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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 pins 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, Aloha Flight 243, a Boeing 737, an upper part of the plane's cabin area rips off in mid-flight. Metal fatigue was the cause of the failure. MAE dept., SJSU

  9. Fracture Surface Characteristics Mode of fracture Typical surface characteristics Ductile Cup and ConeDimplesDull SurfaceInclusion at the bottom of the dimple Brittle Intergranular ShinyGrain Boundary cracking Brittle Transgranular ShinyCleavage fracturesFlat Fatigue BeachmarksStriations (SEM)Initiation sitesPropagation zoneFinal fracture zone MAE dept., SJSU

  10. 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 MAE dept., SJSU

  11. 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

  12. 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

  13. ′ Se Infinite life Finite life S′e = endurance limit of the specimen Fatigue Failure, S-N Curve N > 103 N < 103 MAE dept., SJSU

  14. 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

  15. ′ ′ 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

  16. ′ 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

  17. ′ ′ Se = Cload Csize Csurf Ctemp Crel (Se) Sf = Cload Csize Csurf Ctemp Crel (Sf) 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 or • Load factor, Cload (page 326, Norton’s 3rd ed.) MAE dept., SJSU

  18. 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. 327, Norton’s 3rd 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

  19. A95 d d95 = .95d dequiv = ( )1/2 0.0766 Rectangular parts Solid or hollow non-rotating 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

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

  21. Csurf = A (Sut)b Correction Factors for Specimen’s Endurance Limit • surface factor, Csurf(p. 328-9, Norton’s 3rd 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

  22. Correction Factors for Specimen’s Endurance Limit • Temperature factor, Ctemp(p.331, Norton’s 3rd 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 ≤ 450 oC (840 oF) MAE dept., SJSU

  23. Correction Factors for Specimen’s Endurance Limit • Reliability factor, Crel(p. 331, Norton’s 3rd ed.) The reliability correction factor accounts for the scatter and uncertainty of material properties (endurance limit). MAE dept., SJSU

  24. Notch sensitivity factor Fatigue stress concentration factor Kf=1+ (Kt–1)q (p. 340, Norton’s 3rd 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. The stress concentration factor seems to be sensitive to the notch radius and the ultimate strength of the material. MAE dept., SJSU

  25. Fatigue Stress Concentration Factor, q for Aluminum (p. 341, Norton’s 3rd ed.) MAE dept., SJSU

  26. Use the design equation to calculate the size Se Kf a = n Design process – Fully Reversed Loading for Infinite Life • Determine the maximum alternating applied stress (a )in terms of the size and 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 • Determine the fatigue stress concentration factor, Kf • 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

  27. 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 MAE dept., SJSU

  28. 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” MAE dept., SJSU

  29. 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

  30. 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

  31. 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

  32. 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

  33. 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

  34. 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

  35. 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

  36. 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

  37. 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

  38. 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

  39. 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

  40. 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

  41. Fillet r 4 = .16 = 25 d Kt = 1.76 D 35 = = 1.4 d 25 Example A section of a component is shown. The material is steel with Sut = 620 MPa and a fully corrected endurance limit of Se = 180 MPa. The applied axial load varies from 2,000 to 10,000 N. Use modified Goodman diagram and find the safety factor at the fillet A, groove B and hole C. Which location is likely to fail first? Use Kfm = 1 Pm = (Pmax + Pmin) / 2 = 6000 N Pa = (Pmax – Pmin) / 2 = 4000 N MAE dept., SJSU

  42. Calculate the alternating and the mean stresses, Pa 4000 a= 1.65 = 52.8 MPa Kf = A 25x5 Pm 6000 m= = 48 MPa = A 25x5 Fatigue design equation a m 1 Infinite life = + 1 n Sut 52.8 48 Se n = 2.7 → = + n 180 620 Example Using r = 4and Sut = 620 MPa, q (notch sensitivity) = .85 Kf= 1 + (Kt – 1)q = 1 + .85(1.76 – 1) = 1.65 MAE dept., SJSU

  43. d 5 → Kt= 2.6 = = .143 w 35 Using r = 2.5and Sut = 620 MPa, q (notch sensitivity) = .82 Calculate the alternating and the mean stresses, Pm 6000 m= = 40 MPa = A 30x5 Pa 4000 a= 2.3 = 61.33 MPa Kf = A (35-5)5 1 61.33 40 n = 2.5 → = + n 180 620 Example Hole Kf= 1 + (Kt – 1)q = 1 + .82(2.6 – 1) = 2.3 MAE dept., SJSU

  44. Using r = 3and Sut = 620 MPa, q (notch sensitivity) = .83 Calculate the alternating and the mean stresses, Pm 6000 m= = 41.4 MPa = A 29x5 Pa 4000 a= 2.1 = 58.0 MPa Kf = A (35-6)5 1 58.0 41.4 n = 2.57 → = + n 180 620 Example Groove r 3 = .103 = 29 d → Kt= 2.33 D 35 = = 1.2 d 29 Kf= 1 + (Kt – 1)q = 1 + .83(2.33 – 1) = 2.1 The part is likely to fail at the hole, has the lowest safety factor MAE dept., SJSU

  45. Fa= (Fmax – Fmin) / 2=7.5 lb. Fm= (Fmax + Fmin) / 2=22.5 lb. Ma=7.5 x 16 = 120 in - lb Mm=22.5 x 16 = 360 in - lb 32Ma 32(120) Mc a= = = 23178.6 psi = I πd 3 π(.375)3 32Mm 32(360) Mc m= = 69536 psi = = I πd 3 π(.375)3 Example The figure shows a formed round wire cantilever spring subjected to a varying force F. The wire is made of steel with Sut = 150 ksi. The mounting detail is such that the stress concentration could be neglected. A visual inspection of the spring indicates that the surface finish corresponds closely to a hot-rolled finish. For a reliability of 99%, what number of load applications is likely to cause failure. MAE dept., SJSU

  46. A95= .010462d 2 (non-rotating round section) dequiv= √ A95 / .0766 = .37d = .37 x.375 = .14 1 23178.6 69536 n = .7< 1 → a → m = + 1 n 24077 150000 = + Finite life n Sut Se Find Sn, strength for finite number of cycle a m 23178.6 69536 Sn = 43207 psi → 1 → = 1 + = + Sut Sn Sn 150000 Example Calculate the endurance limit Csurf = A (Sut)b = 14.4(150)-.718 = .394 Cload = 1 (pure bending) Ctemp= 1 (room temp) Crel= .814 (99% reliability) dequiv= .14 < .3 → Csize = 1.0 Se = Cload Csize Csurf Ctemp Crel (Se) = (.394)(.814)(.5x150) = 24.077 ksi MAE dept., SJSU

  47. 24.077 ( ) log ⅓ .9x150 N ( ) 24077 = → 43207 106 Se ( ) .9Sut log ⅓ N ( ) Se Sn = 106 Example N = 96,000 cycles MAE dept., SJSU

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