Tuesday May 25 2021 Chapter 6 Fatigue Failure

Tuesday, May 25, 2021 Chapter 6 Fatigue Failure Resulting from Variable Loading 1 Dr. Mohammad Suliman Abuhaiba, PE

2 Tuesday, May 25, 2021 Chapter Outline q Introduction to Fatigue in Metals q Approach to Fatigue Failure in Analysis and Design q Fatigue-Life Methods ü The Stress-Life Method ü The Strain-Life Method ü The Linear-Elastic Fracture Mechanics Method q The Endurance Limit q Fatigue Strength q Endurance Limit Modifying Factors q Stress Concentration and Notch Sensitivity q Characterizing Fluctuating Stresses q Fatigue Failure Criteria for Fluctuating Stress q Torsional Fatigue Strength under Fluctuating Stresses q Combinations of Loading Modes q Varying, Fluctuating Stresses; Cumulative Fatigue Damage q Surface Fatigue Strength q Stochastic Analysis q Road Maps and Important Design Equations for the Stress-Life Method Dr. Mohammad Suliman Abuhaiba, PE

3 Tuesday, May 25, 2021 Introduction to Fatigue in Metals q Loading produces stresses that are variable, repeated, alternating, or fluctuating q Maximum stresses well below yield strength q Failure occurs after many stress cycles q Failure is by sudden ultimate fracture q No visible warning in advance of failure Dr. Mohammad Suliman Abuhaiba, PE

4 Tuesday, May 25, 2021 Stages of Fatigue Failure q Stage I: Initiation of microcrack due to cyclic plastic deformation q Stage II: Progresses to macro-crack that repeatedly opens & closes, creating bands called beach marks q Stage III: Crack has propagated far enough that remaining material is insufficient to carry the load, and fails by simple ultimate failure Dr. Mohammad Suliman Abuhaiba, PE

Fatigue Fracture Example AISI 4320 drive shaft q B: crack initiation at stress concentration in keyway q C: Final brittle failure q Dr. Mohammad Suliman Abuhaiba, PE

6 Tuesday, May 25, 2021 Fatigue-Life Methods q. Three major fatigue life models q. Methods predict life in number of cycles to failure, N, for a specific level of loading Dr. Mohammad Suliman Abuhaiba, PE

7 Tuesday, May 25, 2021 Fatigue-Life Methods 1. 2. 3. Stress-life method Strain-life method Linear-elastic fracture method mechanics Dr. Mohammad Suliman Abuhaiba, PE

8 Tuesday, May 25, 2021 1. Stress-Life Method Test specimens subjected to repeated stress while counting cycles to failure q Pure bending with no transverse shear q completely reversed stress cycling q Dr. Mohammad Suliman Abuhaiba, PE

9 Tuesday, May 25, 2021 S-N Diagram Dr. Mohammad Suliman Abuhaiba, PE

10 Tuesday, May 25, 2021 S-N Diagram for Steel q Stress levels below Se predict infinite life q 103 to 106 cycles: finite life q Below 103 cycles: low cycle ü quasi-static ü Yielding usually occurs before fatigue Dr. Mohammad Suliman Abuhaiba, PE

11 Tuesday, May 25, 2021 S-N Diagram for Nonferrous Metals no endurance limit Fatigue strength Sf S-N diagram for aluminums Dr. Mohammad Suliman Abuhaiba, PE

12 Tuesday, May 25, 2021 2. Strain-Life Method Detailed analysis of plastic deformation at localized regions q Compounding of several idealizations leads to significant uncertainties in numerical results q Useful for explaining nature of fatigue q Dr. Mohammad Suliman Abuhaiba, PE

13 Tuesday, May 25, 2021 2. Strain-Life Method Fatigue failure begins at a local discontinuity q When stress at discontinuity exceeds elastic limit, plastic strain occurs q Cyclic plastic strain can change elastic limit, leading to fatigue q Fig. 6– 12 Dr. Mohammad Suliman Abuhaiba, PE

