2014 L2m - Fatigue I

7/27/2019 2014 L2m - Fatigue I

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ENS6104 Advanced Mechanical Design

Lecture 2 –

Fatigue I

Dr. Ferdinando Guzzomi

[email protected]

Image Source: http://www.prlog.org/11380026-mechanical-design-services-mechanical-engineering-drawing-services.html

mailto:[email protected]

mailto:[email protected]

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Case Study: De Havilland Comet

Image source: http://en.wikipedia.org/wiki/De_Havilland_Comet

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De Havilland Comet:

http://www.youtube.com/watch?v=QFKP7xCsGsc

Disaster & Progression



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Fatigue Failure: Evidence

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Design for Strength or Life?

σ σ

t t

Failure

Failure

Design for Strength Design for Life

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Stage I – Micro-cracks

Stage II – Macro-cracks

Stage III – Minimised area (Brittle or yield failure)

Fatigue Failure: Fracture Surfaces

Image source: http://www.twi-global.com/technical-knowledge/job-knowledge/fatigue-testing-078/

Crack initiation due to: Rapid cross section changes

(design)

Pitting/Spalling due to contact

loads

Construction marks / FabricationFaults

Material composition due to

processing

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Fatigue Failure: Fracture Surfaces

Image source: Shigley’s Mechanical Engineering Design

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Fatigue Fracture: Worked Example

Stage I – Micro-cracks

Stage II – Macro-cracks

Stage III – Minimised area (Brittle or yield failure)

Initial crack

Crack

propagation

(Beach marks)

Final fast

failure (Roughsurface)

Low nominal stressNo stress concentration

Rotational bending

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Fatigue Fracture: Class Example

Image source: www.uh.edu and http://www.doobybrain.com/2009/03/02/giant-crankset-on-a-bicycle/

High Nominal Stress

Unidirectional bending

Severe Stress Concentration

High Nominal Stress?

What type of loading conditions produced this fatigue failure surface?

Application?

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S-N Diagram: Basics

Image Source: http://www.accutektesting.com/testing-services/mechanical-testing/rotating-beam/

Fatigue life can be predicted using one of the

following methods:

Stress Life

Strain Life

Linear-Elastic Fracture Mechanics

Sut

High Cycle

Finite Life Infinite Life

106

Number of stress cycles, N

Log(N)

Fatigue Strength, SfLog(Sf)

Low

Cycle

103

Se

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S-N Diagram: Steel Example

Sut

Se

Low Cycle High Cycle

Finite Life Infinite Life

106


Log(N)

Fatigue Strength, Sf

Log(Sf)

103

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S-N Diagram: Aluminium Example

Sut


106

Log(N)

Log(S)

Note:• Only some materials show endurance strengths

• Materials like Aluminium have no infinite life

Steel

Aluminium

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Endurance Strength

Ferrite Pearlite Martensite

Carbon Steel 0.57 - 0.63 0.38 - 0.41 …

Alloy Steel … … 0.23 - 0.47

A table for the endurance limit ratio S’e/Sut is shown below.

In practice when using simple relationships, without testing it is worth being

conservative:

MPaS MPaS

MPaS S S

ut e

ut ut e

1400700

14005.0'

'

The above is a reasonable approach for steel. Cast irons generally have a lower

endurance limit ratio and a lower peak endurance strength, defined by:

MPaS MPaS

MPaS S S

ut e

ut ut e

600275

60045.0

'

'

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High-Cycle Fatigue

Low cycle fatigue (< 10 3 cycles):

Life prediction from Strain-Life method

High cycle fatigue (> 10 3 cycles):

Life prediction from Stress-Life method

Identified estimate of mean endurance limit for steel is either half the tensile

strength or 700MPa (whichever is lower) at > 10 6 cycles

However, is it possible to

estimate the fatigue strength

within the high cycle region?

Sut

Se


106


Log(N) F a t i g u e

S t r e n g t h ,

S f

L o g ( S f )

103

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S-N curve: Estimating Fatigue Strength

Traditionally, for wrought steels data suggests that at 1000 cycles (the start of high-cycle

fatigue) the mean fatigue strength is:

)65.0 vely,conservati(or80.0 33 10,10, ut f ut f S S S S

In the absence of actual fatigue testing , a fatigue strength fraction, f, is estimated from the

graph below to determine the fatigue strength (as a fraction of Sut) at 1000 cycles (Sf,103).

Conservatively, f = 0.9 for Sut < 490MPa

The following equation is used to estimate thehigh-cycle fatigue strength (103 < N < 106 ):

: whereb

f N aS

e

ut

S

S f a

2)(

e

ut

S

S f b

log

3

1

490 560 630 700 770 840 910 980 1050 1120 1190 1260 1330 140

Sut [MPa]

0.76

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For a completely reversed stress, Sf can be set to the reversed stress and

rearranging the equation above can provide the expected life (103 N 106 cycles):

b

f N aS

Calculates: Mean Fatigue Strength Sf for given

cycles, N

Calculates: Cycles N, for given Mean Fatigue

Strength Sf (or Reversed stress a)

S-N curve: Finding Fatigue Strength or Cycles

b

aS

N f

/1

High-cycle (between 103 and 106 cycles) fatigue strength can be estimated using

the following equation:

3log

f

ut f N S S

Although a crude approximation, the following equation can estimate the low-cycle

fatigue strength:

Estimates: Mean Fatigue Strength Sf for given

cycles, N (where 1 N 1000)

W k d E l F i S h

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Worked Example: Fatigue Strength

The endurance limit of a thoroughly tested (in fatigue) wrought steel

member is 112MPa and the tensile strength is 385MPa. What is the

fatigue strength corresponding to a life of 70x103 cycles?

Example source: Shigley’s Mechanical Engineering Design

ℎ , , = 0.8 = 0.8 385 = 308

= (308)

112 = 847 =

1

3log

308

112 = 0.146

Now finding the equation coefficients:

The high-cycle fatigue strength can then be calculated using the followingequation:

= = 847 70 × 10 −. = 166

W k d E l F ti St th

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Worked Example: Fatigue Strength

For a hot-rolled 1050 steel sample with Tensile Strength of 620MPa, calculate the following:

I. The rotating-beam endurance limit at 106 cycles,

II. The endurance strength of a polished rotating-beam specimen corresponding to 10

4

cycles to failure,

III. The expected life a polished rotating-beam specimen under a completely reversed

stress of 385MPa.

Example source: Shigley’s Mechanical Engineering Design

I. Sut = 620MPa < 1400MPa, therefore: ′ = 0.5 ⟹ ′ = 0.5 620 = 310

II. For Sut = 620MPa, f = 0.86

= (533)

310 = 916 =

1

3log

533

310 = 0.0785

= = 916 1 × 10 −. = 445

⇒ ,= () = 0.86 620 = 533

III. For a = 385MPa, =

⇒ = 385

916

−.

= 62.4 × 10

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2014 L2m - Fatigue I