Quantifying Fatigue Failurepkwon/me471/Lect 6.3.pdf7! Quantifying Fatigue Failure:" Stress-Life...
Transcript of Quantifying Fatigue Failurepkwon/me471/Lect 6.3.pdf7! Quantifying Fatigue Failure:" Stress-Life...
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Quantifying Fatigue Failure:���Stress-Life Method for���Non-Zero Mean Stress ���
(Failure Surface = “Goodman Diagram”) Variation in cyclic stress.
2minmax σσ
σ+
=m
RRA
m
a
+
−=
+
−==
11
minmax
minmax
σσσσ
σσ
2minmax σσ
σ−
=a
Mean stress:
max
min
σσ
=R
Stress Amplitude:
Stress ratio:
Amplitude ratio:
6-11 Characterizing Fluctuating Stresses
2minmax FFFm
+=
2minmax FFFa
−=
Terms for Stress Cycling
Stre
ss a
t 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 <∞
Nf = ∞
Nf < ∞
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
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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
Nf < ∞
Nf = ∞
1=+u
m
e
a
SS
SSEquation of Goodman line:
1=+u
m
e
a
SS
SS“Goodman Criterion” for ∞-‐life:
Sm Su
(Se = Sa|Nf =∞)
Sf = σa|Nf <∞
“Goodman Criterion” for finite life:
1=+u
m
f
a
SS
SS
( ) ( )1=+
nSnS u
m
f
a σσor
( ) ( )1=+
nSnS u
m
e
a σσor
↑ failure surface ↑ design equation (i.e., shrink failure surface until you hit service loads)
↑ failure surface ↑ design equation
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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
Nf < ∞
Nf = ∞
1=+u
m
e
a
SS
SS
1=+u
m
e
a
SS
SS“Goodman Criterion” for ∞-‐life:
Sm Su
(Se = Sa|Nf =∞)
Sf (Nf <∞)
“Goodman Criterion” for finite life:
1=+u
m
f
a
SS
SS
1=+u
m
f
a
Sn
Sn σσ
or 1=+u
m
e
a
Sn
Sn σσor
↑ 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
Nf < ∞
Nf = ∞
1=+us
ms
es
as
SS
SS“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
SS
SS
nSS us
m
fs
a 1=+
ττor
nSS us
m
es
a 1=+
ττor
experimental correlation
1=+us
ms
es
as
SS
SS
↑ 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
( )MPa66.63MPa9.15
636620201.02
mN100mN25kN2kN5.0
m 05.0
34
≤≤
⋅====
⋅≤≤⋅⇒≤≤
⋅=⋅=
τππ
τ TT
cTc
JTc
TFFdFT
MPa9.92MPa2.23or
)66.63(46.1)9.15(46.1
toe
toe
≤≤
≤≤⇒
ττ
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:
( )11 −+= tssfs KqK
( ) ( ) 46.116.1775.0111 =−+=−+= tssfs KqK
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
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( ) MPa4.21432067.067.0 === utus SS
39.114.2141.58
9.779.34
Goodman =⇒=+=+ nnSS us
m
es
a ττ
08.582
21.2394.92,86.342
21.2394.92=
+==
−= ma ττ
E.g. 3. Non-Zero Mean Stress
Convert ultimate tensile to ultimate shear:
Convert ideal to actual endurance limit:
( ) MPa1603205.05.0 ===ʹ′ ute SSApproximate ideal endurance limit:
nSS us
m
es
a 1=+
ττSafety factor based on Goodman criterion:
Stress amplitude and mid-range stress:
(kc converts tensile endurance to shear endurance!)
