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Chapter 6
Fatigue Failure Resulting
from Variable Loading
Lecture Slides
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Chapter Outline
6–11 Characterizing Fluctuating Stresses
325
6–1
Introduction to Fatigue 286
6–2
Chapter Overview 287
6–3
Crack Nucleation and Propagation
288
6–12 The Fluctuating-Stress Diagram 327
6–4
Fatigue-Life Methods 294
6–14 Constant-Life Curves 342
6–5
The Linear-Elastic Fracture
Mechanics Method 295
6–15 Fatigue Failure Criterion for Brittle
Materials 345
6–6
The Strain-Life Method 299
6–7
The Stress-Life Method and the S-N
Diagram 302
6–16 Combinations of Loading Modes
347
6–8
6–9
The Idealized S-N Diagram for Steels
304
Endurance Limit Modifying Factors
309
6–13 Fatigue Failure Criteria 333
6–17 Cumulative Fatigue Damage 351
6–18 Surface Fatigue Strength 356
6–19 Road Maps and Important Design
Equations for the Stress-Life
Method 359
6–10 Stress Concentration and Notch
Sensitivity 320
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2
Introduction to Fatigue in Metals 1
Prior to the nineteenth century, engineering design was based primarily
on static loading.
Speeds were relatively slow, loads were light, and factors of safety were
large.
With the development of engines capable of higher speeds, and materials
capable of higher loads, parts began to be subject to significantly higher
cycles at high stress.
Though these stresses were well below the yield strength, an increase in
sudden ultimate fractures occurred.
The most distinguishing feature of the failures was a large number of
cycles.
This led to the notion that the part had simply become “tired” from
repeated cycling, hence the origin of the term fatigue failure.
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3
Introduction to Fatigue in Metals 2
Testing proved that the material properties had not changed.
Fatigue failure is due to a crack initiating and growing when
subjected to many repeated cycles.
Some of the first notable fatigue failures involved railroad axles in
the mid-1800s.
Albert Wöhler is credited with deliberately studying and
articulating some of the basic principles of fatigue failure.
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4
Introduction to Fatigue in Metals 3
Some notable examples of dramatic fatigue failures (look them up for
some interesting reading).
• Versailles railroad axle (1842).
• Liberty ships (1943).
• multiple de Havilland Comet crashes (1954).
• Kielland oil platform collapse (1980).
• Aloha B737 accident (1988).
• DC10 Sioux City accident (1989).
• MD-88 Pensacola engine failure (1996).
• Eschede railway accident (1998).
• GE CF6 engine failure (2016).
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5
Chapter Overview
Crack Nucleation and Propagation (Section 6–3).
Fatigue-Life Methods (Section 6–4).
The Linear-Elastic Fracture Mechanics Method (Section 6–5).
The Strain-Life Method (Section 6–6).
The Stress-Life Method in Detail (Section 6–7 through 6–17).
• Completely Reversed Loading (Sections 6–7 through 6–10).
• Fluctuating Loading (Section 6–11 through 6–15).
• Combinations of Loading Modes (Section 6–16).
• Cumulative Fatigue Damage (Section 6–17).
Surface Fatigue (Section 6–18).
Road Maps and Important Design Equations for the Stress-Life Method
(Section 6–19).
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6
Crack Nucleation and Propagation
Fatigue failure is due to crack nucleation and propagation.
A fatigue crack will initiate at a location that experiences repeated
applications of locally high stress (and thus high strain).
The locally high stress is often at a discontinuity.
• Geometric changes, for example, keyways, holes.
• Manufacturing imperfections, for example, stamp marks,
scratches.
• Composition of the material, for example, from rolling, forging,
casting, heat treatment, inclusions, voids.
© McGraw Hill
7
Stages of Fatigue Failure
Stage I – Initiation of microcrack due to cyclic plastic
deformation.
Stage II – Progresses to
macro-crack that repeatedly
opens and closes, creating
bands called beach marks.
Stage III – Crack has
propagated far enough that
remaining material is
insufficient to carry the load,
and fails by simple ultimate
failure.
© McGraw Hill
Fig. 6–1
From ASM Handbook, Vol. 12: Fractography, 2nd printing, 1992, ASM International, Materials Park, OH 44073-0002, fig. 50, p. 120. Reprinted by permission of ASM International®, www.asminternational.org.
8
Schematics of Fatigue Fracture Surfaces
Fig. 6–2
Access the text alternative for slide images.
© McGraw Hill
From ASM Metals Handbook, Vol. 11: Failure Analysis and Prevention, 1986, ASM International, Materials Park, OH 44073-0002, fig. 18, p. 111. Reprinted by permission of ASM International®, www.asminternational.org.
9
Fatigue Fracture Examples 1
AISI 4320 drive shaft.
B– crack initiation at
stress concentration in
keyway.
C– Final brittle failure.
Fig. 6–3
© McGraw Hill
(From ASM Handbook, Vol. 12: Fractography, 2nd printing, 1992, ASM International, Materials Park, OH 44073-0002, fig. 51, p. 120. Reprinted by permission of ASM International®, www.asminternational.org.
10
Fatigue Fracture Examples 2
Fatigue failure
initiating at
mismatched grease
holes.
Sharp corners (at
arrows) provided
stress concentrations.
Fig. 6–4
© McGraw Hill
From ASM Handbook, Vol. 12: Fractography, 2nd printing, 1992, ASM International, Materials Park, OH 44073-0002, fig. 520, p. 331. Reprinted by permission of ASM International®, www.asminternational.org.
11
Fatigue Fracture Examples 3
Fatigue failure of
forged connecting rod.
Crack initiated at flash
line of the forging at
the left edge of picture.
Beach marks show
crack propagation
halfway around the
hole before ultimate
fracture.
Fig. 6–5
© McGraw Hill
From ASM Handbook, Vol. 12: Fractography, 2nd printing, 1992, ASM International, Materials Park, OH 44073-0002, fig. 523, p. 332. Reprinted by permission of ASM International®, www.asminternational.org.
12
Fatigue Fracture Examples 4
Fatigue failure of a
200-mm diameter
piston rod of an alloy
steel steam hammer.
Loaded axially.
Crack initiated at a
forging flake internal to
the part.
Internal crack grew
outward symmetrically.
Fig. 6–6
© McGraw Hill
From ASM Handbook, Vol. 12: Fractography, 2nd printing, 1992, ASM International, Materials Park, OH 44073-0002, fig. 570, p. 342. Reprinted by permission of ASM International®, www.asminternational.org.
13
Fatigue Fracture Examples 5
Double-flange trailer wheel.
Cracks initiated at stamp marks.
Fig. 6–7
Access the text alternative for slide images.
© McGraw Hill
From ASM Metals Handbook, Vol. 11: Failure Analysis and Prevention, 1986, ASM International, Materials Park, OH 44073-0002, fig. 51, p. 130. Reprinted by permission of ASM International®, www.asminternational.org.
14
Fatigue Fracture Examples 6
Aluminum allow landing-gear torque-arm assembly redesign to
eliminate fatigue fracture at lubrication hole
Fig. 6–8
Access the text alternative for slide images.
© McGraw Hill
Photo: From ASM Metals Handbook, Vol. 11: Failure Analysis and Prevention, 1986, ASM International, Materials Park, OH 44073-0002, fig 23, p. 114. Reprinted by permission of ASM International®, www.asminternational.org.
15
Crack Nucleation 1
Crack nucleation occurs in the presence of localized plastic strain.
Plastic strain involves breaking of a limited number of atomic bonds, forming
slip planes, in which atoms in crystal planes slip past one another.
The slip planes prefer movement within a grain of the material in a direction
requiring the least energy.
The preferential orientation is usually along the plane of maximum shear stress,
at 45° to the loading direction. (Fig. 6–9a)
Fig. 6–9
Access the text alternative for slide images.
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16
Crack Nucleation 2
Slip planes tend to be parallel to one another, and bunch together to
form slip bands. (Fig. 6–9b)
When the slip bands reach the edge of a grain, and especially at the
surface of the material, they extrude very slightly, and are called
persistent slip bands. (Fig. 6–9c)
Fig. 6–9
© McGraw Hill
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Crack Nucleation 3
Continued cyclic loading of sufficient level eventually causes further
sliding of the persistent slip bands.
Extrusions and intrusions are formed at the grain boundaries, on the order
of 1 to 10 microns. (Fig. 6–9d)
These tiny steps in the surface act as stress concentrations, which locally
accelerates the process, tending to nucleate a microcrack.
Fig. 6–9
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Crack Nucleation 4
Microcrack nucleation is much more likely at the free surface of a part, where
• Stresses are often highest.
• Stress concentrations often exist.
• Surface roughness exists.
• Oxidation and corrosion accelerate the process.
• There is less resistance to plastic deformation.
Fig. 6–9
© McGraw Hill
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Crack Propagation 1
Stage I crack growth (shear mode).
• Continued cycling
progressively breaks bonds
between slip planes across a
single grain.
• The growth rate is very slow,
on the order of 1 nm per cycle.
• At the grain boundary, the
crack may slow or halt.
• Eventually, the crack may
propagate into the next grain,
especially if the grain is
preferentially oriented with
shear planes near 45° from the
loading direction.
Fig. 6–10
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20
Crack Propagation 2
Stage II crack growth (tensile mode).
• When the crack has grown across
approximately 3 to 10 grains, it is
sufficiently large to form a stress
concentration at its tip that forms
a tensile plastic zone.
• Several microcracks in near
vicinity may join, increasing the
size of the tensile plastic zone.
• The crack is now vulnerable to
being “opened” by a tensile
normal stress.
Fig. 6–10
© McGraw Hill
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Crack Propagation 3
Stage II crack growth (tensile mode).
• The “opened” crack now starts
Stage II crack growth by growing
perpendicular to the applied load.
• The crack grows particularly
when opened in tension.
• Compressive stress does not tend
to open the crack, and therefore
contributes little to crack growth.
Fig. 6–10
© McGraw Hill
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Crack Nucleation and Propagation 1
Crack nucleation and growth as a portion of total fatigue life is shown.
At higher stress levels, a crack initiates quickly, and most of the fatigue life is
growing a crack.
• This is well modeled by methods of fracture mechanics.
At lower stress levels, a large fraction of the fatigue life is spent to nucleate a
crack, followed by a quick crack growth.
Fig. 6–11
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23
Crack Nucleation and Propagation 2
If the stress level is low enough, it is possible that a crack never nucleates, or that
a nucleated crack never grows to fracture.
• This phenomenon is one of the early discoveries by Wöhler.
• It is significant in that it predicts the possibility of designing for long or
infinite life.
Fig. 6–11
© McGraw Hill
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Crack Nucleation and Propagation 3
High-cycle fatigue domain deals with long fatigue life (say, greater than 10000
cycles) due to low loads, elastic stresses and strains.
Low-cycle fatigue domain deals with short fatigue life, due to high loads, mostly
plastic stresses and strains.
Fig. 6–11
© McGraw Hill
25
Fatigue-Life Methods
Fatigue-Life Methods predict life in number of cycles to failure, N, for a specific
level of loading.
There are three major fatigue life methods in use.
Strain-life method.
• Focuses on crack nucleation (Stage I).
• Detailed analysis of plastic deformation at localized regions.
Linear-elastic fracture mechanics (LEFM) method.
• Focuses on crack propagation (Stage II).
• Predicts crack growth with respect to stress intensity.
Stress-life method.
• Estimates life to fracture, ignoring details of crack nucleation and propagation.
• Based on comparison to experimental test specimens.
© McGraw Hill
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Fatigue-Life Methods (Big Picture) 1
Stress-life method.
• Based on nominal stresses, applying stress concentrations at notches, with no
accounting for local plastic strain.
• Consequently, not useful for condition with high stresses, plastic strains, and
low cycles (that is, the low-cycle fatigue domain).
• Based on empirical data with little theoretical basis.
• Least accurate method, but most traditional.
• Easiest to implement for rough approximations.
• Represents high-cycle fatigue domain adequately.
• Good for observing the relative impact of factors that affect fatigue life.
• Good starting point.
© McGraw Hill
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Fatigue-Life Methods (Big Picture) 2
Strain-life method.
• Detailed analysis of the plastic deformation at localized regions where both
elastic and plastic strains are considered.
• Compares to test specimens that are strain-based, taking into account the
cyclic material properties at the localized level.
• Requires material properties from cyclic stress-strain curves and strain-life
curves.
• Especially suited for low-cycle fatigue domain where the strains are high, but
also works for high-cycle domain.
• Widely viewed as the best method to predict fatigue life with reasonable
reliability.
• But, requires high learning curve and more material properties.
© McGraw Hill
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Fatigue-Life Methods (Big Picture) 3
Linear-elastic fracture mechanics method.
• Assumes a crack exists.
• Predicts crack growth with respect to stress intensity.
• The only method that actually tracks the crack growth, rather
than just estimating cycles to fracture.
• Particularly useful when the stress levels are high and a large
fraction of the fatigue life is spent in the slow growth of a crack.
© McGraw Hill
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Fatigue-Life Methods (Big Picture) 4
All three methods have a place in fatigue design.
For monitoring the actual growth rate of a crack, LEFM is the
prime tool.
For low-cycle domain in the presence of a notch, strain-life is
optimal.
For high-cycle domain, both strain-life and stress-life are
applicable. Strain-life is more accurate, but requires significantly
more overhead.
Stress-life is great for beginning engineers, occasional fatigue
analysis, rough estimates, and observing the impact of various
factors on the fatigue life.
© McGraw Hill
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Fatigue Design Criteria
Four design philosophies have evolved to provide strategies for safe designs.
o Infinite-life design.
• Design for infinite life by keeping the stresses below the level for crack initiation.
o Safe-life design.
• Design for a finite life, for applications subject to a limited number of cycles.
• Due to the large scatter in actual fatigue lives under similar conditions, large safety
factors are used.
o Fail-safe design.
• Incorporates an overall design such that if one part fails, the system does not fail.
• Uses load paths, crack stoppers, and scheduled inspections.
• For applications with high consequences for failure, but need low factors of safety,
such as aircraft industry.
o Damage-tolerant design.
• Assumes existence of a crack, and uses LEFM to predict the growth, in order to
dictate inspection and replacement schedule.
• Best for materials that exhibit slow crack growth and high fracture toughness.
© McGraw Hill
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Linear-Elastic Fracture Mechanics Method (A bit more detail)
Fracture mechanics is the field of mechanics that studies the propagation of
cracks.
Linear-elastic fracture mechanics is an analytical approach to evaluating the
stress field at the tip of a crack.
Assumes the material is isotropic and linear elastic.
Assumes plastic deformation at the tip of a crack is small compared to the size of
the crack.
The stress field is evaluated at the crack tip using the theory of elasticity.
When the stresses near the crack tip exceed the material fracture toughness, the
crack is predicted to grow.
See Section 5–12 for basics of fracture mechanics, as applied to quasi-static
loading and brittle materials.
Concepts are extended here for dynamically loaded applications.
© McGraw Hill
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Linear-Elastic Fracture Mechanics Method
Useful for studying and understanding the fracture mechanism.
For a certain class of problems, it is effective in predicting fatigue
life.
• High stresses.
• Crack either exists from the start, or is expected to nucleate
quickly.
• Most of the fatigue life consists of a slow propagation of a crack.
Often used for damage-tolerant design criterion.
Prominent in the aircraft industry.
© McGraw Hill
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Crack Growth 1
Fatigue cracks nucleate and grow when stresses vary and there is
some tension in each stress cycle.
Consider a stress fluctuating in the stress range
   max   min
The stress intensity factor is defined by
K I    a
(5 - 37)
For a stress range , the stress intensity range per cycle is
K I    max   min   a    a
© McGraw Hill
(6 - 1)
34
Crack Growth 2
Testing specimens at various levels of  provides plots of crack length vs. stress
cycles.
A higher stress range produces a longer crack at a particular cycle count.
Note the slope is the rate of crack growth per cycle da/dN.
Fig. 6–12
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© McGraw Hill
35
Crack Growth 3
Log-log plot of rate
of crack growth,
da/dN, shows all
three stages of
growth.
Stage II data are
linear on log-log
scale.
Similar curves can be
generated by
changing the stress
ratio R = σmin / σmax.
Fig. 6–13
Access the text alternative for slide images.
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36
Crack Growth 4
Three unique regions
of crack development
are observed.
In Region I, below
the threshold value
(ΔKI)th, the crack
does not grow.
Above the threshold,
the crack growth rate
is still small, but
increasing rapidly.
Fig. 6–13
© McGraw Hill
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Crack Growth 5
Region II has stable crack
growth.
Linear relationship between
crack growth rate and stress
intensity factor range.
In Region III, the crack growth
rate is very high and rapidly
accelerates to instability.
When ΔKI exceeds the critical
stress intensity factor ΔKIc (also
known as the fracture
toughness), the remaining
cross section suddenly and
completely fractures.
Fig. 6–13
© McGraw Hill
38
Crack Growth 6
Region II is stable
enough to allow
estimation of the
remaining life of a part
after a crack is
discovered.
Fig. 6–13
© McGraw Hill
39
Crack Growth 7
Crack growth in Region II is approximated by the Paris equation
da
m
 C  K I 
(6 - 2)
dN
C and m are empirical material constants. Conservative representative values are
shown in Table 6–1.
Table 6–1 Conservative Values of Factor C and Exponent m in Equation (6–2)
for Various Forms of Steel (R = σmin∕σmax ≈ 0)
C,
m cycle
C,
Material

