Chapter 3: Design of
Mechanical Failure Prevention 2: Fatigue Failure
DR. AMIR PUTRA BIN MD SAAD
C24-322
[email protected] | [email protected]
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
β’ Let π be the number of cycles to fatigue for a specified level of loading
- For 1 β€ π β€ 103, generally classified as low-cycle fatigue
- For π > 103, generally classified as high-cycle fatigue
β’ 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
β’ The most widely used fatigue-testing device is the R. R. Moore high-speedrotating-beam machine.
β’ Specimens in R.R. Moore machines are subjected to pure bending by means ofadded weights.
β’ 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
β’ In R. R. Moore machine tests, a constant bending load is applied, and thenumber of revolutions of the beam required for failure is recorded.
β’ Tests at various bending stress levels are conducted.
β’ These results are plotted as an S-N diagram.
β’ Log plot is generally used to emphasize the bend in the S-N curve.
β’ Ordinate of S-N curve is fatigue strength, ππ, 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, ππ
β’ Low-cycle fatigue considers the range from π = 1 to about 1000 cycles.
β’ In this region, the fatigue strength, ππ is only slightly smaller than the tensile
strength, ππ’π‘.
β’ High-cycle fatigue domain extends from 103 to the endurance limit life (106
to 107 cycles).
β’ 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, ππ
For line b-c:
ππ = πππ
where,
π =(π3)
2
ππβ²
π = β1
3log
π3ππβ²
π =ππππ£π
1π
π3 = 0.75ππ’π‘ 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, ππ
3.8 CHARACTERISTICS OF S-N CURVES FOR METALS
β’ 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 ππ
β² .
β’ Non-ferrous metals and alloys do not have an endurance limit, since their S-N curve never become horizontal.
β’ For materials with no endurance limit, the fatigue strength is normally reported at π = 5 Γ 108
β’ π = 1/2 is the simple tension test.
Aluminum Alloy
3.14 BASIC TYPES OF FLUCTUATING STRESS
β’ Example of loading history on the wind blades.
3.14 BASIC TYPES OF FLUCTUATING STRESS
β’ Fluctuating stresses often of sinusoidal patterns due to the nature of somerotating machinery.
β’ The peaks of the wave are more important than its shape.β’ 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
πΉπ =πΉπππ₯ + πΉπππ
2
ππ =ππππ₯ β ππππ
2
πΉπ =πΉπππ₯ β πΉπππ
2
ππ =ππππ₯ + ππππ
2
ππππ₯ = maximum stress
ππππ = minimum stress
ππ = amplitude stress
ππ = midrange stress
3.14 STRESS COMPONENTS IN A
CYCLIC STRESS
πππ =ππππ₯π β πππππ
2; πππ =
ππππ₯π + πππππ
2
πΉππ =πΉπππ₯π β πΉππππ
2; πΉππ=
πΉπππ₯π + πΉππππ
2
πππ =Tmaxo β Tmino
2; πππ =
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
πππ =ππππ₯π β πππππ
2; πππ =
ππππ₯π + πππππ
2
πππ =ππππ₯π β πππππ
2; πππ =
ππππ₯π + πππππ
2
πππ =ππππ₯π β πππππ
2; πππ =
ππππ₯π + πππππ
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, ππβ² FOR
STEELS
β’ For steels, the endurance limit relates directly to the minimum tensile strength as observed in experimental measurements.
β’ From the observations, the endurance of steels can be estimated as
ππβ² = 0.5ππ’π‘
β’ with the prime mark on the endurance limit referring to the rotating-beam specimen.
