1. Hasmukh Goswami College Of Engineering
DME
Submitted by:
Enrolment no:-
SEM. 5 Mechanical Engineering

2. Fatigue Failure
It 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 Fracture

3. VW crank shaft – fatigue failure due to cyclic bending and torsional stresses
Propagation zone, striations
Crack initiation site
Fracture area

4. 928 Porsche timing pulley
Crack started at the fillet

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.

6. Crank shaft
Gear tooth failure
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.
MAE dept., SJSU

8. Fracture Surface Characteristics
Mode of fracture
Typical surface characteristics
Ductile
Cup and Cone Dimples Dull Surface Inclusion at the bottom of the dimple
Brittle Intergranular
Shiny Grain Boundary cracking
Brittle Transgranular
Shiny Cleavage fractures Flat
Fatigue
Beachmarks Striations (SEM) Initiation sites Propagation zone Final fracture zone

9. Fatigue Failure – Type of Fluctuating Stresses
max
min 2
Alternating stress
Mean stress
m =
max
min 2 +

10. Fatigue Failure, S-N Curve
Test specimen geometry for R.R. Moore rotating beam machine. The surface is polished in the axial direction. A constant bending load is applied.
Motor
Load
Rotating beam machine – applies fully reverse bending stress
Typical testing apparatus, pure bending

11. Fatigue Failure, S-N Curve
Finite life
Infinite life
S′e
= endurance limit of the specimen Se ′

12. Relationship Between Endurance Limit and Ultimate Strength
Se = ′
0.5Sut
100 ksi
700 MPa
Sut ≤ 200 ksi (1400 MPa)
Sut > 200 ksi
Sut > 1400 MPa
Steel
Steel
0.4Sut
Se = ′
Sut < 60 ksi (400 MPa)
Sut ≥ 60 ksi
24 ksi
160 MPa
Sut < 400 MPa
Cast iron

13. Relationship Between Endurance Limit and Ultimate Strength
Aluminum
Aluminum alloys
Se = ′
0.4Sut
19 ksi
130 MPa
Sut < 48 ksi (330 MPa)
Sut ≥ 48 ksi
Sut ≥ 330 MPa
For N = 5x108 cycle
Copper alloys
Se = ′
0.4Sut
14 ksi
100 MPa
Sut < 40 ksi (280 MPa)
Sut ≥ 40 ksi
Sut ≥ 280 MPa
For N = 5x108 cycle

14. Correction Factors for Specimen’s Endurance Limit
For materials exhibiting a knee in the S-N curve at 106 cycles
= endurance limit of the specimen (infinite life > 106) Se ′
= endurance limit of the actual component (infinite life > 106) Se N S
106
103
= fatigue strength of the specimen (infinite life > 5x108) Sf ′
= fatigue strength of the actual component (infinite life > 5x108)
For materials that do not exhibit a knee in the S-N curve, the infinite life taken at 5x108 cycles N S
5x108
103

15. Correction Factors for Specimen’s Endurance Limit
Se = Cload Csize Csurf Ctemp Crel (Se) ′
Load factor, Cload
Pure bending
Cload = 1
Pure axial
Cload = 0.7
Pure torsion
Cload = 1 if von Mises stress is used, use if von Mises stress is NOT used.
Combined loading
Cload = 1

16. Correction Factors for Specimen’s Endurance Limit
Size factor, Csize
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 solid round cross section
d ≤ 0.3 in. (8 mm)
Csize = 1
0.3 in. < d ≤ 10 in.
Csize = .869(d)-0.097
8 mm < d ≤ 250 mm
Csize = 1.189(d)-0.097
If the component is larger than 10 in., use Csize = .6

17. Correction Factors for Specimen’s Endurance Limit
For 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
dequiv = (
d95 d
)1/2
0.0766

18. Correction Factors for Specimen’s Endurance Limit
surface factor, Csurf
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

19. Correction Factors for Specimen’s Endurance Limit
Temperature factor, Ctemp
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 = for
T ≤ 450 oC (840 oF)

20. Correction Factors for Specimen’s Endurance Limit
Reliability factor, Crel
The reliability correction factor accounts for the scatter and uncertainty of material properties (endurance limit).

21. Fatigue Stress Concentration Factor, Kf
Experimental 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.
Kf = 1 + (Kt – 1)q
Notch sensitivity factor

22. Fatigue Stress Concentration Factor, Kf for Aluminum

23. Design process – Fully Reversed Loading for Infinite Life
Determine the maximum alternating applied stress, a, in terms of the size and cross sectional profile
Select material → Sy, Sut
Choose a safety factor → n
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 a = 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 factor

24. Design for Finite Life Sn = a (N)b equation of the fatigue lineSe
106
103 A B N S Sf
5x108
103 A B
Point A
Sn = .9Sut
N = 103
Point A
Sn = .9Sut
N = 103
Point B
Sn = Se
N = 106
Point B
Sn = Sf
N = 5x108