14 Tuesday, May 25, 2021 Relation of Fatigue Life to Strain q q q Figure 6– 13: relationship of fatigue life to true -strain amplitude Fatigue ductility coefficient e'F = true strain at which fracture occurs in one reversal (point A in Fig. 6– 12) Fatigue strength coefficient s'F = true stress corresponding to fracture in one reversal (point A in Fig. 6– 12) Dr. Mohammad Suliman Abuhaiba, PE

15 Tuesday, May 25, 2021 Relation of Fatigue Life to Strain Fig. 6– 13 Dr. Mohammad Suliman Abuhaiba, PE

16 Tuesday, May 25, 2021 Relation of Fatigue Life to Strain q Equation of plastic-strain line in Fig. 6– 13 q Equation of elastic strain line in Fig. 6– 13 Dr. Mohammad Suliman Abuhaiba, PE

17 Tuesday, May 25, 2021 Relation of Fatigue Life to Strain q q q Fatigue ductility exponent c = slope of plastic -strain line 2 N stress reversals = N cycles Fatigue strength exponent b = slope of elastic -strain line Dr. Mohammad Suliman Abuhaiba, PE

18 Tuesday, May 25, 2021 Relation of Fatigue Life to Strain q Manson-Coffin: relationship between fatigue life and total strain q Table A– 23: values of coefficients & exponents q Equation has limited use for design ü Values for total strain at discontinuities are not readily available Dr. Mohammad Suliman Abuhaiba, PE

19 Tuesday, May 25, 2021 The Endurance Limit Fig. 6– 17 Dr. Mohammad Suliman Abuhaiba, PE

20 Tuesday, May 25, 2021 The Endurance Limit Simplified estimate of endurance limit for steels for the rotating-beam specimen, S'e Dr. Mohammad Suliman Abuhaiba, PE

21 Fatigue Strength Tuesday, May 25, 2021 Dr. Mohammad Suliman Abuhaiba, PE q For design, an approximation of idealized S-N diagram is desirable. q To estimate fatigue strength at 103 cycles, start with Eq. (6 -2) q Define specimen fatigue strength at a specific number of cycles as

22 Tuesday, May 25, 2021 Fatigue Strength q At 103 cycles, q f = fraction of Sut represented by q SAE approximation for steels with HB ≤ 500 may be used. Dr. Mohammad Suliman Abuhaiba, PE

23 Tuesday, May 25, 2021 Fatigue Strength q To find b, substitute endurance strength and corresponding cycles into Eq. (6– 9) and solve for b Dr. Mohammad Suliman Abuhaiba, PE

24 Tuesday, May 25, 2021 Fatigue Strength Substitute Eqs. 6– 11 & 6– 12 into Eqs. 6– 9 and 6 – 10 to obtain expressions for S'f and f Dr. Mohammad Suliman Abuhaiba, PE

25 Tuesday, May 25, 2021 Fatigue Strength Fraction f q q q Plot Eq. (6– 10) for the fatigue strength fraction f of Sut at 103 cycles Use f from plot for S'f = f Sut at 103 cycles on S-N diagram Assume Se = S'e= 0. 5 Sut at 106 cycles Fig. 6– 18 Dr. Mohammad Suliman Abuhaiba, PE

Equations for S-N Diagram Fig. 6– 10 Dr. Mohammad Suliman Abuhaiba, PE

Equations for S-N Diagram Write equation for S-N line from 103 to 106 cycles q Two known points ü At N =103 cycles, Sf = f Sut ü At N =106 cycles, Sf = Se q Equations for line: q Dr. Mohammad Suliman Abuhaiba, PE

Equations for S-N Diagram q If a completely reversed stress srev is given, setting Sf = srev in Eq. (6– 13) and solving for N gives, Typical S-N diagram is only applicable for completely reversed stresses q For other stress situations, a completely reversed stress with the same life expectancy must be used on the S-N diagram q Dr. Mohammad Suliman Abuhaiba, PE

