f2012 posted notes for advanced topics mec531

8/13/2019 F2012 Posted Notes for Advanced Topics Mec531

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Introduction to advanced topicsin mechanics of solids

F2012

I. Inelastic material behavior and failurecriteria

II. Stress concentration

III. Materials fatigue failure

Iv. Fracture mechanics and materialsfatigue fracture failure

I. Inelastic Behavior of Solids


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Introduction

Materials elastic deformation is recoverable upon removal of external load

Opposite is inelastic behaviorViscoelastic, Visco-plastic and plastic

To predict failure with measured quantities like yield strength needs acriterion

Ductile materials fail by shear stress on planes of maximum shear stress

Brittle materials by direct tensile loading without much yielding

Other factors - Temperature- Rate of loading- Loading/ Unloading cycles

-20

-2

16

34

52

70

S t

r e s s

( k s

i )

0 0.05 0.1 0.15 0.2 0.25 0.3Strain (in/in)

Test results plotted for 1018 steel

Stress-strain curve for a structural steel


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Structural steels

All dim. in mm

?

Ideal Stress Strain Curves(simplified for modeling)


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Y

Y

E = tan

= tan / tan

Bi-linear stress strain curve: A simplified elastic-strain hardening model


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Models for Uniaxial stress-strain

All constitutive equations are models that are supposed to

represent the physical behavior as described by experimentalstress-strain response

Experimental Stress strain curves Idealized stress strain curvesElastic- perfectly plastic response

Models for Uniaxial stress-strain contd.

. Linear elastic response Elastic strain hardening response


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Models for Uni-axial stress-strain contd.

.

Rigid models

Rigid- perfectly plasticresponse

Rigid- strain hardening plasticresponse

Stress-strain curves

Cast Iron(brittle)

Copper(ductile)

Difference between tension andcompression

Weaker under tension


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Youngs Modulus Proportional limit

Yield Strength

Pl = proportional limit

Material parameters

Tangent Modulus Slope of a line tangent to thestress-strain curve at the pointof interest. It is used todescribe the stiffness of amaterial in the plastic range

Secant Modulus Slope of a line from the origin ofthe stress-strain diagram andintersecting the curve at thepoint of interest. It is also usedto describe the stiffness of amaterial in the plastic range


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Poissons ratio

Poissons ratio can be determined indirectly from stress -strain curve byknowing the change in the cross-sectional area of the specimen at a pointalong the elastic region of the stress-strain curve.

Definition: the strain ratio for a sample loaded inuniaxial stress

=

for most metals: 0.25 ~ 0.35

Rubber: ~ 0.5 (0.5 means incompressible!)

Cork: ~ zero. (good for making bottle

stoppers - an axially-loaded cork will notswell laterally to resist plug insertion.)

theoretical limits: [-1, 0.5] (for engineeringmaterials negative is rare )

Some interesting points aboutPoissons ratio


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- materials soften with increasing temperature

Temperature Effecton Inelastic Behavior of Solids

Typical stress-strain curves for polycrystalline aluminum and semi-crystalline polyethylene

Comparisons: Metallic and plastic materials

Both materials show necking but for polyethylene, the molecule chainalignment results in stiffening before failure.


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Creep / Stress Relaxation ExperimentDead weight creep machine for constant stress

Effects of Stress and Temperature on Strain Rate


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General Yielding Failure of a material happens when the structure cannot support the

intended function In some cases, load continues to increase beyond initial yielding load,

and part of the member still in elastic range General yielding occurs as entire member reaches the inelastic range

and the load cannot increase anymore

2

,6Y Y

bh P Ybh M Y

P Y P Ybh P

2

1.54 P Y

bh M Y M

Example:A rectangular cross-sectionsmember loaded in axial tensionor pure bending

Py load limit at initial yielding, P p load limit at generalyielding for axially loaded barMY load limit at initial yielding M pload limit at generalyielding for bending beam

Models for uniaxial stress-strain contd.

.