Ferritic-pearlitic steels
6.89(10−12)
3.60(10−10)
3.00
Martensitic steels
1.36(10−10)
6.60(10−9)
2.25
Austenitic stainless steels
5.61(10−12)
3.00(10−10)
3.25
MPa m

m

in cycle
kpsi in

m
m
Source: Barsom, J. M. and Rolfe, S. T., Fatigue and Fracture Control in Structures, 2nd ed., Prentice Hall, Upper Saddle
River, NJ, 1987, 288–291.
© McGraw Hill
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Crack Growth 8
Substituting Eq. (6–1) into Eq. (6–2) and integrating,

Nf
0
1 af
da
dN  N f  
C ai   a m


(6 - 3)
ai is the initial crack length.
af is the final crack length corresponding to failure.
Nf is the estimated number of cycles to produce a failure after the
initial crack is formed.
© McGraw Hill
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Example 6–1 (1)
The bar shown in Figure 6–14 is subjected to a repeated moment 0 ≤ M ≤ 1200
lbf · in. The bar is AISI 4430 steel with Sut = 185 kpsi, Sy = 170 kpsi, and
K lc  73 kpsi in. Material tests on various specimens of this material with identical
heat treatment indicate worst-case constants of C  3.8(10 11 ) (in/cycle) (kpsi in ) m
and m = 3.0. As shown, a nick of size 0.004 in has been discovered on the bottom
of the bar. Estimate the number of cycles of life remaining.
Fig. 6–14
© McGraw Hill
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42
Example 6–1 (2)
Solution
The stress range Δσ is always computed by using the nominal (uncracked) area. Thus
I bh 2 0.25  0.5


 0.01042 in 3
c
6
6
Therefore, before the crack initiates, the stress range is
M
1200
 

 115.2 10 3 psi  115.2 kpsi
I c 0.01042
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 Equation (5–37) with KI = KIc, we
approximate af as
2
2
1  K Ic 
1  73 
af  
 
  0.1278 in
   max 
  115.2 
2
 
© McGraw Hill
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Example 6–1 (3)
We compute the ratio af ∕ h as
af
0.1278

 0.256
h
0.5
Access the text alternative for slide images.
© McGraw Hill
Fig. 5–27
44
Example 6–1 (4)
Thus, af ∕ h varies from near zero to approximately 0.256. From
Figure 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
2

1
73
af  
 0.112 in

  1.07 115.2 
Thus, from Equation (6–3), the estimated remaining life is
0.112
1 af
da
1
da
N


Answer f
C ai   a m 3.8 10 11 0.004 1.07 115.2   a  3




© McGraw Hill
 
5.047 103
a



 
3

65
10
cycles
0.004
0.112
45
Strain-Life Method (A bit more detail)
When the stress at a local discontinuity exceeds the elastic limit,
plastic strain occurs.
Fatigue fractures occur in the presence of cyclic plastic strains.
Unlike fracture mechanics approach, strain-life method does not
specifically analyze the crack growth.
Rather, it predicts life based on comparison to experimentally
measured behavior of test specimens.
Assumes life of notched part will equal life of an unnotched
specimen cycled to the same strain levels.
Customarily counts load reversals, which is twice the number of
complete tension-compression cycles.
© McGraw Hill
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Strain-Life Method The Stable Cyclic Hysteresis Loop 1
Cycling between
constant magnitudes of
tensile strain and
compressive strain may
lead to cyclic hardening
or softening (See
Section 2–4).
After a few cycles, the
material settles into a
stable cyclic hysteresis
loop.
Fig. 6–15
Access the text alternative for slide images.
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47
Strain-Life Method The Stable Cyclic Hysteresis Loop 2
The total true strain
amplitude is the sum of
the elastic and plastic
components
  e  p


+
2
2
2
(6 - 6)
A strain-life fatigue test
produces two data
points, elastic and
plastic, for each strain
reversal
Fig. 6–15
© McGraw Hill
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Strain-Life Method Log-Log Plot
The strain amplitudes are plotted on a log-log scale versus the
number of strain reversals 2N.
Both elastic and plastic components have a linear relationship.
Fig. 6–16
Access the text alternative for slide images.
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49
Strain-Life Method Manson-Coffin and Basquin Equations
The plastic and elastic lines are well known and named.
 p
c
Plastic-strain Manson-Coffin equation:
 f   2 N 
2
Elastic-strain Basquin equation:
 e f
b

2 N 
2
E
(6 - 4)
(6 - 5)
Fig. 6–16
© McGraw Hill
50
Strain-Life Method Slopes and Intercepts
Fatigue ductility coefficient ε′f is the ordinate intercept (at 1 reversal, 2N = 1) of the
plastic-strain line. It is approximately equal to the true fracture strain.
Fatigue strength coefficient σ′f is approximately equal to the true fracture strength. σ′f ̸ E
is the ordinate intercept of the elastic-strain line.
Fatigue ductility exponent c is the slope of the plastic-strain line.
Fatigue strength exponent b is the slope of the elastic-strain line.
Fig. 6–16
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Strain-Life Method Cyclic Strain Material Properties
Fatigue ductility coefficient ε′f is the ordinate intercept (at 1 reversal, 2N = 1) of
the plastic-strain line. It is approximately equal to the true fracture strain.
Fatigue strength coefficient σ′f is approximately equal to the true fracture
strength. σ′f ̸ E is the ordinate intercept of the elastic-strain line.
Fatigue ductility exponent c is the slope of the plastic-strain line.
Fatigue strength exponent b is the slope of the elastic-strain line.
These four parameters are considered empirical material properties.
See Table A–23 for representative values.
Cyclic strain testing is carried out to obtain stable cyclic hysteresis
loops, over a broad range of strain amplitudes.
From this, these four material properties are obtained.
© McGraw Hill
52
Strain-Life Method The strain-life relation 1
The total strain amplitude is the sum of the elastic and plastic components.
  e  p


(6 - 6)
2
2
2
Therefore, the total strain amplitude is given by the strain-life relation
  f
b
c

