Example 1
For the following R=-1 AISI 1090 steel test data, plot two S-N curves, one using loglinear coordinates and the other using log-log coordinates. a) Use linear regression to
estimate the best fit curves for the two coordinate systems. Show these lines on the
same graphs with the data b) Estimate the fatigue limit for the material, c) estimate the
fatigue strength at 5x105 cycles to failure, d) estimate the expected fatigue life at 200
MPa, e) Sketch upper and lower bounds for the data and comment on the observed
scatter.
Stress amplitude (MPa)
130
130
130
130
130
150
150
150
150
170
170
170
170
185
185
185
185
220
220
220
220
220
245
245
245
245
Nf
1x107 (no failure)
1750000
1600000
1930000
1x107 (no failure)
1860000
1100000
601300
485000
190567
465000
153140
311250
144430
152060
176960
116430
46240
52020
62500
95000
65300
30100
38500
26300
29600
Solution
The equations of the least square method for fitting of the straight line y
B
n6xiyi 6xi6yi
n6x 2 i (6xi)
2
A
A B
are,
6yi
6xi
B
, where
n
n
n is number of data points and xi , y are values of data. Least square method with respect
to log y is exploited as follows:
The power function y=AxB is by taking logarithms transformed to log y = log A+Blog x and
the previously given equations for A and B are used. Hence, in the diagram with log-log
scales, y=AxB becomes a straight line. In this case, n = 24 because the two non-failed data
points should be excluded from the analysis. We compute the needed values and sums to
obtain the constants A and B. It MUST be noted that in linear regression, the dependent
variable y is Log(cycles to failure)
log-log
y=AxB
coordinates
Nf=A(Va)B
Log(Nf)=A’+BLog(Va)
log linear
y=A+Bx
coordinates
Log(Nf)=A+BVa
A’=19.94
A=10A’= 5.5x1019
B=-6.48
A = 8.11
B = -0.0152
2 run-out data points are
excluded from linear regression
Y=19.94 -6.48X
Y = 8.11 - 0.0152X
2 run-out data points are
excluded from linear regression
These values are transformed to represent the traditional plotting method with Log(Nf) on
the horizontal axis
log-log
Va =C(Nf)D
coordinates Log (Va)=C’+DLog(Nf)
log linear
Va = C+DLog(Nf)
coordinates
C’ = - A’/B = 3.08
C =10C’= 1202
D =1/B = -0.154
C = -A/B = 536
D =1/B = -66.0
Va
Nf
Linear-log plot with regression lines for Va =1202(Nf)-0.154 (green line) and Va = 536+(66.0)Log(Nf)(violet line)
'()
220
200
180
Va
160
140
120
&f
Log-log plot with regression lines for Va =1202(Nf)-0.154 (green line) and Va = 536+(66.0)Log(Nf)(violet line)
b) Estimate the fatigue limit for the material
The exact fatigue limit cannot analytically be defined; it is, however, Va § 130 MPa,
because two of five specimens did not fail with this stress.
c) estimate the fatigue strength at 5x105 cycles to failure,
Using Va =1202(Nf)-0.154 the strength is Va = 159 MPa
Using Va =536+(-66.0)Log(Nf) the strength is Va = 160 MPa
The figures show that these curves nearly intersect at 500 000 cycles
d d Using (Va/1202) (1/-0.154) =Nf the life is Nf = 114000 cycles
Using Log-1((536-Va ) /66.0) = Nf the life is Nf = 123000 cycles
e ! "d #$! %#"d #! data and comment on the observed scatter.
Scatter was found especially at longer fatigue lives. A standard deviation for all data could
be estimated.
Example 2
E
Example 2
An unnotched member fabricated from AISI 4142 steel (see table) is subjected to the
load history shown below. Use the Gerber mean stress correction equation and a)
perform a rainflow count of the load history, b) estimate the number of cycles to failure
and c) estimate the number of (blocks) repetitions to failure.
Rainflow counting yield 3 large amplitude cycles and 200 small amplitude cycles
Use the Gerber equation:
2
Va §Vm ·
¸ ,
1
¨
V ar ¨© V u ¸¹
§ § V ·2 ·
¨1 ¨ m ¸ ¸V
¨ ¨© V u ¸¹ ¸ ar
©
¹
Va
§ § V ·2 ·
¨1 ¨ m ¸ ¸ AN b
f
¨ ¨© V u ¸¹ ¸
©
¹
From table for 4142 steel:
A
1
MPa
m
= 600 MPa.
