KTU S3 ME - 2019 SCHEME - METALLURGY & MATERIAL SCIENCE(MET 205 )
MODULE 1 - IMPORTANT TOPCS
Areas Discussed:
Crystallography: - SC, BCC, FCC, HCP structures, APF - theoretical density simple problems - Miller
Indices: - crystal plane and direction - Modes of plastic deformation: - Slip and twinning -Schmid's
law - Crystallization: Effects of grain size, Hall - Petch theory, simple problems.
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MET 205 METALLURGY & MATERIAL SCIENCE
1. (MET 205)
METALLURGY &
MATERIAL SCIENCE
MODULE I
Mr. Joe Jeba Rajan K
Assistant Professor
Department of Mechanical Engineering
Rajadhani Institute of Engineering & Technology, Nagaroor
3. A lattice is a regular repeated three-dimensional arrangement of atoms, ions, or molecules in a metal or other
crystalline solid.
LATTICE
JOE JEBA RAJAN, AST PROF - MECH, RIET
4. The smallest group of atoms or the smallest unit of lattice which has the overall symmetry of a crystal, and
from which the entire lattice can be built up by repetition in three dimensions
UNIT CELL
JOE JEBA RAJAN, AST PROF - MECH, RIET
5. Coordination number is the number of atoms or ions immediately surrounding a central atom in a crystal
structure.
COORDINATION NUMBER
Atomic packing factor (APF), packing efficiency or packing fraction is the fraction of volume in a crystal
structure that is occupied by constituent atoms. It is dimensionless and always less than unity.
ATOMIC PACKING FACTOR
𝐀𝐏𝐅 =
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐚𝐭𝐨𝐦𝐬 𝐢𝐧 𝐮𝐧𝐢𝐭 𝐜𝐞𝐥𝐥
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐮𝐧𝐢𝐭 𝐜𝐞𝐥𝐥
JOE JEBA RAJAN, AST PROF - MECH, RIET
6. 1. Simple Cubic (SC) structure
2. Body Centre Cubic (BCC) structure
3. Face Centre Cubic (FCC) structure
4. Hexagonal Closed pack (HCP) structure
IMPORTANT CRYSTALLINE UNIT CELLS
1. Simple Cubic (SC) structure
a) Effective Number of Atoms
b) Coordination Number
c) Atomic Packing Factor
JOE JEBA RAJAN, AST PROF - MECH, RIET
7. 1. Simple Cubic (SC) structure
a) Effective Number of Atoms = (1/8)*8 =1
b) Coordination Number = 6
c) Atomic Packing Factor
𝐀𝐏𝐅 =
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐚𝐭𝐨𝐦𝐬 𝐢𝐧 𝐮𝐧𝐢𝐭 𝐜𝐞𝐥𝐥
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐮𝐧𝐢𝐭 𝐜𝐞𝐥𝐥
APF for a simple cubic structure = 0.52
JOE JEBA RAJAN, AST PROF - MECH, RIET
8. 2. Body Centre Cubic (BCC) structure
a) Effective Number of Atoms = (1/8)*8+1 =1+1 = 2
b) Coordination Number =8
c) Atomic Packing Factor
APF for a body-cantered cubic structure = 0.68
Close-packed directions: length = 4R =√3 a
JOE JEBA RAJAN, AST PROF - MECH, RIET
9. 3. Face Centered Cubic Structure (FCC)
a) Effective Number of Atoms = (1/8)*8+(1/2)*6 =1+3 = 4 atoms/unit
cell
b) Coordination Number =12
c) Atomic Packing Factor
Close-packed directions: length = 4R =√2a
APF for a face-centered cubic structure = 0.74
JOE JEBA RAJAN, AST PROF - MECH, RIET
10. 4. Hexagonal Closed pack (HCP) structure
a) Effective Number of Atoms = (12 corner atoms)*1/6 + 2(Face center atoms)*1/2+3(Mid plane
atoms)*1
= 6 atoms/unit cell
a) Coordination Number =12
b) Atomic Packing Factor
JOE JEBA RAJAN, AST PROF - MECH, RIET
12. 4. Hexagonal Closed pack (HCP) structure
a) Effective Number of Atoms =6 atoms/unit cell
b) Coordination Number =12
c) Atomic Packing Factor
JOE JEBA RAJAN, AST PROF - MECH, RIET
13. 4. Hexagonal Closed pack (HCP) structure
Atomic Packing Factor
a
JOE JEBA RAJAN, AST PROF - MECH, RIET
14. 4. Hexagonal Closed pack (HCP) structure
