Chapter 3: Design of

Mechanical Failure Prevention 2: Fatigue Failure

DR. AMIR PUTRA BIN MD SAAD

C24-322

amirputra@utm.my | amirputra@mail.fkm.utm.my

mech.utm.my/amirputra

3.1 INTRODUCTION

3.1 INTRODUCTION

i. Static conditions : (a) Loads are applied gradually, to give sufficient time for the strain to fully develop. OR (b) Abuse load are applied rapidly.

ii. Variable conditions : Stresses vary with time or fluctuate between different levels, also called repeated, alternating, or fluctuating stresses.

iii. When machine members are found to have failed under fluctuating stresses, the actual maximum stresses were well below the ultimate strength of the material, even below yielding strength.

iv. Since these failures are due to stresses repeating for a large number of times, they are called fatigue failures.

v. When machine parts fails statically, the usually develop a very large deflection, thus visible warning can be observed in advance; a fatigue failure gives no warning!

3.2 FATIGUE FAILURE IN METALS

A fatigue failure arises from three stages of development:

- Stage I : initiation of microcracks due to cyclic plastic deformation (these cracks are not usually visible to the naked eyes).

- Stage II : propagation of microcracks to macrocracksforming parallel plateau like fracture surfaces separated by longitudinal ridges (in the form of dark and light bands referred to as beach marks).

- Stage III : fracture when the remaining material cannot support the loads.

Initiation

Propagation

Fracture

3.3 FRACTURE PATTERNS OF FATIGUE FAILURE

3.4 FATIGUE LIFE METHODS IN FATIGUE

FAILURE ANALYSIS

• Let 𝑁 be the number of cycles to fatigue for a specified level of loading

- For 1 ≤ 𝑁 ≤ 103, generally classified as low-cycle fatigue

- For 𝑁 > 103, generally classified as high-cycle fatigue

• Three fatigue life methods used in design and analysis are

1. Stress-Life Method : is based on stress only, least accurate especially for low-cyclefatigue; however, it is the most traditional and easiest to implement for a widerange of applications.

2. Strain-Life Method : involves more detailed analysis, especially good for low-cyclefatigue; however, idealizations in the methods make it less practical whenuncertainties are present.

3. Linear-Elastic Fracture Mechanics Method : assumes a crack is already present.Practical with computer codes in predicting in crack growth with respect to stressintensity factor

3.5 STRESS-LIFE METHOD: R. R. MOORE

• The most widely used fatigue-testing device is the R. R. Moore high-speedrotating-beam machine.

• Specimens in R.R. Moore machines are subjected to pure bending by means ofadded weights.

• Other fatigue-testing machines are available for applying fluctuating orreversed axial stresses, torsional stresses, or combined stresses to the testspecimens.

3.6 S-N CURVE

• In R. R. Moore machine tests, a constant bending load is applied, and thenumber of revolutions of the beam required for failure is recorded.

• Tests at various bending stress levels are conducted.

• These results are plotted as an S-N diagram.

• Log plot is generally used to emphasize the bend in the S-N curve.

• Ordinate of S-N curve is fatigue strength, 𝑆𝑓, at a specific number of cycles

3.6 S-N CURVE

S-N diagram from the results of completely reversed axial fatigue test. Material : UNS G41300 steel.

3.7 FATIGUE STRENGTH, 𝑆𝑓

• Low-cycle fatigue considers the range from 𝑁 = 1 to about 1000 cycles.

• In this region, the fatigue strength, 𝑆𝑓 is only slightly smaller than the tensile

strength, 𝑆𝑢𝑡.

• High-cycle fatigue domain extends from 103 to the endurance limit life (106

to 107 cycles).

• Experience has shown that high-cycle fatigue data are rectified by alogarithmic transform to both stress and cycles-to-failure.

