Design Against Fatigue

Failure

2/3/2015 1

Fatigue is the failure of a mechanical element by the

growth of a crack within a material under variable,

repeated, alternating, or fluctuating stresses.

Generally, fatigue crack growth occurs under stresses

below the ultimate tensile strength (σut), yield strength

(σy) and critical stress for the original crack (σc).

A fatigue failure has an appearance similar to a brittle

fracture, as the fracture surfaces are flat and

perpendicular to the stress axis with the absence of

necking.

The fatigue of material is effected by the size of the

component, relative magnitude of static and variable

loads and the number of load reversals.

2/3/2015 2

It is observed that about 80% of failure of mechanical components are due to “Fatigue Failure” resulting from fluctuating stresses.

NOTES

The stresses which vary from a minimum value to a maximum value of the same nature, (i.e. tensile or compressive) are called fluctuating stresses.

The stresses which vary from zero to a certain maximum value are called repeated stresses.

The stresses which vary from a minimum value to a maximum value of the opposite nature (i.e. from a certain minimum compressive to a certain maximum tensile or from a minimum tensile to a maximum compressive) are called alternating stresses.

The stresses which vary from one value of compressive to the same value of tensile or vice versa, are known as completely reversed or cyclic stresses.

2/3/2015 3

Alternating stress Repeated stress

Completely reversed stress Fluctuating stress

σMax= Maximum stress σa ,σv=Alternating stress

σMin= Minimum stress σr = Range of stress

σm= Mean or average stress

2/3/2015 4

Fatigue failure is characterized by three stages

I. Crack Initiation

II. Crack propagation

III. Final fracture

Fracture zone Propagation zone, striation

Crack initiation site

VW crank shaft – fatigue failure due to cyclic bending and torsional stresses

2/3/2015 5

Jack hammer component,

shows no yielding before

fracture.

Crack initiation site

Propagation zone, striation

Fracture zone

2/3/2015 6

Fatigue-Life Methods

The three major fatigue life methods used in design and

analysis are

1. The stress-life method.

2. The strain-life method.

3. The linear-elastic fracture mechanics method.

• These methods attempt to predict the life in number of

cycles to failure (N) for a specific level of loading.

• This cycles number (N) is generally classified as

1. low-cycle fatigue , where 1 ≤ N ≤ 103

2. high-cycle fatigue , where N > 103.

2/3/2015 7

The Stress-Life Method (SLM)

The stress-life method, based on stress levels only, is the

least accurate approach, especially for low-cycle

applications.

To determine the strength of materials under the action

of fatigue loads, specimens are subjected to repeated or

varying forces of specified magnitudes while the cycles

or stress reversals are counted to destruction.

The most widely used fatigue-testing device is the R.R.

Moore high-speed rotating-beam machine. This machine

subjects the specimen to pure bending (no transverse

shear) by means of weights.

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Test-specimen geometry for the R. R. Moore rotating beam machine.

Motor

Load

Rotating beam machine – applies fully reverse bending stress

Typical testing apparatus (pure bending)

The specimen

is polished

2/3/2015 9

The S-N Diagram • The S-N diagram is the graphical representation of

stress amplitude (Sf) verses the number of stress cycles (N) before the fatigue failure on semilog paper or on log-log paper.

• To establish the fatigue strength of a material, quite a number of tests are necessary because of the statistical nature of fatigue. For the rotating-beam test, a constant bending load is applied, and the number of revolutions (stress reversals) of the beam required for failure is recorded.

• The first test is made at a stress that is somewhat under the ultimate strength of the material. The second test is made at a stress that is less than that used in the first. This process is continued, and the results are plotted as S-N diagram.

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In the case of the steels as shown in Fig.(A), a knee

occurs in the graph, and beyond this knee failure will not

occur, no matter how great the number of cycles. The

strength corresponding to the knee is called the

endurance limit or the fatigue limit.

Aluminium alloys does not have an endurance limit as

shown in Fig.(B), normally the fatigue strength Sf is

reported at a specific number of cycles, normally N =

5(108) cycles of reversed stress.

Endurance or fatigue limit stress (Se) is defined as the

maximum amplitude of completely reversed stress that

the standard specimen can sustain for an unlimited

number of cycles without fatigue failure.

2/3/2015 11

Fig.(A) S-N diagram plotted from the results of completely reversed axial

fatigue tests (Steel-normalized σu=125kpsi) 2/3/2015 12

Fig.(B) S-N bands for representative aluminium alloys, excluding

wrought alloys (Sut <38 kpsi).

2/3/2015 13

We also distinguish a finite-life region and an infinite-

life region in S-N diagram as shown in Fig.(A).

The boundary between these regions cannot be clearly

defined except for a specific material; but it lies

somewhere between 106 and 107 cycles for steels, as

shown in Fig.(A).

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The Strain-Life Method (ELM)

The Strain-Life Method is The best approach yet advanced to explain the nature of fatigue failure which is based on the investigation of behavior of materials subject to cyclic deformation.

Generally, the strength decreases with stress repetitions, while other materials may be strengthened, instead, by cyclic stress reversals (i.e. the elastic limits of annealed steels are likely to increase when subjected to cycles of stress reversals, while cold-drawn steels exhibit a decreasing elastic limit).

Fig.(C) has been constructed to show the general appearance of these plots for the first few cycles of controlled cyclic strain. In this case the strength decreases with stress repetitions, as evidenced by the fact that the reversals occur at ever-smaller stress levels.

2/3/2015 15

Fig. (D)A log-log plot showing how the

fatigue life is related to the true-strain

amplitude for hot-rolled SAE 1020 steel.

Fig.(C) True stress–true strain

hysteresis loops showing the

first five stress reversals of a

cyclic-softening material.

