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Fatigue FailureIt has been recognized that a metal subjected to a repetitive or fluctuating stress will fail at a stress much lower than that required to cause failure on a single application of load. Failures occurring under conditions of dynamic loading are called fatigue failures.Fatigue failure is characterized by three stages
Crack Initiation Crack Propagation Final FractureMAE dept., SJSU
Ken Youssefi
1
Jack hammer component, shows no yielding before fracture.
Crack initiation site
Fracture zone Propagation zone, striationKen Youssefi MAE dept., SJSU
2
VW crank shaft fatigue failure due to cyclic bending and torsional stresses
Propagation zone, striations
Crack initiation site
Fracture area
Ken Youssefi
MAE dept., SJSU
3
928 Porsche timing pulley
Ken Youssefi
MAE dept., SJSU
Crack started at the fillet
4
Fracture surface of a failed bolt. The fracture surface exhibited beach marks, which is characteristic of a fatigue failure.
1.0-in. diameter steel pins from agricultural equipment. Material; AISI/SAE 4140 low allow carbon steel
Ken Youssefi
MAE dept., SJSU
5
bicycle crank spider arm
This long term fatigue crack in a high quality component took a considerable time to nucleate from a machining mark between the spider arms on this highly stressed surface. However once initiated propagation was rapid and accelerating as shown in the increased spacing of the 'beach marks' on the surface caused by the advancing fatigue crack.Ken Youssefi MAE dept., SJSU
6
Crank shaft
Gear tooth failure
Ken Youssefi
MAE dept., SJSU
7
Hawaii, Aloha Flight 243, a Boeing 737, an upper part of the plane's cabin area rips off in mid-flight. Metal fatigue was the cause of the failure.
Ken Youssefi
MAE dept., SJSU
8
Fracture Surface Characteristics Mode of fracture Ductile Typical surface characteristics Cup and Cone Dimples Dull Surface Inclusion at the bottom of the dimple Shiny Grain Boundary cracking Shiny Cleavage fractures Flat Beachmarks Striations (SEM) Initiation sites Propagation zone Final fracture zoneMAE dept., SJSU
Brittle Intergranular
Brittle Transgranular
Fatigue
Ken Youssefi
9
Fatigue Failure Type of Fluctuating Stresses Wa = Wmax Wmax = - Wmin
Alternating stress
Wa = Wmin = 0
Wmax Wmin2
Mean stress
Wa = Wm = Wmax / 2 Wm=Ken Youssefi MAE dept., SJSU
Wmax + Wmin210
Fatigue Failure, S-N CurveTest specimen geometry for R.R. Moore rotating beam machine. The surface is polished in the axial direction. A constant bending load is applied.
Typical testing apparatus, pure bendingMotor
Load
Rotating beam machine applies fully reverse bending stressKen Youssefi MAE dept., SJSU
11
Fatigue Failure, S-N CurveN < 103 N > 103
Finite life
Infinite life
Se
e
= endurance limit of the specimen
Ken Youssefi
MAE dept., SJSU
12
Relationship Between Endurance Limit and Ultimate StrengthSteel Steel 0.5Sute=
Sut 200 ksi (1400 MPa) Sut > 200 ksi
100 ksi
700 MPa Sut > 1400 MPa
Cast iron 0.4Sut
Cast iron
Sut < 60 ksi (400 MPa) Sut 60 ksi
Se =
24 ksi
160 MPa Sut < 400 MPaKen Youssefi MAE dept., SJSU
13
Relationship Between Endurance Limit and Ultimate StrengthAluminumAluminum alloys
0.4Sut
Sut < 48 ksi (330 MPa) Sut 48 ksi
Se =
19 ksi
130 MPa Sut 330 MPa For N = 5x108 cycle
Copper alloysCopper alloys
0.4Sut
Sut < 40 ksi (280 MPa) Sut 40 ksi
Se =
14 ksi
100 MPa Sut 280 MPa For N = 5x108 cycleKen Youssefi MAE dept., SJSU
14
Correction Factors for Specimens Endurance LimitFor materials exhibiting a knee in the S-N curve at 106 cycles
Se = endurance limit of the specimen (infinite life > 106) Se = endurance limit of the actual component (infinite life > 106)S103
Se106
N
For materials that do not exhibit a knee in the S-N curve, the infinite life taken at 5x108 cycles
Sf = fatigue strength of the specimen (infinite life > 5x108) Sf = fatigue strength of the actual component (infinite life > 5x108)S103
Sf5x108
Ken Youssefi
MAE dept., SJSU
N
15
Correction Factors for Specimens Endurance Limit Se = Cload Csize Csurf Ctemp Crel (Se)or
Sf = Cload Csize Csurf Ctemp Crel (Sf)(page 326, Nortons 3rd ed.)