103

Number of Stress Cycles, N

Fatigue Strength, Sf (MPa)

Sut

f.Sut or S3

100 106

a

b

c

Finite Life Infinite Life

Low Cycle Fatigue High Cycle Fatigue

High Nominal Stress

Low Nominal Stress

Valid Zone for S-N Method

Sn

Components in S-N Curve:

3.7 FATIGUE STRENGTH, 𝑆𝑓

For line b-c:

𝑆𝑓 = 𝑎𝑁𝑏

where,

𝑎 =(𝑆3)

2

𝑆𝑛′

𝑏 = −1

3log

𝑆3𝑆𝑛′

𝑁 =𝜎𝑟𝑒𝑣𝑎

1𝑏

𝑆3 = 0.75𝑆𝑢𝑡 for steels

103

Number of Stress Cycles,

N

Fatigue Strength, Sf (MPa)

Sn

Sut

f.Sut or S3

100 106

a

b

c

Valid Zone for S-N Method

3.7 FATIGUE STRENGTH, 𝑆𝑓

3.8 CHARACTERISTICS OF S-N CURVES FOR METALS

• In the case of steels, a knee occurs in the graph, and beyond this knee failure will not occur, no matter how great the number of cycles - this knee is called the endurance limit, denoted as 𝑆𝑛

′ .

• Non-ferrous metals and alloys do not have an endurance limit, since their S-N curve never become horizontal.

• For materials with no endurance limit, the fatigue strength is normally reported at 𝑁 = 5 × 108

• 𝑁 = 1/2 is the simple tension test.

Aluminum Alloy

3.14 BASIC TYPES OF FLUCTUATING STRESS

• Example of loading history on the wind blades.

3.14 BASIC TYPES OF FLUCTUATING STRESS

• Fluctuating stresses often of sinusoidal patterns due to the nature of somerotating machinery.

• The peaks of the wave are more important than its shape.• Fluctuating stresses are described using a steady component and an alternating

component.

3.14 BASIC TYPES OF FLUCTUATING STRESS

Tension-Compression or completely reversed stress

Tension-Zero stress or Repeated Stress

Tension-Tension stress or Fluctuating Stress

Three types of fluctuating stress:

3.14 STRESS COMPONENTS IN A

CYCLIC STRESS

𝐹𝑚 =𝐹𝑚𝑎𝑥 + 𝐹𝑚𝑖𝑛

2

𝜎𝑎 =𝜎𝑚𝑎𝑥 − 𝜎𝑚𝑖𝑛

2

𝐹𝑎 =𝐹𝑚𝑎𝑥 − 𝐹𝑚𝑖𝑛

2

𝜎𝑚 =𝜎𝑚𝑎𝑥 + 𝜎𝑚𝑖𝑛

2

𝜎𝑚𝑎𝑥 = maximum stress

𝜎𝑚𝑖𝑛 = minimum stress

𝜎𝑎 = amplitude stress

𝜎𝑚 = midrange stress

3.14 STRESS COMPONENTS IN A

CYCLIC STRESS

𝑀𝑎𝑜 =𝑀𝑚𝑎𝑥𝑜 − 𝑀𝑚𝑖𝑛𝑜

2; 𝑀𝑚𝑜 =

𝑀𝑚𝑎𝑥𝑜 + 𝑀𝑚𝑖𝑛𝑜

2

𝐹𝑎𝑜 =𝐹𝑚𝑎𝑥𝑜 − 𝐹𝑚𝑖𝑛𝑜

2; 𝐹𝑚𝑜=

𝐹𝑚𝑎𝑥𝑜 + 𝐹𝑚𝑖𝑛𝑜

2

𝑇𝑎𝑜 =Tmaxo − Tmino

2; 𝑇𝑚𝑜 =

Tmaxo + Tmino

2

Alternating Part Steady Part

Note: Subscript with o mean without consideration of stress concentration effect.

