Fatigue resistance

(1)

Fatigue resistance

Contents

• Fatigue failure phenomena

• Fluctuating loads / stress -- > models

• Fatigue resistance determination:

• Life calculation, endurance limit, endurance diagram

• Fatigue safety factor

• Load/stress spectrum – cumulative damage theory

• 1^stapproach on cracking

• Cracks, types, propagation,

• Stress intensity factor

(2)

Fatigue failure phenomena

Fatigue failure - History

• Observed phenomena around 1800 on the train car axles after a limited service life

• Introduction of the notion of fluctuating moments/stresses (rotary bending…..)

• August Wöhler published in 1870 his research founding's:

• Definition of :

• Endurance limit vs. Number of Cycles,

• Diagram (σ-N)

• Fatigue : term used by Poncelet in 1839

(3)

Fatigue failure - Introduction

• A part subjected to a variable loading breaks at a stress level below the yeld/ultimate stress of the material: fatigue

phenomena

• 90% of failures in operation are due to fatigue

• The demands for increasing performance and reducing environmental impacts imply to design systems by

considering fatigue either to ligthen them, and to extend their service life

The 3 phases of failure

1 : ignition 2 : Propagation

2 : Propagation 2 : Final failure

Any surface discontinuity contributes to the crack

ignition

(4)

Fatigue failure – Wöhler experiments

Probabilistic behavior: for a given

stress amplitude several life !

Line of 50%

Modeling Wöhler

experiments

(5)

Rotary bending

L1 L1

R1 2L2

F F

R2

A

C

B D

Force equilibrium Moment equilibrium in A

Rotary bending

L1 + L2 F

F F

F

A

B C

D 0

Cohesion wrench in 0 Global basis

Constant in the global basis

(6)

Rotary bending

In the local basis rotating at ω

y z

L₁F

ωt

Variable in the local basis P

Stress in P (x2, x3)

Stress in Q (d/2, 0) Q

Cohesion wrench in 0

Fatigue failure – Wöhler experiments

σ

moy

= 0

Temps σ a et σ

(7)

Sinusoidal stress

time

σ^a σ^m

σ^max

σ^min

σ _{1 cycle} σm is the mean or average stress:

σais the amplitude stress:

fully reversed stress: σm= 0, σa≠0 alternated stress: σm≠0, σa≠0 Repeated stress: σmin= 0 ou σmax= 0

Variable stress: σmax< 0 in compression, σmin> 0 intraction

Wöhler experiments

Modeling the Wöhler experimental results

10² 10³

10 10⁴ 10⁵ 10⁶ 10⁷ 10⁸ 10⁹ 10¹⁰

Time defined by Number of cycles (Log. scale) Possible fatigue failure

Fatigue test

Nb of cycles for 50% of failure

Equi-probability curve for 50 % Stress, σa (withσmoy= 0)

(8)

Using the Wöhler curve

10² 10³

10 10⁴ 10⁵ 10⁶ 10⁷ 10⁸ 10⁹ 10¹⁰

Oligocyclic

Fatigue Finite life (endurance) 0,9 σu

Prob.50%

Fatigue Limit à 10⁷Cycles (andProb.50%) σ_D(10⁷,σ_moy= 0)

Stress, σa (withσmoy= 0)

Time defined by Number of cycles (Log. scale)

Endurance limitσ_Dfor steels

10² 10³

10 10⁴ 10⁵ 10⁶ 10⁷ 10⁸ 10⁹ 10¹⁰

Infinite life σ_D

Prob.50%

Prob.99%

Endurance limit

Time defined by Number of cycles (Log. scale) Stress, σa (withσmoy= 0)

(9)

Dimensioning based on Wöhler criteria

• For steels, one can use the endurance limit for dimensioning : thanks M.Wöhler !

• This is valid only for: σ_a≠ 0 et σ_m= 0, which is very limitating in terms of applications…

• This valid only for materials that have a constant endurance limit when N tends to infinity. NOT THE CASE OF ALUMINIUMS….

