2 Fatigue Failure 2.1 Endurance Limit The R. R. Moore’s high-speed rotating beam machine is used to performing a fatigue test. The specimen, shown in Fig. 1(a), is rotating with a constant angular speed. The specimen is subjected to pure bending by means of weights. The intensity of the reversed stress causing failure after a given test specimen d motor (a) n M M weights Weights mg σ Se (b) O t Figure 1: R. R. Moore’s high-speed rotating beam machine number of cycles is the fatigue strength corresponding to that number of loading cycles. The fatigue strengths considered for each test are plotted against the corresponding number of revolutions. The resulting chart is called the strength-life, S-N, diagram. The diagram depicts the fatigue strength versus cycle life N of a part. The S-N curves are plotted on log-log coordinates. For ferrous materials the endurance limit of the test specimen is defined as the highest level of alternating stress that can be withstood indefinitely by 1 Fatigue strength S (log) a test specimen without failure, Fig. 1(b). The symbol for endurance limit is Se0 . The S-N diagram becomes horizontal at the endurance limit as shown in Fig. 2. S Se 10 3 10 4 10 5 106 107 N Life N (cycles) (log) Figure 2: S-N diagram The endurance limit is related to the ultimate tensile strength through the relationships [5]: 0.50 Su for Su ≤ 200 kpsi, 0 Se = (2.1) 100 kpsi for Su > 200 kpsi, or Se0 = 0.50 Su 700 MPa for Su ≤ 1400 MPa, for Su > 1400 MPa. (2.2) The endurance limit of the test specimen, Se0 , can be different from the endurance limit, Se , of the machine part subjected to different loadings. The endurance limit Se is affected by a number of factors called modifying factors. Some of these factors are the surface factor kS , the gradient (size) factor kG , the load factor kL , the temperature factor kT , and the reliability factor kR . Thus, the endurance limit of a member can be related to the endurance limit of the test specimen by the following relationship: Se = kS kG kL kT kR Se0 . Surface factor, kS The surface modification factor, kS , depends on the tensile strength of the material, Sut and on the part surface finish. The surface factor formula 2 is given by [5] b . kS = a Sut (2.3) The factor a and the exponent b depends upon the quality of the surface finish [5]: 1.58 4.51 for (MPa) a = 57.7 272 for for for for ground surface finish, machined or cold-drawn surface finish, hot-rolled surface finish, forged surface finish, (2.4) 1.34 2.70 for (kpsi) a = 14.4 39.9 for for for for ground surface finish, machined or cold-drawn surface finish, hot-rolled surface finish, forged surface finish, (2.5) −0.085 −0.256 b= −0.718 −0.995 for for for for ground surface finish, machined or cold-drawn surface finish, hot-rolled surface finish, forged surface finish. (2.6) Gradient (size) factor, kG To evaluate the size factor the following relations are used [5]: For d < 0.4 in. or d < 10 mm, where d is the diameter of the test bar 0.7 − 0.9 for axial loading, kG = (2.7) 1 for bending and torsion. For 0.4 in. < d < 2 in. or 10 mm < d < 50 mm 0.7 − 0.9 for axial loading, kG = 0.9 for bending and torsion. (2.8) For 2 in. < d < 4 in. or 50 mm < d < 100 mm the factors are reduced by 0.1. For 4 in. < d < 6 in. or 100 mm < d < 150 mm the factors are reduced by 0.2. 