( )
( )
MPa9.77)160)(59.0)(9.0(917.0
59.0
9.02024.1
917.03207.57
59.0
107.0
718.0
===
=
==
===
ʹ′=
=
−
−
ckees
c
b
buta
efedcbae
SS
kk
aSk
SkkkkkkS
(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
nSS y
m
e
a 1=+
σσ
12
=⎟⎟⎠
⎞⎜⎜⎝
⎛+
ut
m
e
a
Sn
Sn σσ
12
=⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟⎠
⎞⎜⎜⎝
⎛
y
m
e
a
Sn
Sn σσ
nSS ut
m
e
a 1=+
σσ
ASME-elliptic
Langer Static Yield
nSy
ma =+σσ
“First-Cycle Yield”: Goodman Fatigue Line and Langer Yield Line
Se
Sa
(Se = Sa|Nf =∞)
1=+y
m
y
a
SS
SS
ModifiedGoodman 1=+
u
m
e
a
SS
SS
Sm Su
Sy
Sy
Langer
for this lower value of r, yield happens before fatigue!
generic load line r = Sa /Sm
highe
r valu
e of 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
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( ) MPa4.21432067.067.0 === utus SS
MPa9.77)160)(59.0)(9.0(917.059.0
==
ʹ′=== efedcbakees SkkkkkkSS
c
08.582
21.2394.92
86.342
21.2394.92
=+
=
=−
=
m
a
τ
τ
Revisit E.g. 3, Non-Zero Mean Stress, Test for First-Cycle Yield
6.008.5886.34
===m
arττ
Ses = 77.9
Sa
Goodman
Sm
Sus = 214.4
Sys = 103.9
103.9
Langer (yield)
load line r = 0.6
(not to scale)
8.80Goodman
=+
=esus
usesa
SrSSS
rS
5.48line load
Goodman =+
=∩ esus
usesa SrS
SrSS
98.381lineload
Langer =+
=∩ r
rSS ya
4.671Langer
=+
=r
SrS ya
MPa9.1033MPa 180
3=== yS
ysS
Material YIELDS first!
( ) MPa4.21432067.067.0 === utus SS
MPa9.77)160)(59.0)(9.0(917.059.0
==
ʹ′=== efedcbakees SkkkkkkSS
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)
MPa9.1033MPa 180
3=== yS
ysS
39.114.2141.58
9.779.34
GoodmanGoodman
=⇒=+=+ nnSS us
m
es
a ττ
12.119.1031.58
9.1039.34
LangerLanger
=⇒=+=+ nnSS ys
m
ys
a ττ
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
Sn
Sn σσ
Modified-Goodman design eqn
12
=⎟⎟⎠
⎞⎜⎜⎝
⎛ ʹ′+
ʹ′
u
m
e
a
Sn
Sn σσ
Gerber design eqn
122
=⎟⎟⎠
⎞⎜⎜⎝
⎛ ʹ′+⎟⎟
⎠
⎞⎜⎜⎝
⎛ ʹ′
u
m
e
a
Sn
Sn σσ
ASME-ellipse design eqn
1=ʹ′
+ʹ′
y
m
e
a
Sn
Sn σσ
Main Idea to Combined Loading:���Finding “Effective” Stresses σa and σm for The Many Fatigue Criteria
Soderberg design eqn
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 σ’a only:
Combined Loading: Another Way to View Effective Stress Amplitude
kc,axial ≈ (kc,torsion)−2
( )( )
( ) ( ) ( ) ( )2/12
torsion,
torsiontorsion
2
axial ,
axialaxial
bending ,
bendingbending
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎟⎟⎠
⎞⎜⎜⎝
⎛⋅+⎥
⎦
⎤⎢⎣
⎡⋅+⋅≈ʹ′
c
afs
c
af
c
afa k
τK
kσ
Kkσ
Kσ
⎪⎪⎩
⎪⎪⎨
⎧
=
torsionalpure
axial pure
bending rotating
59.0
85.0
1
ckwhere 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 utsy
utsu
SSSS
577.067.0
=
=Note
In Goodman diagram
!! a = K f( )Bending ! a( )Bending + K f( )Axial! a( )Axial0.85
"
#$
%
&'
2
+3 K fs( )torsion " a( )Torsion"#
