Fatigue IIFatigue II
Lecture 11Lecture 11
Engineering 473Engineering 473Machine DesignMachine Design
Finite Life EstimatesFinite Life Estimates
mσ Stress,Mean
aσ Stress,gAlternatin
utS
eS
ytS
ytS
How can the life of a part be estimated if the mean stress-alternating stress pair lie above the Goodman line?
Goodman DiagramGoodman Diagram
Infinite LifeStress State
Finite Life(Cycles to failure?)
SS--N CurveN Curve
Shigley, Fig. 7-6
Completely reversed cyclic stress, UNS G41200 steel
The S-N curve gives the cycles to failure for a completely reversed (R=-1) uniaxial stress state.
What do you do if the stress state is not completely reversed?
DefinitionsDefinitions
2σσσ
2σσσ
σσσ
minmaxm
minmaxa
minmaxr
+=
−=
−=Stress RangeStress Range
Alternating StressAlternating Stress
Mean StressMean Stress
max
min
σσR =
m
a
σσA =
Stress RatioStress Ratio Amplitude RatioAmplitude RatioNote that R=-1 for a completely reversed stress state with zero mean stress.
FluctuatingFluctuating--Stress Failure Stress Failure Interaction CurvesInteraction Curves
Shigley, Fig. 7-16
The interaction curves provide relationships between alternating stress and mean stress.
When the mean stress is zero, the alternating component is equal to the endurance limit.
The interaction curves are for infinite life or a large number of cycles.
Goodman Interaction LineGoodman Interaction Line
1SS
SSk
ut
m
e
af =+Any combination of mean and alternating stress that lies on or below Goodman line will have infinite life.
Factor of Safety FormatFactor of Safety Format
fut
m
e
af
N1
SS
SSk =+
Note that the fatigue stress concentration factor is applied only to the alternating component.
Master Fatigue PlotMaster Fatigue Plot
Shigley, Fig. 7-15
Constant cycles till failure interaction curves.
Equivalent Alternating StressEquivalent Alternating Stress
mσ Stress,Mean
aσ Stress,gAlternatin
utS
eS
ytS
ytS0σa
mσ
=Alternating stress at zero mean stress that fails the part in the same number of cycles as the original stress state.
The red and blue lines are estimated fatigue interaction curves associated with a specific number of cycles to failure.
cycles 106
cycles 105
Number of Cycles to Number of Cycles to FailureFailure
Once the equivalent alternating stress is found, the S-N curve may be used to find the number of cycles to failure.
Equivalent Alternating Stress Equivalent Alternating Stress FormulaFormula
ut
m
f
af0σa
fut
m
0σa
af
fut
m
e
af
Sσ
N1σkσ
N1
Sσ
σσk
N1
Sσ
Sσk
m
m
−=
=+
=+
=
=
Goodman LineGoodman Line
pair. σ and σ original theas cycles ofnumber same in the failure
fatigue causes that stress -1)(R
reversed completely Equivalentσ
ma
0σam
=
≡=
ExampleExample
1 2
1.5 in. dia. 0.875 in. dia.0.125 in. rad.
Material UNS G41200 Steel
lb 2000Plb 0030P
min
max
==5 in 5 in
Notch sensitivityq=0.3
( )
( ) 4442
4441
in 088.0875.064πD
64πI
in 249.05.164πD
64πI
2
1
===
===
34
2
22
34
1
11
in 0.201in 0.438in 0.088
cIS
in 0.332in 0.75in 0.249
cIS
===
===
ExampleExample(Continued)(Continued)
1 2
1.5 in. dia. 0.875 in. dia.0.125 in. rad.
Material UNS G41200 Steel
lb 2000Plb 0003P
min
max
==5 in 5 in
Notch sensitivityq=0.3
( )1kq1k1k1kq
tf
t
f
−+=−−= ( )
18.1)161.1(3.011kq1k
61.1k
tf
t
=−+=−+=
=
0.143875.0125.0
dr
71.1in 0.875
in 1.5dD
==
==
Ref. Peterson
ExampleExample(Continued)(Continued)
1 2
1.5 in. dia. 0.875 in. dia.0.125 in. rad.
Material UNS G41200 Steel
lb 2000Plb 0003P
min
max
==
5 in 5 in
Notch sensitivityq=0.3
( )( )
( )( ) ksi 2.60in 332.0
in 01lb 2000SMσ
ksi 4.90in 332.0
in 10lb 3000SMσ
31
1min
31
1max
===
===
Section 1 (Base)Section 1 (Base)
ksi 3.752σσσ
ksi 1.152σσσ
minmaxm
minmaxa
=+=
=−=
ExampleExample(Continued)(Continued)
1 2
1.5 in. dia. 0.875 in. dia.0.125 in. rad.
