1. SUMMARY DUCTILE VS BRITTLE FRACTURE:
DUCTILE
BRITTLE
Fracture surface? Dimpled Flat
Deformation? Lots of plastic Not
appreciable
Crack propagation? Slow Rapid
Type of Failure? Gradual
Catastrophic
2. As observed during previous lectures, materials possess low fracture strengths
relative to their theoretical capacity because:
1. Most materials deform plastically at much lower stress levels & eventually fail
by accumulation of this microdamage.
2. Materials contain microstructural and/or manufacturing defects.
If we therefore assume that materials fail as a consequence of pre-existent
defects, then the problem reduces to one of statistics:
• How many defects are present in the component, how big are they and where
are they located with respect to the highly stressed regions of the component?
• What is the shape and orientation of these defects with respect to the loading
direction?
• What is the volume of the component – larger components more likely to
contain more and larger defects than small ones as established by Leonardo da
Vinci about 500 years ago.
FRACTURE MECHANICS
Introduction
3. a
b
c
d
Discuss how the through
thickness defects a, b, c and d
will contribute to the failure of
the component under
uniformly distributed load P
P
P
Component
with defects
It will be shown later that only
defect ‘b’ will lead to failure when
b reaches a critical value –
assuming no interaction.
Since defect a<b, failure will not occur due to defect ‘a’ .
Failure will also not occur due to defects c & d since they are parallel
to the direction of stress. This is so even though defect d is larger
than defect b!!
4. The application of materials in engineering design has posed difficult
problems throughout history.
This has unfortunately been manifested as numerous accidents with
significant loss of lives and money.
Some of these accidents have been due to poor design but it has
gradually been discovered that material deficiencies in form of pre-
existing flaws can initiate cracks leading to fracture.
Introduction of all-welded structures brought it with accident prone
design structures such as the Liberty ships built during WW II.
Several of them broke into two while an even larger number
experienced serious failures. Many bridges and other structures
have encountered similar fate!
5. Extensive investigations were initiated in many countries such the
US.
The work revealed that flaws and stress concentrations & to a certain
extent internal stresses were responsible for the brittle failures.
It turned out that the brittle fracture of steel was promoted by low
temperatures and by conditions of the triaxial stress that exist at a
flaw.
After WW II, use of high strength materials has increased
tremendously and continues to do so due to weight saving
considerations.
Stress analysis methods have also advanced enabling reliable
determination of local stresses such that low margins of safety can
be comfortably adopted.
6. Unfortunately, with the low margins of safety, structures designed in
high strength materials can experience service stresses that may be
enough to induce cracks.
This is particularly so in aggressive environments and if pre-existing
flaws or stress concentrations (e.g., at notches) are present.
Appreciation of this problem is what led to the development of
LINEAR ELASTIC FRACTURE MECHANICS (ELFM).
Now, many structures are designed to carry service loads that are
high enough to initiate cracks, particularly when pre-existing flaws or
stress concentrations are present.
The designer has to anticipate this possibility of cracking and
therefore accept a certain risk that the structure will fail but should be
able predict how fast the cracks will grow and how fast the residual
strength will decrease. (See Figure in next slide)
7. The object of LEFM is to develop methods to make these predictions.
LEFM should answer:
• What is the residual strength as a function of crack size?
• What crack size can be tolerated at expected service load?
• How long will it take the crack to grow from initial size to this critical size?
• What size of pre-existing flaw can be permitted when the structure is put in service. Use of
NDT!
• How often should the structure be inspected? Use of NDT!
Crack size
Time
Residual strength
Failure may
occur
Design strength
Expected highest service load
Normal service load
Failure
Load cycles
Time
Crack size, a
ai
8. Inglis analyzed a plate containing an elliptical hole & showed that the
applied stress a was magnified at the ends of the major axis of the
ellipse:
a
a
2b
2a
;
2
1
max
b
a
a
max = Maximum stress at the
end of the major axis.
a = Applied stress normal
to the major axis.
a = Half major axis.
b = Half minor axis.
Where,
a
b
But
2
,
The term
;
2
1
max
a
a
a
a
2
max since a>>
a
2 is defined as the stress concentration factor, kt
9. Kt describes the effect of crack geometry on the local crack-tip stress
level.
Many textbooks and handbooks describe stress concentrations in
components with a wide range of crack configurations.
It is clear that Kt increases with increasing crack length and
decreasing crack-tip radius.
Cracks should be kept as small as possible if present. Periodic
inspection should be carried out so as to replace components once
the cracks therein reach dangerous levels.
