上海交通大学：《Design and Manufacturing》课程教学资源（参考文献）Shigley's Mechanical Engineering Design ch6 fatigue failures resulting from variable loading.pdf_P16-P20

Budynas-Nisbett:Shigley's ll.Failure Prevention 6.Fatigue Failure Resulting ©The McGraw-Hil 275 Mechanical Engineering from Variable Loading Companies,2008 Design,Eighth Edition 272 Mechanical Engineering Design Figure 6-15 Los贵 When da/dN is measured in Fig.6-14 and plotted on Region I RegionⅡ loglog coordinates,the data Crack for different stress ranges propagation RegionⅢl superpose,giving rise to a Crack sigmoid curve as shown. Increasing unstable (AK th is the threshold value stress ratio of AKI,below which a crack does not grow.From threshold to rupture an aluminum alloy will spend 85-90 percent of life in region I,5-8 percent in region I,and 1-2 percent in region IIl. (△K) Log△K Table 6-1 m/cycle in/cycle Conservative Values of C Material MPa√mm (kpsivin" m Factor C and Exponent m in Eg.(6-5)for Ferritic-pearlitic steels 6.8910-121 3.6010-101 3.00 Various Forms of Steel Martensitic steels 1.36(10-101 6.60(10-91 2.25 (R=O) Austenitic stainless steels 5.6110-121 3.0010-101 3.25 From JM.Barsom and S.T.Rolfe,Fatigue and Frocture Control in Strctures,2nded.Prentice Hall,Upper Saddle River,NJ,1987 pp.8oprightASTMIntio Reprinted wth peision. conditions prevail.Assuming a crack is discovered early in stage II,the crack growth in region II of Fig.6-15 can be approximated by the Paris equarion,which is of the form aN =C(AK1)m da (6-5) where C and mn are empirical material constants and AK is given by Eq.(6-4). Representative,but conservative,values of C and m for various classes of steels are listed in Table 6-1.Substituting Eq.(6-4)and integrating gives dN-N- da (B△o√πa)m (661 Here a;is the initial crack length,at is the final crack length corresponding to failure, and Nf is the estimated number of cycles to produce a failure after the initial crack is formed.Note that B may vary in the integration variable (e.g.see Figs.5-25 to 5-30). SRecommended references are:Dowling,op.cit.;J.A.Collins,Failure of Materials in Mechanical Design, John Wiley Sons,New York,1981:H.O.Fuchs and R.I.Stephens,Metal Fatigue in Engineering.John Wiley Sons.New York,1980:and Harold S.Reemsnyder,"Constant Amplitude Fatigue Life Assessment Models,"SAE Trans.820688.vol.91,Nov.1983

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition II. Failure Prevention 6. Fatigue Failure Resulting from Variable Loading © The McGraw−Hill 275 Companies, 2008 272 Mechanical Engineering Design conditions prevail.8 Assuming a crack is discovered early in stage II, the crack growth in region II of Fig. 6–15 can be approximated by the Paris equation, which is of the form da d N = C(KI) m (6–5) where C and m are empirical material constants and KI is given by Eq. (6–4). Representative, but conservative, values of C and m for various classes of steels are listed in Table 6–1. Substituting Eq. (6–4) and integrating gives Nf 0 d N = Nf = 1 C af ai da (βσ√πa)m (6–6) Here ai is the initial crack length, af is the final crack length corresponding to failure, and Nf is the estimated number of cycles to produce a failure after the initial crack is formed. Note that β may vary in the integration variable (e.g., see Figs. 5–25 to 5–30). Log ΔK Log da dN Increasing stress ratio R Crack propagation Region II Crack initiation Region I Crack unstable Region III (ΔK)th Kc Figure 6–15 When da/dN is measured in Fig. 6–14 and plotted on loglog coordinates, the data for different stress ranges superpose, giving rise to a sigmoid curve as shown. (K I) th is the threshold value of K I, below which a crack does not grow. From threshold to rupture an aluminum alloy will spend 85--90 percent of life in region I, 5--8 percent in region II, and 1--2 percent in region III. Table 6–1 Conservative Values of Factor C and Exponent m in Eq. (6–5) for Various Forms of Steel (R . = 0) Material C, m/cycle MPa√mm C, in/cycle kpsi√ inm m Ferritic-pearlitic steels 6.89(10−12) 3.60(10−10) 3.00 Martensitic steels 1.36(10−10) 6.60(10−9) 2.25 Austenitic stainless steels 5.61(10−12) 3.00(10−10) 3.25 From J.M. Barsom and S.T. Rolfe, Fatigue and Fracture Control in Structures, 2nd ed., Prentice Hall, Upper Saddle River, NJ, 1987, pp. 288–291, Copyright ASTM International. Reprinted with permission. 8 Recommended references are: Dowling, op. cit.; J. A. Collins, Failure of Materials in Mechanical Design, John Wiley & Sons, New York, 1981; H. O. Fuchs and R. I. Stephens, Metal Fatigue in Engineering, John Wiley & Sons, New York, 1980; and Harold S. Reemsnyder, “Constant Amplitude Fatigue Life Assessment Models,” SAE Trans. 820688, vol. 91, Nov. 1983.