2/3/2015 16

Note that the slope of the line AB is the modulus of

elasticity E.

∆σ is the stress range.

∆εp is the plastic-strain range.

∆εe is the elastic strain range.

∆ε is the total-strain range.

∆ε =∆εe + ∆εp

To explain the Fig.(D), we first define the following

terms:

Fatigue ductility coefficient (ε′F) is the true strain

corresponding to fracture in one reversal (point A in

Fig.(C)). The plastic-strain line begins at this point in

Fig.(D).

2/3/2015 17

Fatigue strength coefficient (σ′F) is the true stress

corresponding to fracture in one reversal (point A in

Fig.(C). Note in Fig.(D) that the elastic-strain line

begins at σ′F/E.

Fatigue ductility exponent (c) is the slope of the

plastic-strain line in Fig.(D) and is the power to which

the life 2N must be raised to be proportional to the

true plastic-strain amplitude. If the number of stress

reversals is 2N, then N is the number of cycles.

Fatigue strength exponent (b) is the slope of the

elastic-strain line, and is the power to which the life

2N must be raised to be proportional to the true-stress

amplitude.

2/3/2015 18

Now, from Fig.(D), we see that the total strain is the sum

of the elastic and plastic components. Therefore, the total

strain amplitude is half the total strain range.

The equation of the plastic-strain line is

The equation of the elastic strain line is

∴ The total-strain amplitude is

222

pe

)2('

2N

c

F

p

)2('

2N

bFe

E

)2()2('

'

2NN

bF

c

F E

2/3/2015 19

The Linear-Elastic Fracture Mechanics

Method (LEFM)

The fracture mechanics method assumes a crack is already

present and detected which is employed to predict crack

growth with respect to stress intensity.

Including three stages,

1) The first phase of fatigue cracking is designated as

stage I fatigue.

2) The second phase, that of crack extension, is called

stage II fatigue.

3) Final fracture is occurs during stage III fatigue which

is associated with rapid acceleration of crack growth

then fracture.

2/3/2015 20

Crack Growth

Fatigue cracks nucleate and grow when stresses vary

and there is some tension in each stress cycle.

There are three modes of crack propagation,

I. The opening crack propagation mode.

II. The sliding crack propagation mode.

III. The tearing crack propagation mode.

2/3/2015 21

Consider the stress to be variable between the limits of

σmin and σmax, where the stress range is defined as

∆σ = σmax− σmin.

Consider the stress intensity factor (K) which is used in

fracture mechanics to predict the stress state ("stress

intensity") near the tip of a crack caused by a remote

load or residual stresses.

Thus, for ∆σ, the stress intensity

range per cycle is

aK I

aMinMaxIK )(

aK I

2/3/2015 22

Where,

β: The stress intensity modification factor.

a: The length of crack.

KI: The stress intensity

factor for first mode.

Chart of Thin plate in tension

or simple compression with a

certain crack to select proper β

for mode I crack propagation..

2/3/2015 23

To develop fatigue strength data, a number of specimens

of the same material are tested at various levels of ∆σ.

Cracks nucleate at or very near a free surface or large

discontinuity.

Fig.(E) The increase in crack length (a) from an initial length of (ai)

as a function of cycle count for three stress ranges,

(∆ σ)3 > (∆σ)2 > (∆ σ)1. 2/3/2015 24

Fig.(F) When da/dN is measured in Fig.(E) and plotted on log-log

coordinates, the data for different stress ranges superpose, giving rise to a

sigmoid curve as shown. (∆KI)th is the threshold value of ∆KI, below

which a crack does not grow. From threshold to rupture an aluminum

alloy will spend 85–90 percent of life in region I, 5–8 percent in region II,

and 1–2 percent in region III. Where, Kc is critical stress intensity factor

and R = σmin/σmax 2/3/2015 25

Here we present a simplified procedure for estimating

the remaining life of a cyclically stressed part after

discovery of a crack. This requires the assumption that

plane strain conditions prevail.

Assuming a crack is discovered early in stage II, the

crack growth in region II of Fig.(F) can be approximated

by the Paris equation, which is of the form

where C and m are empirical material constants.

2/3/2015 26

Substituting in above equation and

integrating gives,

ai: The initial crack length

af: The final crack length corresponding to failure

Nf: The estimated number of cycles to produce a failure

after the initial crack is formed.

aK I

2/3/2015 27

Example: The bar shown in figure below is subjected to a

repeated moment 0 ≤ M ≤ 1200 lbf·in. The bar is AISI

4430 steel with Sut= 185 kpsi, Sy= 170 kpsi, and KIc = 73

kpsi√in. Material tests on various specimens of this

material with identical heat treatment indicate worst-

case constants of C = 3.8(10−11)(in/cycle)/(kpsi√in)m and

m = 3.0. As shown, a nick of size 0.004in has been

discovered on the bottom of the bar. Estimate the

number of cycles of life remaining.

L=4in

2/3/2015 28

Solution:

This value is below the yield strength. If β = 1.03, we

approximate af as

kpsi

cI

M

inbh

c

I

20.11501042.0

1200

01042.06

)5.0(25.0

6

322

ina

Ka

f

Icf

1205.0)2.115*03.1

73(

1

)(1

2

2

max

2/3/2015 29

From this figure, we compute

Thus af/h varies from near zero

to approximately 0.241. From

shown Figure, for this range β

Is nearly constant at approxima-

tely 1.07. We will assume it to be

so, and re-evaluate af as

241.05.0

1205.0

h

a f

ina

Ka

f

Icf

112.0)2.115*07.1

73(

1

)(1

2

2

max

2/3/2015 30

Thus the estimated remaining life is

cyclesNf

aa

daN

a

daN

a

da

CN

f

f

a

a

mf

f

i

35.05.03

112.0

004.0

5.03112.0

004.0 3

3

112.0

004.0 311

10*6782.64]004.0112.0[10*0438.5

)](2[10*5219.210*5219.2

])2.115(07.1[10*8.3

1

)(

1

2/3/2015 31

The Endurance Limit

Fig.1 Graph of endurance limits versus tensile strengths from actual test results

for a large number of wrought irons and steels. Ratios of S′e /Sut of 0.60, 0.50,

and 0.40 are shown by the solid and dashed lines. Note also the horizontal

dashed line for S′e= 105 kpsi.

