School of Engineering – EN334
M. Daniel K. Wood
DESIGN FOR FATIGUE STRENGTH
• In real life situation, mechanical elements are not only loaded
statically, but they are also loaded in such a manner that the
stresses in the members vary from a maximum value to a minimum
value during infinite number of cycles.
• A shock absorber in a car is a typical example where the springs are
loaded cyclically as the car is driven through a rough road that has
several holes. The springs are repeatedly loaded by forces that are
at one time at a maximum value and at the other time at a minimum
value.
• The same can be said of a rotating shaft that experiences bending
moments. The effect of this is that at one time some fibres on the
surface experience compressive stress and at other times they
experience tensile stress. This variation between compressive and
tensile stresses may be repeated several times within one minute,
depending on the speed of rotation.
School of Engineering – EN334
M. Daniel K. Wood
• Stresses of this nature are referred to as fluctuating stresses and
they result in mechanical members failing under fatigue failure
mode.
• In fatigue failure, ten million or 107 cycles are referred to as an
infinite life. What this means is that if a shaft rotates ten million
times, then it is assumed that it has attained its design life.
• Fatigue failure mode is very dangerous to mechanical parts because
the stress required to cause it, is normally below the ultimate
strength and the yield strength of the material.
• The designer should be familiar with this kind of failure mode so that
care should be taken to design a machine part that is resistant to the
failure mode.
• In fact, stress concentration factor is linked with fatigue failure. A
small crack developed on a turbine blade is dangerous and it can
indeed cause a major failure. This is because a small crack can
propagate very easily under fluctuating stresses and can very easily
lead to a catastrophic failure of the engine.
School of Engineering – EN334
M. Daniel K. Wood
• The concept of fluctuating stresses is shown in Figure 4.1. For
fluctuating stresses the stress at a point, is given by a range stress,
r, and a mean stress, m. These stresses, are functions of the
maximum stress and minimum stress respectively.

a
m
max
min
t
Figure 4.1 Range and mean stresses
School of Engineering – EN334
M. Daniel K. Wood
•
The maximum stress is given as:
 max   min
  max   min 
m  
   min 
2
2


•
(4.1)
The stress range is given as
 r   max   min
a 
 max   min
2
(4.2)
School of Engineering – EN334
M. Daniel K. Wood
Endurance Limit
• Engineering parts, which fail under fatigue loading, experience
extreme stresses, σmax and σmin. Such parts are more likely to fail
than those parts that experience only maximum stress, σmax. The
stress amplitude, at which the machine member will fail after a given
number of stress-cycle is known as the fatigue strength, designated
as Sn. As n approaches ∞ the fatigue strength approaches the
endurance limit, designated as S’n, as shown in Figure 4.2.
Sn
Figure 4.2
S n
n (stress cycle)
School of Engineering – EN334
M. Daniel K. Wood
• However, surface imperfections due to manufacturing processes
and environmental affects negatively impacts the endurance limit,
and results in lower values compared to those published. Therefore
a modified endurance limit is given as
S e  k sf k r k s kt k m S n
– where
ksf = surface finish factor
kr = reliability factor
ks = size factor
kt = temperature factor
km = stress concentration modifying factor
= inverse of fatigue stress concentration factor
(km = 1/kf)
ksf = surface finish factor
S n = half of the ultimate strength of the material (Su/2)
School of Engineering – EN334
M. Daniel K. Wood
•
•
ksf : The surface finish factor depends on the quality of the finish and the
tensile strength of the material. Graphs of these are normally available.
kr : The reliability factor can be obtained from the Table 4.1.
Table 4.1 Reliability factors
Reliability Reliability factor, kr
•
50%
1
90%
0.897
95%
0.868
99%
0.814
ks : The size factor is obtained from the following relations:
1 if d  8m m


k s  0.85 if 8m m  d  50m m
 0.75 if d  50m m

School of Engineering – EN334
M. Daniel K. Wood
• kt : The temperature factor is obtained from the relation
620