 2 N   f  2 N 
2
E
(6 - 7)
Fig. 6–16
© McGraw Hill
53
Strain-Life Method The strain-life relation 2
The strain-life relation includes both elastic and plastic influences on the fatigue life.
For high strain amplitudes, the strain-life curve approaches the plastic-strain MansonCoffin line.
For low strain amplitudes, the curve approaches the elastic-strain Basquin equation.
The Basquin equation is essentially the same as an elastic stress-life line used in the
stress-life method.
Fig. 6–16
© McGraw Hill
54
Stress-Life Method
The Stress-Life Method relies on studies of test specimens
subjected to controlled cycling between two stress levels, while
counting cycles to ultimate fracture.
Known as constant amplitude loading.
Reasonable model for many real situations, such as rotating
equipment.
Provides a controlled environment to study the nature of fatigue.
© McGraw Hill
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Constant Amplitude Stress Terminology 1
σmin = minimum stress.
σmax = maximum stress.
σm = mean stress, or midrange stress.
σa = alternating stress, or stress amplitude.
σr = stress range.
a 
m 
 max   min
2
 max   min
2
Fig. 6–17a
(6 - 8)
(6 - 9)
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© McGraw Hill
56
Constant Amplitude Stress Terminology 2
Mean stress can be positive or negative.
Alternating stress is always a positive magnitude of fluctuation
around the mean stress.
Fig. 6–17a
© McGraw Hill
57
Constant Amplitude Stress Terminology 3
Special Case: Repeated Stress.
Stress cycles from zero to a maximum.
Fig. 6–17b
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© McGraw Hill
58
Constant Amplitude Stress Terminology 4
Special Case: Completely Reversed Stress.
Stress cycles with equal magnitudes of tension and compression
around a mean stress of zero.
An r subscript (for reversed) may be added to the alternating
component when it is desired to clarify that it is completely
reversed, that is σar.
Fig. 6–17c
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© McGraw Hill
59
Completely Reversed Stress Testing
Most stress-life fatigue testing is done with completely reversed stresses.
Then the modifying effect of nonzero mean stress is considered separately.
A common test machine is R. R. Moore high-speed rotating-beam machine.
Subjects specimen to pure bending with no transverse shear.
Each rotation subjects a stress element on the surface to a completely reversed
bending stress cycle.
Specimen is carefully machined and polished.
Fig. 6–18
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60
The S-N Diagram
Number of cycles to failure at varying stress levels is plotted on log-log scale.
Known as Wöhler curve, or stress-life diagram, or S-N diagram.
Fig. 6–19
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© McGraw Hill
61
Low-Cycle Fatigue
Fatigue failure with less than 1000 cycles is known as low-cycle fatigue, and is often
considered quasi-static.
Yielding usually occurs before fatigue in this zone, minimizing the need for fatigue analysis.
Low-cycle fatigue often includes plastic strain, and is better modeled with strain-life
method.
Fig. 6–19
© McGraw Hill
62
The Endurance Limit
Ferrous metals usually exhibit a bend, or “knee”, in the S-N diagram where it flattens.
The fatigue strength corresponding to the knee is called the endurance limit Se.
Stress levels below Se predict infinite life.
This is an important phenomenon for designers to use.
Fig. 6–19
© McGraw Hill
63
S-N Diagram for Nonferrous Metals
Nonferrous metals and plastics often do not have an endurance limit.
Fatigue strength Sf is reported at a specific number of cycles.
Figure 6–20 shows typical S-N diagram for aluminums.
Fig. 6–20
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64
The Idealized S-N Diagram for Steels 1
For steels, an idealized S-N diagram can be represented by three lines, representing the
median of the failure data.
Fig. 6–21
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65
The Idealized S-N Diagram for Steels 2
Particular attention is given to the line between 103 and 106 cycles, where finite life is
predicted.
Two points are needed: f Sut at 103 cycles, and Se at 106 cycles
Fig. 6–21
© McGraw Hill
66
Estimating the Endurance Limit 1
The endurance limit for steels has been experimentally found to have a reasonably strong
correlation to the ultimate strength
Fig. 6–22
Access the text alternative for slide images.
© McGraw Hill
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.)
67
Estimating the Endurance Limit 2
Simplified estimate of endurance limit for steels for the rotating-beam specimen, S'e
Sut  200 kpsi 1400 MPa 
 0.5 Sut

Se   100 kpsi
Sut  200 kpsi
(6 - 10)
700 MPa
Sut  1400 MPa

Fig. 6–22
© McGraw Hill
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.)
68
Estimating the Fatigue Strength at 103 Cycles
The point f Sut at 103 cycles is
needed.
f is experimentally determined,
or often simply estimated to be
between about 0.8 and 0.9.
For steels, the elastic strain line
in the strain-life approach
indicates f should be related to
Sut as shown in the plot, and
expressed by the curve-fit
equations.
  S  6.9 10  S
f  1.06  4.110  S  1.5 10  S
f  1.06  2.8 10
3
6
2
ut
70  S ut  200 kpsi
7
2
ut
500  S ut  1400 MPa
ut
4
ut
Fig. 6–23
(6 - 11)
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© McGraw Hill
69
Equations for High-Cycle S-N Line 1
Write equation for S-N
line from 103 to 106 cycles
Two known points
At N =103 cycles,
Sf = f Sut
At N =106 cycles,
Sf = Se
Sf is the fatigue strength
correlating to a life N
Fig. 6–21
© McGraw Hill
70
Equations for High-Cycle S-N Line 2
General form of the linear relationship on the log-log scale can be represented by
a power function known as the Basquin equation
S f  aN b
(6 - 12)
a and b are the ordinate intercept and the slope of the line in log-log coordinates.
Fig. 6–21
© McGraw Hill
71
Equations for High-Cycle S-N Line 3
To obtain a and b, substitute into Eq. (6–12) the two known points.
S f  aN b
a
 f Sut 
(6 - 12)
2
Se
 f S ut 
1
b   log 
3
 S e 
(6 - 13)
(6 - 14)
These equations can be used to estimate a fatigue strength Sf
correlating to a life N when 103 < N < 106
© McGraw Hill
72
Equations for High-Cycle S-N Line 4
If a completely reversed stress σar is given, setting Sf = σar in
Eq. (6–12) and solving for N gives,
S f  aN b
 
N   ar 
 a 
(6 - 12)
1b
(6 - 15)
We replace Sf with σar to be very clear that the alternating stress for
the S-N diagram must be completely reversed.
For other stress situations, a completely reversed stress with the
same life expectancy must be used on the S-N diagram.
The effect of mean stress is covered in Section 6-11.
© McGraw Hill
73
Basquin’s Equation
Basquin’s equation is commonly encountered in research literature
as an alternate version of Eq. (6–12).
It is usually expressed in terms of load reversals (two reversals per
cycle).
 ar  f  2 N 
b
(6 - 16)
b is the fatigue strength exponent, and is the slope of the line.
σ'f is the fatigue strength coefficient.
These parameters are empirically determined material properties.
This equation is equivalent to the strain-based version of Eq. (6–5)
used in the strain-life method.
This equation is more accurate, but requires the material parameters
to be obtained.
© McGraw Hill
74
Example 6–2 (1)
Given a 1050 HR steel, estimate
(a) the rotating-beam endurance limit at 106 cycles.
(b) the endurance strength of a polished rotating-beam specimen corresponding to 104 cycles to failure.
(c) the expected life of a polished rotating-beam specimen under a completely reversed stress of 55
kpsi.
Solution
(a) From Table A–20, Sut = 90 kpsi. From Equation (6–10),
Answer
S e  0.5  90  45 kpsi
(b) From Figure 6–23, or Equation (6–11), for Sut = 90 kpsi, f ≈ 0.86. From Equation (6–13),
 0.86  90 
a
 133.1 kpsi
45
2
From Equation (6–14),
Thus, Equation (6–12) is
© McGraw Hill
 0.86  90 
1
b   log 
 0.0785

3
 45 
S f  133.1N 0.0785
75
Example 6–2 (2)
For 104 cycles to failure, Sf = 133.1(104)−0.0785 = 65 kpsi
(c) From Equation (6–15), with σar = 55 kpsi,
Answer
 55 
N 
 133.1
1/ 0.0785
 
 77 500  7.8 10 4 cycles
Keep in mind that these are only estimates, thus the rounding of the
results to fewer significant figures.
© McGraw Hill
76
Endurance Limit Modifying Factors
Endurance limit S'e is for carefully prepared and tested specimen.
If warranted, Se is obtained from testing of actual parts.
When testing of actual parts is not practical, a set of Marin factors are used to
adjust the endurance limit.
S e  k a kb k c k d k e Se
(6 - 17)
where ka = surface factor.
kb = size factor.
kc = load factor.
kd = temperature factor.
ke = reliability factor.
S′e = rotary-beam test specimen endurance limit.
Se = endurance limit at the critical location of a machine part in the
geometry and condition of use.
© McGraw Hill
77
Surface Factor ka 1
Stresses tend to be high at the surface.
Surface finish has an impact on initiation of cracks at localized
stress concentrations, due to plastic strain at the roots of surface
imperfections.
From a practical perspective, surface roughness is difficult to
separate from other stress raisers present due to such things as
metallurgical treatment, cold working, residual stresses from
manufacturing operations.
© McGraw Hill
78
Surface Factor ka 2
Lipson and Noll collected
data from many studies,
organizing them into
several common
commercial surface
finishes.
Clearly, surface effect is
significant.
Higher strengths are more
sensitive to rough
surfaces.
Access the text alternative for slide images.
© McGraw Hill
Fig. 6–24
Generated from data from C. J. Noll and C. Lipson, “Allowable Working Stresses,” Society for Experimental Stress Analysis, vol. 3, no. 2, 1946, p. 29.
79
Surface Factor ka 3
Polished category.
• Matches the test specimen, so by definition has a value of unity.
Ground category.
• Includes ground, honed, and lapped finishes.
• Test data is scattered and limited for higher strengths.
Machined or cold-drawn category.
• Includes rough and finish machining operations.
• Includes unmachined cold-drawn surfaces.
• Test data is limited above 160 kpsi, and is extrapolated.
© McGraw Hill
80
Surface Factor ka 4
Hot-rolled category.
• Represents surfaces typical of hot-rolled manufacturing processes.
• The data includes metallurgical and processing conditions, such as scale
defects, oxide, and partial surface decarburization.
• Not strictly a surface finish factor.
As-forged category.
• Heavily influenced by metallurgical conditions.
• Includes effects of significant decarburization.
• McKelvey and others note that forging processes are significantly improved
since the Lipson and Noll data from the 1940s. They recommend using hotrolled curve even for the as-forged condition.
© McGraw Hill
81
Surface Factor ka 5
The Lipson and Noll curves are only intended to capture the broad
tendencies.
The data came from many studies under a variety of conditions.
In general, the curves are thought to represent the lower bounds of
the spread of the data, and are therefore likely to be conservative
compared to testing of a specific part.
© McGraw Hill
82
Surface Factor ka 6
For convenience, the curves are fitted with a power curve equation.
k a  aS but
(6 - 18)
Factor a
Factor a
Sut , kpsi
Sut , MPa
Exponent b
Ground
1.21
1.38
−0.067
Machined or cold-drawn
2.00
3.04
−0.217
Hot-rolled
11.0
38.6
−0.650
As-forged
12.7
54.9
−0.758
Surface Finish
Table 6–2
© McGraw Hill
83
Example 6–3
A steel has a minimum ultimate strength of 520 MPa and a
machined surface. Estimate ka.
Solution
From Table 6–2, a = 3.04 and b = −0.217. Then, from Equation
(6–18)
Answer
© McGraw Hill
k a  3.04  520
0.217
 0.78
84
Size Factor kb 1
The endurance limit of specimens loaded in bending and torsion has been
observed to decrease slightly as the size increases.
Larger parts have greater surface area at high stress levels, thus a higher
probability of a crack initiating.
Size factor is obtained from experimental data with wide scatter.
For bending and torsion or round rotating bars, the trend of the size factor data is
given by
 d 0.3 0.107  0.879 d 0.107

