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Chapter 6: Fatigue Failure (Review)I)II)III)IV)V)VI)Sec 6-17 Road Maps and Important EquationsLoadingSimple loading- Axial loading- Torsion- BendingCombined LoadingCharacterizing Stress (Fig. 6-23d, e, f)Completely reversed stressRepeated stressFluctuating stressEndurance Limit πΊβ²π (Eq. 6-8)Correction Factors: ππ ππ ππ ππ ππ (Sec. 6-9)Corrected Endurance Limit πΊπ (Sec. 6-18)Equivalent StressesTheoretical stress concentration factors: ππ πππ (A-15)North sensitivity: π ππππππ (Sec. 6-10)Fatigue stress concentration factors: ππ πππ (Eq. 6-32)Von Mises Stress for alternating comp. (Eq. 6-55)Von Mises Stress for mid-range comp. (Eq. 6-56)πβ²π π (πΊππ. π π)If πβ²π πΊπ ; infinite life and factor of safety is:πΊππ β²ππElse finite life and number of stress cycles π΅ is by (Eq. 6-16) where ππππ is set to πβ²ππβ²π π (πΊππ. π ππ)Decide what to use, Soderberg? Modified Goodman? Gerber? ASME Elliptic?(Eq. 6-45 to Eq. 6-48) for factor of safety π.If π π, finite life.Else finite life and π΅ is by (Eq. 6-16) but ππππ is per Step 4 on pp. 340.
Figure 1: Fluctuating StressFigure 2: Repeated StressFigure 3: Completely Reversed Stress
Example (1): A low carbon steel stock is lathe-turned to have a diameter of 1β The stock has ππ’π‘ 100 ππ π, ππ¦ 76 ππ π. Axial load varies 10 50 ππππ . Fatigue stress concentration factor is πΎπ 1.3.Find factor of safety π if infinite life, or number of cycles π if infinite life. Assume room temperature and99% reliability.Case: Simple loading, fluctuating stressπΉπππ 10 ππππ πΉπππ₯ 50 ππππ ππ πΉπππ 16.552 ππ ππ΄ππ πΉπππ₯ 82.761 ππ ππ΄ππππ ππππ₯ππ 33.105 ππ πππ 49.657 ππ πππβ² 0.5ππ’π‘ 50 ππ πππ 0.797ππ 1ππ 0.85ππ 1ππ 0.814ππ 27.572 ππ πCriterionSoderbergModified GoodmanGerberASME-elliptic πππππ‘π 470.540.54ππππ£ , οΏ½οΏ½ 0.845π 258.97 ππ ππ 0.16213CriterionSoderbergModified GoodmanGerberASME-elliptic(see pp. 314 for procedure)
Chapter 7: Shafts and Shaft ComponentsPart 1(7-1): Introduction(7-2): Shaft Materials(7-4): Design for StressPart 2(7-5): Deflection Calculations(7-6): Critical Speeds for ShaftsPart 3(7-3): Shaft Layout(7-7): Misc. Shaft Components(7-8): Limits and Fits(7-1): IntroductionShaft Loading Power transmission shafting is to transmit power/motion from an input source (e.g. motors,engines) to an output work site. Shafts are supported by bearings, and loaded torsionally, transversely, and/or axially as themachine operates. Shafts can be solid or hollow, and are often stepped. They are widely required by virtually all types of machinery and mechanical systems.New Shaft Design Procedure Conceptual sketch for shaft layout, based on functional spec. and system config (7-3) Shaft materials (7-2) An appropriate design factor Support reactions, bending moment diagrams (in tow planes, as well as resultant/combined),and torque diagram, critical cross-sections. Shaft diameters based on strength requirement (7-4) Slopes and Deflections at locations of interest in order to select bearings, couplings, etc.: or toensure proper functioning of bearings, couplings, gears, etc. (7-5) Critical Speeds and other vibration characteristics (7-6)Calculations Needed Shaft diameters based on strength requirement by ANSI/ASME standard B106-1M-1985βDesign for Transmission Shaftingβ, or by other practices. Slopes and deflections simplified / approximate geometry, numerical, graphical, FEA; Critical speed and other vibration characteristics specific deflections of shaft.(7-2): Shaft Materials Requiring generally/typically high strength and high modulus of elasticity Typical selection: low carbon steel (cold-drawn or hot-rolled) such as ANSI 1020-1050 steels If higher strength is required, alloy steels plus heat treatment such as ANSI 1340-50, 4140, 4340,5140, 8650.