%&2
!! m = K f( )Bending ! m( )Bending + K f( )Axial ! m( )Axial"#
%&2+3 K fs( )torsion "m( )Torsion"#
%&2
( ) ( ) 84.21392.0111 =−+=−+= tf KqK ( ) ( ) 74.118.193.0111 =−+=−+= tssfs KqK
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 MPa1000MPa;800 == uty SS
13
( ) ( ) 84.21392.0111 =−+=−+= tf KqK
( )
( )( )
PJTcKτττ
Pτττ
Pπ
PJTcKττ
PPdDfPT
PrπPKσσσ
Pσσσ
Pπ
PrπPKσσ
fsm
a
fs
fm
a
f
2216021
2
22162
443132030.0015.00135.074.10
0135.0403.015.03.0
4
2009021
2
20092
4018015.0
84.20
minmax
minmax
4minmax
2minmax
minmax
22minmax
=⎟⎠⎞
⎜⎝⎛ −=
+=
=−
=
==−=−
=⎟⎠⎞
⎜⎝⎛ +
=⎟⎠⎞
⎜⎝⎛ +
=
−=⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛−+=
+=
=−
=
==−−=−
( ) ( ) 74.118.193.0111 =−+=−+= tssfs KqK
For the present r = 3-mm fillet, Kt=3 and Kts=1.8,
0 < Axial load < P
Problem 5. Clutch MPa1000MPa;800 == uty SS
( )
[ ]MPa4508
2216385.0
20093185.0
2222
'
P
PPτσσ aaa
⋅=
+⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟
⎠⎞
⎜⎝⎛=
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.
( ) ( )[ ]MPa4332
2216320093 2222'
PPPτσσ mmm
⋅=
+−=+=
E.g. 4. Clutch, continued
( )
( )
( )( ) MPa312500862.0723.0
1862.03024.1
723.0100051.4
MPa500)1000(5.05.0
'
107.0
265.0
'
'
===
=
=⋅=
===
=
===
−
−
efedcbae
c
b
buta
efedcbae
ute
SkkkkkkS
kk
aSk
SkkkkkkS
SS
Newtons 53.3e3Newtons .0533e60
11000
4332312
45081''
nnP
nPP
nSσ
Sσ
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
[ ]MPa45083185.0
22' Pτσσ aaa ⋅=⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟
⎠⎞
⎜⎝⎛= [ ]MPa4332
31
22' Pτσσ mmm ⋅=⎟⎟
⎠
⎞⎜⎜⎝
⎛+=
Quantifying Fatigue Failure:���Stress-Life Method for���
Stress Cycles of Unequal Amplitudes
• Palmgren-Miner Cyclic ratio summation rule Miner’s rule (Linear damage Rule)
6-15 Varying, Fluctuating Stresses; Cumulative Fatigue Damage
1=∑i
i
Nn
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
14
Problem 5: Miner’s 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.
S a1
S a3
S a2
N 1
N 3 = ∞
N 2
Solution:
a =fSut( )2
Se=0.843(100)( )2
50=142.13 (Fig. 6-18)
b = !13log 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 =(cycle0.2N11683
+0.5N283509
+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
K f =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 ! =
McI=14750 1.625 / 2( )! 1.6254( ) 64
= 35ksi
Problem 7: cont Eq. (6-19) Eq. (6-20) Eq. (6-26) Eq. (6-18)
ka = aSutb = 2.70(85)!0.265 = 0.832
kb = 0.879d!0.107 = 0.879(1.625)!0.107 = 0.835
kc =1Se = kakbkcS 'e = 0.832( ) 0.835( ) 1( ) 42.5( ) = 29.5ksi
nf =SeK f!
=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 = !13log fSut
Se
"
#$
%
&'= !
13log 0.867(85)
29.5"
#$
%
&'= !0.1325
N =17.2(35)184.1
"
#$
%
&'
1!0.1325
= 4611cycles