3.14 STRESS COMPONENTS IN A

CYCLIC STRESS

Alternating Part Steady Part

𝜎𝑎𝑜 =𝜎𝑚𝑎𝑥𝑜 − 𝜎𝑚𝑖𝑛𝑜

2; 𝜎𝑚𝑜 =

𝜎𝑚𝑎𝑥𝑜 + 𝜎𝑚𝑖𝑛𝑜

2

𝜎𝑎𝑜 =𝜎𝑚𝑎𝑥𝑜 − 𝜎𝑚𝑖𝑛𝑜

2; 𝜎𝑚𝑜 =

𝜎𝑚𝑎𝑥𝑜 + 𝜎𝑚𝑖𝑛𝑜

2

𝜏𝑎𝑜 =𝜏𝑚𝑎𝑥𝑜 − 𝜏𝑚𝑖𝑛𝑜

2; 𝜏𝑚𝑜 =

𝜏𝑚𝑎𝑥𝑜 + 𝜏𝑚𝑖𝑛𝑜

2

Note: Subscript with o mean without consideration of stress concentration effect.

3.14 STRESS COMPONENTS IN A

CYCLIC STRESS

Figure 3.1 shows a sander machine. The most severe loading occurs when an object isheld near the periphery of the sander disk (100-mm radius at black dot as shown inFigure 3.1) with sufficient force to develop a friction torque of 12 Nm. Assume acoefficient of friction of 0.6 between the object and the disk.

3.15 SAMPLE PROBLEM

Figure 3.1: Sander Machine

Touch Point

(i) Show all the acting loads on the sander disk.(ii) Draw free body diagram of the sander shaft. (iii) Identify the critical point.(iv) Draw all the cyclic stress.

3.15 SAMPLE PROBLEM

Figure 3.1: Sander Machine

Touch Point

Sander Disk

Sander Shaft

3.9 ENDURANCE LIMIT, 𝑆𝑛′ FOR

STEELS

• For steels, the endurance limit relates directly to the minimum tensile strength as observed in experimental measurements.

• From the observations, the endurance of steels can be estimated as

𝑆𝑛′ = 0.5𝑆𝑢𝑡

• with the prime mark on the endurance limit referring to the rotating-beam specimen.

3.10 MODIFICATION FACTORS ON

ENDURANCE LIMIT

where,

i. 𝐶𝐿 = Load Factor

ii. 𝐶𝐺 = Size or Gradient Factor

iii. 𝐶𝑆 = Surface Condition Factor

iv. 𝐶𝑇 = Temperature Factor

v. 𝐶𝑅 = Reliability Factor

vi. 𝑆𝑛′ = Rotary-Beam Test Specimen Endurance Limit

𝑆𝑛 = 𝑆𝑛′ 𝐶𝐿𝐶𝐺𝐶𝑆𝐶𝑇𝐶𝑅

modification factors (Empirical Data)

3.10.1 LOAD FACTOR, 𝐶𝐿

Load Factor, 𝐶𝐿 :

3.10.2 GRADIENT FACTOR, 𝐶𝐺

Gradient Factor, 𝐶𝐺 :

3.10.3 SURFACE FACTOR, 𝐶𝐿

Surface Factor, 𝐶𝐿 :

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.

3.10.4 TEMPERATURE FACTOR, 𝐶𝑇

Temperature Factor, 𝐶𝑇 :

3.10.5 RELIABILITY FACTOR, 𝐶𝑅

Reliability Factor, 𝐶𝑅 :

• The fatigue stress concentration factor from the existence of irregularities or discontinuities in materials is defined a

where,

𝐾𝑡 is the static stress concentration factor.𝐾𝑓 is the fatigue stress concentration.

𝑞 is the notch sensitivity

3.11 FATIGUE STRESS CONCENTRATION

FACTOR, 𝐾𝑓 AND 𝐾𝑓𝑠

𝐾𝑓 = 1 + 𝑞(𝐾𝑡 − 1)

𝐾𝑓𝑠 = 1 + 𝑞𝑠(𝐾𝑡𝑠 − 1)

Bending or Axial

Torsion

3.11 FATIGUE STRESS CONCENTRATION

FACTOR, 𝐾𝑓

3.12 NOTCH SENSITIVITY FACTOR, 𝑞 or 𝑞𝑠

The ratio of Fatigue Stress Concentration Factor, Kf over Static StressConcentration Factor, Kt gives property of notch sensitivity, q.

