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Fatigue Failure

Fatigue failure is characterized by three stages. Crack Initiation. Crack Propagation. Final Fracture. Fatigue Failure.

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Fatigue Failure

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  1. Fatigue failure is characterized by three stages • Crack Initiation • Crack Propagation • Final Fracture Fatigue Failure It has been recognized that a metal subjected to a repetitive or fluctuating stress will fail at a stress much lower than that required to cause failure on a single application of load. Failures occurring under conditions of dynamic loading are called fatigue failures. MAE dept., SJSU

  2. Crack initiation site Fracture zone Propagation zone, striation Jack hammer component, shows no yielding before fracture. MAE dept., SJSU

  3. VW crank shaft – fatigue failure due to cyclic bending and torsional stresses Propagation zone, striations Crack initiation site Fracture area MAE dept., SJSU

  4. 928 Porsche timing pulley Crack started at the fillet MAE dept., SJSU

  5. Fracture surface of a failed bolt. The fracture surface exhibited beach marks, which is characteristic of a fatigue failure. 1.0-in. diameter steel pin from agricultural equipment. Material; AISI/SAE 4140 low allow carbon steel MAE dept., SJSU

  6. bicycle crank spider arm This long term fatigue crack in a high quality component took a considerable time to nucleate from a machining mark between the spider arms on this highly stressed surface. However once initiated propagation was rapid and accelerating as shown in the increased spacing of the 'beach marks' on the surface caused by the advancing fatigue crack. MAE dept., SJSU

  7. Crank shaft Gear tooth failure MAE dept., SJSU

  8. Hawaii (1988), Aloha Flight 243, a Boeing 737, an upper part of the plane's cabin area rips off in mid-flight. Metal fatigue compounded by corrosion MAE dept., SJSU

  9. Fracture Surface Characteristics Mode of fracture Typical surface characteristics Ductile Cup and ConeDimplesDull SurfaceInclusion at the bottom of the dimple MAE dept., SJSU

  10. Brittle Fracture The cracks usually travel so fast that you can't tell when the material is about to break. In other words, there is very little plastic deformation before failure occurs. In most cases, this is the worst type of fracture because you can't repair visible damage in a part or structure before it breaks. Cleavage fractures Grain Boundary cracking MAE dept., SJSU

  11. Brittle fracture surfaces slightly bumpy crack surface Chevron Fracture Surface Brittle fracture in a mild steel (shinny surface) MAE dept., SJSU

  12. MAE dept., SJSU

  13. Low temperatures can severely embrittle steels. The Liberty ships, produced in great numbers during the WWII were the first all-welded ships. A significant number of ships failed by catastrophic fracture. Fatigue cracks nucleated at the corners of square hatches and propagated rapidly by brittle fracture. MAE dept., SJSU

  14. Striations (beach markesInitiation sitesPropagation zoneFinal fracture zone Fatigue MAE dept., SJSU

  15. max a = - min max = Alternating stress min max a = 2 min = 0 Mean stress m = a = max / 2 min max + m= 2 Fatigue Failure – Type of Fluctuating Stresses a = Alternating Stress m = Mean Stress MAE dept., SJSU

  16. Typical testing apparatus, pure bending Motor Load Rotating beam machine – applies fully reverse bending stress Fatigue Failure, S-N Curve Test specimen geometry for R.R. Moore rotating beam machine. The surface is polished in the axial direction. A constant bending load is applied. MAE dept., SJSU

  17. The standard machine operates at an adjustable speed of 500 RPM to 10,000 RPM. At the nominal rate of 10,000 RPM, the R. R. Moore machine completes 600,000 cycles per hour, 14,400,000 cycles per day. Bending moment capacity 20 in-lb to 200 in-lb MAE dept., SJSU

  18. ′ Se Finite life Infinite life S′e = endurance limit of the specimen Fatigue Failure, S-N Curve (Strength vs. # of cycles to failure) N > 103 N < 103 MAE dept., SJSU

  19. Steel ′ ′ Se = Se = Sut ≤ 200 ksi (1400 MPa) 0.5Sut Sut> 200 ksi 100 ksi Sut> 1400 MPa 700 MPa Cast iron Cast iron 0.4Sut Sut< 60 ksi (400 MPa) Sut ≥60 ksi 24 ksi Sut< 400 MPa 160 MPa Relationship Between Endurance Limit and Ultimate Strength Steel MAE dept., SJSU