Cold-drawn steel is used for diameters under 3 inches; machining is not needed where there isno fitting with other components.Hot-rolled steels should be machined all over.Stainless steels when environment is corrosive, for example.Equations for the Fatigue Failure Criteria(6-40) Soderbeg(6-41) Modified Goodman(6-42) Gerber(6-43) ASME-EllipticFor factor of safety:Replace ππ with πππReplace ππ with πππResulting in Equations (6 45) (6 48)For ππππ£ (which is needed for number of cycles π):Replace ππ with ππReplace ππ with ππReplace ππ with ππππ£and solving for ππππ£Modified-Goodman line β too dangerous, goes directly to ππ’π‘Soderberg β too conservativeMost of the time, weβll be using ASME-elliptic line, notExample (2): A non-rotating shaft is lathe-turned to have a 1β-diameter. The shaft is subject to a torquethat varies 0 ππππ₯ (ππ ππ ππ). Determine ππππ₯ such that the shaft will have an infinite life with afactor of safety of 1.8. The shaftβs material has ππ’π‘ 100 ππ π and ππ¦ 76 ππ π. Assume roomtemperature and 99% reliability. Fatigue stress concentration factor is πππ 1.6.
Answer: Based on simple loading and ASME-elliptic criterion, ππππ₯ 2480 ππ ππDesign Factor vs. Factor of Safety (not from the text) Design factor is to indicate the level of overload that the part/component is required/intended towithstand Safety factor indicated how much overload the designed part will actually be able to withstand. Design factor is chosen, generally in advance and often set by regulatory code or an industryβsgeneral practice. Safety factor is obtained from design calculations.The following are some recommended values of design factor based on strength considerations. Theyare valid for general applications. 1.25 - 1.5: for reliable materials under controlled conditions subjected to loads and stresses knownwith certainty;1.5 β 2: for well-known materials under reasonably constant environmental conditions subjected toknown loads and stresses;2 β 2.5: for average materials subjected to known loads and stresses;2.5 β 3: for less well-known materials under average conditions of load, stress and environment;3 β 4: for untried materials under average conditions of load, stress and environment;3 β 4: for well-known materials under uncertain conditions of load, stress and environment.Courtesy of Mechanical Design, 2nd Edition, P. Childs, Elsevier Ltd. (p. 95)(7-4): Shaft Design for StressLoads on a Shaft The primary function of a shaft is to transmit torque, typically through only a portion of the shaft; Due to the means of torque transmission, shafts are subject to transverse loads in two planes suchthat the shear and bending moment diagrams are needed in two planes.Shaft Stress from Fatigue Perspective In 1985, ASME published ANSI/ASME Standard B106.1M-1985 βDesign for Transmission Shaftingβ.However, it was withdrawn in 1994.o Little or no axial loado Fully reversed bending and steady (or constant) torsion General case: Fluctuating bending and fluctuating torsiono As considered by the textFully reversed bending and steady torsion is in fact special cases of the general one. In reality, thegeneral case is not as common as the typical case of fully reversed bending and steady torsion.Critical LocationsStresses are evaluated at the critical locations. Look or where: Bending moment is largeThere is torqueThere is stress concentration
Factor of Safety or Required Shaft Diameter β General CaseGenerally, ππππππππ ππ and ππAnd ππ‘πππ πππ ππ and ππAssuming negligible axial load, due to fluctuating bending and fluctuating torsion, the amplitude andmean stresses are given by (Eq. 7-1) and (Eq. 7-2):For solid shaft, stresses can be written in terms of π, the diameter, see (Eq. 7-3) and (Eq. 7-4):Next, the equivalent (von Mises) amplitude and mean stresses are determined, resulting in (Eq. 7-5) and(Eq. 7-6):Now, define terms A and B (pp. 361):Finally, the pair of equations determining factor of safety and diameter, are (DE stands for distortionenergy): For DE-Goodman: (Eq. 7-7), (Eq. 7-8)
For DE-Gerber: (Eq. 7-9), (Eq-7-10) For DE-ASME-Elliptic: (Eq. 7-11), (Eq. 7-12) For DE-Soderberg: (Eq. 7-13), (Eq. 7-14)
