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Lec2 Fatigurefailure

The document discusses fatigue failure and cyclic loading. It provides background on fatigue failure including its causes and types. It then discusses some basic concepts associated with fatigue including crack initiation and propagation. The document also reviews the history of fatigue studies and concepts. It examines fatigue testing methods and diagrams including stress cycles and S-N diagrams.

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0% found this document useful (0 votes)
34 views34 pages

Lec2 Fatigurefailure

The document discusses fatigue failure and cyclic loading. It provides background on fatigue failure including its causes and types. It then discusses some basic concepts associated with fatigue including crack initiation and propagation. The document also reviews the history of fatigue studies and concepts. It examines fatigue testing methods and diagrams including stress cycles and S-N diagrams.

Uploaded by

elwakilwkw
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Failure Prediction for Cyclic

Loading
Dr. Tarek Elmelegy
Faculty of Engineering, Room 05
Email: tarek.elmelegy@bue.edu.eg
What is Fatigue Failure ?

◼ Fatigue: from Latin word for “to tire”


◼ Engineering Fatigue: damage and failure of
materials under repeated loading
◼ Fatigue Failure: common failure mode in most
materials (90% of field failures)
◼ Causes: repeated loads (or stresses) on a member
for certain number of cycles
◼ Example: bending a paper clip repeatedly to
failure
◼ Characteristics: occurs at low stress levels, looks
brittle even in ductile metals
2
What is Fatigue Failure ?

◼ Types of Fatigue
◼ Mechanical
◼ Fluctuations in applied stresses or strains
◼ Thermomechanical
◼ Fluctuations in Temperature
◼ Corrosion
◼ Acceleration of crack initiation
◼ Wear
◼ Acceleration of crack initiation
3
On the Bridge!

“On the Bridge”, an illustration from


Punch magazine in 1891 warning the
populace that death was waiting for
them on the next bridge. Note the
cracks in the iron bridge.

I-35W Mississippi River bridge Collapse, August, 2007


Fatigue disasters

Alexander L. Kielland oil platform 1980

Versailles train 1842

De Havilland Comet plane 1954 Eschede train disaster 1998


Some of the Basic Concepts
Associated with Fatigue
◼ Fatigue is a complex phenomenon with no universal
theory
◼ There are many theories for different materials
◼ Most of the design experience is based on carbon steel
experiments
◼ Fatigue involves damage accumulation and crack
growth with each stress cycle
◼ Fatigue cracks usually start at a surface and
propagate through the bulk
◼ Surface quality is important for fatigue resistance
◼ Fatigue cracks compete for dominance and growth rate
◼ Fatigue testing is essential for safe design
6
History of Fatigue
◼ Albert was the first to publish fatigue test results in 1837
◼ WÖhler measured service loads of railway axles and
suggested design for finite fatigue life in 1858
◼ WÖhler published fatigue test results of railway axles and
discussed the influence of mean stress and the difference
between finite and infinite fatigue life in 1860
◼ WÖhler suggested larger safety factors for fatigue design
and observed fatigue crack propagation and its rate for
casted steel
◼ The WÖhler-curve was plotted by Spangenberg, WÖhler’s
successor, and represented by Basquin with the
equation:a=CRn in 1920
◼ The first stress concentration factor was calculated by
Kirsch in 1889
7
History of Fatigue

◼ Notches reduce the fatigue strength of high-strength materials


◼ Shot-peening improves the fatigue strength by inducing
compressive stresses
◼ Damage accumulation theories were developed by Palmgren,
Serensen, and Miner
◼ Corrosion fatigue, initiation and propagation stages,
and stress concentration factors were introduced between 1920-
1945
◼ Stress concentration factors are less severe for fatigue than for
static loading (i.e. Kt>Kf ).
◼ De Havilland Comet crashes in 1954 led to extensive research and
testing on fatigue failure propagating from a window cutout (which
were rectangular and not oval as they are today, thus giving rise to
large stress concentrations).
◼ Low-Cycle-Fatigue (LCF) was established as a new field by Manson
and Coffin in 1954 8
Microstructure of Fatigue Failures

◼ Features of fatigue fracture


surface
◼ Origin region is smooth due to

slow crack growth and sliding


between surfaces
◼ Final fracture region has

visible striations or beachmarks


due to rapid crack growth and
arrest Cross-section of a fatigued section,
showing fatigue striations or beachmarks
◼ Intermediate region has originating from a fatigue crack
microscopic striations
◼ Final fracture surface looks
rough and is indicative of brittle
fracture, but can also appear
ductile depending on the
material
Methods to Maximize Design Life

produce smooth surfaces by


minimize initial flaws grinding or polishing and protect
them from damage.

maximize crack initiation impart or relieve surface residual


stresses by shot peening,
time burnishing, or other treatments.

maximize crack use materials with elongated grains


transverse to the crack direction,
propagation time such as cold-worked components.

