MECH 401 Mechanical Design ApplicationsDr. M. O’Malley– Master Notes
Spring 2006Dr. D. M. McStravickRice University
Nature of fatigue failure
Starts with a crack Usually at a stress concentration
Crack propagates until the material fractures suddenlyFatigue failure is typically sudden and complete, and doesn’t give warning
Fatigue Failure Examples
Door Stop [Text fig 6-12]Rocker Arm [Text fig. 6-9]Auto Door Handle [Text fig. 6-7]Large Flanged Connection FailureJacob Neu chair failure
Fatigue
Fatigue strength and endurance limitEstimating FS and ELModifying factors
Thus far we’ve studied static failure of machine elementsThe second major class of component failure is due to dynamic loading
Variable stressesRepeated stressesAlternating stressesFluctuating stresses
The ultimate strength of a material (Su) is the maximum stress a material can sustain before failure assuming the load is applied only once and heldA material can also fail by being loaded repeatedly to a stress level that is LESS than Su
Fatigue failure
Approach to fatigue failure in analysis and design
Fatigue-life methods (7-3 to 7-6)Stress-life method (rest of chapter 7)
Fatigue-life methods
Three major methodsStress-lifeStrain-lifeLinear-elastic fracture mechanics
Each predict life in number of cycles to failure, N, for a specified level of loadingLow-cycle fatigue
1≤N≤103 cyclesHigh-cycle fatigue
N>103 cycles
The 3 major methods
Stress-lifeBased on stress levels onlyLeast accurate for low-cycle fatigueMost traditional
Easiest to implementAmple supporting dataRepresents high-cycle applications adequately
Strain-lifeMore detailed analysis of plastic deformation at localized regionsGood for low-cycle fatigue applicationsSome uncertainties exist in the results
Linear-elastic fracture mechanicsAssumes crack is already present and detectedPredicts crack growth with respect to stress intensityPractical when applied to large structures in conjunction with computer codes and periodic inspection
Strain-life method
Fatigue failure almost always begins at local discontinuity
Notch, crack or other SCWhen stress at discontinuity > elastic limit, plastic strain occursFatigue fracture occurs for cyclic plastic strains
Can find fatigue life given strain and other cyclic characteristics
Often the designer does not have these a priori
Linear-elastic fracture mechanics method
Stage 1 – crystal slip through several contiguous grainsStage 2 – crack extensionStage 3 – fractureMethod involves
Determining stress intensity as function of crack lengthFrom here, determine life
In reality, computer programs are used to calculate fatigue crack growth and therefore onset of failure
Fatigue analysis
Always good engineering practice to conduct a testing program on the materials to be employed in design and manufactureIs actually a requirement in guarding against possibility of fatigue failureBecause of this necessity, it would really be unnecessary for usto proceed in study of fatigue failure except for one important reason:
The desire to know why fatigue failures occur so that the most effective method or methods can be used to improve fatigue strength
Stress-life method Least accurate for low-cycleMost traditionalWe will come back to this method later
Fatigue analysis
2 primary classifications of fatigue
Alternating – no DC component
Fluctuating – non-zero DC component
Analysis of alternating stresses
As the number of cycles increases, the fatigue strength Sf (the point of failure due to fatigue loading) decreasesFor steel and titanium, this fatigue strength is never less than the endurance limit, SeOur design criteria is:
As the number of cycles approaches infinity (N ∞), Sf(N) = Se (for iron or Steel)
a
f NSσ
η′
=)(
Method of calculating fatigue strength
Seems like we should be able to use graphs like this to calculate our fatigue strength if we know the material and the number of cyclesWe could use our factor of safety equation as our design equation
But there are a couple of problems with this approach
S-N information is difficult to obtain and thus is much more scarce than σ−ε informationS-N diagram is created for a lab specimen
SmoothCircularIdeal conditions
Therefore, we need analytical methods for estimating Sf(N) and Se
a
f NSσ
η′
=)(
Terminology and notationInfinite life versus finite life
Infinite life Implies N ∞Use endurance limit (Se) of materialLowest value for strength
