Jariguthundi dii so) at all AAP so so

1. 7
Quantifying Fatigue Failure:
Stress-Life Method for
Non-Zero Mean Stress
(Failure Surface = “Goodman Diagram”)
Variation in cyclic stress.
2
min
max σ
σ
σ
+
=
m
R
R
A
m
a
+
−
=
+
−
=
=
1
1
min
max
min
max
σ
σ
σ
σ
σ
σ
2
min
max σ
σ
σ
−
=
a
Mean stress:
max
min
σ
σ
=
R
Stress Amplitude:
Stress ratio:
Amplitude ratio:
6-11 Characterizing Fluctuating Stresses
2
min
max F
F
Fm
+
=
2
min
max F
F
Fa
−
=
Terms for Stress Cycling
Stress
at
A
t
mid-range stress, σm
load reversals (2 reversals = 1 cycle)
stress amplitude, σa
maximum stress, σmax
minimum stress, σmin
Follow a material point A
on the outer fiber
Schematic: Effect of Midrange Stress Sm = σm|Nf =∞
Se
Su
Sf|Nf ∞
N
f = ∞
N
f
∞
Sa
Sm
(Se = Sf|Nf =∞)
Alternating stress
amplitude (σa ) at failure,
or fatigue strength Sa
Mid-range stress (σm )
at failure, or mid-range
strength Sm
Sut
103
106
N
Se
S-N diagram for
Sm = 0
Sf = Sa|Sm=0

2. 8
Data: Effect of Midrange Stress σm
That is: the endurance
limit drops as σm|Nf
= Sm
increases
Simple Model to Account for Midrange Stress σm
That is: the endurance
limit drops as σm|Nf
= Sm
increases
Goodman Diagram for Fatigue
This line is known as
the Goodman diagram
Goodman diagram ≈
“endurance limit as a
function of mean
stress”
Goodman diagram =
drop in Se for rise in
tensile Sm
Mean stress, σm , as mid-range
strength Sm
Alternating stress amplitude, σa ,
as fatigue strength Sa
Goodman Diagram: Fatigue Failure with σm ≠ 0
Se
Sa
N
f ∞
N
f = ∞
1
=
+
u
m
e
a
S
S
S
S
Equation of Goodman line:
1
=
+
u
m
e
a
S
S
S
S
“Goodman Criterion” for ∞-­‐life:
Sm
Su
(Se = Sa|Nf =∞)
Sf = σa|Nf ∞
“Goodman Criterion” for finite
life:
1
=
+
u
m
f
a
S
S
S
S
( ) ( )
1
=
+
n
S
n
S u
m
f
a σ
σ
or
( ) ( )
1
=
+
n
S
n
S u
m
e
a σ
σ
or
↑ failure surface
↑ design equation
(i.e., shrink failure
surface until you hit
service loads)
↑ failure surface
↑ design equation

3. 9
Mean stress, σm , as mid-range
strength Sm
Alternating stress amplitude, σa ,
as fatigue strength Sa
Goodman Diagram: Fatigue Failure with σm ≠ 0
Se
Sa
N
f ∞
N
f = ∞
1
=
+
u
m
e
a
S
S
S
S
1
=
+
u
m
e
a
S
S
S
S
“Goodman Criterion” for ∞-­‐life:
Sm
Su
(Se = Sa|Nf =∞)
Sf (Nf ∞)
“Goodman Criterion” for finite
life:
1
=
+
u
m
f
a
S
S
S
S
1
=
+
u
m
f
a
S
n
S
n σ
σ
or
1
=
+
u
m
e
a
S
n
S
n σ
σ
or
↑ failure surface
↑ design equation
(equivalently, expand
service loads until you
hit the failure surface)
↑ failure surface
↑ design equation
Equation of Goodman line:
Mean stress, σm , as mid-range
strength Sm
Alternating stress amplitude, τa ,
as fatigue strength Sas
Goodman Diagram for Torsion: Failure with τm ≠ 0
Ses
Sas
N
f ∞
N
f = ∞
1
=
+
us
ms
es
as
S
S
S
S
“Goodman Criterion” for ∞-­‐life:
For pure torsion, use Se,shear =
kc·∙Se,tension
Su,shear = 0.67 Su,tension
Sm
Su
(Ses = Sas|Nf =∞)
Sfs (Nf ∞)
“Goodman Criterion” for finite
life:
1
=
+
us
ms
fs
as
S
S
S
S
n
S
S us
m
fs
a 1
=
+
τ
τ
or
n
S
S us
m
es
a 1
=
+
τ
τ
or
experimental correlation
1
=
+
us
ms
es
as
S
S
S
S
↑ failure surface
↑ design equation
↑ failure surface
↑ design equation
Equation of Goodman line:
Sa-Intercept When There Is No Clear Endurance Limit
[Elements of the Mechanical Behavior of Solids, by N. P. Suh and Arthur P.