0.157
0.91d
kb  
0.107
d
7.62
 1.24 d 0.107


1.51d 0.157

0.3  d  2 in
2  d  10 in
7.62  d  51 mm
(6 - 19)
51  d  254 mm
For d less than 0.3 inches (7.62 mm), kb = 1 is recommended.
For axial load, there is no size effect, so kb = 1.
© McGraw Hill
85
Size Factor kb 2
For parts that are not round and rotating, an equivalent round rotating
diameter is obtained.
Equate the volume of material stressed at and above 95% of the
maximum stress to the same volume in the rotating-beam specimen.
Lengths cancel, so equate the areas.
For a rotating round section, the 95% stress area is the area of a ring,
A0.95 
 2
d   0.95d    0.0766 d 2

4
2
(6 - 21)
Equate 95% stress area for other conditions to Eq. (6–21) and solve for d
as the equivalent round rotating diameter
© McGraw Hill
86
Size Factor kb 3
For non-rotating round,
A0.95  0.01046 d 2
(6 - 22)
Equating to Eq. (6–21) and solving for equivalent diameter,
d e  0.370 d
(6 - 23)
Similarly, for rectangular section h x b, A95 = 0.05 hb. Equating to
Eq. (6–21),
d e  0.808  hb 
12
(6 - 24)
Other common cross sections are given in Table 6–3.
© McGraw Hill
87
Size Factor kb 4
Table 6–3
A0.95σ for common
non-rotating
structural shapes
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© McGraw Hill
88
Example 6–4
A steel shaft loaded in bending is 32 mm in diameter, abutting a filleted shoulder 38 mm
in diameter. Estimate the Marin size factor kb if the shaft is used in
(a) A rotating mode.
(b) A nonrotating mode.
Solution
(a) From Equation (6–19)
Answer
 d 
kb  
 7.62 
0.107
 32 

 7.62 
0.107
 0.86
(b) From Table 6–3,
d e  0.37 d  0.37  32   11.84 mm
From Equation (6–19),
Answer
© McGraw Hill
 11.84 
kb  
 7.62 
0.107
 0.95
89
Loading Factor kc
Estimates for endurance limit are typically obtained from testing
with completely reversed loading.
Fatigue tests indicate axial and torsional loading results in different
relationship of endurance limit and ultimate strength.
The loading factor accounts for changes in endurance limit for
different types of fatigue loading.
Only to be used for single load types. Use Combination Loading
method (Sec. 6–16) when more than one load type is present.
bending
1

kc  0.85 axial
0.59 torsion

© McGraw Hill
(6 - 25)
90
Temperature Factor kd 1
Fatigue life predictions can be complicated at temperatures
significantly below or above room temperature, due to complex
interactions between a variety of other time-dependent and
material-dependent processes.
See the discussion on pages 314-316, 11th edition for details.
© McGraw Hill
91
Temperature Factor kd 2
For steels operating in steady
temperatures in the range 20°C
(70°F) to 380°C (720°F), the
primary fatigue life effect is
probably just the temperature effect
on the ultimate strength.
This relation is depicted graphically
in Fig. (2–17).
The ultimate strength relation can
be obtained from curve-fit
polynomials.
 
 
 0.99  5.9 10  T  2.110  T
ST S RT  0.98  3.5 10 4 TF  6.3 10 7 TF2
ST S RT
4
6
C
2
C
(6 - 26)
Fig. 2–17
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© McGraw Hill
92
Temperature Factor kd 3
If ultimate strength is known for operating temperature, then just
use that strength. Let kd = 1 and proceed as usual.
If ultimate strength is known only at room temperature, then use
Eq. 6–26 to estimate ultimate strength at operating temperature.
With that strength, let kd = 1 and proceed as usual.
Alternatively, use ultimate strength at room temperature and apply
temperature factor Eq. (6–27) to the endurance limit.
k d  ST S RT
© McGraw Hill
 6 - 27
93
Example 6–5
A 1035 steel has a tensile strength of 80 kpsi and is to be used for a part that operates in a
steady temperature of 750°F. Estimate the endurance limit at the operating temperature if
(a) only the tensile strength at room temperature is known.
(b) the room-temperature endurance limit for the material is found by test to be (S′e)70° =
39 kpsi.
Solution
(a) Estimate the tensile strength at the operating temperature from Equation (6–26),
 ST S RT  750°  0.98  3.5 10 4  750  6.3 10 7  750 2  0.89
Thus,
 Sut  750°   ST S RT  750°  Sut 70°  0.89  80  71.2 kpsi
From Equation (6–10),

Answer



 Se  750°  0.5  Sut 750°  0.5  71.2  35.6 kpsi
and use kd = 1 since this is already adjusted for the operating temperature.
(b) Since the endurance limit is known at room temperature, apply the temperature factor
to adjust it to the operating temperature. From Equation (6–27),
k d   ST S RT  750°  0.89
Answer
© McGraw Hill
 Se  750°  kd  Se  70°  0.89  39  35 kpsi
94
Reliability Factor ke 1
From Fig. 6–22, S'e = 0.5 Sut is typical of the data and represents 50% reliability.
Reliability factor adjusts to other reliabilities.
Only adjusts Fig. 6–22 assumption. Does not imply overall reliability.
Fig. 6–22
© McGraw Hill
95
Reliability Factor ke 2
Data analysis indicates standard deviations of endurance strengths of less than 8
percent.
Thus the reliability factor to account for this can be written as
ke  1  0.08 z a
(6 - 28)
The transformation variate is defined by Eq. (1–5) and values are available from
Table A–10, with a few values given in Table 6–4.
Or, simply obtain ke for desired reliability from Table 6–4.
Reliability, %
50
90
95
99
99.9
99.99
Transformation Variate za
0
1.288
1.645
2.326
3.091
3.719
Reliability Factor ke
1.000
0.897
0.868
0.814
0.753
0.702
Table 6–4
© McGraw Hill
96
Miscellaneous Effects
Reminder to consider other possible factors.
• Residual stresses
• Directional characteristics from cold working
• Case hardening
• Corrosion
• Surface conditioning, for example, electrolytic plating and metal
spraying
• Cyclic Frequency
• Fretting Corrosion
Limited data is available.
May require research or testing.
More discussion on pages 317-319, 11th edition
© McGraw Hill
97
Example 6–6 (1)
A 1080 hot-rolled steel bar has been machined to a diameter of 1 in. It is to be placed in
reversed axial loading for 70 000 cycles to failure in an operating environment of 650°F.
Using ASTM minimum properties, and a reliability for the endurance limit estimate of 99
percent, estimate the endurance limit and fatigue strength at 70 000 cycles.
Solution
From Table A–20, Sut = 112 kpsi at 70°F. Since the rotating-beam specimen endurance
limit is not known at room temperature, we determine the ultimate strength at the elevated
temperature first, using Equation (6–26),
 ST S RT  650°  0.98  3.5 10 4   650  6.3 10 7   650 2  0.94
The ultimate strength at 650°F is then
 Sut  650°   ST S RT  650°  Sut 70°  0.94 112  105 kpsi
The rotating-beam specimen endurance limit at 650°F is then estimated from Equation
(6–10) as
Se  0.5 105  52.5 kpsi
Next, we determine the Marin factors. For the machined surface, Equation (6–18) with
Table 6–2 gives
0.217
k a  aS but  2.0 105
 0.73
© McGraw Hill
98
Example 6–6 (2)
For axial loading, from Equation (6–20), the size factor kb = 1, and from Equation (6–25) the loading
factor is kc = 0.85. The temperature factor kd = 1, since we accounted for the temperature in modifying
the ultimate strength and consequently the endurance limit. For 99 percent reliability, from Table 6–4,
ke = 0.814. The endurance limit for the part is estimated by Equation (6–17) as
Answer
S e  k a kb kc k d k e Se
 0.73 1 0.851 0.814  52.5  26.5 kpsi
For the fatigue strength at 70 000 cycles we need to construct the S-N equation. From Equation (6–11),
or we could use Figure 6–23,




f  1.06  2.8 10 3 105  6.9 10 6 105  0.84
From Equation (6–13),
a
and Equation (6–14)
 f Sut 
Se
2
2
 0.84 105 

 293.6 kpsi
26.5
2
 fS 
 0.84 105 
1
1
b   log  ut    log 
 0.1741

3
3
 Se 
 26.5 
Finally, for the fatigue strength at 70 000 cycles, Equation (6–12) gives
Answer
© McGraw Hill
S f  a N b  293.6  70 000 
0.1741
 42.1 kpsi
99
Fatigue Stress-Concentration Factor 1
The stress-concentration factor Kt represents the local increase in
stress near a discontinuity, for static loading conditions.
For dynamic loading, it turns out that the fatigue strength of a
notched specimen does not experience the full amount of Kt.
• For discussion of potential explanations, see the last paragraph
on p. 320, 11th edition.
Consequently, for fatigue purposes, a fatigue stress-concentration
factor Kf is defined.
Kf 
© McGraw Hill
Fatigue strength of notch-free specimen
Fatigue strength of notched specimen
(6 - 29)
100
Fatigue Stress-Concentration Factor 2
Kf is a reduced version of Kt, taking into account the sensitivity of
the actual part to the stress concentrating effects in a fatigue
situation.
Kf is used in place of Kt to increase the nominal stress.
 max  K f  0
© McGraw Hill
or
 max  K fs 0
(6 - 30)
101
Notch Sensitivity 1
To quantify the sensitivity of materials to notches, a notch
sensitivity q is defined.
q
K f 1
Kt  1
or
qs 
K fs 1
K ts  1
(6 - 31)
q is between zero and unity.
When the notch sensitivity is zero, Kf = 1, and the material has no
sensitivity to notches.
When the notch sensitivity is one, the Kf = Kt, and the material is
fully sensitive to notches.
Solve Eq. (6–31) for Kf.
K f  1  q  K t  1
© McGraw Hill
or
K fs  1  qs  K ts  1
(6 - 32)
102
Notch Sensitivity 2
Notch sensitivities for specific materials are obtained experimentally.
For steels and aluminum, obtain q for bending or axial loading from Fig. 6–26.
Then get Kf from Eq. (6–32):
Kf = 1 + q( Kt – 1)
Fig. 6–26
Access the text alternative for slide images.
© McGraw Hill
Source: Sines, George and Waisman, J. L. (eds.), Metal Fatigue, McGraw-Hill, New York, 1969.
103
Notch Sensitivity 3
Obtain qs for torsional loading from Fig. 6–27.
Then get Kfs from Eq. (6–32):
Kfs = 1 + qs( Kts – 1)
Fig. 6–27
Access the text alternative for slide images.
© McGraw Hill
104
Notch Sensitivity 4
Alternatively, can use curve fit equations for Figs. 6–26 and 6–27
to get notch sensitivity, or go directly to Kf .
1
q
(6 - 33)
a
1
r
K f  1
Kt  1
1 a
r
(6 - 34)
r is the notch radius
a is a material characteristic length, roughly several times the
microstructure grain size, and can be thought of as near the size of
the material’s natural internal imperfections.
It is often shown in the form of the Neuber constant a
© McGraw Hill
105
Notch Sensitivity 5
For steels, the Neuber constant can be obtained from curve-fit
equations.
Bending or axial:






a  1.24  2.25 10  S  1.60 10  S  4.1110  S
a  0.246  3.08 10 3 S ut  1.51 10 5 Sut2  2.67 10 8 S ut3 50  Sut  250 kpsi
3
6
ut
2
ut
10
3
ut
340  Sut  1700 MPa
(6 - 35)
Torsion:
 