After the above fatigue-based calculations, it is customary to check against static failure. The factor of safety against yielding in the first loading cycle is (Eq. 7-16):β²Where the equivalent maximum stress is ππππ₯by (Eq. 7-15) A quick but conservative check against yielding in the first loading cycle is, see the paragraphfollowing (Eq. 7-16):ππ¦ ππβ²ππ¦β² ππFactor of Safety or Required Shaft Diameter β Typical CasesThe typical case is defined as, fully reversed bending and steady (or constant) torsion. Therefore, settingππ 0 and ππ 0 in above equations will result in what are needed.ASME-Elliptic:Setting ππ 0 and ππ 0, (Eq. 7-5) and (Eq. 7-6) and the A and B terms become:ππβ² ππ2 (πΎπ32ππ 2 πΎπ 32ππ) ππ3ππ3
β²ππ 3πΎππ 16ππππ 3π΄ 2πΎπ πππ΅ 3πΎππ ππ(Eq. 7-11) and (Eq. 7-12) become,1π 2πβ²πβ² ( π ) ( π )ππππ¦2ππ 3161π΄ 2π΅ 2 ( ) ( )ππππ¦216π΄ 2π΅ π ( ) ( )3ππππππ¦3And (Eq. 7-15) simplified to:β²ππππ₯ 16 π΄2 π΅2ππ3Estimating πΊπ Shaft design equations involve ππ which in turn involves five modification factors. Therefore, it requires knowing the material, its surface condition, the size and geometry (stressraisers), and the level of reliability. Material and surface condition can be decided before the analysis. The sizes and geometry are however unknown in the preliminary stage of design. For size factor, a diameter may be determined from safety against yielding, or use 0.9 as theestimate of size factor. Reliability is typically set at 90%. Stress concentration factors πΎπ‘ and πΎπ‘π for first iteration are given in (Table 7-1, pp. 365)After that (For values not in table - second round), we would have to use (A-15)
Example 7-1Loading and design details are given. We are asked to, (1) determine π using DE-Goodman, DE-Gerber,DE-ASME, and DE-Soderberg; and (2) check against yielding failure by evaluating ππ¦Example 7-2Countershaft AB carries two spur gears at G and J; is supported by two bearings at A and B. Its layout isshown in Figure 7-10. Gear loads areππ23 197 πππ‘π23 540 ππππ54 885 πππ‘π54 2431 ππWe are to select appropriate materials and/or diameters at various cross sections, based on fatigue withinfinite life. Design factor is 1.5.The text starts with cross-section πΌ where there are torque and bending moment, and a shoulder forstress concentration. Generous shoulder fillet (r/d 0.1) is assumed DE-Goodman is used to determinediameters.ExampleA critical cross-section of a shat is subject to a combined bending moment of 63-lb-in and a torque of 74lb-in. The cross-section is the seat of a rolling-element bearing, and sharp shoulder fillet is expected.Shaft material is SAE 1040 CD. Estimate the shaftβs diameter at the cross-section for infinite life withπ? 1.5. Base calculations on ASME-Elliptic criterion. Operating conditions are typical.
Solution(1) First iterationππ 63 ππ ππππ 74 ππ ππππ’π‘ 85 ππ πππ¦ 71 ππ πππβ² 42.5 ππ πππ 2.7(85) 0.265 0.832ππ 0.9ππ 1ππ 1ππ 0.897ππ 28.55 ππ πππ‘ 2.7ππ‘π 2.2Set π ππ βπππ 1, so that πΎπ πΎπ‘ 2.7 and πΎππ πΎπ‘π 2.2, andπ΄ 2πΎπ ππ 340.2 ππ πππ΅ 3πΎππ ππ 282.0 ππ ππ216ππ΄ 2π΅ π ( ) ( ) 0.458"πππππ¦3Round off π 12 ππ 0.472"(2) Second iterationππ 63 ππ ππππ 74 ππ ππππ’π‘ 85 ππ πππ¦ 71 ππ πππβ² 42.5 ππ πππ 2.7(85) 0.265 0.832ππ 0.879(0.472) 0.107 0.953ππ 1ππ 1ππ 0.897ππ 30.23 ππ πππ 0.02, then π 0.009β , and π 0.57, ππ βπππ 0.6, so that πΎπ 1.91 andπΎππ 1.66Finally,π΄ 2πΎπ ππ 240.7 ππ πππ΅ 3πΎππ ππ 212.8 ππ ππ
ππ31π ()( 2.41622π΄π΅ ( ) ( )ππππ¦16 π΄2 π΅2 15.56 ππ πππ3ππ¦ππ¦ β² 4.6ππππ₯β²ππππ₯ Therefore, π 12 ππ or 0.472" is sufficient, giving a factor of safety against fatigue at 2.4 and a factorof safety against yielding at 4.6.(7-5) Deflection ConsiderationWhy deflection considerations? Shaft deflects transversely like a beam. Shaft also has torsional deflection like a torsion bar. Excessive deflections affect the proper functioning of gears and bearings, for example. Permissibleslopes and transverse deflections are listed in Table 7-2. To minimize deflections, keep shaft short, and avoid cantilever or overhang.How to Determine Beam Deflection AnalyticallyClosed-form solutions (4-4)Superposition (4-5)Singularity Functions (4-6)Strain Energy (4-7)Castiglianoβs 2nd Theorem (4-8)Statically indeterminate beams (4-10) .and so on.Limitations: effective when EI const. (but a stepped shaft does not have constant I) Numerical integrationSimplified geometry; for example, small shoulders (diameter-wise and length-wise), fillets, keyways,notches, etc., can be omitted.May be tedious.