(10)

Fatigue resistance calculation for a part (fatigue limit)

Recherche d’une loi représentative des essais

Real curve

Wöhler model

Basquin model

* Basquin : model the most used

Fatigue resistance for σ

_m

= 0

(11)

Fatigue resistance

endurance of a part (fatigue limit) (for σm=0)

• The Wölher curve and its models permits determining:

• The endurance limit σ D:

Largest amplitude of σ_afor which no failure occurs (only for Ferrous metals)

• The fatigue limit σ D(N), σ D(N_i) :

Amplitude of σafor which it is observed 50% of survive with σm=0, after N_igiven cycles of loading. >> obtained from Wöhler curves

For steels, one consider that for N_i≥ 10⁷cycles σD(N_i)=σD

• The endurance ratio:

σ_a< σ_D

Use of Basquin model

log Ncycles

logσ ^a

σ m=0

10³ 10⁴ 10⁵ 10⁶ Log 0.9.Rm

10⁷

B

C

logσ^D^(N) logσD(Ni)

Ni

From the garph (Thalès) Fatigue resistance

(12)

Use of Basquin model

Sample of steel ac: _σ_u=920 MPa; σ_D=400 MPa σD(10⁵)=?

N_i if σD(N_i)=600 MPa?

• Endurance limit σ D, approximations:

• Steels N=10⁷

• σD(N)=O.5*σu for σu<1300 Mpa

• σD(N)=600 MPa for σu>1300 Mpa

• Founts N=10⁷

• σ^D^(N)=O.4*^σ^u

• Aluminium alloys N=5*10⁶

• σD(N)=O.4*σu

(13)

Correction factors for σ_D(pour σ_m=0)

• Tests are done with samples !! -> σD

• ?? Endurance limit of a real part : σD*=(k_ak_bk_ck_dk_ek_f).σD

k_a surface finishing k_b size of the part k_c reliability expected

k_d temperature of operation k_e stress concentration k_f side effects ….

Correction factors for σD(pour σm=0)

• k_a , surface finishing

(14)

• k_bsize of the part

Relative volume criteria

V, volume of the part

V₀, reference volume (test sample)

Characteristic dimension criteria

K_b=1 for d≤7.6 mm

K_b=0.85 for 7.6mm<d≤50 mm K_b=0.75 for d>50 mm

Reliability (R) Reliability factor (k_C)

0,5 1

0,9 0,897

0,95 0,868

0,95 0,814

0,99 0,753

0,999 0,702

0,9999 0,659

0,99999 0,620

0,999999 0,584

k_c Reliability factor

(15)

Correction factors for σ_D(pour σ_m=0)

• k_dtemperature factor

1

0,5

71 100 150 200 250

k_d

• k_e stress concentration factor

Discontinuity in cross sections In Statics -- >(factor K_t).

With variable loading, one must account for cross section discontinuities since cracks often ignitiate there

Stress concentration factor in fatigue is given by:

It depends on the part geometry and on the material loading

q: factor related to the discontinuity, see after

(16)

k_e stress concentration_in fatigue

_ .

And when σ

_m

≠ 0 ??

-> endurance diagram, etc …

(17)

Load/stress fluctuation definition

Random stress

Non centered sinusoidal stress

contrainte

temps

σ^a σ^m

σ^max

σ^min

σ _{1 cycle}

σ

moy

= 0

Temps σ a et σ

Centered sinusoidal stress: Wöhler

Sinusoidal stresses

Non centered sinusoidal stress

stress

time

σ^a σ^m

σ^max

σ^min

σ _{1 cycle} σm , mean stress:

σ^aamplitude stress (alternated) :

fully reversed stress: σm= 0, σa≠0 alternated stress: σm≠0, σa≠0 Repeated stress: σmin= 0 ou σmax= 0

Variable stress: σmax< 0 in compression, σmin> 0 intraction

(18)

Sinusoidal stresses

• Example

V _B

A

C

M ωt

V

x y

z y

Case 1

L

∅d

Case 2 M

ωt

V

z

y F

σ_a< σ_D ????

Calcul de résistance à la fatigue pour σ

_m

≠ 0

stress

time

σ^a σ^m

σ^max

σ^min

σ _{1 cycle}

(19)

Endurance limit of a part (fatigue limit) for σm≠0

σm

Endurance limit for σm≠0

Endurance diagram, HAIGH diagram

Endurance diagrams (abscissa σ_met ordinate σ_a) are deduced from Wölher curves. They defines σ^D(N)as a function of the mean stress for a given number of cycle N

Number of cycles σD2 (10⁶;σ_m2)

σD1 (10⁶;σ_m1) σD3(10⁶;σ_m3)

σD2 (10⁸;σ_m2) σD1 (10⁸;σ_m1) σD3(10⁸;σ_m3)

Wöhler curves with Prob. Of failureα%

10⁵ 10⁶ 10⁷ 10⁸ 10⁹

σa

σm= 0 σ_{m 1}

σm 2

σ_{m 3}

σD

0 < σm1< σm 2< σm 3 σD (N,σ_moy=0)

σR σmoy Haigh diagram

(pour N cycles)

σm 1 σm2 σm 3

0

Possible failure zone

σD2 (N , σ_m2) σD1 (N , σ_m1)

σD3(N, σ_m3) σD (N , σ_m=0)

σa

(20)

Do not forget the probabilistic aspect !!!

