Material Strength Griffiths early approach

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• Material Strength Griffiths early approach
• Homogeneous and isotropic solid
• Goal to calculate the strength of the solid and
  compare with experimental results.
• The application of a tensile stress s causes an
  increase in the interatomic distance, which
  itself is accompanied by an increase in the
  interatomic potential energy The application of
  a stress causes an increase in the energy of the
  system.
• Around the equilibrium point (the minimum of the
  potential energy), the stress varies linearly
  with distance ( Hookes behavior).
• However, for larger distances, the stress reaches
  a maximum at the point of inflection of the
  energy-separation curve.

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smax
a0
l/2
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Orowan (1949 Polanyi (1921)

• Theoretical strength is seen to increase if
  interatomic spacing decreases, and if Youngs
  modulus and fracture energy increase. Thus, when
  looking for strong solids, atoms with small ionic
  cores are preferred Beryllium, Boron, Carbon,
  Nitrogen, Oxygen, Aluminum, Silicon etc - the
  strongest materials always contain one of these
  elements.
• Using the O-P expression above, it is found that
  the theoretical strength can be approximated by
  (for most solids)

In practice, the tensile strength is much less
than E/10 because of the omnipresence of defects.
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Materials strength is critically sensitive to
defects

• Example surface cracks
• What is the weakening effect of a surface
  defect on the fiber strength?
• Without any defect, the measured (applied)
  strength s0 would equal the theoretical strength
• Case 1 Circular defect at fiber surface
• For a semi-circular defect, we use the
  analytical solution of Inglis

s0
s0
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INGLIS, 1913
(b)
Note Purely geometric effect!
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If x a (point A), then slocal 3s0 If the
local stress reaches the theoretical strength,
then the applied stress is s0 sth/3 But with
sth E/10, we get s0 E/30 (instead of the
theoretical E/10) A more realistic situation is
that of a sharper crack
A
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• Case 2 Sharper (elliptical) defect at fiber
  surface
• Inglis result in this case is, at point A
• a crack length
• r radius of curvature at A.

s0
A
a
s0
So again, if , and a
1 micron, and r 20 Å, then And thus
s0 E/460 (instead of the theoretical E/10)
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Therefore, defects are indeed a major source of
material weakness

• Defects are the major players for strength
• Removal of surface cracks/defects to improve
  strength (example etching of glass by HF)
• Defects should be made as small as possible, as
  round as possible
• Griffiths experiments and model are the
  historical basis of the fracture mechanics
  approach

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Griffiths theory

• Brittle solids contain defects
• Griffiths original query
• What is the strength of a brittle solid if a
  realistic, sharp crack of length a is present?
• Alternatively At which applied stress will a
  crack of length a start to propagate?
• Remember the prediction

This gave too high a prediction because no defects
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Assumptions Sharp crack in infinite, thin plate
(thickness t) Self-similar crack propagation
Solution A balance must be struck between the
decrease in potential energy and the increase in
surface energy resulting from the presence of a
crack. (The surface energy arises from the fact
that there is a non-equilibrium configuration of
nearest neighbor atoms at any surface in a
solid).
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Basic idea around the crack 2a, a volume
approximately equal to a circular cylinder
carries no stress and thus there is a reduction
in strain energy
s

2a
s
In fact, Griffith showed that the actual strain
energy reduction is
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There exists an energy balance between (1) strain
energy decrease as the crack extends (negative),
and (2) surface energy increase necessary for the
formation of the new crack surfaces
The total energy balance is thus
The following plot shows the roles of the
conflicting energetic contributions
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Plot of total energy UT against crack length
Below the critical crack length, the crack is
stable and does not spontaneously grow. Beyond
the critical crack length (at equilibrium), the
crack propagates spontaneously without limit.
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The condition for spontaneous propagation is
(Griffith, 1921)
This is a (thermodynamics-based) necessary
condition for fracture in solids. Note the math
is similar to nucleation in phase transition Note
that if the Orowan-Polanyi expression
is combined with Inglis expression for an
ellipse
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We obtain
Griffith
Correction factor
The correction factor amounts to about 0.6 if r
a0, and to about 0.8 if r 2a0. This gives
confidence in the result.
In 1930, Obreimoff carried out an experiment on
the cleavage of mica, using a different
experimental configuration. Contrasting with the
Griffith experiment, the equilibrium
configuration used by Obreimoff proves to be
stable
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Griffiths model clearly demonstrates that large
cracks or defects lower the strength of a material