ExampleA stepped shaft is shown below, which is supported by ball bearings at A and F. Determine its maximum(in magnitude) lateral deflection, and the slopes at A and F. Use πΈ 30 πππ π.1) Plot bending moment π(π₯)2) Plot π/π464π (ππΈ ) π ππ/π4πΈπΌπ43) Integrateπ "π ππππ"π44) Integrate "π ππππ"π "πππππππ‘πππ"π45) Baseline6) Deflection7) Slope
To obtain deflection curve of step (6), at any cross-section, subtract value obtained in step (4) frombaseline value.For example, at B, step (4) has 5107.0;baseline value is 26,123; 21,106 26,123 5107.0 will be used in step (6)
117,552 lb/inπ ππππ ππ πππ πππππ π 117,552 ππ( ) 3265.3 ππ/ππ236"ππ7) Subtract the slope of the baseline from value obtained in step (3)
64ππ2 0.6791 10 6 ( )ππΈππππππ2 πΏπππ₯ (26,113 ) (0.679 10 6)ππππ 0.0177 ππ ππππ2ππ΄ (3265.3 2 ) (0.679 10 6)ππππ 0.00222 πππ πππ·π ππ’ππ£πππΉ 0.00181 πππ πππ·π ππ’ππ£ππ Comments regarding the numerical integration method:Applicable to simple supports as outlined;Fixed supports?More divisions for better accuracy;Vertical plane, horizontal plane, and vector sum.An exact numerical method for determining the bending deflection and slope of stepped shafts, C.R.Mischke, in Advanced in reliability and stress analysis, ASME winter annual meeting, December 1978Mechanical Design of Machine Elements and Machines, J.A. Collins, John Wiley & Sons, 2003 (Sec. 8.5)Example (7-3)By the end of Example 7-2, diameters π·1 through π·7 were determined. The layout is shown below(Figure 7-10). Here we are to evaluate the slopes and deflections at key locations.The text uses βBeam 2D Stress Analysisβ (a software with FEA-core) for the evaluation.The results are verified by the above numerical integration method implemented with MATLAB.
DiameterExample 7-2π·1 π·71.0Point of InterestSlope, left bearing (A)Slope, right bearing (B)Slope , left gear (G)Slope, right gear (J)Deflection, left gear (G)Deflection, right gear (J)π·2 π·61.4Example 7-30.000501 rad0.001095 rad0.000414 rad0.000426 rad0.0009155 in0.0017567 inπ·3 π·51.625π·42.0Numerical Integration0.000507 rad0.001090 rad0.000416 rad0.000423 rad0.0009201 in0.0017691 inHow to Determine Torsional (Angular) Deflection Important for shafts carrying components that are required to function in sync with each other; forexample, cam shafts; For a stepped shaft with individual cylinder length ππ , torque ππ and material πΊπ the angulardeflection is,7-6 Critical Speeds for ShaftsIt is about applying knowledge of vibrations and deflections of shafts in the design of shafts.
