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Chapter 8 (and some of 6)Fracture and Failure Analysis
LEARNING OBJECTIVES…
• Define fracture toughness in terms of a brief statement and an equation.
• Define the conditions for fracture.
El b t th l ti hi b t i t i t• Elaborate upon the relationship between impact resistance, tensile strength, tensile ductility, and fracture toughness.
• Define hardness and relate it to strength.
FRACTUREThe separation or fragmentation of a solid body into two or more pieces under an applied stress
Figure 8.1 (a) Highly ductile fracture in which the specimen necks down to a
“Fracture is when something breaks into pieces!”
which the specimen necks down to a point. (b) Moderately ductile fracture after some necking. (c) Brittle fracture without any plastic deformation. [Callister, 7th Ed.]
• Steps in fracture:– Crack initiation (formation)– Crack propagation (growth)
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Modes of Fracture• Depends upon the ability of the material to undergo plastic deformation prior
to fracture.
• Fracture modes:
– DUCTILE: most metallic materials, polymers• Extensive plastic deformation in front of crack• Extensive plastic deformation in front of crack.• This allows the crack to resist further extension unless applied stress is
increased.
– BRITTLE: ceramics, ice, cold steel• Very little plastic deformation• Cracks move rapidly without an increase in applied stress.
No fracture is most desirable.When fracture does occur, a ductile fracture is generally better.
With a ductile fracture, the component may maintain some structural capacity before catastrophic failure.
• Ductile Materials:
– Undergo extensive plastic deformation prior to failure
• Brittle Materials:
– Undergo little or no plastic deformation prior to failure.
Figure 6.13 Schematic representations of tensile stress‐strain behavior for brittle and ductile materials loaded to fracture. [Callister, 7th Ed.]
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Brittle vs. Ductile Fracture
a) Very ductile, soft metals (e.g., Au and Pb) at room temperature. Other metals, polymers, and glasses at high temperatures.
b) Moderately ductile fracture, typical for ductile metals (e.g., Al, Ni).
c) Brittle fracture, typical of cold ferrous metals and ceramics.
Ductile FractureCrack grows 90° to applied stress
Figure 8.2 Stages in the cup‐and‐cone fracture. (a) Initial necking. (b) Small cavity formation. (c) Coalescence of cavities to form a crack. (d) Crack propagation. (e) Final shear fracture at a 45° angle relative to the tensile direction (From Ralls Courtney and
a) Necking
b) Cavity formation
tensile direction. (From Ralls, Courtney and Wulff, Introduction to Materials Science and Engineering, p. 468, Copyright 1976, Wiley, New York). [Callister, 7th Ed.]
) y
c) Cavity coalescence to form cracks
d) Crack propagation (growth)
e) Fracture
Maximum shear stress
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DUCTILE FRACTURE
• Dimples on fracture surface correspond to microcavities that initiate crack formation.
• Picture at left is a typical “cup and cone” fracture.
BRITTLE FRACTURE
• No appreciable plastic deformation.
• Crack propagates very fast; nearly perpendicular to applied stress.
• Cracks often propagate along • Cracks often propagate along specific crystal planes or boundaries.
Transgranular fracture Intergranular fracture
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Transgranular Fracture
Transgranular fracture
Cracks pass through grains, often along specific crystal planes. Fracture surfaces have faceted texture because of
different orientation of cleavage planes in grains.
Transgranular fracture
Intergranular Fracture
Intergranular fracture
Cracks propagate along grain boundaries.
Intergranular fracture
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What is the significance of fracture?
• Many high strength materials such as ceramics fracture before yielding.
• Result is that engineering structures may fail at stresses well below the designed load
icapacity.
• Why???
Fracture of a “perfect” solid
Fracture plane
• Simultaneous rupture of all atomic bonds
A solid that contains no defects
Fracture plane
• Theoretical strength,th E/10
E/10Material GPa psi Au 7.8 1,080,000Cu 12.1 1,680,000
Far greater than experimental values
Thus, fracture does not occur in this fashionWhy is fracture stress lower than theoretical?
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Why/How do materials fail?
Are typical loading conditions severe enough to break all interatomicbonds?
NO!
Since we know the stress that is required to break bonds, why do materials fail in service?
DEFECTS!
They concentrate stress locally to levels high enough to rupture bonds
Wh b i l h f ?What about materials that are perfect?
NO MATERIAL IS PERFECT!
There is ALWAYS some statistical distribution of flaws or defects
P
Defect free solid
PP
Defect free solidFlaws concentrate
applied stress!area = P/A
P = load
PPP
P
2a
P
2a
P
Defect in solid
PP
Defect in solid
P
2b
P
2b
b/a 0, local
A1 >> A2
1 >> 2
PPP
1
2
<<
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a
As we noted earlierFracture occurs by the nucleation and growth of cracks!
Fundamental ways that loads can operate on cracks
TENSILE SLIDING TEARING
Figure 8.10 The three modes of crack surface
displacement. (a) Mode I, opening; (b) mode II, sliding; and (c) mode III,
tearing mode.[Callister, 7th Ed.]
