Dynamic Loading _ Failure-final
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Transcript of Dynamic Loading _ Failure-final
I/C: KALLURI VINAYAK
Variable Loading• Variable loading results when the applied load or
the induced stress on a component is not constant but changes with time
• In reality most mechanical components experience variable loading due to
-Change in the magnitude of applied load
Example: Extrusion process
-Change in direction of load application
Example: a connecting rod
-Change in point of load application
Example: a rotating shaft
Fatigue • Fatigue is a phenomenon associated with variable
loading or more precisely to cyclic stressing or
straining of a material
• ASTM Definition of fatigue
– The process of progressive localized permanent
structural changes occurring in a material subjected
to conditions that produce fluctuating stresses at
some point or points and that may culminate in
cracks or complete fracture after a sufficient number
of fluctuations.
Fatigue Failure- Mechanism
• Three stages are involved in fatigue failure
-Crack initiation
-Crack propagation
-Fracture / Rupture
Crack initiation, propagation and rupture in a shaft subjected to repeated bending
Introduction to Fatigue in Metals
Crack initiation at
the outer surface
Beach marks
showing the
nature of crack
propagation
Final rupture occurs
over a limited area,
characterizing a very
small load required to
cause it
Crack initiation at
the root of keyway
at B
Final failure over
the small area at
C due to sudden
rupture
Crack
propagation
occurs over a
time period
Connecting rod failed by fatigue failure
The crack got initiated at the flash line of forging.
Flash
line of
forging
Fatigue failure of a steam engine connecting rod due to PURE TENSION load.
No surface crack.
Crack may initiate
anywhere that is
the weakest or
unknown source
of weakness.
In this rod, the crack
initiated due to
forging flake slightly
below the centre line.The crack propagated radially outward until some
time after which the sudden rupture occurred.
Radial direction of
crack propagation
Scope of this Topic: Approach to Fatigue Failure in Analysis and Design
• Fatigue life methods
• Fatigue strength and endurance limit
• Endurance limit modifying factors
• Stress concentration and notch sensitivity
• Fluctuating stresses
• Combination of loading modes
• Variable, fluctuating stresses, cumulative fatigue
damage
Fatigue Life Methods: Objective is to predict the failure in number of cycles
N to failure for a specific type of loading
• Stress life methods
– Based on stress levels only
– Least accurate of the three, particularly for LCF
– It is the most traditional because easiest to implement for a wide range of applications
– Has ample supporting data
– Represents high cycle fatigue adequately
• Strain life methods
– Involves more detailed analysis of plastic deformation at localized regions
– Good for LCF
– Some uncertainties may exist in results because several idealizations get compounded
– Hence normally not used in regular practice but only for completeness and special occasions
• Linear elastic fracture mechanics methods (LEFM)
– Assumes that crack is already present and detected
– The crack location is then employed to predict crack growth and sudden rupture with respect to the stress nature and intensity
– Most practical when applied to large structures in conjunction with computer codes and periodic inspection
33 10 :(HCF) fatigue cycleHigh ;101:(LCF) fatigue cycle Low >≤≤ NN
Stress Life Method: Determination of the strength of materials under
action of fatigue loads
R. R. Moore high-speed rotating beam machine.
Pure bending by means of weights and no transverse shear.
The specimen shown is very carefully machined and polished with a final polishing
in the axial direction to void circumferential scratches.
Number of revolutions of the specimen required for failure are recorded.
The first test is made at a stress that is some what under the ultimate strength of
the material.
ext, the test is repeated for a lower load, and so on.
The results are plotted in the S-N diagram, which is either semi-log or log-log.
Specimen preparation for R. R.
Moore Method
• The specimen can be machined on lathe
using formed tool of radius
and workpiece of length
inch8
79
inch10
73
How to apply pure reversed bending without transverse shear?
SFD
BMD
Mb
( )FaFaFxFxM
axFFxM
b
b
=+−=
−−=
The S-N Diagram for steel (UNS G41300), normalized, Sut=812 MPa.