2/3/2015 32

Therefore, after simplifying the observation, the estimation

of the endurance limit is

The prime mark on S′e in this equation refers to the rotating

beam specimen itself.

The unprimed symbol Se for the endurance limit of an

actual machine element subjected to any kind of loading.

2/3/2015 33

Fatigue Strength

is the highest stress that a material can withstand for a

given number of cycles without breaking. In other words,

it is the maximum stress that can be applied for a certain

number of cycles without fracture.

It is affected by environmental factors such as corrosion.

To approximate relationship between S-N during high

cycle, two stages approximation are required which are

I. Approximation of the fatigue strength fraction (f ) at

N= 103.

II. Approximation of the S–N during high cycle at

N>103.(For an actual mechanical component)

2/3/2015 34

I. Approximation of the fatigue strength fraction (f )

at N= 103.

◦ Defining the specimen fatigue strength at a specific number of

cycles as (S′f )N = E∆εe/2,

Since,

At 103 cycles,

)2('

2N

bFe

E

)( NS b

FNf *)2(')'(

b

FfS )10*2(')'( 3

103

ut

f

S

Sf

310)'(

b

ut

F

Sf )10*2(

'3

2/3/2015 35

If this true-stress–true-strain equation (σ′F= σ0εm, with

ε=ε′F) is not known, the SAE approximation for steels

with HB ≤ 500 may be used:

To find b, substitute the endurance strength and

corresponding cycles, S′e and Ne, respectively into Eq.

(*) and solving for b

MpaS' or kpsi S utFutF 34550'

)2log(

)'/'log(

e

eF

N

Sb

2/3/2015 36

Above figure is a plot of f for 70 ≤ Sut ≤ 200 kpsi. To be conservative, for Sut < 70 kpsi, let f = 0.9.

Fig.2 Fatigue strength fraction, f, of Sut at 103 cycles for S′e = Se= 0.5Sut at 106

cycles.

2/3/2015 37

II. Approximation of the S–N during high cycle at N>103

Sf = a Nb

Where N is cycles to failure and the constants a and b are

defined by the points 103, Sf103 and 106, Se with Sf10

3= f Sut.

Substituting these two points in above equation gives

If a completely reversed stress σrev is given, setting Sf =

σrev. the number of cycles-to-failure can be expressed as

)log()( 2

e

ut

e

ut

S

Sf

3

1b and

S

Sfa

brev

aN

1

)(

2/3/2015 38

Example: Given a 1050 HR steel, estimate

(a) the rotating-beam endurance limit at 106 cycles.

(b) the endurance strength of a polished rotating-beam

specimen corresponding to 104 cycles to failure

(c) the expected life of a polished rotating-beam specimen

under a completely reversed stress of 55 kpsi.

Solution:

(a) From Table A–20, Sut = 90 kpsi.

Owing of Sut ≤ 200 kpsi, the endurance limit is

kpsiS

SS

e

ute

45'

)90(5.05.0'

2/3/2015 39

2/3/2015 40

(b) From Fig.2, for Sut = 90 kpsi, f =0.86.

For 104 cycles to failure,

S′f = aNb= 133.1(104)-0.0785

S′f = 64.6 kpsi

(c) For σrev= 55 kpsi

kpsi S

Sfa

e

ut 1.13345

)90*86.0(

'

)( 22

0785.0)45

)90(86.0log(

3

1)

'log(

e

ut

S

Sf

3

1b

cyclesN

aN brev

4

0785.011

10*75.7

)1.133

55()(

2/3/2015 41

Endurance Limit Modifying Factors

The experiments results of endurance limit in laboratory

are not matched with a mechanical or structural member.

Some difference include

o Material: composition, basis of failure, variability.

o Manufacturing: method, heat treatment, fretting

corrosion, surface condition, stress concentration.

o Environment: corrosion, temperature, stress state,

relaxation times.

o Design: size, shape, life, stress state, speed, fretting,

galling

2/3/2015 42

A Marin equation is therefore written as

Where,

ka: surface condition modification factor

kb: size modification factor

kc: load modification factor

kd: temperature modification factor

ke: reliability factor

kf: miscellaneous-effects modification factor (In this course, Kf will not be included)

S′e: rotary-beam test specimen endurance limit

Se: endurance limit at the critical location of a machine part in the geometry and condition of use. (Se < S′e)

When endurance tests of parts are not available, estimations are made by applying Marin factors to the endurance limit.

'SkkkkkkS efedcbae

2/3/2015 43

Surface Factor ka

Where, Sut is the minimum tensile strength and a and b are to be found in

below Table 6–2.

Example: A 1030 CD steel has a machined surface. Estimate

surface factor ka.

Solution: From Table 6–2, a = 4.51 and b =−0.265.

b

uta aSK

858.0)520(51.4aSK 265.0b

uta

2/3/2015 44

Size Factor kb

For axial loading there is no size effect,

For bending and torsion,

Rotating

Nonrotating

It’s depending on an equivalent diameter de ... Table 6–3

1bK

2/3/2015 45

Table 6–3

2/3/2015 46

Example: A steel shaft loaded in bending is 32 mm in

diameter, abutting a filleted shoulder 38 mm in diameter.

