'08 DMcSLectureNotes-Chapter6abWeb

MECH 401 Mechanical Design Applications Dr. M. OMalley Master NotesSpring 2008Dr. D. M. McStravickRice University

ReadingChapter 6HomeworkHW 4 available, due 2-7TestsFundamentals Exam will be in class on 2-21

Nature of fatigue failureStarts with a crack Usually at a stress concentrationCrack propagates until the material fractures suddenlyFatigue failure is typically sudden and complete, and doesnt give warning

Fatigue Failure Examples

Various Fatigue Crack Surfaces [Text fig. 6-2]Bolt Fatigue Failure [Text fig. 6-1]Drive Shaft [Text fig. 6-3]AISI 8640 Pin [Text fig. 6-4]Steam Hammer Piston Rod [Text fig. 6-6]Jacob Neu chair failure (in this classroom)

Fatigue Example 1

Fatigue Failure Example

Fatigue Failure Example

Fatigue Failure Example

Stamping Fatigue Failure Example

Schematic of Various Fatigue Failure

Jim Neu Chair Failure (Pedestal)

Fatigue Failure of Chair Shaft

Seat Fatigue Failure

FatigueFatigue strength and endurance limitEstimating FS and ELModifying factors

Thus far weve studied static failure of machine elementsThe second major class of component failure is due to dynamic loadingVariable stressesRepeated stressesAlternating stressesFluctuating stressesThe 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 SuFatigue failure

More Fatigue Failure Examples (ASM)

More Fatigue Failure Examples (ASM)

More Fatigue Failure Examples

The 3 major methodsStress-lifeBased on stress levels onlyLeast accurate for low-cycle fatigueMost traditionalEasiest to implementAmple supporting dataRepresents high-cycle applications adequatelyStrain-lifeMore detailed analysis of plastic deformation at localized regionsGood for low-cycle fatigue applicationsSome uncertainties exist in the resultsLinear-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

Fatigue analysis2 primary classifications of fatigueAlternating no DC component

Fluctuating non-zero DC component

Analysis of alternating stressesAs 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)

Method of calculating fatigue strengthSeems 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 approachS-N information is difficult to obtain and thus is much more scarce than s-e informationS-N diagram is created for a lab specimenSmoothCircularIdeal conditionsTherefore, we need analytical methods for estimating Sf(N) and Se

Terminology and notationInfinite life versus finite lifeInfinite life Implies N Use endurance limit (Se) of materialLowest value for strengthFinite lifeImplies we know a value of N (number of cycles)Use fatigue strength (Sf) of the material (higher than Se)

Prime () versus no primeStrength variable with a (Se)Implies that the value of that strength (endurance limit) applies to a LAB SPECIMEN in controlled conditionsVariables without a (Se, Sf)Implies that the value of that strength applies to an actual caseFirst 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 wont be able to account for safety factors until we have calculated Se and Se

S-N Plot with Endurance Limit

Estimating Se Aluminum and Copper AlloysFor aluminum and copper alloys, there is no endurance limit Eventually, 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 Sutfor Sut < 48 ksi (330 MPa)= 19 ksi (130 MPa) for all other values of SutCopper alloysSe (Sf at 108 cycles) = 0.4 Sutfor Sut < 35 ksi (250 MPa)= 14 ksi (100 MPa) for all other values of Sut

Constructing an estimated S-N diagramNote 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 diagramsOne for steel and ironOne for aluminum and copperWe 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 [so f=0.9]

Estimating Sf (N)For steel and ironFor f=0.9

For aluminum and copper

For 103 < N < 106For N < 108Where Se is the value of Sf at N = 1085.7

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 factorsStrength reduction factorsMarin modification factorsThese will account for differences between an ideal lab specimen and real lifeSe = ka kb kc kd ke kf Seka surface factorkb size factorkc load factorkd temperature factorke reliability factorKf miscellaneous-effects factorModification factors have been found empirically and are described in section 6-9 of Shigley-Mischke-Budynas (see examples)If calculating fatigue strength for finite life, (Sf), use equations on previous slide

Surface Finish Effect on Se

Temperature Effect on Se

Reliability Factor, ke

Steel Endurance Limit vs. Tensile Strength

Compressive Residual Stresses

Now what?Now that we know the strength of our part under non-laboratory conditions how do we use it?Choose a failure criterionPredict failurePart will fail if:s > Sf(N) Factor of safety or Life of the part:h = Sf(N) / sWhere b = - 1/3 log (0.9 Sut / Se)log (a) = log (0.9 Sut) - 3b

Example Homework Problem 6-9A solid rod cantilevered at one end. The rod is 0.8 m long and supports a completely reversing transverse load at the other end of +/- 1 kN. The material is AISI 1045 hot-rolled steel. If the rod must support this load for 104 cycles with a factor of safety of 1.5, what dimension should the square cross section have? Neglect any stress concentrations at the support end and assume f= 0.9.