• But then, what is the physical meaning of the
  strength
• of a material?? Is this parameter meaningful at
  all?
• The strength is not a material constant !
• Deeper insight is necessary to quantify fracture
  in a more universal way
• This leads to the field of Linear Elastic
  Fracture Mechanics (or LEFM), developed in the
  1950s by Irwin and others

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FIRST WAY OF REWRITING GRIFFITHS EQUATION
Two variables s and a
Only material properties This is a constant
Fracture criterion Fracture occurs when the
left-hand side product reaches the critical value
given in the right-hand side expression, which is
a new material constant with non-intuitive units
(MPa m1/2)
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The Stress intensity factor (a variable)
The fracture toughness (a constant)
Fracture occurs when K becomes equal to Kc
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SECOND WAY OF REWRITING GRIFFITHS EQUATION
Two variables s and a
Only material property This is a constant
Fracture criterion Fracture occurs when the
left-hand side product reaches the critical value
given in the right-hand side expression, which is
a new material constant with energy units
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The energy release rate Or Crack driving force (a
variable)
The materials resistance to crack extension (a
constant)
Fracture occurs when G becomes equal to Gc
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IMPERFECTIONS IN CRYSTALSIntroduction to
dislocations

• Many of the important physical properties of
  crystalline materials are determined by the
  various types of imperfections and defects
  present in them.
• The imperfections are disruptions in the space
  lattice and are called lattice imperfections
• The imperfections can be characterized
  geometrically according to whether the center of
  disruption is at a point, along a line, or over a
  surface.

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Point imperfections in simple crystalline solid
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Point imperfections in ionic crystalline solid
Schottky
Frenkel
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Line imperfections in crystalline solids

• DEFINITION Dislocations are line imperfections
  in crystals that produce lattice distortions
  centered about a line
• IMPORTANCE In three different aspects of the
  mechanical properties of materials
• Plastic deformations In most cases, plastic
  deformations and dislocation movement are
  mutually inclusive (one does not occur without
  the other)
• Strength Materials will support static loads
  without undergoing permanent deflection only if
  dislocations are prevented from moving
• Fracture Groups of dislocations can provide
  high stress concentration to form a crack

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Definition
Slice along ABCD Displace atoms on top side by
a distance equal to the lattice parameter along a
main axis Stick back together. The distorted
crystal along AB is the dislocation line
But there are other ways too
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EDGE (movement perp to AB)
SCREW (movement parallel to AB)
MIXED
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Positive edge dislocation
Extra-plane
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Screw dislocation
Does not have an extra plane associated with it
the atomic planes form a spiral
May be right-handed or left-handed
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Mixed dislocation
EDGE
SCREW
ATOMS BELOW THE PLANE OF SLIDE
ATOMS ABOVE THE PLANE OF SLIDE
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Dislocation can be a loop
Dislocations move in a slip plane The motion of
the dislocation causes the crystal to change its
shape permanent deformation
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• The lattice is severely distorted in the
  immediate vicinity of the dislocation line (thus,
  there is a stress field) but away from the line
  perfect crystal
• Example In a positive edge dislocation there is
  compression above the slip plane, and tension
  below the slip plane.

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Dislocations can attract or repel each other
REPULSION
ATTRACTION
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In a screw dislocation, there is only a shear
stress field BURGERS VECTOR (b) Specifies the
direction and amount of slip associated with a
dislocation BURGERS CIRCUIT
EDGE
(b perpendicular to dislocation line)
SCREW (b
parallel to dislocation line)
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Observations

• In a perfect crystal (no dislocation), the
  circuit is closed
• If a dislocation is present, the circuit is open
  and a nonzero Burgers vector exists
• HOWEVER sometimes the Burgers circuit is open
  BUT it does not reveal the presence of
  dislocations. Example

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Density of dislocations
Length of dislocation lines per unit volume
cm/cm3 EXAMPLE Silicon in solid-state devices
disloc density 1 10 cm/cm3 EXAMPLE
High-strength steel disloc density 1012 cm/cm3
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WHY ARE DISLOCATIONS NEEDED AT ALL?