Acier AISI 4340 N = 2,5 . 10⁶cycles (r = σ_a/ σ_m)

138 276

414 552

Mean Stress σ_m (MPa)

138 276 414 552 690 828 966 1104 1242

σ_D(N , σ_m=0)

Modeling of HAIGH’s diagram

138 276

414 552

Mean stress σ_m (MPa)

138 276 414 552 690 828 966 1104 1242

GERBER parabola

> Too complex for simple use

σa

σD (N , σ_m=0)

σ_u

(21)

138 276

414 552

138 276 414 552 690 828 966 1104 1242

GOODMAN ‘s line

> Most used

σa

σD (N , σ_m=0)

Modeling of HAIGH’s diagram

σ_u

Modélisons le diagramme de HAIGH

138 276

414 552

138 276 414 552 690 828 966 1104 1242

SODERBERG line: simple , accounts for elastic limit

> To be used in addition to GOODMAN’s criteria

σa

σD (N , σ_m=0)

σE σ_u

(22)

Recommended model

138 276

414 552

138 276 414 552 690 828 966 1104 1242

The modified GOODMAN line :

• If, σ

a + σ

m < σ

E

• 1for

• Else, σ_a+ σ_m= σ_E

σa

σD (N , σ_m=0)

σR

Endurance limit for σm≠0 σD, σE,Yand σuare known: plot of the red line,

for compression stress one considers thatσmhas no influence d’influence on σa

Zone 1(white): fatigue failure for nb cycles smaller than Ni (point B)

Zone 2(Blue): life between Ni (point C’) and 10⁷(point A’)

Zone 3(green):

Infinite life (point A)

σu

σE

σ_D σ_D(Ni)

σ_a

σ_m Re

Rm

σ_m>0 σ_m<0

A

C

σ_a(A)

σ_m(A) B

A’

C’ Modified Goodman

(23)

Safety factor in fatigue α

σ_D σ_D(Ni)

σ_a

σm

Re Rm

A

C

σ_a(A)

σ_m(A) σ_D/α

σE/α α= OC/OA,

This is equivalent as

considering a material with characteristics:

σ_D/α

σ_e/αet σ_m/α

O σE σu

σ_D σ_D(Ni)

σ_a

σ_m Re

Rm

A

C

σ_a(A)

σ_m(A) σ_D/α

Modified goodman criteria

• If σ_a+ σ_m < σ_E/α

• Else

σ

_a

+ σ

_m

= σ

_E

/ α

For a given safety factor α

σu

σE

σE/α

(24)

σ_D σ_D(Ni)

σ_a

σ_m Re

Rm

A

C

σ_a(A)

σ_m(A) _σ

E/α σ_D/α

Using Soderberg criteria

σu

σE

With Soderberg criteria

138 276

414 552

138 276 414 552 690 828 966 1104 1242

σa

σD (N , σ_m=0)

(25)

Example

correction factor of σD

Steel shaftcold-drawnof diameter 40mm: σE=490MPa, σu=590MPa.

Inititial axial load F₀=70 kN, variable load F_v: 0 to 100kN.

A stress concentration factor in statics K_t=2.02 for a groove r=5mm.

Determine the safety factor for an unlimited life and a reliability of 90%

Material endurance limit:

σ‘_D(N)=O.5*σu= 295MPa as σu<1300 Mpa Endurance limit for the part:

σ^D^=ka.k_b.k_c.k_d.k_e.k_f.σ^D’

k_a=0.76 k_b==0.85 k_c=0.897 k_d=1

k_e=1/K_foù K_f= q(K_t-1)+1 q=0.86,

K_f= 0.86(2.02-1)+1)=1.877 k_e=0.532 k_f= 1 (no correction)

σ^D=0.76*0.85*0.897*1*0.532*1*295=90.9 MPa

Fi + Fv(t)