 
 
a  0.958  1.83 10  S  1.43 10  S  4.1110  S 340  S  1500 MPa
a  0.190  2.51 10 3 S ut  1.35 10 5 S ut2  2.67 10 8 S ut3 50  Sut  220 kpsi
3
6
ut
© McGraw Hill
2
ut
10
3
ut
(6 - 36)
ut
106
Notch Sensitivity for Cast Irons
Cast irons are already full of discontinuities, which are included in
the strengths.
Additional notches do not add much additional harm.
Recommended to use q = 0.2 for cast irons.
© McGraw Hill
107
Example 6–7
A steel shaft in bending has an ultimate strength of 690 MPa and a shoulder with a fillet
radius of 3 mm connecting a 32-mm diameter with a 38-mm diameter. Estimate Kf using:
(a) Figure 6–26.
(b) Equations (6–34) and (6–35).
Solution
From Figure A–15–9, using D ∕d = 38 ∕ 32 = 1.1875, r ∕ d = 3 ∕ 32 = 0.093 75, we read the
graph to find Kt = 1.65.
(a) From Figure 6–26, for Sut = 690 MPa and r = 3 mm, q = 0.84. Thus, from Equation
(6–32)
K f  1  q  K t  1  1  0.84 1.65  1  1.55
Answer
(b) From Equation (6–35) with Sut = 690 MPa,
Equation (6–34) with r = 3 mm gives
Answer
© McGraw Hill
K f  1
Kt  1
1 a r
a  0.314 mm. Substituting this into
 1
1.65  1
 1.55
0.314
1
3
108
Example 6–8 (1)
Figure 6–28a shows a rotating shaft simply supported in ball bearings at A and D and
loaded by a nonrotating force F of 6.8 kN. The shaft is machined from AISI 1050 colddrawn steel. Estimate the life of the part.
Solution
From Figure 6–28b 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.
Fig. 6–28
(a) Shaft drawing showing all
dimensions in millimeters; all
fillets 3-mm radius.
(b) Bending-moment diagram.
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© McGraw Hill
109
Example 6–8 (2)
We shall solve the problem by first estimating the strength at point B 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 S′e is estimated as
Se  0.5  690  345 MPa
From Equation (6–18) and Table 6–2,
k a  3.04  690
0.217
 0.74
0.107
 0.86
From Equation (6–19),
kb   32 7.62
Since kc = kd = ke = 1,
© McGraw Hill
S e  0.74  0.86  345  220 MPa
110
Example 6–8 (3)
To find the geometric stress-concentration factor Kt we enter Figure A–15–9 with D ∕d =
38 ∕ 32 = 1.1875 and r ∕d = 3 ∕ 32 = 0.093 75 and read Kt = 1.65. From Equation (6–35a)
with Sut = 690 MPa, a  0.314 mm. Substituting this into Equation (6–34) gives
K f  1
Kt  1
1 a r
 1
1.65  1
1  0.314
3
 1.55
The next step is to estimate the bending stress at point B. The bending moment is
M B  R1 x 
225  6.8
225 F
250 
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,
 
MB
695.5
6
 ar  K f
1.55

10  335.1 10 6 Pa  335.1 MPa
=1.55
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.
© McGraw Hill
111
Example 6–8 (4)
For finite life, we will need to use Equation (6–15). The ultimate strength, Sut
= 690 MPa. From Figure 6–23, f = 0.85. From Equation (6–13)
a
 f Sut 
Se
2
 0.85  690  

 1564 MPa
220
2
and from Equation (6–14)
 f Sut 
 0.85  690 
1
1
b   log 
  log 
 0.1419


3
3
 Se 
 220 
From Equation (6–15),
Answer
© McGraw Hill
  ar 
N 
 a 
1/ b
 335.1

 1564 
1/0.1419
 
 52 103 cycles
112
Characterizing Fluctuating Stresses 1
The S-N diagram is applicable for completely reversed stresses.
Other fluctuating stresses exist.
Sinusoidal loading patterns are common, but not necessary.
Fig. 6–29
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© McGraw Hill
113
Characterizing Fluctuating Stresses 2
Fluctuating stresses can often
be characterized simply by the
minimum and maximum
stresses, σmin and σmax.
Define σm as mean steady
component of stress (sometimes
called midrange stress) and σa
as amplitude of alternating
component of stress.
a 
m 
 max   min
2
 max   min
2
(6 - 8)
(6 - 9)
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114
Characterizing Fluctuating Stresses 3
The stress ratio is defined as
 min
R
 max
(6 - 37)
It has values between –1 and +1, and is commonly used to
represent with a single value the nature of the stress pattern.
R = –1 is completely reversed
R = 1 is steady
© McGraw Hill
115
Application of Kf for Fluctuating Stresses
For fluctuating loads at points with stress concentration, the best
approach when using the stress-life approach is to design to avoid
all localized plastic strain.
In this case, Kf should be applied to both alternating and midrange
stress components.
© McGraw Hill
 a  K f  a0
(6 - 38)
 m  K f  m0
(6 - 39)
116
Characteristic Family of S-N curves
It is possible to generate S-N diagrams with increasing levels of mean
stress.
Fig. 6–30
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© McGraw Hill
117
Fluctuating-Stress Diagram 1
Historically, there have been many ways of plotting the data for
general fluctuating stress.
Includes Goodman diagram, modified Goodman diagram, master
fatigue diagram, and Haigh diagram.
Probably most common and simple to use is the plot of σa versus σm
which we shall call the fluctuating-stress diagram.
© McGraw Hill
118
Fluctuating-Stress Diagram 2
From the family of S-N curves in Fig. 6–30, take sets of points correlating to the
same value of life.
With these points, plot constant-life curves on a fluctuating stress diagram
(Fig. 6–31).
Fig. 6–30
© McGraw Hill
Fig. 6–31
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119
Fluctuating-Stress Diagram 3
Now, from Fig. 6–31, focus on the data points correlating to 106 cycles.
This is shown in Fig. 6–32 with many more data points to indicate the scatter of
data.
Fig. 6–32
Fig. 6–31
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120
Fluctuating-Stress Diagram 4
With steels, with an idealized assumption that the endurance limit corresponds to a life of
106 cycles, these data points represent the boundary between finite life and infinite life.
The modified-Goodman line, or simply Goodman line, between Se and Sut represents a
conservative boundary for infinite life.
The equation for the Goodman line is
a
Se

m
Sut
1
(6 - 40)
Fig. 6–32
© McGraw Hill
121
Fluctuating-Stress Diagram 5
Assuming the stress at point A
would increase along the load line
from the origin (for example, the
ratio of σa /σm remains constant), a
factor of safety with respect to
infinite life can be defined.
nf = strength/stress = OB/OA
Or, applying a design factor to the
stresses, and solving for the
factor,
n    n    1
f
Se
a
f
m
© McGraw Hill
(a )
Sut

 
nf   a  m 
 S e Sut 
Fig. 6–32
1
m  0
(6 - 41)
122
Fluctuating-Stress Diagram 6
Experimental data on normalized plot of σa versus σm.
Demonstrates little detrimental effect of negative mean stress.
Fatigue factor of safety for negative mean stress, based on horizontal line is
S
nf  e
m  0
(6 - 42)
a
Fig. 6–33
Access the text alternative for slide images.
© McGraw Hill
Data source: Thomas J. Dolan, “Stress Range,” Section 6.2 in O. J. Horger (ed.), ASME Handbook—Metals Engineering Design, McGraw-Hill, New York, 1953.
123
Fluctuating-Stress Diagram 7
To consider first-cycle yielding on the fluctuating-stress diagram,
Sy
Sy
ny 

(6 - 43)
 max  a   m
The absolute value allows the equation to be used for both positive
and negative mean stress.
It is helpful to plot the yield condition on the fluctuating stressdiagram to compare to the fatigue criterion.
Setting ny = 1, results in a linear equation representing two lines,
known as Langer lines.
a  m  Sy
© McGraw Hill
124
Fluctuating-Stress Diagram 8
Plotting the two fatigue lines and two yield lines defines a design space
with zones for infinite life, finite life, and first-cycle yielding.
Access the text alternative for slide images.
© McGraw Hill
Fig. 6–34
125
Example 6–9 (1)
A steel bar undergoes cyclic loading such that at the critical notch location the nominal stress
cycles between σmax = 40 kpsi and σmin = 20 kpsi, and a fatigue stress-concentration factor is
applicable with Kf = 1.2. For the material, Sut = 100 kpsi, Sy = 85 kpsi, and a fully corrected
endurance limit of Se = 40 kpsi. Estimate
(a) the fatigue factor of safety based on achieving infinite life according to the Goodman line.
(b) the yielding factor of safety.
Solution
(a) From Equations (6–8) and (6–9),
40  20
40  20
 a0 
 10 kpsi
 m0 
 30 kpsi
2
2
Applying Equations (6–38) and (6–39),
 a  K f  a 0  1.2 10  12 kpsi
 m  K f  m 0  1.2  30  36 kpsi
For a positive mean stress, apply Equation (6–41),
Answer
Infinite life is predicted.
© McGraw Hill

 
nf   a  m 
S 
S
e
ut
1
 12 36 


 40 100 
1
 1.52
126
Example 6–9 (2)
(b) To avoid even localized yielding at the notch, keep Kf applied to the stresses for the yield
check. Using Equation (6–43),
Sy
85
ny 

 1.8
Answer
 a   m 12  36
No yielding is predicted at the notch at the first stress cycle. Of course, realize that with
continued cycling, at the grain level the cyclic stress will eventually lead to very localized
plastic strain (see Section 6–3). If there were truly no plastic strain, there would be no fatigue.
Fig. 6–35
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© McGraw Hill
127
Example 6–10 (1)
Repeat Example 6–9, except for a nominal stress that cycles between σmax = 60 kpsi and σmin = −20 kpsi.
Solution
60   20
(a) Equations (6–8), (6–9):
 a0 
Equations (6–38), (6–39):
 a  K f  a 0  1.2  40  48 kpsi
Answer

 
nf   a  m 
S 
S
Equation (6–41):
2
e
ut
 40 kpsi
1
 m0 
 48 24 


 40 100 
60   20
 20 kpsi
2
 m  K f  m 0  1.2  20   24 kpsi
1
 0.69
Infinite life is not predicted. In Example 6–15 this problem will be revisited to estimate the predicted
finite life.
Fig. 6–35
© McGraw Hill
128
Example 6–10 (2)
Sy
85
 1.2
 a   m 48  24
No yielding is predicted at the notch at the first stress cycle. The stress point,
fatigue line intercept, and yield line intercept are plotted as A′, B′, and C′,
respectively, on the fluctuating-stress diagram of Figure 6–35.
(b) Equation (6–43):
ny 

Fig. 6–35
© McGraw Hill
129
Example 6–11 (1)
Repeat Example 6–9, except for a nominal stress that cycles between σmax = −20 kpsi and σmin = −40
kpsi,
Solution
(a) Equations (6–8), (6–9):
Equations (6–38), (6–39):
 a0 
20   40
 10 kpsi
2
 a  k f  a 0  1.2 10  12 kpsi
 m0 
20   40
2
 30 kpsi
 m  k f  m 0  1.2  30   36 kpsi
For a negative mean stress, apply Equation (6–42),
S e 40
n


 3.3
f
Answer
 a 12
Infinite life is predicted, but with a factor of safety more than double the similar problem in Example
6–9, with the only difference being the negative mean stress.
Fig. 6–35
© McGraw Hill
130
Example 6–11 (2)
(b) Equation (6–43):
ny 
Sy
a  m