Endurance Limit,
It is the stress at
which the
component can
sustain infinite
number of cycles
Endurance limit, not applicable for non-
ferrous metals and alloys
• The plot in the S-N diagram never
becomes horizontal for non-ferrous metals
and alloys
• Hence there is no endurance limit for non-
ferrous metals and alloys
• Fatigue strength (Se) is used instead which
is specified, normally, as fatigue strength
at 5*108 cycles
For different aluminium alloys (which is non-ferrous)
For non-ferrous metals and alloys, as can be seen here, the S-N diagram never
becomes horizontal and hence they do not have endurance limit. Hence, a
stress at a specific number of cycles, normally at 5*108 cycles, must be used as
fatigue strength
Quick Estimation of Endurance Limit
• Instead of referring to experimental data-bank each time, it should be possible to quickly estimate the value of endurance limit using some kind of formula
• To enable that, data has been generated for different types of steels, for endurance limit with respect to the ultimate tensile strength
• This plot seemed to closely follow a combination of two straight lines, of which the second being almost horizontal at Sut=1460 MPa
For steels, Endurance limit is estimated as:
conditions loading actual in thelimit Endurance
bending reversein obtainedlimit Endurance
1460740
14605040
'
'
=
=
>
≤=
e
e
ut
utut
e
S
S
MPa Sfor MPa
MPaSfor S.S
Stress concentration
• The single most influential factor leading to
high possibility of crack initiation
• Stress concentration can be due to
– Function of geometry (sudden change in
size/diameter; holes in the structure etc.
– and surface texture (surface finish, presence
of disintegrations etc.)
What is Kt?Kt=Theoretical stress concentration factor
stress Nominal
stress Maximum=tK
( )
FEM assuch simulation numerical
or sexperiment through Determined
stress Nominal
max
=
×=
−=
t
nomt
K
K
tdw
P
σσ
dw
What is Kt?: Determination from FEA
Determination of Kt through FEM
stress Nominal
stress Maximum=tK
Actual stress concentration factor, Kf
• Also called as fatigue strength reduction factor
( ) ( )
tables)from factor, geometric(or
factorion concentrat stress lTheoretica
21)-6& 20-6 Fig. (from y valuesensitivitnotch
1111
=
=
−+=−+=
t
tsshearfstf
K
q
KqKorKqK
Notch Sensitivity plot for normal stress
Fig: 6-20
Notch Sensitivity plot for shear stress
Fig: 6-21
Endurance limit ≠ Endurance strength
• Endurance limit (S’e) is only for rotational
bending of round bar
• Endurance strength (Se) is for all other
types of loading, geometry and operating
conditions
Endurance limit modifying factors
'
eedcbae SkkkkkS =
b
uta
a
aSk
k
=
= factoron modificaticondition surface
Table 6.2
Size factor, kb
( )
1. effect, size no loading axialFor
25451000837.0859.0
5179.224.162.7/
:only torsion and bendingin bars CScircular rotatingFor
factor modifying size
107.0107.0
=
≤≤−
≤≤==
=
−−
b
b
b
k
mmdifd
mmdifddk
k
etc.?section channel
section,-I r,rectangula circular, rotating-Non
:are that barsabout What
Kb for non-conforming situations:
�Effective dimension is used
�Effective dimension “de” obtained by equating the
volume of material stressed at and above 95 percent of
the maximum stress to the same volume in the rotating-
beam specimen
( )[ ]
dd
dA
dddA
Case
e
eee
37.0(2) and (1)Equation
bars CScircular rotating-nonfor ),2(01046.0
i.e 0.95d of spacing a having chords parallel twoof
outside area the twiceis area stresspercent 95 therounds, hollowor solid gnonrotatinFor
bars CS hollowcircular rotatingfor ),1(0766.095.04
bars CScircular rotating-Non:1
2
95.0
222
95.0
=
=
=−=
K
K
σ
σ
π
Kb for non-conforming situations:
Table 6-3
Load modification factor, kc
=
torsion
axial
bending
k c
,59.0
,85.0
,1
Actually the kc is dependent on the Sut of the material.
Tables 6-11 to 6-14 (page no. 325) in Text Book give the
details. The above values are average values.
Temperature modifying factor, kd( ) ( ) ( ) ( )
FT
where
TTTTk
o
F
FFFFd
100070
10595.010104.010115.010432.0975.0 41238253
≥≤
−+−+= −−−−
Reliability factor, ke
ae zk 08.01−=R za R za
50% 0 99.9% 3.031
90% 1.288 99.99% 3.719
95% 1.645 99.999% 4.265
99% 2.326 99.9999% 4.753
• Accounts for
– Corrosion
– Coating failure
– Spraying etc.
Miscellaneous effects factor, kf
Four specific types of cyclic loading identified in mechanical
systems:
• Reversed (completely reversed) – mean stress is zero; equal reversals on both sides
• Repeated – minimum stress is zero; mean stress equal to half of the range stress
• Fluctuating – maximum, minimum and mean stress are all non-zero and arbitrary
• Alternating – minimum stress is zero; mean stress is always compressive and is equal in magnitude to range stress
Pictorial depiction of various types of cyclic loading
Two important of those four types of cyclic (fatigue) loading
• Completely reversed cyclic loading
– The mean load is zero
– Normally has a well defined mathematical variation such harmonic, square etc.