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition II. Failure Prevention 6. Fatigue Failure Resulting from Variable Loading 276 © The McGraw−Hill Companies, 2008 Fatigue Failure Resulting from Variable Loading 273 If this should happen, then Reemsnyder9 suggests the use of numerical integration employing the algorithm δaj = C(KI) m j (δN)j aj+1 = aj + δaj Nj+1 = Nj + δNj (6–7) Nf = δNj Here δaj and δNj are increments of the crack length and the number of cycles. The procedure is to select a value of δNj , using ai determine β and compute KI, determine δaj , and then find the next value of a. Repeat the procedure until a = af . The following example is highly simplified with β constant in order to give some understanding of the procedure. Normally, one uses fatigue crack growth computer programs such as NASA/FLAGRO 2.0 with more comprehensive theoretical models to solve these problems. EXAMPLE 6–1 The bar shown in Fig. 6–16 is subjected to a repeated moment 0 ≤ M ≤ 1200 lbf · in. The bar is AISI 4430 steel with Sut = 185 kpsi, Sy = 170 kpsi, and KIc = 73 kpsi√in. Material tests on various specimens of this material with identical heat treatment indicate worst-case constants of C = 3.8(10−11)(in/cycle)(kpsi√in)m and m = 3.0. As shown, a nick of size 0.004 in has been discovered on the bottom of the bar. Estimate the number of cycles of life remaining. Solution The stress range σ is always computed by using the nominal (uncracked) area. Thus I c = bh2 6 = 0.25(0.5)2 6 = 0.010 42 in3 Therefore, before the crack initiates, the stress range is σ = M I/c = 1200 0.010 42 = 115.2(103 ) psi = 115.2 kpsi which is below the yield strength. As the crack grows, it will eventually become long enough such that the bar will completely yield or undergo a brittle fracture. For the ratio of Sy/Sut it is highly unlikely that the bar will reach complete yield. For brittle fracture, designate the crack length as af . If β = 1, then from Eq. (5–37) with KI = KIc , we approximate af as af = 1 π KIc βσmax 2 . = 1 π 73 115.2 2 = 0.1278 in Figure 6–16 M M Nick in 1 2 in 1 4 9 Op. cit.

Budynas-Nisbett:Shigley's ll.Failure Prevention 6.Fatigue Failure Resulting T©The McGraw-Hill Mechanical Engineering from Variable Loading Companies,2008 Design,Eighth Edition 274 Mechanical Engineering Design From Fig.5-27,we compute the ratio a/h as =01278 =0.256 h 0.5 Thus af/h varies from near zero to approximately 0.256.From Fig.5-27,for this range B is nearly constant at approximately 1.07.We will assume it to be so,and re-evaluate af as 1 73 2 af= 元1.07115.2) =0.112in Thus,from Eq.(6-6),the estimated remaining life is da N好=7 1 r0.112 da (BAo√ram= 3.8(10-)J0.004[1.07115.2)√ma 5.047(103)10.112 =64.7(103)cycles Va 10.004 6-7 The Endurance Limit The determination of endurance limits by fatigue testing is now routine,though a lengthy procedure.Generally,stress testing is preferred to strain testing for endurance limits. For preliminary and prototype design and for some failure analysis as well,a quick method of estimating endurance limits is needed.There are great quantities of data in the literature on the results of rotating-beam tests and simple tension tests of specimens taken from the same bar or ingot.By plotting these as in Fig.6-17,it is possible to see whether there is any correlation between the two sets of results.The graph appears to suggest that the endurance limit ranges from about 40 to 60 percent of the tensile strength for steels up to about 210 kpsi (1450 MPa).Beginning at about S=210 kpsi (1450 MPa),the scatter appears to increase,but the trend seems to level off,as sug- gested by the dashed horizontal line at S=105 kpsi. We wish now to present a method for estimating endurance limits.Note that esti- mates obtained from quantities of data obtained from many sources probably have a large spread and might deviate significantly from the results of actual laboratory tests of the mechanical properties of specimens obtained through strict purchase-order specifi- cations.Since the area of uncertainty is greater,compensation must be made by employ- ing larger design factors than would be used for static design. For steels,simplifying our observation of Fig.6-17,we will estimate the endurance limit as 0.5St Sir 200 kpsi (1400 MPa) 100 kpsi Sur 200 kpsi (6-8) 700 MPa Sur 1400 MPa where S is the minimum tensile strength.The prime mark on Sin this equation refers to the rotating-beam specimen itself.We wish to reserve the unprimed symbol S.for the endurance limit of any particular machine element subjected to any kind of loading. Soon we shall learn that the two strengths may be quite different