3.10 MODIFICATION FACTORS ON
ENDURANCE LIMIT
where,
i. πΆπΏ = Load Factor
ii. πΆπΊ = Size or Gradient Factor
iii. πΆπ = Surface Condition Factor
iv. πΆπ = Temperature Factor
v. πΆπ = Reliability Factor
vi. ππβ² = Rotary-Beam Test Specimen Endurance Limit
ππ = ππβ² πΆπΏπΆπΊπΆππΆππΆπ
modification factors (Empirical Data)
3.10.1 LOAD FACTOR, πΆπΏ
Load Factor, πΆπΏ :
3.10.2 GRADIENT FACTOR, πΆπΊ
Gradient Factor, πΆπΊ :
3.10.3 SURFACE FACTOR, πΆπΏ
Surface Factor, πΆπΏ :
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, πΆπ
Temperature Factor, πΆπ :
3.10.5 RELIABILITY FACTOR, πΆπ
Reliability Factor, πΆπ :
β’ The fatigue stress concentration factor from the existence of irregularities or discontinuities in materials is defined a
where,
πΎπ‘ is the static stress concentration factor.πΎπ is the fatigue stress concentration.
π is the notch sensitivity
3.11 FATIGUE STRESS CONCENTRATION
FACTOR, πΎπ AND πΎππ
πΎπ = 1 + π(πΎπ‘ β 1)
πΎππ = 1 + ππ (πΎπ‘π β 1)
Bending or Axial
Torsion
3.11 FATIGUE STRESS CONCENTRATION
FACTOR, πΎπ
3.12 NOTCH SENSITIVITY FACTOR, π or ππ
The ratio of Fatigue Stress Concentration Factor, Kf over Static StressConcentration Factor, Kt gives property of notch sensitivity, q.
π =πΎπ β 1
πΎπ‘ β 1
ππ =πΎππ β 1
πΎπ‘π β 1
Bending or Axial
Torsion
* for reversed bending and reversed axial loadings
3.12 NOTCH SENSITIVITY FACTOR, π or ππ
3.13 STATIC STRESS CONCENTRATION
FACTOR, πΎπ‘
r/d: 2/40 = 0.05D/d: 60/40 = 1.5Kt = 2.38
3.13 STATIC STRESS CONCENTRATION
FACTOR, πΎπ‘
r/d: 2 /40 = 0.05D/d: 60/40 = 1.5Kts = 1.66
If ππ’π‘ = 900 MPa and ππ¦ = 750 MPa.
(v) Determine the endurance limit, ππβ² .
(vi) Determine all the modification factors (πΆπΏ, πΆπΊ , πΆπ, πΆπ and πΆπ )(vii) Determine the corrected endurance limits, ππ.(viii) Determine the stress concentration factor/s (πΎπ or/and πΎπ π).
3.16 SAMPLE PROBLEM
3.17 MASTER FATIGUE DIAGRAM
π΄ =ππππ
π =ππππ
ππππ₯
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, ππ’π‘, and are very convenient. For
infinite life design strategies, the methods use various curves to connect the
Endurance Limit, ππ, on the alternating stress axis to either the yield stress, ππ¦ or
ultimate strength, ππ’π‘ 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
ππππ
+ππππ’π‘
=1
π
ππππ¦
+ππππ¦
=1
π
3.18 FATIGUE FAILURE CRITERION
Goodman Line
Langer Line (Yield Line)ππ¦
ππ¦ ππ’π‘
ππ
Midrange Stress, ππ
Alt
ern
atin
g St
ress
, ππ
π¦
π+π₯
π= 1
3.18 FATIGUE FAILURE CRITERION
y
x
a
b
β π = safety factorππππ
+ππππ’π‘
=1
π
3.18 FATIGUE FAILURE CRITERION
Goodman Line
ππ’π‘
ππ
Midrange Stress, ππ
Alt
ern
atin
g St
ress
, ππ
π > 1
π < 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 limitβthemean stress is zero and the alternatingstress Sn. Fluctuation b involves a tensilemean stress. In order to have an equal (inthis case, βinfiniteβ) 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
ππ
ππ’π‘
ππ
ππ
π = 106ππ¦ππππ
π = 105ππ¦ππππ
π = 104ππ¦ππππ ππ
πππ = 103ππ¦ππππ
ππ
number of cycles getting smaller
Constant Life Line
3.18 FATIGUE FAILURE CRITERION
Constant-life fatigue diagram for ductile materials.