3 Load factor, kL For the load factor the following 0.85 1 kL = 0.59 average values are used [5]: for axial, for bending, for pure torsion (2.9) For shear the load factor is kL = 0.577 according to [30]. Temperature factor, kT The temperature factor for axial loading, bending, and torsion, for steel, is given by [10] 1 for T ≤ 840 ◦ F, kT = (2.10) 1 − (0.0032 T − 2.688) for 840 ◦ F < T ≤ 1020 ◦ F. Reliability factor, kR The reliability factor is calculated with by [10] 1 for 50% reliability, 0.897 for 90% reliability, 0.868 for 95% reliability, kR = 0.814 for 99% reliability, 0.753 for 99.9% reliability. (2.11) Juvinall and Marshek present a summary of all modifying factors for bending, axial loading, and torsion used for fatigue of ductile materials [10]. For 103 -cycle strength: for bending loads, 0.9 kT Su 0.75 kT Su for axial loads, Sf = (2.12) 0.9 kT Sus for torsional loads. The ultimate shear strength, Sus , is 0.8 Su for steel, Sus ≈ 0.7 Su for other ductile materials. (2.13) For 106 -cycle strength (endurance limit) for axial, bending, and torsional loads: Se = kS kG kL kT kR Se0 . 4 (2.14) 2.2 Fluctuating Stresses The components of the time varying stresses are depicted in Fig. 3, where σmin is minimum stress, σmax the maximum stress, σa the stress amplitude or the alternating stress, σm the midrange or the mean stress, and σr the stress range. The mean stress and the alternating stress are σmax + σmin , (2.15) σm = 2 σmax − σmin . (2.16) σa = 2 The stress ratios are defined by σmin σa R= and A = . (2.17) σmax σm The steady stress or static stress, σs , has any value between σmin and σmax and exists because of a fixed load. Figure 3(a) shows a sinusoidal fluctuating stress, Fig. 3(b) represents a repeated stress, and Fig. 3(c) is a completely reversed sinusoidal stress. A fluctuating stress is a combination of static plus completely reversed stress. 2.3 Constant Life Fatigue Diagram [10] A graphical representation of mean and alternating stress in relation to yielding and various fatigue life is shown in Fig. 4. The horizontal axis, σa = 0, corresponds to static loading. The vertical axis is given by σm = 0. The points on the vertical axis, ordinate axis, (0, Se ), (0, S103 ), (0, S104 ), (0, S105 ), ... are selected from the S-N diagram. The point (Su , 0) represents the ultimate tensile strength. The lines (0, Se )−(Su , 0), (0, S105 )−(Su , 0), (0, S104 )−(Su , 0) and (0, S103 ) − (Su , 0) are the estimated lines of constant life. The diagram, σm − σa , is called the constant life fatigue diagram because it has lines corresponding to a constant 106 cycle or “infinite” life, constant 105 cycle, constant 104 cycle, and so forth. These lines are called the Goodman lines. The point (Sy , 0) represents the yield strength and for ductile materials the point (−Sy , 0) represents the compressive yield strength. At the point (0, Sy ) the stress fluctuates between +Sy and −Sy . The points on line (Sy , 0) − (0, Sy ) correspond to fluctuations having a tensile peak of Sy . The points on line (−Sy , 0) − (0, Sy ) correspond to fluctuations having a compressive peak of Sy . Within the area defined by the lines (Sy , 0) − (0, Sy ), (−Sy , 0) − (0, Sy ) there is no yielding. 5 fluctuating stress σ σa σr (a) σa σmax σm σmin O σ t repeated stress σa (b) σmax σr σa σm t O σmin = 0 σ reversed stress σmax σa (c) O σr σm = 0 t σmin σa Figure 3: Time varying stresses: a) sinusoidal fluctuating stress; (b) repeated stress; and (c) reversed sinusoidal stress. 6 Peak alternating stress S (log) S S-N diagram S103 S104 S105 Se 10 3 10 4 10 5 10 6 N 10 7 Life N (cycles (log)) Constant-Life Curves S103 103 S104 S105 Sy S106 = Se 104 Se 105 106 σa σm −Sy σm (compression) Sy σm (tension) O Su Figure 4: Constant life fatigue diagram The shaded area in Fig. 5(a) corresponds to a life of at least 106 cycles and no yielding. The shaded area in Fig. 5(b) corresponds to a life of at least 106 cycles and yielding. The shaded area in Fig. 5(c) corresponds to less than 106 cycles of life and no yielding. For torsional loads Ssy and Sus are used instead of Sy and Su with Ssy ≈ 0.58 Sy and Sus ≈ 0.8 Su 7 for