Design Against Fatigue Failure 2/3/2015 1

Transcription

1 Design Aginst Ftigue Filure /3/015 1

2 Ftigue is the filure of mechnicl element by the growth of crck within mteril under vrible, repeted, lternting, or fluctuting stresses. Generlly, ftigue crck growth occurs under stresses below the ultimte tensile strength (σ ut ), yield strength (σ y ) nd criticl stress for the originl crck (σ c ). A ftigue filure hs n ppernce similr to brittle frcture, s the frcture surfces re flt nd perpendiculr to the stress xis with the bsence of necking. The ftigue of mteril is effected by the size of the component, reltive mgnitude of sttic nd vrible lods nd the number of lod reversls. /3/015

3 It is observed tht bout 80% of filure of mechnicl components re due to Ftigue Filure resulting from fluctuting stresses. NOTE The stresses which vry from minimum vlue to mximum vlue of the sme nture, (i.e. tensile or compressive) re clled fluctuting stresses. The stresses which vry from zero to certin mximum vlue re clled repeted stresses. The stresses which vry from minimum vlue to mximum vlue of the opposite nture (i.e. from certin minimum compressive to certin mximum tensile or from minimum tensile to mximum compressive) re clled lternting stresses. The stresses which vry from one vlue of compressive to the sme vlue of tensile or vice vers, re known s completely reversed or cyclic stresses. /3/015 3

4 Alternting stress Repeted stress Fluctuting stress Completely reversed stress σ Mx = Mximum stress σ Min = Minimum stress σ m = Men or verge stress σ,σ v =Alternting stress σ r = Rnge of stress /3/015 4

5 Ftigue filure is chrcterized by three stges I. Crck Initition II. Crck propgtion III. Finl frcture VW crnk shft ftigue filure due to cyclic bending nd torsionl stresses Crck initition site Propgtion zone, strition Frcture zone /3/015 5

6 Jck hmmer component, shows no yielding before frcture. Crck initition site Propgtion zone, strition Frcture zone /3/015 6

7 Ftigue-Life Methods The three mjor ftigue life methods used in design nd nlysis re 1. The stress-life method.. The strin-life method. 3. The liner-elstic frcture mechnics method. These methods ttempt to predict the life in number of cycles to filure (N) for specific level of loding. This cycles number (N) is generlly clssified s 1. low-cycle ftigue, where 1 N high-cycle ftigue, where N > /3/015 7

8 The tress-life Method (LM) The stress-life method, bsed on stress levels only, is the lest ccurte pproch, especilly for low-cycle pplictions. To determine the strength of mterils under the ction of ftigue lods, specimens re subjected to repeted or vrying forces of specified mgnitudes while the cycles or stress reversls re counted to destruction. The most widely used ftigue-testing device is the R.R. Moore high-speed rotting-bem mchine. This mchine subjects the specimen to pure bending (no trnsverse sher) by mens of weights. /3/015 8

9 Typicl testing pprtus (pure bending) Motor Lod Rotting bem mchine pplies fully reverse bending stress The specimen is polished Test-specimen geometry for the R. R. Moore rotting bem mchine. /3/015 9

10 The -N Digrm The -N digrm is the grphicl representtion of stress mplitude ( f ) verses the number of stress cycles (N) before the ftigue filure on semilog pper or on log-log pper. To estblish the ftigue strength of mteril, quite number of tests re necessry becuse of the sttisticl nture of ftigue. For the rotting-bem test, constnt bending lod is pplied, nd the number of revolutions (stress reversls) of the bem required for filure is recorded. The first test is mde t stress tht is somewht under the ultimte strength of the mteril. The second test is mde t stress tht is less thn tht used in the first. This process is continued, nd the results re plotted s -N digrm. /3/015 10

11 In the cse of the steels s shown in Fig.(A), knee occurs in the grph, nd beyond this knee filure will not occur, no mtter how gret the number of cycles. The strength corresponding to the knee is clled the endurnce limit or the ftigue limit. Aluminium lloys does not hve n endurnce limit s shown in Fig.(B), normlly the ftigue strength f is reported t specific number of cycles, normlly N = 5(10 8 ) cycles of reversed stress. Endurnce or ftigue limit stress ( e ) is defined s the mximum mplitude of completely reversed stress tht the stndrd specimen cn sustin for n unlimited number of cycles without ftigue filure. /3/015 11

12 Fig.(A) -N digrm plotted from the results of completely reversed xil ftigue tests (teel-normlized σ u =15kpsi) /3/015 1

13 Fig.(B) -N bnds for representtive luminium lloys, excluding wrought lloys ( ut <38 kpsi). /3/015 13

14 We lso distinguish finite-life region nd n infinitelife region in -N digrm s shown in Fig.(A). The boundry between these regions cnnot be clerly defined except for specific mteril; but it lies somewhere between 10 6 nd 10 7 cycles for steels, s shown in Fig.(A). /3/015 14

15 The trin-life Method (ELM) The trin-life Method is The best pproch yet dvnced to explin the nture of ftigue filure which is bsed on the investigtion of behvior of mterils subject to cyclic deformtion. Generlly, the strength decreses with stress repetitions, while other mterils my be strengthened, insted, by cyclic stress reversls (i.e. the elstic limits of nneled steels re likely to increse when subjected to cycles of stress reversls, while cold-drwn steels exhibit decresing elstic limit). Fig.(C) hs been constructed to show the generl ppernce of these plots for the first few cycles of controlled cyclic strin. In this cse the strength decreses with stress repetitions, s evidenced by the fct tht the reversls occur t ever-smller stress levels. /3/015 15