L. Turner, McGraw-Hill, 1975, p. 471]
Set Se = Sa|
Nf=10e8
E.g. 3. Goodman Diagram for Non-Zero Mean Stress
( )
MPa
66
.
63
MPa
9
.
15
636620
2
01
.
0
2
m
N
100
m
N
25
kN
2
kN
5
.
0
m
05
.
0
3
4
≤
≤
⋅
=
=
=
=
⋅
≤
≤
⋅
⇒
≤
≤
⋅
=
⋅
=
τ
π
π
τ T
T
c
Tc
J
Tc
T
F
F
d
F
T
MPa
9
.
92
MPa
2
.
23
or
)
66
.
63
(
46
.
1
)
9
.
15
(
46
.
1
toe
toe
≤
≤
≤
≤
⇒
τ
τ
Static shear-stress concentration is Kts=1.6
at the toe of the 3-mm fillet: τtoe → Ktsτnom
Calculate the shear-stress concentration factor
with Kts=1.6 at the toe of the 3-mm fillet:
( )
1
1 −
+
= ts
s
fs K
q
K
( ) ( ) 46
.
1
1
6
.
1
775
.
0
1
1
1 =
−
+
=
−
+
= ts
s
fs K
q
K
Shaft A: 1010 hot-rolled steel
Kts= 1.6 at the 3-mm fillet
F = 0.5
to 2 kN cycle
Sut = 320 MPa
Find safety factor against failure of shaft A.
Sut = 320 MPa too low
to register on Fig. 6-21.
Use Eqn. (6-35b)
toe of fillet

4. 10
( ) MPa
4
.
214
320
67
.
0
67
.
0 =
=
= ut
us S
S
39
.
1
1
4
.
214
1
.
58
9
.
77
9
.
34
Goodman =
⇒
=
+
=
+ n
n
S
S us
m
es
a τ
τ
08
.
58
2
21
.
23
94
.
92
,
86
.
34
2
21
.
23
94
.
92
=
+
=
=
−
= m
a τ
τ
E.g. 3. Non-Zero Mean Stress
Convert ultimate tensile to ultimate shear:
Convert ideal to actual endurance limit:
( ) MPa
160
320
5
.
0
5
.
0 =
=
=
ʹ′ ut
e S
S
Approximate ideal endurance limit:
n
S
S us
m
es
a 1
=
+
τ
τ
Safety factor based on Goodman criterion:
Stress amplitude and mid-range stress:
(kc converts tensile endurance to shear endurance!)
( )
( )
MPa
9
.
77
)
160
)(
59
.
0
)(
9
.
0
(
917
.
0
59
.
0
9
.
0
20
24
.
1
917
.
0
320
7
.
57
59
.
0
107
.
0
718
.
0
=
=
=
=
=
=
=
=
=
ʹ′
=
=
−
−
c
k
e
es
c
b
b
ut
a
e
f
e
d
c
b
a
e
S
S
k
k
aS
k
S
k
k
k
k
k
k
S
(95% stress area is the same as R. R. Moore specimen)
Recall that for τm = 0 (E.g. 2),
nfully-reversed = (23.2/77.9)−1 = 3.36,
so the mid-range stress reduces the
fatigue resistance of a material!
Various Criteria of failure
Gerber
Mod. Goodman
Soderberg
6-12 Fatigue Failure Criteria
n
S
S y
m
e
a 1
=
+
σ
σ
1
2
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
ut
m
e
a
S
n
S
n σ
σ
1
2
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
y
m
e
a
S
n
S
n σ
σ
n
S
S ut
m
e
a 1
=
+
σ
σ
ASME-elliptic
Langer StaticYield
n
Sy
m
a =
+σ
σ
“First-Cycle Yield”: Goodman Fatigue Line and Langer Yield Line
Se
Sa
(Se = Sa|Nf =∞)
1
=
+
y
m
y
a
S
S
S
S
Modified
Goodman
1
=
+
u
m
e
a
S
S
S
S
Sm
Su
Sy
Sy
Langer
for this lower value of
r, yield happens
before fatigue!
generic load
line r = Sa /Sm
h
i
g
h
e
r
v
a
l
u
e
o
f
r
lower value of r
Table 6-6 lists the coordinates on the Sa-Sm diagram where the
load line intersects the Langer and the Goodman lines
(Goodman)
These are two
eqns in two
unknowns Sa and
Sm, as r = σa/σm is
considered
known)
These are two
eqns in two
unknowns Sa and
Sm, as r = σa/σm is
considered
known)
Two eqns in two
unknowns Sa
and Sm
“First-Cycle Yield”: Goodman Fatigue Line and Langer Yield Line

5. 11
( ) MPa
4
.
