Dislocation and strengthening mechanisms

1. DISLOCATION AND
STRENGTHENING MECHANISMS
PRESENTED BY: YOMNA M. IBRAHIM

2. OUTLINE
1- Dislocations
• Edge dislocation
• Screw dislocation
• Mixed dislocation
2- Slip
3- Twinning
4- Strengthening mechanisms
• Decreasing grain size
• Solid solution strengthening
• Strain hardening
• Precipitation strengthening

3. DID YOU EVER THINK WHY?

4. WHAT IS A DISLOCATION?
• Dislocations are line defects or
imperfections in crystals (areas that are
out of position).
• They are introduced during solidification
(cooling) or deformation (stress).
• It occurs in crystalline structures (metals,
ceramics and some polymers) but do not
occur in amorphous structures.

5. TYPES OF
DISLOCATION
1- Edge dislocation
2- Screw dislocation
3- Mixed dislocation

6. 1- EDGE
DISLOCATIONS
• The edge defect can be easily
visualized as an extra half-
plane of atoms in a lattice.
• The interatomic bonds are
significantly distorted only in
the immediate vicinity of the
dislocation line.

7. • The lattice is cut with an extra half
plane of atoms. The bottom edge
of this inserted plane represents
the edge dislocation which is
represented by an inverted T.

8. BURGER VECTOR (b)
• If we draw a clockwise loop
around the edge dislocation,
starting at point (x) and traveling
an equal number of atom spacings
in each direction, we finish at
point (y) one atom spacing from
the starting point.
• If an edge dislocation was not
present, the loop would close.
• The vector required to complete the
loop is Burger’s vector.
X
Y

9. IN THIS CASE, THE BURGERS VECTOR IS
PERPENDICULAR TO THE DISLOCATION LINE.

10. 2- SCREW DISLOCATIONS
• The screw dislocation can be
illustrated by cutting partway through
a perfect crystal and then skewing the
crystal by one atom spacing (shear
stress).

11. BURGER VECTOR (b)
• If we follow a crystallographic plane one
revolution around the axis on which the
crystal was skewed, starting at point (x) and
traveling equal atom spacings in each
direction, we finish at point (y) one atom
spacing below our starting point.
• If a screw dislocation was not present, the
loop would close. The vector required to
complete the loop is the Burger’s vector.
• If we continued our rotation, we would trace
out a spiral path. The axis, or line around
which we trace out this path, is the screw
dislocation line.
• In this case, the Burgers vector is parallel to
the dislocation line.
X
Y

12. A screw dislocation is named after screw spirals as the starting
point is a unit cell higher than the ending point

13. 3- MIXED DISLOCATIONS
• Mixed dislocations have both
edge and screw components,
with a transition region
between them. The Burger’s
vector, however, remains the
same for all portions of the
mixed dislocation.

14. DISLOCATION MOTION (SLIP)
• When a sufficiently large stress is applied to a
crystal containing a dislocation, the dislocation
can move through a process known as slip.
• A plane at which the slip occurs is known as a
slip plane.
• A direction at which the slip occurs is known as
a slip direction.
• Slip system= slip planes and slip directions
• This motion explains the plastic deformation.

15. 1- SLIP OF EDGE DISLOCATION
• The slip plane is parallel to
burger’s vector and
perpendicular to the
dislocation line.

16. • The bonds across the slip plane between the atoms in the column to the right of the dislocation are
broken.
• The atoms in the column to the right of the dislocation below the slip plane are shifted slightly so
that they establish bonds with the atoms of the edge dislocation. In this way, the dislocation has
shifted to the right.
• If this process continues, the dislocation moves through the crystal until it produces a step on the
exterior of the crystal in the slip direction.

17. • Dislocation motion is analogous to movement of a caterpillar. The caterpillar would have to exert a
large force to move its entire body at once. Instead it moves the rear portion of its body forward a
small amount and creates a hump. The hump then moves forward and eventually moves all the body
forward by a small amount.
• Movement in this manner requires a much smaller force than breaking all the bonds across the
middle plane simultaneously as dislocations allow deformation to occur at much lower stress than in
a perfect crystal. The dislocation moves in a slip system that requires the least expenditure of energy.

18. • By introducing the dislocation,
the atoms above the dislocation
line are squeezed too closely
together (compression), while
the atoms below the dislocation
are stretched too far apart
(tension). The surrounding
region of the crystal has been
disturbed by the presence of the
dislocation (shear).

19. LATTICE STRAIN
ANNIHILATIONS
• Areas of compression repel each
other as well as areas of tension.
• Compression and tension would
attract each other and annihilate
the defects forming a perfect
orientation again.

20. The slip plane is parallel to
Burger’s vector and parallel
to the dislocation line.
2- SLIP OF SCREW DISLOCATION

21. 3- SLIP OF MIXED DISLOCATION

22. • Also, the net plastic deformation of both edge and screw dislocations is the
same.