Material UNS G41200 Steel
lb 2000Plb 0003P
min
max
==5 in 5 in
Notch sensitivityq=0.3
( )( )
( )( ) ksi 8.49in 201.0
in 5lb 2000SMσ
ksi 6.74in 201.0
in 5lb 3000SMσ
31
1min
31
1max
===
===
Section 2 (Fillet)Section 2 (Fillet)
ksi 2.622σσσ
ksi 4.122σσσ
minmaxm
minmaxa
=+=
=−=
ExampleExample(Continued)(Continued)
Completely reversed cyclic stress, UNS G41200 steel
Shigley, Fig. 7-6
ee
ut
Sksi 30Sksi 116S
==′=
fult
m
e
af
N1
Sσ
Sσk =+
ExampleExample(Continued)(Continued)
ee
ut
Sksi 30Sksi 116S
==′=
fult
m
e
af
N1
Sσ
Sσk =+
( ) 15.1ksi 116ksi 75.3
ksi 30ksi 15.10.1
1Nf
=+
=
Section 1 (Base)Section 1 (Base)
( )( )
( )( ) ksi 2.60in 332.0
in 01lb 2000SMσ
ksi 4.90in 332.0
in 10lb 3000SMσ
31
1min
31
1max
===
===
ksi 3.752σσσ
ksi 1.152σσσ
minmaxm
minmaxa
=+=
=−=
Part has finite life at base.Part has finite life at base.
ExampleExample(Continued)(Continued)
Section 2 (Fillet)Section 2 (Fillet)
ee
ut
Sksi 30Sksi 116S
==′=
fult
m
e
af
N1
Sσ
Sσk =+
( ) 02.1ksi 116ksi 62.2
ksi 30ksi 12.41.18
1Nf
=+
=
Part has finite life.Part has finite life.
( )( )
( )( ) ksi 8.49in 201.0
in 5lb 2000SMσ
ksi 6.74in 201.0
in 5lb 3000SMσ
31
1min
31
1max
===
===
ksi 2.622σσσ
ksi 4.122σσσ
minmaxm
minmaxa
=+=
=−=
Calculation of Equivalent Calculation of Equivalent Alternating StressAlternating Stress
ut
m
f
af0σa
Sσ
N1σkσ
m
−=
=
ksi 0.4311675.3
1.01(1.0)15.1σ
0σam
=
−=
=
ksi 5.3111662.2
1.01
(1.18)12.4σ0σa
m
=
−=
=
ksi 3.75σksi 1.15σ
m
a
==
ksi 2.62σksi 4.12σ
m
a
==
BaseBase FilletFillet
Cycles to Failure EstimateCycles to Failure Estimate
10
20
30
50
70
90
Base
Fillet
MultiMulti--axis Fluctuating axis Fluctuating Stress StatesStress States
Everything presented on fatigue has been based on experiments involving a single stress component.
What do you do for problems in which there are more than one stress component?
Marin Load Factor, Marin Load Factor, kkcc
The endurance limit is a function of the load/stress component used in the test.
edcbae SkkkkS ′⋅⋅⋅⋅=
��
�
��
�
�
>≤
=
shear andTorsion 0.577Bending1
MPa) (1520 ksi 220Sloading Axial1MPa) (1520 ksi 220Sloading Axial0.923
k ut
ut
c
Alternating and Mean Von Alternating and Mean Von Mises Mises StressesStresses
1. Increase the stress caused by an axial force by 1/kc.
2. Multiply each stress component by the appropriate fatigue stress concentration factor.
3. Compute the maximum and minimum von Mises stresses.
4. Compute the alternating and mean stresses based on the maximum and minimum values of the von Mises stress.
5. Use the Goodman alternating and mean stress interaction curve and S-N curve to estimate the number of cycles to failure. Use the reversed bending endurance limit.
Complex LoadsComplex Loads
t, time
Fσ,
cycles, nfor σ
cycles, nfor σcycles, nfor σcycles, nfor σfollows as stresses
reversed completely tosubjected ispart A
mm
33
22
11
�
What is the cumulative effect of these different load cycles?
1σ2σ
3σ
Minor’s RuleMinor’s Rule
CNn
Nn
Nn
Nn
m
m
3
3
2
2
1
1 =++++ �
Cumulative Damage LawCumulative Damage Law
2.2. to0.7 from rangingConstant Ci level stressat failure tocyclesN
i level stressfor cycles ofnumber n
i
i
≡≡≡
C is usually taken as 1.0C is usually taken as 1.0Minor’s Rule is the simplest and most widely used Minor’s Rule is the simplest and most widely used
Cumulative Damage LawCumulative Damage Law
ExampleExample
StressState
1
Cycles(n)
Life(N) N
n
23
1,000
5,000
10,000
2,000
10,000
100,000
0.5
0.5
0.1
1.1 Part will fail
AssignmentAssignment(Problem No. 1)(Problem No. 1)
A rotating shaft is made of 42 x 4 mm AISI 1020 cold-drawn steel tubing and has a 6-mm diameter hole drilled transversely through it. Estimate the factor of safety guarding against fatigue failure when 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. Use the stress concentration factor tables found in the appendices, and estimate the Marin factors using information in the body of the text.
AssignmentAssignment(Problem No. 2)(Problem No. 2)
A solid circular bar with a 5/8 inch diameter is subjected to a reversed bending moment of 1200 in-lb for 2000 cycles, 1000 in-lb for 100,000 cycles and 900 in-lb for 10,000 cycles. Use the S-N curve used in this lecture. Determine whether the bar will fail due to fatigue. Assume all Marin factors are equal to 1.0.
AssignmentAssignment(Problem No. 3)(Problem No. 3)
Same as Problem No. 2 except there is a constant axial force of 5,000 lb acting on the bar in addition to the completely reversed bending moment.