450◦ (Sut )70◦ = 1.007(70) = 70.5 kpsi

From Eq. (6–8) then,

(Se)450◦ = 0.5 (Sut )450◦ = 0.5(70.5) = 35.2 kpsi
Part a gives the better estimate due to actual testing of the particular material.

Reliability Factor ke

The discussion presented here accounts for the scatter Soe/f Sduat t=a. such as shown in
Fig. 6–17 where the mean endurance limit is shown to be 0.5, or as given by

Eq. (6–8). Most endurance strength data are reported as mean values. Data presented

by Haugen and Wirching19 show standard deviations of endurance strengths of less than

8 percent. Thus the reliability modification factor to account for this can be written as

ke = 1 − 0.08 za (6–29)

where za is defined by Eq. (20–16) and values for any desired reliability can be deter-
mined from Table A–10. Table 6–5 gives reliability factors for some standard specified
reliabilities.

For a more comprehensive approach to reliability, see Sec. 6–17.

19E. B. Haugen and P. H. Wirsching, “Probabilistic Design,” Machine Design, vol. 47, no. 12, 1975,
pp. 10–14.

Fatigue Failure Resulting from Variable Loading 285

Table 6–5 Reliability, % Transformation Variate za Reliability Factor ke

Reliability Factors ke 50 0 1.000
Corresponding to 90 1.288 0.897
8 Percent Standard 95 1.645 0.868
Deviation of the 99 2.326 0.814
Endurance Limit 99.9 3.091 0.753
99.99 3.719 0.702
99.999 4.265 0.659
99.9999 4.753 0.620

Figure 6–19 Se (case) Case
␴ or ␶ Core
The failure of a case-hardened
part in bending or torsion. In Se (core)
this example, failure occurs in
the core.

Miscellaneous-Effects Factor kf

Though the factor kf is intended to account for the reduction in endurance limit due to
all other effects, it is really intended as a reminder that these must be accounted for,
because actual values of kf are not always available.

Residual stresses may either improve the endurance limit or affect it adversely.
Generally, if the residual stress in the surface of the part is compression, the endurance
limit is improved. Fatigue failures appear to be tensile failures, or at least to be caused
by tensile stress, and so anything that reduces tensile stress will also reduce the possi-
bility of a fatigue failure. Operations such as shot peening, hammering, and cold rolling
build compressive stresses into the surface of the part and improve the endurance limit
significantly. Of course, the material must not be worked to exhaustion.

The endurance limits of parts that are made from rolled or drawn sheets or bars,
as well as parts that are forged, may be affected by the so-called directional character-
istics of the operation. Rolled or drawn parts, for example, have an endurance limit
in the transverse direction that may be 10 to 20 percent less than the endurance limit in
the longitudinal direction.

Parts that are case-hardened may fail at the surface or at the maximum core radius,
depending upon the stress gradient. Figure 6–19 shows the typical triangular stress dis-
tribution of a bar under bending or torsion. Also plotted as a heavy line in this figure are
the endurance limits Se for the case and core. For this example the endurance limit of the
core rules the design because the figure shows that the stress σ or τ, whichever applies,
at the outer core radius, is appreciably larger than the core endurance limit.

286 Mechanical Engineering Design

Of course, if stress concentration is also present, the stress gradient is much
steeper, and hence failure in the core is unlikely.

Corrosion
It is to be expected that parts that operate in a corrosive atmosphere will have a lowered
fatigue resistance. This is, of course, true, and it is due to the roughening or pitting of
the surface by the corrosive material. But the problem is not so simple as the one of
finding the endurance limit of a specimen that has been corroded. The reason for this is
that the corrosion and the stressing occur at the same time. Basically, this means that in
time any part will fail when subjected to repeated stressing in a corrosive atmosphere.
There is no fatigue limit. Thus the designer’s problem is to attempt to minimize the fac-
tors that affect the fatigue life; these are:

• Mean or static stress
• Alternating stress
• Electrolyte concentration
• Dissolved oxygen in electrolyte
• Material properties and composition
• Temperature
• Cyclic frequency
• Fluid flow rate around specimen
• Local crevices

Electrolytic Plating
Metallic coatings, such as chromium plating, nickel plating, or cadmium plating, reduce
the endurance limit by as much as 50 percent. In some cases the reduction by coatings
has been so severe that it has been necessary to eliminate the plating process. Zinc
plating does not affect the fatigue strength. Anodic oxidation of light alloys reduces
bending endurance limits by as much as 39 percent but has no effect on the torsional
endurance limit.

Metal Spraying
Metal spraying results in surface imperfections that can initiate cracks. Limited tests
show reductions of 14 percent in the fatigue strength.

Cyclic Frequency
If, for any reason, the fatigue process becomes time-dependent, then it also becomes
frequency-dependent. Under normal conditions, fatigue failure is independent of fre-
quency. But when corrosion or high temperatures, or both, are encountered, the cyclic
rate becomes important. The slower the frequency and the higher the temperature, the
higher the crack propagation rate and the shorter the life at a given stress level.

Frettage Corrosion
The phenomenon of frettage corrosion is the result of microscopic motions of tightly
fitting parts or structures. Bolted joints, bearing-race fits, wheel hubs, and any set of
tightly fitted parts are examples. The process involves surface discoloration, pitting, and
eventual fatigue. The frettage factor kf depends upon the material of the mating pairs
and ranges from 0.24 to 0.90.