  20. ′ ′ Se = Se = Sut < 40 ksi (280 MPa) Sut < 48 ksi (330 MPa) 0.4Sut 0.4Sut Sut ≥ 40 ksi Sut ≥ 48 ksi 14 ksi 19 ksi Sut ≥ 330 MPa Sut ≥ 280 MPa 130 MPa 100 MPa Copper alloys Copper alloys For N = 5x108 cycle Relationship Between Endurance Limit and Ultimate Strength Aluminum Aluminum alloys For N = 5x108 cycle MAE dept., SJSU

  21. ′ Se = endurance limit of the specimen (infinite life > 106) Se = endurance limit of the actual component (infinite life > 106) Sf = fatigue strength of the actual component (infinite life > 5x108) S Se ′ Sf = fatigue strength of the specimen (infinite life > 5x108) N 106 103 For materials that do not exhibit a knee in the S-N curve, the infinite life taken at 5x108 cycles S Sf N 5x108 103 Correction Factors for Specimen’s Endurance Limit For materials exhibiting a knee in the S-N curve at 106 cycles MAE dept., SJSU

  22. Pure bending Cload = 1 Pure axial Cload = 0.7 Pure torsion Cload = 1 if von Mises stress is used, use 0.577 if von Mises stress is NOT used. Combined loading Cload = 1 Correction Factors for Specimen’s Endurance Limit Material exhibits a knee in S-N curve, infinite life at 106cycle Se = Cload Csize Csurf Ctemp Crel (Se)’ Material does not exhibit a knee in S-N curve, infinite life at 5x108 cycle Sf = Cload Csize Csurf Ctemp Crel (Sf) • Load factor, Cload (page 330, Norton’s 4th ed. or page 362 in 5th ed.) MAE dept., SJSU

  23. For rotating solid round cross section d ≤ 0.3 in. (8 mm) Csize = 1 0.3 in. < d ≤ 10 in. Csize = .869(d)-0.097 8 mm < d ≤ 250 mm Csize = 1.189(d)-0.097 Correction Factors for Specimen’s Endurance Limit • Size factor, Csize(p. 331, Norton’s 4th ed. or page 363 in 5th ed.) Larger parts fail at lower stresses than smaller parts. This is mainly due to the higher probability of flaws being present in larger components. If the component is larger than 10 in., use Csize = .6 MAE dept., SJSU

  24. A95 d d95 = .95d dequiv = ( )1/2 0.0766 Solid or hollow round non-rotating parts Rectangular parts dequiv = .37d dequiv = .808 (bh)1/2 Correction Factors for Specimen’s Endurance Limit For non rotating components, use the 95% area approach to calculate the equivalent diameter. Then use this equivalent diameter in the previous equations to calculate the size factor. A95 = (π/4)[d2 – (.95d)2] = .0766 d2 MAE dept., SJSU

  25. Correction Factors for Specimen’s Endurance Limit I beams and C channels MAE dept., SJSU

  26. Csurf = A (Sut)b Correction Factors for Specimen’s Endurance Limit • surface factor, Csurf(p. 332/3, Norton’s 4th ed. or page 363/4 in 5th ed.) The rotating beam test specimen has a polished surface. Most components do not have a polished surface. Scratches and imperfections on the surface act like a stress raisers and reduce the fatigue life of a part. Use either the graph or the equation with the table shown below. MAE dept., SJSU

  27. Correction Factors for Specimen’s Endurance Limit • Temperature factor, Ctemp(p.335, Norton’s 4th ed. or page 367 in 5th ed.)) High temperatures reduce the fatigue life of a component. For accurate results, use an environmental chamber and obtain the endurance limit experimentally at the desired temperature. For operating temperature below 450 oC (840 oF) the temperature factor should be taken as one. Ctemp = 1 for T ≤ 840 oF (450 oC) Ctemp = 1 – 0.0032(T – 840) for 840 oF < T ≤ 1020 oF MAE dept., SJSU

  28. Correction Factors for Specimen’s Endurance Limit • Reliability factor, Crel(p. 335, Norton’s 4th ed. or page 367 in 5th ed.) The reliability correction factor accounts for the scatter and uncertainty of material properties (endurance limit). MAE dept., SJSU

  29. Notch sensitivity factor Fatigue stress concentration factor Kf=1+ (Kt–1)q (p. 344, Norton’s 4th ed. or p. 376 in 5th ed.) Steel Fatigue Stress Concentration Factor, Kf Experimental data shows that the actual stress concentration factor is not as high as indicated by the theoretical value, Kt (depends on the geometry only). The stress concentration factor seems to be sensitive to the notch radius and the ultimate strength of the material. Theoretical stress concentration factor, Kt MAE dept., SJSU