The shaft material has a mean ultimate tensile strength of

690 MPa. Estimate the Marin size factor kb if the shaft is

used in (a) A rotating mode, (b) A nonrotating mode.

Solution:

(a) Since, 2.79 ≤ (d=32) ≤ 51 mm

(b) From Table 6–3,

858.0)62.7

32()

62.7( 107.0107.0 d

Kb

954.0)62.7

84.11(

mm 51 11.84)d( 2.79,84.11)32(37.037.0

107.0

e

b

e

K

mmdd

2/3/2015 47

Loading Factor kc

Temperature Factor kd

Where,

torsion

axial

bending

59.0

85.0

1

Kc

RT

Td

S

SK ST = tensile strength at operating temperature

SRT = tensile strength at room temperature

F1000T 70 F

T)10(595.0T)10(104.0

T)10(115.0T)10(432.0975.0K

4

F

123

F

8

2

F

5

F

3

d

2/3/2015 48

2/3/2015 49

Table 6–4

Example: A 1035 steel has a tensile strength of 70 kpsi

and is to be used for a part that sees 450°F in service.

Estimate the Marin temperature modification factor and

(Se)450◦ if

(a) The room-temperature endurance limit by test is

(S′e)70◦ = 39.0kpsi.

(b) Only the tensile strength at room temperature is

known.

Solution:

(a)

2/3/2015 50

4

F

123

F

8

2

F

5

F

3

d

T)10(595.0T)10(104.0

T)10(115.0T)10(432.0975.0K ,Using

007.1)450)(10(595.0)450)(10(104.0

)450)(10(115.0)450)(10(432.0975.0K

41238

253

d

2/3/2015 51

kpsi 3.39)39(007.1)'S(K)S( Thus,70ed450e

(b) Interpolating from Table 6–4 gives,

Thus, the tensile strength at 450°F is estimated as

kpsi 49.70)70(007.1)S()S/S()S(70ut450RTT450ut

kpsi245.35)S(

)49.70(5.0)S(5.0)S( Then,

450e

450ut450e

007.1)S/S(400500

400450

018.1995.0

018.1)S/S(

450RTT450RTT

Reliability Factor ke

2/3/2015 52

ae z08.01K

Table 6–5

Example: A 1015 hot-rolled steel bar has been machined to a

diameter of 1 in. It is to be placed in reversed axial loading

for 7(104) cycles to failure in an operating environment of

550°F. Using ASTM minimum properties, and a reliability

of 99 percent, estimate the endurance limit and fatigue

strength at 7(104) cycles.

Solution:

From Table A–20, Sut = 50 kpsi at 70°F.

2/3/2015 53

4

F

123

F

8

2

F

5

F

3

d

T)10(595.0T)10(104.0

T)10(115.0T)10(432.0975.0K ,Using

98.0)550)(10(595.0)550)(10(104.0

)550)(10(115.0)550)(10(432.0975.0K

41238

253

d

The ultimate strength at 550°F is then

The rotating-beam specimen endurance limit at 550°F is then estimated as

Next, we determine the Marin factors. For the machined surface, Eq. (6–19) with Table 6–2 gives,

The size factor kb = 1.

The loading factor is kc = 0.85.

The temperature factor kd = 1, since we accounted for the temperature in modifying the ultimate strength and consequently the endurance limit.

2/3/2015 54

kpsi 0.49)50(98.0)S(K)S(70utd550ut

kpsi 5.24)49(5.0)S(5.0)'S(550ut550e

963.0)49(70.2aSK 265.0b

uta

For 99 percent reliability, from Table 6–5, ke = 0.814.

The endurance limit for the part is estimated by Eq. (6–18)

as

For the fatigue strength at 7(104) cycles,

(Sut)550 ̊ = 49 < 70 kpsi, then f = 0.9.

2/3/2015 55

kpsi 16.3 814)24.5.85)(1)(0.0.963(1)(0

)'S(kkkkk)S( 550eedcba550e

kpsi3.1193.16

)49*9.0(

S

) Sf(a

2

e

2

ut

1441.0)3.16

)49(9.0log(

3

1)

S

Sflog(

3

1b

e

ut

Sf = aNb= 119.3(7*104)-0.1441 = 23.9 kpsi

Stress Concentration and Notch Sensitivity

Where, Kf is a reduced value of Kt and σ0 is the nominal

stress. The factor Kf is commonly called a fatigue stress-

concentration factor, and hence the subscript f.

Notch sensitivity q

2/3/2015 56

specimen free-notchin stress

specimen notchedin stress maximumfK

0fsmax0fmax KK or

stress nominalover stress al theoreticof increase

stress nominalover stress actual of increaseq

Where, q is usually between zero and unity.

NOTES

When the material has no sensitivity to notches,

q = 0 and Kf= 1

When the material is fully sensitivity to notches,

q = 1 and Kf= Kt

q = 0.20 for all grades of cast iron

2/3/2015 57

1K

1Kq

K

Kq

ts

fs

shear

00ts

00fs

shear

1K

1Kq

K

Kq

t

f

00t

00f

)1K(q1K)1K(q1K tsshearfstf or

2/3/2015 58

Fig. 6–20 Notch-sensitivity charts for steels and wrought aluminum alloys

subjected to reversed bending or reversed axial loads.

2/3/2015 59

Figure 6–21 Notch-sensitivity curves for materials in reversed torsion

There is another way to estimate notch sensitivity q, which is by Neuber’s equation

For bending or axial

For torsion

o Where √a is defined as the Neuber constant and is a material constant. r is notch radius.

NOTE: the equations apply to steel and Sut is in kpsi.