214
320
67
.
0
67
.
0 =
=
= ut
us S
S
MPa
9
.
77
)
160
)(
59
.
0
)(
9
.
0
(
917
.
0
59
.
0
=
=
ʹ′
=
= = e
f
e
d
c
b
a
k
e
es S
k
k
k
k
k
k
S
S c
08
.
58
2
21
.
23
94
.
92
86
.
34
2
21
.
23
94
.
92
=
+
=
=
−
=
m
a
τ
τ
Revisit E.g. 3, Non-Zero Mean Stress, Test for First-Cycle Yield
6
.
0
08
.
58
86
.
34
=
=
=
m
a
r
τ
τ
Ses = 77.9
Sa
Goodman
Sm
Sus = 214.4
Sys = 103.9
103.9
Langer
(yield)
load line r
= 0.6
(not to
scale)
8
.
80
Goodman
=
+
=
es
us
us
es
a
S
rS
S
S
r
S
5
.
48
line
load
Goodman =
+
=
∩
es
us
us
es
a
S
rS
S
rS
S
98
.
38
1
line
load
Langer =
+
=
∩ r
rS
S
y
a
4
.
67
1
Langer
=
+
=
r
S
r
S y
a
MPa
9
.
103
3
MPa
180
3
=
=
= y
S
ys
S
Material YIELDS first!
( ) MPa
4
.
214
320
67
.
0
67
.
0 =
=
= ut
us S
S
MPa
9
.
77
)
160
)(
59
.
0
)(
9
.
0
(
917
.
0
59
.
0
=
=
ʹ′
=
= = e
f
e
d
c
b
a
k
e
es S
k
k
k
k
k
k
S
S c
Revisit E.g. 3, Non-Zero Mean Stress, Test for First-Cycle Yield
Ses = 77.9
Sa
Goodman
Sm
Sus = 214.4
Sys = 103.9
103.9 = Sys
Langer
(yield)
load line r
= 0.6
(not to
scale)
MPa
9
.
103
3
MPa
180
3
=
=
= y
S
ys
S
39
.
1
1
4
.
214
1
.
58
9
.
77
9
.
34
Goodman
Goodman
=
⇒
=
+
=
+ n
n
S
S us
m
es
a τ
τ
12
.
1
1
9
.
103
1
.
58
9
.
103
9
.
34
Langer
Langer
=
⇒
=
+
=
+ n
n
S
S ys
m
ys
a τ
τ
Material YIELDS first!
Suggested alternative calculation that avoids Table 6-6:
τm = 58.1
τa = 34.9
Quantifying Fatigue Failure:
Stress-Life Method for
Non-Zero Mean Stress and
Combined Loading
'
'
1
=
ʹ′
+
ʹ′
u
m
e
a
S
n
S
n σ
σ
Modified-Goodman design eqn
1
2
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ ʹ′
+
ʹ′
u
m
e
a
S
n
S
n σ
σ
Gerber design eqn
1
2
2
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ ʹ′
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ ʹ′
u
m
e
a
S
n
S
n σ
σ
ASME-ellipse design eqn
1
=
ʹ′
+
ʹ′
y
m
e
a
S
n
S
n σ
σ
Main Idea to Combined Loading:
Finding “Effective” Stresses σá and σḿ for The Many Fatigue Criteria
Soderberg design eqn

6. 12
Combined Loading: Choice of Von Mises as “Effective” Stress
kc,axial
1. Stresses are higher on average due to stress concentrations; mid-range stress is adjusted:
adjusts for effects of axial
misalignment on fatigue
Adapt for fatigue in tension/compression, bending and torsion of shafts:
2. Axial misalignment affects the axial stress alternating about the mean, and hence applied to σʼa only:
Combined Loading: Another Way to View Effective Stress Amplitude
kc,axial
≈ (kc,torsion)−2
( )
( )
( )
( )
( )
( )
2
/
1
2
torsion
,
torsion
torsion
2
axial
,
axial
axial
bending
,
bending
bending
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⋅
+
⎥
⎦
⎤
⎢
⎣
⎡
⋅
+
⋅
≈
ʹ′
c
a
fs
c
a
f
c
a
f
a
k
τ
K
k
σ
K
k
σ
K
σ
⎪
⎪
⎩
⎪
⎪
⎨
⎧
=
torsional
pure
axial
pure
bending
rotating
59
.