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition II. Failure Prevention 6. Fatigue Failure Resulting from Variable Loading © The McGraw−Hill 277 Companies, 2008 274 Mechanical Engineering Design From Fig. 5–27, we compute the ratio af /h as af h = 0.1278 0.5 = 0.256 Thus af /h varies from near zero to approximately 0.256. From Fig. 5–27, for this range β is nearly constant at approximately 1.07. We will assume it to be so, and re-evaluate af as af = 1 π 73 1.07(115.2) 2 = 0.112 in Thus, from Eq. (6–6), the estimated remaining life is Nf = 1 C af ai da (βσ√πa)m = 1 3.8(10−11) 0.112 0.004 da [1.07(115.2) √πa]3 = −5.047(103) √a 0.112 0.004 = 64.7 (103 ) cycles 6–7 The Endurance Limit The determination of endurance limits by fatigue testing is now routine, though a lengthy procedure. Generally, stress testing is preferred to strain testing for endurance limits. For preliminary and prototype design and for some failure analysis as well, a quick method of estimating endurance limits is needed. There are great quantities of data in the literature on the results of rotating-beam tests and simple tension tests of specimens taken from the same bar or ingot. By plotting these as in Fig. 6–17, it is possible to see whether there is any correlation between the two sets of results. The graph appears to suggest that the endurance limit ranges from about 40 to 60 percent of the tensile strength for steels up to about 210 kpsi (1450 MPa). Beginning at about Sut = 210 kpsi (1450 MPa), the scatter appears to increase, but the trend seems to level off, as suggested by the dashed horizontal line at S e = 105 kpsi. We wish now to present a method for estimating endurance limits. Note that estimates obtained from quantities of data obtained from many sources probably have a large spread and might deviate significantly from the results of actual laboratory tests of the mechanical properties of specimens obtained through strict purchase-order specifi- cations. Since the area of uncertainty is greater, compensation must be made by employing larger design factors than would be used for static design. For steels, simplifying our observation of Fig. 6–17, we will estimate the endurance limit as S e = ⎧ ⎨ ⎩ 0.5Sut Sut ≤ 200 kpsi (1400 MPa) 100 kpsi Sut > 200 kpsi 700 MPa Sut > 1400 MPa (6–8) where Sut is the minimum tensile strength. The prime mark on S e in this equation refers to the rotating-beam specimen itself. We wish to reserve the unprimed symbol Se for the endurance limit of any particular machine element subjected to any kind of loading. Soon we shall learn that the two strengths may be quite different.

278 Budynas-Nisbett:Shigley's ll.Failure Prevention 6.Fatigue Failure Resulting ©The McGraw-Hill Mechanical Engineering from Variable Loading Companies,2008 Design,Eighth Edition Fatigue Failure Resulting from Variable Loading 275 140 0 120 O Carbon steels ·Alloy stcels 8 04 +Wrought irons 6 o 105 kpsi 0 60 8.8。 0 40 20 020 40 60 80 100120140160180200220240260280300 Tensile strength Skpsi Figure 6-17 Graph of endurance limits versus tensile strengths from actual test results for a large number of wrought irons and steels.Ratios of S/Sur of 0.60,0.50,and 0.40 are shown by the solid and dashed lines. Note also the horizontal dashed line for S=105 kpsi.Points shown having a tensile strength greater than 210 kpsi have a mean endurance limit of S=105 kpsi and a standard deviation of 13.5 kpsi. (Collated from data compiled by H.J.Grover,S.A.Gordon,and L.R.Jackson in Fatigue of Metals and Structures,Bureau of Naval Weapons Document NAWWEPS 00-25-534,1960;and from Fatigue Design Handbook,SAE,1968,p.42.) Steels treated to give different microstructures have different S/S ratios.It appears that the more ductile microstructures have a higher ratio.Martensite has a very brittle nature and is highly susceptible to fatigue-induced cracking;thus the ratio is low. When designs include detailed heat-treating specifications to obtain specific microstructures,it is possible to use an estimate of the endurance limit based on test data for the particular microstructure;such estimates are much more reliable and indeed should be used The endurance limits for various classes of cast irons,polished or machined,are given in Table A-24.Aluminum alloys do not have an endurance limit.The fatigue strengths of some aluminum alloys at 5(10)cycles of reversed stress are given in Table A-24. 6-8 Fatigue Strength As shown in Fig.6-10,a region of low-cycle fatigue extends from N =1 to about 103 cycles.In this region the fatigue strength St is only slightly smaller than the ten- sile strength St.An analytical approach has been given by Mischkelo for both 0.E.Shigley.C.R.Mischke,andT.H.Brown,Jr Standard Handbook of Machine Design.3rd ed.. McGraw-Hill,New York,2004.pp.29.25-29.27

