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
1
Fatigue Strength
2
Column Design
3
Column Design
4
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
5
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.

6
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.

7
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
8
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

9
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.
10
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
11
Fatigue Strength
1
a
Fatigue strength fraction

12
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?
13
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
15
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).
16
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 103  T  0.115 105  T 2  0.104 108  T 3  0.595 1012  T 4

17
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
18
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
19
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).
20
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
21
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
22
High Cycle Fatigue
23
Column Design