16 Fig. (D)A log-log plot showing how the ftigue life is relted to the true-strin mplitude for hot-rolled AE 100 steel. Fig.(C) True stress true strin hysteresis loops showing the first five stress reversls of cyclic-softening mteril. /3/015 16

17 Note tht the slope of the line AB is the modulus of elsticity E. σ is the stress rnge. ε p is the plstic-strin rnge. ε e is the elstic strin rnge. ε is the totl-strin rnge. ε = ε e + ε p To explin the Fig.(D), we first define the following terms: Ftigue ductility coefficient (ε F ) is the true strin corresponding to frcture in one reversl (point A in Fig.(C)). The plstic-strin line begins t this point in Fig.(D). /3/015 17

18 Ftigue strength coefficient (σ F ) is the true stress corresponding to frcture in one reversl (point A in Fig.(C). Note in Fig.(D) tht the elstic-strin line begins t σ F /E. Ftigue ductility exponent (c) is the slope of the plstic-strin line in Fig.(D) nd is the power to which the life N must be rised to be proportionl to the true plstic-strin mplitude. If the number of stress reversls is N, then N is the number of cycles. Ftigue strength exponent (b) is the slope of the elstic-strin line, nd is the power to which the life N must be rised to be proportionl to the true-stress mplitude. /3/015 18

19 Now, from Fig.(D), we see tht the totl strin is the sum of the elstic nd plstic components. Therefore, the totl strin mplitude is hlf the totl strin rnge. e The eqution of the plstic-strin line is The eqution of the elstic strin line is The totl-strin mplitude is p p ' (N) c F ' e F E ' F(N) (N) b c ' F E (N) b /3/015 19

20 The Liner-Elstic Frcture Mechnics Method (LEFM) The frcture mechnics method ssumes crck is lredy present nd detected which is employed to predict crck growth with respect to stress intensity. Including three stges, 1) The first phse of ftigue crcking is designted s stge I ftigue. ) The second phse, tht of crck extension, is clled stge II ftigue. 3) Finl frcture is occurs during stge III ftigue which is ssocited with rpid ccelertion of crck growth then frcture. /3/015 0

21 Crck Growth Ftigue crcks nuclete nd grow when stresses vry nd there is some tension in ech stress cycle. There re three modes of crck propgtion, I. The opening crck propgtion mode. II. The sliding crck propgtion mode. III. The tering crck propgtion mode. /3/015 1

22 Consider the stress to be vrible between the limits of σ min nd σ mx, where the stress rnge is defined s σ = σ mx σ min. Consider the stress intensity fctor (K) which is used in frcture mechnics to predict the stress stte ("stress intensity") ner the tip of crck cused by remote lod or residul stresses. Thus, for σ, the stress intensity rnge per cycle is K I ( K I K I ) Mx Min /3/015

23 Where, β: The stress intensity modifiction fctor. : The length of crck. K I : The stress intensity fctor for first mode. Chrt of Thin plte in tension or simple compression with certin crck to select proper β for mode I crck propgtion.. /3/015 3

24 To develop ftigue strength dt, number of specimens of the sme mteril re tested t vrious levels of σ. Crcks nuclete t or very ner free surfce or lrge discontinuity. Fig.(E) The increse in crck length () from n initil length of ( i ) s function of cycle count for three stress rnges, ( σ) 3 > ( σ) > ( σ) 1. /3/015 4

25 Fig.(F) When d/dn is mesured in Fig.(E) nd plotted on log-log coordintes, the dt for different stress rnges superpose, giving rise to sigmoid curve s shown. ( K I ) th is the threshold vlue of K I, below which crck does not grow. From threshold to rupture n luminum lloy will spend percent of life in region I, 5 8 percent in region II, nd 1 percent in region III. Where, K c is criticl stress intensity fctor nd R = σ min /σ mx /3/015 5

26 Here we present simplified procedure for estimting the remining life of cycliclly stressed prt fter discovery of crck. This requires the ssumption tht plne strin conditions previl. Assuming crck is discovered erly in stge II, the crck growth in region II of Fig.(F) cn be pproximted by the Pris eqution, which is of the form where C nd m re empiricl mteril constnts. /3/015 6

27 ubstituting K I integrting gives, in bove eqution nd i : The initil crck length f : The finl crck length corresponding to filure N f : The estimted number of cycles to produce filure fter the initil crck is formed. /3/015 7

28 Exmple: The br shown in figure below is subjected to repeted moment 0 M 100 lbf in. The br is AII 4430 steel with ut = 185 kpsi, y = 170 kpsi, nd K Ic = 73 kpsi in. Mteril tests on vrious specimens of this mteril with identicl het tretment indicte worstcse constnts of C = 3.8(10 11 )(in/cycle)/(kpsi in) m nd m = 3.0. As shown, nick of size 0.004in hs been discovered on the bottom of the br. Estimte the number of cycles of life remining. L=4in /3/015 8