29 Tuesday, May 25, 2021 Low-cycle Fatigue q q 1 ≤ N ≤ 103 On the idealized S-N diagram on a log-log scale, failure is predicted by a straight line between two points (103, f Sut) and (1, Sut) Dr. Mohammad Suliman Abuhaiba, PE

Example 6 -2 Given a 1050 HR steel, estimate a. the rotating-beam endurance limit at 106 cycles. b. the endurance strength of a polished rotating -beam specimen corresponding to 104 cycles to failure c. the expected life of a polished rotating-beam specimen under a completely reversed stress of 55 kpsi. Dr. Mohammad Suliman Abuhaiba, PE

31 Tuesday, May 25, 2021 Endurance Limit Modifying Factors q q q Endurance limit S'e is for carefully prepared and tested specimen If warranted, Se is obtained from testing of actual parts When testing of actual parts is not practical, a set of Marin factors are used to adjust the endurance limit Dr. Mohammad Suliman Abuhaiba, PE

32 Tuesday, May 25, 2021 Endurance Limit Modifying Factors Dr. Mohammad Suliman Abuhaiba, PE

33 Tuesday, May 25, 2021 Surface Factor ka q Surface factor is a function of ultimate strength q Higher strengths are more sensitive to rough surfaces Dr. Mohammad Suliman Abuhaiba, PE

Example 6 -3 A steel has a min ultimate strength of 520 MPa and a machined surface. Estimate ka. Dr. Mohammad Suliman Abuhaiba, PE

35 Tuesday, May 25, 2021 Size Factor kb rotating & Round q Larger parts have greater surface area at high stress levels q Likelihood of crack initiation is higher For bending and torsion loads, the size factor is given by Dr. Mohammad Suliman Abuhaiba, PE

36 Tuesday, May 25, 2021 Size Factor kb rotating & Round q Applies only for round, rotating diameter q For axial load, there is no size effect, ü kb = 1 Dr. Mohammad Suliman Abuhaiba, PE

37 Tuesday, May 25, 2021 Size Factor kb not round & rotating q An equivalent round rotating diameter is obtained. q Volume of material stressed at and above 95% of max stress = same volume in rotating-beam specimen. q Lengths cancel, so equate areas. Dr. Mohammad Suliman Abuhaiba, PE

38 Tuesday, May 25, 2021 Size Factor kb not round & rotating q For a rotating round section, the 95% stress area is the area of a ring, q Equate 95% stress area for other conditions to Eq. (6– 22) and solve for d as the equivalent round rotating diameter Dr. Mohammad Suliman Abuhaiba, PE

39 Tuesday, May 25, 2021 Size Factor kb round & not rotating q For non-rotating round, q Equating to Eq. (6 -22) and solving for equivalent diameter, Dr. Mohammad Suliman Abuhaiba, PE

40 Tuesday, May 25, 2021 Size Factor kb not round & not rotating q For rectangular section h x b, A 95 s = 0. 05 hb. Equating to Eq. (6– 22), Dr. Mohammad Suliman Abuhaiba, PE

41 Tuesday, May 25, 2021 Size Factor kb Table 6– 3: A 95 s for common non-rotating structural shapes undergoing bending Dr. Mohammad Suliman Abuhaiba, PE

42 Tuesday, May 25, 2021 Size Factor kb Table 6– 3: A 95 s for common non-rotating structural shapes undergoing bending Dr. Mohammad Suliman Abuhaiba, PE

Example 6 -4 A steel shaft loaded in bending is 32 mm in diameter, abutting a filleted shoulder 38 mm in diameter. The shaft material has a mean ultimate tensile strength of 690 MPa. Estimate the Marin size factor kb if the shaft is used in a. A rotating mode. b. A nonrotating mode. Dr. Mohammad Suliman Abuhaiba, PE

44 Tuesday, May 25, 2021 Loading Factor kc q q q Accounts for changes in endurance limit for different types of fatigue loading. Only to be used for single load types. Use Combination Loading method (Sec. 6– 14) when more than one load type is present. Dr. Mohammad Suliman Abuhaiba, PE

