# Fatigue Rob

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Fatigue Failure
2.1
Endurance Limit
The R. R. Moore’s high-speed rotating beam machine is used to performing
a fatigue test. The specimen, shown in Fig. 1(a), is rotating with a constant angular speed. The specimen is subjected to pure bending by means
of weights. The intensity of the reversed stress causing failure after a given
test specimen
d
motor
(a)
n
M
M
weights
Weights
mg
σ
Se
(b)
O
t
Figure 1: R. R. Moore’s high-speed rotating beam machine
number of cycles is the fatigue strength corresponding to that number of loading cycles. The fatigue strengths considered for each test are plotted against
the corresponding number of revolutions. The resulting chart is called the
strength-life, S-N, diagram. The diagram depicts the fatigue strength versus
cycle life N of a part. The S-N curves are plotted on log-log coordinates.
For ferrous materials the endurance limit of the test specimen is defined
as the highest level of alternating stress that can be withstood indefinitely by
1
Fatigue strength S (log)
a test specimen without failure, Fig. 1(b). The symbol for endurance limit
is Se0 . The S-N diagram becomes horizontal at the endurance limit as shown
in Fig. 2.
S
Se
10
3
10
4
10
5
106
107
N
Life N (cycles) (log)
Figure 2: S-N diagram
The endurance limit is related to the ultimate tensile strength through
the relationships [5]:
0.50 Su
for Su ≤ 200 kpsi,
0
Se =
(2.1)
100 kpsi for Su &gt; 200 kpsi,
or
Se0
=
0.50 Su
700 MPa
for Su ≤ 1400 MPa,
for Su &gt; 1400 MPa.
(2.2)
The endurance limit of the test specimen, Se0 , can be different from the
endurance limit, Se , of the machine part subjected to different loadings. The
endurance limit Se is affected by a number of factors called modifying factors.
Some of these factors are the surface factor kS , the gradient (size) factor kG ,
the load factor kL , the temperature factor kT , and the reliability factor kR .
Thus, the endurance limit of a member can be related to the endurance limit
of the test specimen by the following relationship: Se = kS kG kL kT kR Se0 .
Surface factor, kS
The surface modification factor, kS , depends on the tensile strength of
the material, Sut and on the part surface finish. The surface factor formula
2
is given by [5]
b
.
kS = a Sut
(2.3)
The factor a and the exponent b depends upon the quality of the surface
finish [5]:

1.58



4.51
for (MPa) a =
57.7



272
for
for
for
for
ground surface finish,
machined or cold-drawn surface finish,
hot-rolled surface finish,
forged surface finish,
(2.4)

1.34



2.70
for (kpsi) a =
 14.4


39.9
for
for
for
for
ground surface finish,
machined or cold-drawn surface finish,
hot-rolled surface finish,
forged surface finish,
(2.5)

−0.085



−0.256
b=
−0.718



−0.995
for
for
for
for
ground surface finish,
machined or cold-drawn surface finish,
hot-rolled surface finish,
forged surface finish.
(2.6)
To evaluate the size factor the following relations are used [5]:
For d &lt; 0.4 in. or d &lt; 10 mm, where d is the diameter of the test bar
kG =
(2.7)
1
for bending and torsion.
For 0.4 in. &lt; d &lt; 2 in. or 10 mm &lt; d &lt; 50 mm
kG =
0.9
for bending and torsion.
(2.8)
For 2 in. &lt; d &lt; 4 in. or 50 mm &lt; d &lt; 100 mm the factors are reduced by
0.1.
For 4 in. &lt; d &lt; 6 in. or 100 mm &lt; d &lt; 150 mm the factors are reduced
by 0.2.
3
For the load factor the following

 0.85
1
kL =

0.59
average values are used [5]:
for axial,
for bending,
for pure torsion
(2.9)
For shear the load factor is kL = 0.577 according to [30].
Temperature factor, kT
is given by [10]
1
for T ≤ 840 ◦ F,
kT =
(2.10)
1 − (0.0032 T − 2.688) for 840 ◦ F &lt; T ≤ 1020 ◦ F.
Reliability factor, kR
The reliability factor is calculated with by [10]

1
for 50% reliability,




0.897
for 90% reliability,

0.868 for 95% reliability,
kR =


0.814 for 99% reliability,



0.753 for 99.9% reliability.
(2.11)
Juvinall and Marshek present a summary of all modifying factors for
bending, axial loading, and torsion used for fatigue of ductile materials [10].
For 103 -cycle strength:

 0.9 kT Su
0.75 kT Su for axial loads,
Sf =
(2.12)

