The fracture of a Al bicycle crank arm.
Failure
• In chapters 6 and 7 we examined the elastic
and plastic behaviour of materials.
• We learned how the motion of dislocations
makes it possible for the material to deform.
• We know from our daily experiences that
there is a limit on how much we can deform
a material. Materials eventually fail.
Another Mystery
• When a material splits in two, all the bonds that used to
hold the two parts together have been broken.
• For years people were puzzled because the force/stress
applied to the material is rarely large enough to break
all the bonds….
• Once again, we ask, what is missing?
Ship: cyclic loading from waves.
3
From Fig. 9.0, Callister (original Neil Boenzi, The New York Times.)
Stress Concentration
• Measured fracture strengths are less than
theoretical values because of microscopic flaws
which always exist under normal conditions.
• An applied stress is amplified at the tip of a crack
• Magnitude of amplification is dependent on:
– Crack orientation
– Crack geometry
• Stress concentration factor (Kt): a measure of the
degree to which an external stress is amplified at the
tip of a crack
 max
a
Kt 
2
0
t
Effect of Flaws
• When the max stress exceeds the yield strength in a
ductile material, plastic deformation occurs. In brittle
materials, cleavage/fracture occurs.
• Consider an elliptical hole in a plate:
5
Magnitude of
stress diminishes
with distance
from the crack tip
Demo: Influence of Notch
Sharpness
J. Hiscocks
2003
As you can see, both notches have the same length
(a), but different rt.
What will happen if we pull the card?
6
Demo: Influence of Notch Length
What happens if we have two cracks with the same
rt but different lengths?
J. Hiscocks,
2003
7
Is this Crack Dangerous?
http://www.newser.com/story/115634/fedsorder-emergency-737-checks-after-crack.html

Griffith Criterion:
• The stress at the tip of the crack is large
enough to break the bonds in the solid.
• Whether the crack propagates or not,
depends on a simple energy balance:
W  U el  GcA
work-done by the load if
the crack grows
Elastic energy released
when the crack grows
Energy needed to expand the
crack (Gc is the surface energy per
unit area of crack).
 0
Griffith Criterion:
Example 1: The balloon
• Let’s introduce a “crack” into the balloon.
• Inflate the balloon to a diameter of 10cm
and ask if
W  U el  GcA
Let’s ignore this term
because d is fixed.
If the balloon blows up, we’d
release all of the energy
stored in the walls.
Energy needed to expand the
crack and cause the balloon to
blow up.
?

Griffith Criterion :
Example 2: Real crack with fixed displacement
 0
• Let’s imagine that the crack got bigger by a
small amount δa.
• As before, we ask the question, is
2a
W  U el  GcA ?

W=0 (displacement of
the block is zero).
Elastic energy released
when the crack grows:
  
2aat 
( 12  )V   12
E


 0
2
Energy needed to expand
the crack:
Gc dA  2Gcta
a

Griffith Criterion :
Example 2: Real crack with fixed displacement
 0
• The crack will grow if:
EGc   2a
or
2a

EGc   a
Gc : Toughness[J/m2 ]
K   a : stress intensity factor [N/m3/2 ]
Kc  EGc : critical stress intensity factor
or fracture toughness[N/m3/2 or Pa m ]
Griffith Criterion: K  Kc
 0
a
Fracture Toughness
• Fracture toughness (Kc): a property that is a
measure of a material’s resistance to brittle fracture
when a crack is present
• when specimen thickness >>crack dimensions, Kc
becomes independent of thickness.
Kc  Y c a
A unit-less parameter
dependent on :
•Crack geometry
•Specimen size
•Manner of load application
KIc (plane strain fracture toughness)
decreases with increasing strain rate and
decreasing temperature
Units:
MPa m
increasing
Fracture Toughness Data
Based on data in Table B5,
Callister 6e.
Composite reinforcement geometry
is: f = fibers; sf = short fibers; w =
whiskers; p = particles. Addition data
as noted (vol. fraction of
reinforcement):
14
1. (55vol%) ASM Handbook, Vol. 21, ASM
Int., Materials Park, OH (2001) p. 606.
2. (55 vol%) Courtesy J. Cornie, MMC, Inc.,
Waltham, MA.
3. (30 vol%) P.F. Becher et al., Fracture
Mechanics of Ceramics, Vol. 7, Plenum
Press (1986). pp. 61-73.
4. Courtesy CoorsTek, Golden, CO.
5. (30 vol%) S.T. Buljan et al., "Development
of Ceramic Matrix Composites for Application
in Technology for Advanced Engines
Program", ORNL/Sub/85-22011/2, ORNL,
1992.
6. (20vol%) F.D. Gace et al., Ceram. Eng.
Sci. Proc., Vol. 7 (1986) pp. 978-82.
How Large a Flaw is Critical?
• A typical steel structure has a fracture toughness of about
KIc = 75 MPa-m1/2
If the structure is designed to withstand a stress of 750 MPa,
how big do flaws in the material have to be to pose a threat?
(Assume that Y = 1)
• Answer:
2
2
1  K IC  1  75 
1
ac 
 

