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Design Against Fatigue
Failure
2/3/2015 1
Fatigue is the failure of a mechanical element by the
growth of a crack within a material under variable,
repeated, alternating, or fluctuating stresses.
Generally, fatigue crack growth occurs under stresses
below the ultimate tensile strength (σut), yield strength
(σy) and critical stress for the original crack (σc).
A fatigue failure has an appearance similar to a brittle
fracture, as the fracture surfaces are flat and
perpendicular to the stress axis with the absence of
necking.
The fatigue of material is effected by the size of the
component, relative magnitude of static and variable
loads and the number of load reversals.
2/3/2015 2
It is observed that about 80% of failure of mechanical components are due to “Fatigue Failure” resulting from fluctuating stresses.
NOTES
The stresses which vary from a minimum value to a maximum value of the same nature, (i.e. tensile or compressive) are called fluctuating stresses.
The stresses which vary from zero to a certain maximum value are called repeated stresses.
The stresses which vary from a minimum value to a maximum value of the opposite nature (i.e. from a certain minimum compressive to a certain maximum tensile or from a minimum tensile to a maximum compressive) are called alternating stresses.
The stresses which vary from one value of compressive to the same value of tensile or vice versa, are known as completely reversed or cyclic stresses.
2/3/2015 3
Alternating stress Repeated stress
Completely reversed stress Fluctuating stress
σMax= Maximum stress σa ,σv=Alternating stress
σMin= Minimum stress σr = Range of stress
σm= Mean or average stress
2/3/2015 4
Fatigue failure is characterized by three stages
I. Crack Initiation
II. Crack propagation
III. Final fracture
Fracture zone Propagation zone, striation
Crack initiation site
VW crank shaft – fatigue failure due to cyclic bending and torsional stresses
2/3/2015 5
Jack hammer component,
shows no yielding before
fracture.
Crack initiation site
Propagation zone, striation
Fracture zone
2/3/2015 6
Fatigue-Life Methods
The three major fatigue life methods used in design and
analysis are
1. The stress-life method.
2. The strain-life method.
3. The linear-elastic fracture mechanics method.
• These methods attempt to predict the life in number of
cycles to failure (N) for a specific level of loading.
• This cycles number (N) is generally classified as
1. low-cycle fatigue , where 1 ≤ N ≤ 103
2. high-cycle fatigue , where N > 103.
2/3/2015 7
The Stress-Life Method (SLM)
The stress-life method, based on stress levels only, is the
least accurate approach, especially for low-cycle
applications.
To determine the strength of materials under the action
of fatigue loads, specimens are subjected to repeated or
varying forces of specified magnitudes while the cycles
or stress reversals are counted to destruction.
The most widely used fatigue-testing device is the R.R.
Moore high-speed rotating-beam machine. This machine
subjects the specimen to pure bending (no transverse
shear) by means of weights.
2/3/2015 8
Test-specimen geometry for the R. R. Moore rotating beam machine.
Motor
Load
Rotating beam machine – applies fully reverse bending stress
Typical testing apparatus (pure bending)
The specimen
is polished
2/3/2015 9
The S-N Diagram • The S-N diagram is the graphical representation of
stress amplitude (Sf) verses the number of stress cycles (N) before the fatigue failure on semilog paper or on log-log paper.
• To establish the fatigue strength of a material, quite a number of tests are necessary because of the statistical nature of fatigue. For the rotating-beam test, a constant bending load is applied, and the number of revolutions (stress reversals) of the beam required for failure is recorded.
• The first test is made at a stress that is somewhat under the ultimate strength of the material. The second test is made at a stress that is less than that used in the first. This process is continued, and the results are plotted as S-N diagram.
2/3/2015 10
In the case of the steels as shown in Fig.(A), a knee
occurs in the graph, and beyond this knee failure will not
occur, no matter how great the number of cycles. The
strength corresponding to the knee is called the
endurance limit or the fatigue limit.
Aluminium alloys does not have an endurance limit as
shown in Fig.(B), normally the fatigue strength Sf is
reported at a specific number of cycles, normally N =
5(108) cycles of reversed stress.
Endurance or fatigue limit stress (Se) is defined as the
maximum amplitude of completely reversed stress that
the standard specimen can sustain for an unlimited
number of cycles without fatigue failure.
2/3/2015 11
Fig.(A) S-N diagram plotted from the results of completely reversed axial
fatigue tests (Steel-normalized σu=125kpsi) 2/3/2015 12
Fig.(B) S-N bands for representative aluminium alloys, excluding
wrought alloys (Sut <38 kpsi).
2/3/2015 13
We also distinguish a finite-life region and an infinite-
life region in S-N diagram as shown in Fig.(A).
The boundary between these regions cannot be clearly
defined except for a specific material; but it lies
somewhere between 106 and 107 cycles for steels, as
shown in Fig.(A).
2/3/2015 14
The Strain-Life Method (ELM)
The Strain-Life Method is The best approach yet advanced to explain the nature of fatigue failure which is based on the investigation of behavior of materials subject to cyclic deformation.
Generally, the strength decreases with stress repetitions, while other materials may be strengthened, instead, by cyclic stress reversals (i.e. the elastic limits of annealed steels are likely to increase when subjected to cycles of stress reversals, while cold-drawn steels exhibit a decreasing elastic limit).
Fig.(C) has been constructed to show the general appearance of these plots for the first few cycles of controlled cyclic strain. In this case the strength decreases with stress repetitions, as evidenced by the fact that the reversals occur at ever-smaller stress levels.