The members AD and CF are made ofelastic- perfectly plastic structural steel, and member BEis made of 7075 T6 Aluminum alloy. The memberseach have a cross-sectional area of 100 mm 2.Determinethe load P= P Y that initiates yield of the structure andthe fully plastic load P P for which all the members yield.

Soln:

Contd..


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Models for uniaxial stress-strain contd.

Materials Mechanical Behavior Failure modes and Failure criteria

How does the concept ofstress matter for them?


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Plane Stress

Plane Stress The state of stresswhen we analyzed bars in tension andcompression, shafts in torsion, andbeams in bending.

The left figure shows a general 3dimensional stress element

For material is in plane stress in the xy

plane

Only the x and y faces of theelement are subjected to stresses

All stresses act parallel to the xand y axis

Review:

Stresses on Inclined Planes

knowing x y and xy at a point

Consider a new stress elementrotated by

Find x, y, and xy,

at the same point in the materialas the original element, but isrotated about the z axis

x and y axis rotated through anangle


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Stresses on Inclined Planes Construct a FBD showing all the

forces acting on the faces The sectioned face is A.

Then the normal and shear forcescan be represented on the FBD.

Summing forces in the x and ydirections and remembering trigidentities, we get:

2cos2sin2

2sin2cos22

xy y x

y x

xy y x y x

x

2D Stress Transformation

The transformation equations for plane stress. transfer the stress component form one set of axes to

another

The state of stress remains unchanged

Based only on force equilibrium, independent ofmaterial properties or geometry There are Strain Transformation equations that can be

obtained based solely on the geometry of deformation.


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2cos2sin2

2sin2cos22

2sin2cos22

'

xy

y x

y x

xy y x y x

y

xy y x y x

x

Formulas for 2D stress transformation

Special cases for simple stress states:

Uniaxial stress- y = xy = 0 Pure Shear - x = y = 0

Biaxial stress - xy = 0

Transformation equations are simplifiedaccordingly.

2D Stress Transformation


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Principal Stress & Maximum Shear Stress

Structural members can fail due to excessive normalstress or shear stress

Failure prediction needs to knowmaximum normal and maximum shear stressesplane (orientation) of the maximum and stresses

Why are they so important for failure analysis?

Principal stresses

Two values of the angle 2 p are obtained from theequation tan(2 p)=

xyx y

.

One value 0 o-180 o, other 180 o-360 o Therefore p has two values 0 o-90 o & 90 o-180 o

Values are called Principal Angles. For one angle x is maximum, the other x is minimum.

Therefore: Principal stresses occur on mutuallyperpendicular planes.


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Principal Stresses Consider the right triangle Using the trig from the triangle

and substituting into thetransformation equation fornormal stress, we get

Formula for principal stresses.

22

2,1 22 xy y x y x

The Third Principal Stress

In x-y plane, rotating stress element about z-axis to obtain in-plane principal stresses

What about the stress element is 3D and has3 principal stresses?


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Maximum In-Plane Shear Stress

Consider the maximum shear stress and theplane in which it exists

Obtain by the transformation equations The planes is at 45 degree to the principal

planes/directions

Maximum shear stress is equal to the

difference of the principal stresses

Average Normal Stress In the planes of maximum shear stress normal

stress is not necessarily zero

Normal stresses acting on the planes of maximumshear stress is equal to half of the sum of theprincipal stresses, called the average normal stress

The sum of the normal stresses for an stresselements in any orientation is constant, calledstress invariant


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Some Important Concepts

The principal stresses are the max and min normalstress at a point

When the state of stress is represented by the principalstresses, no shear stress acts on the element

The state of stress at the point can also be representedin terms of max in-plane shear stress . In this case anaverage normal stress also acts on the element

The element in max in-plane shear stress is oriented45 from the element in principal stresses.

7 - 38

Transformation of Plane Strain Plane strain - deformations of the material

take place in parallel planes and are thesame in each of those planes.

Example: Consider a long bar subjectedto uniformly distributed transverse loads.State of plane stress exists in anytransverse section not located too close tothe ends of the bar.