85
 1.8
12  36
This is the same as in Example 6–9, though it is with regard to compressive
yielding in this case. The stress point, fatigue line intercept, and yield line
intercept are plotted as A″, B″, and C″, respectively, on the fluctuating-stress
diagram of Figure 6–35. Note that the load lines for fatigue and yielding are
different this time.
Fig. 6–35
© McGraw Hill
131
Fatigue Failure Criteria 1
Several fatigue failure criteria that are well known, each providing
options for various purposes.
Fig. 6–36
Access the text alternative for slide images.
© McGraw Hill
132
Fatigue Failure Criteria 2
Goodman.
• Simple, linear.
• To the conservative side
of the data, so good for
design purposes, but not
typical of the data.
• Only for positive mean
stress.
Failure criterion:
Design equation:
© McGraw Hill
a
Se

m
Sut
Fig. 6–36
1

 
nf   a  m 
 S e Sut 
(6 - 40)
1
m  0
(6 - 41)
133
Fatigue Failure Criteria 3
Morrow.
• Replaces Sut with true fracture
strength or the fatigue strength
coefficient, which are not
always readily available.
• Simple, linear.
• More typical of data than
Goodman.
• Reasonable fit of data for both
positive and negative mean
stress.
Failure criterion:
Design equation:
© McGraw Hill
a
Se

Fig. 6–36
m
 1 or
f
 a m 
nf  


S
 e f 
a
Se

m
1
f 
1
or
 a m 
nf  


S
 e f 
(6 - 45)
1
(6 - 46)
134
Fatigue Failure Criteria 4
Morrow.
• For steels (HB<500) a very crude estimate of the fatigue
strength coefficient is given by SAE as
 f  Sut  50 kpsi
© McGraw Hill
or
f  Sut  345 MPa
(6 - 44)
135
Fatigue Failure Criteria 5
Gerber.
• Parabolic.
• Historically known to provide
typical curve through the data,
though other curves actually
fit better.
• Tends to be non-conservative
near the ordinate axis.
• Only for positive mean stress.
 
  m  1
S e  Sut 
2
2 

 2 m S e 
1  Sut    a  

Design equation: n f  
1  1  




2   m   Se  
 Sut  a  


Failure criterion:
© McGraw Hill
a
Fig. 6–36
2
(6 - 47)
m  0
(6 - 48)
136
Fatigue Failure Criteria 6
Soderberg.
• Replaces Sut in Goodman
with Sy.
• Simple, ultra
conservative.
• Provides a simple check
for fatigue and yielding
with a single criterion.
Failure criterion:
a
Se

m
Sy
Fig. 6–36
1
 a m 
Design equation: n  


S
 e Sy 
© McGraw Hill
(6 - 49)
1
m  0
(6 - 50)
137
Fatigue Failure Criteria 7
ASME-Elliptic.
• Elliptic equation.
• Mixes qualities of Gerber and
Soderberg, that is, fit fatigue
data and check yielding.
• Sometimes conservative,
sometimes not.
• Primary recognition is by
ASME standard for
transmission shafting.
Failure criterion:
   2    2 
Design equation: n f   a    m  
 S e   S y  


© McGraw Hill
Fig. 6–36
2
 a   m 
 S    S   1
 y
e
2
(6 - 51)
1 2
m  0
(6 - 52)
138
Fatigue Failure Criteria 8
Smith-Watson-Topper (SWT).
• Relatively more recent
(1970s).
• Gained traction as a good
criterion, based on theory
rather than simply fitting the
data.
• Primarily known in the plastic
strain method, but can be put
into terms of stress.
Fig. 6–36
Failure criterion:
Design equation: n f 
© McGraw Hill
S e   max a 
Se
 m   a   a
 m   a   a
(6 - 53)
(6 - 54)
139
Fatigue Failure Criteria 9
Smith-Watson-Topper (SWT).
• Theorizes that the critical
parameters are σmax and σa .
• Not a function of any strength,
so its curve does not intersect
the mean stress abscissa.
• Acceptable for positive and
negative mean stress, with
range limited by the yield line.
Fig. 6–36
Failure criterion:
Design equation: n f 
© McGraw Hill
S e   max a 
Se
 m   a   a
 m   a   a
(6 - 53)
(6 - 54)
140
Fatigue Failure Criteria 10
Smith-Watson-Topper (SWT).
• Commonly used for predicting
an equivalent completely
reversed stress for a
fluctuating-stress state that
does not predict infinite life.
• Not usually used to predict
fatigue factor of safety.
• Particularly good fit for
aluminum; usually reasonable
for steel.
Failure criterion:
Design equation: n f 
© McGraw Hill
S e   max a 
Se
 m   a   a
Fig. 6–36
 m   a   a
(6 - 53)
(6 - 54)
141
Fatigue Failure Criteria 11
Walker.
•
Generalized version of SWT in
which the square root is replaced
by fitting parameter γ .
•
γ = 0.5 for special case of SWT.
•
γ is determined by experiment for
each material by testing at
multiple values of mean stress.
•
γ essentially shifts weighting
between σmax and σa to better fit
behavior of each material.
Failure criterion:
Design equation: n f 
© McGraw Hill
Fig. 6–36
1 
S e   max
 a   m   a 
1 
Se
 m   a 
1 

a
 a
(6 - 55)
(6 - 56)
142
Fatigue Failure Criteria 12
Walker
• For steels, an approximate relationship between γ and Sut is
experimentally found to be
  0.0002 Sut  0.8818
  0.0014 Sut  0.8818
© McGraw Hill
 Sut in MPa 
 Sut in kpsi 
(6 - 57)
143
Application to a Pure Shear Case
If the fluctuating stresses are entirely shear stresses, the fluctuating-stress
diagram can be adapted with the following adjustments:
• Replace normal stress σm and σa with shear stresses τm and τa.
• Apply the load factor kc = 0.59 to the endurance limit.
• Replace Sy with Ssy = 0.577 Sy, based on the relationship predicted by
the distortion energy theory.
• Replace Sut with Ssu.
For most materials, Ssu ranges from 65 to 80 percent of ultimate strength.
Lacking specific information, use the conservative estimate
S su  0.67 Sut ,
(6 - 58)
Alternatively, pure shear cases can also use the methods of Section 6–16.
© McGraw Hill
144
Example 6–12 (1)
For the part shown in Figure 6–37, the 3-in diameter end is firmly clamped. A force F is
repeatedly applied to deflect the tip until it touches the rigid stop, then released. The part
is machined from AISI 4130 quenched and tempered to a hardness of approximately 250
HB. Use Table A–23 for material properties. Estimate the fatigue factor of safety based on
achieving infinite life, using each of the following criteria. Compare the results.
(a) Goodman
(b) Morrow
(c) Gerber
Fig. 6–37
Access the text alternative for slide images.
© McGraw Hill
145
Example 6–12 (2)
The critical stress location is readily identified as at the fillet radius, on the bottom, where it
experiences repeated bending stress in tension. We shall first find the fully modified endurance limit,
then the stresses. From Table A–23, the closest material option has Sut = 130 kpsi.
k a  a  Sut   2.0 130
b
0.217
 0.70
Equation (6–18):
Machined
Equation (6–23):
Nonrotating round d e  0.37 d  0.37 1  0.37
Equation (6–19):
kb  0.879 d 0.107  0.879  0.37 
0.107
 0.98
Equations (6–10) and (6–17): S e   0.70 0.98 0.5130   45 kpsi
4
I   d 4 64   1 64  0.04909 in 4
3  30 10 6   0.04909 
3 EI
Fmax  ymax 3  0.125
 135 lbf
Table A–9–1:
l
163
Fmin  0
 max  My I  135 16 0.5 0.04909  22.0 kpsi
Equations (6–8), (6–9):  m 0   max   min  22.0 2  11.0 kpsi   a 0
2
Figure A–15–9:
r ∕ d = 0.1, D ∕ d = 3 ∕ 1 = 3, Kt = 1.8
Figure 6–26 or Equation (6–34): q = 0.9
Equation (6–32): K f  1  q  K t  1  1  0.9 1.8  1  1.7
Equations (6–38), (6–39):  m   a  K f  0  1.7 11  18.7 kpsi
© McGraw Hill
146
Example 6–12 (3)
(a) Goodman
Answer Equation (6–41):
 a m 
nf  

S 
S
e
1
ut
 18.7 18.7 


 45 130 
1
 1.8
(b) Morrow
From Table A–23, σ′f = 185 kpsi. Note that if σ′f had not been available for this
material, the estimate for steel in Equation 6–44 would have predicted a value of
180, which would have been quite acceptable to use.
Answer Equation (6–46):
(c) Gerber
Equation (6–48):
Answer
© McGraw Hill

 
nf   a  m 
 S e f 
1
 18.7 18.7 


 45 185 
1
 1.9
2
2 







S

2

S
1
n f   ut   a   1  1   m e  
2   m   Se  
 Sut  a  


2 
2

 2(18.7)(45) 
1  130   18.7  
  2.2
 

1

1






2  18.7   45  
 130(18.7)  


147
Example 6–12 (4)
A criterion that predicts a lower factor of safety is predicting that the
stress is closer to failure, which sends a message to be careful. From a
design perspective, then, a predicted lower factor of safety is more
conservative. Ranking the results in order of most conservative to least
conservative, gives
Goodman
1.8
Morrow
1.9
Gerber
2.2
Because we had a tabulated value for σ′f for the material, the Morrow
result is probably the most typical of reality. As expected, Goodman is on
the conservative side. Gerber is sometimes typical, but for lower mean
stress it has a tendency to be a bit nonconservative.
© McGraw Hill
148
Example 6–13 (1)
For the problem in Example 6–10, estimate the infinite-life fatigue factor of safety for
each of the failure criteria defined in this section. Compare the results.
Solution
From Example 6–10,  a  48 kpsi,  m  24 kpsi, Sut  100 kpsi, and S e  40 kpsi.
Goodman:
1
1
 a m 
 48 24 
Equation (6–41) n f  



  0.69


40 100
 S e S ut 
Morrow:
Lacking specific material properties, use the estimate for steel in Equation (6–44),
f   Sut  50 kpsi  150 kpsi
Equation (6–46)
Gerber:
Equation (6–48)
© McGraw Hill

 
nf   a  m 
 S e f  
1
 48 24 


 40 150 
1
 0.74
2
2 

 2 m S e 
1  S ut    a  

nf  
1  1  




2   m   Se  
 S ut a  


2 
2

 2(24)(40) 
1  100   48  
  0.80
 
   1  1  


2 24
40 
 100(48)  


149
Example 6–13 (2)
Soderberg:
Equation (6–50)
 a m 
nf  


S
 e Sy 
1
 48 24 


 40 85 
ASME-Elliptic:
      
n f   a    m  
 S e   S y  


2
Equation (6–52)
SWT:
Equation (6–54)
nf 
Se
 m   a   a
2

1 2
1
 0.67
 48  2  24  2 
      
 40   85  
40
 24  48 48
1 2
 0.81
 0.68
Walker:
Lacking specific material test data, use the estimate in Equation (6–57),
  0.0014 Sut  0.8818  0.0014 100   0.8818  0.74
Se
40

 0.75
Equation (6–56) n f 
1 
1 0.74
0.74

 m   a   a  24  48 48
© McGraw Hill
150
Example 6–13 (3)
For comparison, sort all of the results in order of most conservative to least conservative.
Soderberg
SWT
Goodman
Morrow
Walker
Gerber
ASME-Elliptic
0.67
0.68
0.69
0.74
0.75
0.80
0.81
Probably, Walker and Morrow are the most accurate. Gerber and ASME-Elliptic are
nonconservative, which they tend to be with low mean stress. Goodman is conservative as
expected. SWT is about the same as Goodman, not being a particularly good match for
the fitting parameter γ needed for this material (as estimated by Walker). Soderberg is, as
always, conservative, though not as much for low mean stress where its line is not much
different from Goodman. Which is best? It depends on the goal. Probably, in this case,
Walker or Morrow were equally good for a result typical of the data, and Goodman is a
good choice for a reasonably predictable amount of conservativeness.
© McGraw Hill
151
Example 6–14 (1)
A flat-leaf spring is used to retain an oscillating flat-faced follower in contact with a plate cam. The
follower range of motion is 2 in and fixed, so the alternating component of force, bending moment,
and stress are 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 14 in thick, as seen in Figure 6–38a. 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 failure criterion curve with the load line.
(b) What are the strength factors of safety corresponding to 2 in and 5 in preload?
Fig. 6–38a
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© McGraw Hill
152
Example 6–14 (2)
Solution
A unique aspect of this problem is that due to the nature of the cam’s motion, the
alternating force is very defined. The preload can change the mean stress. An
appropriate load line is a horizontal line to reflect a steady value of alternating
stress. We begin with preliminaries. The second area moment of the cantilever
cross section is
3
3
2
0.25
bh
   0.00260 in 4
I