Atomic Packing Factor
a) Effective Number of Atoms =6 atoms/unit cell
b) Coordination Number =12
JOE JEBA RAJAN, AST PROF - MECH, RIET
15. Crystal Structure SC BCC FCC HCP
Edge Length a = 2R a = 4R/√3 a = 2R√2
a = 2R,
c = 1.633a
c/a Ratio 1 1 1 1.633
Atomic Packing Factor
(APF)
52% 68% 74% 74%
Coordination Number
(CN)
6 8 12 12
Comparison of SC, BCC, FCC, and HCP
Crystal Structures
JOE JEBA RAJAN, AST PROF - MECH, RIET
16. CLOSE PACKING
Close-packing of atoms means the dense arrangement of atoms in an infinite, regular arrangement or
lattice.
The most efficient way to fill space with spheres
• First layer is considered to be of A type,
• Subsequent layers is also of A type
highest packing density - greatest fraction of space occupied - by a close packing is 0.74
JOE JEBA RAJAN, AST PROF - MECH, RIET
17. • In HCP structure a second layer of spheres is placed in the indentations left by the first layer.
• Space is trapped between the layers that is not filled by the spheres is called voids
The voids in close packing can be called as holes or interstitial sites.
Holes or Interstitial Sites.
JOE JEBA RAJAN, AST PROF - MECH, RIET
18. Copper has FCC crystal structure. Determine its density if the atomic weight is 63.54 and atomic radius is 1.287A
Theoretical Density
A - Atomic Weight
n - No. of atoms in unit cell
NA - Avogadro's Number
a - Lattice parameter or Edge Length
EX:1
To Find
a
FCC
JOE JEBA RAJAN, AST PROF - MECH, RIET
19. Co ordination Number and Radius Ratio
• Ceramic materials are characterized by the presence of more than one type of atomic bond in a
single material.
• These compounds are formed predominantly by ionic bonds
• metallic ions or cations are positively charged
• non-metallic ions or anions are negatively charged
• ratio rc / ra is known as radius ratio
• Each cation likes to have as many nearest anion neighbours as possible.
• The anions also prefer to have maximum number of cations as nearest neighbours.
• At the same time, the anions around a cation need to be in contact with the cation, so that a stable crystal
structure is obtained.
• The ionic radius increases as the co-ordination number increases.
• When an electron is removed from an atom or from an ion, the remaining electron8 are tightly bound to the
nucleus and hence decreasing the ionic radius.
rc is the radius of cation
ra is that radius of anion
• metallic elements shed electrons to become cations so cations are smaller than anions - radius ratio < 1
JOE JEBA RAJAN, AST PROF - MECH, RIET
20. Miller Indices are a symbolic vector representation for the orientation of an atomic plane in a crystal lattice
and are defined as the reciprocals of the fractional intercepts which the plane makes with the
crystallographic axes.
MILLER INDICES
Series of steps that will lead you to its notation using Miller indices for
planes.
1. fractional intercepts that the plane/face makes with the
crystallographic axes.
2. Taking the reciprocal of the fractional intercept of each unit length for
each axis.
3. Fractions are cleared (i.e., make 1 as the common denominator).
4. These integer numbers are then parenthetically enclosed In
the example the Miller indices for the plane is (111)
JOE JEBA RAJAN, AST PROF - MECH, RIET
24. MILLER INDICES
Steps for determination of the Miller indices of a direction
1. Determinate the co-ordinates of two points that lie on the direction, using a right handed co-ordinate
system.