Finite lifeImplies we know a value of N (number of cycles)Use fatigue strength (Sf) of the material (higher than Se)
Prime (‘) versus no primeVariables with a ‘ (Se’)
Implies that the value of that strength (endurance limit) applies to a LAB SPECIMEN in controlled conditions
Variables without a ‘ (Se, Sf)Implies that the value of that strength applies to an actual case
First we find the prime value for our situation (Se’)Then we will modify this value to account for differences between a lab specimen and our actual situationThis will give us Se (depending on whether we are considering infinite life or finite life)Note that our design equation uses Sf, so we won’t be able to account for safety factors until we have calculated Se’ and Se
a
f NSσ
η′
=)(
Estimating Se’ – Steel and Iron
For steels and irons, we can estimate the endurance limit (Se’) based on the ultimate strength of the material (Sut)
SteelSe’ = 0.504 Sut for Sut < 212 ksi (1460 MPa)
= 107 ksi (740 MPa) for all other values of Sut
IronSe’ = 0.4 Sut for Sut < 60 ksi (400 MPa)
= 24 ksi (160 MPa) for all other values of Sut
Estimating Se’ – Aluminum and Copper Alloys
For aluminum and copper alloys, there is no endurance limitEventually, these materials will fail due to repeated loadingTo come up with an “equivalent” endurance limit, designers typically use the value of the fatigue strength (Sf’) at 108 cycles
Aluminum alloysSe’ (Sf at 108 cycles) = 0.4 Sut for Sut < 48 ksi (330 MPa)
= 19 ksi (130 MPa) for all other values of Sut
Copper alloysSe’ (Sf at 108 cycles) = 0.4 Sut for Sut < 40 ksi (280 MPa)
= 14 ksi (100 MPa) for all other values of Sut
Constructing an estimated S-N diagram
Note that Se’ is going to be our material strength due to “infinite”loading
We can estimate an S-N diagram and see the difference in fatigue strength after repeated loading
For steel and iron, note that the fatigue strength (Sf) is never less than the endurance limit (Se’)
For aluminum and copper, note that the fatigue strength (Sf) eventually goes to zero (failure!), but we will use the value of Sf at 108 cycles as our endurance limit (Se’) for these materials
Estimating the value of Sf
When we are studying a case of fatigue with a known number of cycles (N), we need to calculate the fatigue strength (Sf)We have two S-N diagrams
One for steel and ironOne for aluminum and copper
We will use these diagrams to come up with equations for calculating Sf for a known number of cyclesNote: Book indicates that 0.9 is not actually a constant, and uses the variable f to donate this multiplier. We will in general use 0.9.
Estimating Sf (N)
For steel and iron
For aluminum and copper
( )
( ) ( ) bSaSSb
aNNS
ut
e
ut
bf
39.0loglog
9.0log31
−=
⎟⎟⎠
⎞⎜⎜⎝
⎛′
−=
=
( )
( ) ( ) bSaSSb
aNNS
ut
e
ut
bf
39.0loglog
9.0log7.5
1
−=
⎟⎟⎠
⎞⎜⎜⎝
⎛′
−=
= For N < 108
Where Se’ is the value of Sf at N = 108
For 103 < N < 106
Correction factorsNow we have Se’ (infinite life) We need to account for differences between the lab specimen and a real specimen (material, manufacturing, environment, design)We use correction factors
Strength reduction factorsMarin modification factors
These will account for differences between an ideal lab specimen and real lifeSe = ka kb kc kd ke kf Se’
ka – surface factorkb – size factorkc – load factorkd – temperature factorke – reliability factorKf – miscellaneous-effects factorModification factors have been found empirically and are described in section 7-9 of Shigley-Mischke-Budynas (see examples)
If calculating fatigue strength for finite life, (Sf), use equations on previous slide
Endurance limit modifying factors
Surface (ka)Accounts for different surface finishes
Ground, machined, cold-drawn, hot-rolled, as-forgedSize (kb)
Different factors depending on loadingBending and torsion (see pg. 329)Axial (kb = 1)
Loading (kc)Endurance limits differ with Sut based on fatigue loading (bending, axial, torsion)
Temperature (kd)Accounts for effects of operating temperature
Reliability (ke)Accounts for scatter of data from actual test resultsWe will probably not address ke
Miscellaneous-effects (kf)Accounts for reduction in endurance limit due to all other effectsReminder that these must be accounted for
Residual stressesCorrosionetc
Now what?
Now that we know the strength of our part under non-laboratory conditions…… how do we use it?