Fatigue Failure Resulting from Variable Loading 287

6–10 Stress Concentration and Notch Sensitivity

In Sec. 3–13 it was pointed out that the existence of irregularities or discontinuities,
such as holes, grooves, or notches, in a part increases the theoretical stresses signifi-
cantly in the immediate vicinity of the discontinuity. Equation (3–48) defined a stress
concentration factor Kt (or Kts ), which is used with the nominal stress to obtain the
maximum resulting stress due to the irregularity or defect. It turns out that some mate-
rials are not fully sensitive to the presence of notches and hence, for these, a reduced
value of Kt can be used. For these materials, the maximum stress is, in fact,

σmax = K f σ0 or τmax = K f s τ0 (6–30)

where K f is a reduced value of Kt and σ0 is the nominal stress. The factor K f is com-
monly called a fatigue stress-concentration factor, and hence the subscript f. So it is

convenient to think of Kf as a stress-concentration factor reduced from Kt because of
lessened sensitivity to notches. The resulting factor is defined by the equation

maximum stress in notched specimen (a)
K f = stress in notch-free specimen

Notch sensitivity q is defined by the equation

q = Kf −1 or qshear = Kfs − 1 (6–31)
Kt − 1 Kts − 1

where q is usually between zero and unity. Equation (6–31) shows that if q = 0, then

K f = 1, and the material has no sensitivity to notches at all. On the other hand, if
q = 1, then K f = Kt , and the material has full notch sensitivity. In analysis or design
work, find Kt first, from the geometry of the part. Then specify the material, find q, and
solve for Kf from the equation

K f = 1 + q(Kt − 1) or K f s = 1 + qshear(Kts − 1) (6–32)

For steels and 2024 aluminum alloys, use Fig. 6–20 to find q for bending and axial
loading. For shear loading, use Fig. 6–21. In using these charts it is well to know that
the actual test results from which the curves were derived exhibit a large amount of

Figure 6–20 Notch radius r, mm

Notch-sensitivity charts for 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
steels and UNS A92024-T 1.0
wrought aluminum alloys = 200 kpsi (1.4 GPa)
subjected to reversed bending (1.0)
or reversed axial loads. For S ut
larger notch radii, use the
values of q corresponding 0.8 150 (0.7)
to the r = 0.16-in (4-mm)
ordinate. (From George Sines Notch sensitivity q 100 (0.4)
and J. L. Waisman (eds.),
Metal Fatigue, McGraw-Hill, 0.6 60
New York. Copyright ©
1969 by The McGraw-Hill 0.4
Companies, Inc. Reprinted by Steels
permission.) Alum. alloy

0.2

0
0 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16
Notch radius r, in

288 Mechanical Engineering Design

Figure 6–21 Notch sensitivity qshear 0 Notch radius r, mm
1.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
Notch-sensitivity curves for 0.8
materials in reversed torsion. 0.6 Quenched and drawn steels (Bhn > 200)
For larger notch radii, use Annealed steels (Bhn < 200)
the values of qshear
corresponding to r = 0.16 in
(4 mm).

0.4
Aluminum alloys

0.2

0
0 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16
Notch radius r, in

scatter. Because of this scatter it is always safe to use K f = Kt if there is any doubt
about the true value of q. Also, note that q is not far from unity for large notch radii.

The notch sensitivity of the cast irons is very low, varying from 0 to about 0.20,
depending upon the tensile strength. To be on the conservative side, it is recommended
that the value q = 0.20 be used for all grades of cast iron.

Figure 6–20 has as its basis the Neuber equation, which is given by

Kf =1+ Kt √− 1 (6–33)
1 + a/r
√
where a is defined as the Neuber constant and is a material constant. Equating

Eqs. (6–31) and (6–33) yields the notch sensitivity equation

q = 1√ (6–34)
1 + √a
r

For steel, with Sut in kpsi, the Neuber constant can be approximated by a third-order

polynomial fit of data as

√ = 0.245 799 − 0.307 794(10−2)Sut (6–35)
a

+ 0.150 874(10−4)Su2t − 0.266 978(10−7)Su3t

To use Eq. (6–33) or (6–34) for torsion for low-alloy √steels, increase the ultimate
strength by 20 kpsi in Eq. (6–35) and apply this value of a.

EXAMPLE 6–6 A steel shaft in bending has an ultimate strength of 690 MPa and a shoulder with a fil-
let radius of 3 mm connecting a 32-mm diameter with a 38-mm diameter. Estimate Kf
using:
(a) Figure 6–20.
(b) Equations (6–33) and (6–35).

Fatigue Failure Resulting from Variable Loading 289

Solution From Fig. A–15–9, =u.si1n.g65D. /d = 38/32 = 1.1875, r/d = 3/32 = 0.093 75, we read
Answer the graph to find Kt 690 MPa and r = 3 mm, q =. 0.84. Thus, from Eq. (6–32)
Answer
(a) From Fig. 6–20, for Sut =

K f = 1 + q(Kt − 1) =. 1 + 0.84(1.65 − 1) = 1.55
√ √√

(b) From Eq. (6–35) with Sut = 690 MPa = 100 kpsi, a = 0.0622 in = 0.313 mm.

Substituting this into Eq. (6–33) with r = 3 mm gives

Kf = 1 + Kt √− 1 =. 1 + 1.65 − 1 = 1.55
1 + a/r 1 + 0√.313

3

For simple loading, it is acceptable to reduce the endurance limit by either dividing

the unnotched specimen endurance limit by K f or multiplying the reversing stress by
K f . However, in dealing with combined stress problems that may involve more than one
value of fatigue-concentration factor, the stresses are multiplied by K f .

EXAMPLE 6–7 Consider an unnotched specimen with an endurance limit of 55 kpsi. If the specimen
Solution
Answer was notched such that K f = 1.6, what would be the factor of safety against failure for
Answer N > 106 cycles at a reversing stress of 30 kpsi?

(a) Solve by reducing Se.
(b) Solve by increasing the applied stress.

(a) The endurance limit of the notched specimen is given by

Se = Se = 55 = 34.4 kpsi
Kf 1.6

and the factor of safety is

n = Se = 34.4 = 1.15
σa 30

(b) The maximum stress can be written as

(σa)max = K f σa = 1.6(30) = 48.0 kpsi
and the factor of safety is

n = Se = 55 = 1.15
K f σa 48

Up to this point, examples illustrated each factor in Marin’s equation and stress
concentrations alone. Let us consider a number of factors occurring simultaneously.

