Fatigue Failure Resulting From Variable Loading (1)

7/22/2019 Fatigue Failure Resulting From Variable Loading (1)

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BITS PilaniPilani Campus


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BITS PilaniPilani Campus

F a t i g u e F a i l u r e R e s u l t i n gf r o m Va r i a b l e L o a d i n g


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BITS Pilani, Pilani Campus

Variable Loading Variable loading results when the applied load or

the induced stress on a component is not constantbut changes with time

In reality most mechanical components experience

variable loading due to-Change in the magnitude of applied load

Example: Extrusion process

-Change in direction of load applicationExample: a connecting rod

-Change in point of load application

Example: a rotating shaft


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Fatigue

Fatigue is a phenomenon associated with variable

loading or more precisely to cyclic stressing orstraining of a material

ASTM Definition of fatigue

The process of progressive localized permanent

structural changes occurring in a material subjected

to conditions that produce fluctuating stresses at

some point or points and that may result in cracks or

complete fracture after a sufficient number of

fluctuations.


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Fracture Failure- Mechanism

Three stages are involved in fatigue failure

-Crack initiation-Crack propagation

-Fracture / Rupture


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Fracture Failure- Mechanism

Crack propagation modes.


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fatigue failure of the crankshaft under pure bending load.

Fatigue failure


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Crack initiation, propagation and rupture in a shaft subjected to repeated bending

Introduction to Fatigue in Metals

Crack initiation at

the outer surface

Beach marks

showing the nature

of crack propagation

Final rupture occurs

over a limited area,

characterizing a very

small load required

to cause it


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Crack initiation at the root of

keyway at B

Final failure over the smallarea at C due to sudden

rupture

Crack propagation occurs

over a time period



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Connecting rod failed by fatigue failure

The crack got initiated at the flash line of forging.

Flash line of

forging



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Fatigue failure of a steam engine connecting rod due to PURE TENSION load.

No surface crack.

Crack may initiate

anywhere that is the

weakest or unknown

source of weakness.

In this rod, the crack

initiated due to forging

flake slightly below the

centre line.The crack propagated radially outward

until some time after which the sudden

rupture occurred.

Radial direction of

crack propagation



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Approach to Fatigue Failure in Analysis and Design

Fatigue life methods

Fatigue strength and endurance limit

Endurance limit modifying factors

Stress concentration and notch sensitivity Fluctuating stresses

Combination of loading modes

Variable, fluctuating stresses, cumulative fatigue

damage


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Fatigue Life Methods

predict the failure in number of cycles N to failure for a specific type of loading

33 10:(HCF)fatiguecycleHigh;101:(LCF)fatiguecycleLow > NN Stress life methods

Based on stress levels only

Least accurate of the three, particularly for LCF

It is the most traditional because easiest to implement for a wide range of applications

Has ample supporting data

Represents high cycle fatigue adequately Strain life methods

Involves more detailed analysis of plastic deformation at localized regions

Good for LCF

Some uncertainties may exist in results because several idealizations get compounded

Hence normally not used in regular practice but only for completeness and special occasions

Linear elastic fracture mechanics methods (LEFM) Assumes that crack is already present and detected

The crack location is then employed to predict crack growth and sudden rupture with respect tothe stress nature and intensity

Most practical when applied to large structures in conjunction with computer codes and periodicinspection


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Stress Life Method

R. R. Moore high-speed rotating

beam machine.


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pure reversed bending without transverse shear

SFD

BMD

Mb

Stress Life Method


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Stress Life Method

Pure bending by means of weights and no transverse shear.

The specimen shown is very carefully machined and polished witha final polishing in the axial direction to void circumferential

scratches.

Number of revolutions of the specimen required for failure arerecorded.

The first test is made at a stress that is some what under the

ultimate strength of the material.

Next, the test is repeated for a lower load, and so on.

The results are plotted in the S-N diagram, which is either semi-log

or log-log.


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The S-N Diagram for steel (UNS G41300), normalized, Sut=812 MPa.