23. • When thousands of dislocations move on the same plane, they
produce a large step at the crystal surface forming a slip line (a).
• A group of slip lines is known as a slip band (b).

24. • Dislocations do not move easily in materials such as silicon, which have covalent
bonds because of the strength and directionality of the bonds as well as the low
packing of covalently bonded materials.
• Materials with ionic bonding, including many ceramics such as MgO, are brittle
and resistant to slip. Movement of a dislocation disrupts the charge balance
around the anions and cations so the ionic bond must be broken, and the ion must
move past an area with a repulsive charge in order to get to the next location of
the same charge.

25. SIGNIFICANCE OF SLIP
PHENOMENON
• 1- Dislocation and slip explain plastic
deformation.
• 2- Ductility of metals is controlled by
dislocations. The easier the slip, the more
ductile the material (Aluminum).
• 3- A material can be strengthened by
introducing an obstacle into the crystal to
prevent the dislocation from slipping unless we
apply higher forces (Stainless steel).
• 4- The electrical resistivity of a material
increases with increasing slip and dislocation
(dislocation free silicon and copper).

26. • Dislocation motion is easier with close-
packed planes (greatest planar density or
planar packing) according to the crystal
structure (allow shearing and sliding with
the least energy and least shear stress).
SLIP IN RELATION TO CRYSTAL STRUCTURE

27. PLANAR PACKING FACTOR
• PPF (planar density)=
Area of atoms in a plane
Area of the plane
PPF= 0.785 PPF= 0.555 PPF= 0.906

28. Families are non-parallel directions or planes that overlap when the
crystal is rotated 90 degrees around x, y or z axes.
Families of directions <100>
Families of planes {111}

29. 1- FCC METALS
• FCC metals contain four non-parallel close-
packed planes of the form {111} and three
close-packed directions of the form <110>
within each plane, giving a total of twelve slip
systems.

30. • BCC metals have as many as 48 slip systems
that are nearly close-packed. Several slip
systems are always properly oriented for slip
to occur (12) while the others require high
forces to be activated.
2- BCC METALS

31. 3- HCP METALS
• HCP metals have only 1 primary slip plane which is
the hexagonal base of the unit cell. There are 3 slip
directions for this slip plane, so that the hcp-lattice
has a total of 3 slip systems. Additional slip planes
can be activated by greater force or high
temperatures.

32. Slip systems for FCC, BCC
and HCP

33. • FCC has high ductility and low strength (wrought metals)
• BCC and HCP have low ductility and high strength (cast metals)

34. • Due to the same number of slip systems,
one could prematurely conclude that both
FCC and BCC lattices are equally
deformable. In practice, however, it is
found that the FCC lattice shows a
significantly higher ductility.
• The close packing is more efficient in the
FCC crystals than BCC ones.
• In addition to the main criterion of the
number of slip systems (quantity), the
quality of the slip planes also plays a role
in ductility.

35. SCHMID’S LAW

36. • Only forces in slip direction are relevant for shifting of lattice
planes.
• In order for the dislocation to move in its slip system, a shear
force acting in the slip direction must be produced by the
applied force.
• The resolved shear force is:

37. • Area of the slip plane (A) is given as:
• The shear stress (τr) acting on the plane can be calculated by dividing
resolved shear force (Fr) by the area of the slip plane (A)
• Schmid’s law is then deduced to be:
• Schmid factor (m):

38. CRITICAL RESOLVED SHEAR STRESS (τcrss)
•τcrss represents the minimum resolved shear stress required to initiate slip.
• A high τcrss means that high shear stress is needed in order to plastically
deform the material and that this material has high yield strength (if τcrss
is not exceeded, elastic deformation occurs).

39. • Crystals have many available slip
systems each with its own τcrss.
• As the tensile load increases, the τcrss
is reached first on one of these systems
which is then called primary slip
system.
• As the load is increased further, τcrss
may be reached on other slip systems;
these then begin to operate.
• From Schmid's Law, the primary slip
system is the system with the greatest
Schmid factor.

40. • When the slip system is aligned
perpendicular to the pulling axis,
cos(Φ) =1 for Φ = 0 and in this
case λ = 90° and thus cos(λ) =
0.
• The Schmid factor is thus zero
and no shear stress acts (no slip
occurs).

41. • The critical resolved shear stress is
reached first when the slip system is
inclined such that Φ = 45° and λ = 45°.
• The Schmid factor reaches 0.5 in this
ideal case. This corresponds to the
maximum possible value of Schmid’s
orientation factor:

42. •The critical resolved shear stresses are thus ideally half
the size of the externally applied normal stresses
(observed yield stress).