Load factor, Cload
Pure bending Pure axial Pure torsion Combined loading
Cload = 1 Cload = 0.7 Cload = 1 if von Mises stress is used, use 0.577 if von Mises stress is NOT used. Cload = 1
Ken Youssefi
MAE dept., SJSU
16
Correction Factors for Specimens Endurance Limit Size factor, Csize(p. 327, Nortons 3rd ed.)
Larger parts fail at lower stresses than smaller parts. This is mainly due to the higher probability of flaws being present in larger components.
For rotating solid round cross section
d 0.3 in. (8 mm)0.3 in. < d 10 in. 8 mm < d 250 mm
Csize = 1 Csize = .869(d)-0.097 Csize = 1.189(d)-0.097
If the component is larger than 10 in., use Csize = .6
Ken Youssefi
MAE dept., SJSU
17
Correction Factors for Specimens Endurance LimitFor non rotating components, use the 95% area approach to calculate the equivalent diameter. Then use this equivalent diameter in the previous equations to calculate the size factor. A95 = (/4)[d2 (.95d)2] = .0766 d2 d95 = .95d d
dequiv = (
A95 0.0766
)1/2
Solid or hollow non-rotating parts
Rectangular parts
dequiv = .37dKen Youssefi MAE dept., SJSU
dequiv = .808 (bh)1/2
18
Correction Factors for Specimens Endurance LimitI beams and C channels
Ken Youssefi
MAE dept., SJSU
19
Correction Factors for Specimens Endurance Limit surface factor, Csurf(p. 328-9, Nortons 3rd ed.)
The rotating beam test specimen has a polished surface. Most components do not have a polished surface. Scratches and imperfections on the surface act like a stress raisers and reduce the fatigue life of a part. Use either the graph or the equation with the table shown below.
Csurf = A (Sut)b
Ken Youssefi
MAE dept., SJSU
20
Correction Factors for Specimens Endurance Limit Temperature factor, Ctemp(p.331, Nortons 3rd ed.)
High temperatures reduce the fatigue life of a component. For accurate results, use an environmental chamber and obtain the endurance limit experimentally at the desired temperature. For operating temperature below 450 oC (840 oF) the temperature factor should be taken as one.
Ctemp = 1
for T 450 oC (840 oF)
Ken Youssefi
MAE dept., SJSU
21
Correction Factors for Specimens Endurance Limit Reliability factor, Crel (p. 331, Nortons 3rd ed.)The reliability correction factor accounts for the scatter and uncertainty of material properties (endurance limit).
Ken Youssefi
MAE dept., SJSU
22
Fatigue Stress Concentration Factor, KfExperimental data shows that the actual stress concentration factor is not as high as indicated by the theoretical value, Kt. The stress concentration factor seems to be sensitive to the notch radius and the ultimate strength of the material. Notch sensitivity factor Fatigue stress Kf = 1 + (Kt 1)q concentration factor rd(p. 340, Nortons 3 ed.)
Steel
Ken Youssefi
MAE dept., SJSU
23
Fatigue Stress Concentration Factor, Kf for Aluminum(p. 341, Nortons 3rd ed.)