290 Mechanical Engineering Design

EXAMPLE 6–8 A 1015 hot-rolled steel bar has been machined to a diameter of 1 in. It is to be placed
Solution in reversed axial loading for 70 000 cycles to failure in an operating environment of
550°F. Using ASTM minimum properties, and a reliability of 99 percent, estimate the
endurance limit and fatigue strength at 70 000 cycles.

From Table A–20, Sut = 50 kpsi at 70°F. Since the rotating-beam specimen endurance
limit is not known at room temperature, we determine the ultimate strength at the ele-
vated temperature first, using Table 6–4. From Table 6–4,

ST = 0.995 + 0.963 = 0.979
SRT 550◦ 2

The ultimate strength at 550°F is then

(Sut )550◦ = (ST /SRT )550◦ (Sut )70◦ = 0.979(50) = 49.0 kpsi

The rotating-beam specimen endurance limit at 550°F is then estimated from Eq. (6–8)
as

Se = 0.5(49) = 24.5 kpsi

Next, we determine the Marin factors. For the machined surface, Eq. (6–19) with
Table 6–2 gives

ka = a Subt = 2.70(49−0.265) = 0.963

For axial loading, from Eq. (6–21), the size factor kb = 1, and from Eq. (6–26) the load-
ing factor is kc = 0.85. The temperature factor kd = 1, since we accounted for the tem-
perature in modifying the ultimate strength and consequently the endurance limit. For
99 percent reliability, from Table 6–5, ke = 0.814. Finally, since no other conditions
were given, the miscellaneous factor is kf = 1. The endurance limit for the part is esti-
mated by Eq. (6–18) as

Answer Se = kakbkckd kek f Se
Answer = 0.963(1)(0.85)(1)(0.814)(1)24.5 = 16.3 kpsi

For the fatigue strength at 70 000 cycles we need to construct the S-N equation. From
p. 277, since Sut = 49 < 70 kpsi, then f ϭ 0.9. From Eq. (6–14)

a = ( f Sut )2 = [0.9(49)]2 = 119.3 kpsi
Se 16.3

and Eq. (6–15)

b = − 1 log f Sut = − 1 log 0.9(49) = −0.1441
3 Se 3 16.3

Finally, for the fatigue strength at 70 000 cycles, Eq. (6–13) gives

Sf = a N b = 119.3(70 000)−0.1441 = 23.9 kpsi

Fatigue Failure Resulting from Variable Loading 291

EXAMPLE 6–9 Figure 6–22a shows a rotating shaft simply supported in ball bearings at A and D and
Solution loaded by a nonrotating force F of 6.8 kN. Using ASTM “minimum” strengths, estimate
the life of the part.

From Fig. 6–22b we learn that failure will probably occur at B rather than at C or at the
point of maximum moment. Point B has a smaller cross section, a higher bending
moment, and a higher stress-concentration factor than C, and the location of maximum
moment has a larger size and no stress-concentration factor.

We shall solve the problem by first estimating the strength at point B, since the strength
will be different elsewhere, and comparing this strength with the stress at the same point.

From Table A–20 we find Sut = 690 MPa and Sy = 580 MPa. The endurance limit
Se is estimated as

Se = 0.5(690) = 345 MPa

From Eq. (6–19) and Table 6–2,

ka = 4.51(690)−0.265 = 0.798

From Eq. (6–20),

kb = (32/7.62)−0.107 = 0.858

Since kc = kd = ke = kf = 1,

Se = 0.798(0.858)345 = 236 MPa

To find the geometric stress-concentration factor Kt we enter FKitg=..A√1–.1655–. 9 with D/d =
38/32 = 1.1875 and r/d = 3/32 = 0.093 75 and√read Substi√tuting

Sut = 690/6.89 = 100 kpsi into Eq. (6–35) yields a = 0.0622 in = 0.313 mm.

Substituting this into Eq. (6–33) gives

Kf = 1 + Kt √− 1 = 1 + 1.65 − 1√ = 1.55
1 + a/r 1 + 0.313/ 3

Figure 6–22 A B 6.8 kN C D
10 125
(a) Shaft drawing showing all 250 75 100
dimensions in millimeters; all 10
fillets 3-mm radius. The shaft
rotates and the load is 30 32 38 35
stationary; material is R1 (a) 30
machined from AISI 1050
cold-drawn steel. (b) Bending- R2
moment diagram.

Mmax
MB

MC

A B CD
(b)

292 Mechanical Engineering Design

The next step is to estimate the bending stress at point B. The bending moment
is

MB = R1 x = 225F 250 = 225(6.8) 250 = 695.5 N ·m
550 550

Just to the left of B the section modulus is I /c = πd3/32 = π323/32 = 3.217 (103)mm3.
The reversing bending stress is, assuming infinite life,

σ = Kf MB = 1.55 695.5 (10)−6 = 335.1(106) Pa = 335.1 MPa
I /c 3.217

This stress is greater than Se and less than Sy. This means we have both finite life and
no yielding on the first cycle.