2/3/2015 60

r

a1

1q

r/a1

1K1K t

f

3

ut

82

ut

5

ut

3 S)10(67.2S)10(51.1S)10(08.3246.0a

3

ut

82

ut

5

ut

3 S)10(67.2S)10(35.1S)10(51.2190.0a

Example: A steel shaft in bending has an ultimate strength

of 690 MPa and a shoulder with a fillet radius of 3 mm

connecting a 32-mm diameter with a 38-mm diameter.

Estimate Kf using:

(a) Figure 6–20.

(b) Equations.

Solution:

From Fig. A–15–9, using

D/d = 38/32 = 1.1875,

r/d = 3/32 = 0.093 75,

we read the graph to find

Kt =1.65.

2/3/2015 61

(a) From Fig. 6–20, for Sut = 690 MPa and r = 3 mm,

q .= 0.84.

Thus,

(b) Converting Sut= 690 Mpa= 100kpsi.

2/3/2015 62

55.1)165.1(84.01)1K(q1K tf

55.1

3

313.01

165.11

r/a1

1k1K

mm313.0in0622.0

)100)(10(67.2)100)(10(51.1)100)(10(08.3246.0

S)10(67.2S)10(51.1S)10(08.3246.0a

tf

38253

3

ut

82

ut

5

ut

3

Thus,

Example: For the step-shaft of previous example, it is

determined that the fully corrected endurance limit is

Se = 280 MPa. Consider the shaft undergoes a fully

reversing nominal stress in the fillet of (σrev)nom = 260

MPa. Estimate the number of cycles to failure.

Solution:

From previous example, Kf = 1.55, and the ultimate

strength is Sut = 690 MPa = 100 kpsi. The maximum

reversing stress is

From Fig.2, f = 0.845.

2/3/2015 63

Mpa403)260(55.1)(K)( nomrevfmaxrev

2/3/2015 64

Mpa1214280

)690*845.0(

S

) Sf(a

2

e

2

ut

1062.0)280

)690(845.0log(

3

1)

S

Sflog(

3

1b

e

ut

cycles10*3.32N

)1214

403(

)a

(N

3

1062.01

b1

rev

Modifying Factor to Account for fatigue stress

concentration

The endurance limit is reduced due to fatigue stress

concentration. The fatigue stress concentration factor is

less than stress concentration factor due to notches

sensitivity of the material.

To apply the effect of fatigue stress concentration, the

designer can (1) either reduce the endurance limit by

dividing it by Kf or (2) increase the nominal stress

amplitude by multiplied it by Kf .

o In this course, we will consider (2) which increases the

nominal stress amplitude by multiplied it by Kf

2/3/2015 65

Example: Figure 6–22a shows a rotating shaft simply

supported in ball bearings at A and D and loaded by a

nonrotating force F of 6.8 kN. Using ASTM “minimum”

strengths, estimate the life of the part.

Solution:

We will solve the problem by

first estimating the strength at

point B,

2/3/2015 66

Fig.6-22(a) Shaft drawing showing all

dimensions in millimeters; all fillets 3-mm

radius. The shaft rotates and the load is

stationary; material is machined from AISI

1050 cold-drawn steel.

From Table A–20 we find Sut = 690 MPa and Sy = 580

MPa. The endurance limit S′e is estimated as

Se′= 0.5(690) = 345 MPa

Estimating Marin’s factors,

2/3/2015 67

798.0)690(51.4aSKfactor, Surface 265.0b

uta

MPa236

8)3450.798(0.85 S

1 k k k Since,

858.0)62.7

32(

)62.7

d(K ,factor Size

e

edc

107.0

107.0

b

To find the geometric stress

-concentration factor Kt,

with D/d = 38/32 = 1.1875

and r/d = 3/32 = 0.093 75

and read Kt = 1.65.

2/3/2015 68

To find the fatigue stress-concentration factor kf,

Substituting Sut = 690/6.89 = 100 kpsi into Eq. (6–35a)

gives, 33)–(6 Eq. into thisngSubstituti ,mm313.0in0622.0

100)10(67.2100)10(51.1100)10(08.3246.0a 38253

55.1r/a1

1K1K t

f

30.3131

1-1.651

The next step is to estimate the bending stress at point B.

The bending moment is

The reversing bending stress is,

2/3/2015 69

(b) Bending moment diagram. m.N5.695

250*550

)8.6(225

250*550

F225xRM 1B

3333 mm )(10 3.217 32/32π 32/πd I/c

MPa335.1 Pa )335.1(10

)(103.217

695.51.55

I/c

MK σ

6

6Bfrev

This stress is greater than Se and less than Sy. This means

we have both finite life and no yielding on the first

cycle.

The life of the part,

The ultimate strength, Sut = 690MPa = 100 kpsi. From Fig.

6–18, we will get f = 0.844

2/3/2015 70

Mpa1437236

)690*844.0(

S

) Sf(a

2

e

2

ut

1308.0236

)690(844.0log

3

1)

S

Sflog(

3

1b

e

ut

cycles10*68)1437

1.335()

a(N 31308.0

1b

1rev

Characterizing Fluctuating Stresses

It has been found that in periodic patterns exhibiting a

single maximum and a single minimum of force, the

shape of the wave is not important, but the peaks on

both the high side (maximum) and the low side

(minimum) are important.

Thus Fmax and Fmin in a cycle of force can be used to

characterize the force pattern.

Fm: the midrange steady component of force, and

Fa: the amplitude of the alternating component of force.

2/3/2015 71

2

FFF and

2

FFF minmax

aminmax

m

2/3/2015 72

2 and

2

minmaxa

minmaxm

m

a

max

min A and R

σa = amplitude component σr = range of stress

σm = midrange component A= the amplitude ratio

R= the stress ratio

σa = σao and σm= σmo → in the absence of a notch

σa = kf σao and σm= kf σmo → with a notch

2/3/2015 73

Fatigue Failure Criteria for Fluctuating Stress

we want to vary both the midrange stress and the stress amplitude, or alternating component, to learn something about the fatigue resistance of parts when subjected to such situations.