2/3/2015 15
Fig. (D)A log-log plot showing how the
fatigue life is related to the true-strain
amplitude for hot-rolled SAE 1020 steel.
Fig.(C) True stress–true strain
hysteresis loops showing the
first five stress reversals of a
cyclic-softening material.
2/3/2015 16
Note that the slope of the line AB is the modulus of
elasticity E.
∆σ is the stress range.
∆εp is the plastic-strain range.
∆εe is the elastic strain range.
∆ε is the total-strain range.
∆ε =∆εe + ∆εp
To explain the Fig.(D), we first define the following
terms:
Fatigue ductility coefficient (ε′F) is the true strain
corresponding to fracture in one reversal (point A in
Fig.(C)). The plastic-strain line begins at this point in
Fig.(D).
2/3/2015 17
Fatigue strength coefficient (σ′F) is the true stress
corresponding to fracture in one reversal (point A in
Fig.(C). Note in Fig.(D) that the elastic-strain line
begins at σ′F/E.
Fatigue ductility exponent (c) is the slope of the
plastic-strain line in Fig.(D) and is the power to which
the life 2N must be raised to be proportional to the
true plastic-strain amplitude. If the number of stress
reversals is 2N, then N is the number of cycles.
Fatigue strength exponent (b) is the slope of the
elastic-strain line, and is the power to which the life
2N must be raised to be proportional to the true-stress
amplitude.
2/3/2015 18
Now, from Fig.(D), we see that the total strain is the sum
of the elastic and plastic components. Therefore, the total
strain amplitude is half the total strain range.
The equation of the plastic-strain line is
The equation of the elastic strain line is
∴ The total-strain amplitude is
222
pe
)2('
2N
c
F
p
)2('
2N
bFe
E
)2()2('
'
2NN
bF
c
F E
2/3/2015 19
The Linear-Elastic Fracture Mechanics
Method (LEFM)
The fracture mechanics method assumes a crack is already
present and detected which is employed to predict crack
growth with respect to stress intensity.
Including three stages,
1) The first phase of fatigue cracking is designated as
stage I fatigue.
2) The second phase, that of crack extension, is called
stage II fatigue.
3) Final fracture is occurs during stage III fatigue which
is associated with rapid acceleration of crack growth
then fracture.
2/3/2015 20
Crack Growth
Fatigue cracks nucleate and grow when stresses vary
and there is some tension in each stress cycle.
There are three modes of crack propagation,
I. The opening crack propagation mode.
II. The sliding crack propagation mode.
III. The tearing crack propagation mode.
2/3/2015 21
Consider the stress to be variable between the limits of
σmin and σmax, where the stress range is defined as
∆σ = σmax− σmin.
Consider the stress intensity factor (K) which is used in
fracture mechanics to predict the stress state ("stress
intensity") near the tip of a crack caused by a remote
load or residual stresses.
Thus, for ∆σ, the stress intensity
range per cycle is
aK I
aMinMaxIK )(
aK I
2/3/2015 22
Where,
β: The stress intensity modification factor.
a: The length of crack.
KI: The stress intensity
factor for first mode.
Chart of Thin plate in tension
or simple compression with a
certain crack to select proper β
for mode I crack propagation..
2/3/2015 23
To develop fatigue strength data, a number of specimens
of the same material are tested at various levels of ∆σ.
Cracks nucleate at or very near a free surface or large
discontinuity.
Fig.(E) The increase in crack length (a) from an initial length of (ai)
as a function of cycle count for three stress ranges,
(∆ σ)3 > (∆σ)2 > (∆ σ)1. 2/3/2015 24
Fig.(F) When da/dN is measured in Fig.(E) and plotted on log-log
coordinates, the data for different stress ranges superpose, giving rise to a
sigmoid curve as shown. (∆KI)th is the threshold value of ∆KI, below
which a crack does not grow. From threshold to rupture an aluminum
alloy will spend 85–90 percent of life in region I, 5–8 percent in region II,
and 1–2 percent in region III. Where, Kc is critical stress intensity factor
and R = σmin/σmax 2/3/2015 25
Here we present a simplified procedure for estimating
the remaining life of a cyclically stressed part after
discovery of a crack. This requires the assumption that
plane strain conditions prevail.
Assuming a crack is discovered early in stage II, the
crack growth in region II of Fig.(F) can be approximated
by the Paris equation, which is of the form
where C and m are empirical material constants.
2/3/2015 26
Substituting in above equation and
integrating gives,
ai: The initial crack length
af: The final crack length corresponding to failure
Nf: The estimated number of cycles to produce a failure
after the initial crack is formed.
aK I
2/3/2015 27
Example: The bar shown in figure below 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 KIc = 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 0.004in has been
discovered on the bottom of the bar. Estimate the
number of cycles of life remaining.