0
85
.
0
1
c
k
where loading factor
In other words,
effective 1-D fatigue
stress amplitude
Note: When 1/kc is applied to amplify the service load in combined loading, we do
not apply kc to reduce the strength in the Marin equation (See Sections 6-9 and 6-14)
6-13,14 (Summary) Torsional Fatigue Strength
under fluctuating Stresses Combine loading
• Torsional Fatigue
Strength under
fluctuating stresses
• Combine loading
ut
sy
ut
su
S
S
S
S
577
.
0
67
.
0
=
=
Note
In Goodman diagram
!
!a = Kf
( )Bending
!a
( )Bending
+ Kf
( )Axial
!a
( )Axial
0.85
#
$
%
'
2
+3 Kfs
( )torsion
a
( )Torsion
#
%
2
!
!m = Kf
( )Bending
!m
( )Bending
+ Kf
( )Axial
!m
( )Axial
#
%
2
+3 Kfs
( )torsion
m
( )Torsion
#
%
2
( ) ( ) 84
.
2
1
3
92
.
0
1
1
1 =
−
+
=
−
+
= t
f K
q
K ( ) ( ) 74
.
1
1
8
.
1
93
.
0
1
1
1 =
−
+
=
−
+
= ts
s
fs K
q
K
Given: friction coefficient f = 0.3, and for the present r = 3-mm fillet, Kt=3 and Kts=1.8
0 Axial load P
The steel shaft rotates at a constant speed ω
while the axial load is applied linearly from zero
to P and then released. The cycle is repeated.
Given that T = fP(D+d)/4, find the maximum
allowable load, P, for infinite life of the shaft.
E.g. 4. Clutch
MPa
1000
MPa;
800 =
= ut
y S
S

7. 13
( ) ( ) 84
.
2
1
3
92
.
0
1
1
1 =
−
+
=
−
+
= t
f K
q
K
( )
( )
( )
P
J
Tc
K
τ
τ
τ
P
τ
τ
τ
P
π
P
J
Tc
K
τ
τ
P
P
d
D
fP
T
P
r
π
P
K
σ
σ
σ
P
σ
σ
σ
P
π
P
r
π
P
K
σ
σ
fs
m
a
fs
f
m
a
f
2216
0
2
1
2
2216
2
4431
32
030
.
0
015
.
0
0135
.
0
74
.
1
0
0135
.
0
4
03
.
0
15
.
0
3
.
0
4
2009
0
2
1
2
2009
2
4018
015
.
0
84
.
2
0
min
max
min
max
4
min
max
2
min
max
min
max
2
2
min
max
=
⎟
⎠
⎞
⎜
⎝
⎛ −
=
+
=
=
−
=
=
=
−
=
−
=
⎟
⎠
⎞
⎜
⎝
⎛ +
=
⎟
⎠
⎞
⎜
⎝
⎛ +
=
−
=
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛−
+
=
+
=
=
−
=
=
=
−
−
=
−
( ) ( ) 74
.
1
1
8
.
1
93
.
0
1
1
1 =
−
+
=
−
+
= ts
s
fs K
q
K
For the present r = 3-mm fillet, Kt=3 and Kts=1.8,
0 Axial load P
Problem 5. Clutch
MPa
1000
MPa;
800 =
= ut
y S
S
( )
[ ]
MPa
4508
2216
3
85
.
0
2009
3
1
85
.
0
2
2
2
2
'
P
P
P
τ
σ
σ a
a
a
⋅
=
+
⎟
⎠
⎞
⎜
⎝
⎛
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
⎟
⎠
⎞
⎜
⎝
⎛
=
Fatigue effective stress amplitude:
The steel shaft rotates at a constant speed ω
while the axial load is applied linearly from zero
to P and then released. The cycle is repeated.
Given that T = fP(D+d)/4, find the maximum
allowable load, P, for infinite life of the shaft.
( ) ( )
[ ]
MPa
4332
2216
3
2009
3
2
2
2
2
'
P
P
P
τ
σ
σ m
m
m
⋅
=
+
−
=
+
=
E.g. 4. Clutch, continued
( )
( )
( )( ) MPa
312
500
862
.
0
723
.
