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Fatigue of Materials
Dr. Richard Chung
Department of Chemical and Materials Engineering
San Jose State University
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Learning Objectives• Explain why the fatigue problems are more profound in
polymers and metals than ceramics and composites• Describe three stages of a fatigue process in a material (crack
nucleation, crack growth and crack propagation)• Discuss how the fatigue crack propagation is determined by
the relationship between dC/dN (crack advance rate) and ΔK (cyclic stress intensity factor)
• Design and use a material having ΔK is less than ΔKth (fail-safe failure mode)
• Examine and discuss the physical meaning of striation formed on the fractured surface and determine the crack advance between cycles
• Determine the conditions of the slow crack growth region and fast crack growth region of the fractured surface
• Find the relationship between cyclic stress (or strain) amplitude and number of cycles to help design fatigue resistant material or applications
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What is fatigue?
An engineering structure is often subjected to the repeated application of a stress below its yield strength of the material.
This cyclical stress may occur in the form of rotation, bending, or vibration.
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Fatigue Testing
A common test to measure a material’s fatigue properties is to use a rotating cantilever beam.
A cylindrical beam is mounted in a motor-driven chuck with a load applied from the opposite end.
A fatigue mode (a sinusoidal cycle) of C-0-T-0 is repeatedly applied to the beam.
The maximum stress acting on the beam is governed by the following equation:
where l is the length of the beam, P is the load, and d is the diameter of the beam.
3
18.10
d
P
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Example 1: A solid tool-steel shaft must be 96 inch long and must survive continuous operation for one year with an applied force of 12,500 pounds. The shaft is rotating one revolution per minute during operation. Design a shaft that will meet these requirements.
Solution:
No. of cycles= (1 cycle/min)(60 x 24 x 365min) = 5.256 x 105 cycles/yr.
From figure 6-19, the applied stress is around 72,000psi
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Example 1: A solid tool-steel shaft must be 96 inch long and must survive continuous operation for one year with an applied force of 12,500 pounds. The shaft is rotating one revolution per minute during operation. Design a shaft that will meet these requirements.
Solution:No. of cycles= (1 cycle/min)(60 x 24 x 365min) = 5.256
x 105 cycles/yr.From figure 6-19, the applied stress is around 72,000psi
d = 5.54 inches Add a safety factor to the system:
d = 5.54 x1.05 = 5.82 inches
3
)12500.)(96)(18.10(000,72
d
lbsinpsi
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TerminologyEndurance limit: is the stress below which that
failure by fatigue will never occur, this is a preferred design criterion.
Fatigue life: indicates how long (no. of cycles) a component survives a particular stress.
Fatigue strength: is applicable to a component has No endurance limit. It is the maximum stress for which fatigue will not occur at a particular number of cycles, in general, 500 million cycles.
Endurance ratio: the endurance limit is approximately ½ the tensile strength.
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Definitions
max
minRatio Stress
R
R
RA
m
a
1
1Ratio Amplitude
A
AR
1
1
R
RA
1
1
22Stress gAlternatin minmax
Ra
2StressMean minmax
m
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The difference between Point Stresses () and Nominal Stresses (S)
For simple axial loading, = S For bending, S
S=Mc/I; where M =bending moment, c = the distance from neutral axis to edge, and I = the area moment of inertia about the axis.
For notched specimen (No yielding), = KtS
where Kt = elastic stress concentration factor.
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Stress vs. Life (S-N) Curves
An equation can be derived to represent an S-N curve
fa NDC logDfa NC log
DfC
Df
C
Ne
e
Ne
a
a
Bfa AN
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bffa N )2('
A = 2b ’f B = b
Bfa AN
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Plotting in Linear vs. Logarithmic scales
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Three factors are necessary to cause fatigue failure:
A maximum tensile stress of sufficiently high value
A large enough variation or fluctuation in the applied stress
A sufficiently large number of cycles of the applied stress
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Variables Affecting Fatigue In A Material
Stress concentration Corrosion (Environment) Temperature Overload Metallurgical structure (Microstructure) Residual stress (shot peening, presetting) Combined stress Surface condition
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Stress Amplitude versus Mean Stress
Mean stress effects can be plotted in a diagram using stress amplitude versus mean stress.
