피로 파괴 (Fatigue Failure)

1. Fatigue Failure
It has been recognized that a metal subjected to a repetitive or fluctuating stresswill 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.

2. Fatigue failure is characterized by three stages
1) Crack Initiation
2) Crack Propagation
3) Final Fracture

3. Fatigue failure examples
1) Jack hammer component, shows no yielding before fracture. 2)VW crank shaft – fatigue failure due to cyclic bending and torsional stresses

3) 928 Porsche timing pulley

4) 1.0-in. diameter steel pins from agricultural equipment. Material; AISI/SAE 4140 low allow carbon steel	5) Fracture surface of a failed bolt. The fracture surface exhibited beach marks, which is characteristic of a fatigue failure.

6) 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.

7) Gear tooth failure

5. Fatigue Failure – Type of Fluctuating Stresses

(a) Full reversed σa = σmax, σa = σmin	(b) Repeated σa = σm = σmax / 2.0 , σmin = 0	(c) Fluctuating σa = (σmax - σmin) / 2.0 σm = (σmax + σmin) / 2.0

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

6.2 Fatigue Failure, S-N Curve

Se = endurance limit of the specimen

7. Relationship Between Endurance Limit and Ultimate Strength

7.1 Steel Se' = 0.5 ksi Sut ≤ 200 ksi (1400 MPa) Se' = 100 ksi Sut > 200 ksi Se' = 700 MPaSut > 1400 MPa
7.2 Cast iron Se' = 0.4 Sut Sut ≤ 60 ksi (400 MPa) Se' = 24 ksi Sut > 60 ksi Se' = 160 MPa Sut > 400 MPa
7.3 Aluminum Se' = 0.4 Sut Sut ≤ 48 ksi (330 MPa) Se' = 19 ksi Sut > 48 ksi Se' = 130 MPa Sut > 330 MPa   For N = 5 x 108cycle
7.4 Copper alloys Se' = 0.4 Sut Sut ≤ 40 ksi (330 MPa) Se' = 19 ksi Sut > 40 ksi Se' = 100 MPa Sut > 280 MPa   For N = 5 x 108cycle

Correction Factors for Specimen's Endurance Limit
For materials exhibiting a knee in the S-N curve at 10⁶cycles

8. Correction Factors for Specimen’s Endurance Limit

8.1 For materials exhibiting a knee in the S-N curve at 106cycles   Se'= endurance limit of the specimen   (infinite life > 106)   Se= endurance limit of the actual component   (infinite life > 106) 8.2 For materials that do not exhibit a knee in the S-N curve, the infinite life taken at 5 x 108cycles   Sf'= fatigue strength of the specimen   (infinite life > 5 x 108)   Sf = fatigue strength of the actual component   (infinite life > 5 x 108)

9. Correction Factors for Specimen's Endurance Limit
S_e = C_load · C_size · C_surf · C_temp · C_rel · (S_e')
S_f = C_load · C_size · C_surf · C_temp · C_rel · (S_f')

9.1 Load factor, C_load (page 326, Norton's 3^rd ed.)
  1) Pure bending   C_load = 1.0
  2) Pure axial   C_load = 0.7
  3) Pure torsion   C_load = 1.0
   if von Mises stress is used, use 0.577 if von Mises stress is NOT used.
  4) Combined loading   C_load = 1.0

9.2 Size factor, C_size (p. 327, Norton's 3^rd 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.
1) For rotating solid round cross section
   d ≤ 0.3 in. (8 mm)   C_size = 1.0
   0.3 in. < d ≤ 10 in.   C_size = 0.869 (d)^-0.097
   8 mm < d ≤ 250 mm   C_size = 1.189 (d)^-0.097
   If the component is larger than 10 in., use C_size = 0.6

2) For noncircular section
A₉₅ = (π / 4) [ d² - (0.95 d)²] = 0.766 d²
d_equiv = (A / 0.0766)^1/2

9.3 surface factor, C_surf (p. 328-9, Norton's 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.
C_surf= A?(S_ut)^b ≤ 1

Table. Coefficients for the surface-factor Equation
Source : Shigley and Mischke, Mechanical engineering Design. 5th ed. McGraw-Hill,
New York, 1989, p 283 permission

9.4 Temperature factor, C_temp (p.331, Norton's 3^rd 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.
Temperature factor (For Steel)

T ≤ 450 °C   :   C_temp = 1.0
450 °C T ≤ 550 °C   :   C_temp = 1.0 - 0.0058 (T-450)

9.5 Reliability factor, C_rel (p. 331, Norton's 3^rd ed.)
The reliability correction factor accounts for the scatter and uncertainty of material properties
(endurance limit).