• Early computation of theoretical strengths of
  perfect crystals led to values that are many
  times higher than those measured
• Example The theoretical shear strength of a
  crystal is G/2p, where G shear modulus. This
  is far above the experimental values (similar as
  for tensile strength).

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• Discrepancies between theory and observations
  could only be explained by the presence of linear
  crystalline defects
• However direct observations in the 1950s only

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• The original concept of such linear defects was
  introduced independently- in the 1930s by
  Taylor, Orowan, and Polanyi.
• The slip/breakage of cubic lattice occurs
  through a consecutive rather than simultaneous
  movement.
• The result of both movements is the same BUT
  much less energy is involved in consecutive slip.
• Simple-minded analogies

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Caterpillar movement
Worm movement
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energy-saving devices
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IMPORTANT Slip is favored on close-packed planes
because less force is required to move the atoms
from one position to the next closest one. Each
crystal structure (FCC, BCC, etc) has its
preferred slip planes and directions !
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ENERGY OF A DISLOCATION

• Dislocations have a stress/strain field
  surrounding the line because of the severe
  distorsion
• There is elastic energy stored in the field
• Screw dislocation as an example we calculate the
  strain energy stored in an annular ring of radius
  r, thickness dr, unit length, centered on the
  dislocation

b total elastic displacement
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Shear strain (elastic displacement)/(distance
over which it occurs) b/2pr Thus, the strain
energy per unit volume is (1/2)(stress)(strain)
(1/2)G(strain)2 (1/2)G(b/2pr)2
Thus Elastic strain energy per unit length of
dislocation is
(r gt r0)
Which leads to
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(Below r0, the strains are inelastic
dislocation core region) THUS The dislocation
energy is proportional to the square of the
Burgers vector We must expect dislocations to
have the smallest Burgers vector possible (energy
minimization) EXAMPLE Assume a dislocation with
a Burgers vector 2b. Its energy is G 4Gb2.
On the other hand, two individual dislocations
with a Burgers vector b have a total energy G
Gb2 Gb2 2Gb2. CONCLUSION Such a
dislocation can reduce its energy by splitting
into two individual dislocations.
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DISLOCATION SOURCES

• Looking at the previous Zn crystal photograph,
  the magnitude of slip steps indicate that
  thousands of dislocations have passed through the
  crystal
• But if we examine the crystal prior to
  deformation, we find only a few dislocations
• Dislocations are therefore created within the
  crystal, already under low stresses. How?
• Take a cube-shaped crystal with an edge
  dislocation, apply a shear stress t. The
  dislocation moves a distance x, the displacement
  is (x/L)b, where L is the cube edge.

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The force on the top surface due to stress t is
tL2, and the work done is tL2(xb/L) txbL The
force acting on the dislocation is
d(energy)/d(position) tbL Therefore, the force
per unit length is simply tb, and it acts
perpendicular to the dislocation line (or
parallel to the direction of motion. Now,
remember the dislocation loop
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• Apply a shear force to the top and bottom of the
  cube two forces are conflicting
• Strain energy of dislocation line decreases if
  length decreases, which makes the loop want to
  collapse
• Applied force tends to expand the loop

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If these forces are equated, we get
collapse
expansion
Thus
In other words, the shear force necessary to
expand the loop is proportional to 1/r
This has a direct application the mechanism of
generation of dislocations in a crystal
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No stress. A and B are pinning points for the
dislocation line
l
A
B
tbL
Under stress t. The dislocation line bows out of
the pinning points
q
A
B
T
T
r l/2, the line behaves as a loop with rl/2
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If tgt(2Gb/l), the segment bows out, and we are
left with 2 portions of dislocations with the
same Burgers vector of opposite signs They will
anihilate each other! What then remains is a
loop and a central dislocation segment
Frank-Read source of dislocations! If the stress
level above is maintained, the emission of
dislocation loops continues
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