Plane strain occurs in a plate subjectedalong its edges to a uniformly distributedload and restrained from expanding orcontracting laterally by smooth, rigid andfixed supports

0 :strainof components

x zy zx z xy y


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2

7 - 39

Transformation of Plane Strain

State of strain at the point Q results indifferent strain components with respectto the xy and xy reference frames.

y xOB xy

xy y xOB

xy y x

2

45

cossinsincos

21

22

2cos2

2sin22

2sin2

2cos22

2sin2

2cos22

xy y x y x

xy y x y x y

xy y x y x x

Applying the trigonometric relationsused for the transformation of stress,

7 - 40

Mohrs Circle for Plane Strain The equations for the transformation of

plane strain are of the same form as theequations for the transformation of planestress - Mohrs circle techniques apply .

Abscissa for the center C and radius R ,22

222

xy y x y xave R

Principal axes of strain and principal strains,

R R aveave

y x

xy p

minmax

2tan

22max 2 xy y x R Maximum in-plane shearing strain,


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7 - 41

Three-Dimensional Analysis of Strain

Previously demonstrated that three principalaxes exist such that the perpendicularelement faces are free of shearing stresses.

By Hookes Law, it follows that theshearing strains are zero as well and thatthe principal planes of stress are also the

principal planes of strain.

Rotation about the principal axes may berepresented by Mohrs circles.

7 - 42

Three-Dimensional Analysis of Strain by Mohr Circle For the case of plane strain where the x and y

axes are in the plane of strain,- the z axis is also a principal axis- the corresponding principal normal strain

is represented by the point Z = 0 or theorigin.

If the points A and B lie on opposite sidesof the origin, the maximum shearing strainis the maximum in-plane shearing strain, D and E .

If the points A and B lie on the same side ofthe origin, the maximum shearing strain isout of the plane of strain and is represented

by the points D and E .


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7 - 43

Three-Dimensional Analysis of Strain

Consider the case of plane stress,0 z b ya x

Corresponding normal strains,

babac

bab

baa

E

E E

E E

1

If B is located between A and C on theMohr-circle diagram, the maximumshearing strain is equal to the diameter CA.

Strain perpendicular to the plane of stressis not zero.

7 - 44

Measurements of Strain: Strain Rosette

Strain gages indicate normal strain throughchanges in resistance.

y xOB xy 2

With a 45 o rosette, x and y are measureddirectly. xy is obtained indirectly with,

3332

32

3

2222

22

2

1112

12

1

cossinsincos

cossinsincos

cossinsincos

xy y x

xy y x

xy y x

Normal and shearing strains may beobtained from normal strains in any threedirections,


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7 - 45

Mohrs Circle for 3D Analysis of 2D Stress (review)

If A and B (principal stresses)are on the

same side of the origin (i.e., have the samesign)

c) planes of maximum shearing stress areat 45 degrees to the plane of stress

b) maximum shearing stress for theelement is equal to half of themaximum stress

a) the circle defining max , min , and max for the element is not the circle

corresponding to transformations withinthe plane of stress

7 - 46

Stresses in Thin-Walled Pressure Vessels

Cylindrical vessel 1 = hoop stress 2 = longitudinal stress

t

pr

xr p xt F z

1

1 220

Hoop stress:

21

2

22

22

20

t pr

r prt F x

Longitudinal stress:

Example


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2

7 - 47


Points A and B correspond to hoop stress,1, and longitudinal stress, 2

Maximum in-plane shearing stress:

t pr 42

12) planeinmax(

Maximum out-of-plane shearing stresscorresponds to a 45 o rotation of the planestress element around a longitudinal axis

t pr 22max

7 - 48


Spherical pressurevessel:

t pr 221

Mohrs circle for in -planetransformations reduces to a

point0

constant

plane)-max(in

21

Maximum out-of-planeshearing stress

t

pr 4

121

max


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49

Failure Modes Failure Theory for Ductile Materials Failure Theory for Brittle Materials Safety Factor Selection of Failure Criteria

Materials Failure Theory

50

Failure Mode Ductile Failure:

Failure strain > 5% Fails by shear stress characterized w. sloped break surface Materials: steel, aluminum, copper,

Brittle Failure: Failure strain < 5% Fails by normal stress characterized w. flat break surface Materials: ceramics, glass, steel (low temp.),

Ductile failure Brittle failure


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Failure Mode

Failure Criteria:

d o

l o

x

y

y

y

xzx

xy

yx

xz

yz

zyz

f S y 1D

2D

3D

Material property test

Stress condition in atensile testing sample

The maximum shear stress when yield

happens is Y = Y/2


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Yield criteria for ductile materialsunder plane stress

Material property testingto obtain yield limit Y

Actual stress state in abeam

How to use Yinpredicting ductile

failure of an actualmember ??

7 - 54

Yield Criteria for Ductile Materials Under Plane Stress Failure of a machine component

subjected to uniaxial stress is directly predicted from an equivalent tensile test

Failure of a machine componentsubjected to plane stress cannot bedirectly predicted from the uniaxial stateof stress in a tensile test specimen

It is convenient to determine the principal stresses and to base the failurecriteria on the corresponding biaxialstress state

Failure criteria are based on themechanism of failure. Allowscomparison of the failure conditions fora uniaxial stress test and biaxialcomponent loading


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Materials failure testing subjects a materialcoupon in simpler stress (1D)

But in general, mechanical components aresubjected to complicated stress state (2D, 3D)

How to use the testing obtained materialsfailure data (such as yield stress) to predictfailure of a component under arbitrary stresscondition? ------ need failure theory and criteria

The necessity of Failure Criteria

Materials failure: Ductile and Brittle Fracture

Ductile

Brittle


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Materials failure process -Crack initiation and propagation

Common failure modes of a structural member:

1. Failure by excessive deflection

2. Failure by general yielding

3. Failure by fracture

4. Failure by instability (buckling)

Elastic deflection

Deflection by creep

Sudden (brittle) fracture

Fracture of cracked member

Progressive fracture, Fatigue


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Failure and limits on design: (isotropic material)

For predicting materials failure, need stress based criteria

MaterialType Failure Theories

Ductile Maximum shear stress criterion, von Mises criterion

Brittle Maximum normal stress criterion, Mohr's theory

Or non-Stress based criteria

Stiffness, vibrational characteristics, fatigueresistance, creep resistance etc.

For stress analysis for isotropic materials under elastic deformation, only two independent elastic constants are needed describing the stress-strain

relationship, i.e., Hooke's Law / = E

7 - 60

Failure Criteria for Brittle Materials Under Plane Stress

Maximum normal stress criterionStructural component is safe as long as themaximum normal stress is less than the ultimatestrength of a tensile test specimen.

U b

U a

Brittle materials fail suddenly throughrupture or fracture in a tensile test. Thefailure condition is characterized by theultimate strength U.


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61

Failure Theory for Brittle Materials

Max. Normal Stress Criterion Brittle material fails @ the max. normal stress

uc

ut

S

or S

3

1

crack 900

Example:

Maximum principal stress criterion ( Rankines criterion)for 3D

Yield begins at a point in a member where themaximum principal stress reaches a value equal tothe tensile/compressive yield stress Y .

1 = Y

The criterion ignores the existence of 2 and 3Effective stress

e = max (l 1 l, l 2 l, l 3 l)

Yield function f = max (l 1 l, l 2 l, l 3 l) - Y

Since ductile materials fail due to shear. This criteriononly applies to brittle materials whose failure is due totensile normal stress.

1

2

3

Y

Y

-Y

-Y

Y

-Y

Yield surface


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Maximum principal strain criterion (St. Venants criterion)

Yield starts when the maximum principal strain at a point reaches a valueequal to the yield strain

y= Y / E

This criterion takes into consideration of other principal stresses since 1=1/ E ( 1- 2- 3).

Similar to maximum principal stress criterion, this one also applies tobrittle materials for predicting fracture failure.