12
12
Since, from Table A–9, beam 1, force F and deflection y in a cantilever are
related by F = 3EI y ∕ l 3, then stress σ and deflection y are related by
Mc 32 Fc 32  3 EIy  c 96 Ecy




 Ky
I
I
l3
I
l3


96 Ec 96 30 10 0.125
3
where K  3 

10.99
10
psi/in  10.99 kpsi/in
3
l
32
© McGraw Hill
6
 
153
Example 6–14 (3)
Now the minimums and maximums of y and σ can be defined by
ymin  
ymax  2  
 min  K 
 max  K  2   
The stress components are thus
a 
m 
© McGraw Hill
K  2     K
2
 K  10.99 kpsi
K  2     K
2
 K 1     10.99 1   
For   0,
 a   m  10.99  11 kpsi
For   2 in,
 a  11 kpsi,
 m  10.99 1  2   33 kpsi
For   5 in,
 a  11 kpsi,
 m  10.99 1  5  65.9 kpsi
154
Example 6–14 (4)
(a) A plot of the Gerber criterion is shown in Figure 6–38b. 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 design equation of Equation
(6–48) was derived for the load line from the origin, so it is not applicable in this case. The
intersection point (Sm, Sa) between the Gerber line and the load line is found from solving
Equation (6–47) for Sm and substituting 11 kpsi for Sa:
S m  Sut 1 
Sa
11
 150 1 
 116.9 kpsi
Se
28
Fig. 6–38b
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© McGraw Hill
155
Example 6–14 (5)
(b) The factor of safety is found as the proportion of the distance along the load
line toward the failure point that the stress point has come. For δ = 2 in,
Answer
nf 
and for δ = 5 in,
Answer
nf 
Sm
m

116.9
 3.54
33
116.9
 1.77
65.9
The problem statement didn’t ask for it, but yielding should also be checked in a
similar fashion, using the load line and the Langer yield line.
Fig. 6–38b
© McGraw Hill
156
Constant-Life Curves 1
When a fluctuating stress is predicted
to have finite life, it is often desirable
to estimate the life.
The concept of constant-life curves
was previously introduced with Fig.
6–31.
A constant-life curve that passes
through a finite-life point of interest
will have an ordinate intercept which
is a completely reversed stress with
the same predicted life.
This equivalent completely reversed
stress can be evaluated on an S-N
diagram to estimate the life.
Fig. 6–31
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© McGraw Hill
157
Constant-Life Curves 2
Any of the fatigue failure
criterion can be adapted to
model a constant-life curve.
The better fatigue failure
criteria for this purpose will be
the ones that are typical of the
data, and that don’t attempt to
check yielding simultaneously.
In particular, good options are
Goodman, Morrow, SWT, and
Walker.
© McGraw Hill
Fig. 6–31
158
Constant-Life Curves 3
Goodman.
• The criterion is conservative with respect to the data, so it is
expectedly inaccurate in predicting an equivalent completely reversed
stress.
• Its inaccuracy is in the very conservative direction, in that it predicts a
higher equivalent completely reversed stress than is validated by
experiment.
• The criterion is historically commonly used, and is acceptable when
excessive design conservatism is warranted.
• In the equation for the Goodman line, Eq. (6–40), substituting the
equivalent completely reversed stress for the endurance limit gives
 ar 
© McGraw Hill
a
1   m Sut
(6 - 59)
159
Constant-Life Curves 4
Morrow.
• The criterion is generally more typical of the data than Goodman.
• Its weakness is that the material properties are not always readily
available.
• The SAE estimate can be used for steels.
f  Sut  50 kpsi
or f  Sut  345 MPa
(6 - 44)
• Even with the estimate, the Morrow criterion is probably as good or
better than Goodman for estimating an equivalent completely reversed
stress.
• The equivalent completely reversed stress is obtained from Eq. (6–45).
 ar 
© McGraw Hill
a
1 m  f
or  ar 
a
1   m f
(6 - 60)
160
Constant-Life Curves 5
Smith-Watson-Topper (SWT).
• This criterion is very good for aluminum and usually reasonably
good for steels.
• An advantage is only needing the two stresses, and no strengths.
• From Eq. (6–53),
 ar   max a 
© McGraw Hill
 m   a   a
(6 - 61)
161
Constant-Life Curves 6
Walker.
• This criterion is considered the best match to experimental predictions
when the material fitting parameter  is known.
• From Eq. (6–55),

 ar   1max
 a  ( m   a )1  a
(6 - 62)
• For aluminum,  is close to 0.5, which is equivalent to SWT.
• For steels, the estimate can be used:
  0.0002 Sut  0.8818  Sut in MPa 
  0.0014 Sut  0.8818  Sut in kpsi 
(6 - 57)
• With this, the Walker method is very usable for steels and is as good or
better than any other method.
© McGraw Hill
162
Example 6–15 (1)
For the problem defined in Example 6–10 and extended in Example 6–13, all of the fatigue
criteria predicted finite life. For the fatigue criteria of Goodman, Morrow, SWT, and
Walker, estimate the equivalent completely reversed stress and the predicted life. Compare
the results.
Solution
 a  48 kpsi,  m  24 kpsi, Sut  100 kpsi, and S e  40 kpsi.
From Example 6–10,
a
48

 63 kpsi
Equation (6–59): Goodman  ar 
1   m Sut 1  24 100
Equation (6–44):
f  Sut  50 kpsi  150 kpsi
Equation (6–60): Morrow
 ar 
a
48

 57 kpsi
1   m f 1  24 150
Equation (6–61): SWT
 ar 
 m   a   a   24  48 48  59 kpsi
From Example 6–13, γ = 0.74
Equation (6–62): Walker
© McGraw Hill

 ar   1max
 a   m   a 
1
 a   24  48
1 0.74
480.74  53 kpsi
163
Example 6–15 (2)
Use the S-N diagram equations with these equivalent completely reversed stresses to estimate the life
based on each criterion.
Figure 6–23:
f  0.84
 f Sut    0.84100   176.4
a
2
Equation (6–13):
Equation (6–14):
Equation (6–15):
2
Se
40
 fS 
  0.84100 
1
1
b   log  ut    log 
  0.1074
3
3
40

 Se 
 
N   ar 
 a 
1b
Calculate the estimated life for each criterion. Results are reported in order of lowest life to highest life.
σar = 63 kpsi
N = 15 000 cycles
Answer Goodman
SWT
σar = 59 kpsi
N = 27 000 cycles
Morrow
σar = 57 kpsi
N = 37 000 cycles
Walker
σar = 53 kpsi
N = 73 000 cycles
A higher equivalent completely reversed stress on an S-N diagram will predict a shorter life. From a
design perspective, a prediction of a shorter life is more conservative. The Walker estimate is
expected to be the most typical of reality. The Morrow result is an improvement on Goodman, even
with just the estimate for σ′f . As expected, Goodman is conservative. Though Goodman is in the
correct order of magnitude, it is substantially less than the presumably better value from Walker.
© McGraw Hill
164
Fatigue Failure Criterion for Brittle Materials
For many brittle materials, the first quadrant fatigue failure criteria
follows a concave upward Smith-Dolan locus,
S a 1  S m S ut

S e 1  S m S ut
(6 - 63a )
Or as a design equation,
n a 1  n m Sut

Se
1  n m S ut
(6 - 63b )
For a radial load line of slope r, the intersection point is
rS ut  S e 
4 rS ut S e 

Sa 
  1  1 
2
2

 rSut  Se  
(6 - 64)
In the second quadrant,
S

S a  S e   e  1 S m
 Sut 
© McGraw Hill
 S ut  S m  0
 for cast iron 
(6 - 65)
165
Fatigue Criteria for Brittle Materials
Table A–24 gives properties of gray cast iron, including endurance
limit.
The endurance limit already includes ka and kb .
The average kc for axial and torsional is 0.9.
© McGraw Hill
166
Example 6–16 (1)
A grade 30 gray cast iron is subjected to a load F applied to a 1 by 83 -in cross-section link
with a 14 -in-diameter hole drilled in the center as depicted in Figure 6–39a. The surfaces
are machined. In the neighborhood of the hole, what is the factor of safety guarding
against failure under the following conditions:
(a) The load F = 1000 lbf tensile, steady.
(b) The load is 1000 lbf repeatedly applied.
(c) The load fluctuates between −1000 lbf and
300 lbf without column action.
Use the Smith-Dolan fatigue locus.
Fig. 6–39a
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© McGraw Hill
167
Example 6–16 (2)
Solution
Some preparatory work is needed. From Table A–24, Sut = 31 kpsi, Suc = 109 kpsi, kakbS′e = 14 kpsi.
Since kc for axial loading is 0.9, then Se = (kakbS′e )kc = 14(0.9) = 12.6 kpsi. From Table A–15–1,
A = t(w − d) = 0.375(1 − 0.25) = 0.281 in2, d ∕ w = 0.25 ∕ 1 = 0.25, and Kt = 2.45. The notch sensitivity for
cast iron is 0.20 (see Section 6–10), so
Kf = 1 + q(Kt − 1) = 1 + 0.20(2.45 − 1) = 1.29
(a) Since the load is steady, σa = 0, the load is static. Based on the discussion of cast iron in Section 5–2,
Kt, and consequently Kf, need not be applied. Thus, σm = Fm ∕A = 1000(10−3) ∕ 0.281 = 3.56 kpsi, and
n
Answer
Sut
m

31.0
 8.71
3.56
Fig. 6–39a
© McGraw Hill
168
Example 6–16 (3)
(b)
F 1000

 500 lbf
2
2
K f Fa 1.29  500
a  m 

10 3  2.30 kpsi
A
0.281
Fa  Fm 


a
r
1
m
From Equation (6–64),
1 31  12.6 
4 1 3112.6  

 1  1 
  7.63 kpsi
Sa 
2
2

1 31  12.6  

Answer
© McGraw Hill
n
Sa
a

7.63
 3.32
2.30
169
Example 6–16 (4)
(c)
1
300   1000  650 lbf
2
1
Fm  300   1000   350 lbf
2
a 
Fa 
r
m 
1.29  650
10   2.98 kpsi

0.281
1.29  350
0.281
3
10   1.61 kpsi
3
a
3.0

 1.86
 m 1.61
From Equation (6–65), S a  Se   Se Sut  1 S m and S m  S a r . It follows that
Sa 
Answer
© McGraw Hill
Se
12.6

 18.5 kpsi
1
12.6



1 S
 1
1   e  1 1 


1.86 31
r  Sut 
n
Sa
a

18.5
 6.20
2.98
170
Example 6–16 (5)
Figure 6–39b shows the portion of the designer’s fatigue diagram that was constructed.
Fig. 6–39b
Access the text alternative for slide images.
© McGraw Hill
171
Combinations of Loading Modes 1
When more than one type of loading (bending, axial, torsion)
exists, use the Distortion Energy theory to combine them.
Obtain von Mises stresses for both mean and alternating
components.
Apply appropriate Kf to each type of stress.
For load factor, use kc = 1. The torsional load factor (kc = 0.59) is
inherently included in the von Mises equations.
If needed, axial load factor can be divided into the axial stress.
© McGraw Hill
172
Combinations of Loading Modes 2
For case of a shaft with bending stresses, torsional shear stresses,
and axial stress, the von Mises stress is of the form

    x2  3 xy2

12
The von Mises stresses for alternating and mean stress elements are
12
2
2




a    K f


K


3
K

 a 0  bending
f axial  a 0  axial 
bending
 fs torsion  a 0  torsion  


 
 