§ § 600 · 2 ·
¨1 ¨
¸1837 N f 0.0762 , which gives Nf =
¨ © 1757 ¸¹ ¸
©
¹
600
For
0.0
1757 MPa
u
For
b
m
= 880 MPa.
§ § 880 · 2 ·
¨1 ¨
¸1837 N f 0.0762 , which gives Nf =
¨ © 1757 ¸¹ ¸
©
¹
320
Stress amplitude:
a
max
min
Mean stress:
m max min
The stress history Rainflow Table:
a
N fj
N j / N fj
600
600
4,68E+05
6,41E-06
860
340
1,13E+08
1,77E-06
j
Nj
!
1
3
0
1200
2
200
520
1200
Ȉ
203
8,18E-06
The Palmgren-Miner rule:
j
¦N
Nj
j
"
Bf
k 1
fj
Bf
§ j Nj
¨¦
¨k 1 N
fj
©
¦N
k 1
1
Nj
1,
where B f is the number of repetitions.
fj
·
¸ | #$$%%% repetitions = 24 800 000 cycles
¸
¹
Example 2b
An aircraft component is fabricated from Ti-6Al-4V solution treated and aged titanium
alloy (see table with Problem 3). At the critical point in the component, the stress history is
determined to be
Develop a rainflow table of for the history shown. Estimate the number of repetitions of the
given history necessary to cause fatigue failure.
The Palmgren-Miner rule:
j
¦N
Nj
j
1
Bf
fj
¦N
Nj
k 1
fj
Stress amplitude:
a
k 1
1,
where B f is the number of repetitions.
max
min
2
Equivalent completely reversed stress for the cases where mean levels are involved:
V ar
Nf
V maxV a
V ´f 2 N f
1 §¨ V max V a
2 ¨© V 'f
b
, V ar
V maxV a is known as SWT-equation (Smith, Watson & Topper).
1
·b
¸
¸
¹
For Ti-6Al-4V, b = -0.104 and
The stress history:
´f
= 2030 MPa.
Rainflow Table:
j
Nj
m
m
N j
N / N 1
400
200
800
300
4,32E+05
9,26E-04
2
8000
-400
200
300
3,39E+08
2,36E-05
3
1
-400
1000
700
2,51E+03
3,98E-04
Ȉ
Bf
§
¨
¨
©
j
¦
k 1
Nj
N fj
·
¸
¸
¹
1
| 7 repetitions.
1,35E-03
Example 3
A stepped shaft has a diameter D = 50mm, d = 30mm and r = 3mm. determine and
compare Kt for a) axial loading and b) bending and c) torsion. Estimate the fatigue
notch factor, Kf for R=-1 axial loading if the shaft is made of the following steels: 1)
normalized 1038 (HB 163), 2) quenched and tempered 1038 (HB 195), 3) hot rolled
4340 (HB 243)
and 4) quenched and tempered 4340 (HB 409).
r
d
3
30
0.1
D
d
50
1.6
30
The elastic stress concentration factor Kt for axial loading is defined from Figure:
K t | 1.9
Kt for bending loading is defined from Figure:
K t | 1.7
Kt for torsion loading is defined from Figure:
K t | 1.4
A value of fatigue notch factor k f could be obtained according to Peterson for R = -1 loading from
kf
1
kt 1
1
a
, where a 0.0254§¨ 2070 ·¸
© Su ¹
1.8
(for steels).
r
1.8
§ 2070 ·
The value of a can be obtained for 1038 (HB 163), Su = 582 MPa, 0.0254¨
¸
© 582 ¹
We obtain k f
1
1.90 1
0.249
1
3
1.
0.2 .
1.8
§ 2070 ·
For 1038 (HB 195), Su = 649 MPa, 0.0254¨
¸
© 649 ¹
We obtain k f
1
1.90 1
0.205
1
3
0. .
1.
1.8
§ 2070 ·
For 4340 (HB 243) Su = 827 MPa, 0.0254¨
¸
© 827 ¹
We obtain k f
1
1.90 1
0.132
1
3
0. .
1.
1.8
§ 2070 ·
For 4340 (HB 409), Su = 1468 MPa, 0.0254¨
¸
© 1468 ¹
We obtain k f
1
1.90 1
0.047
1
3
0.0 .