The organization of this section:(Eq. 7-22): Exact solution of critical speed for a simply supported shaft with uniform cross section andmaterial(eq. 7-23): Rayleighβs method to estimate critical speed.(Eq. 7-24): Through (Eq. 7-32) Derivation of Dunkerleyβs method to estimate critical speed.(see pp. 376-377)
Example (7-5)Notes: (a) Rayleighβs and Dunkerleyβs methods only give rise to estimates; (b) They yield the upper andlower bound solutions, respectively. That is, π1(π·π’πππππππ¦) π1 π1(π ππ¦ππππβ) ; (c) The methods andtheir derivations fall under vibrations/dynamics of continuum by energy method.Critical Speeds Critical speeds refer to speeds at which the shaft becomes unstable, such that deflections (due tobending or torsion) increase without bound. A critical speed corresponds to the fundamental natural frequency of the shaft in a particularvibration mode. Three shaft vibration modes are to be concerned: lateral, vibration, shaft whirling and torsionalvibration. Critical speeds for lateral vibration and shaft whirling are identical. Numerically speaking, two critical speeds can be determined, one for lateral vibration or shaftwhirling, and another for torsional vibration. Focus will be the critical speed for lateral vibration or shaft whirling. Regarding critical speed fortorsional vibration, one can reference βrotor dynamicsβ and the transfer matrix method. If the critical speed is π1 , it is required that the operating speed π be:If the shaft is rigid (shafts in heavy machinery):π 3π1If the shaft is flexible (shafts that are long with small-diameters):π 1/3π1The text recommends:π 1/2π1Exact solution, (Eq. 7-22)Simply supported, uniform cross-section and material.
Rayleighβs Method for Critical Speed, (Eq. 7-23)Shaft is considered massless and flexible; Components such as gears, pulleys, flywheels, and so on, aretreated as lumped masses; The weight of the shaft, if significant, will be lumped as a mass or masses.Textbook equation should be updated to include the following absolute symbols:π1 π π€π π¦π π€π π¦π2Where:π€π weight of mass ππ€π should be treated as a force with a magnitude equation the weight of mass π;Forces π€π (π 1, ) should be applied in such a way that the deflection curve resembles thefundamental mode shape of lateral vibration.π¦π lateral deflection at location π (where π€π is applied) and caused by all forces.Dunkerleyβs Method for Critical Speed, (Eq. 7-32)The model for Dunkerleyβs method is the same as that for Rayleighβs.πand πππ π¦ππ Where π¦ππ is the lateral deflection at location π and caused by π€π only. πππ represents the critical speedwith only π€π on the shaft.
Deflections ππ πππ and πΉπππ¦ππ is the deflection at location π and due to a load applied at location π. When the load at location π is aunit load, then π¦ππ is denoted by πΏππ . πΏππ is also known as the influence coefficient.π¦π is the deflection at location π and caused by all applied loads. Therefore,π¦π π¦πππunit load, then Type equation here.unit load, then π¦ππ is denoted at πΏππ . πΏππProblem SolvingClosed-form solutions (Sec. 4-4 and Table A-9).Superposition (Section 4-5);Indexes π and πeach runs 1 through the number of masses/forces;π¦ππ are signed numbers;Example:Evaluate the range of the shaftβs critical speed corresponding to its lateral vibration, in terms of πΈπΌwhere πΈπΌ ππππ π‘πππ‘.π 1,2π 1,2