 0.003m  3mm



   designY    750   (100)
– The largest tolerable flaws in many engineering structures are of
this dimension
15
Types of Fracture
• Fracture: crack formation and
propagation in response to an imposed
stress
• 2 Fracture Modes (classified based on
the ability of the material to experience
plastic deformation)
– Ductile (stable)
• Exhibit substantial plastic deformation
large % RA and % elongation (often >
20%)
• High energy absorption
Ductile Material
– Brittle (unstable)
• Little or no plastic deformation low %
elongation (often ≤ 1%)
• Low energy absorption
• Ductility is a function of:
– Temperature
– Strain rate
– Stress rate
Brittle Material
Ductile Fracture
• Ductile fracture is induced by plastic
deformation.
• It’s caused by damage accumulation.
– nucleation, growth, coalescence of voids
• After tensile instability starts, the damage
is concentrated in the neck.
• Note the importance of particles.
Indicates plastic
deformation
(fibrous)
17
Fracture Process
a) Initial Necking
b) Microvoids form in the
interior of cross section
c) Microvoids enlarge and
coalesce to form an elliptical
crack (long axis
perpendicular to stress
direction)
d) Crack propagation
e) Rapid propagation of crack
around outer perimeter of
neck  shear fracture at 45˚
relative to tensile direction
Brittle Fracture
• Occurs by rapid crack
propagation
• Fracture surfaces will have
their own distinctive
patterns; for example:
– Steel: chevron markings
– Lines or ridges originating
from near the center of the
cross section
– Amorphous materials: shiny,
smooth surface
• Cleavage: crack propagation
corresponding to the
successive and repeated
breaking of atomic bonds
along specific
crystallographic planes
Brittle transcrystalline steel fracture
http://www.tescan.com/gallery-gallery.php?obr=40&menu=2
Failure Origins
• What do you see on the mild steel specimen that
suggests why it failed at this location?
Defects are the
key to brittle
fracture
Fig. 9.3(b) Callister
20
An Example: Failure of a
Pressurized Pipe
• Failure may be…
– … ductile
• specimen still a
single piece
• extensive
deformation
– … brittle
• many pieces
• little
deformation
21
Figures from V.J. Colangelo and F.A.
Heiser, Analysis of Metallurgical Failures
(2nd ed.), Fig. 4.1(a) and (b), p. 66 John
Wiley and Sons, Inc., 1987. Used with
permission.
Impact Fracture
Testing
Impact Testing Techniques
• 2 standardized tests used to
measure impact energy (notch
toughness):
– Charpy
– Izod
Charpy
• The difference between the tests
lies in the way the specimen is
supported
Izod
http://met-tech.com/mechanical_testing.html
Ductile-to-Brittle Transition
• Occurs when the
temperature of a material
drops
• Related to the temperature
dependence of the
measured impact energy
absorption
A283 Steel
Photograph of fracture
surfaces of A36 steel
Charpy V-notch specimens
tested at the indicated
temperatures
Ductile to Brittle Transition
• Many materials become brittle at low temperatures.
– BCC materials (like steel) fail by cleavage along the <100> plane
• Material exhibiting a ductile to brittle transition should
only be used above the transition temperature so as to
avoid brittle failure
Impact energy vs. Temperature Plot
 3 general behaviours
Influence of carbon content on the
Charpy energy-temperature
behaviour for steel
Chapter 8 Review
What did you learn?
•
•
•
•
•
•
26
Chapter 8
Practice Problems
Practice Problems
1. A thick glass baking pan is removed from the oven at
200°C and placed in cold water at 20°C. Upon being
placed in cold water the pan breaks due to thermal
shock. Estimate the minimum crack length present at
the surface of the pan given that the Elastic modulus of
the glass is 100 GPa, the coefficient of thermal
expansion is 7 x 10-6 /K, the fracture toughness is 5
MPa.sqrt(m) and Y =1.
a)
b)
c)
d)
a min = 0.1 mm
a min = 0.5 mm
a min = 1 mm
a min = 2 mm
Practice Problems
Answer:
First, solve for  design using the known data :
 design  El T  126MPa
Now, we can solve for ac :
2
1  Kc 
1  5MPa m 
ac 
 


   design  Y    126MPa 
ac  0.5 mm or b)
2
Practice Problems
2. A specimen with a surface crack size of 2.0
mm fractures at a stress of 80 MPa. Estimate
the fracture toughness, K, assuming that Y=1.
a)
b)
c)
d)
6.3 MPa.sqrt(m)
12 MPa.sqrt(m)
3 MPa.sqrt(m)
0.3 MPa.sqrt(m)
Practice Problems
Answer:
Note: Be careful of units
a  2 103 m
Kc  Y y a  1 80MPa   2 10 m
3
Kc  6.3MPa m or a)
Practice Problems
3.
A structural component is fabricated from an alloy that has a
plane strain fracture toughness of 48.6 MPa-m1/2. It has been
determined that this component fails at a stress of 178 MPa when
the maximum length of a surface crack is 1.08 mm. For a stress of
178 MPa, what is the maximum allowable surface crack length
without fracture for this same component that is made from
another alloy that has a plane strain fracture toughness of 47.5
MPa-m1/2?
a)
b)
c)
d)
e)
0.57 mm
1.03 mm
1.08 mm
2.09 mm
3.12 mm
Practice Problems
Answer:
Using the design stress equation :
K
 c  IC
Y a
Create two equations.Solve for a c2 :
K IC1
ac

1
K IC2
ac
2
1  K IC2    ac1
ac2  
 
K IC1

ac2  1.03 mm or b)
2

 47.5MPa m   1.08103 m 
1
  


  
48.6MPa m



2