For finite life, we will need to use Eq. (6–16). The ultimate strength, Sut = 690
MPa = 100 kpsi. From Fig. 6–18, f = 0.844. From Eq. (6–14)

a = ( f Sut )2 = [0.844(690)]2 = 1437 MPa
Se 236

and from Eq. (6–15)

b = − 1 log f Sut = − 1 log 0.844(690) = −0.1308
3 Se 3 236

From Eq. (6–16),

Answer N= σa 1/b 335.1 −1/0.1308

= = 68(103) cycles
a 1437

6–11 Characterizing Fluctuating Stresses

Fluctuating stresses in machinery often take the form of a sinusoidal pattern because
of the nature of some rotating machinery. However, other patterns, some quite irreg-
ular, do occur. It has been found that in periodic patterns exhibiting a single maxi-
mum and a single minimum of force, the shape of the wave is not important, but the
peaks on both the high side (maximum) and the low side (minimum) are important.
Thus Fmax and Fmin in a cycle of force can be used to characterize the force pattern.
It is also true that ranging above and below some baseline can be equally effective
in characterizing the force pattern. If the largest force is Fmax and the smallest force
is Fmin, then a steady component and an alternating component can be constructed
as follows:

Fm = Fmax + Fmin Fa = Fmax − Fmin
2 2

where Fm is the midrange steady component of force, and Fa is the amplitude of the
alternating component of force.

Fatigue Failure Resulting from Variable Loading 293

Figure 6–23 Stress Stress ␴max ␴a ␴r
␴a ␴m
Some stress-time relations: Time
(a) fluctuating stress with high- ␴min
frequency ripple; (b and c)
nonsinusoidal fluctuating (a)
stress; (d) sinusoidal fluctuating
stress; (e) repeated stress; O Time
(f) completely reversed
sinusoidal stress. (d )

Stress Time Stress ␴a
␴a
␴max ␴r
␴m

(b) O ␴min = 0 Time

(e)

+

Stress Time Stress O ␴a Time ␴r
␴a

␴m = 0

(c) ( f )

Figure 6–23 illustrates some of the various stress-time traces that occur. The com-
ponents of stress, some of which are shown in Fig. 6–23d, are

σmin = minimum stress σm = midrange component
σmax = maximum stress σr = range of stress
σa = amplitude component σs = static or steady stress

The steady, or static, stress is not the same as the midrange stress; in fact, it may have
any value between σmin and σmax. The steady stress exists because of a fixed load or pre-
load applied to the part, and it is usually independent of the varying portion of the load.
A helical compression spring, for example, is always loaded into a space shorter than
the free length of the spring. The stress created by this initial compression is called the
steady, or static, component of the stress. It is not the same as the midrange stress.

We shall have occasion to apply the subscripts of these components to shear stress-
es as well as normal stresses.

The following relations are evident from Fig. 6–23:

σm = σmax + σmin
2
(6–36)
σmax − σmin
σa = 2

294 Mechanical Engineering Design

In addition to Eq. (6–36), the stress ratio

R = σmin (6–37)
σmax

and the amplitude ratio

A = σa (6–38)
σm

are also defined and used in connection with fluctuating stresses.
Equations (6–36) utilize symbols σa and σm as the stress components at the loca-

tion under scrutiny. This means, in the absence of a notch, σa and σm are equal to the
nominal stresses σao and σmo induced by loads Fa and Fm , respectively; in the presence
of a notch they are K f σao and K f σmo, respectively, as long as the material remains
without plastic strain. In other words, the fatigue stress concentration factor K f is
applied to both components.

When the steady stress component is high enough to induce localized notch yield-

ing, the designer has a problem. The first-cycle local yielding produces plastic strain

and strain-strengthening. This is occurring at the location where fatigue crack nucle-

ation and growth are most likely. The material properties (Sy and Sut ) are new and dif-
ficult to quantify. The prudent engineer controls the concept, material and condition of

use, and geometry so that no plastic strain occurs. There are discussions concerning

possible ways of quantifying what is occurring under localized and general yielding in

the presence of a notch, referred to as the nominal mean stress method, residual stress
method, and the like.20 The nominal mean stress method (set σa = K f σao and
σm = σmo) gives roughly comparable results to the residual stress method, but both are
approximations.

There is the method of Dowling21 for ductile materials, which, for materials with a

pronounced yield point and approximated by an elastic–perfectly plastic behavior

model, quantitatively expresses the steady stress component stress-concentration factor

K f m as

Kfm = Kf K f |σmax,o| < Sy
K f |σmax,o| > Sy
Kfm = Sy − K f σao K f |σmax,o − σmin,o| > 2Sy (6–39)
|σmo|

Kfm = 0

For the purposes of this book, for ductile materials in fatigue,

• Avoid localized plastic strain at a notch. Set σa = K f σa,o and σm = K f σmo.

• When plastic strain at a notch cannot be avoided, use Eqs. (6–39); or conservatively,
set σa = K f σao and use K f m = 1, that is, σm = σmo.

20R. C. Juvinall, Stress, Strain, and Strength, McGraw-Hill, New York, 1967, articles 14.9–14.12; R. C.
Juvinall and K. M. Marshek, Fundamentals of Machine Component Design, 4th ed., Wiley, New York, 2006,
Sec. 8.11; M. E. Dowling, Mechanical Behavior of Materials, 2nd ed., Prentice Hall, Englewood Cliffs,
N.J., 1999, Secs. 10.3–10.5.
21Dowling, op. cit., p. 437–438.

Fatigue Failure Resulting from Variable Loading 295

6–12 Fatigue Failure Criteria for Fluctuating Stress

Now that we have defined the various components of stress associated with a part sub-
jected to fluctuating stress, we want to vary both the midrange stress and the stress
amplitude, or alternating component, to learn something about the fatigue resistance of
parts when subjected to such situations. Three methods of plotting the results of such
tests are in general use and are shown in Figs. 6–24, 6–25, and 6–26.

The modified Goodman diagram of Fig. 6–24 has the midrange stress plotted along
the abscissa and all other components of stress plotted on the ordinate, with tension in
the positive direction. The endurance limit, fatigue strength, or finite-life strength,
whichever applies, is plotted on the ordinate above and below the origin. The midrange-
stress line is a 45◦ line from the origin to the tensile strength of the part. The modified
Goodman diagram consists of the lines constructed to Se (or Sf ) above and below the
origin. Note that the yield strength is also plotted on both axes, because yielding would
be the criterion of failure if σmax exceeded Sy.