  30. Fatigue Stress Concentration Factor, q for Aluminum (p. 345, Norton’s 4th ed. or p. 376 in 5th ed.) Steel q = .6 for r =.04 and Sut = 50,000 psi Aluminum – heat treated q = .52 for r =.04 and Sut = 50,000 psi MAE dept., SJSU

  31. Use the design equation to calculate the size for infinite life Se Kf a = n Design process – Fully Reversed Loading for Infinite Life Mean stress = 0 • Determine the maximum alternating applied stress (a )in terms of the size for the selected cross sectional profile • Select material → Sy, Sut • Choose a safety factor → n • Determine all modifying factors and calculate the endurance limit of the component → Se (infinite life 106) or, Sf (infinite life 5x108) • Determine the fatigue stress concentration factor, Kf , q and Kt Sf Kf a = or n • Investigate different cross sections (profiles), optimize for size or weight • You may also assume a profile and size, calculate the alternating stress and determine the safety factor. Iterate until you obtain the desired safety factor MAE dept., SJSU

  32. ′ Se Finite life Infinite life S′e = endurance limit of the specimen Fatigue Failure, S-N Curve (Strength vs. # of cycles to failure) N > 103 N < 103 MAE dept., SJSU

  33. Sn = a (N)b equation of the fatigue line A A S S B B Sf Se Sn = .9Sut N N 106 5x108 103 103 Point A N = 103 Sn = .9Sut Sn = Se Sn = Sf Point A Point B Point B N = 103 N = 106 N = 5x108 Design for Finite Life Strength for finite number of cycle MAE dept., SJSU

  34. log .9Sut = loga + blog103 logSe = loga + blog106 (.9Sut)2 Se .9Sut ( ) a 1 = b log .9Sut = Se log ⅓ Se 3 N ( ) Se Sn = 106 Sn Kf a = n CalculateSnand replace Sein the design equation Design equation Design for Finite Life Sn = a (N)b log Sn = log a + b log N Apply boundary conditions for point A and B to find the two constants “a” and “b” or Kf (na) = Sn MAE dept., SJSU

  35. Norton’s book (Machine Design) notations MAE dept., SJSU

  36. Norton’s book (Machine Design) notations N = N1 = 103 For all materials N = N2 = 106 For materials showing the knee N = N2 For materials that exhibit no knee. Select N2 and look up the value for z from the table MAE dept., SJSU

  37. (.9Sut)2 (.9Sut)2 .9Sut .9Sut a a 1 1 = = b b log log = = Se Se Se Se 3 5.699 For material exhibiting an endurance-limit knee N1 = 103 and N = N2 = 106 Bending and combined load For material exhibiting no endurance-limit knee If infinite life is taken as 5x108 then, and N = N2 = 5x108 N1 = 103 MAE dept., SJSU

  38. Review of fatigue strength notations ′ Se= endurance limit of a specimen, material showing a knee Se= corrected endurance limit, material showing a knee ′ Sf= endurance limit of a specimen, material showing NO knee Sf= corrected endurance limit, material showing No knee Sn= Fatigue strength for finite life Sm= Fatigue strength at N=1000 cycle MAE dept., SJSU

  39. a Gerber curve Se Alternating stress Goodman line m Sut Sy Soderberg line Mean stress The Effect of Mean Stress on Fatigue Life Mean stress exist if the loading is of a repeating or fluctuating type. Mean stress is not zero MAE dept., SJSU

  40. Yield line a C Safe zone Alternating stress m The Effect of Mean Stress on Fatigue Life Modified Goodman Diagram Sy Se Goodman line Sut Sy Mean stress MAE dept., SJSU

  41. Safe zone Goodman line - Syc Sut The Effect of Mean Stress on Fatigue Life Modified Goodman Diagram a Sy Yield line Se C Safe zone +m Sy - m MAE dept., SJSU

  42. m > 0 m≤0 Fatigue, Fatigue, Infinite life Se a m 1 a= Finite life = + nf nf a m Se Sut 1 = + Yield Sy Syc Sn Sut a+ m= a+ m= Yield ny ny The Effect of Mean Stress on Fatigue Life Modified Goodman Diagram a Se C Safe zone Safe zone +m Sy Sut - m - Syc MAE dept., SJSU

  43. Calculate the stress concentration factor for the mean stress using the following equation, Kfa Sy Kfm= m Fatigue design equation Kf a Kfmm 1 Infinite life = + nf Se Sut Applying Stress Concentration factor to Alternating and Mean Components of Stress • Determine the fatigue stress concentration factor, Kf, apply directly to the alternating stress → Kfa • If Kf max < Sy then there is no yielding at the notch, use Kfm =Kf and multiply the mean stress by Kfm → Kfmm • If Kf max > Sy then there is local yielding at the notch, material at the notch is strain-hardened. The effect of stress concentration is reduced. MAE dept., SJSU