𝑞 =𝐾𝑓 − 1

𝐾𝑡 − 1

𝑞𝑠 =𝐾𝑓𝑠 − 1

𝐾𝑡𝑠 − 1

Bending or Axial

Torsion

* for reversed bending and reversed axial loadings

3.12 NOTCH SENSITIVITY FACTOR, 𝑞 or 𝑞𝑠

3.13 STATIC STRESS CONCENTRATION

FACTOR, 𝐾𝑡

r/d: 2/40 = 0.05D/d: 60/40 = 1.5Kt = 2.38

3.13 STATIC STRESS CONCENTRATION

FACTOR, 𝐾𝑡

r/d: 2 /40 = 0.05D/d: 60/40 = 1.5Kts = 1.66

If 𝑆𝑢𝑡 = 900 MPa and 𝑆𝑦 = 750 MPa.

(v) Determine the endurance limit, 𝑆𝑛′ .

(vi) Determine all the modification factors (𝐶𝐿, 𝐶𝐺 , 𝐶𝑆, 𝐶𝑇 and 𝐶𝑅)(vii) Determine the corrected endurance limits, 𝑆𝑛.(viii) Determine the stress concentration factor/s (𝐾𝑠 or/and 𝐾𝑠𝑓).

3.16 SAMPLE PROBLEM

3.17 MASTER FATIGUE DIAGRAM

𝐴 =𝜎𝑎𝜎𝑚

𝑅 =𝜎𝑚𝑖𝑛

𝜎𝑚𝑎𝑥

Since the tests required to generate a Haigh or Master Diagram are quite

expensive, several empirical relationships which relate alternating stress

amplitude to mean stress have been developed. These relationships characterize

a material through its ultimate tensile strength, 𝑆𝑢𝑡, and are very convenient. For

infinite life design strategies, the methods use various curves to connect the

Endurance Limit, 𝑆𝑛, on the alternating stress axis to either the yield stress, 𝑆𝑦 or

ultimate strength, 𝑆𝑢𝑡 on the mean stress axis. Of all the proposed relationships,

three have been most widely accepted (i.e., those of Goodman, Soderberg or

Gerber).

3.18 FATIGUE FAILURE CRITERION

Goodman

Langer

𝜎𝑎𝑆𝑛

+𝜎𝑚𝑆𝑢𝑡

=1

𝑛

𝜎𝑎𝑆𝑦

+𝜎𝑚𝑆𝑦

=1

𝑛

3.18 FATIGUE FAILURE CRITERION

Goodman Line

Langer Line (Yield Line)𝑆𝑦

𝑆𝑦 𝑆𝑢𝑡

𝑆𝑛

Midrange Stress, 𝜎𝑚

Alt

ern

atin

g St

ress

, 𝜎𝑎

𝑦

𝑎+𝑥

𝑏= 1

3.18 FATIGUE FAILURE CRITERION

y

x

a

b

∗ 𝑛 = safety factor𝜎𝑎𝑆𝑛

+𝜎𝑚𝑆𝑢𝑡

=1

𝑛

3.18 FATIGUE FAILURE CRITERION

Goodman Line

𝑆𝑢𝑡

𝑆𝑛

Midrange Stress, 𝜎𝑚

Alt

ern

atin

g St

ress

, 𝜎𝑎

𝑛 > 1

𝑛 < 1

Goodman

SAFE

FAIL

Infinite Life

Finite Life

Also called as Constant-Life Fatigue Diagram

Various fluctuating uniaxial stresses,all of which correspond to equalfatigue life.

The existence of a static tensile stressreduces the amplitude of reversed stressthat can be superimposed. Figure on theright illustrates this concept. Fluctuation ais a completely reversed stresscorresponding to the endurance limit—themean stress is zero and the alternatingstress Sn. Fluctuation b involves a tensilemean stress. In order to have an equal (inthis case, “infinite”) fatigue life, thealternating stress must be less than Sn. Ingoing from b to c, d, e, and f, the meanstress continually increases; hence, thealternating stress must correspondinglydecrease.