From previous example:Table A 9{Case 10π¦12 π¦π΄π΅ π₯ 0.45 ( 700)(0.225)(0.45)(0.92 0452 )(6 0.9)πΈπΌ7.9738πΈπΌπ¦22 π¦π‘ ( 700)(0.225)2 (0.9 0.225) 13.289 3πΈπΌπΈπΌTable A 9{Case 5π¦11 π¦πππ₯ π¦π΄π΅ (900)(0.9)3(13.669) 48πΈπΌπΈπΌπΉπ₯(4π₯ 2 3π 2 )48πΈπΌπππ΄π΅πΉ(4π₯ 2 π 2 ) ; ππ₯16πΈπΌ(900)( 0.92 )45.563ππ΄ 16πΈπΌπΈπΌ10.252π¦21 ( ππ΄ )(0.225) πΈπΌOr:10.252π¦21 (ππΆ )(0.225) πΈπΌππ΄π΅ 0 π₯ π221.643πΈπΌ23.541 πΈπΌ π¦1 π¦11 π¦12 π¦2 π¦21 π¦22Rayleighβs Method:π( π€π π¦π ) π1 2( π€π π¦π )21.64323.541) (700) ()πΈπΌπΈπΌ (9.81)21.643 223.541 2(900) () (700) ()πΈπΌπΈπΌ 0.66011 πΈπΌ(900) (Dunkerleyβs Method:13.669π¦11 πΈπΌ13.289π¦22 πΈπΌπ2π11 0.71768 πΈπΌ π¦11
2π22 π 0.73820 πΈπΌ π¦22 1112.74802 2 2 πΈπΌπ1 π11 π22πΈπΌπ1 0.60324 πΈπΌ2.7480Take πΈ 200 πΊππ, π 25 ππ πΈπΌ 1261.6 (π π 2 ) π1(π·π’πππππππ¦) (0.60324)(1261.6) 761.0 πππ/π ππ π1(π·π’πππππππ¦) 7267 ππππ΄ππ π π1(π ππ¦ππππβ) 832.8 πππ/π ππ π1(π ππ¦ππππβ) 7952 πππ ππππππ‘πππ πππππ 1π3 1ππ π 2422 πππ7-3 Shaft Layout Between a shaft and its components (e.g., gears, bearings, pulleys, etc.), the latter must be locatedaxially and circumferentially. Means to provide for torque transmissionKeysSplinesSetscrewsPinsPress/shrink fitsTapered fitsetc. Means to provide for axial location β large axial load shouldersShouldersRetaining ringssleevesCollarsetc. Means to provide for axial location β small axial loadPress/shrink fitsSetscrewsetc. Locating rolling element bearingsSee Ch. 11
7-7 Miscellaneous Shaft ComponentsIncludes:- Setscrews- Keys and pins- Retaining ringsFocus:KeysExample 7-67-8 Limits and Fits Fits (clearance, transition, and interference) are to ensure that a shaft and itscomponents/attachments will function as intended. Preferred fits are listed in Table 7-9 Medium drive fit and force fit will give rise to the press/shrink fits. Press-fit is typically for small hubs; Shrink fit (or expansion fit) it used with larger hubs How much diameter interference to have?0.001β for up to 1β of diameter0.002β for diameter 1β to 4β Press/shrink fits can be designed to transfer torque and axial load. Press/shrink fits are known to be associated with fretting corrosion (loss of material from theinterface) (Section 3-14) for stress distributions in thick-walled cylinder under pressures. (Section 3-16) for stresses developed in the shaft and hub due to pressure induced by a press/shrinkfit; or (Eq 7-39) to (Eq. 7-47) Axial load and torque capacities: (Eq. 7-48) and (Eq. 7-49) Radial interference versus diametral interference (not necessarily the same thing).
Chapter 11: Rolling-Contact BearingsPart 1: (Introduction)11-1 Bearing TypesPart 2: (The Basics)11-2 Bearing Life11-3 Bearing Life at Rated Reliability11- 4Reliability versus Life β The Weibull Distribution11-5 Relating Bearing Load, Life and ReliabilityPart 3: (Selection of Bearing)11-6 Combined Radial and Thrust Bearing11-8 Selection of Ball and Roller Bearings11-7 Variable Loading11-10 Design AssessmentPart 4: Others11-12 Mounting and Enclosure11-11 Lubrication11-9 Selection of Tapered Roller Bearings11-1 Bearing TypesNomenclatureSee Figure 11-1Classifications By shape of rolling elements (sphere, cylinder, tapered, etc.) By type of loads taken (radial only, axial only, combination) By permissible slope (Self-aligning, non-self-aligning) Sealed? Shielded? Figures 11-2, 11-3
11-2 Bearing LifeWhy Bearing Life? Bearings are under cyclic contact stresses (compressive as well as shear). As a result, they mayexperience crack, putting, spalling, fretting, excessive noise, and vibration. Common life measures are:o Number of revolutions of the inner ring (outer ring stationary) until the first evidence offailure; ando Number of hours of use at standard angular speed until the first evidence of fatigue Number of revolutions is more common