maximize the critical use materials with high fracture


toughness.
crack length
Stress Cycle
◼ Cyclic stress is a function of time, but the
variation is such that the stress sequence repeats
itself.
◼ Mean Stress: the average of the maximum and
minimum stresses in the cycle
 max +  min
m =
2
◼ The range stress: is the difference between the
maximum and minimum stresses
Variation in nonzero cyclic mean stress.
 r =  max −  min
◼ The Stress amplitude is one half of the stress
range The most frequently used patterns of
  −
a = r
= max min constant amplitude cyclic stress are:
2 2 1. Completely reserved (m=0, Rs=-1,
◼ The Stress ratio: is the ratio of the minimum to
the maximum stress amplitudes Aa=∞)
 min
Rs = 2. Non zero mean
 max 3. Released tension (min=0, Rs=0,
◼ The Amplitude ratio: is the ratio of the stress Aa=1, m=max/2 )
amplitude to the mean stress
  −  min 1 − Rs 4. Released compression (max=0,
Aa = a = max = Rs=∞, Aa=-1, m=min/2 )
 m  max +  min 1 + Rs
Fatigue Testing: S-N Diagram

R.R. Moore machine fatigue test


specimen. Dimensions in inches.

◼ Rotating 4-point bending


◼ Constant F and constant M
◼ Allows for un-notched and
notched specimen options
◼ Ra = -1
◼ Sinusoidal wave form
◼ Constant frequency
◼ Not well-adaptable to non-zero
m
Fatigue Testing: S-N Diagram
◼ Endurance limit
Ferrous and titanium alloys exhibit distinct endurance limit.
◼ Nonferrous alloys (e.g Aluminum, copper, and Magnesium) do not
have a significant endurance limit. Their fatigue strength continues to
decrease with increasing cycles. Fatigue will occur regardless of the
stress level.

Fatigue strengths as a function of number of loading cycles. (a) Ferrous alloys, showing clear endurance limit,
(b) Aluminum alloys, with less pronounced knee and no endurance limit
S-N Diagram: Endurance Limit

Bending Se = 0.5Su


Axial Se = 0.45Su
Torsion Se = 0.29Su

Approximate endurance limit for various materials.


High Cycle, Finite Life Fatigue
In applications, such as car door hinge and aircraft body
panels, the number of cycles imposed on the component
are between 103 and 106.
The straight line for the finite life HCF part of the S-N
diagram can be represented by:

log 𝑆𝑓′ = 𝑏𝑠 log 𝑁𝑡′ + 𝐶ሜ


where 𝑏𝑠 = slope
𝐶ሜ = intercept
At the endpoints, the above equation
becomes
log 𝑆𝑙′ = 𝑏𝑠 log 1 03 + 𝐶ሜ = 3𝑏𝑠 + 𝐶ሜ
Then the fatigue strength is given by
log 𝑆𝑒′ = 𝑏𝑠 log 1 03 + 𝐶ሜ = 6𝑏𝑠 + 𝐶ሜ
S f = 10C ( Nt ) s for 103  Nt  106
b
Solving for 𝑏𝑠 and 𝐶ሜ gives:
1 𝑆𝑙′ or
𝑏𝑠 = − log ′
3 𝑆𝑒
( )
1/ bs
N  = S f 10 −C
for 103  Nt  106
′ 2
𝑆𝑙
𝐶ሜ = log
𝑆𝑒′
Endurance Limit Modification Factor

◼ The modification factors presented are for completely


reversed cycles.
◼ The modified endurance limit can be expressed as:
Se = k f ks kr kt km ko Se
where Se = Endurace limit from experimental
apparatus under indealized conditions
k f = Surface finish factor
ks = Size factor
kr = Reliability factor
kt = Temperature factor
km = Miscellaneous factor
ko = Stress concentration factor
Endurance Limit Modification Factor

Surface Finish Factor

Table 7.3 Surface finish factor.

Surface finish factors for steel. (a) As function


of ultimate strength in tension for different
manufacturing processes.
Endurance Limit Modification Factor

Surface Finish Factor


Effect of Surface Roughness

Figure 7.10 Surface finish factors for steel. (b) As function of ultimate strength and
surface roughness as measured with a stylus profilometer.
Endurance Limit Modification Factor

Size Factor
 0.869d −0.112 0.3 in  d  10 in

ks = 1 d  0.3in or d  8 mm
 1.189d −0.112 8 mm  d  250 mm

For components that are not circular, calculate an effective


diameter based on the cross-sectional area of the part
Endurance Limit Modification Factor

Reliability Factor

Reliability factor for six probabilities of survival.


Endurance Limit Modification Factor

Temperature Factor

The temperature factor is given by:

Sut
kt =
Sut ,ref
Where Sut= ultimate tensile strength of material at desired temperature
Where Sut,ref= ultimate tensile strength of material at reference
temperature
Endurance Limit Modification Factor
Miscellaneous Effects
◼ Manufacturing History
◼ Crack propagation is more rapid along the grain boundaries than
through grains.
◼ In design, cautions must be taken to make the critical stress
direction along the elongated grains.
◼ Coatings
◼ Coating layer adds compressive stresses to the surface layer of the
component and hence increase the fatigue strength
◼ If the coating layer is porous, it will be a high potential source for
crack initiation, and hence reducing fatigue strength
◼ Corrosion
◼ Working in corrosive media will cause surface degradation and
sites for crack initiation.
◼ Oxygen and hydrogen embrittlements caused by diffusion and
assisted by large stress at the initiated cracks will cause material
to be brittle and reduce fatigue strength
Endurance Limit Modification Factor
Miscellaneous Effects
Residual Stresses: Shot Peening
Shot peening is a cold working process in which the surface of a part is bombarded with small
spherical media called shot. Each impact leads to plastic deformation at the workpiece surface
leading to compressive residual stress after recovery.