45 Tuesday, May 25, 2021 Temperature Factor kd q Endurance limit appears to maintain same relation to ultimate strength for elevated temperatures as at RT q Table 6– 4: Effect of Operating Temperature on Tensile Strength of Steel. * (ST = tensile strength at operating temperature (OT); SRT = tensile strength at room temperature; 0. 099 ≤ ˆσ ≤ 0. 110) Dr. Mohammad Suliman Abuhaiba, PE

46 Tuesday, May 25, 2021 Temperature Factor kd Table 6– 4 Dr. Mohammad Suliman Abuhaiba, PE

47 Tuesday, May 25, 2021 Temperature Factor kd q q q If ultimate strength is known for OT, then just use that strength. Let kd = 1. If ultimate strength is known only at RT, use Table 6– 4 to estimate ultimate strength at OT. With that strength, let kd = 1. Use ultimate strength at RT and apply kd from Table 6– 4 to the endurance limit. Dr. Mohammad Suliman Abuhaiba, PE

48 Tuesday, May 25, 2021 Temperature Factor kd q A fourth-order polynomial curve fit of the data of Table 6– 4 can be used in place of the table, Dr. Mohammad Suliman Abuhaiba, PE

Example 6 -5 A 1035 steel has a tensile strength of 70 kpsi and is to be used for a part that sees 450°F in service. Estimate the Marin temperature modification factor and (Se)450◦ if a. RT endurance limit by test is (S’e)70◦ = 39. 0 kpsi b. Only the tensile strength at RT is known. Dr. Mohammad Suliman Abuhaiba, PE

50 Tuesday, May 25, 2021 Reliability Factor ke q Fig. 6– 17, S'e = 0. 5 Sut is typical of the data and represents 50% reliability. q Reliability factor adjusts to other reliabilities. Dr. Mohammad Suliman Abuhaiba, PE

51 Reliability Factor ke Tuesday, May 25, 2021 Fig. 6– 17 Dr. Mohammad Suliman Abuhaiba, PE

52 Tuesday, May 25, 2021 Reliability Factor ke Table 6– 5 Dr. Mohammad Suliman Abuhaiba, PE

53 Tuesday, May 25, 2021 Miscellaneous-Effects Factor kf q Consider other possible factors: Residual stresses ü Directional characteristics from cold working ü Case hardening ü Corrosion ü Surface conditioning ü q Limited data is available. q May require research or testing. Dr. Mohammad Suliman Abuhaiba, PE

54 Tuesday, May 25, 2021 Stress Concentration and Notch Sensitivity q Obtain Kt as usual (Appendix A– 15) q Kf = fatigue stress-concentration factor q q = notch sensitivity, ranging from 0 (not sensitive) to 1 (fully sensitive) q For q = 0, Kf = 1 q For q = 1, Kf = Kt Dr. Mohammad Suliman Abuhaiba, PE

55 Tuesday, May 25, 2021 Notch Sensitivity Fig. 6– 20: q for bending or axial loading Kf = 1 + q( Kt – 1) Fig. 6– 20 Dr. Mohammad Suliman Abuhaiba, PE

56 Tuesday, May 25, 2021 Notch Sensitivity Fig. 6– 21: qs for torsional loading Kfs = 1 + qs( Kts – 1) Dr. Mohammad Suliman Abuhaiba, PE

57 Tuesday, May 25, 2021 Notch Sensitivity Use curve fit equations for Figs. 6– 20 & 6– 21 to get notch sensitivity, or go directly to Kf. Bending or axial: Torsion: Dr. Mohammad Suliman Abuhaiba, PE

58 Tuesday, May 25, 2021 Notch Sensitivity for Cast Irons q Cast irons are already full of discontinuities, which are included in the strengths. q Additional notches do not add much additional harm. q Recommended to use q = 0. 2 for cast irons. Dr. Mohammad Suliman Abuhaiba, PE

Example 6 -6 A steel shaft in bending has an ultimate strength of 690 MPa and a shoulder with a fillet radius of 3 mm connecting a 32 -mm diameter with a 38 -mm diameter. Estimate Kf using: a. Figure 6– 20 b. Equations (6– 33) and (6– 35) Dr. Mohammad Suliman Abuhaiba, PE