Rating Life (or Rated Life) This is the terminology used by ABMA (American Bearing Manufacturers Association) and mostbearing manufacturers. It is defined as, of a group of nominally identical bearings, the number of revolutions that 90% of thebearings in the group will achieve or exceed, before failure occurs. It is denoted as πΏ10 or π΅10 life. The typical value for πΏ10 or π΅10 is 1 million. However, a manufacturer can choose its own specific rating life. For example, Timken uses 90 million for tapered roller bearings, but 1 million for its other bearings. Refer to bearings catalog for value(s) of πΏ10 .11-3 Bearing Life at Rated Reliability At rated reliability of 90%, bearing life πΏ relates to bearingβs radial load πΉ by:πΉπΏ1/π ππππ π‘πππ‘Where π 3 for ball bearings and π 10/3 or roller bearings.Figure 11-4 shows the meaning of (Eq. 11-1) o Itβs a straight line on log-log scales;o Points on the line will have the same reliability;o The line corresponding to 90% reliability is called the rated line.To determine the constant on the RHS of (Eq. 11-1), πΏ is set to πΏ10 . The corresponding πΉ isdesignated as πΆ10 .πΆ10 is called the Basic Dynamic Load Rating, or the Basic Dynamic Rated Load It is defined as theradial load that causes 10% of the group of nominally identical bearings to fail at or before πΏ10(1 million, or 90 million, or revs as chosen by a manufacturer).(Equation 11-1) becomes:1/π1/ππΉπ· πΏπ· πΆ10 πΏ10(11 3)
Where the subscript π· means design. (Equation 11-3)* is essentially (Eq. 11-3) of the text. In (Eq. 11-3),the subscript π means rated. Example 11-1: find πΆ10 from known πΏπ· , πΉπ· and πΏ1011-4 Reliability versus Life β The Weibull Distribution11-5 Relating Load, Life and Reliability At 90% reliability, (Eq. 11-3*) forms the basis for selecting a bearing. What if the reliability is not 90%?Figure 11-5 shows the process of going from the rated line to a different line.π΄ π·, with π΅ being the intermediary. π΄ π΅: Load is constant, and life measure (which is a random variable) follows a three-parameterWeibull distributionπ΅ π·: Reliability is constant, and (Eq. 11-3*) is valid; The mathematics is given in (Sec. 11-4) and(Sec. 11-5).Another approach is to use a reliability factor π1 , the value of which depends on π , the reliability.Values of π1 are available from a number of references.R, %909596979899 Reliability Factor, ππ1.000.620.530.440.330.21The factor π1 can be determined by (courtesy of, for example SKF catalog)100 2/3π1 4.48 (ln)π Where π is in %, e.g., π 92.5. Note: π 99.
Timken recommends the following formula for π1100 2/3π1 4.26 (ln) 0.05π Where π 99.9The Basic Bearing EquationThe basic bearing equation can now be obtained by, in (Eq. 11-3), introducing the reliability factor π1and a load-application factor ππ . That is,1/π1/ππΉπ· πΏπ· πΆ10 πΏ10Becomes,πΆ10πΏπ· 1/π ()ππ πΉπ·π1 πΏ10ππ is given in (Table 11-5)This basic equation can be used for bearing selection and for assessment after selection.Example 1A SKF deep-groove ball bearing is subjected to a radial load of 495 lb. The shaft rotates at 300 rpm. Thebearing is expected to last 30,000 hours (continuous operation). Catalog shows a πΆ10 19.5 ππ on thebasis of 106 revs. (1) is the bearing suitable for 90% reliability? (2) Also assess the bearingβs reliability.Set ππ 1.Solution:The basic bearing equation is:πΆ10πΏπ· 1/π ()ππ πΉπ·π1 πΏ10(Where π and πΏ10 are generally set by the manufacturer, π is a variable we can change.)Where:πΆ10 19.5 ππ 4387.5 ππ; πΏ10 106 πππ£π ; ππ 1; π 3Also πΉπ· 495 ππ; and πΏπ· (30,000)(60)(300) (540)(106 ) πππ£π .Assume 90% reliability, then π1 1. From the basic bearing equation,πΆ10πΏπ· 1/π ()ππ πΉπ·π1 πΏ10Substituting values, πΏπ»π 8.92, π π»π 8.14. Therefore, (πΏπ· , πΉπ· ) is not on the rated line.
