1ME311 Machine DesignW Dornfeld06Oct2014Fairfield UniversitySchool of EngineeringLecture 5: Fully-Reversing Fatigue and the S-N DiagramDiscovering FatigueHamrock Section 7.2With the Industrial Revolution came the steam engine (~1760-1775) and then the locomotive (~1830). This began an age of dynamic loading that had never been seen before.Even though equipment was designed with stresses below Yield, failures were still occurring. This was very publicly visible in the failure of bridges and of train wheel axles.2The Versailles Train CrashOne of the worst rail disasters of the 19th century occurred in May 1842 near Versailles, France. Following celebrations at the Palace of Versailles, a train returning to Paris crashed after the leading locomotive broke an axle. The carriages behind piled into the wrecked engines and caught fire. At least 55 passengers were trapped in the carriages and died.Examination of several broken axles from British railway vehicles showed that they had failed by brittle cracking across their diameters (fatigue). The problem of broken axles was widespread on all railways at the time, and continued to occur for many years before engineers developed better axle designs, mostly resulting from improved testing of axles.From Wikipedia.The S-N DiagramA German engineer, August Wöhler, discovered that reversed loading was the cause, and in 1871 he produced the first S-N diagram:The diagram shows the relationship between the stress amplitude of reversing loading and the number of cycles to failure. Note that some materials have an Endurance Limit – a stress level below which the material can take unlimited cycles.SutEndurance Limit, Se-8-6-4-2024680 400TimeStress3What’s up with fatigue?Stage I: Initiation with Small Cracks• Shear driven• Interact with microstructure• Mostly analyzed by continuum mechanics approachesStage II: Propagation with Large Cracks• Tension driven• Fairly insensitive to microstructure• Mostly analyzed by fracture mechanics modelsMicrocracks act as localized stress concentrations to exceed Sy and support local plastic yielding. As the crack grows, the section is reduced, stresses increase, and propagation accelerates until the part fractures – usually rapidly and unexpectedly.From Anders Ekberg, Chalmers U.Crack Growth and FailureThe progression of fatigue cracking is sometimes visible as “beach marks” with each line representing the crack growth due to one load cycle:The left photo shows that the crack grew and the remaining section reduced until P/A exceeded yield and the remainder of the part broke by yielding. From Majid Mirzaei, TMU.4Characterizing FatigueFatigue is very complex, so has been largely driven by test data from rotating beam tests.• Standard R.R.Moore test specimen is 0.3 in. diameter at thinnest section• Specimen is polished• Each revolution gives one fully reversing cycle• At 3600RPM = 60 Hz, get 216,000 cycles per hour, 5.2 million cycles per dayHamrock Section 7.5Ø 0.3”Although moment is uniform, the narrowing specimen has max bending stress at the middle.The S-N DiagramThe S-N diagram has three regions for materials with Endurance Limits: Low Cycle, High Cycle, & Infinite LifeThe Low Cycle Fatigue (LCF) region is from 1 to 1000 cycles.The High Cycle Fatigue (HCF) region is from 1000 to 1 million cycles.The Infinite Life region is above 1 million cycles.Some consider the Infinite Life region to begin at 10 million cycles.SutEndurance Limit, SeS-N Diagram for Steel Specimen in Bending with Sut = 1000204060801001201 10 100 1,000 10,000 100,000 1,000,000 10,000,000 100,000,000No. of Stress Cycles (N)Fatigue Strength (Sf) SutLCFHCF∞∞∞∞Hamrock Section 7.65Constructing our S-N Diagram1. Start with Sut at 1 cycle (Because it can endure 1 cycle to Sut)2. Plot the LCF value SL’ at 1000 cycles3. Plot the Endurance Strength Se’ at 1 million cycles4. Draw the first and third segments as straight lines5. Connect SL’ and Se’ with a curve = aNbS-N Diagram for Steel Specimen in Bending with Sut = 1000204060801001201 10 100 1,000 10,000 100,000 1,000,000 10,000,000 100,000,000No. of Stress Cycles (N)Fatigue Strength (Sf) SutSL' = 0.90 SutSe' = 0.50 SutaNbSL’ and Se’ Depend on the Type of LoadingFind the fatigue strength at the 1000 cycle Low Cycle Fatigue point, SL’For steel, use:Bending: SL’ = 0.90 SutAxial: SL’ = 0.75 Sut (Eqn. 7.8)Torsion: SL’ = 0.72 SutFind the “raw” endurance strength, Se’For steel, use (from Eqn. 7.7):Bending: Se’ = 0.50 Sut up to a max Se’ of 100 ksiAxial: Se’ = 0.45 Sut up to a max Se’ of 90 ksiTorsion: Se’ = 0.29 Sut up to a max Se’ of 58 ksiThis is Se’ for a polished 0.3” diameter bar.This is because Se’ tops out at ~100 ksi. See Fig. 7.86Plotting the 1000 to 1 million CurveConnect (1000, SL’) and (106, Se) with the curvewhere and N = Number of cycles.If you have aor cand b, and want to know the life for an alternating stress σaltbetween Se’ and SL’, computeHamrock uses this.bCfbfNSoraNS 10==−==='log,')(log,')('10312'102'eLeLeLSSbandSSCSSabbCaltaltoraN1110=σσI use this.S-N ExercisePlot the S-N diagram for a steel with Sut = 120ksi under fully reversing Axial loading. Find a & b, and the Sf at 100,000 cycles.S-N Diagram for Steel Specimen with Sut = 1200204060801001201401 10 100 1,000 10,000 100,000 1,000,000 10,000,000 100,000,000No. of Stress Cycles (N)Fatigue Strength (Sf)7Reality Sets InIf your actual product doesn’t happen to be 0.3” diameter polished bars, then you must make some modifications to the Endurance Limit to make this real.The modification (Marin) factors are:kfSurface effect – if not polished ksSize effect – if not < 0.3” diameterkrReliability effect – if other than 50% survivalktTemperature effect – if not at Room TemperaturekmMiscellaneous – Mat’l processing, Residual stress, Coatings, CorrosionThen Se = kfkskrktkmSe’(My Part) (Test Specimen)Another factor is Kf, the fatigue stress concentration factor.Main 3Surface Finish FactorEquation 7.19 giveswhere Sut is the material’s Ultimate Tensile Strength, and e & f are coefficients defined in Table 7.3.Note 1: e is NOT the “exponential”.Note 2: Sut value is entered as MPa or ksi, NOT Pa or psi. In other words, if Sut = 100ksi, you use the value 100.Exercise: What is kffor AISI 1080 steel that has a machined surface?Why do we need to