Fatigue failure.ppt

Ken Youssefi MAE dept., SJSU 1
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
Fatigue failure is characterized by three stages
 Crack Initiation
 Crack Propagation
 Final Fracture

Jack hammer component,
shows no yielding before
fracture.
Crack initiation site
Fracture zone
Propagation zone, striation

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

928 Porsche timing pulley
Crack started at the fillet

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

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.
bicycle crank spider arm

Gear tooth failure
Crank shaft

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.

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

Fatigue Failure – Type of Fluctuating Stresses
a =
max min
2
Alternating stress
Mean stress
m
=
max min
2
+
min = 0
a = max / 2
m =
a = max
max = - min

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

The standard machine operates at an
adjustable speed of 500 RPM to
10,000 RPM. At the nominal rate of
10,000 RPM, the R. R. Moore machine
completes 600,000 cycles per hour,
14,400,000 cycles per day.
Bending moment capacity
20 in-lb to 200 in-lb

Fatigue Failure, S-N Curve
Finite life Infinite life
N < 103
N > 103
S′e
= endurance limit of the specimen
Se
′

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

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

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

Se = Cload Csize Csurf Ctemp Crel (Se)
′
• Load factor, Cload (page 326, Norton’s 3rd ed.)
Pure bending Cload = 1
Pure axial Cload = 0.7
Combined loading Cload = 1
Pure torsion Cload = 1 if von Mises stress is used, use
0.577 if von Mises stress is NOT used.
Sf = Cload Csize Csurf Ctemp Crel (Sf)
′
or

• Size factor, Csize (p. 327, Norton’s 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) 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

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.
dequiv = (
A95
0.0766
)1/2
d
d95 = .95d
A95 = (π/4)[d2
– (.95d)2
] = .0766 d2
dequiv = .37d
Solid or hollow non-rotating parts
dequiv = .808 (bh)1/2
Rectangular parts

I beams and C channels

• surface factor, Csurf (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.
Csurf = A (Sut)b

• Temperature factor, Ctemp (p.331, Norton’s 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 o
C (840 o
F) the
temperature factor should be taken as one.
Ctemp = 1 for T ≤ 450 o
C (840 o
F)

• Reliability factor, Crel (p. 331, Norton’s 3rd ed.)
The reliability correction factor accounts for the scatter and
uncertainty of material properties (endurance limit).

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.
(p. 340, Norton’s 3rd ed.)
Steel
Kf = 1 + (Kt – 1)q
Notch sensitivity factor
Fatigue stress
concentration factor

Fatigue Stress
Concentration Factor,
q for Aluminum
(p. 341, Norton’s 3rd ed.)

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
• Use the design equation to calculate the size
Se
Kf a =
n
• 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
• 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

Design for Finite Life
Sn = a (N)b
equation of the fatigue line
N
S
Se
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 = Sf
N = 5x108
Point B
Sn = Se
N = 106

Design for Finite Life
Sn = a (N)b
log 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 =
(.9Sut)
2
Se
b =
.9Sut
Se
1
3
log
Sn
Kf a =
n
Design equation
Calculate Sn and replace Se in the design equation
Sn = Se (
N
106
)
⅓ (
Se
.9Sut
)
log

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
Se
Sy
Soderberg line
Sut
Goodman line
Gerber curve
Mean stress is not zero

Modified Goodman Diagram
Mean stress
Alternating
stress
m
a
Sut
Goodman line
Sy
Sy
Se
Safe zone
C
Yield line

- Syc
+m
a
Sut
Goodman line
Sy Yield line
Safe zone
- m
C
Sy
Se
Safe zone

+m
a
Sut
Safe zone
- m
C
Sy
Safe zone
Se
- Syc
Finite life
Sn
1
=
Sut
a m
+
Fatigue, m > 0
Fatigue, m ≤ 0
a =
Se
nf
a + m =
Syc
ny
Yield
a + m =
Sy
ny
Yield
nf
Se
1
=
Sut
a m
+ Infinite life

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

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

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

Design Example
R1 R2
10,000 lb.
6˝
6˝
12˝
D = 1.5d
d
r (fillet radius) = .1d
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.
Calculate the support forces, R1 = 2500, R2 = 7500 lb.
A
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
d
= 1.5
Kt = 1.7

Design Example
Assume d = 1.0 in
Using r = .1 and Sut = 120 ksi,
q (notch sensitivity) = .85
Kf = 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)
-.265
= .759
0.3 in. < d ≤ 10 in. Csize = .869(d)-0.097
= .869(1)-0.097
= .869
Se = Cload Csize Csurf Ctemp Crel (Se) = (.759)(.869)(.5x120) = 39.57 ksi
′

