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'08 DMcSLectureNotes-Chapter6abWeb

Jan 17, 2016




MECH 401 Machine Design

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

Approach to fatigue failure in analysis and designFatigue-life methods (6-3 to 6-6)Stress Life Method (Used in this course)Strain Life MethodLinear Elastic Fracture Mechanics MethodStress-life method (rest of chapter 6)Addresses high cycle Fatigue (>103 ) WellNot Accurate for Low Cycle Fatigue ( 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 = 2.14


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

Modification factors

Example, cont.Design criteriaGoodman line:

For arbitrary variation in sa and sm,


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

Less 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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