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition II. Failure Prevention 6. Fatigue Failure Resulting from Variable Loading 278 © The McGraw−Hill Companies, 2008 Fatigue Failure Resulting from Variable Loading 275 Steels treated to give different microstructures have different S e/Sut ratios. It appears that the more ductile microstructures have a higher ratio. Martensite has a very brittle nature and is highly susceptible to fatigue-induced cracking; thus the ratio is low. When designs include detailed heat-treating specifications to obtain specific microstructures, it is possible to use an estimate of the endurance limit based on test data for the particular microstructure; such estimates are much more reliable and indeed should be used. The endurance limits for various classes of cast irons, polished or machined, are given in Table A–24. Aluminum alloys do not have an endurance limit. The fatigue strengths of some aluminum alloys at 5(108 ) cycles of reversed stress are given in Table A–24. 6–8 Fatigue Strength As shown in Fig. 6–10, a region of low-cycle fatigue extends from N = 1 to about 103 cycles. In this region the fatigue strength Sf is only slightly smaller than the tensile strength Sut . An analytical approach has been given by Mischke10 for both 0 20 40 60 80 100 120 140 160 180 200 260 300 220 240 280 Tensile strength Sut, kpsi 0 20 40 60 80 100 120 140 Endurance limit S 'e , kpsi 105 kpsi 0.4 0.5 S 'e Su = 0.6 Carbon steels Alloy steels Wrought irons Figure 6–17 Graph of endurance limits versus tensile strengths from actual test results for a large number of wrought irons and steels. Ratios of S e/Sut of 0.60, 0.50, and 0.40 are shown by the solid and dashed lines. Note also the horizontal dashed line for S e = 105 kpsi. Points shown having a tensile strength greater than 210 kpsi have a mean endurance limit of S e = 105 kpsi and a standard deviation of 13.5 kpsi. (Collated from data compiled by H. J. Grover, S. A. Gordon, and L. R. Jackson in Fatigue of Metals and Structures, Bureau of Naval Weapons Document NAVWEPS 00-25-534, 1960; and from Fatigue Design Handbook, SAE, 1968, p. 42.) 10J. E. Shigley, C. R. Mischke, and T. H. Brown, Jr., Standard Handbook of Machine Design, 3rd ed., McGraw-Hill, New York, 2004, pp. 29.25–29.27.

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition II. Failure Prevention 6. Fatigue Failure Resulting from Variable Loading © The McGraw−Hill 279 Companies, 2008 276 Mechanical Engineering Design high-cycle and low-cycle regions, requiring the parameters of the Manson-Coffin equation plus the strain-strengthening exponent m. Engineers often have to work with less information. Figure 6–10 indicates that the high-cycle fatigue domain extends from 103 cycles for steels to the endurance limit life Ne, which is about 106 to 107 cycles. The purpose of this section is to develop methods of approximation of the S-N diagram in the highcycle region, when information may be as sparse as the results of a simple tension test. Experience has shown high-cycle fatigue data are rectified by a logarithmic transform to both stress and cycles-to-failure. Equation (6–2) can be used to determine the fatigue strength at 103 cycles. Defining the specimen fatigue strength at a specific number of cycles as (S f )N = Eεe/2, write Eq. (6–2) as (S f )N = σ F (2N)b (6–9) At 103 cycles, (S f )103 = σ F (2.103)b = f Sut where f is the fraction of Sut represented by (S f )103 cycles. Solving for f gives f = σ F Sut (2 · 103 ) b (6–10) Now, from Eq. (2–11), σ F = σ0εm , with ε = ε F . If this true-stress–true-strain equation is not known, the SAE approximation11 for steels with HB ≤ 500 may be used: σ F = Sut + 50 kpsi or σ F = Sut + 345 MPa (6–11) To find b, substitute the endurance strength and corresponding cycles, S e and Ne, respectively into Eq. (6–9) and solving for b b = −log σ F /S e log (2Ne) (6–12) Thus, the equation S f = σ F (2N) b is known. For example, if Sut = 105 kpsi and S e = 52.5 kpsi at failure, Eq. (6–11) σ F = 105 + 50 = 155 kpsi Eq. (6–12) b = −log(155/52.5) log 2 · 106 = −0.0746 Eq. (6–10) f = 155 105 2 · 103−0.0746 = 0.837 and for Eq. (6–9), with S f = (S f )N , S f = 155(2N)−0.0746 = 147 N −0.0746 (a) 11Fatigue Design Handbook, vol. 4, Society of Automotive Engineers, New York, 1958, p. 27.