Another way to display test results is shown in Fig. 6–25. Here the abscissa repre-
sents the ratio of the midrange strength Sm to the ultimate strength, with tension plot-
ted to the right and compression to the left. The ordinate is the ratio of the alternating
strength to the endurance limit. The line BC then represents the modified Goodman
criterion of failure. Note that the existence of midrange stress in the compressive region
has little effect on the endurance limit.

The very clever diagram of Fig. 6–26 is unique in that it displays four of the stress
components as well as the two stress ratios. A curve representing the endurance limit
for values of R beginning at R = −1 and ending with R = 1 begins at Se on the σa axis
and ends at Sut on the σm axis. Constant-life curves for N = 105 and N = 104 cycles

Figure 6–24 +
Su
Modified Goodman diagram
showing all the strengths and Stress Sy
the limiting values of all the ␴max
stress components for a Max. stress
particular midrange stress.
Se
strMeisdsrange ␴a
␴min ␴r
45°
␴a
0
␴m Sy
Parallel Midrange stress Su
Min. stress

Se

296 Mechanical Engineering Design

1.2 B
A

1.0

Amplitude ratio Sa/Se 0.8

0.6

0.4

0.2

C

–1.2 –1.0 –0.8 –0.6 –0.4 –0.2 0 0.2 0.4 0.6 0.8 1.0

Compression Sm/Suc Tension Sm/Sut

Midrange ratio

Figure 6–25

Plot of fatigue failures for midrange stresses in both tensile and compressive regions. Normalizing

the data by using the ratio of steady strength component to tensile strength Sm /Sut , steady strength
component to compressive strength Sm /Suc and strength amplitude component to endurance limit
Sa/Se enables a plot of experimental results for a variety of steels. [Data source: Thomas J. Dolan,
“Stress Range,” Sec. 6.2 in O. J. Horger (ed.), ASME Handbook—Metals Engineering Design,

McGraw-Hill, New York, 1953.]

Figure 6–26 4.0 2.33 1.5 A=1 0.67 0.43 0.25 0.11 0
1.0
–0.6 –0.4 –0.2 R = 0 0.2 0.4 0.6 0.8

Master fatigue diagram RA
created for AISI 4340 steel

having Sut = 158 and 180
Sy = 147 kpsi. The stress
components at A are 160 Su t
σmin = 20, σmax = 120,
σm = 70, and σa = 50, all in A=ϱ 104 cycles 140
kpsi. (Source: H. J. Grover, R = –1.0 105
A
Fatigue of Aircraft Structures, 120 120 kpsi120

U.S. Government Printing Maximum stress ␴max, kpsi 100 106 stress10␴0
Office, Washington, D.C., m,
100
1966, pp. 317, 322. See
also J. A. Collins, Failure of 80 60 Midran8g0e
Al8te0rnating stress
Materials in Mechanical 60 Se
Design, Wiley, New York,
60

1981, p. 216.) 40 ␴

, 40 40

a

kpsi

20 20 20

–120 –100 –80 –60 –40 –20 0 20 40 60 80 100 120 140 160 180
Minimum stress ␴min, kpsi

have been drawn too. Any stress state, such as the one at A, can be described by the min-
imum and maximum components, or by the midrange and alternating components. And
safety is indicated whenever the point described by the stress components lies below the
constant-life line.

Fatigue Failure Resulting from Variable Loading 297

Figure 6–27 Sy Yield (Langer) line
Se
Fatigue diagram showing Alternating stress ␴a Sa Gerber line
various criteria of failure. For 0
each criterion, points on or Load line, slope r = Sa/Sm
“above” the respective line 0
indicate failure. Some point A A Modified Goodman line
on the Goodman line, for Soderberg line ASME-elliptic line
example, gives the strength Sm
as the limiting value of σm
corresponding to the strength
Sa, which, paired with σm , is
the limiting value of σa.

Sm Sy Sut
Midrange stress ␴m

When the midrange stress is compression, failure occurs whenever σa = Se or
whenever σmax = Syc, as indicated by the left-hand side of Fig. 6–25. Neither a fatigue
diagram nor any other failure criteria need be developed.

In Fig. 6–27, the tensile side of Fig. 6–25 has been redrawn in terms of strengths,
instead of strength ratios, with the same modified Goodman criterion together with four
additional criteria of failure. Such diagrams are often constructed for analysis and
design purposes; they are easy to use and the results can be scaled off directly.

The early viewpoint expressed on a σaσm diagram was that there existed a locus which
divided safe from unsafe combinations of σa and σm. Ensuing proposals included the
parabola of Gerber (1874), the Goodman (1890)22 (straight) line, and the Soderberg (1930)
(straight) line. As more data were generated it became clear that a fatigue criterion, rather
than being a “fence,” was more like a zone or band wherein the probability of failure could
be estimated. We include the failure criterion of Goodman because

• It is a straight line and the algebra is linear and easy.

• It is easily graphed, every time for every problem.

• It reveals subtleties of insight into fatigue problems.

• Answers can be scaled from the diagrams as a check on the algebra.

We also caution that it is deterministic and the phenomenon is not. It is biased and we
cannot quantify the bias. It is not conservative. It is a stepping-stone to understanding; it
is history; and to read the work of other engineers and to have meaningful oral exchanges
with them, it is necessary that you understand the Goodman approach should it arise.

Either the fatigue limit Se or the finite-life strength Sf is plotted on the ordinate of
Fig. 6–27. These values will have already been corrected using the Marin factors of
Eq. (6–18). Note that the yield strength Sy is plotted on the ordinate too. This serves as
a reminder that first-cycle yielding rather than fatigue might be the criterion of failure.

The midrange-stress axis of Fig. 6–27 has the yield strength Sy and the tensile
strength Sut plotted along it.