60 Tuesday, May 25, 2021 Application of Fatigue Stress Concentration Factor q Use Kf as a multiplier to increase the nominal stress. q Some designers apply 1/Kf as a Marin factor to reduce Se. q For infinite life, either method is equivalent, since q For finite life, increasing stress is more conservative. Decreasing Se applies more to high cycle than low cycle. Dr. Mohammad Suliman Abuhaiba, PE

Example 6 -7 For the step-shaft of Ex. 6– 6, it is determined that the fully corrected endurance limit is Se = 280 MPa. Consider the shaft undergoes a fully reversing nominal stress in the fillet of (σrev)nom = 260 MPa. Estimate the number of cycles to failure. Dr. Mohammad Suliman Abuhaiba, PE

Example 6 -8 A 1015 hot-rolled steel bar has been machined to a diameter of 1 in. It is to be placed in reversed axial loading for 70 000 cycles to failure in an operating environment of 550°F. Using ASTM minimum properties, and a reliability of 99%, estimate the endurance limit and fatigue strength at 70 000 cycles. Dr. Mohammad Suliman Abuhaiba, PE

Example 6 -9 Figure 6– 22 a shows a rotating shaft simply supported in ball bearings at A and D and loaded by a non-rotating force F of 6. 8 k. N. Using ASTM “minimum” strengths, estimate the life of the part. Fig. 6– 22 Dr. Mohammad Suliman Abuhaiba, PE

64 Tuesday, May 25, 2021 Characterizing Fluctuating Stresses q The S-N diagram is applicable for completely reversed stresses q Other fluctuating stresses exist q Sinusoidal loading patterns are common, but not necessary Dr. Mohammad Suliman Abuhaiba, PE

Fluctuating Stresses Figure 6– 23 fluctuating stress with high frequency ripple non-sinusoidal fluctuating stress Dr. Mohammad Suliman Abuhaiba, PE

Fluctuating Stresses Figure 6– 23 General Fluctuating Completely Reversed Repeated Dr. Mohammad Suliman Abuhaiba, PE

67 Tuesday, May 25, 2021 Characterizing Fluctuating Stresses Stress ratio Amplitude ratio Dr. Mohammad Suliman Abuhaiba, PE

68 Tuesday, May 25, 2021 Application of Kf for Fluctuating Stresses q For fluctuating loads at points with stress concentration, the best approach is to design to avoid all localized plastic strain. q In this case, Kf should be applied to both alternating and midrange stress components. Dr. Mohammad Suliman Abuhaiba, PE

69 Tuesday, May 25, 2021 Application of Kf for Fluctuating Stresses q When localized strain does occur, some methods (nominal mean stress method and residual stress method) recommend only applying Kf to the alternating stress. Dr. Mohammad Suliman Abuhaiba, PE

70 Tuesday, May 25, 2021 Application of Kf for Fluctuating Stresses q Dowling method recommends applying Kf to the alternating stress and Kfm to the mid-range stress, where Kfm is Dr. Mohammad Suliman Abuhaiba, PE

71 Tuesday, May 25, 2021 Fatigue Failure for Fluctuating Stresses q Vary sm and sa to learn about the fatigue resistance under fluctuating loading q Three common methods of plotting results follow. Dr. Mohammad Suliman Abuhaiba, PE

72 Tuesday, May 25, 2021 Modified Goodman Diagram Fig. 6– 24 Dr. Mohammad Suliman Abuhaiba, PE

73 Tuesday, May 25, 2021 Plot of Alternating vs Midrange Stress Fig. 6– 25 Little effect of negative midrange stress Dr. Mohammad Suliman Abuhaiba, PE

74 Tuesday, May 25, 2021 Master Fatigue Diagram Fig. 6– 26 Dr. Mohammad Suliman Abuhaiba, PE

75 Tuesday, May 25, 2021 Plot of Alternating vs Midrange Stress q Most common and simple to use q Goodman or Goodman diagram Modified Dr. Mohammad Suliman Abuhaiba, PE

