mech1641 - w2008 lecture 05 - fatigue iv

MECH1641 – Machine Design II

1

Lecture 5: Fatigue IV

Homework Review

Interpreting Constant Life Diagrams

Constructing Constant Life Diagrams

Component Design for Fluctuating Loads

Review

Homework & Reading Assignment


2

Objectives

After this lecture and associated homework, you will be able to:

Interpret a Constant Life diagram.

Calculate the fatigue life for fluctuating stresses using Constant Life diagrams.

Construct a Constant Life diagram from the S-N curve.

Apply Constant Life diagrams to the design of parts with a fluctuating load profile.


3

Homework Review

1. Determine the mean stress and stress amplitude of an application that cycles between -350 and 780 MPa? Sketch the stress vs. time profile.

2. Determine the minimum stress and stress amplitude of an application with a mean stress of 925 MPa and a maximum stress of 1190 MPa. Sketch the stress vs. time profile.


4

Constant Life Diagram Review

The horizontal axis is mean stress σm.

The vertical axis is stress amplitude σa.

Sy

0

-Sy

0

Sy

-Sy

Sy

0

-Sy

Sy

0

-Sy

Sy

0

-Sy

Sy

0

-Sy


5

Designing for Fatigue

An S-N Curve is needed to design parts for fatigue applications involving fully reversing loads.

A Constant Life diagram is needed to design parts for fatigue applications involving fluctuating loads.

102030405060708090

100110120130140150160

1.E+00 1.E+01 1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07 1.E+08 1.E+09

Life N (cycles)

Fatig

ue S

treng

th S

f (ks

i)


6

Fatigue Strength Diagram for 7075-T6 Aluminum

Ref. Juvinall Figure 8.19.


7

Fatigue Strength Diagram for Alloy SteelsApplicable to AISI 4340, 4130, 2330, 8630.

Ref. Juvinall Figure 8.17.


8

Problem 1If a test specimen made from normalized 4130 steel with an ultimate strength of 100 ksi is subject to a maximum bending stress of 90 ksi and a minimum stress of 0, what is the expected fatigue life? Sketch the stress vs. time profile.

4: 90 90% 0 10max u minFind N Given ksi S N cyclesσ σ= = = → =

0

90 ksi


9

Problem 2For an normalized 4130 test specimen with Su=100 ksi, what is the maximum stress that can be applied for 106 cycles if the loading is of the zero-to-max-to-zero type? Sketch the stress vs. time profile.

610 : 0

70% 70max min

max u

Find for N cycles Given

S ksi

σ σ

σ

= =

= =


10

Problem 3For a quenched and tempered 4340 test specimen with an ultimate strength of 180 ksi, what life is expected from a fluctuating load with a mean stress of 90 ksi and a stress amplitude of 80 ksi? Sketch the stress vs. time profile.

4 4

: 90 50% 72 40%

10 3 10m u a uFind N Given ksi S ksi S

N

σ σ= = = =

< < ×


11

Problem 4For a hardened 4340 test specimen with Su=230 ksi what is the highest stress amplitude that can be applied with a superimposedtensile stress of 184 ksi if infinite life is required? Sketch the stress vs. time profile.

7: 184 80% ; 10

10% 23a m u

a u

Find Given ksi S N

S ksi

σ σ

σ

= = >

= =

184 ksi

0


12

Problem 5Use the empirical relationships discussed to construct the S-N Curve and the Constant Fatigue Life diagram in Excel for a precision steel part with an ultimate strength of 150 ksi, a yield strength of 120 ksi, commercially polished surface. All cross-section dimensions are under 2 in, and the loads are axial. Assume that actual fatigue test data is not available.

Ref. Juvinall Sample Problem 8.1.


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Problem 5

, 1000

:150 ; 120

For steel, 0.5For axial loading, 0.75 , and 1

For a ground surface, 0.9For a diameter<2 , axially loaded, 0.9Assume 1; 1

0.5 0.5 150 75

u y

n u

f N u T L

S

G

T R

n u

n n L G

AnalysisS ksi S ksi

S SS S C C

Cin C

C CS S ksi ksi

S S C C

=

= =

′ =

= =

==

= =′ = = × =

′=

, 1000 , 1000

75 1 0.9 0.9 1 1 61

0.75 0.75 150 1 112

S T R n

f N u T f N

C C C ksi S ksi

S S C ksi S ksi= =

= × × × × × ∴ =

= = × × ∴ =

This is all that is needed to draw the predicted S-N curve, which shows fatigue behaviour for fully fully-reversing loads only.


14

102030405060708090

100110120130140150160

1.E+00 1.E+01 1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07 1.E+08 1.E+09

Life N (cycles)

Fatig

ue S

treng

th S

f (ks

i)

Problem 5 ( S-N Curve)

6, 1061nf N

S S ksi=

′= =

3, 10112

f NS ksi

==

, 1 150f N uS S ksi= = =

These values can be read from this S-N curve to construct the Constant Life Diagram.

4, 1094

f NS ksi

==

5, 1077

f NS ksi

==

S-N Curve DataN Sf (ksi)

1.E+00 150 Su

1.E+03 111 Sf,N=1000

1.E+06 61 Sn

1.E+09 61 Sn


15

Problem 5 (Constant Life Diagram )


16

112 N=103

94 N=104

77 N=105

61 N=106

0

20

40

60

80

100

120

140

-120 -100 -80 -60 -40 -20 0 20 40 60 80 100 120 140 160Mean Stress σm (ksi)

Stre

ss A

mpl

itude

σa (

ksi)


σm

N=103 N=104 N=105 N=106

-120 112 94 77 610 112 94 77 61

150 0 0 0 0

σa

Constant Life Line Data


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N=103112

N=10494

N=10577

N=10661

0

20

40

60

80

100

120

140


Stre

ss A

mpl

itude

σa (

ksi)


This line represents combinations of σm & σa that result in a σmax=Sy. I.e., yielding occurs above this loading, no yielding below.

