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.