There are a number of ways to seek the answer.90% reliability, π1 1;The first: similar to the typical calculations done for selecting a bearingSet πΉπ· 495 ππ; πΏπ· (540)(106 ) πππ£π ; and πΏ10 106 πππ£π ; find πΆ10 and check if it is less than the4387.5 ππ that the bearing is capable of providing.From:πΆ10πΏπ· 1/π ()ππ πΉπ·π1 πΏ10Substituting known values:1/3πΆ10(540)(106 ) ()(1)(106 )(1)(495)Resulting in πΆ10 4031 ππSince itβs less than the catalogβs πΆ10 , or 4387.5 ππ, the selected bearing is suitable for 90% reliability.The second: can be used to select a bearing (pre-selecting a bearing, then checking to make sure it issuitable)Set πΏπ· (540)(106 ) πππ£π , find πΉπ· , and check is πΉπ· 495 ππ.From:πΆ10πΏπ· 1/π ()ππ πΉπ·π1 πΏ10Substituting known values:1/3(4387.5)(540)(106 ) ()(1)(106 )(1)πΉπ·Solving gives πΉπ· 538.8 ππ with 90% reliability and a life of (540)(106 ) πππ£π , the bearing can take on a maximum radial load of538.8 ππ. Since the applied radial load is only 495 ππ, the bearing will have better than 90% reliability.The third: similar to post-selection calculation to evaluate the life of the bearing.Set πΉπ· 495 ππ, find πΏπ· , and check if πΏπ· (540)(106 ) πππ£π .From:πΆ10πΏπ· 1/π ()ππ πΉπ·π1 πΏ10Substituting known values:1/3(4387.5)πΏπ· ()(1)(106 )(1)(495)Solving gives πΏπ· (696)(106 ) πππ£π With a radial load at 495 ππ, the bearing has 90% chance proabability to survive at least(696)(106 ) πππ£π . The chance of surviving only (540)(106 ) πππ£π is better than 90%.(2) Set πΉπ· 495 ππ; πΏπ· (540)(106 ) πππ£π To assess reliability means to evaluate π1 . This is typically done after selection.
From:πΆ10πΏπ· 1/π ()ππ πΉπ·π1 πΏ10Substituting known values:1/3(4387.5)(540)(106 ) ()(1)(495)π1 (106 )Solving gives π1 0.775Finally, from:π1 4.26 (ln100 2/3) 0.05π Solving for π results in π 93%.With the radial load at 495 ππ, there is a 7% of chance that the bearing would fail at or before 540millions of revs.It shows that the bearing is more than suitable for 90% reliability.Example 2Select bearings A and B for the shaft of Example 7-2. They are to be used for a minimum of 1,000 hoursof continuous operation. Shaft rpm is 450. Radial loads are, πΉπ΄ 375 ππ and πΉπ΅ 1918 ππ. Shaftdiameter at both locations is 1" (π·1 and π·7 in Figure 7-10). Assume 90% reliability.Solution:Table 11-2Table 11-3Since there is no thrust load, deep-groove ball bearings may
Example 2Select bearings A and B for the shaft of Example 7-2. They are to be used for a minimum of 1,000 hoursof continuous operation. Shaft rpm is 450. Radial loads are, πΉπ΄ 375 ππ and πΉπ΅ 1918 ππ. Shaftdiameter at both locations is 1" (π·1 and π·7 in Figure 7-10). Assume 90% reliability.Solution:Since there is no thrust load, deep-groove ball bearings are first considered. Table 11-2 has a list of 02series deep groove ball bearings.π1 1;ππ 1.2 (πππππ 11 5, ππππππππππ πππππππ, 1.1 1.3);πΏ10 106 πππ£π ;πΏπ· 1000 60 450 27 106 πππ£π ;
Bearing π΅: πΉπ· πΉπ΅ 1918 ππ 8535 π. From:πΆ10πΏπ· 1/π ()ππ πΉπ·π1 πΏ10Itβs found that πΆ10 30,726 π. Note π 3.Table 11-2 Shows that the smallest (in dimensions) bearing meeting the requirements is the one withbore diameter of 40 ππ and its πΆ10 is 30.7 ππ.Switch to roller bearing (Table 11-3). With π 10/3, then πΆ10 27,529 π.From Table 11-3, under 02-series, the bearing with 35-mm bore has πΆ10 31.9 π;Under 03 series, the bearing with 25-mm bore has a πΆ10 28.6 kN.
Select 03-series, ππππ 25 ππ, ππ· 62 ππ, π€πππ‘β 17 ππ, and πΆ10 28.6 ππAssessing reliability: π1 0.88, and π 91.7Bearing π΄: πΉπ· πΉπ΄ 375 ππ 1669 π. Use the same bearings as at B.Assessing reliability: π1 0.00382, π 99.9975. So π 99.The two bearings combined will have a reliability of (0.917) (0.99) 0.9111-6 Combine
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