Ken Youssefi
MAE dept., SJSU
24
Design process Fully Reversed Loading for Infinite Life Determine the maximum alternating applied stress (Wa ) in terms of the size and cross sectional profile Select material Sy, Sut n
Choose a safety factor
Determine all modifying factors and calculate the endurance limit of the component Se Determine the fatigue stress concentration factor, Kf Use the design equation to calculate the size
Se Kf Wa = n Investigate different cross sections (profiles), optimize for size or weight You may also assume a profile and size, calculate the alternating stress and determine the safety factor. Iterate until you obtain the desired safety factorMAE dept., SJSU
Ken Youssefi
25
Design for Finite Life
Sn = a (N)b equation of the fatigue lineA S103
A B Se106
S Sf N103
B N
5x108
Point A
Sn = .9SutN = 103
Point A
Sn = .9SutN = 103
Point BKen Youssefi
Sn = SeN = 106MAE dept., SJSU
Point B
Sn = SfN = 5x10826
Design for Finite LifeSn = a (N)blog Sn = log a + b log N Apply boundary conditions for point A and B to find the two constants a and b
log .9Sut = log a + b log 103 log Se = log a + b log 106
a= b= log (
(.9Sut) Se1 3
2
log
.9Sut Se
Sn = Se ( 106 )Calculate Sn
N
Se ) .9Sut
and replace Se in the design equationSn Kf Wa = nDesign equation27
Ken Youssefi
MAE dept., SJSU
The Effect of Mean Stress on Fatigue LifeMean stress exist if the loading is of a repeating or fluctuating type.
Wa
Mean stress is not zero
Gerber curve Alternating stress Se Goodman line
Soderberg lineKen Youssefi MAE dept., SJSU
Sy Mean stress
Sut
Wm28
The Effect of Mean Stress on Fatigue Life Modified Goodman DiagramWaSy Yield line
Alternating stress
Se Goodman line Safe zoneC
Sy Mean stress
Sut
Wm
Ken Youssefi
MAE dept., SJSU
29
The Effect of Mean Stress on Fatigue Life Modified Goodman Diagram
WaSy Yield line
Se Goodman line Safe zone - Wm - Syc Safe zoneC
Sy
Sut
+Wm
Ken Youssefi
MAE dept., SJSU
30
The Effect of Mean Stress on Fatigue Life Modified Goodman DiagramFatigue,
Wm 0
Fatigue,
Wm > 0 Sut= =1 1 Infinite life
WaSe
WaSe
+ +
Wm WmSut
nfFinite life Yield
Wa = n fYield Se Syc Safe zone - Wm - Syc
WaSn
Wa + Wm = n y
Safe zone
C
Wa + Wm = n ySy Sut +Wm
Sy
Ken Youssefi
MAE dept., SJSU
31
Applying Stress Concentration factor to Alternating and Mean Components of Stress Determine the fatigue stress concentration factor, Kf, apply directly to the alternating stress Kf Wa
If Kf Wmax < Sy then there is no yielding at the notch, use Kfm = Kf and multiply the mean stress by Kfm Kfm Wm
If Kf Wmax > Sy then there is local yielding at the notch, material at the Calculate the stress concentration factor for the mean stress using the following equation, Kfm =
notch is strain-hardened. The effect of stress concentration is reduced.