2. Subtract the co-ordinates of tail point from those of the head point and express the same in terms of
the unit cell dimensions.
3. Multiply or divide these numbers by a common factor to reduce them to the smallest integers.
4. The three indices are enclosed in square bracket is as u v w. A negative integer is represented with a bar
over the number.
EXAMPLE
Determine the Miller indices of direction A
as shown in the cubic unit cells.
1. Points are 1, 0, 0 and 0, 0, 0
2. 1,0, 0 - 0, 0, 0 = 1, 0, 0
3. No fractions to clear
4. [1 0 0]
H T
JOE JEBA RAJAN, AST PROF - MECH, RIET
25. Determine the Miller indices of direction B
as shown in the cubic unit cells.
1. Points are 1, 1, 1 and 0, 0, 0
2. 1, 1, 1 - 0, 0, 0 = 1, 1, 1
3. No fractions to clear
4. [1 1 1]
1. Points are 0, 0, 1 and
1
2
, 1,0
2. 0, 0, 1 -
1
2
, 1, 0 = -
1
2
, -1, 1
3. 2(-
1
2
, -1, 1) = -1, -2, 2
4. [1 2 2]
Direction C
Direction B
X Y Z
STEP 1 Head end 1 1 1
Tail end 0 0 0
STEP 2 H.E – T.E 1 1 1
STEP 3 Fraction
reduction
No fraction
STEP 4 Enclose
within [ ]
[1 1 1]
Direction D 1. Points are 0, 1, 0 and 2, 0, 0
2. 2, 0, 0-0, 1, 0 = 2, -1, 0
3. 2, 0, 0-0, 1, 0 = 2, -1, 0
4. [2 1 0]
JOE JEBA RAJAN, AST PROF - MECH, RIET
26. • [ 1 0 0] is not equal to [1 0 0] since both are representing opposite direction
• A direction and its multiple are identical; [100] is the same direction as [200]
• The groups of equivalent directions can be said as directions of a form or family
• The special brackets < > are used to indicate this collection of directions.
• Can expect the material have the same properties in all this directions
FAMILY OF DIRECTION
JOE JEBA RAJAN, AST PROF - MECH, RIET
27. FAMILY OF PLANES
• In each unit cell, planes of a form or family represent groups of equivalent planes
that have their particular indices because of the orientation of the coordinates.
• We represent these groups of similar planes with the notation {}. The
• planes of the form {110} in cubic systems are
• In cubic systems, a direction that has the same indices as a plane is perpendicular to that plane.
• Planes and their negatives are identical (this was not the case for directions) because they are
parallel.
• Planes and their multiples are not identical. (Planar density)
JOE JEBA RAJAN, AST PROF - MECH, RIET
28. Draw the [121] direction and the plane (201) in a cubic unit cell
Since a negative index is involved, the origin needs to be shifted towards the positive y-
direction.
a) [121]
Y
X
Z
-Y
1. Locate the tail of direction ‘a’ i.e orgin
2. move +1 in x direction,
3. -2 in y-direction the origin,
4. +1 in z-direction to obtain the other point (Head)
5. Join these points to represent [121] direction
- 2
+1
+1
JOE JEBA RAJAN, AST PROF - MECH, RIET
29. As the negative index is in the x-direction, origin needs to be shifted towards positive x-
direction
(b) (2 0 1)
Y
X
Z
Take reciprocals of the indices
The intercepts are x=
1
−2
y =
1
0
= ∞ and z=
1
1
= 1
• Locate the x intercept along x axis
• draw a line parallel to y axis
𝟏
−𝟐
Since plane is parallel to y axis,
• Locate the z intercept along z axis
• Draw a line parallel to y axis
Join the end points of these lines
𝟏
JOE JEBA RAJAN, AST PROF - MECH, RIET
30. MILLER- BRAVAIS INDICES
Planes and directions in hexagonal crystals are defined
• 4 digits instead of 3
• known as Miller- Bravais indices
• Consider the two planes Plane 1 & Plane 2 with
the three co-ordinate axes a1, a2, and c.