There are Five criteria of failure are diagrammed in Fig. 6–27: the Soderberg, the modified Goodman, the Gerber, the ASME-elliptic, and yielding.

The Fig. 6–27 shows that only the Soderberg criterion guards against any yielding, but is biased low.

Considering the modified Goodman line as a criterion, point A represents a limiting point with an alternating strength Sa and midrange strength Sm. The slope of the load line shown is defined as r = Sa/Sm.

2/3/2015 74

2/3/2015 75

Fig. 6–27 Fatigue diagram showing various criteria of failure. For each

criterion, points on or “above” the respective line indicate failure.

Name The criterion equation

Soderberg Fatigue Line

Modified Goodman Fatigue Line

Gerber Fatigue Curve

ASME-elliptic Fatigue Curve

Yield (Langer) line

2/3/2015 76

1S

S

S

S

y

m

e

a

1S

S

S

S

ut

m

e

a

1S

S

S

S2

ut

m

e

a

1S

S

S

S2

y

m

2

e

a

yma S S S

The stresses nσa and nσm can replace Sa and Sm, where n

is the design factor or factor of safety. See table Table *

There are two ways to proceed with a typical analysis.

One method is to assume that fatigue occurs first and

use one of Eqs. (1) to (4) to determine n or size,

depending on the task. Most often fatigue is the

governing failure mode. Then follow with a static check.

If static failure governs then the analysis is repeated

using Eq. (5). See table Table *

2/3/2015 77

2/3/2015 78

No. Name The criterion equation with F.S

1 Soderberg

2 Modified Goodman

3 Gerber

4 ASME-elliptic

5 Langer static yield

n

1

SS y

m

e

a

n

1

SS ut

m

e

a

1S

n

S

n2

ut

m

e

a

1S

n

S

n2

y

m

2

e

a

n

S

y

ma

Table *

2/3/2015 79

Table 6–6

Amplitude and Steady Coordinates of Strength and Important Intersections in

First Quadrant for Modified Goodman and Langer Failure Criteria

2/3/2015 80

Table 6–7

Amplitude and Steady Coordinates of Strength and Important Intersections in

First Quadrant for Gerber and Langer Failure Criteria

2/3/2015 81

Table 6–8

Amplitude and Steady Coordinates of Strength and Important Intersections in

First Quadrant for ASME-elliptic and Langer Failure Criteria

2/3/2015 82

NOTES for the above three tables (6-6) to (6-8):

The first row of each table corresponds to the fatigue

criterion, the second row is the static Langer criterion, and

the third row corresponds to the intersection of the static

and fatigue criteria.

The first column gives the intersecting equations and the

second column the intersection coordinates.

Example: A 1.5-in-diameter bar has been machined from an

AISI 1050 cold-drawn bar. This part is to withstand a

fluctuating tensile load varying from 0 to 16 kip. Because of

the ends, and the fillet radius, a fatigue stress-concentration

factor Kf is 1.85 for 106 or larger life. Find Sa and Sm and

the factor of safety guarding against fatigue and first-cycle

yielding, using (a) the Gerber fatigue line and (b) the

ASME-elliptic fatigue line.

Solution:

A . We start with calculate the axial midrange force

component and amplitude of force component,

The nominal axial stress components σao and σmo are

Applying Kf to both components σao and σmo,

2/3/2015 83

kIP82

016

2

FFF

kiP82

016

2

FFF

minmaxa

minmaxm

kpsi53.44/5.1

8

4/d

F

kpsi53.44/5.1

8

4/d

F

22

aao

22

mmo

maofa σ kpsi 8.38 1.85(4.53) K

B . From Table A–20, Sut = 100 kpsi and Sy = 84 kpsi.

Estimating the endurance limit,

The Marin factors are, deterministically,

ka = 2.70(100)−0.265 = 0.797

kb = 1 (axial loading)

kc = 0.85

kd = ke = 1

2/3/2015 84

kpsi 9.33 50)11)((1)(0.85)(7970.

'SkkkkkS eedcbae

(Ideal) kpsi50 )100(5.0S5.0'S ute

C . Let us calculate the factors of safety first.

(a) From the bottom panel from Table 6–7 the factor of

safety for fatigue is

From Eq. (6–49) the factor of safety guarding against first-

cycle yield is

2/3/2015 85

66.3)38.8(100

)9.33)(38.8(211

9.33

38.8

38.8

100

2

1

S

S211

S

S

2

1n

22

2

aut

em

e

a

2

m

utf

01.538.838.8

84

S n

ma

y

y

For drawing the designer’s diagram,

1. We have to know all the details about the load line

starting from its slope (r),

2. Also, the intersection point (B) with Gerber fatigue

curve

2/3/2015 86

45rTan 138.8

38.8r 1

m

a

kpsi7.30100)1(

)9.33(211

)9.33(2

)100(1

rS

S211

S2

SrS

222

2

ut

e

e

2

ut

2

a

kpsi7.301

7.30

r

SS a

m

3. In this Ex., the intersection point (C) between the load

line and Langer line can be found by continuously

drawing the load line with respect to till

intersects the Langer line

4. Finally, the intersection point (D) between the Gerber

fatigue curve with Langer line.

2/3/2015 87

kpsi0.649.33

841

100

)9.33(211

)9.33(2

100

S

S1

S

S211

S2

SS

22

e

y

2

ut

e

e

2

utm

kpsi 20 6484 S S S mya

45rTan 1

The critical slope is thus

which is less than the actual load line of r = 1. This

indicates that fatigue occurs before first-cycle-yield.

As a check on the previous result, nf = OB/OA = Saf /σa =

Smf /σm = 30.7/8.38 =3.66. ny= OC/OA= Say/σa =Smy/σm =

5.01 and we see total agreement.