1.
The following summary and comparison of the values of Kt shows that axial is the most critical
loading type. Kt for bending loading is only slightly less critical. However for torsion, Kt is only
about 74 % compared to bending load.
Kt
1,90
1,75
1,40
Kt %
100
92
73,7
axial loading
bending
torsion
Kf
1038 (HB 163), Su = 582 MPa
1,83
1038 (HB 195), Su = 649 MPa
1,84
4340 (HB 243), Su = 827 MPa
1,86
4340 (HB 409), Su = 1468 MPa
1,89
a
Example 4
Example 4Exa
The smooth specimen is subjected to cyclic loading. The required fatigue life is
100 000 cycles to failure. Compute the allowable stress amplitude for mean stress 0, 200,
400 and 800 MPa. Use the Goodman mean stress correction equation. Use this data to
construct a constant life diagram for Nf = 1X105. (see the sample constant life diagram
below taken from the lecture slides)
The Goodman’s equation generally gives reasonable results:
1
Va
a
ar
m
,
Vu
§ Vm ·
¨¨1 ¸¸V ar
© Vu ¹
§ Vm ·
¨¨1 ¸¸ A
N f b
© Vu ¹
From table for 4142 steel:
A
u
For
Va
1 MPa
b
1757 MPa
m
= 100 MPa.
V ·
§
0.0762
, at Nf = 100 000 cycles, we get
¨1 m ¸1837 100000 © 1757 ¹
ıa
MPa
Nf cycles
801,19 1
Therefore:
Va
For
Va
0.0
801.19N f
m
0.0648
= 0 MPa.
9382 N f
0.0648
ıa MPa Nf cycles
896,8
1
Therefore:
Va
For
Va
896.8N f
0.0648
= -100 MPa.
938 100 2 N f
ıa
MPa
0.0648
Nf cycles
992,41 1
Therefore:
Va
992.41N f
0.0648
A stepped shaft with a diameter D = 30 mm, d = 24 mm and r = 1.3 mm. The required
fatigue life is 1 000 000 cycle and is loaded in bending. a) Determine Kt and Kf , b)
Compute the allowable stress amplitude for mean stress 0, 200, 400 and 800 MPa and c)
use this data to construct a constant life diagram for Nf = 1X106.
Kt for bending loading is defined from Figure:
D/d = 30/24 = 1.25, r/d = 1.3/24 § 0.054167.
Kt | 1.9
A value of fatigue notch factor k f could be obtained according to Peterson for R = -1 loading from
kf
kt 1
1
1
a
, where a 0.0254§¨ 2070 ·¸
1.8
© Su ¹
(for steels).
r
1 .8
§ 2070 ·
The value of a can be obtained, 0.0254¨
¸
© 1757 ¹
We obtain k f
1
0.0
.
1.98 1
1.
0.0341
1
2
Using the value of k f , the nominal stress amplitude corresponding to the fatigue limit is:
S ar
ar
k
Example 5
For RQC-100 steel and using the Morrow mean stress correction, obtain equations relating
stress amplitude(Va)vs. cycles to failure (Nf) in the cases where the mean stress(Vm)is a)
100 MPa tension, b) zero and c) 100 MPa compression. Plot the curves on the same graph
using log-log coordinates.
The Morrow’s equation generally gives reasonable results for ductile steels:
1
a
ar
m
V'
,
f
V ar
V 'f 2 N f
Va
V 'f
b
V m 2 N f
b
From table for RQC-100:
$ 'f
!"# MPa
b
0.%&()
For
Va
*m
= 100 MPa.
+938 100 , 2 N f
ıa
MPa
0.0648 , by substituting Nf = 1 cycle, we get
Nf cycles
801,19 1
Therefore:
Va
For
Va
0.0648
801.19-N f .
1/
= 0 MPa.
93822 N f 3
0.0648
ıa MPa Nf cycles
896,8
1
Therefore:
Va
For
Va
0.0648
896.84N f 5
1/
= -100 MPa.
