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1、P1: GIG ham65328_Ch07Hamrock-0402TPB466-Hamrock-v28.clsApril 10, 200414:50 CHAPTERCHAPTER 7FAILUREPREDICTION FORCYCLIC ANDIMPACTLOADING Aloha Airlines Flight 243, a Boeing 737-200, taken April 28, 1988. The midfl ight fuselage failure was caused by corrosion-assisted fatigue. (Steven Minkowski/ Gamm
2、a Liaison) All machine and structural designs are problems in fatigue because the forces of Nature are always at work and each object must respond in some fashion. Carl Osgood, Fatigue Design 265 P1: GIG ham65328_Ch07Hamrock-0402TPB466-Hamrock-v28.clsApril 10, 200414:50 266PART 1 Fundamentals SYMBOL
3、S Aarea, m2 Aaamplitude ratio, a/m afatigue strength exponent bwidth, m bsslope Cconstant used in Equation (7.47) C1,C2integration constants Cintercept c distance from neutral axis to outer fi ber of solid, m ddiameter, m Emodulus of elasticity, Pa e, ffactors used in Equation (7.21) fexponent ggrav
4、itational acceleration, 9.807 m/s2 (386.1 in/s2) Hheight including two notch radii, m hheight without two notch radii, m Iarea moment of inertia, m4 Imimpact factor Kstress intensity factor, MPam Kcstress concentration factor Kffatigue stress concentration factor kspring rate, N/m ?Kstress intensity
5、 range, MPam kf surface fi nish factor kmmiscellaneous factor krreliability factor kssize factor kttemperature factor k0stress concentration factor llength, m lccrack length, m Mbending moment, N m mconstant used in Equation (7.47) Nfatigue life in cycles Ncnumber of stress cycles N? number of cycle
6、s to failure at a specifi c stress N? t total number of cycles to failure n? number of cycles at a specifi c stress when n? 1 2 1 da dN (a)(b) Figure 7.4Illustration of fatigue crack growth. (a) Size of a fatigue crack for two different stress ratios as a function of the number of cycles; (b) rate o
7、f crack growth, illustrating three regimes. Source: Suresh (1998). P1: GIG ham65328_Ch07Hamrock-0402TPB466-Hamrock-v28.clsApril 10, 200414:50 CHAPTER 7 Failure Prediction for Cyclic and Impact Loading275 7.5.3MICROSTRUCTURE OFFATIGUEFAILURES As discussed, even the most ductile materials can exhibit
8、brittle behavior in fatigue and will fracture with little or no plastic deformation. The reasons for this are not at all obvious, but an investigation of fatigue fracture microstructure can help explain this behavior. AtypicalfatiguefracturesurfaceisshowninFigure7.5andhasthefollowingfeatures: 1.Near
9、theoriginofthefatiguecrack,thesurfaceisburnished,orverysmooth.Intheearly stagesoffatigue,thecrackgrowsslowlyandelasticdeformationsresultinmicroscopic sliding between the two surfaces, resulting in a rubbing of the surfaces. 2. Nearthefi nalfracturelocation,striationsorbeachmarksareclearlyvisibletoth
10、enaked eye. During the last few cycles of a fatigue failure, the crack growth is very rapid, and these striations are indicative of fast growth and growth-arrest processes. 3.Microscopic striations can exist between these two extremes, as shown in Figure 7.5, and are produced by the slower growth of
11、 fatigure cracks at this location in the part. 4. The fi nal fracture surface often looks rough and is indicative of brittle fracture, but it can also appear ductile depending on the material. Theactualpatternofstriationsdependsontheparticulargeometry,material,andloading (Fig. 7.6) and can require e
12、xperience to evaluate a failure cause. Figure 7.5Cross-section of a fatigued section, showing fatigue striations or beachmarks originating from a fatigue crack at B. (From Rimnac et al, 1986. Copyright American Society for Testing and Materials. Reprinted by permission.) P1: GIG ham65328_Ch07Hamrock