22It is difficult to date Goodman’s work because it went through several modifications and was never
published.

298 Mechanical Engineering Design

Five criteria of failure are diagrammed in Fig. 6–27: the Soderberg, the modified
Goodman, the Gerber, the ASME-elliptic, and yielding. The diagram shows that only
the Soderberg criterion guards against any yielding, but is biased low.

Considering the modified Goodman line as a criterion, point A represents a limit-
ing point with an alternating strength Sa and midrange strength Sm. The slope of the load
line shown is defined as r = Sa/Sm .

The criterion equation for the Soderberg line is

Sa + Sm = 1 (6–40)
Se Sy
Similarly, we find the modified Goodman relation to be

Sa + Sm = 1 (6–41)
Se Sut

Examination of Fig. 6–25 shows that both a parabola and an ellipse have a better
opportunity to pass among the midrange tension data and to permit quantification of the
probability of failure. The Gerber failure criterion is written as

Sa + Sm 2 (6–42)

Se Sut =1

and the ASME-elliptic is written as

Sa 2 Sm 2 (6–43)

+ =1
Se Sy

The Langer first-cycle-yielding criterion is used in connection with the fatigue
curve:

Sa + Sm = Sy (6–44)

The stresses nσa and nσm can replace Sa and Sm , where n is the design factor or factor
of safety. Then, Eq. (6–40), the Soderberg line, becomes

Soderberg σa + σm = 1 (6–45)
Se Sy n

Equation (6–41), the modified Goodman line, becomes

mod-Goodman σa + σm = 1 (6–46)
Se Sut n (6–47)
(6–48)
Equation (6–42), the Gerber line, becomes

Gerber nσa + nσm 2

Se Sut =1

Equation (6–43), the ASME-elliptic line, becomes

ASME-elliptic nσa 2 nσm 2

+ =1
Se Sy

We will emphasize the Gerber and ASME-elliptic for fatigue failure criterion and the
Langer for first-cycle yielding. However, conservative designers often use the modified
Goodman criterion, so we will continue to include it in our discussions. The design
equation for the Langer first-cycle-yielding is

Langer static yield σa + σm = Sy (6–49)
n

Fatigue Failure Resulting from Variable Loading 299

The failure criteria are used in conjunction with a load line, r = Sa/Sm = σa/σm .
Principal intersections are tabulated in Tables 6–6 to 6–8. Formal expressions for
fatigue factor of safety are given in the lower panel of Tables 6–6 to 6–8. The first row
of each table corresponds to the fatigue criterion, the second row is the static Langer
criterion, and the third row corresponds to the intersection of the static and fatigue

Table 6–6 Intersecting Equations Intersection Coordinates

Amplitude and Steady Sa + Sm =1 Sa = r SeSut
Coordinates of Strength Se Sut r Sut + Se
and Important
Intersections in First Load line r = Sa Sm = Sa
Quadrant for Modified Sm r
Goodman and Langer
Failure Criteria Sa + Sm =1 Sa = r Sy
Sy Sy 1+r

Load line r = Sa Sm = Sy
Sm 1+r

Sa + Sm =1 Sm = Sy − Se Sut
Se Su t Sut − Se

Sa + Sm =1 Sa = Sy − Sm , rcrit = Sa/Sm
Sy Sy

Fatigue factor of safety

nf = σa 1 σm
Se + Su t

Table 6–7 Intersecting Equations Intersection Coordinates

Amplitude and Steady ⎡⎤
Coordinates of Strength
and Important Sa + Sm 2 Sa = r 2 Su2t ⎣−1 + 1+ 2Se 2
Intersections in First Se Su t 2Se r Sut
Quadrant for Gerber =1 ⎦
and Langer Failure
Criteria Load line r = Sa Sm = Sa
Sm r

Sa + Sm =1 Sa = r Sy
Sy Sy 1+r

Load line r = Sa Sm = Sy
Sm 1+r

⎡⎤

Sa Sm 2 Su2t 2Se 2 Sy
Se Su t 2Se Su t Se
+ =1 Sm = ⎣1 − 1+ 1− ⎦

Sa + Sm =1 Sa = Sy − Sm , rcrit = Sa/Sm
Sy Sy

Fatigue factor of safety

⎡⎤

1 Su t 2 σa 2σm Se 2
2 σm Se
nf = ⎣−1 + 1+ Sut σa ⎦ σm > 0

300 Mechanical Engineering Design

Table 6–8 Intersecting Equations Intersection Coordinates

Amplitude and Steady Sa 2 Sm 2 Sa = r 2 Se2 Sy2
Coordinates of Strength Se Sy Se2 + r 2 Sy2
and Important + =1
Intersections in First
Quadrant for ASME- Load line r = Sa/Sm Sm = Sa
Elliptic and Langer r
Failure Criteria
Sa + Sm =1 Sa = r Sy
Sy Sy 1+r

Load line r = Sa/Sm Sm = Sy
1+r

Sa 2 Sm 2 Sa = 0, 2Sy Se2
Se Sy Se2 + Sy2
+ =1

Sa + Sm =1 Sm = Sy − Sa, rcrit = Sa/Sm
Sy Sy

Fatigue factor of safety 1
nf = (σa/Se)2 + σm /Sy 2

criteria. The first column gives the intersecting equations and the second column the
intersection coordinates.

There are two ways to proceed with a typical analysis. One method is to assume
that fatigue occurs first and use one of Eqs. (6–45) to (6–48) to determine n or size,
depending on the task. Most often fatigue is the governing failure mode. Then
follow with a static check. If static failure governs then the analysis is repeated using
Eq. (6–49).

Alternatively, one could use the tables. Determine the load line and establish which
criterion the load line intersects first and use the corresponding equations in the tables.

Some examples will help solidify the ideas just discussed.

