Fatigue IIFatigue II

Lecture 11Lecture 11

Engineering 473Engineering 473Machine DesignMachine Design

Finite Life EstimatesFinite Life Estimates

mσ Stress,Mean

aσ Stress,gAlternatin

utS

eS

ytS

ytS

How can the life of a part be estimated if the mean stress-alternating stress pair lie above the Goodman line?

Goodman DiagramGoodman Diagram

Infinite LifeStress State

Finite Life(Cycles to failure?)

SS--N CurveN Curve

Shigley, Fig. 7-6

Completely reversed cyclic stress, UNS G41200 steel

The S-N curve gives the cycles to failure for a completely reversed (R=-1) uniaxial stress state.

What do you do if the stress state is not completely reversed?

DefinitionsDefinitions

2σσσ

2σσσ

σσσ

minmaxm

minmaxa

minmaxr

+=

−=

−=Stress RangeStress Range

Alternating StressAlternating Stress

Mean StressMean Stress

max

min

σσR =

m

a

σσA =

Stress RatioStress Ratio Amplitude RatioAmplitude RatioNote that R=-1 for a completely reversed stress state with zero mean stress.

FluctuatingFluctuating--Stress Failure Stress Failure Interaction CurvesInteraction Curves

Shigley, Fig. 7-16

The interaction curves provide relationships between alternating stress and mean stress.

When the mean stress is zero, the alternating component is equal to the endurance limit.

The interaction curves are for infinite life or a large number of cycles.

Goodman Interaction LineGoodman Interaction Line

1SS

SSk

ut

m

e

af =+Any combination of mean and alternating stress that lies on or below Goodman line will have infinite life.

Factor of Safety FormatFactor of Safety Format

fut

m

e

af

N1

SS

SSk =+

Note that the fatigue stress concentration factor is applied only to the alternating component.

Master Fatigue PlotMaster Fatigue Plot

Shigley, Fig. 7-15

Constant cycles till failure interaction curves.

Equivalent Alternating StressEquivalent Alternating Stress

mσ Stress,Mean

aσ Stress,gAlternatin

utS

eS

ytS

ytS0σa

mσ

=Alternating stress at zero mean stress that fails the part in the same number of cycles as the original stress state.

The red and blue lines are estimated fatigue interaction curves associated with a specific number of cycles to failure.

cycles 106

cycles 105

Number of Cycles to Number of Cycles to FailureFailure

Once the equivalent alternating stress is found, the S-N curve may be used to find the number of cycles to failure.

Equivalent Alternating Stress Equivalent Alternating Stress FormulaFormula

ut

m

f

af0σa

fut

m

0σa

af

fut

m

e

af

Sσ

N1σkσ

N1

Sσ

σσk

N1

Sσ

Sσk

m

m

−=

=+

=+

=

=

Goodman LineGoodman Line

pair. σ and σ original theas cycles ofnumber same in the failure

fatigue causes that stress -1)(R

reversed completely Equivalentσ

ma

0σam

=

≡=

ExampleExample

1 2

1.5 in. dia. 0.875 in. dia.0.125 in. rad.

Material UNS G41200 Steel

lb 2000Plb 0030P

min

max

==5 in 5 in

Notch sensitivityq=0.3

( )

( ) 4442

4441

in 088.0875.064πD

64πI

in 249.05.164πD

64πI

2

1

===

===

34

2

22

34

1

11

in 0.201in 0.438in 0.088

cIS

in 0.332in 0.75in 0.249

cIS

===

===

ExampleExample(Continued)(Continued)

1 2

1.5 in. dia. 0.875 in. dia.0.125 in. rad.

Material UNS G41200 Steel

lb 2000Plb 0003P

min

max

==5 in 5 in

Notch sensitivityq=0.3

( )1kq1k1k1kq

tf

t

f

−+=−−= ( )

18.1)161.1(3.011kq1k

61.1k

tf

t

=−+=−+=

=

0.143875.0125.0

dr

71.1in 0.875

in 1.5dD

==

==

Ref. Peterson

ExampleExample(Continued)(Continued)

1 2

1.5 in. dia. 0.875 in. dia.0.125 in. rad.

Material UNS G41200 Steel

lb 2000Plb 0003P

min

max

==

5 in 5 in

Notch sensitivityq=0.3

( )( )

( )( ) ksi 2.60in 332.0

in 01lb 2000SMσ

ksi 4.90in 332.0

in 10lb 3000SMσ

31

1min

31

1max

===

===

Section 1 (Base)Section 1 (Base)

ksi 3.752σσσ

ksi 1.152σσσ

minmaxm

minmaxa

=+=

=−=

ExampleExample(Continued)(Continued)

1 2

1.5 in. dia. 0.875 in. dia.0.125 in. rad.