6938 100 7 2 N f
0.0648
ıa
MPa
Nf cycles
992,41 1
Therefore:
Va
0.0648
992.41:N f ;
Sm
=
S @ ? ><==
S @ ? <==
Example 6
SometestdatapointsforthestressͲstraincurveof2024ͲT351aluminiumduringaxialstressare
given:
V,MPa
H
317
0.00474
341
0.00607
366
0.00950
390
0.01910
414
0.03290
439
0.05230
TheelasticmodulusforthematerialisgivenasE=73100MPa.
a) determinetheplasticstrain,Hp,foreachpointandplotVvsHponalogͲloggraphand
determinetheconstantsHandnfortheRambergͲOsgoodmaterialmodel(seeslide13
page6fromlectureslideset5)
b) PlottheresultingH=f(V)lineandtestdata(usealinearͲlineargraph).Determinerthe0.2%
offsetyieldstrength.
Foreachpoint,Histhetotalstrain,He=V/EandHp=HͲHe
V,MPa
317
341
366
390
414
439
H
0.00474
0.00607
0.00950
0.01910
0.03290
0.05230
He
0,004337
0,004665
0,005007
0,005335
0,005663
0,006005
Hp
0,000403
0,001405
0,004493
0,013765
0,027237
0,046295
Theresultinglog–logplotisshownandthevaluesHandnfortheequation
ߪ ଵൗ௡
ߝ௣ ൌ ቀ ቁ ‫ܪ‬
arefoundbylinearregression.H=528MPa,n=0.066
1000
stress(MPa)
H=528MPa
1/n
1
0,0001
0,001
0,01
pstrain
0,1
1
totalstrainisfromtheRambergͲOsgoodcurve
ߪ
ߪ ଵൗ௡
ߪ
ߪ ଵൗ଴Ǥ଴଺଺
ߝ ൌ ൅ቀ ቁ ൌ
൅ቀ
ቁ
‫ܧ‬
‫ܪ‬
͹͵ͳͲͲ
ͷʹͺ
Theresultingcurvewiththedatapointsisshown
500
0
450
offsetyield
400
350
stress(MPa)
300
250
200
150
100
50
0
0
0,01
offset
0,02
0,03
0,04
0,05
0,06
s
The0.2%offsetyieldstrengthoccurswhenHp=0.2%=0.002.ThisisV=350.3MPa
Example 7
Anelastic,linearhardeningmaterialhaselasticmodulusE=200GPa,yieldstrengthV0=fy=500
MPaandavalueofG=0.1(seeslide11page5fromlectureslideset5).Assumethatthematerial
behavesaccordingtotheelasticͲlinearhardeningspringͲslidermodel(seelectureslideset4).
Estimate(sketch)thematerialbehaviorfor:
a) completelyreversedcyclicstrainingatHa=0.006
b) strainingfrom H=0toH=0.012followedbydecreasingstraintoH=0.005andthenincreasing
straintoH=0.015.
UsingthespringͲslidermaterialmodelforelasticͲlinearhardeningmaterialandtheparameters
given,theresultingstressͲstrainresponseduringmonotonicloadingwouldbeasshown.
Forcompletelyreversedcyclicloading,thestressstrainresponseisasshown.Notethatoncethe
straincycleisreversed,thematerialyieldonlywhenthechangeinstrainattainstwicetheinitial
yieldstrength.Seethefigure.Forcompletelyreversedyieldingthecycleisalwaysperfectly
symmetric.
b)ThestraincycleforstrainingfromH=0toH=0.012followedbydecreasingstraintoH=0.005and
thenincreasingstraintoH=0.015isshown
Example 8
Anotchedcomponenthaskt=3.Thecomponentisloadeduntilnominalstress,S=200MPa.The
componentisthenunloadedtoanominalstressofS=0.StrainͲlifepropertiesofthematerialare:
E=100GPa,Vf’=1000MPa,Hf’=1.0,b=Ͳ0.08andc=Ͳ0.60.
a) determinethelocalstressandlocalstrainatthenotchatS=200MPa.
b) determinetheresiduallocalstressandlocalstrainatthenotchatS=0MPa
c) usetheNeuberanalysisandMorrowmeanstresscorrectiontoestimatethefatiguelifeof
thecomponent.
Findthefollowingequationsfromthelecturenotesfortheweek6andsolvethemissingconstants
forthecyclicstressͲstraincurve:
H' n'
' f
H ' f b / c b
c
H'
1000MPa
10.08 / 0.60 n'
0.08
0.60
1000MPa
2
1 Neuber´s rule and the cyclic stressͲstrain curve are used to estimate both the maximums and
amplitudes of the local notch stress and strain. For the initial monotonic response, assumed to
followthecyclicstressͲstraincurve,wehave:
H max
f V max 1
V max
§V
· nc
¨ max ¸
E
© Hc ¹
1
V max
§ V max · 2 / 15
¨
¸
100000MPa © 1000MPa ¹
Eliminationof betweenequationsgivesanequationinvolvingstressthatcanbesolvedbytrial
anderrororothernumericalprocedure.However,e.g.acalculatorwithanumericalsolvercanbe
suggestedtobeusedtosolve .