Example

safety factor calulation

Induced stress (by the axial loading) F₀=70kN ; F_v=100kN F_a=F_v/2 F_m=F₀+F_a=120kN

A=πd²/4=1.257.10^-3m² σ_a=F_a/A=39.8 MPa; σm=F_m/A=95.5 MPa

Safety factor calulation σ_D=295 MPa / 90.9 MPa ; σ_E=490 MPa ; σ_u=590 MPa

SODERBERG α=OS/OA=3,03 / 1,58 GOODMAN α=OB/OA=3,37 / 1,66

In statics:

F_a

F₀ F_m

F_v

σ_r

Goodman

σ_D σ_a

σ_m σ_E

A B σ_a

σ_m

C

σ_E

Sollicitation line

S

Fi + Fv(t)

(26)

Application

Application

The beam ,with a circular cross section, is clamped on its left and subjected to a fluctuating load F(F_min=-F₀; F_max=3F₀) on its right.

One aims at determining F₀such that the life of the beam is infinite.

By neglecting the internal shear force, determine the critical section along the beam length. The stress concentration will be neglected in a first analysis.

Then determine F₀such that the life of the beam is infinite.

Material data: σu=560 Mpa; σY,E=480 Mpa; σD=280 Mpa

(27)

• Calculation scheme : Forces/Moments in A Application

A B C

l₁=l

! F

l₂=5l

d₁=2d d₂=d

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*_# 0

+# $ ,# 0 +# 0 -# 6)/

A B C

• Cohesion wrench - stresses Application

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A C

l₁=l

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CC AB !9 79 H_?

> 2 (_@_II ^I₂ !:= _{> 2}^ED

7: 8:

J !1 +7

K77 !1 !⁹ 0

Stresses in section G

(28)

Application

• Critical section

77 () !2( 6/

K:: 2 !9

79 ()

J !1

7: 0

Max on the outer diameter, I₃₃/xG max A1 or B1

Max section B+

• Critical section Application

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4 J9 SL⁹

4 K^::,9 SL^U 64

77 160). / SL^:

79 (4)

SL⁹

7: 0

Resistance criteria to be applied in section B+

(29)

Application

• Fluctuating load Fluctuating stresses

) ) 1 2 sin YZ

77 160. /.) 1 2 sin YZ SL^:

79 [ (4) 1 2 sin YZ SL⁹

160. /. 2. )

SL^: ^\ 160. /. )

SL^: [_\ (4)

SL⁹

[ (8)

SL⁹

• N.A.

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77 160). / SL^:

79 (4)

SL⁹

7: 0

Application

• Fatigue resistance criteria

1- Shear stress and stress concentration factor neglected

1,3. ) _\ 0,65. ) [ (0,016. ) [\ (0,008. )

<

\c 1 d 1,3. )

<

0,65. ) c 1

d ) c1

d . 1 1,3

<

0,65

) c 166 -

d_e4fg 1

σ_D(Ni) σ_a

σ_m σu

σa(A) A σm(A) σE

σ_D

σ_E σu

Soderberg

Goodman σa(C)

σm(C) σD/α C

σ_u/α σ_E/α

Soderberg

<

\ h c1

d

) c1

d 1

1,3

<

0,65

h

) c 171 - 1,3. )

<

0,65. ) c1

d Goodman

Material data: σu=560 Mpa; σY,E=480 Mpa; σD=280 Mpa

(30)

Application

2- stress concentration considered – shear stress neglected

• calculation of Kf

• Soderberg, Goodman

1,3. ) _\ 0,65. ) [ (0,016. ) [\ (0,008. )

′_< k^<_l

′_m,g k^m,g_n

k_l.

<

k_n. _\

m c 1

d k_l.

<

k_n. _\

o c 1

d

Application

• D/d=2, r/d= 0.28

Kt=1.3

q=0.9

(31)

Application

3- Shear stress considered multi-axial stress state

Calculation of an equivalent stress based on Von Mises formula

1,3. ) \ 0,65. ) [ (0,016. ) [_\ (0,008. )

Fatigue resistance in torsion

for τ

_m

≠ 0

(32)

Endurance limit for τm≠0

Haigh’s diagram

Pure torsion

τE

τ_D(Ni) τ_a

τ_m τu

A C

τ_a(A)

τ_m(A)

B

τu

τE

τ_D Some experiments have shown thatthe

mean stress τmhas no effect on the torsion endurance limit τ_D.

