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### Transcript of Quantifying Fatigue Failure pkwon/me471/Lect 6.3.pdf7! Quantifying Fatigue Failure:"...

7

Quantifying Fatigue Failure:Stress-Life Method forNon-Zero Mean Stress

(Failure Surface = Goodman Diagram)

Variation in cyclic stress.

2minmax

+=m

RRA

m

a

+

=

+

==

11

minmax

minmax

2minmax

=a

Mean stress:

max

min

=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, m

load reversals (2 reversals = 1 cycle)

stress amplitude, a

maximum stress, max

minimum stress, min

Follow a material point A on the outer fiber

Schematic: Effect of Midrange Stress Sm = m|Nf =

Se

Su

Sf|Nf

8

Data: Effect of Midrange Stress m

That is: the endurance limit drops as m|Nf = Sm increases

Simple Model to Account for Midrange Stress m

That is: the endurance limit drops as m|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, m , as mid-range strength Sm

Alternating stress amplitude, a ,

as fatigue strength Sa

Goodman Diagram: Fatigue Failure with m 0

Se

Sa

Nf <

Nf =

1=+u

m

e

a

SS

SSEquation of Goodman line:

1=+u

m

e

a

SS

SSGoodman Criterion for -life:

Sm

Su

(Se = Sa|Nf =)

Sf = a|Nf

9

Mean stress, m , as mid-range strength Sm

Alternating stress amplitude, a ,

as fatigue strength Sa

Goodman Diagram: Fatigue Failure with m 0

Se

Sa

Nf <

Nf =

1=+u

m

e

a

SS

SS

1=+u

m

e

a

SS

SSGoodman Criterion for -life:

Sm

Su

(Se = Sa|Nf =)

Sf (Nf

10

( ) MPa4.21432067.067.0 === utus SS

39.114.2141.58

9.779.34

Goodman ==+=+ nnSS usm

es

a

08.582

21.2394.92,86.342

21.2394.92=

+==

= ma

E.g. 3. Non-Zero Mean Stress

Convert ultimate tensile to ultimate shear:

Convert ideal to actual endurance limit:

( ) MPa1603205.05.0 === ute SSApproximate ideal endurance limit:

nSS usm

es

a 1=+ Safety 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

===

=

==

===

=

=

ckees

c

b

buta

efedcbae

SS

kk

SkkkkkkS

(95% stress area is the same as R. R. Moore specimen)

Recall that for m = 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 ym

e

a 1=+

12

=

+

ut

m

e

a

Sn

Sn

12

=

+

y

m

e

a

Sn

Sn

nSS utm

e

a 1=+

ASME-elliptic

Langer Static Yield

nSy

ma =+

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 = a/m is considered known)

These are two eqns in two unknowns Sa and Sm, as r = a/m 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

==

=== efedcbakees SkkkkkkSS c

08.582

21.2394.92

86.342

21.2394.92

=+

=

=

=

m

a

Revisit E.g. 3, Non-Zero Mean Stress, Test for First-Cycle Yield

6.008.5886.34

===m

ar

Ses = 77.9

Sa

Goodman

Sm

Sus = 214.4

Sys = 103.9

103.9

Langer (yield)

(not to scale)

8.80Goodman

=+

=esus

usesa

rS

Goodman =+

= esus

usesa SrS

Langer =+

= r

4.671Langer

=+

=r

SrS ya

MPa9.1033MPa 180

3=== ySysS

Material YIELDS first!

( ) MPa4.21432067.067.0 === utus SS

MPa9.77)160)(59.0)(9.0(917.059.0

==

=== 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)

(not to scale)

MPa9.1033MPa 180

3=== ySysS

39.114.2141.58

9.779.34

GoodmanGoodman

==+=+ nnSS us

m

es

a

12.119.1031.58

9.1039.34

LangerLanger

==+=+ nnSS ys

m

ys

a

Material YIELDS first!

Suggested alternative calculation that avoids Table 6-6:

m = 58.1

a = 34.9

Quantifying Fatigue Failure:Stress-Life Method for

'

'

1=

+

u

m

e

a

Sn

Sn

Modified-Goodman design eqn

12

=

+

u

m

e

a

Sn

Sn

Gerber design eqn

122

=

+

u

m

e

a

Sn

Sn

ASME-ellipse design eqn

1=

+

y

m

e

a

Sn

Sn

Main Idea to Combined Loading:Finding Effective Stresses a and m for The Many Fatigue Criteria

Soderberg design eqn

12

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 a only:

kc,axial

(kc,torsion)2

( )( )

( ) ( ) ( ) ( )2/12

torsion,

torsiontorsion

2

axial ,

axialaxial

bending ,

bendingbending

+

+

c

afs

c

af

c

afa k

K

k

Kk

K

=

torsionalpure

axial pure

bending rotating

59.0

85.0

1

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)

Torsional Fatigue Strength under fluctuating stresses

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 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

( )

( )( )

PJTcK

P

P

PJTcK