k t   460  T
if T  160o F
o

1
if
T

160
F

• km : The stress concentration modifying factor is given as
km  1 k f
• where the fatigue stress concentration factor kf, is given as
k f  1  q(k st  1)
School of Engineering – EN334
M. Daniel K. Wood
• and q is the notch sensitivity usually given graphically and is a
function of the material’s ultimate strength and notch radius of the
part. The stress concentration factor, kst, is also a function of the
part’s geometry. When q and kst have been established, kf can then
be obtained.
• For no notch round shaft, kt = 1, which implies kf = 1, and hence km
= 1. By using these modified factors in equation (4.3), the modified
endurance limit can then be obtained.
School of Engineering – EN334
M. Daniel K. Wood
Fatigue Strength
Figure 4.3.
Goodman diagram
Alternating Stress Component σa
• When the mean stress and the stress range are varied, the fatigue
resistance of parts subjected to these fluctuating stresses can be
studied using Goodman diagram shown in Figure 4.3.
Midrange stress σm
School of Engineering – EN334
M. Daniel K. Wood
• The diagram is drawn by marking out the yield strength of a material
in both the x- and y-axes, Sy.
• The ultimate strength of the material is marked out in the x-axis, Sult,
and this is usually greater than the yield strength .
• The endurance limit is marked out in the y-axis, Se.
• The points corresponding to the yield points on the x-axis and y-axis
are connected by a straight line, so are the points corresponding to
the ultimate strength along the x-axis and the endurance limit along
the y-axis.
• A region is defined by the intersection of the endurance limit-ultimate
strength line and the yield strength-yield strength lines, within which
design is feasible (shaded area).
• The stress range is associated with the y-axis and the mean stress
is associated with the x-axis. By plotting a line defined by the
gradient obtained from the loads related to the stress range and
mean stress respectively, it is possible to locate the stress range
and mean stress values.
School of Engineering – EN334
M. Daniel K. Wood
Figure 4.4
Deriving mean
stress and stress
range
from Goodman
diagram
Alternating Stress Component σa
• As the mean stress increases, the semi-range reduces. Within the
region bounded by the yield strength-yield strength line and ultimate
strength-endurance limit line, the sum of the mean stress and the
stress range is equal to the yield strength.
• The stress range and the mean stress in a machine member can be
estimated if we superimpose a line whose gradient is given by the
ratio of the stress range to mean load as shown in Figure 4.4.
Midrange stress σm
School of Engineering – EN334
M. Daniel K. Wood
• The maximum stress is given as
Wmean KWsr
ˆ
S  S mean  S sr 

A
A
(4.8)
• The mean stress is given as
S mean
Wmean

A
(4.9)
• The stress range is given as
KW sr
S sr 
A
(4.10)
School of Engineering – EN334
M. Daniel K. Wood
• The ratio of the stress range to the mean stress is given as follows:
S sr
KWsr

S mean Wmean
(4.11)
• The ratio of the stress range to the mean stress given in equation
(4.11) is a much quicker way of obtaining the stress range and mean
stress from the Goodman diagram if the associated loads are
known.
School of Engineering – EN334
M. Daniel K. Wood
Example 4.1
• Problem:
An AISI 1020 CR steel bar is to withstand a tensile pre-load of
356kN and a fluctuating tensile load varying from 0 to 71.2kN. The
steel bar has a geometric stress-concentration factor of 1.995
corresponding to a fillet radius of 5mm. Assume the notch sensitivity
is 0.70, and the surface factor ksf = 0.78. Determine the bar
diameter for a factor of safety of 1.5 and an infinite life.
• Solution:
From the appendix (material properties), the yield and ultimate
stresses for AISI 1020 CR steel are Sy = 414 x106 N/m2 and Su =
552 x106 N/m2 respectively.
School of Engineering – EN334
M. Daniel K. Wood
S’n= half of the ultimate strength of the material (Su/2) = 0.5×552 ×106
N/m2 = 276 ×106 N/m2.
• ksf = surface finish factor = 0.78 (given)
• kr = reliability factor = 1 (assume since no information is given)
• ks = size factor = 0.85 (mid-value assumed since initial diameter is
not known)
• kt = temperature factor = 1 (assume since no information is given)
• k f  1  q(k st  1) = fatigue stress concentration factor
= 1 + 0.7×(1.995 – 1) = 1.6965
• km = 1 / kf = the stress concentration modifying factor
= 1/1.6956 = 0.5894
• Therefore, Se  ksf kr ks kt km Sn = 0.78×1×0.85×1×0.5894×276 ×106
= 107.85 ×106 N/m2 (Pa).
School of Engineering – EN334
M. Daniel K. Wood
• The stress induced in the steel bar is:
Fs 35.6 x103 N 142.4 x103 N
s 


2
A
d
d 2
4
• The stress range in the steel bar is:
Fr 71.2 x103 N 284.8 x103 N
r 


2
A
d
d 2
4
School of Engineering – EN334
M. Daniel K. Wood
• The stress amplitude in the steel bar is:
r
142.4 x103 N
a 

2
d 2
• The mean stress in the steel bar is:
142.4 x103 N 142.4 x103 N 284.8 x103 N
m  s a 


2
2
d
d
d 2
• We now use the Goodman diagram to find the stress range as
follows:
School of Engineering – EN334
M. Daniel K. Wood
• The load line can now be drawn with a gradient of  a  m  0.5 .
• Using basic knowledge of intersection of straight line, the points of
intersection can be found from the ultimate strength-endurance limit
line, and the superimposed design line-equations as follows:
• Ultimate strength-endurance limit line:
Y1  
107 .85
X  107 .85  0.195 X  107 .85
552
• Design line equation:
Y2  0.5 X
• Where Y is the alternating stress and X is the midrange stress
School of Engineering – EN334
M. Daniel K. Wood
• At the point of intersection, Y1 = Y2. Solving the simultaneous
equation leads to:
 0.195 X  107 .85  0.5 X
• We find the value of X = 155.179Pa. The value of Y1 is found as
follows:
Y1  0.195X  107.85  0.195155.179 107.85
 77.59x106 N / m2
• Using the relation
a
S sr
 nf
School of Engineering – EN334
M. Daniel K. Wood
142.4 x103 N
1

 1.5
2
6
d
77.59x10 N
• Solving for the diameter, Voila!
142.4 x103
d
 0.0197m (19.7mm)
6
77.59x10 *1.5