Endurance Limit,

It is the stress at which the

component can sustain

infinite number of cycles

Stress Life Method: S-N Diagram


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The plot in the S-N diagram never becomes horizontal

for non-ferrous metals and alloys For non-ferrous metals and alloys, stress at a specific

number of cycles, normally at 5*108 cycles, must be

used as fatigue strength

Endurance limit for non-ferrous metals and alloys


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For different aluminium alloys (which is non-ferrous)

For non-ferrous metals and alloys, the S-N diagram never becomes horizontal and

hence they do not have endurance limit. Therefore, a stress at a specific number

of cycles, normally at 5*108 cycles, must be used as fatigue strength

Stress Life Method: S-N Diagram


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Instead of referring to experimental data-bank each time,it should be possible to quickly estimate the value ofendurance limit using some kind of formula

To enable that, data has been generated for differenttypes of steels, for endurance limit with respect to the

ultimate tensile strength This plot seemed to closely follow a combination of two

straight lines, of which the second being almost

horizontal at Sut=1460 MPa

Estimation of Endurance Limit


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Estimation of Endurance Limit

For steels, Endurance limit :

conditionsloadingactualin thelimitEndurance

bendingreverseinobtainedlimitEndurance

1460700

146050

'

'

=

=

>

=

e

e

ut

utut

e

S

S

MPaSforMPa

MPaSforS.S


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Endurance limit (Se) is only for rotational bending of

round bar at idealistic conditions (prepared very

carefully and tested under closely controlled conditions).

Endurance strength (Se) is for all other types of loading,geometry and operating conditions.

Endurance limit Vs. Endurance strength

Endurance limit Endurance strength


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Endurance limit modifying factors'

eedcbae SkkkkkS =

strengthendurancei.e.useofconditionandgeometry

theinpartmachineaoflocationcriticalat thelimitenduranceS

limitendurancespecimentestbeam-otaryS

factoronmodificatim

factorr

factoronmodificati

factoronmodificatiload

factoronmodificatisize

factoronmodificaticonditionsurface

e

'

e

=

=

=

=

=

=

=

=

r

effectsusiscellaneok

eliabilityk

etemperaturk

k

k

k

f

e

d

c

b

a


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b

uta aSk =

Table 6.2; page:288

Surface condition modification factor (ka)

The surface modification factor depends on the quality of the finish of

the actual part surface and on the tensile strength of the part material.


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Size modification factor, kb

( )

1.effect,sizenoloadingaxialFor

25451000837.0859.0

5179.224.162.7/

:onlytorsionandendinginarsCScircularrotatingFor

107.0107.0

=

==

b

b

k

mmdifd

mmdifddk

etc.sectioncrosschannelsection,-Ir,rectangulacircular,

butrotating-nonarethatbarsaboutWhat


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( )[ ]

dd

dA

dddA

Case

e

eee

37.0(2)and(1)Equation

),2(01046.0

i.e0.95dofspacingahavingchordsparalleltwoof

outsideareathetwiceisareastresspercent95the

rounds,solidgnonrotatinFor

),1(0766.095.04

circularrotatingfor:1

2

95.0

222

95.0

=

=

==

K

K

Kb for non-rotating shapes

Effective dimension is used

Effective dimension de obtained by equating the volume of

material stressed at and above 95 percent of the maximum stressto the same volume in the rotating-beam specimen

Table 6-3; page:290


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Load modification factor, kc

=torsion

axial

bending

kc

,59.0

,85.0

,1

Actually the kc is dependent on

the Sutof the material. Tables 6-

11 to 6-14 (page no. 333) in

Text Book give the details. Theabove values are average

values.


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Temperature modifying factor, kd

When operating temperatures are below room

temperature, brittle fracture is a strong possibility andshould be investigated first.

When the operating temperatures are higher than room

temperature, yielding should be investigated first becausethe yield strength drops off so rapidly with temperature.

Any stress will induce creep in a material operating at

high temperatures


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Temperature modifying factor, kd

FT

where

TTTTk

o

F

FFFFd

100070

10595.010104.010115.010432.0975.041238253

++=

145 tests of 21 different carbon and alloy steels results a

fourth-order polynomial curve fit to the data underlying


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Temperature modifying factor, kd( ) ( ) ( ) ( )

FT

where

TTTTk

o

F

FFFFd

100070

10595.010104.010115.010432.0975.0 41238253

++=

Effect of Operating Temperature on the Tensile Strength of Steel.(ST= tensile strength at operating temperature;SRT= tensile strength at room temperature)

Table 6.4; page :291

RT

Td

S

Sk =


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ae zk 08.01=

Reliability factor, ke

Table 65


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Accounts for

Residual stress due to shot-peening, hammering

etc.