13、-0402TPB466-Hamrock-v28.clsApril 10, 200414:50 276PART 1 Fundamentals High Nominal StressLow Nominal Stress No stress concentration Mild stress concentration Severe stress concentration No stress concentration Mild stress concentration Severe stress concentration Tension-tension or tension-compressi
14、on Unidirectional bending Reversed bending Rotational bending BeachmarksFracture surface Figure 7.6Typical fatigue fracture surfaces of smooth and notched cross-sections under different loading conditions and stress levels. From Metals Handbook, American Society for Metals (1975). 7.5.4SN DIAGRAMS D
15、ata from reversed-bending experiments are plotted as the fatigue strength versus the logarithm of the total number of cycles to failure N? t for each specimen. These plots are called SN diagrams or W ohler diagrams, after August W ohler, a German engineer who publishedhisfatigueresearchin1870.Theyar
16、eastandardmethodofpresentingfatiguedata and are common and informative. Two general patterns for two classes of material, those with and those without endurance limits, emerge when the fatigue strength is plotted versus the logarithm of the number of cycles to failure. Figure 7.7 shows typical resul
17、ts for several materials.Figure7.7(a)presentstestdataforwroughtsteel.Notethelargeamountofscatter inthedata,evenwiththegreatcareinpreparingtestspecimens.Thus,lifepredictionsbased on curves such as those in Figure 7.7 are all subject to a great deal of error. Figure 7.7(a) also shows a common result.
18、For some materials with endurance limits, such as ferrous and titanium alloys, a horizontal straight line occurs at low stress levels, implying that an endurance limit S? e is reached below which failure will not occur. This endurance limit S? e P1: GIG ham65328_Ch07Hamrock-0402TPB466-Hamrock-v28.cl
19、sApril 10, 200414:50 Not broken 1.0 0.9 0.8 0.7 0.6 0.5 0.4 103 103 104103105106107 104105106107108109 104105106107 Number of cycles to failure N Number of cycles to failure N Number of cycles to failure N Fatigue stress ratio Sf /Sut 80 70 60 50 40 35 30 25 20 18 16 14 12 10 8 7 6 5 Alternating str
20、ess sa, ksi Alternating stress sa, MPa Alternating stress sa, psi 00 2 4 6 8 10 20 30 40 50 60 PTFE Phenolic Epoxy Diallyl-phthalate Alkyd Nylon (dry) Polycarbonate Polysulfone (a) (b) (c) Wrought Sand cast Permanent mold cast Figure 7.7Fatigue strengths as a function of number of loading cycles. (a
21、) Ferrous alloys, showing clear endurance limit adapted from Lipson and Juvinall (1963); (b) aluminum alloys, with less pronounced knee and no endurance limit adapted from Juvinall and Marshek (1991); (c) selected properties of assorted polymer classes adapted from Norton (1996). 277 P1: GIG ham6532
22、8_Ch07Hamrock-0402TPB466-Hamrock-v28.clsApril 10, 200414:50 278PART 1 Fundamentals represents the largest fl uctuating stress that will not cause failure for an infi nite number of cycles. For many steels the endurance limit ranges between 35% and 60% of the materials ultimate strength. Most nonferr
23、ous alloys (e.g., aluminum, copper, and magnesium) do not have a signif- icant endurance limit. Their fatigue strength continues to decrease with increasing cycles. Thus, fatigue will occur regardless of the stress amplitude. The fatigue strength for these materials is taken as the stress level at w
24、hich failure will occur for some specifi ed number of cycles (e.g., 107cycles). Determining the endurance limit experimentally is lengthy and expensive. The MansonCoffi nrelationshipEq.(7.6)demonstratesthatthefatiguelifewilldependonthe materials fracture strength during a single load cycle, suggesti
25、ng a possible relationship between static material strength and strength in fatigue. Such a relationship has been noted by several researchers (e.g., Fig. 7.8). The stress endurance limits of steel for three types of loading can be approximated as follows: BendingS? e = 0.5Su AxialS? e = 0.45Su Tors