3.18 FATIGUE FAILURE CRITERION

3.18 FATIGUE FAILURE CRITERION

𝜎𝑎

𝑆𝑢𝑡

𝑆𝑛

𝑆𝑓

𝑁 = 106𝑐𝑦𝑐𝑙𝑒𝑠

𝑁 = 105𝑐𝑦𝑐𝑙𝑒𝑠

𝑁 = 104𝑐𝑦𝑐𝑙𝑒𝑠𝑆𝑓

𝑆𝑓𝑁 = 103𝑐𝑦𝑐𝑙𝑒𝑠

𝜎𝑚

number of cycles getting smaller

Constant Life Line

3.18 FATIGUE FAILURE CRITERION

Constant-life fatigue diagram for ductile materials.

Goodman

Langer

𝜎𝑎𝑆𝑛

+𝜎𝑚𝑆𝑢𝑡

=1

𝑛

𝜎𝑎𝑆𝑦

+𝜎𝑚𝑆𝑦

=1

𝑛

3.16 FATIGUE FAILURE CRITERION

Modified Goodman Line

Langer Line (Yield Line)𝑆𝑦

𝑆𝑦 𝑆𝑢𝑡

𝑆𝑛

Midrange Stress, 𝜎𝑚

Alt

ern

atin

g St

ress

, 𝜎𝑎

(a)

(b)

(c)

(d)

(e)

𝑛 > 1

SAFEInfinite Life

Question: Explain the fatigue condition for all the points (a, b, c, d and e).

3.18 SIMULTANEOUS ACTION OF BENDING,

TORSION AND AXIAL STRESSES WITH 𝐾𝑓

Equivalent Amplitude Stress, 𝜎𝑒𝑎 is calculated using the Distortion Energy Theory:

Equivalent Midrange Stress, 𝜎𝑒𝑚 is calculated using the Maximum Principal Stress. Use Mohr Circle Theory:

𝜎𝑒𝑎 = 𝜎𝑎𝑎𝑜 + 𝜎𝑎𝑏𝑜2 + 3 𝜏𝑎𝑜

2

𝜎𝑒𝑚 =𝜎𝑚𝑎𝑜 + 𝜎𝑚𝑏𝑜

2+ 𝐾𝑓𝑠𝜏𝑚𝑜

2+

𝜎𝑚𝑎𝑜 + 𝜎𝑚𝑏𝑜

2

2

Axial Bending Torsion

Axial Bending Torsion

STRENGTH

YOU ARE ALLOWED TO MODIFIED EITHER ‘STRENGTH’

OR ‘STRESS’ ONLY !

𝑆𝑛 = 𝑆𝑛′ 𝐶𝐿𝐶𝐺𝐶𝑆𝐶𝑇𝐶𝑅

reducing

increasing

OR

STRESS

𝜎𝑎𝑎 = 𝐾𝑓𝑎𝜎𝑎𝑎𝑜

𝜎𝑚𝑎 = 𝐾𝑓𝑎𝜎𝑚𝑎𝑜

𝜏𝑎 = 𝐾𝑓𝑠𝜏𝑎𝑜

𝜏𝑚 = 𝐾𝑓𝑠𝜏𝑚𝑜

AXIAL

𝜎𝑎𝑏 = 𝐾𝑓𝑏𝜎𝑎𝑏𝑜

𝜎𝑚𝑏 = 𝐾𝑓𝑏𝜎𝑚𝑏𝑜

BENDING TORSION

𝜎𝑎 = 𝐾𝑓𝜎𝑎𝑜 𝜎𝑚 = 𝐾𝑓𝜎𝑚𝑜

𝜏𝑎 = 𝐾𝑓𝑠𝜏𝑎𝑜 𝜏𝑚 = 𝐾𝑓𝑠𝜏𝑚𝑜

Note: Subscript with o mean without consideration of stress concentration effect.