 
(6 - 66)
12
2
2




 m    K f
 m 0  bending  K f axial  m 0 axial   3  K fs torsion  m 0  torsion  
bending



 
© McGraw Hill
 
 
(6 - 67)
173
Example 6–17 (1)
A shaft is made of 42- × 4-mm AISI 1018 cold-drawn steel tubing and has a 6-mm-diameter hole
drilled transversely through it. Estimate the factor of safety guarding against fatigue and static
failures using the Goodman and Langer failure criteria for the following loading conditions:
(a) The shaft is rotating and is subjected to a completely reversed torque of 120 N · m in phase with
a completely reversed bending moment of 150 N · m.
(b) The shaft is nonrotating and is subjected to a pulsating torque fluctuating from 20 to 160 N · m
and a steady bending moment of 150 N · m.
Solution
Here we follow the procedure of estimating the strengths and then the stresses, followed by relating
the two.
From Table A–20 we find the minimum strengths to be Sut = 440 MPa and Sy = 370 MPa. The
endurance limit of the rotating-beam specimen is 0.5(440) = 220 MPa. The surface factor, obtained
from Equation (6–18) and Table 6–2, is
0.217
k a  3.04 S ut0.217  3.04  440 
 0.81
From Equation (6–19) the size factor is
0.107
0.107
 d 
 42 
kb  

 0.83
 7.62 
 7.62 
The remaining Marin factors are all unity, so the modified endurance strength Se is
S e  0.81 0.83 220  148 MPa
© McGraw Hill
174
Example 6–17 (2)
(a) Theoretical stress-concentration factors are found from Table A–16. Using a ∕D = 6∕42 = 0.143
and d ∕D = 34∕42 = 0.810, and using linear interpolation, we obtain A = 0.798 and Kt = 2.37 for
bending; and A = 0.89 and Kts = 1.75 for torsion. Thus, for bending,
Z net 
A

32 D

D4  d 4 
  0.798 
32  42
42   34   3.3110 3  mm 3



4
4
and for torsion
J net 
   42  34   155 10 mm
D d 
   

 
32
32 
A
4
4
 0.89
4
4
3
4
Next, using Figures 6–26 and 6–27, with a notch radius of 3 mm we find the notch sensitivities to be
0.78 for bending and 0.81 for torsion. The two corresponding fatigue stress concentration factors are
obtained from Equation (6–32) as
K f  1  q  K t  1  1  0.78  2.37  1  2.07
K fs  1  0.811.75  1  1.61
The alternating bending stress is now found to be
 xa  K f
© McGraw Hill
 
M
150
 2.07
 93.8 10 6 Pa  93.8 MPa
6
Z net
3.31 10


175
Example 6–17 (3)
and the alternating torsional stress is
 
 
120  42 10 3
TD
 xya  K fs
 1.61
 26.2 10 6 Pa  26.2 MPa
9
2 J net
2 155 10
 
The mean von Mises component σ′m is zero. The alternating component σ′a is given by

a    3
2
xa
2
xya

12


12
 93.82  3 26.2 2 
 104 MPa
For this completely reversed loading, the fatigue factor of safety nf is
S e 148
n


 1.42
f
Answer
a 104
The first-cycle yield factor of safety is
Answer
ny 
Sy
a

370
 3.56
104
There is no localized yielding; the threat is from fatigue.
© McGraw Hill
176
Example 6–17 (4)
(b) This part asks us to find the factors of safety when the alternating component is due to pulsating
torsion, and a steady component is due to both torsion and bending. We have Ta = (160 − 20) ∕ 2 =
70 N · m and Tm = (160 + 20) ∕ 2 = 90 N · m. The corresponding amplitude and steady-stress
components are
70  42 10 3 
Ta D
 xya  K fs
 1.61
 15.3 10 6  Pa  15.3 MPa
9
 
90  42 10 
T D
K
 1.61
 19.7 10  Pa  19.7 MPa
2J
2 155 10 
2 155 10
2 J net
3
 xym
6
m
9
fs
net
The steady bending stress component σxm is
 xm  K f
 
Mm
150
 2.07
 93.8 10 6 Pa  93.8 MPa
6
Z net
3.31 10


The von Mises components σ′a and σ′m, from Equations (6–66) and (6–67), are
a  3 15.3 
2 12
 26.5 MPa
m  93.82  3 19.7  
2 12
 99.8 MPa
From Equation (6–41),
Answer
  
nf   a  m 
S 
S
e
© McGraw Hill
ut
1
 26.5 99.8 


 148 440 
1
 2.46
177
Example 6–17 (5)
The first-cycle yield factor of safety ny is
Sy
370
ny 

 2.93
Answer
a  m 26.5  99.8
There is no notch yielding.
© McGraw Hill
178
Varying Fluctuating Stresses
Loading patterns may be
complex.
Simplifications may be
necessary.
Small fluctuations may be
negligible compared to large
cycles.
Access the text alternative for slide images.
© McGraw Hill
Fig. 6–40
179
Cumulative Fatigue Damage
A common situation is to load at σ1 for n1 cycles, then at σ2 for n2 cycles,
etc.
The cycles at each stress level contributes to the fatigue damage.
Accumulation of damage is represented by the Palmgren-Miner cycleratio summation rule, also known as Miner’s rule.
ni
N c
i
(6 - 68)
where ni is the number of cycles at stress level σi and Ni is the number of
cycles to failure at stress level σi .
c is experimentally found to be in the range 0.7 < c < 2.2, with an average
value near unity.
Defining D as the accumulated damage,
D
© McGraw Hill
ni
Ni
(6 - 69)
180
Example 6–18 (1)
Given a steel part with Sut = 151 kpsi and at the critical location of the part, Se = 67.5 kpsi.
For the loading of Figure 6–40, estimate the number of repetitions of the stress-time block
in Figure 6–40 that can be made before failure. Use the Morrow criteria.
Fig. 6–40
© McGraw Hill
181
Example 6–18 (2)
Solution
From Figure 6–23, for Sut = 151 kpsi, f = 0.795. From Equation (6–13),
a
 f Sut 
Se
2
 0.795 151 

 213.5 kpsi
67.5
2
From Equation (6–14),
 f Sut 
 0.795 151 
1
1
b   log 


log
 67.5   0.0833
3
3
 S e 


From Equation (6–15),
  
N   ar 
 213.5 
1 0.0833
(1)
We prepare to add two columns to the previous table. Lacking specific material
information, use the estimate for steel from Equation (6–44),
f   Sut  50 kpsi  151  50  201 kpsi
© McGraw Hill
182
Example 6–18 (3)
Cycle 1: Check the fatigue factor of safety to see if damage is expected.
Equation (6–46):
 a m 
nf  


 S e f 
1
10 
 70


 67.5 201
1
 0.92
Since nf < 1, fatigue damage is predicted from cycle 1. Find the equivalent completely
reversed stress.
Equation (6–60):  ar 
From Equation (1),
a
70

 73.7 kpsi
1   m f 1  10 201
 
N   ar 
 a 
1b
 73.7 

 213.5 
1 0.0833
 
 351 103 cycles
Cycle 2: Repeat the process with the second cycle of stresses.
Equation (6–46):
 a m 
nf  


 S e f 
1
50 
 10


 67.5 201
1
 2.52
Since nf > 1, no fatigue damage is predicted from cycle 2, so infinite life is predicted.
© McGraw Hill
183
Example 6–18 (4)
Cycle 3: This cycle has a negative mean stress. Though the Morrow line can be
continued into the negative mean stress region, it is not necessary, as a quick
check shows that the alternating stress is well below the endurance limit. No
damage is predicted from cycle 3, so infinite life is predicted.
From Equation (6–69) the damage per block is

ni
1
1 1
N
D
N




3
Ni
   351 10 3
 351 10

 
 
Answer
Setting D = 1 yields N = 351(103) cycles.
© McGraw Hill
184
Illustration of Miner’s Rule
Figure 6–41 illustrates effect of Miner’s rule on endurance limit and fatigue failure line.
Note that the damaged material line is predicted to be parallel to original material line.
Fig. 6–41
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© McGraw Hill
185
Weaknesses of Miner’s Rule
Miner’s rule fails to agree with experimental results in two ways.
• It predicts the static strength Sut is damaged.
• It does not account for the order in which the stresses are applied.
© McGraw Hill
186
Manson’s Method
Manson’s method overcomes deficiencies of Miner’s rule.
It assumes all fatigue lines on the S-N diagram converge to a common point at 0.9Sut at 103
cycles.
It requires each line to be constructed in the same historical order in which the stresses
occur.
Fig. 6–42
Access the text alternative for slide images.
© McGraw Hill
187
Surface Fatigue Strength 1
When two surfaces roll or roll and slide against one another, a
pitting failure may occur after a certain number of cycles.
The surface fatigue mechanism is complex and not definitively
understood.
Factors include Hertz stresses, number of cycles, surface finish,
hardness, lubrication, and temperature.
© McGraw Hill
188
Surface Fatigue Strength 2
From Eqs. (3–73) and (3–74), the pressure in contacting cylinders,
b
2 F (1  v12 ) E1  (1  v22 ) E2
l
(1 d1 )  (1 d 2 )
(6 - 70)
2F
 bl
(6 - 71)
Pmax 
Converting to radius r and width w instead of length l,
4 F (1  v12 ) E1  (1  v22 ) E2
b 
w
1 r1  1 r2
2
pmax 
2F
 bw
(6 - 72)
(6 - 73)
Define pmax as surface endurance strength (also called contact strength, contact
fatigue strength, or Hertzian endurance strength)
SC 
© McGraw Hill
2F
 bw
(6 - 74)
189
Surface Fatigue Strength 3
Combining Eqs. (6–72) and (6–74),
2
2


1

v
1

v
F 1 1
2
1
2
    SC 

 K1


w  r1 r2 
E2 
 E1
(6 - 75)
K1 is known as Buckingham’s load-stress factor, or wear factor.
In gear studies, a similar factor is used,
Kg 
K1
sin 
4
(6 - 76)
From Eq. (6–75), with material property terms incorporated into an
elastic coefficient CP
SC  C p
© McGraw Hill
F 1 1
 

w  r1 r2 
(6 - 66)
190
Surface Fatigue Strength 4
Experiments show the following relationships
K1  1 N 1
1 
log  K11 K12 
log  N1 N 2 
b
K g  aN b

log K g 1 K g 2
log  N1 N 2 
SC   N 
   log  S
C1
SC 2 
log  N1 N 2 
(6 - 78)
Data on induction-hardened steel on steel give (SC)107 = 271 kpsi
and (SC)108 = 239 kpsi, so β, from Equation (6–78), is

© McGraw Hill
log  271 239

7
log 10 10
8

 0.055
191
Surface Fatigue Strength 5
A longstanding correlation in steels between SC and HB at 108
cycles is
 0.4 H B  10 kpsi
 SC 108   2.76 H  70 MPa
B

(6 - 79)
AGMA uses
0.99
© McGraw Hill
 SC 10  0.327 H B  26 kpsi
7
(6 - 80)
192
Surface Fatigue Strength 6
Incorporating design factor into Eq. (6–77),
 C  CP
CP
F 1 1
  

wnd  r1 r2 
nd
SC
F 1 1
  

w  r1 r2 
nd
Since this is nonlinear in its stress-load transformation, the
definition of nd depends on whether load or stress is the primary
consideration for failure.
If the loss of function is focused on the load,
nd   SC  C 
2
If the loss of function is focused on the stress,
nd  SC  C
© McGraw Hill
193