26、ionS? e = 0.29Su (7.7) These equations can be used to approximate the endurance limits for other ferrous alloys, but it must be recognized that the limits can differ signifi cantly from experimentally determined endurance limits. Since the ultimate stress and the type of loading are known for variou
27、s materials, their endurance limits can be approximated. 160 103 140 120 100 80 60 40 20 0 020406080100140180220260300 103120160200240280 Se _ Su = 0.6 0.5 0.4 Carbon steels Alloy steels Wrought irons Tensile strength Sut, psi Endurance limit Se, psi 100 103 psi Figure 7.8Endurance limit as function
28、 of ultimate strength for wrought steels. Adapted from Shigley and Mitchell (1983). P1: GIG ham65328_Ch07Hamrock-0402TPB466-Hamrock-v28.clsApril 10, 200414:50 CHAPTER 7 Failure Prediction for Cyclic and Impact Loading279 Table 7.2Approximate endurance limit for various materials MaterialNumber of cy
29、clesRelation Magnesium alloys108S?e= 0.35Su Copper alloys1080.25Su 1 and thus failure should not occur; however because the safety factor is just above 1, the margin of safety is a minimum. Case Study 7.1POWERPRESSBRAKESTUDDESIGNREVIEW Given:An engineer working for a manufacturer of me- chanicalpowe
30、rpressesisassistinginthedevelopmentofa new size of machine. When starting to design a new brake stud, the engineer fi nds the information in Figure 7.19 from an existing computer-assisted drawing (CAD) of a machine with the closest capacity to the new project. The brake stud supports the brake on th
31、e camshaft of the me- chanical power press. The brake actuates with every cycle of the press, stopping the ram at the top-dead-center posi- tion so that an operator can remove a workpiece from the dies and insert another workpiece for the next cycle. If the stud fails, the press could continue to co
32、ast downward and could result in a serious injury. FocusingonsectionA-A,theengineerrecognizesthat nofi lletradiushadbeenspecifi edforthebrakestudforma- chinessoldpreviously.Conversationswithmachinistswho routinely worked on the part led to the conclusion that the common practice was to undercut the
33、fi llet to make sure assembly was complete; the fi llet radius in effect was the radius of the machine tool insert, a value as low as 1 8 in. The immediate concern is whether a product recall is in (continued) P1: GIG ham65328_Ch07Hamrock-0402TPB466-Hamrock-v28.clsApril 10, 200414:50 310PART 1 Funda
34、mentals Case Study(Continued) 2.25 in1 in 1.375 in.3.0625 in R = 0.375 in Machine frame AA P x Shoulder Figure 7.19Dimensions of existing brake stud design. order, since a brake stud failure could result in a machine operators hands being in a die while the machine fails to stop after a cycle. The s
35、upervisor shares the engineers fears but assigns one immediate task. Find:Is the machine safe as manufactured? What con- cerns exist for this machine in service? Is a recall neces- sary? The carbon steel used for the stud has a minimum ultimate strength of 74.5 ksi, and no yield strength was prescri
36、bed in the drawing. Solution: (a)Applied stresses. Figure 7.19 shows that the power press frame and the brake stud share the loading if the stud-retaining nut is tight. If the nut is loose, the stud can conceivably carry the entire loading. Exist- ing brake design calculations reveal that the brake
37、acts over one-quarter of the camshaft revolution and has a relatively constant peak load of almost 1000 lb. Figure 7.20 shows the shear and bending moment di- agrams for the stud and the loading with respect to time. The critical loading is bending (shear must also be considered, but in this case, a