2/3/2015 88

312.064

20

S

Sr

m

acri

Fig. 6–28 Principal points A, B, C, and D on the designer’s diagram drawn for

Gerber, Langer, and load line.

2/3/2015 89

(b) Repeating the same procedure for the ASME-elliptic

line, for fatigue

Again, this is less than ny = 5.01 and fatigue is predicted to

occur first.

For drawing the designer’s diagram,

2/3/2015 90

75.3)84/38.8()9.33/38.8(

1

)S/()S/(

1n

22

2

ym

2

ea

f

1. We have to know all the details about the load

line starting from its slope (r),

45rTan 138.8

38.8r 1

m

a

2/3/2015 91

2. Also, the intersection point (B) with ASME fatigue

curve. We obtain the coordinates Sa and Sm of point B

by using

kpsi4.311

4.31

r

SS

kpsi4.31)84)(1()9.33(

)84)(9.33)(1(

SrS

SSrS

am

222

222

2

y

22

e

2

y

2

e

2

a

3. In this Ex., the intersection point (C) between the load

line and ASME curve can be found by continuously

drawing the load line with respect to till

intersects the ASME-elliptic fatigue curve.

45rTan 1

4. Finally, the intersection point (D) between the ASME-

elliptic fatigue curve with Langer line.

2/3/2015 92

kpsi5.605.2384SSS

kpsi5.23849.33

9.33)84(2

SS

SS2S

aym

22

2

2

y

2

e

2

ey

a

388.05.60

5.23

S

Sr

m

acri

The critical slope is thus

Which is less than the actual load line of r =1. This indicates

that fatigue occurs before first-cycle-yield

As a check on the previous result, nf = OB/OA = Saf /σa =

Smf /σm = 30.7/8.38 =3.75. ny= OC/OA= Say/σa =Smy/σm =

5.01 and we see total agreement.

Therefore, the Gerber and the ASME-elliptic fatigue failure

criteria are very close to each other and are used

interchangeably.

2/3/2015 93

Fig. 6–29 Principal points A, B, C, and D on the designer’s diagram drawn for

ASME-elliptic, Langer, and load line.

Torsional Fatigue Strength under Fluctuating Stresses

For ductile material, polished, notch-free, and cylindrical,

a torsional steady-stress component not more than the

torsional yield strength has no effect on the torsional

endurance limit.

For materials with stress concentration, notches, or

surface imperfections, the torsional fatigue limit decreases

monotonically with torsional steady stress.

The modified Goodman relation for pulsating torsion is

from distortion-energy theory,

2/3/2015 94

utsu 0.67S S

ytsy 0.5777S S

Combinations of Loading Modes

It may be helpful to think of fatigue problems as being in

three categories:

• Completely reversing simple loads (σm = 0).

• Fluctuating simple loads.

• Combinations of loading modes.

Here will be used von-mises theory for combined loads

1. The first step is to generate two stress elements—one

for the alternating stresses and one for the midrange

stresses.

2. Apply the appropriate fatigue stress-concentration

factors to each of the stresses; i.e., apply (Kf )bending for

the bending stresses, (Kf s)torsion for the torsional

stresses, and (Kf )axial for the axial stresses.

2/3/2015 95

3. Next, calculate an equivalent von Mises stress for each

of these two stress elements, σ′a and σ′m.

4. For the endurance limit, Se, use the endurance limit

modifiers, ka, kb, and kc, for bending.

The torsional load factor, kc = 0.59 should not be

applied as it is already accounted for in the von Mises

stress calculation.

The load factor for the axial load can be accounted for

by dividing the alternating axial stress by the axial

load factor of 0.85.

5. Finally, select a fatigue failure criterion (modified

Goodman, Gerber, ASME-elliptic, or Soderberg) to

complete the fatigue analysis.

2/3/2015 96

Simple Example, consider the common case of a shaft with

bending stresses, torsional shear stresses, and axial

stresses.

2/3/2015 97

2/12

torsionalmtorsionalfs

2

axialmaxialfbendingmbendingf

'

m

2/1

2

torsionalatorsionalfs

2

axiala

axialfbendingabendingf

'

a

2/12

xy

2

x

'

K3KK

K385.0

KK

)3(

Example: A rotating shaft is made of 42-×4-mm AISI

1018 cold-drawn steel tubing and has a 6-mm-diameter

hole drilled transversely through it. Estimate the factor

of safety guarding against fatigue and static failures

using the Gerber and Langer failure criteria for the

following loading conditions:

a) The shaft is subjected to a completely reversed

torque of 120 N·m in phase with a completely

reversed bending moment of 150 N·m.

b) The shaft is subjected to a pulsating torque

fluctuating from 20 to 160 N·m and a steady bending

moment of 150 N·m.

2/3/2015 98

Solution:

(a) Theoretical stress-concentration factors are found from

Table A–16. Using a/D = 6/42 = 0.143 and d/D = 34/42 =

0.810, and using linear interpolation, we obtain A = 0.798

and Kt = 2.366 for bending; and A = 0.89 and Kts = 1.75

for torsion.

Next, using Figs. 6–20 and 6–21, pp. 295–296, with a

notch radius of 3 mm we find the notch sensitivities to be

0.78 for bending and 0.81 for torsion.

2/3/2015 99

61.1)175.1(81.01)1K(q1K

07.2)1366.2(78.01)1K(q1K

tsshearfs

tf

434444

net

334444

net

mm )10(155])34()42[(32

)98.0()dD(

32

AJ

mm )10(31.3])34()42[()42(32

)798.0()dD(

D32

AZ

2/3/2015 100

Table A–16

Approximate Stress-

Concentration Factor Kt

for Bending of a Round

Bar or Tube with a

Transverse Round Hole

The alternating bending stress is now found to be

and the alternating torsional stress is

The midrange von Mises component σ′m is zero. The

alternating component σ′a is given by

From Table A–20, Sut = 440 MPa and Sy = 370 MPa.