L=4in
2/3/2015 28
Solution:
This value is below the yield strength. If β = 1.03, we
approximate af as
kpsi
cI
M
inbh
c
I
20.11501042.0
1200
01042.06
)5.0(25.0
6
322
ina
Ka
f
Icf
1205.0)2.115*03.1
73(
1
)(1
2
2
max
2/3/2015 29
From this figure, we compute
Thus af/h varies from near zero
to approximately 0.241. From
shown Figure, for this range β
Is nearly constant at approxima-
tely 1.07. We will assume it to be
so, and re-evaluate af as
241.05.0
1205.0
h
a f
ina
Ka
f
Icf
112.0)2.115*07.1
73(
1
)(1
2
2
max
2/3/2015 30
Thus the estimated remaining life is
cyclesNf
aa
daN
a
daN
a
da
CN
f
f
a
a
mf
f
i
35.05.03
112.0
004.0
5.03112.0
004.0 3
3
112.0
004.0 311
10*6782.64]004.0112.0[10*0438.5
)](2[10*5219.210*5219.2
])2.115(07.1[10*8.3
1
)(
1
2/3/2015 31
The Endurance Limit
Fig.1 Graph of endurance limits versus tensile strengths from actual test results
for a large number of wrought irons and steels. Ratios of S′e /Sut of 0.60, 0.50,
and 0.40 are shown by the solid and dashed lines. Note also the horizontal
dashed line for S′e= 105 kpsi.
2/3/2015 32
Therefore, after simplifying the observation, the estimation
of the endurance limit is
The prime mark on S′e in this equation refers to the rotating
beam specimen itself.
The unprimed symbol Se for the endurance limit of an
actual machine element subjected to any kind of loading.
2/3/2015 33
Fatigue Strength
is the highest stress that a material can withstand for a
given number of cycles without breaking. In other words,
it is the maximum stress that can be applied for a certain
number of cycles without fracture.
It is affected by environmental factors such as corrosion.
To approximate relationship between S-N during high
cycle, two stages approximation are required which are
I. Approximation of the fatigue strength fraction (f ) at
N= 103.
II. Approximation of the S–N during high cycle at
N>103.(For an actual mechanical component)
2/3/2015 34
I. Approximation of the fatigue strength fraction (f )
at N= 103.
◦ Defining the specimen fatigue strength at a specific number of
cycles as (S′f )N = E∆εe/2,
Since,
At 103 cycles,
)2('
2N
bFe
E
)( NS b
FNf *)2(')'(
b
FfS )10*2(')'( 3
103
ut
f
S
Sf
310)'(
b
ut
F
Sf )10*2(
'3
2/3/2015 35
If this true-stress–true-strain equation (σ′F= σ0εm, with
ε=ε′F) is not known, the SAE approximation for steels
with HB ≤ 500 may be used:
To find b, substitute the endurance strength and
corresponding cycles, S′e and Ne, respectively into Eq.
(*) and solving for b
MpaS' or kpsi S utFutF 34550'
)2log(
)'/'log(
e
eF
N
Sb
2/3/2015 36
Above figure is a plot of f for 70 ≤ Sut ≤ 200 kpsi. To be conservative, for Sut < 70 kpsi, let f = 0.9.
Fig.2 Fatigue strength fraction, f, of Sut at 103 cycles for S′e = Se= 0.5Sut at 106
cycles.
2/3/2015 37
II. Approximation of the S–N during high cycle at N>103
Sf = a Nb
Where N is cycles to failure and the constants a and b are
defined by the points 103, Sf103 and 106, Se with Sf10
3= f Sut.
Substituting these two points in above equation gives
If a completely reversed stress σrev is given, setting Sf =
σrev. the number of cycles-to-failure can be expressed as
)log()( 2
e
ut
e
ut
S
Sf
3
1b and
S
Sfa
brev
aN
1
)(
2/3/2015 38
Example: 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.
Owing of Sut ≤ 200 kpsi, the endurance limit is
kpsiS
SS
e
ute
45'
)90(5.05.0'
2/3/2015 39
2/3/2015 40
(b) From Fig.2, for Sut = 90 kpsi, f =0.86.
For 104 cycles to failure,
S′f = aNb= 133.1(104)-0.0785
S′f = 64.6 kpsi
(c) For σrev= 55 kpsi
kpsi S
Sfa
e
ut 1.13345
)90*86.0(
'
)( 22
0785.0)45
)90(86.0log(
3
1)
'log(
e
ut
S
Sf
3
1b
cyclesN
aN brev
4
0785.011
10*75.7
)1.133
55()(
2/3/2015 41
Endurance Limit Modifying Factors
The experiments results of endurance limit in laboratory
are not matched with a mechanical or structural member.
Some difference include
o Material: composition, basis of failure, variability.
o Manufacturing: method, heat treatment, fretting
corrosion, surface condition, stress concentration.
o Environment: corrosion, temperature, stress state,
relaxation times.
o Design: size, shape, life, stress state, speed, fretting,
galling
2/3/2015 42
A Marin equation is therefore written as
Where,
ka: surface condition modification factor
kb: size modification factor
kc: load modification factor
kd: temperature modification factor
ke: reliability factor
kf: miscellaneous-effects modification factor (In this course, Kf will not be included)
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. (Se < S′e)
When endurance tests of parts are not available, estimations are made by applying Marin factors to the endurance limit.
'SkkkkkkS efedcbae
2/3/2015 43
Surface Factor ka
Where, Sut is the minimum tensile strength and a and b are to be found in
below Table 6–2.
Example: A 1030 CD steel has a machined surface. Estimate
surface factor ka.
Solution: From Table 6–2, a = 4.51 and b =−0.265.
b
uta aSK
858.0)520(51.4aSK 265.0b
uta
2/3/2015 44
Size Factor kb
For axial loading there is no size effect,
For bending and torsion,
Rotating
Nonrotating
It’s depending on an equivalent diameter de ... Table 6–3
1bK
2/3/2015 45
Table 6–3
2/3/2015 46
Example: A steel shaft loaded in bending is 32 mm in
diameter, abutting a filleted shoulder 38 mm in diameter.
