7

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

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

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

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

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

10

( ) 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

11

( ) 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