29 olution: I c bh 6 0.5(0.5) in 3 M I c kpsi This vlue is below the yield strength. If β = 1.03, we pproximte f s f 1 K Ic ( f mx ) 1 73 ( ) 1.03* in /3/015 9

30 From this figure, we compute f h Thus f /h vries from ner zero to pproximtely From shown Figure, for this rnge β Is nerly constnt t pproxim- tely We will ssume it to be so, nd re-evlute f s f 1 K Ic ( f mx ) 1 73 ( ) 1.07* in /3/015 30

31 Thus the estimted remining life is N N f f 1 C N f Nf f i ( 1 3.8*10 11 d ) * *10 3 m 0.11 d d [1.07(115.) 0.5 ] [ * ] [ ( 0.5 )] *10 3 cycles /3/015 31

32 The Endurnce Limit Fig.1 Grph of endurnce limits versus tensile strengths from ctul test results for lrge number of wrought irons nd steels. Rtios of e / ut of 0.60, 0.50, nd 0.40 re shown by the solid nd dshed lines. Note lso the horizontl dshed line for e = 105 kpsi. /3/015 3

33 Therefore, fter simplifying the observtion, the estimtion of the endurnce limit is The prime mrk on e in this eqution refers to the rotting bem specimen itself. The unprimed symbol e for the endurnce limit of n ctul mchine element subjected to ny kind of loding. /3/015 33

34 Ftigue trength is the highest stress tht mteril cn withstnd for given number of cycles without breking. In other words, it is the mximum stress tht cn be pplied for certin number of cycles without frcture. It is ffected by environmentl fctors such s corrosion. To pproximte reltionship between -N during high cycle, two stges pproximtion re required which re I. Approximtion of the ftigue strength frction (f ) t N= II. Approximtion of the N during high cycle t N>10 3.(For n ctul mechnicl component) /3/015 34

35 I. Approximtion of the ftigue strength frction (f ) t N= Defining the specimen ftigue strength t specific number of cycles s ( f ) N = E ε e /, ince, ' e F E (N) b b ( ' f ) N ' F (N) (*) At 10 3 cycles, ' ' 3 ( ) (*10 ) f 10 3 F b f ' ( ' ut f 3 ) 10 ut F 3 b f (*10 ) /3/015 35

36 If this true-stress true-strin eqution (σ F = σ 0 ε m, with ε=ε F ) is not known, the AE pproximtion for steels with HB 500 my be used: ' kpsi or ' F ut 50 F ut 345 Mp To find b, substitute the endurnce strength nd corresponding cycles, e nd N e, respectively into Eq. (*) nd solving for b b log( ' / ' F e) log( N ) e /3/015 36

37 Fig. Ftigue strength frction, f, of ut t 10 3 cycles for e = e= 0.5 ut t 10 6 cycles. Above figure is plot of f for 70 ut 00 kpsi. To be conservtive, for ut < 70 kpsi, let f = 0.9. /3/015 37

38 II. Approximtion of the N during high cycle t N>10 3 f = N b Where N is cycles to filure nd the constnts nd b re defined by the points 10 3, f10 3 nd 10 6, e with f10 3= f ut. ubstituting these two points in bove eqution gives ( f ut e ) nd b 1 3 log( f e ut ) If completely reversed stress σ rev is given, setting f = σ rev. the number of cycles-to-filure cn be expressed s N ( rev ) 1 b /3/015 38

39 Exmple: Given 1050 HR steel, estimte () the rotting-bem endurnce limit t 10 6 cycles. (b) the endurnce strength of polished rotting-bem specimen corresponding to 10 4 cycles to filure (c) the expected life of polished rotting-bem specimen under completely reversed stress of 55 kpsi. olution: () From Tble A 0, ut = 90 kpsi. Owing of ut 00 kpsi, the endurnce limit is ' e ' e 0.5 ut 45kpsi 0.5(90) /3/015 39

40 /3/015 40

41 (b) From Fig., for ut = 90 kpsi, f =0.86. b ( f 1 3 ' For 10 4 cycles to filure, (c) For σ rev = 55 kpsi ut e ) log( N N f ' e (0.86*90) 45 ut ) (90) log( ) 45 f = N b = 133.1(10 4 ) f = 64.6 kpsi 1 rev 55 b ( ) ( ) *10 cycles kpsi /3/015 41

42 Endurnce Limit Modifying Fctors The experiments results of endurnce limit in lbortory re not mtched with mechnicl or structurl member. ome difference include o Mteril: composition, bsis of filure, vribility. o Mnufcturing: method, het tretment, fretting corrosion, surfce condition, stress concentrtion. o Environment: corrosion, temperture, stress stte, relxtion times. o Design: size, shpe, life, stress stte, speed, fretting, glling /3/015 4

43 A Mrin eqution is therefore written s e k k b k c k Where, k : surfce condition modifiction fctor k b : size modifiction fctor k c : lod modifiction fctor k d : temperture modifiction fctor k e : relibility fctor k f : miscellneous-effects modifiction fctor (In this course, K f will not be included) e : rotry-bem test specimen endurnce limit e : endurnce limit t the criticl loction of mchine prt in the geometry nd condition of use. (e < e) When endurnce tests of prts re not vilble, estimtions re mde by pplying Mrin fctors to the endurnce limit. d k e k f ' e /3/015 43