  44. Calculate the alternating and mean principal stresses, 1a, 2a=(xa /2)±(xa /2)2+(xya)2 1m, 2m=(xm /2)±(xm /2)2+(xym)2 Combined Loading All four components of stress exist, xaalternating component of normal stress xmmean component of normal stress xyaalternating component of shear stress xymmean component of shear stress MAE dept., SJSU

  45. a′ = (1a+ 2a - 1a2a)1/2 m′ = (1m+ 2m - 1m2m)1/2 2 2 2 2 ′a ′m 1 = + nf Se Sut Fatigue design equation Infinite life Combined Loading Calculate the alternating and mean von Mises stresses, MAE dept., SJSU

  46. Calculate the support forces, R1= 2500, R2= 7500 lb. The critical location is at the fillet, MA= 2500 x 12 = 30,000 lb-in 32M 305577 Mc a= Calculate the alternating stress, = = I πd 3 d 3 Determine the stress concentration factor D r = 1.5 = .1 Kt = 1.7 d d 10,000 lb. Design Example 6˝ 6˝ 12˝ A rotating shaft is carrying 10,000 lb force as shown. The shaft is made of steel with Sut = 120 ksi and Sy = 90 ksi. The shaft is rotating at 1150 rpm and has a machine finish surface. Determine the diameter, d, for 75 minutes life. Use safety factor of 1.6 and 50% reliability. d D = 1.5d A R1 R2 r (fillet radius) = .1d m= 0 MAE dept., SJSU

  47. Using r = .1and Sut = 120 ksi, q (notch sensitivity) = .85 Csurf = A (Sut)b = 2.7(120)-.265 = .759 0.3 in. < d ≤ 10 in. Csize = .869(d)-0.097 = .869(1)-0.097 = .869 Design Example Assume d = 1.0 in Kf= 1 + (Kt – 1)q = 1 + .85(1.7 – 1) = 1.6 Calculate the endurance limit Cload = 1 (pure bending) Crel = 1 (50% rel.) Ctemp= 1 (room temp) ′ Se = Cload Csize Csurf Ctemp Crel (Se)= (.759)(.869)(.5x120) = 39.57 ksi MAE dept., SJSU

  48. 39.57 ( ) log ⅓ .9x120 86250 ( ) Sn 39.57 = = 56.5 ksi 106 Sn 56.5 305577 n = a= = .116 < 1.6 = = 305.577 ksi Kfa 1.6x305.577 d 3 So d = 1.0 in. is too small Se ( ) .9Sut log ⅓ Assume d = 2.5 in N ( ) Se Sn = All factors remain the same except the size factor and notch sensitivity. 106 Using r = .25and Sut = 120 ksi, q (notch sensitivity) = .9 Kf= 1 + (Kt – 1)q = 1 + .9(1.7 – 1) = 1.63 Se =36.2 ksi → Design Example Design life, N = 1150 x 75 = 86250 cycles Csize = .869(d)-0.097 = .869(2.5)-0.097 = .795 MAE dept., SJSU

  49. 36.2 ( ) log ⅓ .9x120 86250 ( ) Sn 36.20 = = 53.35 ksi 305577 a= 106 = 19.55 ksi (2.5)3 Sn 53.35 n = = 1.67 ≈ 1.6 = Kfa 1.63x19.55 d=2.5 in. Check yielding Sy 90 n= 2.8 > 1.6 okay = = Kfmax 1.63x19.55 Design Example Se =36.2 ksi → MAE dept., SJSU

  50. 6˝ 6˝ 12˝ d D = 1.5d A R1 R2 = 7500 Sn r (fillet radius) = .1d 56.5 n = Calculate an approximate diameter = .116 < 1.6 = Kfa 1.6x305.577 Sn 56.5 n = → d = 2.4 in. = = 1.6 So d = 1.0 in. is too small Kfa 1.6x305.577/d 3 Check the location of maximum moment for possible failure Design Example – Observations So, your next guess should be between 2.25 to 2.5 Mmax (under the load) = 7500 x 6 = 45,000 lb-in MA (at the fillet) = 2500 x 12 = 30,000 lb-in But, applying the fatigue stress conc. Factor of 1.63,Kf MA = 1.63x30,000 = 48,900 > 45,000 MAE dept., SJSU

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