Corrosion

Coating failure

Spraying etc.

Miscellaneous effects factor, kf


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Estimate the endurance strength of the given material for the following two cases

1. A32 mm diameter shaft made of hot rolled AISI 1040 steel. The shaft surfaces are grinded. The shaft is

subjected to reverse torque and to be used for a part that sees 250C in service at 99.99% reliability.

2. A solid 20 mm side-square rod is cantilevered at one end. The rod supports a completely reverse axial load at

the other end. The material is AISI 1015 cold-drawn steel. The expected reliability of the rod is 90%.

Problem


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Two steels are being considered for manufacture of forged

connecting rods. One is AISI 4340 Cr-Mo-Ni steel capable of

being heat-treated to a tensile strength of 1820 MPa. The otheris a plain carbon steel AISI 1040 with an attainable Sut of 790

MPa. If each rod is to have a size giving an equivalent diameter

of 20 mm, is there any advantage to using the alloy steel for this

fatigue application?

Problem

For 4340 Cr-Mo-Ni steel, Sut = 1820 MPa For 1040 HR steel, Sut = 790 MPa

899.0)20(24.124.1107.0107.0 === dkb

155.0)1820(272

995.0

===

b

uta Sak 385.0)730(272995.0

===

b

uta Sak

.1;1;1;899.0 ==== edcb kkkk.1;1;1 === edc kkk

MPaSe 54.97)700()1()1()1()899.0()155.0( == MPaSe 75.136)395()1()1()1()899.0()385.0( ==


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The single most influential factor leading to high

possibility of crack initiation

Stress concentration can be due to

Function of geometry (sudden change in

size/diameter; holes in the structure etc.and surface texture (surface finish, presence of

disintegrations etc.)

Stress concentration


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Kt=Theoretical stress concentration factor

Stress concentration (Kt)-revised

stressNominal

stressMaximum=tK

( )

FEMassuchsimulationnumerical

orsexperimentthroughDetermined

stressNominal

max

=

=

=

t

nomt

K

K

tdwP

dw


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Kf is a reduced value ofKt and it is also called as fatigue

strength reduction factor

Actual / Fatigue stress concentration factor, Kf

factor)(geometricfactorionconcentratstresslTheoretica

21)-6&20-6Fig.(fromy valuesensitivitnotch

=

=

tK

q

( ) ( )1111 +=+= tsshearfstf KqKorKqK

specimenfree-notchinstress

specimennotchedinstressmaximum=fK

Stress-concentration factors for a variety of geometries under different

loading conditions can be found in Table A15 (page:1026-1032)


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Notch Sensitivity plot for Steels and UNS A92024-T wrought Al alloys

Fig: 6-20 ; page : 295

(Reverse bending or reverse axial loads)


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BITS Pilani, Pilani CampusFig: 6-21 ; page : 296

Notch Sensitivity plot for Steels and UNS A92024-T wrought Al alloys

(Reversed torsion condition)


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Estimation of KfKf= 1+q(Kt -1).

When q=0, the material has no sensitivity to notches, and henceKf=1.

When q=1, or when notch radius is large for which q is almost equal

to 1, the material has full notch sensitivity, andKf=Kt.

For all grades of cast iron, use q=0.20.Use the different graphs as given to obtain q for bending/axial and

torsional loading.

Whenever the graphs do not give values ofq for certain combinationsof data, use eitherNeuber equation orHeywood equation.


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Use theNeuber equation when the notch is circular/cylindrical.Estimation of Kf

( )

radiusnotchstrength.ultimateoffunctioni.e),(

constantmaterialaisandconstantNeuberisawhere

11

1

1

==

+=

+

=

rSfa

KqKand

ra

q

ut

tf

For steel, withSutin kpsi, the Neuber constant can be approximated

by a third-orderpolynomial fit of data as

38253

38253

)10(67.2)10(35.1)10(51.219.0:

)10(67.2)10(51.1)10(08.3246.0:

ututut

ututut

SSSaTorsion

SSSaaxialorBending

+=

+=


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Use Heywood equation when the notch is NOT circular/cylindrical

but is a tranverse hole or shoulder or groove.