44 urfce Fctor k b K ut Where, ut is the minimum tensile strength nd nd b re to be found in below Tble 6. Exmple: A 1030 CD steel hs mchined surfce. Estimte surfce fctor k. olution: From Tble 6, = 4.51 nd b = K b ut 4.51( 50 ) /3/015 44

45 ize Fctor k b For xil loding there is no size effect, For bending nd torsion, Rotting K b 1 Nonrotting It s depending on n equivlent dimeter d e... Tble 6 3 /3/015 45

46 Tble 6 3 /3/015 46

47 Exmple: A steel shft loded in bending is 3 mm in dimeter, butting filleted shoulder 38 mm in dimeter. The shft mteril hs men ultimte tensile strength of 690 MP. Estimte the Mrin size fctor k b if the shft is used in () A rotting mode, (b) A nonrotting mode. olution: () ince,.79 (d=3) 51 mm K b d ( ) ( 7.6 ) (b) From Tble 6 3, d e K 0.37d 0.37(3) 11.84mm, b ( ) (d e 11.84) 51mm /3/015 47

48 Loding Fctor k c K c bending xil torsion Temperture Fctor k d K d T RT Where, T = tensile strength t operting temperture RT = tensile strength t room temperture K d ( ( 10 8 )T 3 F 3 )T F 0.595( ( 10 1 )T 4 F 5 )T F 70 TF 1000 F /3/015 48

49 Tble 6 4 /3/015 49

50 Exmple: A 1035 steel hs tensile strength of 70 kpsi nd is to be used for prt tht sees 450 F in service. Estimte the Mrin temperture modifiction fctor nd (e) 450 if () The room-temperture endurnce limit by test is ( e) 70 = 39.0kpsi. (b) Only the tensile strength t room temperture is known. olution: () Using, K d ( ( 10 8 )T 3 F 3 )T F 0.595( ( 10 1 )T 4 F 5 )T F K d ( ( )( 450 ) 3 )( 450 ) 0.115( ( )( 450 ) )( 450 ) /3/015 50

51 Thus, ( ) K ( ' e d e ) ( 39 ) 39.3 kpsi (b) Interpolting from Tble 6 4 gives, ( T / RT ) ( T / RT ) Thus, the tensile strength t 450 F is estimted s ( ut ) ( T / RT ) ( ut ) (70 ) kpsi Then, ( ( e e ) ) ( ut ) kpsi 0.5(70.49 ) /3/015 51

52 Relibility Fctor k e K e z Tble 6 5 /3/015 5

53 Exmple: A 1015 hot-rolled steel br hs been mchined to dimeter of 1 in. It is to be plced in reversed xil loding for 7(10 4 ) cycles to filure in n operting environment of 550 F. Using ATM minimum properties, nd relibility of 99 percent, estimte the endurnce limit nd ftigue strength t 7(10 4 ) cycles. olution: From Tble A 0, ut = 50 kpsi t 70 F. Using, K d ( ( 10 8 )T 3 F 3 )T F 0.595( ( 10 1 )T 4 F 5 )T F K d ( ( )( 550 ) 3 )( 550 ) 0.115( ( )( 550 ) )( 550 ) /3/015 53

54 The ultimte strength t 550 F is then ( ut ) Kd( ut ) ( 50 ) 49.0 kpsi The rotting-bem specimen endurnce limit t 550 F is then estimted s ( ' e ) 0.5( ut ) ( 49 ) 4.5 kpsi Next, we determine the Mrin fctors. For the mchined surfce, Eq. (6 19) with Tble 6 gives, K b ut.70( 49 ) The size fctor k b = 1. The loding fctor is k c = The temperture fctor k d = 1, since we ccounted for the temperture in modifying the ultimte strength nd consequently the endurnce limit. /3/015 54

55 For 99 percent relibility, from Tble 6 5, k e = The endurnce limit for the prt is estimted by Eq. (6 18) s ) k k k k k ( ' ) ( e 550 b c d e e (1)(0.85)(1)(0.814) kpsi For the ftigue strength t 7(10 4 ) cycles, ( ut ) 550 = 49 < 70 kpsi, then f = 0.9. b ( f 1 3 ut e log( ) f (0.9* 49 ) 16.3 e ut ) 1 3 log( 119.3kpsi 0.9( 49 ) 16.3 ) f = N b = 119.3(7*10 4 ) = 3.9 kpsi /3/015 55

56 tress Concentrtion nd Notch ensitivity K f mximum stressin notched specimen stressin notch - free specimen mx K f 0 or mx K fs 0 Where, K f is reduced vlue of K t nd σ 0 is the nominl stress. The fctor K f is commonly clled ftigue stressconcentrtion fctor, nd hence the subscript f. Notch sensitivity q q increse of ctul stressover nominl stress increse of theoreticl stressover nominl stress /3/015 56

57 q K f t 0 K q K K f t 1 1 q sher K K fs ts q sher K K fs ts 1 1 K f 1 q( K t 1) or K 1 q Where, q is usully between zero nd unity. fs sher ( K ts 1) NOTE When the mteril hs no sensitivity to notches, q = 0 nd K f = 1 When the mteril is fully sensitivity to notches, q = 1 nd K f = K t q = 0.0 for ll grdes of cst iron /3/015 57

58 Fig. 6 0 Notch-sensitivity chrts for steels nd wrought luminum lloys subjected to reversed bending or reversed xil lods. /3/015 58