EXAMPLE 6–10 A 1.5-in-diameter bar has been machined from an AISI 1050 cold-drawn bar. This part
Solution is to withstand a fluctuating tensile load varying from 0 to 16 kip. Because of the ends,
and the fillet radius, a fatigue stress-concentration factor K f is 1.85 for 106 or larger
life. Find Sa and Sm and the factor of safety guarding against fatigue and first-cycle
yielding, using (a) the Gerber fatigue line and (b) the ASME-elliptic fatigue line.

We begin with some preliminaries. From Table A–20, Sut = 100 kpsi and Sy = 84 kpsi.
Note that Fa = Fm = 8 kip. The Marin factors are, deterministically,

ka = 2.70(100)−0.265 = 0.797: Eq. (6–19), Table 6–2, p. 279

kb = 1 (axial loading, see kc)

Fatigue Failure Resulting from Variable Loading 301

kc = 0.85: Eq. (6–26), p. 282

kd = ke = kf = 1
Se = 0.797(1)0.850(1)(1)(1)0.5(100) = 33.9 kpsi: Eqs. (6–8), (6–18), p. 274, p. 279

The nominal axial stress components σao and σmo are

σao = 4 Fa = 4(8) = 4.53 kpsi σmo = 4 Fm = 4(8) = 4.53 kpsi
πd2 π 1.52 πd2 π 1.52

Applying K f to both components σao and σmo constitutes a prescription of no notch
yielding:

σa = K f σao = 1.85(4.53) = 8.38 kpsi = σm

(a) Let us calculate the factors of safety first. From the bottom panel from Table 6–7 the

factor of safety for fatigue is ⎧ ⎫

Answer nf = 1 100 2 8.38 ⎨ 1+ 2(8.38)33.9 2⎬
Answer 2 8.38 33.9 ⎩−1 + 100(8.38) ⎭ = 3.66

Answer From Eq. (6–49) the factor of safety guarding against first-cycle yield is

ny = σa Sy = 84 = 5.01
+ σm 8.38 + 8.38

Thus, we see that fatigue will occur first and the factor of safety is 3.68. This can be

seen in Fig. 6–28 where the load line intersects the Gerber fatigue curve first at point B.

If the plots are created to true scale it would be seen that n f = O B/O A.

From the first panel of Table 6–7, r = σa/σm = 1,
⎧ 2⎫⎬
⎨ ⎭ = 30.7 kpsi
Sa = (1)21002 ⎩−1 + 1+ 2(33.9)
2(33.9) (1)100

Figure 6–28 100
84
Principal points A, B, C, and
D on the designer’s diagram
drawn for Gerber, Langer, and
load line.

Stress amplitude ␴a , kpsi 50 Load line
42 C
33.9
30.7 Langer line
B
20
A D rcrit
Gerber
8.38 fatigue curve
0
0 8.38 30.7 42 50 64 84 100

Midrange stress ␴m, kpsi

302 Mechanical Engineering Design

Answer Sm = Sa = 30.7 = 30.7 kpsi
Answer r 1

As a check on the previous result, n f = O B/O A = Sa/σa = Sm/σm = 30.7/8.38 =
3.66 and we see total agreement.

We could have detected that fatigue failure would occur first without drawing Fig.

6–28 by calculating rcrit . From the third row third column panel of Table 6–7, the inter-
section point between fatigue and first-cycle yield is

⎡⎤

Sm = 1002 ⎣1 − 1+ 2(33.9) 2 1 − 84 ⎦ = 64.0 kpsi
2(33.9) 100 33.9

Sa = Sy − Sm = 84 − 64 = 20 kpsi

The critical slope is thus

rcrit = Sa = 20 = 0.312
Sm 64

which is less than the actual load line of r = 1. This indicates that fatigue occurs before
first-cycle-yield.
(b) Repeating the same procedure for the ASME-elliptic line, for fatigue

nf = 1 = 3.75
(8.38/33.9)2 + (8.38/84)2

Again, this is less than ny = 5.01 and fatigue is predicted to occur first. From the first
row second column panel of Table 6–8, with r = 1, we obtain the coordinates Sa and
Sm of point B in Fig. 6–29 as

Figure 6–29 100
84
Principal points A, B, C, and
D on the designer’s diagram
drawn for ASME-elliptic,
Langer, and load lines.

Stress amplitude ␴a , kpsi 50 Load line
42 C
31.4
23.5 B Langer line
D
A
8.38 ASME-elliptic line

0 31.4 42 50 60.5 84 100
0 8.38 Midrange stress ␴m, kpsi

Fatigue Failure Resulting from Variable Loading 303

Answer Sa = (1)233.92(84)2 Sm = Sa = 31.4 = 31.4 kpsi
33.92 + (1)2842 = 31.4 kpsi, r 1

To verify the fatigue factor of safety, n f = Sa/σa = 31.4/8.38 = 3.75.
As before, let us calculate rcrit. From the third row second column panel of

Table 6–8,

Sa = 2(84)33.92 = 23.5 kpsi, Sm = Sy − Sa = 84 − 23.5 = 60.5 kpsi
33.92 + 842

rcrit = Sa = 23.5 = 0.388
Sm 60.5

which again is less than r = 1, verifying that fatigue occurs first with n f = 3.75.
The Gerber and the ASME-elliptic fatigue failure criteria are very close to each

other and are used interchangeably. The ANSI/ASME Standard B106.1M–1985 uses
ASME-elliptic for shafting.

EXAMPLE 6–11 A flat-leaf spring is used to retain an oscillating flat-faced follower in contact with a
Solution
plate cam. The follower range of motion is 2 in and fixed, so the alternating component

of force, bending moment, and stress is fixed, too. The spring is preloaded to adjust to

various cam speeds. The preload must be increased to prevent follower float or jump.