25. Design for Finite Life Sn = a (N)b Sn Se ( ) ⅓ Sn n a
log Sn = log a + b log N
Apply conditions for point A and B to find the two constants “a” and “b” a =
(.9Sut)2 Se b
.9Sut 1 3
log
log .9Sut = log a + b log 103
log Se = log a + b log 106 Sn = Se ( N
106 ) ⅓
.9Sut
log Sn
Kf a = n
Design equation
Calculate Sn and replace Se in the design equation

26. The Effect of Mean Stress on Fatigue Life
Mean stress exist if the loading is of a repeating or fluctuating type.
Mean stress
Alternating stress m a Sy
Yield line
Gerber curve Se Sy
Soderberg line
Sut
Goodman line

27. The Effect of Mean Stress on Fatigue Life Modified Goodman Diagram
Alternating stress m a Sy
Yield line Se
Safe zone C
Goodman line Sy
Sut
Mean stress

28. The Effect of Mean Stress on Fatigue Life Modified Goodman DiagramSy
Yield line
- Syc Se
Safe zone
Sut
Goodman line C
Safe zone
- m Sy
+m

29. The Effect of Mean Stress on Fatigue Life Modified Goodman DiagramSe 1 =
Sut a m +
Infinite life
a = Se nf
Finite life Sn 1 =
Sut a m +
a + m = Sy ny
Yield Se
a + m = Sy ny
Yield C
Safe zone
Safe zone
- m
- Syc Sy
Sut
+m

30. 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 a
If Kf max < Sy then there is no yielding at the notch, use Kfm = Kf and multiply the mean stress by Kfm → Kfm m
If Kf max > Sy then there is local yielding at the notch, material at the notch is strain-hardened. The effect of stress concentration is reduced.
Calculate the stress concentration factor for the mean stress using the following equation,
Kfm = Sy
Kf a m nf Se 1 =
Sut
Kf a
Kfmm +
Infinite life
Fatigue design equation

31. xya alternating component of shear stress
Combined Loading
All four components of stress exist,
xa alternating component of normal stress
xm mean component of normal stress
xya alternating component of shear stress
xym mean component of shear stress
Calculate the alternating and mean principal stresses,
1a, 2a = (xa /2) ± (xa /2)2 + (xya)2
1m, 2m = (xm /2) ± (xm /2)2 + (xym)2

32. Combined Loading a′ = (1a + 2a - 1a2a)1/2
Calculate the alternating and mean von Mises stresses,
a′ = (1a + 2a - 1a2a)1/2 2
m′ = (1m + 2m - 1m2m)1/2 2
Fatigue design equation nf Se 1 =
Sut
′a
′m +
Infinite life

33. Design Example m = 0 a = I πd 3 d 3 10,000 lb. 6˝ 6˝ 12˝
A rotating shaft is carrying 10,000 lb force as shown. The shaft is made of steel with 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. d
D = 1.5d A R1 R2
r (fillet radius) = .1d
Calculate the support forces,
R1 = 2500, R2 = 7500 lb.
The critical location is at the fillet,
MA = 2500 x 12 = 30,000 lb-in
a =
Calculate the alternating stress, Mc I =
32M
πd 3
305577
d 3
m = 0
Determine the stress concentration factor r d
= .1 D
= 1.5
Kt = 1.7

34. Design Example
Assume d = 1.0 in
Kf = 1 + (Kt – 1)q = (1.7 – 1) = 1.6
Using r = .1 and Sut = 120 ksi, q (notch sensitivity) = .85
Calculate the endurance limit
Csurf = A (Sut)b = 2.7(120)-.265 = .759
Cload = 1 (pure bending)
Crel = 1 (50% rel.)
Ctemp= 1 (room temp)
0.3 in. < d ≤ 10 in.
Csize = .869(d) = .869(1) = .869
Se = Cload Csize Csurf Ctemp Crel (Se) = (.759)(.869)(.5x120) = ksi ′

35. Design Example Sn Se ( ) ⅓ Sn ( ) ⅓ n = Sn a = log log Kfa d 3
Design life, N = 1150 x 75 = cycles Sn = Se ( N
106 ) ⅓
.9Sut
log Sn =
39.57 (
86250
106 ) ⅓
.9x120
log
= 56.5 ksi
n = Sn
Kfa =
56.5
1.6x
= .116 < 1.6
So d = 1.0 in. is too small
a =
305577
d 3
= ksi
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
Kf = 1 + (Kt – 1)q = (1.7 – 1) = 1.63
Csize = .869(d) = .869(2.5) = .795
Se = ksi →

36. Design Example a = d = 2.5 in. n = Se = 36.2 ksi Sn = 53.35 ksi→
a =
305577
(2.5)3
= ksi
n = Sn
Kfa =
53.35
1.63x19.55
= 1.67 ≈ 1.6
d = 2.5 in.
Check yielding
n = Sy
Kfmax = 90
1.63x19.55
2.8 > okay