59 Figure 6 1 Notch-sensitivity curves for mterils in reversed torsion /3/015 59

60 There is nother wy to estimte notch sensitivity q, which is by Neuber s eqution K q f For bending or xil ( 10 r K t 1 / r 3 ) ut 1.51( 10 5 ) ut.67( 10 8 ) 3 ut For torsion ( 10 3 ) ut 1.35( 10 5 ) ut.67( 10 8 ) 3 ut o Where is defined s the Neuber constnt nd is mteril constnt. r is notch rdius. NOTE: the equtions pply to steel nd ut is in kpsi. /3/015 60

61 Exmple: A steel shft in bending hs n ultimte strength of 690 MP nd shoulder with fillet rdius of 3 mm connecting 3-mm dimeter with 38-mm dimeter. Estimte K f using: () Figure 6 0. (b) Equtions. olution: From Fig. A 15 9, using D/d = 38/3 = , r/d = 3/3 = , we red the grph to find K t =1.65. /3/015 61

62 () From Fig. 6 0, for ut = 690 MP nd r = 3 mm, q.= Thus, K f 1 q( K t 1) 10.84(1.65 1) 1.55 (b) Converting ut = 690 Mp= 100kpsi Thus, K ( 10 f ( 10 k 1 1 t 3 3 ) ut 1.51( 10 5 ) )(100 ) 1.51( 10 in mm / r ut 5.67( 10 )(100 8 ) 3 ut ).67( 10 8 )(100 3 ) /3/015 6

63 Exmple: For the step-shft of previous exmple, it is determined tht the fully corrected endurnce limit is e = 80 MP. Consider the shft undergoes fully reversing nominl stress in the fillet of (σ rev ) nom = 60 MP. Estimte the number of cycles to filure. olution: From previous exmple, K f = 1.55, nd the ultimte strength is ut = 690 MP = 100 kpsi. The mximum reversing stress is ( rev ) mx K f ( rev ) nom From Fig., f = ( 60 ) 403Mp /3/015 63

64 ( f ut e ) (0.845* 690 ) Mp b 1 3 log( f e ut ) 1 3 log( 0.845(690 ) 80 ) N N 1 rev b ( ) ( ) * cycles /3/015 64

65 Modifying Fctor to Account for ftigue stress concentrtion The endurnce limit is reduced due to ftigue stress concentrtion. The ftigue stress concentrtion fctor is less thn stress concentrtion fctor due to notches sensitivity of the mteril. To pply the effect of ftigue stress concentrtion, the designer cn (1) either reduce the endurnce limit by dividing it by K f or () increse the nominl stress mplitude by multiplied it by K f. o In this course, we will consider () which increses the nominl stress mplitude by multiplied it by K f /3/015 65

66 Exmple: Figure 6 shows rotting shft simply supported in bll berings t A nd D nd loded by nonrotting force F of 6.8 kn. Using ATM minimum strengths, estimte the life of the prt. olution: We will solve the problem by first estimting the strength t point B, Fig.6-() hft drwing showing ll dimensions in millimeters; ll fillets 3-mm rdius. The shft rottes nd the lod is sttionry; mteril is mchined from AII 1050 cold-drwn steel. /3/015 66

67 From Tble A 0 we find ut = 690 MP nd y = 580 MP. The endurnce limit e is estimted s e = 0.5(690) = 345 MP Estimting Mrin s fctors, urfce fctor, ize fctor, ince, k c k d K b ut d Kb ( ) ( ) 7.6 k 1 e 4.51(690 ) e 0.798(0.858) MP /3/015 67

68 To find the geometric stress -concentrtion fctor K t, with D/d = 38/3 = nd r/d = 3/3 = nd red K t = To find the ftigue stress-concentrtion fctor k f, ubstituting ut = 690/6.89 = 100 kpsi into Eq. (6 35) ( 10 3 ) ( 10 5 )100.67( 10 8 ) in mm, ubstituting this into Eq. (6 33) gives, K f 1 K 1 t / r /3/015 68

69 The next step is to estimte the bending stress t point B. The bending moment is 5F R x * (6.8 ) * N.m M B 1 (b) Bending moment digrm. The reversing bending stress is, I/c πd 3 / 3 π 3 3 / (10 3 ) mm 3 σ rev M B K f 1.55 (10 ) I/c (10 ) P MP /3/015 69

70 This stress is greter thn e nd less thn y. This mens we hve both finite life nd no yielding on the first cycle. The life of the prt, The ultimte strength, ut = 690MP = 100 kpsi. From Fig. 6 18, we will get f = ( f ut e ) (0.844* 690 ) Mp b 1 3 log( f e ut ) (690 ) log N ( rev ) 1 b ( ) * 10 3 cycles /3/015 70

71 Chrcterizing Fluctuting tresses It hs been found tht in periodic ptterns exhibiting single mximum nd single minimum of force, the shpe of the wve is not importnt, but the peks on both the high side (mximum) nd the low side (minimum) re importnt. Thus F mx nd F min in cycle of force cn be used to chrcterize the force pttern. F m F mx F min nd F F mx F min F m : the midrnge stedy component of force, nd F : the mplitude of the lternting component of force. /3/015 71