Effective stress e = max l i - j- k l, i j k

Yield function f = e Y = max l i - j- k l - Y , i j k

Yield Criteria for ductile materials: A general concept in Plasticity

Three basic components in Plasticity: Yield criterion to define initiation of yield Flow rule to relate plastic strain increments to stress increments Hardening rule predict changes in yield structure

We only focus on yield initiation theory yield criteria Primary objective of yield criteria is to develop the concept of yield

criteria for uni-axial stress states.

Basis of the development is the definition of effective (equivalent) uni-axial stress that is a particular combination of components in multi-axialstress state.

Basic concept is that yielding initiated in a multi-axial stress state whenthe effective stress reaches a limiting value that is assumed to be sameas that in the un-iaxial stress state). This is because that mostparameters for material property are obtained based on uni-axialtesting.


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Commonly used yield criteriafor Ductile Metals

As the evidence of slip lines shown by ductile metals atyielding suggests, most ductile metals failure due toshearing. This comes to the following two criteria forpredicting the failure for ductile metal materials.

Maximum Shear-Stress (Tresca)Criterion

Distortional Energy Density (vonMises) Criterion

Maximum Shear-Stress (Tresca) Criterion

Yield begins at a point in a member when themaximum shear stress reaches a value equal tothe maximum shear stress at yield of a samplein a uni-axial tension or compression test.

For multi-axial stress case max = ( max - min)/2

In uni-axial Y = Y/2

The criterion


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7 - 67

Mohrs Circle for 3D Analysis of 2D Stress (a review)

If A and B (principal stresses)are onthe same side of the origin (i.e., havethe same sign)

c) planes of maximum shearingstress are at 45 degrees to theplane of stress

b)maximum shearing stress forthe element is equal to half of

the maximum stress

a) the circle defining max , min ,and max for the element is notthe circle corresponding totransformations within the planeof stress

7 - 68

Yield Criteria for Ductile Materials Under Plane Stress

(1) Maximum shearing stresscriterion (for 2D stress)Structural component is safe as long as themaximum shearing stress is less than themaximum shearing stress in a tensile testspecimen at yield, i.e.,

2max

Y Y

For a and b with the same sign,

22or

2maxY ba

For a and b with opposite signs,

22maxY ba

whichever is greater


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Failure Theory for Ductile Materials

Max. Shear CriterionFor plane stress

Safety Factor

y

y

y

S

S

S

1

2

21

3 0

S y

S y

-S y

-S y 1

2

factor Safetynn

S y

:

),,max( 313221

Safe

y y S

70

Safety Factor

Concept and Definition

load normal overload designn

or

stresseffective strengthmaterial

n


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Max. Shear Criterion

P

max

90 0

crack

Example

1D stress

Strain energy density - review

The general form for strain energy density

Uo= 1/2 E [ 12 + 22 + 32 -2 ( 1 2 - 2 3- 3 1)]

The strain energy density at yield in uni-axial tension/compression

UoY= Y2/2 E

(2) von Mises condition (maximum distortion energy criterion) forductile materials under plane stress (2D stress)


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If only distortional energy (shear stress and shear strainrelated )is considered being responsible for ductile failure

2D distortional strain energy:

(2) von Mises condition (maximum distortion energy criterion) forDuctile Materials Under Plane Stress (2D stress)

Structural component is safe as long asthe distortion energy per unit volume isless than that occurring in a tensile testspecimen at yield.

222

2222 006

1

6

1

Y bbaa

Y Y bbaa

Y d

GG

uu

where a , b are the principal stresses at a point as variablesand Y is a material constant (yield strength)



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78


Compare Tresca with Von Mises CriterionDifference between Tresca and von Mises: < 15%

S y

S y

-S y

-S y

1

2Tresca Criterion

von Mises

Tresca condition is more conservative than von Mises


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79


Example: von Mises criterion applied to pure shear

1 3

y S

3

0060002

1 21222222 /

For material under 2D pure shear stress condition,yielding will occur at 57.7% of the Yield Strength.

y S 577.0

Distortion Energy (von Mises Criterion) for 3D Materials fail due to the distortional strain energy When principal stresses 1, 2, 3 are known

When normal & shear stresses in arbitrary coordinatesystem are known

n

S factor Safetyn

stress Effective

S

y

y

:

:

][2

1 2/1231

232

221

n

S factor Safetyn

stress Effective

S

y

y xz yz xy z x z y y x

:

:

](6[2

1 2/1222222

z

x

y 2

1

3


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81


ExampleFailure analysisfor an L shaped bar

subjected to a verticalforce at its tip F L

a

?