The shaft material has a mean ultimate tensile strength of
690 MPa. Estimate the Marin size factor kb if the shaft is
used in (a) A rotating mode, (b) A nonrotating mode.
Solution:
(a) Since, 2.79 ≤ (d=32) ≤ 51 mm
(b) From Table 6–3,
858.0)62.7
32()
62.7( 107.0107.0 d
Kb
954.0)62.7
84.11(
mm 51 11.84)d( 2.79,84.11)32(37.037.0
107.0
e
b
e
K
mmdd
2/3/2015 47
Loading Factor kc
Temperature Factor kd
Where,
torsion
axial
bending
59.0
85.0
1
Kc
RT
Td
S
SK ST = tensile strength at operating temperature
SRT = tensile strength at room temperature
F1000T 70 F
T)10(595.0T)10(104.0
T)10(115.0T)10(432.0975.0K
4
F
123
F
8
2
F
5
F
3
d
2/3/2015 48
2/3/2015 49
Table 6–4
Example: A 1035 steel has a tensile strength of 70 kpsi
and is to be used for a part that sees 450°F in service.
Estimate the Marin temperature modification factor and
(Se)450◦ if
(a) The room-temperature endurance limit by test is
(S′e)70◦ = 39.0kpsi.
(b) Only the tensile strength at room temperature is
known.
Solution:
(a)
2/3/2015 50
4
F
123
F
8
2
F
5
F
3
d
T)10(595.0T)10(104.0
T)10(115.0T)10(432.0975.0K ,Using
007.1)450)(10(595.0)450)(10(104.0
)450)(10(115.0)450)(10(432.0975.0K
41238
253
d
2/3/2015 51
kpsi 3.39)39(007.1)'S(K)S( Thus,70ed450e
(b) Interpolating from Table 6–4 gives,
Thus, the tensile strength at 450°F is estimated as
kpsi 49.70)70(007.1)S()S/S()S(70ut450RTT450ut
kpsi245.35)S(
)49.70(5.0)S(5.0)S( Then,
450e
450ut450e
007.1)S/S(400500
400450
018.1995.0
018.1)S/S(
450RTT450RTT
Reliability Factor ke
2/3/2015 52
ae z08.01K
Table 6–5
Example: A 1015 hot-rolled steel bar has been machined to a
diameter of 1 in. It is to be placed in reversed axial loading
for 7(104) cycles to failure in an operating environment of
550°F. Using ASTM minimum properties, and a reliability
of 99 percent, estimate the endurance limit and fatigue
strength at 7(104) cycles.
Solution:
From Table A–20, Sut = 50 kpsi at 70°F.
2/3/2015 53
4
F
123
F
8
2
F
5
F
3
d
T)10(595.0T)10(104.0
T)10(115.0T)10(432.0975.0K ,Using
98.0)550)(10(595.0)550)(10(104.0
)550)(10(115.0)550)(10(432.0975.0K
41238
253
d
The ultimate strength at 550°F is then
The rotating-beam specimen endurance limit at 550°F is then estimated as
Next, we determine the Marin factors. For the machined surface, Eq. (6–19) with Table 6–2 gives,
The size factor kb = 1.
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.
2/3/2015 54
kpsi 0.49)50(98.0)S(K)S(70utd550ut
kpsi 5.24)49(5.0)S(5.0)'S(550ut550e
963.0)49(70.2aSK 265.0b
uta
For 99 percent reliability, from Table 6–5, ke = 0.814.
The endurance limit for the part is estimated by Eq. (6–18)
as
For the fatigue strength at 7(104) cycles,
(Sut)550 ̊ = 49 < 70 kpsi, then f = 0.9.
2/3/2015 55
kpsi 16.3 814)24.5.85)(1)(0.0.963(1)(0
)'S(kkkkk)S( 550eedcba550e
kpsi3.1193.16
)49*9.0(
S
) Sf(a
2
e
2
ut
1441.0)3.16
)49(9.0log(
3
1)
S
Sflog(
3
1b
e
ut
Sf = aNb= 119.3(7*104)-0.1441 = 23.9 kpsi
Stress Concentration and Notch Sensitivity
Where, Kf is a reduced value of Kt and σ0 is the nominal
stress. The factor Kf is commonly called a fatigue stress-
concentration factor, and hence the subscript f.
Notch sensitivity q
2/3/2015 56
specimen free-notchin stress
specimen notchedin stress maximumfK
0fsmax0fmax KK or
stress nominalover stress al theoreticof increase
stress nominalover stress actual of increaseq
Where, q is usually between zero and unity.
NOTES
When the material has no sensitivity to notches,
q = 0 and Kf= 1
When the material is fully sensitivity to notches,
q = 1 and Kf= Kt
q = 0.20 for all grades of cast iron
2/3/2015 57
1K
1Kq
K
Kq
ts
fs
shear
00ts
00fs
shear
1K
1Kq
K
Kq
t
f
00t
00f
)1K(q1K)1K(q1K tsshearfstf or
2/3/2015 58
Fig. 6–20 Notch-sensitivity charts for steels and wrought aluminum alloys
subjected to reversed bending or reversed axial loads.
2/3/2015 59
Figure 6–21 Notch-sensitivity curves for materials in reversed torsion
There is another way to estimate notch sensitivity q, which is by Neuber’s equation
For bending or axial
For torsion
o Where √a is defined as the Neuber constant and is a material constant. r is notch radius.
NOTE: the equations apply to steel and Sut is in kpsi.