Sy
Kf Wa
Wm1 Infinite life32
Fatigue design equation Kf Wa KfmWm = +
Se
Sut
nf
Ken Youssefi
MAE dept., SJSU
Combined LoadingAll four components of stress exist,
Wxa Wxm
alternating component of normal stressmean component of normal stress alternating component of shear stress mean component of shear stress
Xxya Xxym
Calculate the alternating and mean principal stresses,
W1a, W2a = (Wxa /2) W1m, W2m = (Wxm /2)Ken Youssefi
(Wxa /2)2 + (Xxya)2 (Wxm /2)2 + (Xxym)233
MAE dept., SJSU
Combined LoadingCalculate the alternating and mean von Mises stresses,
Wa Wm
= =
(W1a + W2a - W1aW2a)1/2 (W1m + W2m - W1mW2m)1/22 2
2
2
Fatigue design equation
WaSe
+
WmSut
=
1
nf
Infinite life
Ken Youssefi
MAE dept., SJSU
34
Design ExampleA rotating shaft is carrying 10,000 lb force as shown. The shaft is made of steel with 12 d A
10,000 lb. 6 6 D = 1.5d R2
Sut = 120 ksi and Sy = 90 ksi. The shaft is rotating at 1150 rpm and has a machine finish surface. Determine the diameter, d, for 75 minutes life. Use safety factor of 1.6 and 50% reliability.Calculate the support forces,
R1 r (fillet radius) = .1d
R1 = 2500, R2 = 7500 lb. MA = 2500 x 12 = 30,000 lb-in
The critical location is at the fillet, Calculate the alternating stress,
Wa =
Mc
I
=
32M
d
3
=
305577
d
3
Wm = 0
Determine the stress concentration factor d D = 1.5 dKen Youssefi
r = .1
Kt = 1.735
MAE dept., SJSU
Design ExampleAssume d = 1.0 in
Using r = .1 and Sut = 120 ksi, q (notch sensitivity) = .85Kf = 1 + (Kt 1)q = 1 + .85(1.7 1) = 1.6
Calculate the endurance limit
Cload = 1 (pure bending) Crel = 1 (50% rel.) Ctemp= 1 (room temp) Csurf = A (Sut)b = 2.7(120)0.3 in. < d 10 in.-.265
= .759
Csize = .869(d)-0.097 = .869(1)-0.097 = .869ksi36
Se = Cload Csize Csurf Ctemp Crel (Se) = (.759)(.869)(.5x120) = 39.57Ken Youssefi MAE dept., SJSU
Design ExampleDesign life, N = 1150 x 75 = 86250 cycles Se log ( .9S ) N 86250 ut Sn = 39.57 ( 6 Sn = Se ( 6 )10 10
log (
)56.5
39.57 ) .9x120
= 56.5 ksi
Wa =
305577
d
3
= 305.577 ksi
n=
Sn KfWa
=
1.6x305.577
= .116 < 1.6
So d = 1.0 in. is too small Assume d = 2.5 in All factors remain the same except the size factor and notch sensitivity.
Using r = .25 and Sut = 120 ksi, q (notch sensitivity) = .9 Csize = .869(d)-0.097 = .869(2.5)Ken Youssefi
Kf = 1 + (Kt 1)q = 1 + .9(1.7 1) = 1.63
-0.097
= .795
Se = 36.2 ksi37
MAE dept., SJSU
Design ExampleSe = 36.2 ksi
Sn = 36.20 (= 19.55 ksi
86250106
)
36.2 log ( .9x120 )
= 53.35 ksi
Wa =
305577
(2.5)
3
n=
SnKfWa
=
53.35 1.63x19.55
= 1.67 1.6
d = 2.5 in.
Check yielding
n=
Sy 90 = 2.8 > 1.6 okay = KfWmax 1.63x19.55MAE dept., SJSU
Ken Youssefi
38
Design Example Observationsn= Sn KfWa= 56.5 1.6x305.577 = .116 < 1.6 d R1 r (fillet radius) = .1d A 12 6 6 D = 1.5d R2 = 7500
So d = 1.0 in. is too small Calculate an approximate diameter
n=
Sn KfWa
=
56.5 1.6x305.577/d3
= 1.6
d = 2.4 in.