Plane 1 intercepts a1 1 , a2 1 , c∞
Indices value are 1/1 , 1/1 , 1/ ∞
(1 1 0)
Plane 2 intercepts a1 1 , a2 -1/ 2 ,
c ∞
Indices value are 1/1 , 1/(-1/2) , 1/ ∞
(1 2 0)
two planes of
the
same form have
Non- similar
indices
+a1
+a2
+a3
+C
+ C
JOE JEBA RAJAN, AST PROF - MECH, RIET
31. MILLER- BRAVAIS INDICES
Planes and directions in hexagonal crystals are defined
• 4 digits instead of 3
• known as Miller- Bravais indices
• Consider the two planes Plane 1 & Plane 2 with
the three co-ordinate axes a1, a2, and c.
Plane 1 intercepts a1 1 , a2 1 , c∞
Indices value are 1/1 , 1/1 , 1/ ∞
(1 1 0)
Plane 2 intercepts a1 1 , a2 -1/ 2 ,
c ∞
Indices value are 1/1 , 1/(-1/2) , 1/ ∞
(1 2 0)
two planes of
the
same form have
Non- similar
indices
+a1
+a2
+a3
+C
JOE JEBA RAJAN, AST PROF - MECH, RIET
32. MILLER- BRAVAIS INDICES
• Consider the two planes Plane 1 & Plane 2 with the four co-ordinate axes
a1, a2, a3 and c.
Plane 1 intercepts a11 , a2 1 , a3 −
1
2
, c= ∞
Indices value are 1/1 , 1/1 , 1/ (-1/2),1/∞
(1 1 2 0)
Plane 2 intercepts a11 , a2 -1/ 2 , a3 1, c=∞
Indices value are 1/1 , 1/(-1/2) , 1/1 , 1/∞
(1 2 1 0)
+a1
+a2
+a3
two planes of the same form have similar indices
four indices are designated as (h k i 𝑙)
JOE JEBA RAJAN, AST PROF - MECH, RIET
33. MILLER- BRAVAIS INDICES
+a1
+a2
+a3
four indices are designated as (h k i 𝑙)
upper and lower faces of HCP (0001)
Six vertical surfaces of the unit cell are known as prism
planes.
Corresponding indices (1 0 1 0) and similar indices
Form family {1 1 0 0 }
JOE JEBA RAJAN, AST PROF - MECH, RIET
34. INTERPLANAR SPACING
• The distance between two adjacent parallel planes of atoms with the same Miller indices is called the
interplanar spacing (dhkl).
• The interplanar spacing in cubic materials is given by the general equation,
• where a0 is the lattice parameter and h, k, and l represent the Miller indices of the adjacent planes being considered.
JOE JEBA RAJAN, AST PROF - MECH, RIET
35. POLYMORPHISM AND ALLOTROPY
Polymorphism Allotropy
• Ability of a solid material to exist in more than
one form or crystal structure.
• Find this characteristic in any crystalline
material such as polymers, mineral, metal, etc.
• Allotropy is the existence of two or more
different physical forms of a chemical element.
• These forms exist in the same physical state,
mostly in the solid state.
Polymorphism occurs in chemical compounds Allotropy occurs in chemical elements
All allotropes are polymorphs Carbon – diamond, graphite, graphene,
fullerenes, etc.
JOE JEBA RAJAN, AST PROF - MECH, RIET
36. FCC
hexagonal unit cell of
graphite is a rhombic prism
(not a hexagonal prism)
JOE JEBA RAJAN, AST PROF - MECH, RIET
37. DIAMOND AND GRAPHITE ARE ALLOTROPES OF CARBON
These different forms may have different physical
properties because they have different structures and
chemical behavior
One allotrope may convert into another when we change
some factors such as pressure, light, temperature, etc.