Estimates of mean stress effects for un-notched specimens can be determined by Morrow equation or SWT equation (Smith, Watson, and Topper)
where ar= equivalent completely reversed stress f
m
aar
'1
aar max
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The Palmgren-Miner Rule
13
3
2
2
1
1 jf
j
fff N
N
N
N
N
N
N
N
The fatigue failure of a material under a variable (multiple) amplitude loading is expected when such life fractions sum to unity.
In the case of creep-fatigue, a fracture criterion will be defined as:
1if
i
if
i
t
t
N
N
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Initiation of Fatigue CracksF
ract
ure
Tou
ghn
ess
(KIC
)
Yield Strength (y)
Design based on strength
Design based on toughness
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Crack Initiation and Propagation
Max
imu
m C
ycle
Str
ess
Number of Cycles
Ni Np
Ni Nt = Ni +Np
Ni = # of cycles for initiationNp= # of cycles for propagation
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Crack Rate As A Function of The Stress-Intensity Range (K)
• When the length of the crack is small, the growth rate of the crack (a/N)is also small
• As the the length of the crack increases, the growth rate of the crack (a/N) also increases
• Under identical cyclic loading, larger initial cracks propagate to failure in short cycles
•
slope theis consatnt, theis where
)(
mA
KAdN
dc m
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Example: A metal strip (4 inch wide and 0.2 inch thick) is loaded in a cyclic loading ranging from 6,000 to 43,000 lbs. A crack is found located in the center of the strip that extends through the thickness. For c = 0.1 and 0.4 inch , calculate K.
Answer:
Assume the geometry factor F = 1 cFSK i
inKsiKKK
inKsicSK
inKsicSK
c
inKsiKKK
inKsicSK
inKsicSK
c
psix
S
psix
S
9.514.84.60
4.8)4.0(500,7
3.60)4.0(750,53
0.4inchfor nscalculatio Repeat the
9.252.41.30
2.4)1.0(500,7
1.30)1.0(750,53
0.1inchfor rangeintensity stress Compute
500,742.0
000,6
750,5342.0
000,43
min2max22
minmin2
maxmax2
min1max11
minmin1
maxmax1
min
max
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The Walker Equation
1)1(1
)1(1
11
1
1max
max
min
mm and )1(
)1(
)1(
)(
constant material a is γ
)1()1(
1
1
1
1
1
m
m
m
m
m
R
AA
KR
A
dN
dc
R
KA
dN
dc
KAdN
dc
R
KRKK
S
SR
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The Forman Equation
• The R ratio has strong effects on the behavior of slow growth rate
•
))(1(
)(
)1(
)(
max
2222
KKR
KA
KKR
KA
dN
dc
c
m
c
m
thR
KK th
th
1)1(
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Fatigue Life
dcdc
dNN
f
i
c
c
if )(
2)(m )2/1()(
2/12/1
mSFA
ccN
m
mi
mf
if
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Summary• Fatigue failures are often focused on metals and
polymers.• Endurance limit and fatigue life can be used to help
prevent fatigue in materials• When applied stress magnitude increases, the
number of cycles decreases• Materials can fail by fatigue even when they contain
no cracks• A fatigue failure is based on the accumulation of
fatigue cycles used at low and high cyclic stresses• A final fractured surface resembles a ductile failure
pattern which includes three distinct stages: crack nucleation, crack growth and crack propagation.
• The spacing between the beach marks corresponds to the crack advance per cycle in materials.
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Summary (cont’d)• As long as ΔK is less than ΔKth , the crack growth rate is not
going to increase• Using a material at ΔK value (less than ΔKth), a fail safe
fatigue design can be achieved.• The slow-crack growth surface in metals is generally smooth,
unless oxidation or abrasion have already developed.• The fast-crack region on a fractured surface is observed as
dull and fibrous resembling a tensile ductile failure.• Polymers can develop εDCG (shear band cracking
accompanied by crazing) prior to DCG (discontinuous crack growth)
• Temperature will have great effects on polymer and metal fatigue. High frequency of loading could yield hysteric heating and thermal softening which could significantly reduce the fatigue life/endurance limit in a material. Creep-fatigue could be a complicated process for material failure prediction.