steel. (2.18) σa S103 103 S104 S10 5 S106 = Se Sy (a) 104 Se 105 106 Sy −Sy Su O σa S103 103 S104 S10 5 S106 = Se Sy 104 Se −Sy (b) 105 106 Sy O Su σm σa S103 103 S104 S10 5 σm Sy 104 S106 = Se Se (c) 5 10 6 10 Sy −Sy Su O Figure 5: Constant life fatigue diagrams 8 σm 2.4 Arbitrarily Varying Forces and Fatigue Life An arbitrarily varying force acts on a part. The part is subjected to σ1 for n1 cycles, σ2 for n2 cycles, and so forth. The procedure for estimating the fatigue life of the part is the linear cumulative damage rule (or Miner’s rule) and is expressed by the following equation [10, 30]: k X n2 nk nj n1 + + ... + = 1 or =c N1 N2 Nk Nj j=1 (2.19) where n1 , n2 , · · · , nk represent the number of cycles at specific levels of stress σ1 , σ2 , · · · , σk and N1 , N2 , · · · , Nk represent the life (in cycles) at these stress levels, as taken from the appropriate S-N curve. The parameter c has been determined by experiment and 0.7 < c < 2.2 with an average value near 1. Failure results when c = 1. 2.5 Variable Loading Failure Theories [5, 30] There are various techniques for plotting the results of the fatigue failure test of a part subjected to fluctuating stress. A stress state can be described by the mean and the alternating components. This stress state is characterized by a point and if the point is situated below the constant-life line then safety is suggested. A point on the constant-life line gives the strength Sm as the limiting value of σm . In a similar way the corresponding strength Sa is the limiting value of σa , as shown in Fig. 6. The alternating strength Sa as a limiting value of σa is plotted on the ordinate. The mean strength is Sm as a limiting value of σm and is plotted on the abscissa. The fatigue limit Se is plotted on the ordinate. The tensile yield strength Sy is plotted on both coordinate axes. The ultimate tensile strength Su is plotted on the abscissa. Four criteria of failure are shown in the diagram in Fig. 6 that is, Soderberg, the modified Goodman, Gerber, and yielding. The Soderberg criterion protects against yielding. The equation for the Soderberg line is Sm Sa + = 1. Sy Se 9 (2.20) Sy yield line Soderberg line modified Goodman line Gerber line Se Sa d loa e lin O Sm Sy Su Figure 6: Fatigue diagram The equation for the modified Goodman line is Sm Sa + = 1. Su Se (2.21) The yielding line is described by the equation Sm Sa + = 1. Sy Sy (2.22) The Gerber criterion is also called the Gerber parabolic relation because the curve can be modeled by a parabolic equation of the form 2 Sa Sm + = 1. (2.23) Se Su If SF is a design or safety factor then Sa = SF σa and Sm = SF σm . 10 (2.24) The Soderberg equation becomes 1 σa σm + = , Se Sy SF (2.25) the modified Goodman equation becomes σa σm 1 + = , Se Su SF and the Gerber equation becomes 2 SF σm SF σa + = 1. Se Su Some values of the tensile strength are given in Table 2.1[38]. 11 (2.26) (2.27) Table 2.1 Tensile strength Steel (AISI = American Iron and Steel Institute) tensile strength ultimate yield Sut (MPa) Sy (MPa) AISI 1006 Steel, cold drawn 330 285 AISI 1006 Steel, hot rolled bar, 19-32 mm round 295 165 AISI 1006 Steel, cold drawn bar, 19-32 mm round 330 285 AISI 1008 Steel, hot rolled bar, 19-32 mm round 305 170 AISI 1008 Steel, cold drawn bar, 19-32 mm round 340 285 AISI 1010 Steel, cold drawn 365 305 AISI 1010 Steel, hot rolled bar, 19-32 mm round or thickness 325 180 AISI 1010 Steel, cold drawn bar, 19-32 mm round or thickness 365 305 AISI 1012 Steel, cold drawn 370 310 AISI 1012 Steel, hot rolled bar, 19-32 mm round or thickness 330 185 AISI 1012 Steel, cold drawn bar, 19-32 mm round or thickness 370 310 AISI 1015 Steel, cold drawn 385 325 AISI 1015 Steel, cold drawn, 19-32 mm round 385 325 AISI 