Design Example
Design life, N = 1150 x 75 = 86250 cycles
Sn = Se (
N
106
)
⅓ (
Se
.9Sut
)
log
Sn = 39.57 (
86250
106
)
⅓ (
39.57
.9x120
)
log
= 56.5 ksi
a =
305577
d 3
= 305.577 ksi n =
Sn
Kfa
=
56.5
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,
Kf = 1 + (Kt – 1)q = 1 + .9(1.7 – 1) = 1.63
Csize = .869(d)-0.097
= .869(2.5)-0.097
= .795 Se = 36.2 ksi
→

Design Example
a =
305577
(2.5)3
= 19.55 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 > 1.6 okay
Se = 36.2 ksi → Sn = 36.20 (
86250
106
)
⅓ (
36.2
.9x120
)
log
= 53.35 ksi

Design Example – Observations
n =
Sn
Kfa
=
56.5
1.6x305.577
= .116 < 1.6
So d = 1.0 in. is too small
Calculate an approximate diameter
n =
Sn
Kfa
=
56.5
1.6x305.577/d 3 = 1.6 → d = 2.4 in. So, your next guess
should be between
2.25 to 2.5
Mmax (under the load) = 7500 x 6 = 45,000 lb-in
Check the location of maximum moment for possible failure
R1 R2 = 7500
6˝
6˝
12˝
D = 1.5d
d
r (fillet radius) = .1d
A
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

Example
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
Pa = (Pmax – Pmin) / 2 = 4000 N Pm = (Pmax + Pmin) / 2 = 6000 N
Fillet
r
d
= .16
D
d
= 1.4
4
25
=
35
25
=
Kt = 1.76

Example
Kf = 1 + (Kt – 1)q = 1 + .85(1.76 – 1) = 1.65
Calculate the alternating and the
mean stresses,
a =
Pa
A
=
4000
25x5
= 52.8 MPa
Kf 1.65
m =
Pm
A
=
6000
25x5
= 48 MPa
n
Se
1
=
Sut
a m
+ Infinite life
n = 2.7
n
180
1
=
620
52.8 48
+ →
Using r = 4 and Sut = 620 MPa,

Example
Hole
d
w
= .143
5
35
= → Kt = 2.6
Using r = 2.5 and Sut = 620 MPa,
Kf = 1 + (Kt – 1)q = 1 + .82(2.6 – 1) = 2.3
mean stresses,
a =
Pa
A
=
4000
(35-5)5
= 61.33 MPa
Kf 2.3
m =
Pm
A
=
6000
30x5
= 40 MPa
n = 2.5
n
180
1
=
620
61.33 40
+ →

Example
Groove
r
d
= .103
D
d
= 1.2
3
29
=
35
29
=
→ Kt = 2.33
Using r = 3 and Sut = 620 MPa,
Kf = 1 + (Kt – 1)q = 1 + .83(2.33 – 1) = 2.1
(35-6)5
mean stresses,
a =
Pa
A
=
4000
= 58.0 MPa
Kf 2.1
m =
Pm
A
=
6000
29x5
= 41.4 MPa
n = 2.57
n
180
1
=
620
58.0 41.4
+ →
The part is likely to fail at the hole, has the lowest safety factor

Example
Fa = (Fmax – Fmin) / 2 = 7.5 lb. Fm = (Fmax + Fmin) / 2 = 22.5 lb.
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.
Ma = 7.5 x 16 = 120 in - lb Mm = 22.5 x 16 = 360 in - lb
a =
Mc
I
=
32Ma
πd 3 =
32(120)
π (.375)
3 = 23178.6 psi
m =
Mc
I
=
32Mm
πd 3
=
32(360)
π (.375)
3 = 69536 psi

Example
Cload = 1 (pure bending)
Ctemp= 1 (room temp)
Calculate the endurance limit
Crel= .814 (99% reliability)
Csurf = A (Sut)b
= 14.4(150)
-.718
= .394
A95 = .010462 d
2
(non-rotating round section)
dequiv = √ A95 / .0766 = .37d = .37 x.375 = .14
dequiv = .14 < .3 → Csize = 1.0
Se = Cload Csize Csurf Ctemp Crel (Se) = (.394)(.814)(.5x150) = 24.077 ksi
n
Se
1
=
Sut
a m
+ →
n
24077
1
=
150000
23178.6 69536
+ → n = .7 < 1
Finite life
Sn
1
=
Sut
a m
+
Find Sn, strength for finite number of cycle
Sn
1
=
150000
23178.6 69536
+
→ → Sn = 43207 psi

Example
Sn = Se (
N
106
)
⅓ (
Se
.9Sut
)
log
43207 = 24077 (
N
106
)
⅓ (
24.077
.9x150
)
log
→
N = 96,000 cycles

Fatigue failure.ppt

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