76 Tuesday, May 25, 2021 Commonly Used Failure Criteria Fig. 6– 27 Dr. Mohammad Suliman Abuhaiba, PE

77 Tuesday, May 25, 2021 Commonly Used Failure Criteria q Modified Goodman is linear, so simple to use for design. It is more conservative than Gerber. q Soderberg provides a very conservative single check of both fatigue and yielding. q Langer line represents standard yield check. q It is equivalent to comparing maximum stress to yield strength. Dr. Mohammad Suliman Abuhaiba, PE

78 Tuesday, May 25, 2021 Equations for Commonly Used Failure Criteria Dr. Mohammad Suliman Abuhaiba, PE

79 Tuesday, May 25, 2021 Summarizing Tables for Failure Criteria q Tables 6– 6 to 6– 8: equations for Modified Goodman, Gerber, ASME-elliptic, and Langer failure criteria q 1 st row: fatigue criterion q 2 nd row: yield criterion q 3 rd row: intersection of static and fatigue criteria q 4 th row: equation for fatigue factor of safety q 1 st column: intersecting equations q 2 nd column: coordinates of the intersection Dr. Mohammad Suliman Abuhaiba, PE

Table 6– 6: Modified Goodman and Langer Failure Criteria (1 st Quadrant) Dr. Mohammad Suliman Abuhaiba, PE

Table 6– 7: Gerber & Langer Failure Criteria (1 st Quadrant) Dr. Mohammad Suliman Abuhaiba, PE

Table 6– 8: ASME Elliptic and Langer Failure Criteria (1 st Quadrant) Dr. Mohammad Suliman Abuhaiba, PE

Example 6 -10 A 1. 5 -in-diameter bar has been machined from an AISI 1050 cold-drawn bar. This part is to withstand a fluctuating tensile load varying from 0 to 16 kip. Because of the ends, and the fillet radius, a fatigue stress-concentration factor Kf is 1. 85 for 106 or larger life. Find Sa and Sm and the factor of safety guarding against fatigue and firstcycle yielding, using a. Gerber fatigue line b. ASME-elliptic fatigue line. Dr. Mohammad Suliman Abuhaiba, PE

Example 6 -10 Dr. Mohammad Suliman Abuhaiba, PE

Example 6 -10 Dr. Mohammad Suliman Abuhaiba, PE

Example 6 -11 A flat-leaf spring is used to retain an oscillating flat-faced follower in contact with a plate cam. The follower range of motion is 2 in and fixed, so the alternating component of force, bending moment, and stress is fixed, too. The spring is preloaded to adjust to various cam speeds. The preload must be increased to prevent follower float or jump. For lower speeds the preload should be decreased to obtain longer life of cam and follower surfaces. The spring is a steel cantilever 32 in long, 2 in wide, and 1/4 in thick, as seen in Fig. 6– 30 a. The spring strengths are Sut = 150 kpsi, Sy = 127 kpsi, and Se = 28 kpsi fully corrected. The total cam motion is 2 in. The designer wishes to preload the spring by deflecting it 2 in for low speed and 5 in for high speed. Dr. Mohammad Suliman Abuhaiba, PE

Example 6 -11 a. b. Plot Gerber-Langer failure lines with the load line. What are the strength factors of safety corresponding to 2 in and 5 in preload? Dr. Mohammad Suliman Abuhaiba, PE

Example 6 -11 Dr. Mohammad Suliman Abuhaiba, PE

Example 6 -12 A steel bar undergoes cyclic loading such that σmax = 60 kpsi and σmin = − 20 kpsi. For the material, Sut = 80 kpsi, Sy = 65 kpsi, a fully corrected endurance limit of Se = 40 kpsi, and f = 0. 9. Estimate the number of cycles to a fatigue failure using: a. Modified Goodman criterion. b. Gerber criterion. Dr. Mohammad Suliman Abuhaiba, PE