The endurance limit of the rotating-beam specimen is

0.5(440) = 220 MPa.

2/3/2015 101

Mpa8.93)10(31.3

15007.2

Z

MK

6

net

fxa

Mpa2.26)10)(155(2

)10)(42(12061.1

J2

TDK

9

3

net

fsxya

Mpa2.104

)]2.26(38.93[)3(

a'

2/1222/12

xya

2

xaa'

2/3/2015 102

Mpa165220)1)(1)(1)(833.0(899.0S

833.0K

899.0K

e

b

a

Marin factors are

Since Se = Sa , the fatigue factor of safety nf is

58.12.104

165

'

Sn

a

af

The first-cycle yield factor of safety is

50.32.104

370

'

Sn

a

y

y

There is no localized yielding; the threat is from fatigue.

2/3/2015 103

Figure 6–32 Designer’s fatigue diagram

(b) We have Ta = (160 − 20)/2 = 70 N·m

and Tm = (160 + 20)/2 = 90 N·m. The corresponding

amplitude and steady-stress components are

The steady bending stress component σxm is

The von Mises components are

2/3/2015 104

Mpa8.93)10(31.3

15007.2

Z

MK

6

net

mfxm

Mpa7.19)10)(155(2

)10)(42(9061.1

J2

DTK

Mpa3.15)10)(155(2

)10)(42(7061.1

J2

DTK

9

3

net

mfsxym

9

3

net

afsxya

Mpa8.99))7.19(38.93()3(

Mpa5.26)]3.15(3[)3(

2/1222/12

xym

2m

'

2/122/12

xyaa'

xm

From Table 6–7, p. 307, the fatigue factor of safety is

From the same table, with r = σ′a/σ′m= 26.5/99.8 = 0.28,

the strengths can be shown to be Sa = 85.5 MPa and Sm

= 305 MPa. See the plot.

The first-cycle yield factor of safety ny is

There is no notch yielding. The likelihood of failure may

first come from first-cycle yielding at the notch. 2/3/2015 105

12.3)5.26(440

)165)(8.99(211

165

5.26

8.99

440

2

1

S

S211

S

S

2

1n

22

2

a'

ut

em'

e

a'2

m'

utf

93.28.995.26

370

''

Sn

ma

y

y

Varying, Fluctuating Stresses; Cumulative Fatigue

Damage

Instead of a single fully reversed stress history block

composed of n cycles, suppose a machine part, at a

critical location, is subjected to

oA fully reversed stress σ1 for n1 cycles, σ2 for n2

cycles, . . . , or

oA “wiggly” time line of stress exhibiting many and

different peaks and valleys.

The method which will be used here to count the

number of cycles it is called the rain-flow counting.

2/3/2015 106

2/3/2015 107

Figure 6–33 Variable stress diagram prepared for assessing cumulative damage

The Palmgren-Miner cycle-ratio summation rule, also

called Miner’s rule, is written

Where,

ni is the number of cycles of operation at stress level σi.

Ni is the number of cycles to failure at stress level σi.

c is the empirical constant in the range 0.7 < c < 2.2.

Using the deterministic formulation as a linear damage

rule we write

where D is the accumulated damage. When D = c = 1,

failure ensues.

2/3/2015 108

cN

n

i

i

i

i

N

nD

Example: Given a part with Sut = 151 kpsi and at the

critical location of the part, Se = 67.5 kpsi. For the

loading of Figure below, estimate the number of

repetitions of the stress-time block in the figure that can

be made before failure.

2/3/2015 109

Solution:

For the figure, We will start to count the number of cycles

and find the midrange stresses σm and amplitude of

stresses σa for each one to construct the table below:

From Fig. 6–18, for Sut = 151 kpsi, f = 0.795.

2/3/2015 110

2/3/2015 111

kpsi5.2135.67

)151*795.0(

S

) Sf(a

2

e

2

ut

0833.05.67

)151(795.0log

3

1)

S

Sflog(

3

1b

e

ut

)2( 5.213

S

a

SN

0833.01

fb

1

f

So,

We prepare to add two columns to the previous table. Using

the Gerber fatigue criterion, with Se = Sf , and n = 1, we can

write

)3(

0 S

0 )S/(1S

me

m2

utm

a

f

where Sf is the fatigue strength associated with a completely

reversed stress, σrev, equivalent to the fluctuating stresses.

Cycle 1: r = σa/σm = 70/10 = 7, and the strength amplitude

from table 6-7 is

Since σa > Sa, that is, 70 > 67.2, life is reduced. From Eq.(3)

2/3/2015 112

kpsi2.67151)7(

)5.67(211

)5.67(2

)151(7

rS

S211

S2

SrS

222

2

ut

e

e

2

ut

2

a

kpsi3.70)151/10(1

70

)S/(1S

22

utm

af

and from Eq. (2)

Cycle 2: r = 10/50 = 0.2, and the strength amplitude is

Since σa < Sa, that is 10 < 24.2, then Sf = Se and indefinite

life follows. Thus, N→∞.

Cycle 3: r = 10/−30 = −0.333, and since σm < 0, Sf = Se,

indefinite life follows and N→∞

2/3/2015 113

cycles )10(6195.213

3.70N 3

0833.01

kpsi2.24151)2.0(

)5.67(211

)5.67(2

)151(2.0S

222

a

From Eq. (6–58) the damage per block is

2/3/2015 114

)10(619n

1)10(619

nD

11

)10(619

1n

N

nD

3

o

3

o

3o

i

i