3.18 SIMULTANEOUS ACTION OF BENDING,

TORSION AND AXIAL STRESSES WITH 𝐾𝑓

Equivalent Amplitude Stress, 𝜎𝑒𝑎 is calculated using the Distortion Energy Theory:

Equivalent Midrange Stress, 𝜎𝑒𝑚 is calculated using the Maximum Principal Stress. Use Mohr Circle Theory:

𝜎𝑒𝑎 = 𝐾𝑓𝑎𝜎𝑎𝑎𝑜 + 𝐾𝑓𝑏𝜎𝑎𝑏𝑜2+ 3 𝐾𝑓𝑠𝜏𝑎𝑜

2

𝜎𝑒𝑚 =𝐾𝑓𝑎𝜎𝑚𝑎𝑜 + 𝐾𝑓𝑏𝜎𝑚𝑏𝑜

2+ 𝐾𝑓𝑠𝜏𝑚𝑜

2+

𝐾𝑓𝑎𝜎𝑚𝑎𝑜 + 𝐾𝑓𝑏𝜎𝑚𝑏𝑜

2

2

(viii) Determine the alternating part, 𝜎𝑎𝑜 and 𝜏𝑎𝑜.(ix) Determine the steady part, 𝜎𝑚𝑜 and 𝜏𝑚𝑜.(x) Determine the alternating part, 𝜎𝑎 and 𝜏𝑎.(xi) Determine the steady part, 𝜎𝑚 and 𝜏𝑚.(xii) Determine the equivalent alternating part, 𝜎𝑒𝑎.(xiii) Determine the equivalent steady part, 𝜎𝑒𝑚.

3.19 SAMPLE PROBLEM

(ix) Draw Fatigue Failure Diagram. Plot the operating point. (x) Calculate the Langer’s and Goodman’s safety factor.(xi) Determine the number of stress cycle, N.(xii) Give your critical comments on the results.

3.20 SAMPLE PROBLEM

Goodman

Langer

𝜎𝑎𝑆𝑛

+𝜎𝑚𝑆𝑢𝑡

=1

𝑛

𝜎𝑎𝑆𝑦

+𝜎𝑚𝑆𝑦

=1

𝑛

Goodman Line

Langer Line (Yield Line)𝑆𝑦

𝑆𝑦 𝑆𝑢𝑡

𝑆𝑛

Midrange Stress, 𝜎𝑚

Alt

ern

atin

g St

ress

, 𝜎𝑎

Example :

A component is subjected to a maximum cyclic stress of 750 MPa and aminimum of 70 MPa. The steel from which it is manufactured has an ultimatetensile strength, 𝑆𝑢𝑡 of 1050 MPa and a measured endurance limit, 𝑆𝑛 of 400MPa. The fully reversed stress at 1000 cycles is 750 MPa. Using the Goodmanfatigue failure criterion, calculate the component life.

Step 1: Determine the 𝜎𝑎 and 𝜎𝑚:

𝜎𝑎 =𝜎𝑚𝑎𝑥 − 𝜎𝑚𝑖𝑛

2=750 − 70

2= 340 𝑀𝑃𝑎

𝜎𝑚 =𝜎𝑚𝑎𝑥 + 𝜎𝑚𝑖𝑛

2=750 + 70

2= 410 𝑀𝑃𝑎

3.21 SAMPLE PROBLEM

• Step 2: Choose failure criteria and then, determine the 𝜎𝑓 :

Sut = 1050 MPa

Sn = 400MPa

σm = 410 MPa

σa = 340MPa

SAFE

σm

σa

σf

𝜎𝑎𝜎𝑓

+𝜎𝑚𝑆𝑢𝑡

= 1

𝜎𝑎𝜎𝑓

+𝜎𝑚𝑆𝑢𝑡

= 1

340

𝜎𝑓+

410

1050= 1

𝜎𝑓 = 557.8 𝑀𝑃𝑎These two points have the same number of cyclic life

3.17 EXAMPLE

• Step 3: Determine the a and b :

• Step 4: Determine the number of cycles:

𝑆𝑓 = 𝑎𝑁𝑏

𝑏 = −1

3log

𝑆3𝑆𝑛′ = −

1

3log

750

400= −0.091

𝑎 =𝑆3

2

𝑆𝑛′ =

750 2

400= 1406.25

𝑁 =𝜎𝑓

𝑎

1𝑏=

557.8

1406.25

1−0.091

= 2.544 × 104 𝑐𝑦𝑐𝑙𝑒𝑠

3.17 EXAMPLE

Thank You

END