Solution: -- See Board Work--

Stress concentration (SC) and fatigue failureUnlike 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 fatigues = kfsnom+ = kfso

t = kfstnom = kfsto

kf is a reduced value of kT and so is the nominal stress.kf called fatigue stress concentration factorWhy reduced? Some materials are not fully sensitive to the presence of notches (SCs) therefore, depending on the material, we reduce the effect of the SC

Fatigue SC factorkf = [1 + q(kt 1)]kfs = [1 + qshear(kts 1)]kt or kts and nominal stressesTable A-15 & 16 (pages 1006-1013 in Appendix)q and qshearNotch sensitivity factorFind using figures 6-20 and 6-21 in book (Shigley) for steels and aluminumsUse q = 0.20 for cast ironBrittle materials have low sensitivity to notchesAs kf approaches kt, q increasing (sensitivity to notches, SCs)If kf ~ 1, insensitive (q = 0)Property of the material

ExampleAISI 1020 as-rolled steelMachined finishFind Fmax for:h = 1.8Infinite life

Design Equation:h = Se / sSe because infinite life

Example, cont.h = Se / sWhat do we need?SesConsiderations?Infinite life, steelModification factorsStress concentration (hole)Find snom (without SC)

Example, cont.Now add SC factor:

From Fig. 6-20,r = 6 mmSut = 448 MPa = 65.0 ksi q ~ 0.8

Example, cont.From Fig. A-15-1,Unloaded holed/b = 12/60 = 0.2kt ~ 2.5q = 0.8kt = 2.5snom = 2083 F

Example, cont.Now, estimate SeSteel:Se = 0.5 Sut for Sut < 1400 MPa (eqn. 6-8) 700 MPa elseAISI 1020 As-rolledSut = 448 MPaSe = 0.50(448) = 224 MPa

Constructing an estimated S-N diagramNote 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 factorsStrength reduction factorsMarin modification factorsThese will account for differences between an ideal lab specimen and real lifeSe = ka kb kc kd ke kf Seka surface factorkb size factorkc load factorkd temperature factorke reliability factorKf miscellaneous-effects factorModification factors have been found empirically and are described in section 6-9 of Shigley-Mischke-Budynas (see examples)If calculating fatigue strength for finite life, (Sf), use equations on previous slide

Example, cont.Modification factorsSurface:ka = aSutb (Eq. 6-19)a and b from Table 6-2Machined

ka = (4.45)(448)-0.265 = 0.88

Example, cont.Size:kb Axial loadingkb = 1 (Eq. 6-21)Load:kcAxial loadingkc = 0.85 (Eq. 6-26)

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:

Fluctuating Fatigue Failures

Alternating vs. fluctuatingAlternatingFluctuating

Alternating Stressessa characterizes alternating stress

Fluctuating stressesMean Stress

Stress amplitude

Together, sm and sa characterize fluctuating stress

Alternating vs. Fluctuating

Modified Goodman Diagram

Fluctuating Stresses in Compression and Tension

Failure criterion for fluctuating loadingSoderbergModified GoodmanGerberASME-ellipticYielding

Points above the line: failureBook uses Goodman primarilyStraight 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

Gerber Langer Plot for Fluctuating Stresses

Fluctuating stresses, cont.As with alternating stresses, fluctuating stresses have been investigated in an empirical mannerFor sm < 0 (compressive mean stress)sa > Sf FailureSame as with alternating stressesOr,

Static Failure

For sm > 0 (tensile mean stress)Modified Goodman criteria

h < 1 Failure

Modified Goodman Langer Equations

Fluctuating stresses, cont.Relationship is easily seen by plotting:Goodman Line(safe stress line)Safe design region(for arbitrary fluctuationsin sm and sa )Note: sm + sa = smax

sm + sa > Syt (static failure by yielding)Important point: Part can fail because of fluctuations in either sa, sm, or both. Design for prescribed variations in sa and sm to get a more exact solution.

Special cases of fluctuating stressesCase 1: sm fixed

Case 2: sa fixed

Special cases of fluctuating stressesCase 3: sa / sm fixed

Case 4: both vary arbitrarily

ExampleGiven:Sut = 1400 MPaSyt = 950 MPaHeat-treated (as-forged)Fmean = 9.36 kNFmax = 10.67 kNd/w = 0.133; d/h = 0.55Find:h for infinite life, assuming Fmean is constant

Example, cont.Find sm and sa

Stress Concentration Factor

Example, cont.Since this is uniaxial loading, sm = 200 MPasa = 28 MPaWe need to take care of the SC factorsSu = 1400Mpa

kt ~ 2.2 (Figure A15-2)q ~ 0.95 (Figure 7-20)kf = 2.14nominal

Example, cont.Find strengthEqn. 7-8: Se = .504Sut

Modification factors

Example, cont.Design criteriaGoodman line:

For arbitrary variation in sa and sm,

1211400

Example, cont.However, we know that Fmean = constant from problem statementsm = constantLess conservative!

Combined loading and fatigueSize factor depends on loadingSC factors also depend on loadingCould be very complicated calculation to keep track of each load caseAssuming all stress components are completely reversing and are always in time phase with each other,For the strength, use the fully corrected endurance limit for bending, SeApply the appropriate fatigue SC factors to the torsional stress, the bending stress, and the axial stress componentsMultiply any alternating axial stress components by the factor 1/kc,axEnter the resultant stresses into a Mohrs circle analysis to find the principal stressesUsing the results of step 4, find the von Mises alternating stress saCompare sa with Sa to find the factor of safetyAdditional details are in Section 6-14

More Fatigue Failure Examples

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