43. • FCC metals have low critical resolved shear stress (50 to 100 psi) in a
perfect crystal (high ductility).
• BCC crystal structures have high critical resolved shear stress (10,000 psi)
in perfect crystals (low ductility).

44. • HCP metals such as zinc that have a c:a ratio
greater than or equal to 1.633 have a critical
resolved shear stress of less than 100 psi, just as
in FCC metals.
• In HCP titanium, however, the c:a ratio is less
than 1.633; the close-packed planes are spaced
too closely together. Slip now occurs on planes
such as the prism planes or faces of the
hexagon. These planes are activated by proper
alloying or increased temperatures giving more
ductility.

45. TWINNING
• A twin boundary is a plane
across which there is a special
mirror image misorientation of
the crystal structure.
• Twins can be produced when a
shear force, acting along the
twin boundary, causes the
atoms to shift out of position.

46. Slipping Twinning
Takes place at low strain rates Takes place at high strain rates
Atoms move in one plane only
(slip plane)
Atoms move in all planes of the
twin region
Orientation across the slip plane
is the same
Orientation across the twin
plane is different (reoriented)
Atomic movements are equal to
atomic distances
Atomic movements are lesser
than atomic distances
Results in large deformations
Results in small deformations
and increases the strength
Forms a series of steps
Forms small well-defined
deformed regions
Occurs in FCC and BCC due to
large number of slip systems
Occurs more in HCP due to low
number of slip systems

47. • The real importance of twinning lies with the accompanying crystallographic
reorientations; twinning may place new slip systems in orientations that are
favorable relative to the stress axis such that the slip process can now take
place.
• Twinning occurs during deformation or heat treatment of certain metals. The
twin boundaries interfere with the slip process and increase the strength of
the metal.

48. • The movement of a large numbers of
dislocations causes permanent
deformation
• Any mechanism that impedes dislocation
motion (slip) makes a metal stronger and
harder

49. 1- DECREASING GRAIN SIZE
• By reducing the grain size, we increase the number of
grains and increase the amount of grain boundary area.
• Any dislocation moves only a short distance before
encountering a grain boundary where it is blocked or
piles up, and the strength of the metallic material is
increased.
• 1-The dislocation can not change its orientation from one
crystal to another.
• 2-The slip planes will be discontinued due to the
disordered arrangement at grain boundaries.

50. • The Hall-Petch equation relates the grain size to the yield strength
where (y) is the yield strength, (d) is the average diameter of the grains, and
(σ0) and (K) are constants for the metal
• The Hall-Petch equation is not valid for materials with unusually large or
ultrafine grains.

51. • Small-angle grain boundaries
(which occur due to
superimposed dislocations) are
not effective in interfering with
the slip process because of the
slight crystallographic
misalignment across the
boundary.
• On the other hand, twin
boundaries will effectively
block slip and increase the
strength of the material.

52. • The grain boundaries limit the slip of the grains and the final shape
assumed by them after deformation.
• Polycrystalline materials are less ductile than single crystals due to the
presence of grain boundaries (similar to alloys).

53. • When we introduce interstitial or
substitutional atoms (point defects) of a
guest element to another element, we
create a solid solution.
• Point defects disrupt the perfection of
the crystal structure (strain fields). If a
dislocation moves to the right , it
encounters a disturbed crystal caused
by the point defect; higher stresses are
needed to continue slip of the
dislocation.
2- SOLID SOLUTION STRENGTHENING

54. • An impurity atom that is smaller than a host atom for which it substitutes exerts
tensile strains on the surrounding crystal lattice while a larger substitutional atom
imposes compressive strains on neighboring atoms.
• These impurity atoms diffuse and segregate around dislocations in order to
reduce the overall strain energy (cancel some of the strain in the lattice
surrounding a dislocation) by annihilating the dislocation.
• A smaller impurity atom is located where its tensile strain will partially nullify
some of the dislocation’s compressive strain.

55. • Pure gold is an FCC metal with many active slip systems making it weaker
and softer than alloys.
• Increasing the concentration of the impurity element results in an increase in
tensile and yield strengths due to resistance to slip. This mechanism explains
why carbon steel is stronger than pure iron.

56. 3- STRAIN HARDENING (WORK HARDENING, COLD
WORKING)
• A ductile metal can become harder and stronger as it is plastically deformed.
• The deformation takes place at temperatures below the absolute melting
temperature (cold working). Most metals strain harden at room temperature.
• It is sometimes convenient to express the degree of plastic deformation as
percent cold work rather than as strain. Percent cold work (%CW) is defined as
where A0 is the original area of the cross section that experiences deformation and
Ad is the area after deformation.