0.9 kT Sus for torsional loads.
The ultimate shear strength, Sus , is
0.8 Su for steel,
Sus ≈
0.7 Su for other ductile materials.
(2.13)
For 106 -cycle strength (endurance limit) for axial, bending, and torsional
Se = kS kG kL kT kR Se0 .
4
(2.14)
2.2
Fluctuating Stresses
The components of the time varying stresses are depicted in Fig. 3, where
σmin is minimum stress, σmax the maximum stress, σa the stress amplitude
or the alternating stress, σm the midrange or the mean stress, and σr the
stress range. The mean stress and the alternating stress are
σmax + σmin
,
(2.15)
σm =
2
σmax − σmin
.
(2.16)
σa =
2
The stress ratios are defined by
σmin
σa
R=
and A =
.
(2.17)
σmax
σm
The steady stress or static stress, σs , has any value between σmin and σmax
and exists because of a fixed load. Figure 3(a) shows a sinusoidal fluctuating
stress, Fig. 3(b) represents a repeated stress, and Fig. 3(c) is a completely
reversed sinusoidal stress. A fluctuating stress is a combination of static plus
completely reversed stress.
2.3
Constant Life Fatigue Diagram [10]
A graphical representation of mean and alternating stress in relation to yielding and various fatigue life is shown in Fig. 4. The horizontal axis, σa = 0, corresponds to static loading. The vertical axis is given by σm = 0. The points
on the vertical axis, ordinate axis, (0, Se ), (0, S103 ), (0, S104 ), (0, S105 ), ... are
selected from the S-N diagram. The point (Su , 0) represents the ultimate tensile strength. The lines (0, Se )−(Su , 0), (0, S105 )−(Su , 0), (0, S104 )−(Su , 0)
and (0, S103 ) − (Su , 0) are the estimated lines of constant life. The diagram,
σm − σa , is called the constant life fatigue diagram because it has lines corresponding to a constant 106 cycle or “infinite” life, constant 105 cycle, constant
104 cycle, and so forth. These lines are called the Goodman lines.
The point (Sy , 0) represents the yield strength and for ductile materials the point (−Sy , 0) represents the compressive yield strength. At the
point (0, Sy ) the stress fluctuates between +Sy and −Sy . The points on line
(Sy , 0) − (0, Sy ) correspond to fluctuations having a tensile peak of Sy . The
points on line (−Sy , 0) − (0, Sy ) correspond to fluctuations having a compressive peak of Sy . Within the area defined by the lines (Sy , 0) − (0, Sy ),
(−Sy , 0) − (0, Sy ) there is no yielding.
5
fluctuating stress
σ
σa
σr
(a)
σa
σmax
σm
σmin
O
σ
t
repeated stress
σa
(b)
σmax
σr
σa
σm
t
O σmin = 0
σ
reversed stress
σmax σa
(c)
O
σr
σm = 0
t
σmin σa
Figure 3: Time varying stresses: a) sinusoidal fluctuating stress; (b) repeated
stress; and (c) reversed sinusoidal stress.
6
Peak alternating stress S (log)
S
S-N diagram
S103
S104
S105
Se
10 3
10
4
10
5
10
6
N
10
7
Life N (cycles (log))
Constant-Life Curves
S103
103
S104
S105
Sy
S106 = Se
104
Se
105
106
σa
σm
−Sy
σm (compression)
Sy
σm (tension)
O
Su
Figure 4: Constant life fatigue diagram
The shaded area in Fig. 5(a) corresponds to a life of at least 106 cycles
and no yielding. The shaded area in Fig. 5(b) corresponds to a life of at
least 106 cycles and yielding. The shaded area in Fig. 5(c) corresponds to
less than 106 cycles of life and no yielding.
For torsional loads Ssy and Sus are used instead of Sy and Su with
Ssy ≈ 0.58 Sy
and Sus ≈ 0.8 Su
7
for steel.
(2.18)
σa
S103
103
S104
S10 5
S106 = Se
Sy
(a)
104
Se
105
106
Sy
−Sy
Su
O
σa
S103
103
S104
S10 5
S106 = Se
Sy
104
Se
−Sy
(b)
105
106
Sy
O
Su
σm
σa
S103
103
S104
S10 5
σm
Sy
104
S106 = Se
Se
(c)
5
10
6
10
Sy
−Sy
Su
O
Figure 5: Constant life fatigue diagrams
8
σm
2.4
Arbitrarily Varying Forces and Fatigue Life