Some common examples for allotropes are as follows:
• Carbon – diamond, graphite, graphene, fullerenes, etc.
• Phosphorous – white phosphorous, red phosphorous,
di-phosphorous, etc.
• Oxygen – dioxygen, ozone, tetraoxygen, etc.
JOE JEBA RAJAN, AST PROF - MECH, RIET
39. SHORT-RANGE ORDER VS LONG-RANGE ORDER
The regular and predictable arrangement of the atoms/ions over a short distance -
usually one or two atom spacing.
SHORT-RANGE ORDER
A regular repetitive arrangement of atoms/ions in a solid which extends over a very large
distance.
LONG-RANGE ORDER (LRO)
JOE JEBA RAJAN, AST PROF - MECH, RIET
41. PROPERTIES OF CRYSTALLINE SOLIDS
• Particular three dimensional geometrical structure.
• Arrangement order of atoms/ions - long order.
• Strength of all the bonds between different ions, molecules and atoms is equal.
• Melting point is extremely sharp -reason - heating breaks the bonds at the same time.
• Physical properties are different along different directions.
• Most stable solids as compared to other solids.
Physical properties
(thermal conductivity, electrical conductivity, refractive index and mechanical strength etc.)
JOE JEBA RAJAN, AST PROF - MECH, RIET
42. PROPERTIES OF AMORPHOUS SOLIDS
• They have different bonds with different strength
• No regularity in the external structure
• Don’t have sharp melting point, due to the variable strength of bonds
• Reason low strength bonds on heating break at once But strong bonds take some time to break.
• Isotropic in nature – i.e in all the directions their physical properties will remain same.
Physical properties
(thermal conductivity, electrical conductivity, refractive index and mechanical strength etc.)
JOE JEBA RAJAN, AST PROF - MECH, RIET
43. INTERPLANAR SPACING
1. The distance between two adjacent parallel planes of atoms with the same Miller indices is
called the interplanar spacing (dhkl).
a0 lattice parameter
h, k, l Miller indices
JOE JEBA RAJAN, AST PROF - MECH, RIET
44. ➢ Deformation – change in shape that take place in a
material when subjected to an external force or load.
➢ Elastic (temporary) and plastic (permanent)
deformation.
➢ Both take place due to adjustments or
displacement in atomic arrangement.
➢ Applied load may be tensile, compressive or
shear (constant or fluctuating)
➢ Application time – fraction of second or a period
of years
➢ Environmental factors- temperature or corrosive
atmosphere
DEFORMATION OF METALS
JOE JEBA RAJAN, AST PROF - MECH, RIET
45. Types of stress that cause deformation
JOE JEBA RAJAN, AST PROF - MECH, RIET
46. ELASTIC DEFORMATION
• When a material is subjected to an applied force, atoms within the crystal are displaced
from their normal positions of equilibrium - deformation.
• This displacement will be just enough to develop attractive forces between the atoms to
balance the applied load.
• Elastic deformation in a solid bar (loaded axially in tension)
• Elastic deformation – Displacement of atoms is by small amount – removal of load
allows atoms to return to their normal equilibrium positions – ie, REVERSIBLE
JOE JEBA RAJAN, AST PROF - MECH, RIET
48. • As the applied load increases and atoms are displaced further apart – end of elastic range
• Deformation is no longer a simple separation of atoms – inelastic
• Two possible ways by which elastic stage ends:
• by yielding
• by fracture
• BRITTLE –elastic range followed by fracture
ELASTIC DEFORMATION
JOE JEBA RAJAN, AST PROF - MECH, RIET
49. • In most engineering materials, there also exists a time-dependent elastic strain component
• This time-dependent elastic behavior is known as anelasticity
• For metals the anelastic component is normally small and is often neglected
• For some polymeric materials its magnitude is significant; in this case it is termed viscoelastic behavior.
• It is the property of materials that exhibit both viscous and elastic characteristics.