Reliability(%) Crel 50 1.0 90 0.897 99 0.814 99.9 0.753 99.99 0.702 99.999 0.659

10. Fatigue Stress Concentration Factor, Kf
Experimental data shows that the actual stress concentration actor 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.
1) Fatigue stress concentration factor (p. 340, Norton's 3^rd ed.)
Kf = 1 + (Kt - 1)q   , q : Notch sensitivity factor

2) Fatigue Stress Concentration Factor, Kf for Aluminum (p. 341, Norton's 3^rd ed.)

11. Design process – Fully Reversed Loading for Infinite Life
1) Determine the maximum alternating applied stress (σa) in terms
  of the size and cross sectional profile
2) Select material → Sy, Sut
3) Choose a safety factor → n
4) Determine all modifying factors and calculate the endurance limit
  of the component → Se
5) Determine the fatigue stress concentration factor, Kf
6) Use the design equation to calculate the size
  Kf σa = Se / n
7) Investigate different cross sections (profiles), optimize for size or weight
8) 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

12. Design for Finite Life

1) Design for Finite Life
Sn = a (N)^blog Sn = log a + log N
Apply boundary conditions for point A and B to find the two constants “a” and “b”
log 0.9 Sut = log a + b log 10 ³, a = (0.9 Sut)²/ Se
log Se = log a + b log 10⁶, b = (- 1 / 3)log [0.9 Sut / Se]
Calculate Sn and replace Se in the design equation
Kfσa = Sn / n   Design equation

13.The Effect of Mean Stress on Fatigue Life

Mean stress exist if the loading is of a repeating or fluctuating type. Mean stress is not zero

Repeated

Fluctuating

14.The Effect of Mean Stress on Fatigue Life Modified Goodman Diagram

15. Applying Stress Concentration factor to Alternating and Mean Components of Stress
1) Determine the fatigue stress concentration factor, Kf,
 apply directly to the alternating stress --> Kf σa
2) If Kf σmax < Sy then there is no yielding at the notch, use Kfm = Kf
 and multiply the mean stress by Kfm --> Kfm σm
3) If Kf σmax > 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 σa) / σm

 Fatigue design equation
 ( Kf σa) / Se + ( Kfm σm) / Sut = 1 / nf    Infinite life

16. Combined Loading
1) All four components of stress exist,
 σa : alternating component of normal stress
 σxm : mean component of normal stress
 τxya : alternating component of shear stress
 τxym : mean component of shear stress
2) Calculate the alternating and mean principal stresses,
 σ1a, σ2a = (σxa / 2.0 ) ± √ { (σxa / 2) 2 + (τxya) 2 }
 σ1m, σ2m = (σxm / 2.0 ) ± √ { (σxm / 2) 2 + (τxym) 2 }
3) Calculate the alternating and mean von Mises stresses,
 σa′ = √ (σ1a2 + σ2a2 - σ1a σ2a)
 σm′ = √ (σ1m2 + σ2m2 - σ1m σ2m)
4) Fatigue design equation
 σ'a / Se + σ'm / Sut = 1 / nf    Infinite life

Design Example 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.

1) Calculate the support forces,
R1 = 2500, R2 = 7500 lb.

2) The critical location is at the fillet,
Ma = 2500 x 12 = 30,000 lb-in

3) Calculate the alternating stress,
σa = Mc / I = 32 Ma / π d³ = 305,577 / d³?, σm = 0

4) Determine the stress concentration factor
r / d = 0.1 , D /d = 1.5 ---> Kt = 1.7

5) Assume d = 1.0 in , Using r = 0.1 and Sut = 120 ksi,
q (notch sensitivity) = .85
Kf = 1 + (Kt ? 1)q = 1 + 0.85 (1.7 ? 1) = 1.6

6) Calculate the endurance limit
C_load = 1.0    (pure bending)
0.3 in. < d ≤ 10 in.    C_size = 0.869 (d)^-0.097 = 0.869 (1.0)^-0.097 = 0.869

C_surf = A · (S_ut)^b = 2.7 · (120)^-0.265 = 0.759
C_temp = 1.0    (room temp)

C_rel = 1.0    (50% rel.)
Se' = 0.5 Sut   Sut ≤ 200 ksi (1400 MPa)

Se = C_load · C_size · C_surf · C_temp · C_rel · (Se')
  = (1.0) (0.869) (0.759) (1.0) (1.0) (0.5x120) = 39.57 ksi