Estimation of Kf

( )

335page15;-6Tabletheingivenarevalues

121

a

where

r

a

K

K

KK

t

t

tf

+

=

r= hole/ shoulder/groove size


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Fluctuating Stress

2

stresseor variablamplitude

2stressmeanormidrange

stressofrange

stressmaximumstressminimum

minmax

minmax

minmax

max

min

==

+==

==

==

a

m

r


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Reversed: mean stress is zero;equal reversals on both sides;

useful in conducting experiments

Repeated: minimum stress is zero;mean stress equal to half of therange stress

Fluctuating: maximum, minimumand mean stress are all non-zeroand arbitrary

Three specific types of cyclic loading:


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Design for Infinite life under cyclic loading

fyt

m

e

a

nSS

1=+

1=+yt

m

e

a

S

S

S

S

1=+ut

m

e

a

S

S

S

S

1

2

=

+

ut

m

e

a

S

S

S

1

22

=

+

yt

m

e

a

S

S

S

S


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Different fatigue failure models

yielding)staticfor

checkingfor(onlylineLanger1

lineEllipticASME1

lineGerber1

lineGoodmanModified1

lineSoderberg1

222

2

K

K

K

K

K

yyt

m

yt

a

fyt

m

e

a

fut

m

fe

a

fut

m

e

a

fyt

m

e

a

nSS

nSS

nSn

S

nSS

nSS

=+

=

+

=

+

=+

=+

Where aofamofm KandK ==


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Modified Goodman and

Langer Failure Criteria

Amplitude and Steady Coordinates of Strength and Important

Intersections in First Quadrant


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Gerber and Langer

Failure Criteria




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ASME-Elliptic and Langer

Failure Criteria




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Results based on Smiths tests (72 tests on torsion

strength) and confirmation by Joerres of AssociatedSpring-Barnes Group, the following relation is

used to identify the shear ultimate strength from

tensile strength.

From distortion energy theory,

utsu S0.67S =

ytsy S0.577S =

Torsional Fatigue Strength


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An AISI 1020 cold drawn steel shaft of diameter 30

mm is welded to a fixed support at one end and

another end is subjected to a bending moment

varying from 25 Nm to 100 Nm. Fatigue stress

concentration factor under bending (Kf) is 1.6. Findthe factor of safety using the Gerber failure criteria.

Problem:

390470S;SteelCD1020AISIFor MPaSandMPa ==


53/67


( )( )

( )( )

.

38.61

470

73.37

196.8

64.22

1SS

equation;failurefatigueGerber

MPa196.8=Skkk=S

1=k

0.95830)1.24(0.37=)1.24(d=k

drawn)coldfor-0.265=band4.51=(a0.874=aS=k

MPa235=0.5S=S

64.2215.146.1

73.3758.236.115.14

58.23

43.930

10253232

72.3730

101003232

390470S;SteelCD1020AISIFor

2

2

ut

m

e

a

1

ecbae

c

0.107-0.107-

eb

b

uta

ut

1

e

3

3

3

min

3

3

3

max

ut

safetyfactorhighwithdesignedoverismemberThe

nn

n

nn

MPaK

MPaKMPastressbendingVariable

MPastressbendingMean

MPad

MstressbendingMinimum

MPad

M

stressbendingMaximum

MPaSandMPa

f

f

f

f

f

bafa

bmfm

ba

bm

y

==

+

=

+

=

===

=====

==

=

==

=

==

==


54/67


Different types of cyclic loads may be applied incombination, for example, bending, axial and torsional

on machine components

When the loads and in-phase, the maximum values ofloads occurs at the same time and so are the minimum

values.

Hence in such cases, we can estimate the maximum andminimum von-Mises stress values and then estimate the

mean and amplitude von-Mises stresses. Then fatiguecriterion may be applied.