Material UNS G41200 Steel

lb 2000Plb 0003P

min

max

==5 in 5 in

Notch sensitivityq=0.3

( )( )

( )( ) ksi 8.49in 201.0

in 5lb 2000SMσ

ksi 6.74in 201.0

in 5lb 3000SMσ

31

1min

31

1max

===

===

Section 2 (Fillet)Section 2 (Fillet)

ksi 2.622σσσ

ksi 4.122σσσ

minmaxm

minmaxa

=+=

=−=

ExampleExample(Continued)(Continued)

Completely reversed cyclic stress, UNS G41200 steel

Shigley, Fig. 7-6

ee

ut

Sksi 30Sksi 116S

==′=

fult

m

e

af

N1

Sσ

Sσk =+

ExampleExample(Continued)(Continued)

ee

ut

Sksi 30Sksi 116S

==′=

fult

m

e

af

N1

Sσ

Sσk =+

( ) 15.1ksi 116ksi 75.3

ksi 30ksi 15.10.1

1Nf

=+

=

Section 1 (Base)Section 1 (Base)

( )( )

( )( ) ksi 2.60in 332.0

in 01lb 2000SMσ

ksi 4.90in 332.0

in 10lb 3000SMσ

31

1min

31

1max

===

===

ksi 3.752σσσ

ksi 1.152σσσ

minmaxm

minmaxa

=+=

=−=

Part has finite life at base.Part has finite life at base.

ExampleExample(Continued)(Continued)

Section 2 (Fillet)Section 2 (Fillet)

ee

ut

Sksi 30Sksi 116S

==′=

fult

m

e

af

N1

Sσ

Sσk =+

( ) 02.1ksi 116ksi 62.2

ksi 30ksi 12.41.18

1Nf

=+

=

Part has finite life.Part has finite life.

( )( )

( )( ) ksi 8.49in 201.0

in 5lb 2000SMσ

ksi 6.74in 201.0

in 5lb 3000SMσ

31

1min

31

1max

===

===

ksi 2.622σσσ

ksi 4.122σσσ

minmaxm

minmaxa

=+=

=−=

Calculation of Equivalent Calculation of Equivalent Alternating StressAlternating Stress

ut

m

f

af0σa

Sσ

N1σkσ

m

−=

=

ksi 0.4311675.3

1.01(1.0)15.1σ

0σam

=

−=

=

ksi 5.3111662.2

1.01

(1.18)12.4σ0σa

m

=

−=

=

ksi 3.75σksi 1.15σ

m

a

==

ksi 2.62σksi 4.12σ

m

a

==

BaseBase FilletFillet

Cycles to Failure EstimateCycles to Failure Estimate

10

20

30

50

70

90

Base

Fillet

MultiMulti--axis Fluctuating axis Fluctuating Stress StatesStress States

Everything presented on fatigue has been based on experiments involving a single stress component.

What do you do for problems in which there are more than one stress component?

Marin Load Factor, Marin Load Factor, kkcc

The endurance limit is a function of the load/stress component used in the test.

edcbae SkkkkS ′⋅⋅⋅⋅=

��

�

��

�

�

>≤

=

shear andTorsion 0.577Bending1

MPa) (1520 ksi 220Sloading Axial1MPa) (1520 ksi 220Sloading Axial0.923

k ut

ut

c

Alternating and Mean Von Alternating and Mean Von Mises Mises StressesStresses

1. Increase the stress caused by an axial force by 1/kc.

2. Multiply each stress component by the appropriate fatigue stress concentration factor.

3. Compute the maximum and minimum von Mises stresses.

4. Compute the alternating and mean stresses based on the maximum and minimum values of the von Mises stress.

5. Use the Goodman alternating and mean stress interaction curve and S-N curve to estimate the number of cycles to failure. Use the reversed bending endurance limit.

Complex LoadsComplex Loads

t, time

Fσ,

cycles, nfor σ

cycles, nfor σcycles, nfor σcycles, nfor σfollows as stresses

reversed completely tosubjected ispart A

mm

33

22

11

�

What is the cumulative effect of these different load cycles?

1σ2σ

3σ

Minor’s RuleMinor’s Rule

CNn

Nn

Nn

Nn

m

m

3

3

2

2

1

1 =++++ �

Cumulative Damage LawCumulative Damage Law

2.2. to0.7 from rangingConstant Ci level stressat failure tocyclesN

i level stressfor cycles ofnumber n

i

i

≡≡≡

C is usually taken as 1.0C is usually taken as 1.0Minor’s Rule is the simplest and most widely used Minor’s Rule is the simplest and most widely used

Cumulative Damage LawCumulative Damage Law

ExampleExample

StressState

1

Cycles(n)

Life(N) N

n

23

1,000

5,000

10,000

2,000

10,000

100,000

0.5

0.5

0.1

1.1 Part will fail

AssignmentAssignment(Problem No. 1)(Problem No. 1)

A rotating shaft is made of 42 x 4 mm AISI 1020 cold-drawn steel tubing and has a 6-mm diameter hole drilled transversely through it. Estimate the factor of safety guarding against fatigue failure when 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. Use the stress concentration factor tables found in the appendices, and estimate the Marin factors using information in the body of the text.

AssignmentAssignment(Problem No. 2)(Problem No. 2)

A solid circular bar with a 5/8 inch diameter is subjected to a reversed bending moment of 1200 in-lb for 2000 cycles, 1000 in-lb for 100,000 cycles and 900 in-lb for 10,000 cycles. Use the S-N curve used in this lecture. Determine whether the bar will fail due to fatigue. Assume all Marin factors are equal to 1.0.

AssignmentAssignment(Problem No. 3)(Problem No. 3)

Same as Problem No. 2 except there is a constant axial force of 5,000 lb acting on the bar in addition to the completely reversed bending moment.