90 Tuesday, May 25, 2021 Fatigue Criteria for Brittle Materials q First quadrant fatigue failure criteria follows a concave upward Smith-Dolan locus, q Or as a design equation, Dr. Mohammad Suliman Abuhaiba, PE

91 Tuesday, May 25, 2021 Fatigue Criteria for Brittle Materials Dr. Mohammad Suliman Abuhaiba, PE q For q In a radial load line of slope r, the intersection point is the second quadrant, q Table A– 24: properties of gray cast iron, including endurance limit q Endurance limit already includes ka and kb q Average kc for axial and torsional is 0. 9

Example 6 -13 A grade 30 gray cast iron is subjected to a load F applied to a 1 by 3/8 -in cross-section link with a 1/4 -in-diameter hole drilled in the center as depicted in Fig. 6 – 31 a. The surfaces are machined. In the neighborhood of the hole, what is the factor of safety guarding against failure under the following conditions: a. The load F = 1000 lbf tensile, steady. b. The load is 1000 lbf repeatedly applied. c. The load fluctuates between − 1000 lbf and 300 lbf without column action. Use the Smith-Dolan fatigue locus. Dr. Mohammad Suliman Abuhaiba, PE

Example 6 -13 Dr. Mohammad Suliman Abuhaiba, PE

Example 6 -13 Dr. Mohammad Suliman Abuhaiba, PE

95 Tuesday, May 25, 2021 Torsional Fatigue Strength q Testing: steady-stress component has no effect on the endurance limit for torsional loading if the material is : ü ductile, polished, notch-free, and cylindrical. q For less than perfect surfaces, the modified Goodman line is more reasonable. q For pure torsion cases, use kc = 0. 59 to convert normal endurance strength to shear endurance strength. q For shear ultimate strength, recommended to use Dr. Mohammad Suliman Abuhaiba, PE

96 Tuesday, May 25, 2021 Combinations of Loading Modes q For combined loading, use Distortion Energy theory to combine them. q Obtain Von Mises stresses for both midrange and alternating components. q Apply appropriate Kf to each type of stress. q For load factor, use kc = 1. The torsional load factor (kc = 0. 59) is inherently included in the von Mises equations. Dr. Mohammad Suliman Abuhaiba, PE

97 Tuesday, May 25, 2021 Combinations of Loading Modes q If needed, axial load factor can be divided into the axial stress. Dr. Mohammad Suliman Abuhaiba, PE

98 Tuesday, May 25, 2021 Static Check for Combination Loading q Distortion Energy theory still applies for check of static yielding q Obtain Von Mises stress for maximum stresses q Stress concentration factors are not necessary to check for yielding at first cycle Dr. Mohammad Suliman Abuhaiba, PE

99 Tuesday, May 25, 2021 Static Check for Combination Loading q Alternate simple check is to obtain conservative estimate of s'max by summing s'a and s'm q ≈ Dr. Mohammad Suliman Abuhaiba, PE

Example 6 -14 A rotating shaft is made of 42× 4 mm AISI 1018 CD steel tubing and has a 6 -mm-diameter hole drilled transversely through it. Estimate the factor of safety guarding against fatigue and static failures using the Gerber and Langer failure criteria for the following loading conditions: a. The shaft is subjected to a completely reversed torque of 120 N. m in phase with a completely reversed bending moment of 150 N. m. b. The shaft is subjected to a pulsating torque fluctuating from 20 to 160 N. m and a steady bending moment of 150 N. m. Dr. Mohammad Suliman Abuhaiba, PE

Example 6 -14 Dr. Mohammad Suliman Abuhaiba, PE

102 Tuesday, May 25, 2021 Varying Fluctuating Stresses Dr. Mohammad Suliman Abuhaiba, PE

103 Tuesday, May 25, 2021 Cumulative Fatigue Damage A common situation is to load at s 1 for n 1 cycles, then at s 2 for n 2 cycles, etc. q The cycles at each stress level contributes to the fatigue damage q Accumulation of damage is represented by the Palmgren-Miner cycle-ratio summation rule, also known as Miner’s rule q Dr. Mohammad Suliman Abuhaiba, PE