Goodman
Langer
ππππ
+ππππ’π‘
=1
π
ππππ¦
+ππππ¦
=1
π
3.16 FATIGUE FAILURE CRITERION
Modified Goodman Line
Langer Line (Yield Line)ππ¦
ππ¦ ππ’π‘
ππ
Midrange Stress, ππ
Alt
ern
atin
g St
ress
, ππ
(a)
(b)
(c)
(d)
(e)
π > 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 πΎπ
Equivalent Amplitude Stress, πππ is calculated using the Distortion Energy Theory:
Equivalent Midrange Stress, πππ is calculated using the Maximum Principal Stress. Use Mohr Circle Theory:
πππ = ππππ + ππππ2 + 3 πππ
2
πππ =ππππ + ππππ
2+ πΎππ πππ
2+
ππππ + ππππ
2
2
Axial Bending Torsion
Axial Bending Torsion
STRENGTH
YOU ARE ALLOWED TO MODIFIED EITHER βSTRENGTHβ
OR βSTRESSβ ONLY !
ππ = ππβ² πΆπΏπΆπΊπΆππΆππΆπ
reducing
increasing
OR
STRESS
πππ = πΎππππππ
πππ = πΎππππππ
ππ = πΎππ πππ
ππ = πΎππ πππ
AXIAL
πππ = πΎππππππ
πππ = πΎππππππ
BENDING TORSION
ππ = πΎππππ ππ = πΎππππ
ππ = πΎππ πππ ππ = πΎππ πππ
Note: Subscript with o mean without consideration of stress concentration effect.
3.18 SIMULTANEOUS ACTION OF BENDING,
TORSION AND AXIAL STRESSES WITH πΎπ
Equivalent Amplitude Stress, πππ is calculated using the Distortion Energy Theory:
Equivalent Midrange Stress, πππ is calculated using the Maximum Principal Stress. Use Mohr Circle Theory:
πππ = πΎππππππ + πΎππππππ2+ 3 πΎππ πππ
2
πππ =πΎππππππ + πΎππππππ
2+ πΎππ πππ
2+
πΎππππππ + πΎππππππ
2
2
(viii) Determine the alternating part, πππ and πππ.(ix) Determine the steady part, πππ and πππ.(x) Determine the alternating part, ππ and ππ.(xi) Determine the steady part, ππ and ππ.(xii) Determine the equivalent alternating part, πππ.(xiii) Determine the equivalent steady part, πππ.
3.19 SAMPLE PROBLEM
(ix) Draw Fatigue Failure Diagram. Plot the operating point. (x) Calculate the Langerβs and Goodmanβs safety factor.(xi) Determine the number of stress cycle, N.(xii) Give your critical comments on the results.
3.20 SAMPLE PROBLEM
Goodman
Langer
ππππ
+ππππ’π‘
=1
π
ππππ¦
+ππππ¦
=1
π
Goodman Line
Langer Line (Yield Line)ππ¦
ππ¦ ππ’π‘
ππ
Midrange Stress, ππ
Alt
ern
atin
g St
ress
, ππ
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, ππ’π‘ of 1050 MPa and a measured endurance limit, ππ 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 ππ and ππ:
ππ =ππππ₯ β ππππ
2=750 β 70
2= 340 πππ
ππ =ππππ₯ + ππππ
2=750 + 70
2= 410 πππ
3.21 SAMPLE PROBLEM
β’ Step 2: Choose failure criteria and then, determine the ππ :
Sut = 1050 MPa
Sn = 400MPa
Οm = 410 MPa
Οa = 340MPa
SAFE
Οm
Οa
Οf
ππππ
+ππππ’π‘
= 1
ππππ
+ππππ’π‘
= 1
340
ππ+
410
1050= 1
ππ = 557.8 πππThese two points have the same number of cyclic life
3.17 EXAMPLE
β’ Step 3: Determine the a and b :
β’ Step 4: Determine the number of cycles:
ππ = πππ
π = β1
3log
π3ππβ² = β
1
3log
750
400= β0.091
π =π3
2
ππβ² =
750 2
400= 1406.25
π =ππ
π
1π=
557.8
1406.25
1β0.091
= 2.544 Γ 104 ππ¦ππππ
3.17 EXAMPLE
Thank You
END
Top Related