K f - Department of Mechanical and Aerospace Engineering
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Transcript K f - Department of Mechanical and Aerospace Engineering
Fatigue Strength
(6.4, 6.7-6.8, 6.11)
MAE 316 – Strength of Mechanical Components
NC State University Department of Mechanical and Aerospace Engineering
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Fatigue Strength
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Column Design
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Column Design
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Column Design
Fatigue Strength (6.8)
Up to now, we have designed structures for static loads.
t
d
w
P
P
max S y
P
(σmax is also constant)
t
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Fatigue Strength
Fatigue Strength (6.8)
What if loading is not constant?
P
t
Even if σmax ≤ Sy, failure could occur if enough cycles are
applied.
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Fatigue Strength
Fluctuating Stresses (6.11)
σ
σmax
t
σmin
1
2
m mean ( max min )
1
2
a alternating ( max min )
min
R
max
If σmin = - σmax, this is known as “fully-reversed” loading.
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Fatigue Strength
S-N Diagram (6.4)
Sf (fatigue strength) - stress level
at which a corresponding number
of cycles (N) will lead to failure
(crack initiation)
Se (endurance limit) - stress
level below which failure will
never occur
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Fatigue Strength
Endurance Limit (6.7)
The simplest design rule to prevent fatigue failure is
applied max Se
This is a valid concept, but not quite so simple in reality.
Se is determined experimentally.
Simple approximate Se formulas exist for steel, but must be
used carefully – better to have actual data.
Se ' 0.5Sut Sut 200 kpsi (1400 MPa)
Se ' 100 kpsi Sut > 200 kpsi
Se ' 700 MPa Sut > 1400 MPa
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where Sut = ultimate strength and Se’ = unmodified, laboratory
determined value
Fatigue Strength
Endurance Limit (6.7)
For real design we will modify Se’ to account for the surface
finish, stress concentration, temperature, etc.
These effects decrease the effective endurance limit.
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Fatigue Strength
Predicting Fatigue Life (6.8)
High-cycle fatigue life (N > 1000 cycles)
Typical S-N diagram for steel
Equationof a line (y ax c) :
log S f a(log N ) c
log Sl ' a(log103 ) c 3a c
Sl ’
log Se ' a(log106 ) c 6a c
Se ’
(log Sf)
S'
1
a log l
3
Se '
( Sl ' ) 2
c log
Se '
(log N)
S f 10c N a for103 N 106 cycles
or
N (S f 10 )
-c
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Fatigue Strength
1
a
Fatigue strength fraction
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Fatigue strength
Example
Find Sf of 1020 hot-rolled steel if the required life is
250,000 cycles, bending loads.
Given: Sut = 57 ksi for 1020 steel
Note: For steel, Sl’ = 0.9Su (bending), 0.75Su (axial), and 0.72Su
(torsion).
What is the life if Sf = 40 ksi?
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Fatigue Strength
High Cycle Fatigue
(6.9-6.10)
MAE 316 – Strength of Mechanical Components
NC State University Department of Mechanical and Aerospace Engineering
14
High Cycle Fatigue
Modified Endurance Limit (6.9)
Modified endurance limit is defined as
Se ka kb kc kd ke k f Se '
ka = surface finish factor = aSutb
Table 6-2 Surface finish factors ka
a
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b
Surface finish
MPa
(kpsi)
Ground
1.58
(1.34)
-0.085
Machine or cold drawn
4.51
(2.7)
-0.265
Hot rolled
57.7
(14.4)
-0.718
As-Forged
272.0
(39.9)
-0.995
High Cycle Fatigue
Modified Endurance Limit (6.9)
kb = size factor
Axial loading
Bending and torsion
kb = 1
kb = 0.879d-.107 (0.11 in ≤ d ≤ 2 in)
kb = 0.91d-.157 (2 < d < 10 in)
kb = 1.241d-.107 (2.79 ≤ d ≤ 51 mm)
kb = 1.51d-.157 (51 < d < 254 mm)
d is the diameter of the round bar or the equivalent diameter
(de) of a non-rotating or non-circular bar (Table 6-3).
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High Cycle Fatigue
Modified Endurance Limit (6.9)
kc = loading factor
1 (bending)
0.85 (axial)
0.59 (torsion)
kd = temperature factor
ST
(Table 6-4) or use
If endurance limit (Se’) is known, kd
SRT
equation
kd 0.975 0.432 103 T 0.115 105 T 2 0.104 108 T 3 0.595 1012 T 4
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If Se’ is not known, use kd = 1 and temperature-corrected tensile
strength (Sut) (see Example 6-5 in textbook)
High Cycle Fatigue
Modified Endurance Limit (6.9)
ke = reliability factor
Table 6-5 Reliability factors ke
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Survival Rate (%)
ke
50
1.00
90
0.89
95
0.87
98
0.84
99
0.81
99.9
0.75
99.99
0.70
High Cycle Fatigue
Modified Endurance Limit (6.9)
kf = miscellaneous-effects factor
Corrosion
Electrolytic plating
Metal Spraying
Cyclic frequency
Frettage corrosion
If none of the above conditions apply, kf = 1
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High Cycle Fatigue
Fatigue Stress Concentration Factor (6.10)
Kf = fatigue stress concentration factor
Kf = 1 + q(Kt – 1)
q = notch sensitivity
Kt = stress concentration factor
Kf can be used to reduce Se (multiply Se by 1/Kf) or to modify the
nominal stress (σmax = Kfσnom).
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High Cycle Fatigue
Fatigue Stress Concentration Factor (6.10)
Figure 6-20 Notch sensitivity for bending and axial
Figure 6-21 Notch sensitivity for torsion
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High Cycle Fatigue
Example
For the plate shown below, find the maximum allowable load F
for the plate to have infinite life.
Given: 1018 cold-drawn steel, Sy = 373 MPa, Sut = 442 MPa
t = 10 mm
d = 12 mm
w=
60 mm
F
F
F
t
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High Cycle Fatigue
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Column Design