Choose a failure criterionPredict failure
Part will fail if:σ’ > Sf(N)Factor of safety:η = Sf(N) / σ’Life of part
b = - 1/3 log (0.9 Sut / Se) log(a) = log (0.9 Sut) - 3b
b
aN
1
⎟⎠⎞
⎜⎝⎛ ′
=σ
Stress concentrations and fatigue failure
Unlike with static loading, both ductile and brittle materials are significantly affected by stress concentrations for repeated loading casesWe use stress concentration factors to modify the nominal stressSC factor is different for ductile and brittle materials
SC factor – fatigue
σ = kfσnom+ = kfσo
τ = kfsτnom = kfsτo
kf is a reduced value of kT and σo is the nominal stress.kf called fatigue stress concentration factorWhy reduced? Some materials are not fully sensitive to the presence of notches (SC’s) therefore, depending on the material, we reduce the effect of the SC
Fatigue SC factor
kf = [1 + q(kt – 1)]kfs = [1 + qshear(kts – 1)]
kt or kts and nominal stressesPages 982-988
q and qshearNotch sensitivity factorFind using figures 7-20 and 7-21 in book (SMB) for steels and aluminumsUse q = 0.20 for cast iron
Brittle materials have low sensitivity to notchesAs kf approaches kt, q increasing (sensitivity to notches, SC’s)If kf ~ 1, insensitive (q = 0)
Property of the material
Example
AISI 1020 as-rolled steelMachined finishFind Fmax for:
η = 1.8Infinite life
Design Equation:η = Se / σ’
Se because infinite life
Example, cont.
η = Se / σ’What do we need?
Se
σ’Considerations?
Infinite life, steelModification factorsStress concentration (hole)
Find σ’nom (without SC)
( ) ( ) FFhdb
PAP
nom 2083101260
=−
=−
==′σ
Example, cont.
Now add SC factor:
From Fig. 7-20,r = 6 mmSut = 448 MPa = 65.0 ksiq ~ 0.8
( )[ ] nomtnomf kqk σσσ ′−+=′=′ 11
Example, cont.
From Fig. A-15-1,Unloaded holed/b = 12/60 = 0.2kt ~ 2.5
q = 0.8kt = 2.5σ’nom = 2083 F
( )[ ]( )[ ] ( )
( )FF
kq nomt
4583208315.28.01
11
=′−+=′
′−+=′
σσ
σσ
Example, cont.
Now, estimate SeSteel:Se’ = 0.504 Sut for Sut < 1400 MPa (eqn. 7-8)
740 MPa elseAISI 1020 As-rolled
Sut = 448 MPaSe’ = 0.506(448) = 227 MPa
Constructing an estimated S-N diagram
Note that Se’ is going to be our material strength due to “infinite”loading
We can estimate an S-N diagram and see the difference in fatigue strength after repeated loading
For steel and iron, note that the fatigue strength (Sf) is never less than the endurance limit (Se’)
For aluminum and copper, note that the fatigue strength (Sf) eventually goes to zero (failure!), but we will use the value of Sf at 108 cycles as our endurance limit (Se’) for these materials
Correction factorsNow we have Se’ (infinite life) We need to account for differences between the lab specimen and a real specimen (material, manufacturing, environment, design)We use correction factors
Strength reduction factorsMarin modification factors
These will account for differences between an ideal lab specimen and real lifeSe = ka kb kc kd ke kf Se’
ka – surface factorkb – size factorkc – load factorkd – temperature factorke – reliability factorKf – miscellaneous-effects factorModification factors have been found empirically and are described in section 7-9 of Shigley-Mischke-Budynas (see examples)
If calculating fatigue strength for finite life, (Sf), use equations on previous slide
Example, cont.
Modification factorsSurface: ka = aSut
b (Eq. 7-18)a and b from Table 7-4Machined
ka = (4.45)(448)-0.265 = 0.88
Example, cont.
Temperature: kd = 1 (no info given)
Reliability: ke = 1 (no info given)
Miscellaneous:kf = 1
Endurance limit:Se = kakbkckdkekfSe’ = (0.88)(0.85)(227) = 177 MPa
Design Equation:( )
( ) kN 4.218.14583
10x177
8.14583177
6==
==′
=
F
FMPaSe
ση
Fluctuating stresses
Mean Stress
Stress amplitude
Together, σm and σacharacterize fluctuating stress
2minmax σσσ +
=m
2minmax σσσ −
=a
Failure criterion for fluctuating loading
SoderbergModified GoodmanGerberASME-ellipticYielding
Points above the line: failureBook uses Goodman primarily
Straight line, therefore easy algebraEasily graphed, every time, for every problemReveals subtleties of insight into fatigue problemsAnswers can be scaled from the diagrams as a check on the algebra
Fluctuating stresses, cont.