The mode of loading (state of stress) plays a role in how/when a material will fail.
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MATERIALS DO NOT FAIL AT CERTAIN STRESS LEVELS!
All materials fail when the
FRACTURE TOUGHNESS
is exceededis exceeded
Fracture toughness is a characteristic material property that tells you when a material will fail.
K Y
Materials fail when KIC is exceeded
ICK Y a
Toughness Stress Crack Length
Depends on size of flaw and material properties NOT stress
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Values of KICMaterial KIC (MPa·m1/2)
Minerals 0.5 – 1
Concrete 0.35
Metals that exhibit high ductility, exhibit high toughness.Concrete 0.35
SLS Glass 1
Most Polymers 1 – 3
Ceramics 3 – 10
Cast Iron 10 – 40
Aluminum Alloys 20 – 50
Ceramics are very strong, but have low ductility and low toughness.
Polymers are veryAluminum Alloys 20 50
Plane Carbon Steels 30 – 100
Titanium Alloys 30 – 120
TRIP Steel ~200
Polymers are very ductile but are not generally very strong in shear (compared to metals and ceramics). They have low toughness.
Stress–strain curves for dense, polycrystalline Al2O3.
Why are ceramics so much stronger in compression compared to tension?
Tension CompressionTension Compression
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What is Compression?
AAo
P
hho
PA = Aoho/h = V/h
ENGINEERING: S = -P/Ao e = (ho – h)/ho = A – Ao/A
TRUE: = -P/A = ln(ho/h) = ln(A/Ao)
By convention, stress and strain are negative (forces are opposite than tension)
Compression• Friction is important!
– Friction between specimen and press causes barreling (more to come)
– Barreling is where the specimen expands laterally as the top/bottom surfaces are fixed by friction, consequently
• X’section area changes along its height
• Friction dissipates energy
Makes it difficult to get truly indicative properties
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Compression
• In compression, shape of the engineering plastic region different than tensiledifferent than tensile
• Why is the engineering stress higher?– Barreling (x‐section increased, resulting in for more load). Recall S=F/Ao
– Would the true stress‐strain be
S
the same for compression and tensile? Yes, instantaneous area used! e
Back to our question ‐Why are ceramics so much stronger in compression over tension?
Griffith crack modelGriffith crack model
21
2 cm
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In Class Problem:
A glass plate contains an atomic scale crack (tip radius = dia of O‐2 ion). If crack length = 1 m & theoretical strength = 7.0 GPa, calc. the breaking strength, .
IMPACT – Toughness• Short term dynamic stressing
– Car collisions– Bullets
Athl ti i t– Athletic equipment– Etc…
• Ability of the material to absorb energy prior to fracture.
• This is different than toughness.
• Useful in quality control.
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Influence of Loading Rate on Properties
• Increased loading rate...
-- increases y and TS-- decreases %EL
• Why? An increased rategives less time for dislocations to move pastdecreases %EL dislocations to move past obstacles.
y
TS
TS
larger
ll
27
y
smaller
Figure 8.12(a) Specimen used for Charpy and Izod impact tests. (b) A schematic drawing of an impact testing
hi Th
Charpy or Izod test
Strike a notched sample with an anvil.
machine. The hammer is released from a fixed height
and strikes the specimen; the
energy expended to break the specimen is reflected in the
difference between the initial height of the hammer and
Measure how far the anvil travels following impact.
Distance traveled is related to energy required to break the sample.
h h f l dthe swing height.
[adapted from Callister, 7th Ed.]
Very high rate of loading. Makes materials more “brittle.”
Impact energy is analogous to toughness.
High impact energy means high toughness; material resists crack propagation
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Impact Data for Select Materials
No. Alloy Impact energy [J (ft-lb)]
1 1040 carbon steel 180 (133)
2 8630 low-alloy steel 55 (41)
3 410 stainless steel 34 (25)
4 L2 tool steel 26 (19)
5 Ferrous superalloy (410) 34 (25)5 Ferrous superalloy (410) 34 (25)
6 Ductile iron, quenched 9 (7)
7 2048, plate aluminum 10.3 (7.6)
8 AZ31B magnesium 0.8 (0.6)
9 Ti-5Al-2.5Sn 23 (17)
10 Aluminum bronze, 9% (copper alloy) 48 (35)
11 Monel 400 (Ni-Cu alloy) 298 (220)
12 50:50 solder (Pb alloy) 21.6 (15.9)
13 Nb 1Z 174 (128)13 Nb-1Zr 174 (128)
14 Low density polyethylene 22 (16)
15 High density polyethylene 1.4-16 (1-12)
16 PVC 1.4 (1)
17 Epoxy 1.1 (0.8)
18 Teflon 5 (4)
Impact energies can vary wildly within each class of materials
• Fatigue = failure under cyclic stress.
compression on top
countermotorbe i g
specimenAdapted from Fig. 8.16, Callister 6e (Fig 8 16
FATIGUE – its important
tension on bottom
countermotor
flex coupling
bearing bearing
• Stress varies with time.--key parameters are S and m max
i
time
mS
Callister 6e. (Fig. 8.16 is from Materials Science in Engineering, 4/E by Carl. A. Keyser, Pearson Education, Inc., Upper Saddle River, NJ.)
min time• Key points: Fatigue...