– Used for testing and measurement of endurance limit of a given material
• Fluctuating loading
– The mean load is not zero
– The actual loading may not readily be given by a mathematical function but needs to be approximated
– More critical and realistic than completely reversed loading
Different fatigue failure models:
yielding) staticfor
checkingfor (only lineLanger 1
line Elliptic ASME1
lineGerber 1
lineGoodman Modified1
line Soderberg1
22
2
K
K
K
K
K
nSS
S
n
S
n
S
n
S
n
nSS
nSS
yt
m
yt
a
yt
m
e
a
ut
m
e
a
ut
m
e
a
yt
m
e
a
=+
=
+
=
+
=+
=+
σσ
σσ
σσ
σσ
σσ
Where aofamofm KandK σσσσ ==
How to estimate Kf
•Kf = 1+q(Kt -1).
•When q=0, the material has no sensitivity to notches,
and hence Kf=1.
•When q=1, or when notch radius is large for which q
is almost equal to 1, the material has full notch
sensitivity, and hence Kf = Kt.
•For all grades of cast iron, use q=0.20.
•Use the different graphs as given to obtain q for
bending/axial and torsional loading.
How to estimate Kf
Contd.
• Whenever the graphs do not give values of q for
certain combinations of data, use either Neuber
equation or Heywood equation.
How to estimate Kf
Contd.
• Use the Neuber equation when the notch is
circular/cylindrical.
( )
radiusnotch
strength. ultimate offunction i.e ),(
constant material a is andconstant Neuber is a where
11
1
1
=
=
−+=
+
=
r
Sfa
KqKand
r
aq
ut
tf
For steel, with Sut in kpsi, the Neuber constant can be approximated by a third-
order polynomial fit of data as
How to estimate Kf
Contd.
• Use Heywood equation when the notch is NOT
circular/cylindrical but is a tranverse hole or
shoulder or groove.
( )
size esize/groovder size/shoul hole
book.in text 15-6 Table in thegiven are values
121
=
−+
=
r
a
where
r
a
K
K
KK
t
t
tf
How to apply Kf
• If there is no notch, there is also no notch sensitivity, q=0, and Kf=1. Hence σm= σm0 and σa= σa0. In other words no stress concentration needs to be applied.
• When there is notch, 0<q<1, Kf>1, and:
� If localized plastic strain at the notch is to be avoided, then apply Kf to both mean and amplitude stresses.
�σm= Kf σm0 and σa= Kf σa0.
� If localized plastic strain is not a concern or can not be avoided by incorporating Kf, then apply Kf
only to the amplitude stress (conservative).
�σm= σm0 and σa= Kf σa0.
13
Prob 6-17: The cold drawn AISI 1018 steel bar is subjected to an axial load fluctuating between 3.5 kN and 15 kN. Estimate the factor of safety ny and nf
using ASME Elliptic criterion
Prob 6-20: A formed round wire cantilever spring subjected to a varying force. The ultimate strength is 1296 MPa. No stress concentration. Surface finish is hot rolled finish. Find factor of safety by using modified Goodman criteria
Combination of loading modes
• Different types of cyclic loads may be applied in combination, for example, bending, axial and torsional on machine components
• When the loads and in-phase, the maximum values of loads occurs at the same time and so are the minimum values.
• Hence in such cases, we can estimate the maximum and minimum von-Mises stress values and then estimate the mean and amplitude von-Mises stresses. Then fatigue criterion may be applied.
Combined loadingFor the common case of a shaft with bending stresses, torsional shear stresses,
and axial stresses, the von Mises stress is
Considering that the bending, torsional, and axial stresses have alternating and
midrange components, the von Mises stresses for the two stress elements can be
written as
For plane stress
Design for Combined loading
• Calculate von Mises stresses for alternating and
midrange stress states, σ′a and σ′m .
• Apply stresses to fatigue criterion i.e Soderberg,
Modified-Goodman, Gerber’s or ASME Elliptic
criteria by replacing σa and σm with σ′a and σ′mrespectively
• Conservative check for localized yielding using
von Mises stresses i.e
• 6-27 Fig shows clutch testing machine. Axial load applied to the shaft is cycled from 0 to P. Torque is induced as T=0.25fP(D+d). Sy=800 MPa, Sut=1000 MPa, Kta=3, Kts=1.8, f=0.3. Find the maximum value of P such that the shaft will survive 106 cycles with factor of safety of 3 using Goodman criteria.