3

)(16

3264/

)2/()(

22

3

34

F n

S criterion Misesvon

d

FL

J

Tr

d Fa

d d Fa

I Mc

y

Critical site

82

Summary- Failure Theory Selection

For ductile Materials Von Mises criterion preferred Tresca criterion

For brittle Materials Max. normal stress criterion


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4

Causes for stress concentrationin a structure member

Abrupt geometric changes in a section(notches, holes, groves) Surface contact stressMaterial discontinuities (defects,inclusions, voids) Residual stress due to manufacturingCracks

Almost all engineering components and machineshave to incorporate design features whichintroduce changes in cross-section

These changes cause localized stressconcentrations

Severity of concentrations depends on thegeometry of the discontinuity and natureof the material.

Stress concentration factor, Kt = Smax /S avSmax , maximum stress at discontinuitySav, nominal stressKt, value depends only on geometry of the part


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Stress concentration K f , fatigue stressconcentration factor,

endurance limit of notch freeKf =

endurance limit of notched part

Stress Concentration in fatigue

Guidelines for design

Abrupt changes in cross-section should be avoided. Fillet radii or stress-relieving groove should be provided. Slot and grooves should be provided with generous run-out

radii and with fillet radii in all corners. Stress relieving grooves or undercut should be provided at

the end of threads and splines. Sharp internal corners and external edges should be avoided Weakening features like bolt and oil holes, identification

marks, and part number should not be located in highlystressed areas.

Weakening features should be staggered to avoid theaddition of their stress concentration effects.

Stress Concentration


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4

Factors of Stress Concentration S cc


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4

Directions for using Neubers diagram

Neubers diagram


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4

Example for using Neubers diagram


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In simple cases, stress concentrationfactors can be analytically obtainedusing Theory of Elasticity

More complicated cases usingnumerical methods such as FiniteElements simulations


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III. Fatigue Failure

Damage effect on materials subject toalternating force (failure under a live load)

The level of repeated load causing fatiguefailure is below that of a non-repeated failureload.

Approximately 80% 90% of all structuralfailures occur through some sort of fatiguemechanism

Fatigue process


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Examples forfatigue failure


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A variety of analytical tools and techniques are used to identifyfatigue fractures and their root cause. These include macroscopicexamination, microstructural analysis, hardness testing, chemicalanalysis, microprobe chemical analysis and scanning electronmicroscopy (SEM)

Stress amplitude

Range of stress

Mean stress

Materials fatigue failure under repeated cyclic loadingDefinitions of terms used inmodeling prediction

For Reversed cycle fatigueR = -1


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Fatigue Limit

Maximum stress a material can sustain under the action ofan infinite number of reversals of stress

For Metallic Alloys the stress level is usually between one-third and one-half of the static tensile strength

Causes of Fatigue Failure Member is stressed above its fatigue limit such that stress

alternates/fluctuates Presence of high-frequency vibrations that might not be

noticed Crack growth and damage cumulate assisted by the

environment (chemical)

High-Cycle Fatigue

Features:

Deformation is primarily elastic Stress is controlled More than 100,000 cycles Crack has not started


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Fatigue Limit

Stress

Log (Fatigue Life) 10 6 10 7

Steel

Aluminum

S-N curve for High cycle fatigue failure prediction

c f

p N )2(2 '

Low-Cycle Fatigue Stress is high enough for plastic deformation to occur Strain is controlled Failure occurs at less than 1000 cycles

p/2 is the plastic strain amplitude

f is an empirical constant known as the fatigueductility coefficient, the failure strain for a singlereversal

2N is the number of reversals to failure (N cycles)