72 mx m min R mx min σ = mplitude component σ m = midrnge component R= the stress rtio nd nd mx min A m σ r = rnge of stress A= the mplitude rtio /3/015 7

73 σ = σ o nd σ m = σ mo in the bsence of notch σ = k f σ o nd σ m = k f σ mo with notch /3/015 73

74 Ftigue Filure Criteri for Fluctuting tress we wnt to vry both the midrnge stress nd the stress mplitude, or lternting component, to lern something bout the ftigue resistnce of prts when subjected to such situtions. There re Five criteri of filure re digrmmed in Fig. 6 7: the oderberg, the modified Goodmn, the Gerber, the AME-elliptic, nd yielding. The Fig. 6 7 shows tht only the oderberg criterion gurds ginst ny yielding, but is bised low. Considering the modified Goodmn line s criterion, point A represents limiting point with n lternting strength nd midrnge strength m. The slope of the lod line shown is defined s r = / m. /3/015 74

75 Fig. 6 7 Ftigue digrm showing vrious criteri of filure. For ech criterion, points on or bove the respective line indicte filure. /3/015 75

76 Nme The criterion eqution oderberg Ftigue Line Modified Goodmn Ftigue Line Gerber Ftigue Curve AME-elliptic Ftigue Curve Yield (Lnger) line /3/ y m e 1 ut m e 1 ut m e 1 y m e y m

77 The stresses nσ nd nσ m cn replce nd m, where n is the design fctor or fctor of sfety. ee tble Tble * There re two wys to proceed with typicl nlysis. One method is to ssume tht ftigue occurs first nd use one of Eqs. (1) to (4) to determine n or size, depending on the tsk. Most often ftigue is the governing filure mode. Then follow with sttic check. If sttic filure governs then the nlysis is repeted using Eq. (5). ee tble Tble * /3/015 77

78 /3/ No. Nme The criterion eqution with F. 1 oderberg Modified Goodmn 3 Gerber 4 AME-elliptic 5 Lnger sttic yield n 1 y m e n 1 ut m e 1 n n ut m e 1 n n y m e n y m Tble *

79 Tble 6 6 Amplitude nd tedy Coordintes of trength nd Importnt Intersections in First Qudrnt for Modified Goodmn nd Lnger Filure Criteri /3/015 79

80 Tble 6 7 Amplitude nd tedy Coordintes of trength nd Importnt Intersections in First Qudrnt for Gerber nd Lnger Filure Criteri /3/015 80

81 Tble 6 8 Amplitude nd tedy Coordintes of trength nd Importnt Intersections in First Qudrnt for AME-elliptic nd Lnger Filure Criteri /3/015 81

82 NOTE for the bove three tbles (6-6) to (6-8): The first row of ech tble corresponds to the ftigue criterion, the second row is the sttic Lnger criterion, nd the third row corresponds to the intersection of the sttic nd ftigue criteri. The first column gives the intersecting equtions nd the second column the intersection coordintes. Exmple: A 1.5-in-dimeter br hs been mchined from n AII 1050 cold-drwn br. This prt is to withstnd fluctuting tensile lod vrying from 0 to 16 kip. Becuse of the ends, nd the fillet rdius, ftigue stress-concentrtion fctor K f is 1.85 for 10 6 or lrger life. Find nd m nd the fctor of sfety gurding ginst ftigue nd first-cycle yielding, using () the Gerber ftigue line nd (b) the AME-elliptic ftigue line. /3/015 8

83 olution: A. We strt with clculte the xil midrnge force component nd mplitude of force component, F F m F F mx mx F F min min kiP kIP The nominl xil stress components σ o nd σ mo re mo o Fm 8 d / F 8 d / kpsi / kpsi / 4 Applying K f to both components σ o nd σ mo, K f o 1.85(4.53) 8.38 kpsi σ m /3/015 83

84 B. From Tble A 0, ut = 100 kpsi nd y = 84 kpsi. Estimting the endurnce limit, ' (100 ) 50kpsi e The Mrin fctors re, deterministiclly, k =.70(100) 0.65 = k b = 1 (xil loding) k c = 0.85 k d = k e = 1 e k ut k b k c k d k e ' e (Idel) 0.797(1)(0.85)(1)(1) kpsi /3/015 84

85 C. Let us clculte the fctors of sfety first. () From the bottom pnel from Tble 6 7 the fctor of sfety for ftigue is From Eq. (6 49) the fctor of sfety gurding ginst firstcycle yield is /3/ ( 8.38 ) ( 8.38 )( 33.9 ) n ut e m e m ut f n m y y

86 For drwing the designer s digrm, 1. We hve to know ll the detils bout the lod line strting from its slope (r), r m Also, the intersection point (B) with Gerber ftigue curve m r ut e 1 1 (100 ) 1 ( 33.9 ) r r 1 e ut Tn ( 33.9 ) (1) kpsi 1 r 45 /3/ kpsi

87 3. In this Ex., the intersection point (C) between the lod line nd Lnger line cn be found by continuously drwing the lod line with respect to Tn 1 r 45 till intersects the Lnger line 4. Finlly, the intersection point (D) between the Gerber ftigue curve with Lnger line. m ut e 1 1 ut e 1 y e 100 ( 33.9 ) 1 1 ( 33.9 ) kpsi y m kpsi /3/015 87