2/3/2015 60
r
a1
1q
r/a1
1K1K t
f
3
ut
82
ut
5
ut
3 S)10(67.2S)10(51.1S)10(08.3246.0a
3
ut
82
ut
5
ut
3 S)10(67.2S)10(35.1S)10(51.2190.0a
Example: 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 6–20.
(b) Equations.
Solution:
From Fig. 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.
2/3/2015 61
(a) From Fig. 6–20, for Sut = 690 MPa and r = 3 mm,
q .= 0.84.
Thus,
(b) Converting Sut= 690 Mpa= 100kpsi.
2/3/2015 62
55.1)165.1(84.01)1K(q1K tf
55.1
3
313.01
165.11
r/a1
1k1K
mm313.0in0622.0
)100)(10(67.2)100)(10(51.1)100)(10(08.3246.0
S)10(67.2S)10(51.1S)10(08.3246.0a
tf
38253
3
ut
82
ut
5
ut
3
Thus,
Example: For the step-shaft of previous example, it is
determined that the fully corrected endurance limit is
Se = 280 MPa. Consider the shaft undergoes a fully
reversing nominal stress in the fillet of (σrev)nom = 260
MPa. Estimate the number of cycles to failure.
Solution:
From previous example, Kf = 1.55, and the ultimate
strength is Sut = 690 MPa = 100 kpsi. The maximum
reversing stress is
From Fig.2, f = 0.845.
2/3/2015 63
Mpa403)260(55.1)(K)( nomrevfmaxrev
2/3/2015 64
Mpa1214280
)690*845.0(
S
) Sf(a
2
e
2
ut
1062.0)280
)690(845.0log(
3
1)
S
Sflog(
3
1b
e
ut
cycles10*3.32N
)1214
403(
)a
(N
3
1062.01
b1
rev
Modifying Factor to Account for fatigue stress
concentration
The endurance limit is reduced due to fatigue stress
concentration. The fatigue stress concentration factor is
less than stress concentration factor due to notches
sensitivity of the material.
To apply the effect of fatigue stress concentration, the
designer can (1) either reduce the endurance limit by
dividing it by Kf or (2) increase the nominal stress
amplitude by multiplied it by Kf .
o In this course, we will consider (2) which increases the
nominal stress amplitude by multiplied it by Kf
2/3/2015 65
Example: Figure 6–22a shows a rotating shaft simply
supported in ball bearings at A and D and loaded by a
nonrotating force F of 6.8 kN. Using ASTM “minimum”
strengths, estimate the life of the part.
Solution:
We will solve the problem by
first estimating the strength at
point B,
2/3/2015 66
Fig.6-22(a) Shaft drawing showing all
dimensions in millimeters; all fillets 3-mm
radius. The shaft rotates and the load is
stationary; material is machined from AISI
1050 cold-drawn steel.
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
Estimating Marin’s factors,
2/3/2015 67
798.0)690(51.4aSKfactor, Surface 265.0b
uta
MPa236
8)3450.798(0.85 S
1 k k k Since,
858.0)62.7
32(
)62.7
d(K ,factor Size
e
edc
107.0
107.0
b
To find the geometric stress
-concentration factor Kt,
with D/d = 38/32 = 1.1875
and r/d = 3/32 = 0.093 75
and read Kt = 1.65.
2/3/2015 68
To find the fatigue stress-concentration factor kf,
Substituting Sut = 690/6.89 = 100 kpsi into Eq. (6–35a)
gives, 33)–(6 Eq. into thisngSubstituti ,mm313.0in0622.0
100)10(67.2100)10(51.1100)10(08.3246.0a 38253
55.1r/a1
1K1K t
f
30.3131
1-1.651
The next step is to estimate the bending stress at point B.
The bending moment is
The reversing bending stress is,
2/3/2015 69
(b) Bending moment diagram. m.N5.695
250*550
)8.6(225
250*550
F225xRM 1B
3333 mm )(10 3.217 32/32π 32/πd I/c
MPa335.1 Pa )335.1(10
)(103.217
695.51.55
I/c
MK σ
6
6Bfrev
This stress is greater than Se and less than Sy. This means
we have both finite life and no yielding on the first
cycle.
The life of the part,
The ultimate strength, Sut = 690MPa = 100 kpsi. From Fig.
6–18, we will get f = 0.844
2/3/2015 70
Mpa1437236
)690*844.0(
S
) Sf(a
2
e
2
ut
1308.0236
)690(844.0log
3
1)
S
Sflog(
3
1b
e
ut
cycles10*68)1437
1.335()
a(N 31308.0
1b
1rev
Characterizing Fluctuating Stresses
It has been found that in periodic patterns exhibiting a
single maximum and a single minimum of force, the
shape of the wave is not important, but the peaks on
both the high side (maximum) and the low side
(minimum) are important.
Thus Fmax and Fmin in a cycle of force can be used to
characterize the force pattern.
Fm: the midrange steady component of force, and
Fa: the amplitude of the alternating component of force.
2/3/2015 71
2
FFF and
2
FFF minmax
aminmax
m
2/3/2015 72
2 and
2
minmaxa
minmaxm
m
a
max
min A and R
σa = amplitude component σr = range of stress
σm = midrange component A= the amplitude ratio
R= the stress ratio
σa = σao and σm= σmo → in the absence of a notch
σa = kf σao and σm= kf σmo → with a notch
2/3/2015 73
Fatigue Failure Criteria for Fluctuating Stress
we want to vary both the midrange stress and the stress amplitude, or alternating component, to learn something about the fatigue resistance of parts when subjected to such situations.