38、s with most shafts, bending stresses are the critical consideration). The maximum stress in the absence of stress raisers is given by max= Mc I = (1000)(3.0625/2 + 1.375)(0.5) (/4)(0.5)4 psi = 29.6 ksi The loading is not completely reversing; the mean stress is 14.8 ksi, and the stress amplitude is
39、14.8 ksi (Fig. 7.20). The stress raiser cannot be ignored in this case. From Figure 6.5 the theoretical stress concentration P1: GIG ham65328_Ch07Hamrock-0402TPB466-Hamrock-v28.clsApril 10, 200414:50 CHAPTER 7 Failure Prediction for Cyclic and Impact Loading311 Case Study(Continued) P V Sheer force,
40、 lb M Moment, lb.in Px (a)(b) Time Stress, psi sa sm Figure 7.20Press brake stud loading. (a) Shear and bending-moment diagrams for applied load; (b) stress cycle. factor is approximately 1.7. The notch sensitivity is obtained from Figure 7.9 as 0.85. Thus, from Equa- tion (7.19) Kf= 1 + (1.7 1)0.85
41、 = 1.6 Recallthatforductilematerialsastressconcentration factor is not used for static loads, with the rationale that plastic fl ow (even at microscopic scales) will re- lieve stress concentrations. The brake actuates every time the press cy- cles. Some mechanical presses operate at more than 100 cy
42、cles/min, but the design speed for the subject machine is 40 cycles/min. Still, even when used for only one 8-h shift per day, the number of loading cy- cles on the stud would be 107after 20 years. Many mechanical presses have expected lives of twice this. Clearly, the stud must be designed for infi
43、 nite life. (b)Fatigue strength calculation. The steel has an ulti- mate strength of 74.5 ksi, but no yield strength was given. Thus, the Goodman line will be used for fa- tigueanalysis.Althoughthismayseemconservative, if a plastic deformation in the stud were possible, it would manifest itself in t
44、rials conducted on the machines before shipping. Therefore, the design re- viewer feels reasonably confi dent that the main fail- ure mode is fatigue. From Equation (7.7) the approximate value of an unmodifi ed endurance limit is S?e= 0.5Sut= 37.25 ksi. The following correction factors are obtained:
45、 Surface fi nish: A machined fi nish so that from Equation (7.21) kf= 2.70(74.5)0.265= 0.86 Size factor: From Equation (7.22) ks= 0.869(1)0.112= 0.869 Reliability factor (chosen from Table 7.3 for a 99% reliability) kr= 0.82 The endurance limit for the steel brake stud is then given by Equation (7.1
46、6) as Se= (37.25)(0.86)(0.869)(0.82) = 22.8 ksi (c)Fatiguecriterion.UsingtheGoodmanline Eq. (7.28) gives the safety factor as Kfa Se + m Sut = 1.6(14.8) 22.8 + 14.8 74.5 = 1 ns ns= 0.8 (continued) P1: GIG ham65328_Ch07Hamrock-0402TPB466-Hamrock-v28.clsApril 10, 200414:50 312PART 1 Fundamentals Case
47、Study(Concluded) Thus,thedesignisinadequate.However,iftherewere no stress concentration at all, the safety factor would increase to 1.17, certainly a low safety factor regard- less. Discussion:The machine manufacturer had not experi- enced failures in the brake stud, even though the existing design
48、had been on the market for over three decades. Through laboratory testing it was determined that if the brake-stud-retaining bolt were loose, the brake would make a noticeably loud squeal and the stud would clank. Undoubtedly, such problems were promptly identifi ed by maintenance personnel, and the
49、 stud bolt was retightened whenever it became loose in service. As long as the brake stud was snug against the machine support, or stud shoul- der, the safety factor for infi nite life was high. It was ul- timately determined that a recall was not in order, but a servicememorandumwasissuedtoallpreviouscustomers informing them of the need to properly maintain the tight- nessofthestud-retainingbolt.Subsequentservicemanuals also mentioned this important consideration. 7.13SUMMARY Failures in components or structures