88 The criticl slope is thus r cri m which is less thn the ctul lod line of r = 1. This indictes tht ftigue occurs before first-cycle-yield. As check on the previous result, n f = OB/OA = f /σ = mf /σ m = 30.7/8.38 =3.66. n y = OC/OA= y /σ = my /σ m = 5.01 nd we see totl greement. /3/015 88

89 Fig. 6 8 Principl points A, B, C, nd D on the designer s digrm drwn for Gerber, Lnger, nd lod line. /3/015 89

90 (b) Repeting the sme procedure for the AME-elliptic line, for ftigue n f ( / e ) 1 ( 1 ( 8.38 / 33.9 ) ( 8.38 / 84 ) 3.75 Agin, this is less thn n y = 5.01 nd ftigue is predicted to occur first. m / y ) For drwing the designer s digrm, 1. We hve to know ll the detils bout the lod line strting from its slope (r), r 1 Tn r m /3/015 90

91 . Also, the intersection point (B) with AME ftigue curve. We obtin the coordintes nd m of point B by using r e r e y y (1 )( 33.9 )( 84 ) ( 33.9 ) (1 )( 84 ) 31.4kpsi m r kpsi 3. In this Ex., the intersection point (C) between the lod line nd AME curve cn be found by continuously drwing the lod line with respect to Tn 1 r 45 till intersects the AME-elliptic ftigue curve. 4. Finlly, the intersection point (D) between the AMEelliptic ftigue curve with Lnger line. /3/015 91

92 m e y y e y ( 84 ) kpsi kpsi The criticl slope is thus 3.5 rcri m Which is less thn the ctul lod line of r =1. This indictes tht ftigue occurs before first-cycle-yield As check on the previous result, n f = OB/OA = f /σ = mf /σ m = 30.7/8.38 =3.75. n y = OC/OA= y /σ = my /σ m = 5.01 nd we see totl greement. Therefore, the Gerber nd the AME-elliptic ftigue filure criteri re very close to ech other nd re used interchngebly. /3/015 9

93 Fig. 6 9 Principl points A, B, C, nd D on the designer s digrm drwn for AME-elliptic, Lnger, nd lod line. /3/015 93

94 Torsionl Ftigue trength under Fluctuting tresses For ductile mteril, polished, notch-free, nd cylindricl, torsionl stedy-stress component not more thn the torsionl yield strength hs no effect on the torsionl endurnce limit. For mterils with stress concentrtion, notches, or surfce imperfections, the torsionl ftigue limit decreses monotoniclly with torsionl stedy stress. The modified Goodmn reltion for pulsting torsion is su 0.67 ut from distortion-energy theory, sy yt /3/015 94

95 Combintions of Loding Modes It my be helpful to think of ftigue problems s being in three ctegories: Completely reversing simple lods (σ m = 0). Fluctuting simple lods. Combintions of loding modes. Here will be used von-mises theory for combined lods 1. The first step is to generte two stress elements one for the lternting stresses nd one for the midrnge stresses.. Apply the pproprite ftigue stress-concentrtion fctors to ech of the stresses; i.e., pply (K f ) bending for the bending stresses, (K f s ) torsion for the torsionl stresses, nd (K f ) xil for the xil stresses. /3/015 95

96 3. Next, clculte n equivlent von Mises stress for ech of these two stress elements, σ nd σ m. 4. For the endurnce limit, e, use the endurnce limit modifiers, k, k b, nd k c, for bending. The torsionl lod fctor, k c = 0.59 should not be pplied s it is lredy ccounted for in the von Mises stress clcultion. The lod fctor for the xil lod cn be ccounted for by dividing the lternting xil stress by the xil lod fctor of Finlly, select ftigue filure criterion (modified Goodmn, Gerber, AME-elliptic, or oderberg) to complete the ftigue nlysis. /3/015 96

97 imple Exmple, consider the common cse of shft with bending stresses, torsionl sher stresses, nd xil stresses. ' ( x 3 xy ) 1/ ' K K f bending bending f xil xil K fs torsionl torsionl 1/ ' m K K 3 K 1/ f bending m bending f xil m xil fs torsionl m torsionl /3/015 97

98 Exmple: A rotting shft is mde of 4-4-mm AII 1018 cold-drwn steel tubing nd hs 6-mm-dimeter hole drilled trnsversely through it. Estimte the fctor of sfety gurding ginst ftigue nd sttic filures using the Gerber nd Lnger filure criteri for the following loding conditions: ) The shft is subjected to completely reversed torque of 10 N m in phse with completely reversed bending moment of 150 N m. b) The shft is subjected to pulsting torque fluctuting from 0 to 160 N m nd stedy bending moment of 150 N m. /3/015 98

99 olution: () Theoreticl stress-concentrtion fctors re found from Tble A 16. Using /D = 6/4 = nd d/d = 34/4 = 0.810, nd using liner interpoltion, we obtin A = nd K t =.366 for bending; nd A = 0.89 nd K ts = 1.75 for torsion. A (0.798 ) Z J net net Next, using Figs. 6 0 nd 6 1, pp , with notch rdius of 3 mm we find the notch sensitivities to be 0.78 for bending nd 0.81 for torsion. K K ( D 3D A ( D 3 f fs 4 4 d 1 q( 1 q d 4 K 4 t sher ) (0.98 ) ) [( 4 3 1) 10.78(.366 1).07 ( K ts 3( 4 ) [( ) ( 34 ) ) ( 34 ) 4 1) 10.81( ) ] 3.31(10 ] 155( 10 /3/ ) 3 ) mm mm 4 3