--can cause part failure, even though max < c.--causes ~ 90% of mechanical engineering failures.
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• Fatigue limit, Sfat:--no fatigue if S < Sfat
Sfat
case for steel (typ.)unsafe
S = stress amplitude
FATIGUE DESIGN PARAMETERS
• Sometimes, thefatigue limit is zero!
Sfat
N = Cycles to failure103 105 107 109
safe
case for Al (typ )f
S = stress amplitude
Adapted from Fig. 8.17(a), Callister 6e.
Al (typ.)
N = Cycles to failure103 105 107 109
unsafe
safe Adapted from Fig. 8.17(b), Callister 6e.
• Crack grows incrementally
da
dN K m
typ. 1 to 6
FATIGUE MECHANISM
dN
~ a
increase in crack length per loading cycle
• Failed rotating shaft--crack grew even though
Kmax < Kc--crack grows faster if
crack origin
crack grows faster if• increases• crack gets longer• loading freq. increases.
Adapted fromFig. 8.19, Callister 6e. (Fig. 8.19 is from D.J. Wulpi, Understanding How Components Fail, American Society for Metals, Materials Park, OH, 1985.)
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1. Impose a compressivesurface stress(to suppress surface
k f i )
S = stress amplitude
near zero or compressive m
Adapted fromFig. 8.22, Callister 6e.
IMPROVING FATIGUE LIFE
cracks from growing)
--Method 1: shot peening --Method 2: carburizing
C-rich gasput
surface into
i
shot
N = Cycles to failure
moderate tensile mlarger tensile m
p m
2. Remove stressconcentrators.
bad
bad
better
better
compression
Adapted fromFig. 8.23, Callister 6e.
Ductile‐to‐Brittle Transition(DBTT)
DuctileBrittle
Figure 8.16 Influence of carbon content on the
Charpy V‐notch energy‐
Low toughness
High toughness
Charpy V notch energyversus‐temperature behavior for steel.
[from Callister, 7th Ed.]
Dramatic change in impact energy is associated with a change in fracture mode from brittle to ductile.
DBTT occurs in bcc metals but not in fcc metals
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• Increasing temperature...
--increases %EL and Kc
• Ductile-to-Brittle Transition Temperature (DBTT)...
Temperature
BCC metals (e.g., iron at T < 914°C)
pact
Ene
rgy
High strength materials ( > E/150)
polymers
More DuctileBrittle
FCC metals (e.g., Cu, Ni)
Imp
Temperature
High strength materials (y > E/150)
Ductile-to-brittle transition temperature
Adapted from Fig. 8.15, Callister 7e.
• Pre-WWII: The Titanic • WWII: Liberty ships
Design Strategy:Stay Above The DBTT!
Reprinted w/ permission from R.W. Hertzberg, Reprinted w/ permission from R.W. Hertzberg,
• Problem: Used a type of steel with a DBTT ~ Room temp.
"Deformation and Fracture Mechanics of Engineering Materials", (4th ed.) Fig. 7.1(a), p. 262, John Wiley and Sons, Inc., 1996. (Orig. source: Dr. Robert D. Ballard, The Discovery of the Titanic.)
"Deformation and Fracture Mechanics of Engineering Materials", (4th ed.) Fig. 7.1(b), p. 262, John Wiley and Sons, Inc., 1996. (Orig. source: Earl R. Parker, "Behavior of Engineering Structures", Nat. Acad. Sci., Nat. Res. Council, John Wiley and Sons, Inc., NY, 1957.)
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HARDNESSHARDNESS• The resistance of a
material to deformation by indentation.
• Hardness can provide a qualitative assessment of strength.
• Hardness cannot be used to quantitatively infer strength or ductility.
• Useful in quality control
Hardness• Resistance to permanently indenting the surface.• Large hardness means:
--resistance to plastic deformation or cracking incompression.
--better wear properties.
e.g., 10 mm sphere
apply known force measure size of indent after removing load
dDSmaller indents mean larger hardnesshardness.
increasing hardness
most plastics
brasses Al alloys
easy to machine steels file hard
cutting tools
nitrided steels diamond
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HARDNESSHARDNESSHardness is the resistance of a material to deformation by indentation.
Hardness can provide a qualitative assessment of strength.
Table 6.5 Hardness testing techniques
Early hardness tests were based on scratching materials with minerals
Figure 6.18 Comparison of several hardness
scales.[from Callister, 7th Ed.]
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Hardness: Measurement
• RockwellNo major sample damage– No major sample damage
– Each scale runs to 130 but only useful in range 20‐100.
– Minor load 10 kg
– Major load 60 (A), 100 (B) & 150 (C) kg• A = diamond, B = 1/16 in. ball, C = diamond
HB B i ll H d• HB = Brinell Hardness– TS (psi) = 500 x HB
– TS (MPa) = 3.45 x HB