1015 Steel, hot rolled, 19-32 mm round 345 190 AISI 1015 Steel, as rolled 420 315 ◦ ◦ AISI 1015 Steel, normalized at 925 C (1 700 F) 425 325 ◦ ◦ AISI 1015 Steel, annealed at 870 C (1 600 F) 385 285 AISI 1016 Steel, cold drawn, 19-32 mm round 420 350 AISI 1016 Steel, hot rolled, 19-32 mm round 380 205 AISI 1017 Steel, cold drawn 405 340 AISI 1017 Steel, hot rolled, 19-32 mm round 365 200 AISI 1018 Steel, cold drawn 440 370 AISI 1018 Steel, hot rolled, quenched, and tempered 475 275 AISI 1018 Steel, hot rolled, 19-32 mm round 400 220 AISI 1018 Steel, as cold drawn, 16-22 mm round 485 415 AISI 1018 Steel, as cold drawn, 22-32 mm round 450 380 AISI 1018 Steel, as cold drawn, 32-50 mm round 415 345 AISI 1018 Steel, as cold drawn, 50-76 mm round 380 310 AISI 1019 Steel, cold drawn 455 379 AISI 1019 Steel, hot rolled, 19-32 mm round 407 224 AISI 1020 Steel, cold rolled 420 350 Source: MatWeb - Material Property Data, http://www.matweb.com/ 12 References [1] E.A. 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[38] MatWeb - Material Property Data, http://www.matweb.com/ 15 % fatigue % S-N diagram clear all; clc; close all % alternating stresses are torsional % steel bar, d<2in., fine ground surface Su = 110; % ultimate strength (kpsi) Sy = 77; % yield strength (kpsi) % % % % S peak alternating strength 10^3 cycle strenght S torsional load: S = 0.9 Sue; Sue=0.8 Su (for steel); Sue = 0.7 Su (other) = 0.9*0.8*Su; % peak alternating strength % 10^6 cycle strenght (endurance limits) % Se = kL kG kL Sep % endurance limit of test speciment Sep % Sep = 0.5 Su (for Su<1400 MPa) % Sep = 700 MPa (for Su>1400 MPa) % bending, axial, torsion: Sep = 0.5 Su Sep = 0.5*Su; % modifying factors for endurance limit % surface factor kS (bending, axial, torsion) a = 1.34; % fine ground surface b = -0.085; % fine ground surface kS = a*Su^b; % size (gradient) factor kG % bending and torsion: % kG = 0.9; % for 0.4<d<2 in kG = 0.9; % load factor kL kL = 0.577; % temperature factor, kT % kT = 1 for T<840 degF % kT = 1-(0.0032*T-2.688) for 840<T<1020 degF kT = 1; % reliability factor, kR % kR = 1 for 50% reliability kR = 1; % endurance limit Se = kS kG kL kT kR Sep Se = kS*kG*kL*kT*kR*Sep; % Se = 25.667 (kpsi) % S-N diagram LS = log10(S); LSe= log10(Se); % straight line equation y = m x + b % slope m = (y1-y2)/(x1-x2) Lm = (LSe-LS)/(6–3); % y-intercept b = y1 – m x1 Lb = LS-3*Lm; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % print results fprintf('Su = %6.3f (kpsi) \n',Su) fprintf('Sy = %6.3f (kpsi) \n',Sy) fprintf('10^3 cycle strenght S = %6.3f (kpsi) \n',S) fprintf('Sep = %6.3f (kpsi) \n',Sep) fprintf('kS = %6.3f \n',kS) fprintf('kG = %6.3f \n',kG) fprintf('kL = %6.3f \n',kL) fprintf('10^6 cycle strenght Se = %6.3f (kpsi) \n',Se) fprintf('\n') fprintf('S = %6.3f(kpsi) -> log(S) = %6.3f \n',S,LS) fprintf('Se = %6.3f(kpsi) -> log(Se) = %6.3f \n',Se,LSe) % plot the graph syms x % eq. for S-Se ye = Lm*x+Lb; xn = 3:1:6; yn = Lm*xn+Lb; hold on plot(xn,yn,'k-','LineWidth',2) plot([6,7],[LSe,LSe],'b-','LineWidth',2) title('S-N diagram') ylabel('log S','FontSize',14) xlabel('log N','FontSize',14) axis([3 7 1 LS+.25]) grid ax = gca; ax.XTick = [3,4,5,6,7]; ax.YTick = [LSe,LS]; % eof % % % % % % % % % % % Su = 110.000 (kpsi) Sy = 77.000 (kpsi) 10^3 cycle strenght S = 79.200 (kpsi) Sep = 55.000 (kpsi) kS = 0.899 kG = 0.900 kL = 0.577 10^6 cycle strenght Se = 25.667 (kpsi) S = 79.200(kpsi) -> log(S) = 1.899 Se = 25.667(kpsi) -> log(Se) = 1.409 Su = 110.000 (kpsi) Sy = 77.000 (kpsi) 10^3 cycle strenght S = 79.200 (kpsi) Sep = 55.000 (kpsi) kS = 0.899 kG = 0.900 kL = 0.577 10^6 cycle strenght Se = 25.667 (kpsi) S = 79.200(kpsi) -> log(S) = 1.899 Se = 25.667(kpsi) -> log(Se) = 1.409 Published with MATLAB® R2015a