There are Five criteria of failure are diagrammed in Fig. 6–27: the Soderberg, the modified Goodman, the Gerber, the ASME-elliptic, and yielding.
The Fig. 6–27 shows that only the Soderberg criterion guards against any yielding, but is biased low.
Considering the modified Goodman line as a criterion, point A represents a limiting point with an alternating strength Sa and midrange strength Sm. The slope of the load line shown is defined as r = Sa/Sm.
2/3/2015 74
2/3/2015 75
Fig. 6–27 Fatigue diagram showing various criteria of failure. For each
criterion, points on or “above” the respective line indicate failure.
Name The criterion equation
Soderberg Fatigue Line
Modified Goodman Fatigue Line
Gerber Fatigue Curve
ASME-elliptic Fatigue Curve
Yield (Langer) line
2/3/2015 76
1S
S
S
S
y
m
e
a
1S
S
S
S
ut
m
e
a
1S
S
S
S2
ut
m
e
a
1S
S
S
S2
y
m
2
e
a
yma S S S
The stresses nσa and nσm can replace Sa and Sm, where n
is the design factor or factor of safety. See table Table *
There are two ways to proceed with a typical analysis.
One method is to assume that fatigue occurs first and
use one of Eqs. (1) to (4) to determine n or size,
depending on the task. Most often fatigue is the
governing failure mode. Then follow with a static check.
If static failure governs then the analysis is repeated
using Eq. (5). See table Table *
2/3/2015 77
2/3/2015 78
No. Name The criterion equation with F.S
1 Soderberg
2 Modified Goodman
3 Gerber
4 ASME-elliptic
5 Langer static yield
n
1
SS y
m
e
a
n
1
SS ut
m
e
a
1S
n
S
n2
ut
m
e
a
1S
n
S
n2
y
m
2
e
a
n
S
y
ma
Table *
2/3/2015 79
Table 6–6
Amplitude and Steady Coordinates of Strength and Important Intersections in
First Quadrant for Modified Goodman and Langer Failure Criteria
2/3/2015 80
Table 6–7
Amplitude and Steady Coordinates of Strength and Important Intersections in
First Quadrant for Gerber and Langer Failure Criteria
2/3/2015 81
Table 6–8
Amplitude and Steady Coordinates of Strength and Important Intersections in
First Quadrant for ASME-elliptic and Langer Failure Criteria
2/3/2015 82
NOTES for the above three tables (6-6) to (6-8):
The first row of each table corresponds to the fatigue
criterion, the second row is the static Langer criterion, and
the third row corresponds to the intersection of the static
and fatigue criteria.
The first column gives the intersecting equations and the
second column the intersection coordinates.
Example: A 1.5-in-diameter bar has been machined from an
AISI 1050 cold-drawn bar. This part is to withstand a
fluctuating tensile load varying from 0 to 16 kip. Because of
the ends, and the fillet radius, a fatigue stress-concentration
factor Kf is 1.85 for 106 or larger life. Find Sa and Sm and
the factor of safety guarding against fatigue and first-cycle
yielding, using (a) the Gerber fatigue line and (b) the
ASME-elliptic fatigue line.
Solution:
A . We start with calculate the axial midrange force
component and amplitude of force component,
The nominal axial stress components σao and σmo are
Applying Kf to both components σao and σmo,
2/3/2015 83
kIP82
016
2
FFF
kiP82
016
2
FFF
minmaxa
minmaxm
kpsi53.44/5.1
8
4/d
F
kpsi53.44/5.1
8
4/d
F
22
aao
22
mmo
maofa σ kpsi 8.38 1.85(4.53) K
B . From Table A–20, Sut = 100 kpsi and Sy = 84 kpsi.
Estimating the endurance limit,
The Marin factors are, deterministically,
ka = 2.70(100)−0.265 = 0.797
kb = 1 (axial loading)
kc = 0.85
kd = ke = 1
2/3/2015 84
kpsi 9.33 50)11)((1)(0.85)(7970.
'SkkkkkS eedcbae
(Ideal) kpsi50 )100(5.0S5.0'S ute
C . Let us calculate the factors of safety first.
(a) From the bottom panel from Table 6–7 the factor of
safety for fatigue is
From Eq. (6–49) the factor of safety guarding against first-
cycle yield is
2/3/2015 85
66.3)38.8(100
)9.33)(38.8(211
9.33
38.8
38.8
100
2
1
S
S211
S
S
2
1n
22
2
aut
em
e
a
2
m
utf
01.538.838.8
84
S n
ma
y
y
For drawing the designer’s diagram,
1. We have to know all the details about the load line
starting from its slope (r),
2. Also, the intersection point (B) with Gerber fatigue
curve
2/3/2015 86
45rTan 138.8
38.8r 1
m
a
kpsi7.30100)1(
)9.33(211
)9.33(2
)100(1
rS
S211
S2
SrS
222
2
ut
e
e
2
ut
2
a
kpsi7.301
7.30
r
SS a
m
3. In this Ex., the intersection point (C) between the load
line and Langer line can be found by continuously
drawing the load line with respect to till
intersects the Langer line
4. Finally, the intersection point (D) between the Gerber
fatigue curve with Langer line.
2/3/2015 87
kpsi0.649.33
841
100
)9.33(211
)9.33(2
100
S
S1
S
S211
S2
SS
22
e
y
2
ut
e
e
2
utm
kpsi 20 6484 S S S mya
45rTan 1
The critical slope is thus
which is less than the actual load line of r = 1. This
indicates that fatigue occurs before first-cycle-yield.