1
(k t S max ) 2
§ V · nc
V 2 V ˜ E¨ ¸
© H c ¹ ,Substituting intooriginalequationgives .
1
3 ˜ 200MPa 2
V
§
· 2 / 15
V 2 V ˜ 100000 MPa ¨
¸
© 1000 MPa ¹
max
463.54166 ...M max
| 463.5M
then
1
15
·2
¸
¹
§ 463 5
¨
© 1000
463 5
100000
D D
3 3
0 0
1
c
D D
3 3
0 0
·
§V
¨
¸
c
¹
©
Q
[
D
P+
[
D(
P
[
D
P
I
V
[
D
P
H
V
7 77 ˜ 10 3 Forcyclicloading,itisconvenienttoworkinthetermsofamplitudesusingthecyclicNeuber`srule
andthesamestressͲstraincurve,butevaluatethesefor:
200
D
3
Q
L
P
[
D
P
'
Theneededequationsare:
·2
¸
¹
1
'V
§
2¨
© 2 ˜ 1000
D
3
0
'V
100000
D
3
0
1
c
Q
+
§ 'V ·
2¨
¸
© 2 c¹
(
'H
'V
and
3 0 ˜ 200
D D
3 3
0 0
2
(
W
'
6
.
'V'H
2
100000
Solvingthesegives:
'H
590
D
3
'
.
3 ,
6 11 ˜ 10 126
'H
464 590
[
D
P
H
'
D
3
[
D
P
Q
L
P
H
Q
L
P
D
3
0
Theminimumvaluesforcyclicloadingcanbeobtained:
7 77 ˜ 103 6 11 ˜ 103 | 1 66 ˜ 10 ,andfor1(b) "
$ #126
!
D
3
!
Q
L
P
| 464
D
3
[
D
P
Hence,thesolutionfor1(a)is "
.
TheestimatedstressͲstrainresponseisshowninthefollowingfigure.Thiswasplottedbyequating
'&
2 f ('% / 2) forthebothbranchesofthehysteresisloop.
Morrow'smeanstresscorrectionestimateisknownas:
'H
2
,a
+ ' f + m
E
(2 N f *b H ' f (2 N f *c
wehave:
1
I
1
I
100000
I
2 0 08 1 02 0 60
| 4 4 ˜ 10 5 cycles/
1
D
3 D
0 3
0
-1000 168.
3 06 ˜ 10 3
Exmpale 9
UsingthematerialpropertiesgiveninTableA.2,ConstructthemonotonicandcyclicstressͲsrain
curvesforRQCͲ100hotͲrolledsteelforstrainbetween0and2%.Doesthismaterialcyclically
hardenorsoften?
fromtable
‫ ܧ‬ൌ ʹͲ͹ͲͲͲ‫ܽܲܯ‬
݊ ൌ ͲǤͲ͸
݊Ԣ ൌ ͲǤͳͶ
‫ ܭ‬ൌ ͳͳ͹ʹ‫ܽܲܯ‬
‫ܭ‬Ԣ ൌ ͳͶ͵Ͷ‫ܽܲܯ‬
ߪ
ߪ ଵൗ௡
ߪ
ߪ ଵൗ଴Ǥ଴଺
ߝ ൌ ൅ቀ ቁ ൌ
൅ቀ
ቁ
‫ܧ‬
‫ܭ‬
ʹͲ͹ͲͲͲ
ͳͳ͹ʹ
ߪ௔
ߪ௔ ଵൗ௡ᇱ
ߪ௔
ߪ௔ ଵൗ଴Ǥଵସ
ߝ௔ ൌ
൅ቀ ቁ
ൌ
൅ቀ
ቁ
‫ܧ‬
‫ܭ‬Ԣ
ʹͲ͹ͲͲͲ
ͳͶ͵Ͷ
m
cyclic
V / Va
H / a
Overthisrangeofstrain,themonotoniccurveisabovethecycliccurve,thereforethematerial
softens.