Figure 7.11 The use of shot peening to improve fatigue properties. (a)
Fatigue strength at two million cycles for high strength steel as a function of
ultimate strength; (b) typical S-N curves for nonferrous metals.
Endurance Limit Modification Factor
Stress Concentration Effects
Notch Sensitivity
Usage:

1
ko =
Kf
Where
Kf = the fatigue stress
concentration factor
qn = Notch sensitivity factor
ko = Endurance limit modification
Notch sensitivity as function of notch radius for several factor
materials and types of loading.
Example: 1
Corresponding to a reliability of 99%, estimate the endurance limit of a
round cold drawn BS 070M20 (Sut=430 MPa) steel shaft 30 mm in
diameter

Se = k f ks kr kt km ko Se
where Se = 0.5 Sut = 215 MPa
k f = 0.82 (for cold drawn bar)
ks = 1.189d −0.112 = 1.189(30)−0.112 = 0.812
kr = 0.82
kt = 1.0
km = 1.0
ko = 1.0
Se = 0.82*0.812*0.82*123 = 117 MPa
Example: 2
The Figure below shows a rotating shaft supported on ball bearings at A
& D and loaded by the non-rotating force F. Estimate the shaft life, if it is
made from BS.080M40 steel with an ultimate tensile strength of 690
MPa, yield strength of 385 MPa, and HB of 200 MPa. All fillets are 3 mm
in radius and the shaft is machined finish.
6.8 kN
Example: 2
Point (B) has a smaller cross section, a higher
moment, and higher stress concentration factor
than point (C). Failure most likely to initiate at
this point. Therefore, it is important to calculate
the strength at this point an compare with the
applied stress. First we need to determine th
endurance limit of the shaft at point (B)
Se = k f ks kr kt km ko Se
where Se = 0.5 Sut = 0.5*690 = 345 MPa
k f = 0.73 (for machined surfaces)
ks = 1.189d −0.112 = 1.189(32)−0.112 = 0.806
kr = kt = km = 1.0
For D = 38 / 32 = 1.1875, r = 3/ 32 = 0.09375 get KC = 1.65
d d
for Sut = 690 MPa, qn = 0.83
K f = 1 + ( KC − 1) qn = 1.54
ko = 1.0 = 0.649
Kf
Se = 0.73*0.806*0.649*345 = 131 MPa
Example: 2

Now, determine the stress at point (B). The


bending moment is given by

250*225*6800
MB = = 695 N .m
550

=
My
=
695 x103 32
( )
2 = 216 MPa
I  ( 32 ) 4
64
  Se → Finite Life
To determine (N), we need to find bs & C

1 S 1 0.8Sut
bs = − log l = − log = −0.1996
3 
Se 3 Se
( 0.8Sut )
2

C = log = 3.341
Se
→ N = 111, 000 cycles
Fatigue Failure Theories:
Influence of Non-Zero Mean Stress
Gerber Line

Goodman Line

Soderberg Line

Influence of nonzero mean stress on fatigue life


for tensile loading as estimated by four
empirical relationships.
Fatigue Failure Theories:
Influence of Non-Zero Mean Stress

Modified Goodman
Diagram

Complete modified Goodman diagram, plotting


stress as ordinate and mean stress as abscissa.
Fatigue Failure Theories:
Influence of Non-Zero Mean Stress

Modified Goodman
Criterion
Equations and range of
applicability for construction
of complete modified
Goodman diagram.
Fatigue Failure Theories:
Influence of Non-Zero Mean Stress

Modified Goodman Criterion

Table 7.6 Failure equations and validity limits of equations for four regions
of complete modified Goodman diagram.
Fatigue Failure Theories:
Failure of Brittle Materials

For an single normal stress with a mean and alternating stress components,
the safety factors are given by:
Sut
ns static =
Kc m
Se
ns fatigue =
Kc a

For an single shear stress with a steady and alternating shear stress
components, the safety factors are given by:

Sut
ns =
static
K cs m (1 + Sut Suc )
Se
ns =
fatigue
K cs a (1 + Sut Suc )
Fatigue Failure Theories:
Multiaxial Fatigue

For the case when the stresses are alternating in more than one direction, the
equivalent mean and alternating stresses can be calculated as follows

( a,x −  a, y ) + ( a, y −  a, z ) + ( a, z −  a, x ) + 6 ( a2, xy +  a2, yz +  a2, zx )


2 2 2

 a =
2

( −  m, y ) + ( m, y −  m, z ) + ( m, z −  m, x ) + 6 ( m2 , xy +  m2 , yz +  m2 , zx )
2 2 2

 m =
m, x

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