7) Design life, N = 1150 x 75 = 86250 cycles

8) Assume d = 2.5 in
All factors remain the same except the size factor and notch sensitivity.
Using r = 0.25 and Sut = 120 ksi, q (notch sensitivity) = 9
Kf = 1 + (Kt ? 1)q = 1 + 0.9 (1.7 ? 1) = 1.63
C_load = 1.0    (pure bending)
0.3 in. < d ≤ ≤ in.    C_size = 0.869 (d)^-0.097= 0.869 (2.5)^-0.097= 0.795
Se = C_load · C_size · C_surf · C_temp · C_rel · (Se')
  = (1.0) (0.795) (0.759) (1.0) (1.0) (0.5x120) = 36.2 ksi

9) Design Example ? Observations
n = Sn / ( Kf σa) = 56.5 / (1.6 x 305,577) = 0.116 < 1.6
So d = 1.0 in. is too small
Calculate an approximate diameter
n = Sn / ( Kf σa) = 56.5 / (1.6 x 305,577 / 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 > 45,000

Design Example b)
A 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

1) Pa = (Pmax ? Pmin) / 2 = 4000 N , Pm = (Pmax + Pmin) / 2 = 6000 N

2) Fillet

3) Using r = 4 and Sut = 620 MPa,
q (notch sensitivity) = .85
Kf = 1 + (Kt ? 1)q = 1 + .85(1.76 ? 1) = 1.65

4) Calculate the alternating and the mean stresses,
σa = Kf ( Pa / A) = 1.65 [ 4000 / (25 x 5)] = 52.8 MPa
σm= Pm / A = 6000 (25 x 5) = 48 MPa

5) Fatigue design equation
σa / Se + σm / Sut = 1 / n
Infinite life
52.8 / 180 + 48 / 620 = 1 / n ---> n = 2.7

Hole
d / w = 5 / 35 = 0.143 ⇨ Kf = 2.6

7)Using r = 2.5 and Sut = 620 MPa,
q (notch sensitivity)= 0.82
Kf = 1 + (Kt ? 1)q = 1 + 0.82 (2.6 ? 1) = 2.3

8) Calculate the alternating and the mean stresses,
σa = Kf ( Pa / A) = 2.3 [ 4000 / { (35 - 5) x 5}] = 61.33 MPa
σm= Pm / A = 6000 (30 x 5) = 40 MPa

10) Fatigue design equation
σa / Se + σm / Sut = 1 / n
Infinite life
61.33 / 180 + 40 / 620 = 1 / n --> n = 2.5

11) Groove
r / d = 3 / 29 = 0.103 , d / D = 35 / 29 = 1.2

12) Using r = 3 and Sut = 620 MPa,
q (notch sensitivity) = .83
Kf = 1 + (Kt ? 1)q = 1 + .83(2.33 ? 1) = 2.1

13) Calculate the alternating and the mean stresses
σa = Kf ( Pa / A) = 2.1 [ 4000 / { (35 - 6) x 5}] = 58.0 MPa
σm= Pm / A = 6000 (29 x 5) = 41.4 MPa

14) Fatigue design equation
σa / Se + σm / Sut = 1 / n
Infinite life
58.0 / 180 + 41.4 / 620 = 1 / n --> n = 2.57

15) The part is likely to fail at the hole, has the lowest safety factor

Example c)
The 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.

1) 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

2) Calculate the alternating stress,
σa = Mc / I = 32 Ma / π d³ = 32 (120) / π (0.375) ³= 23178.6 psi
σm= Mc / I = 32 Mm / π d³= 32 (360) / π (0.375) ³= 69536 psi

3) Calculate the endurance limit
C_load = 1.0   (pure bending)
C_surf = A · (S_ut)^b = 14.4 · (150)^-0.718 = 0.394
C_temp = 1.0 (room temp)
C_rel = 0.814 (99% reliability.)

A₉₅ = 0.10462 d²   (non-rotating round section)

d_equiv = (A95 / 0.0766)^1/2 = 0.37 d = 0.37 x 0.375 = 0.14

S_e = C_load · C_size · C_surf · C_temp · C_rel · (Se')
  = (1.0) (1.0) (0.394) (1.0) (0.814) (0.5x150) = 24.077 ksi

σa / Se + σm / Sut = 1 /n
Infinite life 23178.6 / 24077 + 69536 / 150000 = 1 /n ⇨ n = 0.7 < 1

4) Find Sn, strength for finite number of cycle
σa / Sn + σm / Sut = 1
23178.6 / Sn + 69536 / 150000 = 1 ⇨ Sn = 43207 psi

N = 96,000 cycles