An arbitrarily varying force acts on a part. The part is subjected to σ1 for
n1 cycles, σ2 for n2 cycles, and so forth. The procedure for estimating the
fatigue life of the part is the linear cumulative damage rule (or Miner’s rule)
and is expressed by the following equation [10, 30]:
k
X
n2
nk
nj
n1
+
+ ... +
= 1 or
=c
N1 N2
Nk
Nj
j=1
(2.19)
where n1 , n2 , &middot; &middot; &middot; , nk represent the number of cycles at specific levels of stress
σ1 , σ2 , &middot; &middot; &middot; , σk and N1 , N2 , &middot; &middot; &middot; , Nk represent the life (in cycles) at these stress
levels, as taken from the appropriate S-N curve. The parameter c has been
determined by experiment and 0.7 &lt; c &lt; 2.2 with an average value near 1.
Failure results when c = 1.
2.5
There are various techniques for plotting the results of the fatigue failure test
of a part subjected to fluctuating stress.
A stress state can be described by the mean and the alternating components. This stress state is characterized by a point and if the point is
situated below the constant-life line then safety is suggested. A point on the
constant-life line gives the strength Sm as the limiting value of σm . In a similar way the corresponding strength Sa is the limiting value of σa , as shown
in Fig. 6. The alternating strength Sa as a limiting value of σa is plotted
on the ordinate. The mean strength is Sm as a limiting value of σm and is
plotted on the abscissa. The fatigue limit Se is plotted on the ordinate. The
tensile yield strength Sy is plotted on both coordinate axes. The ultimate
tensile strength Su is plotted on the abscissa.
Four criteria of failure are shown in the diagram in Fig. 6 that is, Soderberg, the modified Goodman, Gerber, and yielding. The Soderberg criterion
protects against yielding.
The equation for the Soderberg line is
Sm Sa
+
= 1.
Sy
Se
9
(2.20)
Sy
yield line
Soderberg line
modified Goodman line
Gerber line
Se
Sa
d
loa
e
lin
O
Sm
Sy
Su
Figure 6: Fatigue diagram
The equation for the modified Goodman line is
Sm Sa
+
= 1.
Su
Se
(2.21)
The yielding line is described by the equation
Sm Sa
+
= 1.
Sy
Sy
(2.22)
The Gerber criterion is also called the Gerber parabolic relation because the
curve can be modeled by a parabolic equation of the form
2
Sa
Sm
+
= 1.
(2.23)
Se
Su
If SF is a design or safety factor then
Sa = SF σa
and Sm = SF σm .
10
(2.24)
The Soderberg equation becomes
1
σa σm
+
=
,
Se
Sy
SF
(2.25)
the modified Goodman equation becomes
σa σm
1
+
=
,
Se
Su
SF
and the Gerber equation becomes
2
SF σm
SF σa
+
= 1.
Se
Su
Some values of the tensile strength are given in Table 2.1[38].
11
(2.26)
(2.27)
Table 2.1 Tensile strength
Steel
(AISI = American Iron and Steel Institute)
tensile strength
ultimate
yield
Sut (MPa) Sy (MPa)
AISI 1006 Steel, cold drawn
330
285
AISI 1006 Steel, hot rolled bar, 19-32 mm round
295
165
AISI 1006 Steel, cold drawn bar, 19-32 mm round
330
285
AISI 1008 Steel, hot rolled bar, 19-32 mm round
305
170
AISI 1008 Steel, cold drawn bar, 19-32 mm round
340
285
AISI 1010 Steel, cold drawn
365
305
AISI 1010 Steel, hot rolled bar, 19-32 mm round or thickness
325
180
AISI 1010 Steel, cold drawn bar, 19-32 mm round or thickness
365
305
AISI 1012 Steel, cold drawn
370
310
AISI 1012 Steel, hot rolled bar, 19-32 mm round or thickness
330
185
AISI 1012 Steel, cold drawn bar, 19-32 mm round or thickness
370
310
AISI 1015 Steel, cold drawn
385
325
AISI 1015 Steel, cold drawn, 19-32 mm round
385
325
AISI 1015 Steel, hot rolled, 19-32 mm round
345
190
AISI 1015 Steel, as rolled
420
315
◦
◦
AISI 1015 Steel, normalized at 925 C (1 700 F)
425
325
◦
◦