So, your next guess should be between 2.25 to 2.5
Check the location of maximum moment for possible failure Mmax (under the load) = 7500 x 6 = 45,000 lb-in MA (at the fillet) = 2500 x 12 = 30,000 lb-in But, applying the fatigue stress conc. Factor of 1.63, Kf MA = 1.63x30,000 = 48,900 > 45,000Ken Youssefi MAE dept., SJSU
39
ExampleA section of a component is shown. The material is steel with Sut = 620 MPa and a fully corrected endurance limit of Se = 180 MPa. The applied axial load varies from 2,000 to 10,000 N. Use modified Goodman diagram and find the safety factor at the fillet A, groove B and hole C. Which location is likely to fail first? Use Kfm = 1 Pa = (Pmax Pmin) / 2 = 4000 N Pm = (Pmax + Pmin) / 2 = 6000 N
Fillet
r
d D 35 = = 1.4 d 25
=
4 = .16 25
Kt = 1.76
Ken Youssefi
MAE dept., SJSU
40
ExampleUsing r = 4 and Sut = 620 MPa, q (notch sensitivity) = .85Kf = 1 + (Kt 1)q = 1 + .85(1.76 1) = 1.65 Calculate the alternating and the mean stresses, Pa 4000 = 52.8 MPa = 1.65 Wa = Kf A 25x5 Pm 6000 = 48 MPa = Wm = A 25x5 Fatigue design equation
WaSe+
WmSut
=
1
n
Infinite life
52.8 48 = + 180 620
1
n
n = 2.7
Ken Youssefi
MAE dept., SJSU
41
Hole d 5 = .143 = w 35 Kt = 2.6
Example
Using r = 2.5 and Sut = 620 MPa, q (notch sensitivity) = .82Kf = 1 + (Kt 1)q = 1 + .82(2.6 1) = 2.3 Calculate the alternating and the mean stresses, Pa 4000 = 61.33 MPa = 2.3 Kf Wa = A (35-5)5 Pm 6000 = 40 MPa = Wm = A 30x5 61.33 40 = + 180 620 1
n
n = 2.542
Ken Youssefi
MAE dept., SJSU
ExampleGroove
r
d Kt = 2.33 D 35 = 1.2 = d 29 Using r = 3 and Sut = 620 MPa, q (notch sensitivity) = .83 Kf = 1 + (Kt 1)q = 1 + .83(2.33 1) = 2.1 Calculate the alternating and the mean stresses, Pa 4000 = 58.0 MPa = 2.1 Wa = Kf A (35-6)5
=
3 = .103 29
Wm =
Pm A
=
6000
29x51
= 41.4 MPa
58.0 41.4 = + 180 620Ken Youssefi
n
n = 2.57MAE dept., SJSU
The part is likely to fail at the hole, has the lowest safety factor
43
ExampleThe figure shows a formed round wire cantilever spring subjected to a varying force F. The wire is made of steel with Sut = 150 ksi. The mounting detail is such that the stress concentration could be neglected. A visual inspection of the spring indicates that the surface finish corresponds closely to a hot-rolled finish. For a reliability of 99%, what number of load applications is likely to cause failure. Fa = (Fmax Fmin) / 2 = 7.5 lb. Fm = (Fmax + Fmin) / 2 = 22.5 lb. Ma = 7.5 x 16 = 120 in - lb Mm = 22.5 x 16 = 360 in - lb = 23178.6 psi = 69536 psi
Wa = Wm =
Mc
IMc
=
32Ma
d d
3
= =
32(120)
(.375)
3
I
=
32Mm3
32(360)
(.375)
3
Ken Youssefi
MAE dept., SJSU
44
Calculate the endurance limit
ExampleCsurf = A (Sut)b = 14.4(150)2
Cload = 1 (pure bending) Ctemp= 1 (room temp) Crel= .814 (99% reliability)
-.718
= .394
A95 = .010462 d (non-rotating round section)
dequiv = A95 / .0766 = .37d = .37 x.375 = .14 dequiv = .14 < .3 Csize = 1.0ksi
Se = Cload Csize Csurf Ctemp Crel (Se) = (.394)(.814)(.5x150) = 24.077 WaSe+
WmSut
=
1
n
23178.6 69536 = + 24077 150000
1
n
n = .7 < 1Finite life
Find Sn, strength for finite number of cycle
WaSnKen Youssefi
Wm+
Sut
=1
23178.6
Sn
+
69536 =1 150000
Sn = 43207 psi
MAE dept., SJSU
45
ExampleSe log ( .9S ) ut
Sn = Se ( 106 )
N
43207 = 24077 (
N 106
)
log ( .9x150 )
24.077
N = 96,000 cycles
Ken Youssefi
MAE dept., SJSU
46
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