104 Tuesday, May 25, 2021 Cumulative Fatigue Damage ni = number of cycles at stress level si q Ni = number of cycles to failure at stress level si q c = experimentally found to be in the range 0. 7 < c < 2. 2, with an average value near unity q D = the accumulated damage, q Dr. Mohammad Suliman Abuhaiba, PE

Example 6 -15 Fig. 6– 33 Given a part with Sut = 151 kpsi and at the critical location of the part, Se = 67. 5 kpsi. For the loading of Fig. 6– 33, estimate the number of repetitions of the stress-time block in Fig. 6– 33 that can be made before failure. Dr. Mohammad Suliman Abuhaiba, PE

106 Tuesday, May 25, 2021 Illustration of Miner’s Rule q q Figure 6– 34: effect of Miner’s rule on endurance limit and fatigue failure line. Damaged material line is predicted to be parallel to original material line. Dr. Mohammad Suliman Abuhaiba, PE

107 Tuesday, May 25, 2021 Weaknesses of Miner’s Rule Miner’s rule fails to agree with experimental results in two ways 1. It predicts the static strength Sut is damaged. 2. It does not account for the order in which the stresses are applied Dr. Mohammad Suliman Abuhaiba, PE

108 Tuesday, May 25, 2021 Manson’s Method q Manson’s rule. method overcomes deficiencies of Miner’s q. All fatigue lines on S-N diagram converge to a common point at 0. 9 Sut at 103 cycles. q. It requires each line to be constructed in the same historical order in which the stresses occur. Fig. 6– 35 Dr. Mohammad Suliman Abuhaiba, PE

109 Tuesday, May 25, 2021 Surface Fatigue Strength q When two surfaces roll or roll and slide against one another, a pitting failure may occur after a certain number of cycles. q The surface fatigue mechanism is complex and not definitively understood. q Factors include Hertz stresses, number of cycles, surface finish, hardness, lubrication, and temperature Dr. Mohammad Suliman Abuhaiba, PE

110 Tuesday, May 25, 2021 Surface Fatigue Strength q From Eqs. (3– 73) and (3– 74), the pressure in contacting cylinders, Dr. Mohammad Suliman Abuhaiba, PE

111 Tuesday, May 25, 2021 Surface Fatigue Strength q Converting to radius r and width w instead of length l, q pmax = surface endurance strength (contact strength, contact fatigue strength, or Hertzian endurance strength) Dr. Mohammad Suliman Abuhaiba, PE

112 Tuesday, May 25, 2021 Surface Fatigue Strength q Combining Eqs. (6– 61) and (6– 63), q K 1 = Buckingham’s load-stress factor, or wear factor q In gear studies, a similar factor is used, Dr. Mohammad Suliman Abuhaiba, PE

113 Tuesday, May 25, 2021 Surface Fatigue Strength q From Eq. (6– 64), with material property terms incorporated into an elastic coefficient CP Dr. Mohammad Suliman Abuhaiba, PE

114 Tuesday, May 25, 2021 Surface Fatigue Strength q Experiments show the following relationships q Data on induction-hardened steel give (SC)107 = 271 kpsi and (SC)108 = 239 kpsi, so β, from Eq. (6– 67), is Dr. Mohammad Suliman Abuhaiba, PE

115 Tuesday, May 25, 2021 Surface Fatigue Strength q. A long standing correlation in steels between SC and HB at 108 cycles is q AGMA uses Dr. Mohammad Suliman Abuhaiba, PE

116 Tuesday, May 25, 2021 Surface Fatigue Strength q Incorporating design factor into Eq. (6– 66), Dr. Mohammad Suliman Abuhaiba, PE

117 Tuesday, May 25, 2021 Surface Fatigue Strength q Since this is nonlinear in its stress-load transformation, the definition of nd depends on whether load or stress is the primary consideration for failure. q If loss of function is focused on the load, q If loss of function is focused on the stress, Dr. Mohammad Suliman Abuhaiba, PE