100 Tble A 16 Approximte tress- Concentrtion Fctor K t for Bending of Round Br or Tube with Trnsverse Round Hole /3/

101 The lternting bending stress is now found to be x M Z ( 10 K f 6 net nd the lternting torsionl stress is xy TD J 10( 4 )( (155 )(10 K fs 9 net 93.8Mp ) 6.Mp The midrnge von Mises component σ m is zero. The lternting component σ is given by ' ' ( x 3 xy 104.Mp ) 1/ From Tble A 0, ut = 440 MP nd y = 370 MP. The endurnce limit of the rotting-bem specimen is 0.5(440) = 0 MP. 3 [ 93.8 ) ) 3( 6. )] 1/ /3/

102 Mrin fctors re K K e b (0.833 )(1)(1)(1)0 165Mp ince e =, the ftigue fctor of sfety n f is n f ' The first-cycle yield fctor of sfety is n y y ' There is no loclized yielding; the thret is from ftigue. /3/015 10

103 Figure 6 3 Designer s ftigue digrm /3/

104 (b) We hve T = (160 0)/ = 70 N m nd T m = ( )/ = 90 N m. The corresponding mplitude nd stedy-stress components re xy xym K K fs fs T D J net Tm D J net 70( 4 )( (155 )(10 90( 4 )( (155 )( ) 15.3Mp ) ) 19.7Mp ) The stedy bending stress component σ xm is xm M Z ( 10 m K f 6 net The von Mises components re ' ' m ( 3 ( xy xm ) 1/ 3 xym [ 3(15.3 ) 1/ )] ( / 93.8Mp ) 6.5Mp 3( 19.7 )) 1/ 99.8Mp /3/

105 From Tble 6 7, p. 307, the ftigue fctor of sfety is From the sme tble, with r = σ /σ m = 6.5/99.8 = 0.8, the strengths cn be shown to be = 85.5 MP nd m = 305 MP. ee the plot. The first-cycle yield fctor of sfety n y is There is no notch yielding. The likelihood of filure my first come from first-cycle yielding t the notch. /3/ ( 6.5 ) ( 99.8 )(165 ) n ' ut e m ' e ' m ' ut f ' ' n m y y

106 Vrying, Fluctuting tresses; Cumultive Ftigue Dmge Insted of single fully reversed stress history block composed of n cycles, suppose mchine prt, t criticl loction, is subjected to o A fully reversed stress σ 1 for n 1 cycles, σ for n cycles,..., or o A wiggly time line of stress exhibiting mny nd different peks nd vlleys. The method which will be used here to count the number of cycles it is clled the rin-flow counting. /3/

107 Figure 6 33 Vrible stress digrm prepred for ssessing cumultive dmge /3/

108 The Plmgren-Miner cycle-rtio summtion rule, lso clled Miner s rule, is written Where, n N i i c n i is the number of cycles of opertion t stress level σ i. N i is the number of cycles to filure t stress level σ i. c is the empiricl constnt in the rnge 0.7 < c <.. Using the deterministic formultion s liner dmge rule we write ni D N where D is the ccumulted dmge. When D = c = 1, filure ensues. i /3/

109 Exmple: Given prt with ut = 151 kpsi nd t the criticl loction of the prt, e = 67.5 kpsi. For the loding of Figure below, estimte the number of repetitions of the stress-time block in the figure tht cn be mde before filure. /3/

110 olution: For the figure, We will strt to count the number of cycles nd find the midrnge stresses σ m nd mplitude of stresses σ for ech one to construct the tble below: From Fig. 6 18, for ut = 151 kpsi, f = /3/

111 ( f ut e ) (0.795* 151) kpsi b 1 3 log( f e ut ) (151) log o, N f 1 b f ( ) We prepre to dd two columns to the previous tble. Using the Gerber ftigue criterion, with e = f, nd n = 1, we cn write m 0 1 ( / ) f m ut ( 3 ) e m 0 /3/

112 where f is the ftigue strength ssocited with completely reversed stress, σ rev, equivlent to the fluctuting stresses. Cycle 1: r = σ /σ m = 70/10 = 7, nd the strength mplitude from tble 6-7 is r ut e 1 1 r e ut 7 (151 ) 1 (67.5 ) 1 (67.5 ) (7 ) kpsi ince σ >, tht is, 70 > 67., life is reduced. From Eq.(3) 70 f 70.3kpsi 1( / ) 1(10 / 151) m ut /3/015 11

113 nd from Eq. () N ( 10 ) cycles Cycle : r = 10/50 = 0., nd the strength mplitude is 0. (151 (67.5 ) ) 1 1 ince σ <, tht is 10 < 4., then f = e nd indefinite life follows. Thus, N. 3 (67.5 ) (0. )151 4.kpsi Cycle 3: r = 10/ 30 = 0.333, nd since σ m < 0, f = e, indefinite life follows nd N /3/

114 From Eq. (6 58) the dmge per block is /3/ ) 619( 10 n 1 ) 619( 10 n D 1 1 ) 619( 10 1 n N n D 3 o 3 o 3 o i i