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.
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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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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,
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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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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.
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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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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.
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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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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
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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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Stresses in Thin-Walled Pressure Vessels
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
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Stresses in Thin-Walled Pressure Vessels
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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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 ??
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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
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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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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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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
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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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Failure Theory for Ductile Materials
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)
Failure Theory for Ductile Materials
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Failure Theory for Ductile Materials
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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Failure Theory for Ductile Materials
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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Failure Theory for Ductile Materials
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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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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Factors of Stress Concentration S cc
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Directions for using Neubers diagram
Neubers diagram
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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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