As a check on the previous result, nf = OB/OA = Saf /σa =
Smf /σm = 30.7/8.38 =3.66. ny= OC/OA= Say/σa =Smy/σm =
5.01 and we see total agreement.
2/3/2015 88
312.064
20
S
Sr
m
acri
Fig. 6–28 Principal points A, B, C, and D on the designer’s diagram drawn for
Gerber, Langer, and load line.
2/3/2015 89
(b) Repeating the same procedure for the ASME-elliptic
line, for fatigue
Again, this is less than ny = 5.01 and fatigue is predicted to
occur first.
For drawing the designer’s diagram,
2/3/2015 90
75.3)84/38.8()9.33/38.8(
1
)S/()S/(
1n
22
2
ym
2
ea
f
1. We have to know all the details about the load
line starting from its slope (r),
45rTan 138.8
38.8r 1
m
a
2/3/2015 91
2. Also, the intersection point (B) with ASME fatigue
curve. We obtain the coordinates Sa and Sm of point B
by using
kpsi4.311
4.31
r
SS
kpsi4.31)84)(1()9.33(
)84)(9.33)(1(
SrS
SSrS
am
222
222
2
y
22
e
2
y
2
e
2
a
3. In this Ex., the intersection point (C) between the load
line and ASME curve can be found by continuously
drawing the load line with respect to till
intersects the ASME-elliptic fatigue curve.
45rTan 1
4. Finally, the intersection point (D) between the ASME-
elliptic fatigue curve with Langer line.
2/3/2015 92
kpsi5.605.2384SSS
kpsi5.23849.33
9.33)84(2
SS
SS2S
aym
22
2
2
y
2
e
2
ey
a
388.05.60
5.23
S
Sr
m
acri
The critical slope is thus
Which is less than the actual load line of r =1. This indicates
that fatigue occurs before first-cycle-yield
As a check on the previous result, nf = OB/OA = Saf /σa =
Smf /σm = 30.7/8.38 =3.75. ny= OC/OA= Say/σa =Smy/σm =
5.01 and we see total agreement.
Therefore, the Gerber and the ASME-elliptic fatigue failure
criteria are very close to each other and are used
interchangeably.
2/3/2015 93
Fig. 6–29 Principal points A, B, C, and D on the designer’s diagram drawn for
ASME-elliptic, Langer, and load line.
Torsional Fatigue Strength under Fluctuating Stresses
For ductile material, polished, notch-free, and cylindrical,
a torsional steady-stress component not more than the
torsional yield strength has no effect on the torsional
endurance limit.
For materials with stress concentration, notches, or
surface imperfections, the torsional fatigue limit decreases
monotonically with torsional steady stress.
The modified Goodman relation for pulsating torsion is
from distortion-energy theory,
2/3/2015 94
utsu 0.67S S
ytsy 0.5777S S
Combinations of Loading Modes
It may be helpful to think of fatigue problems as being in
three categories:
• Completely reversing simple loads (σm = 0).
• Fluctuating simple loads.
• Combinations of loading modes.
Here will be used von-mises theory for combined loads
1. The first step is to generate two stress elements—one
for the alternating stresses and one for the midrange
stresses.
2. Apply the appropriate fatigue stress-concentration
factors to each of the stresses; i.e., apply (Kf )bending for
the bending stresses, (Kf s)torsion for the torsional
stresses, and (Kf )axial for the axial stresses.
2/3/2015 95
3. Next, calculate an equivalent von Mises stress for each
of these two stress elements, σ′a and σ′m.
4. For the endurance limit, Se, use the endurance limit
modifiers, ka, kb, and kc, for bending.
The torsional load factor, kc = 0.59 should not be
applied as it is already accounted for in the von Mises
stress calculation.
The load factor for the axial load can be accounted for
by dividing the alternating axial stress by the axial
load factor of 0.85.
5. Finally, select a fatigue failure criterion (modified
Goodman, Gerber, ASME-elliptic, or Soderberg) to
complete the fatigue analysis.
2/3/2015 96
Simple Example, consider the common case of a shaft with
bending stresses, torsional shear stresses, and axial
stresses.
2/3/2015 97
2/12
torsionalmtorsionalfs
2
axialmaxialfbendingmbendingf
'
m
2/1
2
torsionalatorsionalfs
2
axiala
axialfbendingabendingf
'
a
2/12
xy
2
x
'
K3KK
K385.0
KK
)3(
Example: A rotating 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 Gerber and Langer failure criteria for the
following loading conditions:
a) The shaft 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 subjected to a pulsating torque
fluctuating from 20 to 160 N·m and a steady bending
moment of 150 N·m.
2/3/2015 98
Solution:
(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 = 2.366 for bending; and A = 0.89 and Kts = 1.75
for torsion.
Next, using Figs. 6–20 and 6–21, pp. 295–296, with a
notch radius of 3 mm we find the notch sensitivities to be
0.78 for bending and 0.81 for torsion.
2/3/2015 99
61.1)175.1(81.01)1K(q1K
07.2)1366.2(78.01)1K(q1K
tsshearfs
tf
434444
net
334444
net
mm )10(155])34()42[(32
)98.0()dD(
32
AJ
mm )10(31.3])34()42[()42(32
)798.0()dD(
D32
AZ
2/3/2015 100
Table A–16
Approximate Stress-
Concentration Factor Kt
for Bending of a Round
Bar or Tube with a
Transverse Round Hole
The alternating bending stress is now found to be
and the alternating torsional stress is
The midrange von Mises component σ′m is zero. The
alternating component σ′a is given by
From Table A–20, Sut = 440 MPa and Sy = 370 MPa.