0
1
862
.
0
30
24
.
1
723
.
0
1000
51
.
4
MPa
500
)
1000
(
5
.
0
5
.
0
'
107
.
0
265
.
0
'
'
=
=
=
=
=
⋅
=
=
=
=
=
=
=
=
−
−
e
f
e
d
c
b
a
e
c
b
b
ut
a
e
f
e
d
c
b
a
e
ut
e
S
k
k
k
k
k
k
S
k
k
aS
k
S
k
k
k
k
k
k
S
S
S
Newtons
53.3e3
Newtons
.0533e6
0
1
1000
4332
312
4508
1
'
'
n
n
P
n
P
P
n
S
σ
S
σ
ut
m
e
a
=
=
=
+
⇒
=
+
The max. load for
fatigue failure (n=1)
is 53.3 kN.
Note: 1/kc has been applied to amplify the service load already
[ ]
MPa
4508
3
1
85
.
0
2
2
'
P
τ
σ
σ a
a
a ⋅
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
⎟
⎠
⎞
⎜
⎝
⎛
= [ ]
MPa
4332
3
1
2
2
'
P
τ
σ
σ m
m
m ⋅
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
Quantifying Fatigue Failure:
Stress-Life Method for
Stress Cycles of Unequal
Amplitudes
• Palmgren-Miner Cyclic ratio summation rule
Minerʼs rule (Linear damage Rule)
6-15 Varying, Fluctuating Stresses;
Cumulative Fatigue Damage
1
=
∑
i
i
N
n
Assumption:
The stress sequence does not matter and the rate of damage
accumulation at a particular stress level is independent of the
stress history.
where ni is the number of cycle at stress level σi and
Ni is the number of cycle to failure at stress level σi

8. 14
Problem 5: Minerʼs Rule
A rotating beam specimen (4130 steel) with an endurance limit of 50 kpsi and an ultimate
strength of 125 kpsi is cycled 20% of the time at 70 kpsi, 50% at 55 kpsi and 30% at 40
kpsi. Estimate the number of cycle to failure.
Sa1
Sa3
Sa2
N1
N3 = ∞
N2
Solution:
a =
fSut
( )
2
Se
=
0.843(100)
( )
2
50
=142.13 (Fig. 6-18)
b = !
1
3
log
84.3
50
#
$
%
' = !0.07562
!1 = 70ksi N1 =
70
142.13
#
$
%
'
1/!0.07562
=11683cycles
!2 = 55ksi N2 =
55
142.13
#
$
%
'
1/!0.07562
= 283509cycles
!3 = 40ksi N3 = (cycle
0.2N
11683
+
0.5N
283509
+
0.3N
(
=1
N = 52959cycles
Problem 7: Shaft
This stress is far below the yield strength
of 71 kpsi, so yielding is not predicted.
Fig. A-15-9: r/d = 0.0625/1.625 = 0.04, D/
d = 1.875/1.625 = 1.15, Kt =1.95
Get the notch sensitivity either from Fig.
6-20, or from the curve-fit Eqs. (6-34) and
(6-35a). We will use Fig. 6-20.
Eq. (6-32):
Eq. (6-8): Se’=0.5Sut=0.5(85)=42.5 kpsi
Kf =1+ q Kt !1
( )=1+ 0.76(1.95!1) =1.72
q = 0.76
1040 Steels (Sut=85Ksi) rpm=1600rpm, F1=2500lbf and F2=1000lbf
n of infinite life or N
! =
Mc
I
=
14750 1.625 / 2
( )
! 1.6254
( ) 64
= 35ksi
Problem 7: cont
Eq. (6-19)
Eq. (6-20)
Eq. (6-26)
Eq. (6-18)
ka = aSut
b
= 2.70(85)!0.265
= 0.832
kb = 0.879d!0.107
= 0.879(1.625)!0.107
= 0.835
kc =1
Se = kakbkcS'e = 0.832
( ) 0.835
( ) 1
( ) 42.5
( )= 29.5ksi
nf =
Se
Kf!
=
29.5
1.72 35
( )
= 0.49
No infinite life
Fig. 6-18: f=0.867
Eq. (6-14)
Eq. (6-15)
Eq. (6-16)
a =
fSut
( )
2
Se
=
0.867(85)
( )
2
29.5
=184.1
b = !
1
3
log
fSut
Se
#
$
%
' = !
1
3
log
0.867(85)
29.5
#
$
%
' = !0.1325
N =
17.2(35)
184.1
#
$
%
'
1
!0.1325
= 4611cycles