57. • The dislocation density in a metal increases
with deformation or cold work, because of
dislocation multiplication or the formation of
new dislocations.
• Consequently, the average distance of
separation between dislocations decreases
and the dislocations are positioned closer
together (entanglement).
• Mostly, dislocation strain interactions are
repulsive. The motion of a dislocation is
hindered by the presence of other
dislocations.
• So the stress required to cause dislocation
motion increases.

58. • Strain hardening is demonstrated in a stress–strain diagram. The metal with
yield strength (y0) is plastically deformed to point D. The stress is released,
then reapplied with a resultant new yield strength (yi). The metal has thus
become stronger during the process because (yi) is greater than (y0).
• The parameter (n) is called the strain-hardening exponent, which is a measure
of the ability of a metal to strain harden; the larger its magnitude, the
greater the strain hardening for a given amount of plastic strain.

59. Forging
Rolling
Extrusion Drawing

60. • Steel, brass and copper increase in yield and tensile strength with
increasing cold work. The price for this enhancement of hardness and
strength is in the ductility of the metal. The ductility, in percent elongation,
experiences a reduction with increasing percent cold work for the same
three alloys.

61. • Some fraction of the energy expended in
deformation is stored in the metal as strain
energy around the newly created dislocations.
• Dislocation densities and the strain energy
around them can be reduced substantially by
heating a metallic material to a relatively high
temperature (below the melting temperature)
and holding it there for a long period of time.
This heat treatment is known as annealing and
is used to impart ductility to metallic materials
(opposite to strain hardening).
• Annealing is composed of recovery,
recrystallisation and grain growth.

62. ANNEALING STAGES
• 1- Recovery
During recovery, some of the stored internal strain energy is relieved due to
dislocation motion as a result of enhanced atomic diffusion at the elevated
temperature.
2- Recrystallization
Recrystallization is the formation of a new set of strain-free and equally
dimensioned grains that have low dislocation densities and are
characteristic of the pre cold-worked condition (soft, weak, ductile metals).
The new grains form very small nuclei and grow until they completely
consume the parent material.

63. a) Cold-worked brass
b) Nuclei formation
(580ºC)
c) Partial replacement
(580ºC)
d) Complete
recrystallization (580ºC)

64. • Recrystallization is a process which depends
on both time and temperature.
• The degree of recrystallization increases
with time.
• The recrystallization behavior of a metal or
alloy is sometimes specified in terms of
recrystallization temperature which is the
temperature at which recrystallization just
reaches completion in 1h (Brass = 580ºC).
This temperature is typically third or half the
melting temperature.

65. 3- Grain growth
• The strain-free grains will continue to grow if
the metal specimen is left at the elevated
temperature.
• As grains increase in size, the total boundary
area decreases. Thus, the average grain size
increases with time, and at any particular
instant there will exist a range of grain sizes.
• Grain growth proceeds more rapidly as
temperature increases due to the enhancement
of diffusion rate with rising temperature.

66. • The mechanical properties at room temperature of a
fine-grained metal are superior to those of coarse-
grained ones. If the grain structure of a single-phase
alloy is coarser than that desired, refinement may be
accomplished by plastically deforming the material,
then subjecting it to a recrystallization heat treatment.
• It should also be noted that the strengthening effects
due to grain size reduction and strain hardening can be
eliminated or at least reduced by an elevated
temperature heat treatment.
• Conversely, solid-solution strengthening is unaffected
by heat treatment.

67. • Precipitation hardening or age hardening
produces uniformly dispersed particles within a
metal's grain structure that help hinder
dislocation motion.
• This technique is used with aluminum,
magnesium and nickel strengthened by copper.
4- PRECIPITATION STRENGTHENING

68. • Solution Treatment: the metal is
heated to a high temperature and
treated with the solution of the other
metal.
• Quenching: the solution is cooled
down suddenly
• Aging: the metal is heated to a
medium temperature and then
quickly cooled.
• Aging determines the size and
distribution of the particles.

69. • The degree of strengthening from second phase particles depends on
particle distribution and particle size in the matrix.
• Too small or too large particles will allow dislocations to pass through at
low stresses due to large interspacing between them.

70. TANTALUM
• Ductile but strong
• Fast deformation causes
deformation twinning
and twisting of crystals
to stop dislocation
motion.
• Slow deformation
causes slip and ductility.

71. REFERENCES
1- Material science and engineering, 8th edition,
2010, Wiley
2- The science and engineering of materials, 6th
edition, 2010 Cengage Learning Inc.
3- Material science for dentistry, 9th edition,
2009, Woodhead publishing limited.
4- Materials Science and Engineering, University
of Texas, YouTube channel
5- Metallurgy data, YouTube channel
6- Materials behavior, GIT, Coursera online
course
7- Solving mysteries of metal hardening, Allen
Chen, S &TR, September 2018
8- https://www.tec-science.com/material-
science/ductility-of-metals/influence-of-the-
lattice-structure-on-the-ductility/