The endurance limit of the rotating-beam specimen is
0.5(440) = 220 MPa.
2/3/2015 101
Mpa8.93)10(31.3
15007.2
Z
MK
6
net
fxa
Mpa2.26)10)(155(2
)10)(42(12061.1
J2
TDK
9
3
net
fsxya
Mpa2.104
)]2.26(38.93[)3(
a'
2/1222/12
xya
2
xaa'
2/3/2015 102
Mpa165220)1)(1)(1)(833.0(899.0S
833.0K
899.0K
e
b
a
Marin factors are
Since Se = Sa , the fatigue factor of safety nf is
58.12.104
165
'
Sn
a
af
The first-cycle yield factor of safety is
50.32.104
370
'
Sn
a
y
y
There is no localized yielding; the threat is from fatigue.
2/3/2015 103
Figure 6–32 Designer’s fatigue diagram
(b) 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
The steady bending stress component σxm is
The von Mises components are
2/3/2015 104
Mpa8.93)10(31.3
15007.2
Z
MK
6
net
mfxm
Mpa7.19)10)(155(2
)10)(42(9061.1
J2
DTK
Mpa3.15)10)(155(2
)10)(42(7061.1
J2
DTK
9
3
net
mfsxym
9
3
net
afsxya
Mpa8.99))7.19(38.93()3(
Mpa5.26)]3.15(3[)3(
2/1222/12
xym
2m
'
2/122/12
xyaa'
xm
From Table 6–7, p. 307, the fatigue factor of safety is
From the same table, with r = σ′a/σ′m= 26.5/99.8 = 0.28,
the strengths can be shown to be Sa = 85.5 MPa and Sm
= 305 MPa. See the plot.
The first-cycle yield factor of safety ny is
There is no notch yielding. The likelihood of failure may
first come from first-cycle yielding at the notch. 2/3/2015 105
12.3)5.26(440
)165)(8.99(211
165
5.26
8.99
440
2
1
S
S211
S
S
2
1n
22
2
a'
ut
em'
e
a'2
m'
utf
93.28.995.26
370
''
Sn
ma
y
y
Varying, Fluctuating Stresses; Cumulative Fatigue
Damage
Instead of a single fully reversed stress history block
composed of n cycles, suppose a machine part, at a
critical location, is subjected to
oA fully reversed stress σ1 for n1 cycles, σ2 for n2
cycles, . . . , or
oA “wiggly” time line of stress exhibiting many and
different peaks and valleys.
The method which will be used here to count the
number of cycles it is called the rain-flow counting.
2/3/2015 106
2/3/2015 107
Figure 6–33 Variable stress diagram prepared for assessing cumulative damage
The Palmgren-Miner cycle-ratio summation rule, also
called Miner’s rule, is written
Where,
ni is the number of cycles of operation at stress level σi.
Ni is the number of cycles to failure at stress level σi.
c is the empirical constant in the range 0.7 < c < 2.2.
Using the deterministic formulation as a linear damage
rule we write
where D is the accumulated damage. When D = c = 1,
failure ensues.
2/3/2015 108
cN
n
i
i
i
i
N
nD
Example: Given a part with Sut = 151 kpsi and at the
critical location of the part, Se = 67.5 kpsi. For the
loading of Figure below, estimate the number of
repetitions of the stress-time block in the figure that can
be made before failure.
2/3/2015 109
Solution:
For the figure, We will start to count the number of cycles
and find the midrange stresses σm and amplitude of
stresses σa for each one to construct the table below:
From Fig. 6–18, for Sut = 151 kpsi, f = 0.795.
2/3/2015 110
2/3/2015 111
kpsi5.2135.67
)151*795.0(
S
) Sf(a
2
e
2
ut
0833.05.67
)151(795.0log
3
1)
S
Sflog(
3
1b
e
ut
)2( 5.213
S
a
SN
0833.01
fb
1
f
So,
We prepare to add two columns to the previous table. Using
the Gerber fatigue criterion, with Se = Sf , and n = 1, we can
write
)3(
0 S
0 )S/(1S
me
m2
utm
a
f
where Sf is the fatigue strength associated with a completely
reversed stress, σrev, equivalent to the fluctuating stresses.
Cycle 1: r = σa/σm = 70/10 = 7, and the strength amplitude
from table 6-7 is
Since σa > Sa, that is, 70 > 67.2, life is reduced. From Eq.(3)
2/3/2015 112
kpsi2.67151)7(
)5.67(211
)5.67(2
)151(7
rS
S211
S2
SrS
222
2
ut
e
e
2
ut
2
a
kpsi3.70)151/10(1
70
)S/(1S
22
utm
af
and from Eq. (2)
Cycle 2: r = 10/50 = 0.2, and the strength amplitude is
Since σa < Sa, that is 10 < 24.2, then Sf = Se and indefinite
life follows. Thus, N→∞.
Cycle 3: r = 10/−30 = −0.333, and since σm < 0, Sf = Se,
indefinite life follows and N→∞
2/3/2015 113
cycles )10(6195.213
3.70N 3
0833.01
kpsi2.24151)2.0(
)5.67(211
)5.67(2
)151(2.0S
222
a
From Eq. (6–58) the damage per block is
2/3/2015 114
)10(619n
1)10(619
nD
11
)10(619
1n
N
nD
3
o
3
o
3o
i
i
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