For lower speeds the preload should be decreased to obtain longer life of cam and

follower surfaces. The spring is a steel cantilever 32 in long, 2 in wide, and 1 in thick,
4
as seen in Fig. 6–30a. The spring strengths are Sut = 150 kpsi, Sy = 127 kpsi, and Se =

28 kpsi fully corrected. The total cam motion is 2 in. The designer wishes to preload

the spring by deflecting it 2 in for low speed and 5 in for high speed.

(a) Plot the Gerber-Langer failure lines with the load line.

(b) What are the strength factors of safety corresponding to 2 in and 5 in preload?

We begin with preliminaries. The second area moment of the cantilever cross section is

I = bh3 = 2(0.25)3 = 0.00260 in4
12 12

Since, from Table A–9, beam 1, force F and deflection y in a cantilever are related by
F = 3E I y/l3, then stress σ and deflection y are related by

σ = Mc = 32 F c = 32(3E I y) c = 96 E cy = Ky
I I l3 I l3

where K = 96 E c = 96(30 · 106)0.125 = 10.99(103) psi/in = 10.99 kpsi/in
l3 323

Now the minimums and maximums of y and σ can be defined by

ymin = δ ymax = 2 + δ

σmin = K δ σmax = K (2 + δ)

304 Mechanical Engineering Design 2 in 1 in +
4
Figure 6–30 ␦ = 2 in
32 in +
Cam follower retaining spring.
(a) Geometry; (b) designer’s ␦ = 5 in
fatigue diagram for Ex. 6–11. +

␦ = 2 in preload

␦ = 5 in preload (a)
150

Amplitude stress component ␴a, kpsi 100
Langer line

50

A Gerber line A"
0 A'

11 33 50 65.9 100 115.6 127 150

Steady stress component ␴m, kpsi

(b)

The stress components are thus

σa = K (2 + δ) − Kδ = K = 10.99 kpsi
2

σm = K (2 + δ) + Kδ = K (1 + δ) = 10.99(1 + δ)
2

For δ = 0, σa = σm = 10.99 = 11 kpsi

Fatigue Failure Resulting from Variable Loading 305

For δ = 2 in, σa = 11 kpsi, σm = 10.99(1 + 2) = 33 kpsi
For δ = 5 in, σa = 11 kpsi, σm = 10.99(1 + 5) = 65.9 kpsi

(a) A plot of the Gerber and Langer criteria is shown in Fig. 6–30b. The three preload
deflections of 0, 2, and 5 in are shown as points A, A , and A . Note that since σa is
constant at 11 kpsi, the load line is horizontal and does not contain the origin. The
intersection between the Gerber line and the load line is found from solving Eq. (6–42)
for Sm and substituting 11 kpsi for Sa:

Sm = Sut 1 − Sa = 150 1 − 11 = 116.9 kpsi
Se 28

The intersection of the Langer line and the load line is found from solving Eq. (6–44)
for Sm and substituting 11 kpsi for Sa:

Sm = Sy − Sa = 127 − 11 = 116 kpsi

The threats from fatigue and first-cycle yielding are approximately equal.
(b) For δ = 2 in,

Answer nf = Sm = 116.9 = 3.54 ny = 116 = 3.52
Answer σm 33 33

and for δ = 5 in,

nf = 116.9 = 1.77 ny = 116 = 1.76
65.9 65.9

EXAMPLE 6–12 A steel bar undergoes cyclic loading such that σmax = 60 kpsi and σmin = −20 kpsi. For
Solution the material, Sut = 80 kpsi, Sy = 65 kpsi, a fully corrected endurance limit of Se =
40 kpsi, and f = 0.9. Estimate the number of cycles to a fatigue failure using:

(a) Modified Goodman criterion.

(b) Gerber criterion.

From the given stresses,

σa = 60 − (−20) = 40 kpsi σm = 60 + (−20) = 20 kpsi
2 2

From the material properties, Eqs. (6–14) to (6–16), p. 277, give

a = ( f Sut )2 = [0.9(80)]2 = 129.6 kpsi
Se 40

b = − 1 log f Sut = − 1 log 0.9(80) = −0.0851
3 Se 3 40

Sf 1/b Sf −1/0.0851

N= = (1)
a 129.6

where Sf replaced σa in Eq. (6–16).

306 Mechanical Engineering Design

(a) The modified Goodman line is given by Eq. (6–46), p. 298, where the endurance
limit Se is used for infinite life. For finite life at Sf > Se, replace Se with Sf in Eq.
(6–46) and rearrange giving

Sf = σa = 40 = 53.3 kpsi
1 − σm 1 − 20
Sut 80

Substituting this into Eq. (1) yields

Answer N= 53.3 −1/0.0851 =. 3.4(104) cycles
Answer 129.6

(b) For Gerber, similar to part (a), from Eq. (6–47),

Sf = σa 2 = 40 2 = 42.7 kpsi
σm 20
1− 1−
Sut 80

Again, from Eq. (1),

N = 42.7 −1/0.0851 =. 4.6(105) cycles
129.6

Comparing the answers, we see a large difference in the results. Again, the modified
Goodman criterion is conservative as compared to Gerber for which the moderate dif-
ference in Sf is then magnified by a logarithmic S, N relationship.

For many brittle materials, the first quadrant fatigue failure criteria follows a con-
cave upward Smith-Dolan locus represented by

Sa = 1 − Sm /Sut (6–50)
Se 1 + Sm /Sut

or as a design equation,

nσa = 1 − nσm/Sut (6–51)
Se 1 + nσm/Sut

For a radial load line of slope r, we substitute Sa/r for Sm in Eq. (6–50) and solve for
Sa, obtaining

Sa = r Sut + Se −1 + 1 + 4r Sut Se (6–52)
2 Sut + Se)2
(r

The fatigue diagram for a brittle material differs markedly from that of a ductile material
because:

• Yielding is not involved since the material may not have a yield strength.

• Characteristically, the compressive ultimate strength exceeds the ultimate tensile
strength severalfold.