fatigue strength (6.4, 6.7-6.8, 6.11)
DESCRIPTION
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. Fatigue Strength (6.8). Up to now, we have designed structures for static loads. t. w. P. P. P. ( σ max is also constant). d. t. - PowerPoint PPT PresentationTRANSCRIPT
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Fatigue Strength(6.4, 6.7-6.8, 6.11)
MAE 316 – Strength of Mechanical ComponentsNC State University Department of Mechanical and Aerospace Engineering
Fatigue Strength1
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Column Design2
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Column Design3
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Column Design4
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Fatigue Strength (6.8)
Fatigue Strength5
Up to now, we have designed structures for static loads.
PP wd
t
ySmax
P
t
(σmax is also constant)
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Fatigue Strength (6.8)
Fatigue Strength6
What if loading is not constant?
Even if σmax ≤ Sy, failure could occur if enough cycles are applied.
P
t
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Fluctuating Stresses (6.11)
If σmin = - σmax, this is known as “fully-reversed” loading.
Fatigue Strength7
σ
t
σmax
σmin
max
min
minmax
minmax
)(2
1
)(2
1
R
galternatina
meanm
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S-N Diagram (6.4)
Fatigue Strength8
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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Endurance Limit (6.7)
The simplest design rule to prevent fatigue failure is
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.
where Sut = ultimate strength and Se’ = unmodified, laboratory determined value
Fatigue Strength9
eapplied S max
' 0.5 200 kpsi (1400 MPa)
' 100 kpsi > 200 kpsi
' 700 MPa > 1400 MPa
e ut ut
e ut
e ut
S S S
S S
S S
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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.
Fatigue Strength10
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High-cycle fatigue life (N > 1000 cycles) Typical S-N diagram for steel
Predicting Fatigue Life (6.8)
Fatigue Strength11
cNaS f
)(loglog
:c)ax(y line a ofEquation
cacaSl 3)10(log'log 3
cacaSe 6)10(log'log 6
'
)'(log
'
'log
3
1
2
e
l
e
l
S
Sc
S
Sa
(log Sf)
(log N)
Se’
Sl’
a-cf
acf
)(SN
NNS
1
63
10
or
cycles 1010for 10
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Fatigue strength fraction
Fatigue strength12
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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?
Fatigue Strength13
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High Cycle Fatigue(6.9-6.10)
MAE 316 – Strength of Mechanical ComponentsNC State University Department of Mechanical and Aerospace Engineering
High Cycle Fatigue14
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Modified Endurance Limit (6.9)
Modified endurance limit is defined as
ka = surface finish factor = aSutb
High Cycle Fatigue15
'e a b c d e f eS k k k k k k S
a 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
Table 6-2 Surface finish factors ka
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Modified Endurance Limit (6.9)
kb = size factor Axial loading
kb = 1
Bending and torsion 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).
High Cycle Fatigue16
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Modified Endurance Limit (6.9)
kc = loading factor 1 (bending) 0.85 (axial) 0.59 (torsion)
kd = temperature factor If endurance limit (Se’) is known, or use
equation
If Se’ is not known, use kd = 1 and temperature-corrected tensile strength (Sut) (see Example 6-5 in textbook)
High Cycle Fatigue17
3 5 2 8 3 12 40.975 0.432 10 0.115 10 0.104 10 0.595 10dk T T T T
(Table 6-4)Td
RT
Sk
S
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Modified Endurance Limit (6.9)
ke = reliability factor
High Cycle Fatigue18
Table 6-5 Reliability factors ke
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
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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
High Cycle Fatigue19
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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).
High Cycle Fatigue20
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Fatigue Stress Concentration Factor (6.10)
High Cycle Fatigue21
Figure 6-20 Notch sensitivity for bending and axial
Figure 6-21 Notch sensitivity for torsion
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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
High Cycle Fatigue22
FF w =60 mm
d = 12 mm
t = 10 mm
F
t
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Column Design23