The relative severity of the stress concentration can also reduced by
drilling a hole through the crack-tip. This should be done with caution,
the crack is still present and can re-sharpen by re-initiating and then
continue growing!! Continued inspection is therefore important.
10. Kt may also be defined for component configuration changes, e.g.,
section size changes.
Kt increases for several design configurations whenever there is a
large change in cross-sectional area and/or where the associated fillet
radius is small. (See below & next slide)
12. DEFINITION OF LEFM
• Why do we get stress concentrations at
notches ?
• If we consider lines of force: the
presence of a crack leads to locally
raised stresses, which diminish with
distance in an inversely proportional
manner.
• It is the local stresses at the crack tip
which cause failure, these are difficult
to calculate.
• But Griffith developed a global energy
balance approach to describe failure in
brittle materials which can be used to
calculate the fracture stress, f.
app
A B
app
distance, r
app
x
14. Two opposing energetic considerations:
Griffiths used the stress analysis by
Inglis to compute the stored strain
energy (stored potential energy), U,
around the crack as:
The surface energy, Us , of the new
surfaces formed after cracking is given
by,
Semi-infinite plate of
unit thickness
Surface energy
per unit area
Stored strain energy
around crack
Surface energy
of new crack surfaces
2a
sapp
sapp
E
a
U
app
2
2
s
s
s a
a
U
4
)
)(
2
)(
2
(
A negative sign is
used because growth
of the crack releases
elastic strain energy
15. The total change in potential energy resulting from the creation of the
crack is:
According to Griffith, the crack will propagate under constant stress
pp (i.e., f) if an incremental increase in crack length produces no
change in the total energy of the system. i.e., the increased surface
energy is compensated by a decrease in elastic strain energy.
s
app
s a
E
a
U
U
U
4
2
2
a
E
a
E
E
a
E
a s
f
s
f
f
app
s
2
;
4
2
;
2
2
4 2
2
2
Accurate for brittle materials (e.g. glass). Thus, correction required for metals to
allow for plastic deformation at crack tip
s
app
a
E
a
da
d
da
U
d
4
0
2
2
16. E
a
f
s
2
2
E = Young’s modulus; s = surface energy per unit volume
Also called the strain
energy release rate, ‘G’
(drives crack growth)
The surface energy
increase with crack
growth (resists crack
growth)
Can also be rearranged as:
.
2 const
material
E
a s
f
It is clear that as the crack, a, increases, the stress, f, required for crack
extension decreases (i.e., for failure to occur). Note that this only applies to purely
brittle materials!
20. app
app
Thus, failure is associated with the
concentration of stress in the vicinity
of a flaw. And fracture mechanics
describes the theory used to calculate
these crack tip stresses and strains.
For the special case shown of RHS,
where K = the stress intensity factor:
K = sapp√(pa)
i.e. can be calculated from applied
stress and crack length and describes
the magnitude of the local crack tip
stress field
slocal
r, distance
ahead of
crack tip
local
K
2r
infinite
plate
assumption
21. Fracture toughness
As applied stress increases K (and
therefore the local crack tip stress) will
also increase until the local stress
required for fracture is reached.
We define this critical stress intensity
factor value as: KC the fracture
toughness value.
slocal
r, distance
ahead of
crack tip
Increasing
applied stress
KC = sf√(pac)
where sf = the stress applied at fracture ac = critical crack length
More generally: KC = Qsf√(pac) where Q (shape factor) allows for
various types of crack and specimen geometries)
22. How does the Griffith energy balance approach relate to K?
Consider fracture toughness:
Griffith approach says:
Hence KC is a materials constant, as long as LINEAR ELASTIC
ASSUMPTIONS HOLD, i.e. material behaviour must be well
approximated by elastic continuum assumptions.
f a 2E KC
f
2E
a
KC f a
materials constants
29. Thickness
Fracture
toughness
KIc
plane strain
plane stress
mixed mode
USES OF FRACTURE MECHANICS
K takes into account the crack
shape and component geometry
and allows an assessment of the
local crack tip stresses. It can also
take into account the applied
loading (e.g. opening, shearing,
twisting)
KIC is the plane strain fracture
toughness and is found to be a
materials property - i.e. linear
elastic assumptions hold
30. USES OF FRACTURE MECHANICS
Stress intensity analysis can help in a number of safety and lifing
assessments.
EXAMPLE:
A nuclear pressure vessel has a KIC of 40MPa√m and you know
that the manufacturing process has left quench cracks of ~3mm.
What stresses will it be safe to allow the pressure vessel to
experience??
KIC = sf√(pac) =>
PLUS SAFETY FACTOR!!
f
KIC
a
40 106
3 103
4.12 108
Pa