Given the stress loading line, the part resistance is limited by :

• the horizontal line τ_D

• The elasticity limit line Safety factor

α=τ_D/τ_a α=τE/(τa+τm)

Stress loading lines The fatigue resistance in shear (torsion) is determined by analogy to the yeld stress from the distorsion energy principle:

Factors influencing the endurance limitσD(pour σm=0)

Type of loading (bending, traction, torsion, etc.)

σ_m, τ_m

σ_a, τ_a ^{2D Bending}

Rotary bending Traction

Torsion

(33)

Fatigue resistance for: multi- axial fluctuating stresses

Fatigue resistance for: multi-axial fluctuating stresses

• Context–equivalent stress

• Very often, parts of machines are subjected to multi axial and fluctuating stresses (2D or 3D)

• Calculation of equivalent mean and amplitude stresses:

σ_{a_equi}et σ_{m_equi}

σ_z τ_zx τ_zy

σ_y τ_yx

τ_yz σ_x

τ_xz τ_xy σ_y

τ_yx

σ_x τ_xy

σ_y τ_yx

σ_x τ_xy

(34)

• σ_{a_equi} and σ_{m_equi}

τ_yx σ_x τ_xy

σ_y τ_yx σ_x

τ_xy

σ_y

σ_z τ_zx τ_zy

σ_y τ_yx

τ_yz σ_x

τ_xz τ_xy

• σ_{a_equi} and σ_{m_equi}

Then, use of the endurance diagrams

_pqh r r

(35)

Application

3-shear stress accounted for multi-axial stress state

• equivalent mean and amplitude stresses

• Resistance criteria:

• Soderberg, Goodman

In this case, the shear stress has no effect, it can be neglected

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\_4sgt \9 3[\9 0,652. )

_4sgt 9 3[⁹ 1,304. )

) c 166 -

d 1

) c 171 -

Fatigue resistance of a transmission shaft

Method :

• Calculation of the stresses

• Equivalent stresses

• Application of the modified Goodman criteria

σ_u

Modified Goodman

σ_Dou σ_D(Ni) σ_a

σ_m σ_E

A B

σ_{a_equi}

σ_{m_equi} C

σ_E

Stress line

Modified Goodman criteria:

If σa_equi+ σm_equi< σE/α

_pqh _pqh

h

else

σ_{a_equi}+ σ_{m_equi}= σ_E/α

(36)

Cohesion wrench constant in the global frame

y z

Mf ωt Mt Q

Rotating shaft bending/ torsion

Cohesion wrench variable in the local frame

Stresses in Q

• Equivalent stresses

\_4sgt \9 3[_\⁹ 3.[_\ _{_4sgt} ⁹ 3[⁹

3. 16. +₅

S. L^:. _g 32. +

S. L^:. _< 1 d

L d_e4fg

S 3. 16. +₅

g

32. +

<

C

(37)

Synthesis , fatigue resistance, 1 single stress level

Uniaxial

σ_m=0 σ_m≠0

Haigh

(endurance) diagram - Soderberg model - Goodmann model - Goodmann model

Multi axial

Direct use of Wölher diagram

- Wolher model -Basquin model

Tensile strength Yeld stress

Gerber model

σE

σu

σE σu

Example: shaft of a starter-alternator

• Calculation scheme

• Bearing centers

• Pulley center

• Winding center

L

x z

F_A d

e

C_m A

D B

e/2 C y

-C_m

F_B F_D

L e d F_A C_m

150 mm 100 mm 30 mm -4000 N 80 N.m

σr σD σE

700 MPa 300 MPa 600 MPa

(38)

Example: shaft of a starter-alternator

Example: shaft of a starter-alternator, stress state in statics

• Stress in section B+ ⁷⁷ ^-_J ⁺_K₉₉⁹^!^:⁽⁺_K_::^:^!⁹ ^{u` ∗ 64}_SL^U ^!⁹

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B ^/⁷ !

/9

/:

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7:

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O+7 $ 0

L/20

x,yG_I,G_?,G_Cz

77 50 ∗ 64 SL^U L

2 50 ∗ 32 SL^:

79 (2000 SL⁹

7: 80 ∗ 16 SL^:

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[= ₇₉₉ _7:9

77

{ 779 3 799 7:9

L | m

+9,:∗ 32 S

9

3 +7∗ 16 S

9 7:

α=1 d=11,3 mm α=2 d= 14,3 mm