C is a constant called the fatigue ductility exponent(-0.5 to -0.7) for metals

Fatigue prediction:


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Types of cyclic loading

Stress terms in relation to fatigue


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One of failure analysis goals = prediction of fatigue life of componentknowing service constraint and conducting Lab tests

Ignores crack initiationand fracture times

IV. Fracture mechanics, fatigue &fracture fatigue of engineering metals

Modes of fracture depending on types of load


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FRAC

Fracture is the separation of abody into two or more pieces dueto stress

Modes of fracture differ by ductileand brittle fracture

Ductile fracture in coppernucleating around inclusions


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Cup and Cone fracture Brittle fracture

Example of a fatigue fracture in a bicycle


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A close look of Cracks developed and propagated in a metal

FRACTOGRAPHY Pictures taken by scanning electron microscope (SEM) for microscopic

examination of the fracture surface OF ALUMINUM ALLOY 2024-T3


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Replicas of cracktip profiles and grid distortion at various loads

Stress concentrationfactor vs. specimengeometry/configuration

Question:what will happen ifa crack intersects ahole?


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Griffith Energy criterionCrack propagates wheresituation is favorable forreducing the materials strain

energy

Material behindcrack-tip is stress-relaxed so that theElastic strain energy is released

Crack surfaces havesurface energy ( s)

a

Fracture Mechanics

Classical Description of Strength of Materials Based on - curves obtained in uni-axial testing E , and G for linear elastic materials, plus Y and u for elasto-plastic

materials

Simple but powerful failure theories Widely used in the industry

But has some major shortcomings in handling sharp notches or cracks

Fracture Mechanics Quantifies crack-induced effects Predicts failure load accurately Safer design


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Modes of Crack Loading

Mode 1 Opening Mode

Mode 2 Sliding (In-Plane Shear)Mode

Mode 3 Tearing (Anti-Plane Shear)Mode

Linear Elastic Fracture Mechanics (LEFM)


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ij III III ij II II ij I I ij f K f K f K r

1

Linear Elastic Fracture Mechanics (LEFM)

Stress distribution near tip of a crack

Stress intensity factor


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Fracture toughness K c The stress intensity factor K was initially

used to quantify crack-tip stress distribution

Fracture occurs when the crack-tip stressintensity factor reached a critical value Kc

Kc is found to be independent of crack sizeor applied stress

Kc is called fracture toughness, a parameterfor material property

Metal Fatigue is a process which causes premature failure or damageof a component subjected to repeated load variation/cycle

Fatigue life of an initially defect-free structure can be written as thesum

NT = NI + NP

whereNI corresponds to a period for crack initiationNP is crack propagation period, including stable as well as acceleratedstages of fatigue crack growth

Fatigue and fatigue life for a metal


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Features of Materials Fatigue Failure

Damage effect on materials subject to alternatingforce (failure under a live load)

The level of repeated load causing fatigue failure isbelow that of a non-repeated failure load.

Approximately 80 90 per cent of all structuralfailures occur through some sort of fatiguemechanism

Ductile fracture processDuctile process normally develops in five stages:

1) Necking2) Formation of small cavities or micro-voids in interior of cross

section3) Growth of micro-voids in size and coalesce to form elliptical

shaped crack with long axis perpendicular to the stress direction4) Crack growth in direction parallel to major axis of the ellipsis5) Failure takes place by rapid propagation of a crack around the

outer perimeter of the neck due to shear deformation along 45 o degree or maximum shear stress direction


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Fatigue fracture of metals -Crack growth under repeated loads

Stress intensity factor K initially used to quantifycrack-tip damage for fracture scenarios

Fracture occur when the crack-tip stress intensityfactor reached a critical value, Kc,1 independentof crack size or net applied stress

K is related to fatigue crack growth rates d a /dN, orincrement of crack growth per load cycle

K range of variation of SIFfor a crack under cyclic loading


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Fatigue crack growth behavior

Plastic zone

For ductile materials :consider y (i.e. y means direction not yield)


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f2012 posted notes for advanced topics mec531

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