As with alternating stresses, fluctuating stresses have been investigated in an empirical mannerFor σm < 0 (compressive mean stress)
σa > Sf FailureSame as with alternating stressesOr,
Static Failure
For σm > 0 (tensile mean stress)Modified Goodman criteria
η < 1 Failure
)S(or max ucycam S≥−= σσσ
ησσ 1
=+ut
m
f
a
SS
Fluctuating stresses, cont.
Relationship is easily seen by plotting:
Goodman LineSafe design region(for arbitrary fluctuationsin σm and σa )
(safe stress line)
1=+ut
m
f
a
SSσσ
ησσ 1
=+ut
m
f
a
SS
Note: σm + σa = σmax
σm + σa > Syt (static failure by yielding)
Important point:Important point: Part can fail because of fluctuations in either σa, σm, or both.Design for prescribed variations in σa and σm to get a more exact solution.
Special cases of fluctuating stresses
Case 3: σa / σm fixed
Case 4: both vary arbitrarily
m
m
a
a SSσσ
η ==
ut
m
f
a
SSσσ
η+=
1
Example
Given:Sut = 1400 MPaSyt = 950 MPaHeat-treated (as-forged)Fmean = 9.36 kNFmax = 10.67 kNd/w = 0.133; d/h = 0.55
Find:η for infinite life, assuming Fmean is constant
Example, cont.
Find σm and σa( ) ( )
( )( )
( )( )
MPa28
MPa228
MPa200
Nm 8003.0100.67x141
41
22
Nm 7023.010x36.941
41
22
m 009.02
mx1016.3181075121
121
121
max
maxmaxmax
max
3max
maxmax
3
max
48333
=−=
==
==
===⎟⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛=
===⎟⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛=
==
=−=−==
=
−
ma
mm
mm
m
IyM
IyM
LFLFM
LFLFM
hy
hdwbhI
IMy
σσσ
σ
σ
σ
Example, cont.
Since this is uniaxialloading, σm = 200 MPaσa = 28 MPa
We need to take care of the SC factors
Su = 1400Mpa
kt ~ 2.2 (Figure A15-2)q ~ 0.95 (Table 7-20)kf = 2.14
( )11 −+= tf kqk
nominal
( )( )( )( ) MPa 42820014.2
MPa 602814.2
===′=
===′=
nom
nom
mfmm
afaa
k
k
σσσ
σσσ
Example, cont.
Find strengthEqn. 7-8: S’e = .504Sut
Modification factors
( )
86.0
24.1
808.0
mm 51d2.8 :19)-(7Equation
:Size
107.0
21
eq
=
=
=
≤≤
−
b
eqb
eq
k
dk
hbd
MPa1400S since MPa 700~ ut =′eS
201.0995.0
271
:Surface
=−=
==
a
buta
kba
aSk25)-7 (Eq. 1
Bending:Load
=ck
( )( )( ) MPa12170086.0201.0 ==eS
Example, cont.
Design criteriaGoodman line:
For arbitrary variation in σa and σm,
1=+ut
m
e
a
SSσσ
121
1400
11400121
=+ ma σσ
25.11400428
121601
11400121
=
+=
=+
ηη
ησσ ma
Example, cont.
However, we know that Fmean = constant from problem statement
σm = constant
4.16084
MPa84
11400428
121
1
===
=
=+
=+
a
a
a
a
ut
m
e
a
SS
SSS
S
ση
σ
Less conservative!
Combined loading and fatigueSize factor depends on loadingSC factors also depend on loadingCould be very complicated calculation to keep track of each loadcaseAssuming all stress components are completely reversing and are always in time phase with each other,1. For the strength, use the fully corrected endurance limit for
bending, Se2. Apply the appropriate fatigue SC factors to the torsional stress,
the bending stress, and the axial stress components3. Multiply any alternating axial stress components by the factor
1/kc,ax4. Enter the resultant stresses into a Mohr’s circle analysis to find
the principal stresses5. Using the results of step 4, find the von Mises alternating stress
σa’6. Compare σa’ with Sa to find the factor of safety
Additional details are in Section 7-14