AISI 1015 Steel, annealed at 870 C (1 600 F)
385
285
AISI 1016 Steel, cold drawn, 19-32 mm round
420
350
AISI 1016 Steel, hot rolled, 19-32 mm round
380
205
AISI 1017 Steel, cold drawn
405
340
AISI 1017 Steel, hot rolled, 19-32 mm round
365
200
AISI 1018 Steel, cold drawn
440
370
AISI 1018 Steel, hot rolled, quenched, and tempered
475
275
AISI 1018 Steel, hot rolled, 19-32 mm round
400
220
AISI 1018 Steel, as cold drawn, 16-22 mm round
485
415
AISI 1018 Steel, as cold drawn, 22-32 mm round
450
380
AISI 1018 Steel, as cold drawn, 32-50 mm round
415
345
AISI 1018 Steel, as cold drawn, 50-76 mm round
380
310
AISI 1019 Steel, cold drawn
455
379
AISI 1019 Steel, hot rolled, 19-32 mm round
407
224
AISI 1020 Steel, cold rolled
420
350
Source: MatWeb - Material Property Data, http://www.matweb.com/
12
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14
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15
% fatigue
% S-N diagram
clear all; clc; close all
% alternating stresses are torsional
% steel bar, d&lt;2in., fine ground surface
Su = 110; % ultimate strength (kpsi)
Sy = 77; % yield strength
(kpsi)
%
%
%
%
S
peak alternating strength
10^3 cycle strenght S
torsional load: S = 0.9 Sue;
Sue=0.8 Su (for steel); Sue = 0.7 Su (other)
= 0.9*0.8*Su;
% peak alternating strength
% 10^6 cycle strenght (endurance limits)
% Se = kL kG kL Sep
% endurance limit of test speciment Sep
% Sep = 0.5 Su (for Su&lt;1400 MPa)
% Sep = 700 MPa (for Su&gt;1400 MPa)
% bending, axial, torsion: Sep = 0.5 Su
Sep = 0.5*Su;
% modifying factors for endurance limit
% surface factor kS (bending, axial, torsion)
a = 1.34;
% fine ground surface
b = -0.085; % fine ground surface
kS = a*Su^b;
% bending and torsion:
% kG = 0.9; % for 0.4&lt;d&lt;2 in
kG = 0.9;
kL = 0.577;
% temperature factor, kT
% kT = 1 for T&lt;840 degF
% kT = 1-(0.0032*T-2.688) for 840&lt;T&lt;1020 degF
kT = 1;
% reliability factor, kR
% kR = 1 for 50% reliability
kR = 1;
% endurance limit Se = kS kG kL kT kR Sep
Se = kS*kG*kL*kT*kR*Sep;
% Se = 25.667 (kpsi)
% S-N diagram
LS = log10(S);
LSe= log10(Se);
% straight line equation y = m x + b
% slope m = (y1-y2)/(x1-x2)
Lm = (LSe-LS)/(6–3);
% y-intercept b = y1 – m x1
Lb = LS-3*Lm;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% print results
fprintf('Su = %6.3f (kpsi) \n',Su)
fprintf('Sy = %6.3f (kpsi) \n',Sy)
fprintf('10^3 cycle strenght S = %6.3f (kpsi) \n',S)
fprintf('Sep = %6.3f (kpsi) \n',Sep)
fprintf('kS = %6.3f \n',kS)
fprintf('kG = %6.3f \n',kG)
fprintf('kL = %6.3f \n',kL)
fprintf('10^6 cycle strenght Se = %6.3f (kpsi) \n',Se)
fprintf('\n')
fprintf('S = %6.3f(kpsi) -&gt; log(S) = %6.3f \n',S,LS)
fprintf('Se = %6.3f(kpsi) -&gt; log(Se) = %6.3f \n',Se,LSe)
% plot the graph
syms x
% eq. for S-Se
ye = Lm*x+Lb;
xn = 3:1:6;
yn = Lm*xn+Lb;
hold on
plot(xn,yn,'k-','LineWidth',2)
plot([6,7],[LSe,LSe],'b-','LineWidth',2)
title('S-N diagram')
ylabel('log S','FontSize',14)
xlabel('log N','FontSize',14)
axis([3 7 1 LS+.25])
grid
ax = gca;
ax.XTick = [3,4,5,6,7];
ax.YTick = [LSe,LS];
% eof
%
%
%
%
%
%
%
%
%
%
%
Su = 110.000 (kpsi)
Sy = 77.000 (kpsi)
10^3 cycle strenght S = 79.200 (kpsi)
Sep = 55.000 (kpsi)
kS = 0.899
kG = 0.900
kL = 0.577
10^6 cycle strenght Se = 25.667 (kpsi)
S = 79.200(kpsi) -&gt; log(S) = 1.899
Se = 25.667(kpsi) -&gt; log(Se) = 1.409
Su = 110.000 (kpsi)
Sy = 77.000 (kpsi)
10^3 cycle strenght S = 79.200 (kpsi)
Sep = 55.000 (kpsi)
kS = 0.899
kG = 0.900
kL = 0.577
10^6 cycle strenght Se = 25.667 (kpsi)
S = 79.200(kpsi) -&gt; log(S) = 1.899
Se = 25.667(kpsi) -&gt; log(Se) = 1.409
Published with MATLAB&reg; R2015a
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