Combination of loading modes


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Combined loadingFor the common case of a shaft with bending stresses, torsional shear

stresses, and axial stresses, the von Mises stress is

Considering that the bending, torsional, and axial stresses have

alternating and midrange components, the von Mises stresses for the

two stress elements can be written as

For plane stress

[ ]21

222' 3 xyyyxx ++=[ ]2

122' 3 xyx +=

( ) ( ) ( )

( )

( ) ( )[ ]( ) ( ) ( ) ( )[ ] ( ) ( )[ ]{ } 2/122'

2/1

2

2

'

3

385.0

torsionmotorsionfsaxialmoaxialfbendingmobendingfm

torsionaotorsionfs

axialao

axialfbendingaobendingfa

KKK

KKK

++=

+

+=


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Calculate von Mises stresses for alternating and

midrange stress states, a and m .

Apply stresses to fatigue criterion i.e Soderberg,

Modified-Goodman, Gerbers or ASME Elliptic criteria

by replacing a and m with a and m respectively

Conservative check for localized yielding using von

Mises stresses i.e

Design for Combined loading


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Fig shows clutch-testing machine. The steel shaft rotates at a constant speed

. An axial load is applied to the shaft and is cycled from zero to P. The

torque T induced by the clutch face onto the shaft

where D and d are defined in the figure and f is the coefficient of friction of

the clutch face. The shaft is machined with Sy = 800 MPa and Sut = 1000

MPa. The theoretical stress concentration factors for the fillet are 3.0 and 1.8for the axial and torsional loading, respectively.

Assume the load variation P is synchronous

with shaft rotation. With f = 0.3, find themaximum allowable load P such that the shaft

will survive a minimum of 108 cycles with a

factor of safety of 3. Use the modified

Goodman criterion.

Prob:6-57


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Solution


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Solution


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Practice problemIn the figure shaft A, made of AISI 1010 hot-rolled steel, is welded to afixed support and is subjected to loading by equal and opposite forces F via

shaft B. In addition, the shaft is also subjected to a study compressive force

of P. The length of shaft A from the fixed support to the connection at shaftB is 1 m. If the load F cycles from 0.5 to 2kN and P= 1 kN, For shaft A find

the factor of safety for infinite life using (a) the modified Goodman fatigue

failure criterion. (b) Gerber fatigue failure criterion.


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Strain-Life method is used to estimate fatigue

strength for high cycle fatigue (i.e. N>103 cycles)

Fatigue strength

If a completely reversed stress(rev

) in given, the

number of cycles to failure can be expressed as

( )

==

=

e

ut

e

ut

b

f

SfSb

SfSa

whereNaS

log31&

2

Fatigue strength calculation / Design for finite life

brev

a

N

1

=

Where f is fatigue strengthfraction depends on Sut from

Figure 6-18; page:285


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Fatigue strength fraction

Figure 6-18; page:285


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A 25-mm-diameter solid round bar has a groove 2.5-mm

deep with a 2.5-mm radius machined into it. The bar is

made of AISI 1020 CD steel and is subjected to a purely

reversing torque of 200 N-m. For the S-N curve of this

material, letf= 0.9.

(a) Estimate the number of cycles to failure.

(b) If the bar is also placed in an environment with a

temperature of 4500F, estimate the number of cycles to

failure.

Problem : 6-12

Solution


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D= 25 mm ; d= 20 mm ; r= 2.5 mm

r/d =2.5/20 = 0.125; D/d= 1.25From figure: Kts= 1.4

Solution

Sol tion


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MPaSandkpsiMPa y 390)68(470S;SteelCD1020AISIFor ut ==

Solution


66/67


( )

cyclesa

N

S

fSb

S

fSa

MPa

ba

se

su

se

su

22472

169.0log3

1

94.913

315S0.67S

MPa87.94=Skkk=S

0.59=k0.7260)21.24(=)1.24(d=k

drawn)coldfor-0.265=band4.51=(a0.874=aS=k

MPa235=0.5S=S

1

2utsu

1

ecbae

c

0.107-0.107-

eb

b

uta

ut

1

e

=

=

=

=

==

==

=

Solution


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Documents

Fatigue Failure Resulting From Variable Loading (1)