Yield Envelopeσm σa

-120 0 (-Sy,0)0 120 (0,+Sy)

120 0 (+Sy,0)


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Problem 6 – Designing a Part for Fluctuating Loads

A round tensile link is subjected to fluctuating load profile between 1000 lb and 5000 lb. It is a precision member, so CG=0.9 can be used, with commercially-polished surfaces. The material is steel, Su=150 ksi, Sy=120 ksi, so that the S-N curve and Constant Life diagram from the previous question can be used. A safety factor of 2 is required.

Find the diameter required for infinite life, and the diameter required for a life of 1000 cycles. No yielding should be permitted.



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Problem 6 (Constant Life Diagram)


112 N=103

94 N=104

77 N=105

61 N=106

0

20

40

60

80

100

120

140


Stre

ss A

mpl

itude

σa (

ksi)


20

Problem 6 Analysis


First, identify , :5000

The maximum stress in the part

1000The minimum stress in the part

5000 10003000

2 25000 1000

2 2

m a

fmaxmax

fminmin

f f

fmax minm m

f f

max mina

n lbnFA A

n lbnFA A

n lb n lbn lbA A

An lb n lb

A A

σ σ

σ

σ

σ σσ σ

σ σσ

×= =

×= =

× ×+ ×+

= = ∴ =

× ×−−

= =

23

2000

So the ratio 0.666, for whatever the area must be.Draw a line with this slope on the Constant Life diagram.Such a line represents all possibile combinations ofmean stress and stress amplitude for the required loa

am

fa

n lbA

σσ

σ×

∴ =

= =

2

d profile.

3000 2000 12,000 8000Also Area

4f f f f

m a m a

n lb n lb n lb n lbdA dπσ σ πσ πσ

× × × ×= = = ∴ = =


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N=103112

N=10494

N=10577

N=10661

σa/σm=0.667

0

20

40

60

80

100

120

140


Stre

ss A

mpl

itude

σa (

ksi)

Problem 6 (Infinite Life)


This line represents the combinations of σm & σa that describe the given load profile, from 1000 lb to 5000 lb.

σa/σm=0.667σm σa

0 0 σa=0.667σm

150 100 σa=0.667σm


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N=103112

N=10494

N=10577

N=10661

σa/σm=0.667

0

20

40

60

80

100

120

140


Stre

ss A

mpl

itude

σa (

ksi)



Small diameters will be associated with large stresses, on this line far away from the origin. Large diameters will result in small stresses, on this line near the origin.


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N=103112

N=10494

N=10577

N=10661

σa/σm=0.667

0

20

40

60

80

100

120

140


Stre

ss A

mpl

itude

σa (

ksi)



σm=56 ksi & σa =38 ksi will cause yielding, and produce a life of about 106 cycles. d=0.369 in

212,000 12,00021 54,000

0.369

f f

m f

n lb lb indlb

d in

πσ π×

= = × ×

∴ =


24

N=103112

N=10494

N=10577

N=10661

σa/σm=0.667

0

20

40

60

80

100

120

140


Stre

ss A

mpl

itude

σa (

ksi)

Problem 6 (No Yielding)


σm=71 ksi & σa =48 ksi will cause yielding, and produce a life of about 104 cycles. d=0.328 in

212,000 12,00021 71,000

0.328

f f

m f

n lb lb indlb

d in

πσ π×

= = × ×

∴ =


25

N=103112

N=10494

N=10577

N=10661

σa/σm=0.667

0

20

40

60

80

100

120

140


Stre

ss A

mpl

itude

σa (

ksi)

Problem 6 (N=1000)


σm=80 ksi & σa =53 ksi will produce a life of 103

cycles, but with considerable yielding. d=0.309 in

212,000 12,00021 80,000

0.309

f f

m f

n lb lb indlb

d in

πσ π×

= = × ×

∴ =


26

Summary: S-N Curves

An S-N Curve is needed to design parts for fatigue applications involving fully reversing loads.

An estimated S-N Curve can be constructed from ultimate strength.

It is more accurate to use real fatigue test data.

102030405060708090

100110120130140150160

1.E+00 1.E+01 1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07 1.E+08 1.E+09

Life N (cycles)

Fatig

ue S

tren

gth

Sf (

ksi)

S-N Curve Estimated from Ultimate Strength(better than nothing; used for prototypes)

S-N Curve Based on test data(preferred for serious design)


27

Summary: Constant-Life Diagrams

A Constant Life diagram is needed to design parts for fatigue applications involving fluctuating loads.

An estimated Constant Life diagram can be constructed from the S-N curve.

It is more accurate to use real fatigue test data.

Constant Life diagram constructed from the S-N curve(better than nothing; used for prototypes)

Constant Life diagram based on test data(preferred for serious design)


28

Homework and Reading AssignmentFor next class, review these notes then answer the questions below:

1. For a 4340 test specimen with Su=150 ksi, what is the predicted life for a fully reversing stress of 90 ksi?

2. What is the maximum stress that will provide the same life if the load profile is changed to a zero-to-max-to-zero type?

3. What stress amplitude will provide the same life if the load profile is fluctuating with a maximum stress of 105 ksi?

4. Also, do these problems from Juvinall, chapter 8, by creating Excel S-N curves: 8.19, 8.20, 8.23, 8.25 (Excelconstant-life diagram required for this last problem.)


29

Closing Notes

Quiz 2 (4%) is on Friday January 25.

Test 1 (20%) is on Friday February 22.

mech1641 - w2008 lecture 05 - fatigue iv

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