ANELASTIC AND VISCOELASTIC BEHAVIOUR
• result of stretching of atomic bonds along crystallographic planes in an ordered solid.
Elasticity
• result of the diffusion of atoms or molecules inside an amorphous material
Viscosity
Viscoelastic materials have elements of both these properties
JOE JEBA RAJAN, AST PROF - MECH, RIET
51. PLASTIC DEFORMATION
Due to yielding, some of the atoms move to new equilibrium positions
Atoms form new bonds in their new positions; hence material is not weakened and do not
return to original positions
Thus the deformation is inelastic or irrecoverable
Such a deformation is permanent, known as plastic deformation.
Two common mechanisms:
1. by Slip
2. by Twinning
JOE JEBA RAJAN, AST PROF - MECH, RIET
52. PLASTIC DEFORMATION BY SLIP
Metals are significantly weaker in shear
Under the influence of a shear force atoms move relative to each other on certain planes from one position of
equilibrium to another causing a permanent deformation
Some atoms slip past the other atoms on a certain plane in a certain direction. This process is called slip
SLIP PLANE: the plane on which slip takes place
SLIP DIRECTION: directions along which atoms move
SLIP SYSTEM: set of favoured plane and direction
JOE JEBA RAJAN, AST PROF - MECH, RIET
57. SCHMID’SLAW
Shear Force, Fr = F cos α
Area of slip plane, A =
Ao
COS β
Resolved shear stress, τr =
Fr
A
=
F cos α cos β
A0
τr = σ cos α cos β ( Schmid’s Law)
F/A0 = σ , the normal stress applied to the bar
JOE JEBA RAJAN, AST PROF - MECH, RIET
58. SCHMID’SLAW
When β = 0 the slip plane is parallel with force applied
When α = 0 the slip plane is perpendicular to force applied
β
90o
JOE JEBA RAJAN, AST PROF - MECH, RIET
61. PLASTIC DEFORMATION BY TWINNING
In certain materials, when a shear stress is applied, planes of atoms in the lattice move parallel to
a specific plane so that the lattice is divided into two symmetrical parts which are differently
oriented. This phenomenon is known as Twinning
Twinning plane: the planes parallel to which atomic movement has taken place
Twinned region: differently oriented region within the crystal and the twinning planes
Twinning is produced suddenly and is accompanied with sound
JOE JEBA RAJAN, AST PROF - MECH, RIET
62. ➢ In both slip and twinning, lattice is sheared
➢ In slip, shear is uniformly distributed over a volume rather than localized on a discrete number of
slip planes
➢ Twinning – atoms move only a fraction of interatomic distance relative to each other
➢ The amount of movement of each plane of atoms in the twinned region is proportional to its
distance from the twinning plane –Mirror image
➢ Twinning is common in HCP crystals.
➢ During twinning the lattice over the twinned regions may get favorably oriented for slip w.r.t the
applied load.
➢ Thus under certain conditions a twinned metal can be more easily deformed by slip
JOE JEBA RAJAN, AST PROF - MECH, RIET
66. Brittleness of BCC, HCP and ductility of FCC
{111} planes and <110> directions have maximum atomic density and there are
12 easy slip systems - well distributed – minimum one slip system oriented in a
favorable position for slip to take place
Hence, slip and thereby plastic deformation is initiated easily in FCC crystals.
Materials having FCC structure are therefore ductile in nature.
Also have 12 easy slip systems with the {110} planes and <111> directions.
But, the {110} planes of BCC have lesser atomic density
So stress required to cause unit atomic displacement will be more
So less plastic (or less ductile) compared to FCC.
FCC Crystals
BCC crystals
JOE JEBA RAJAN, AST PROF - MECH, RIET
67. • Single crystals exist in nature but are very rare - can be grown artificially
• All familiar crystalline solids are composed of a collection of many small
crystals known as grains - polycrystalline.
• Solidification of pure metals, also known as crystallization i.e transition
from liquid state to solid state.
• In the liquid state - atoms will be continuously moving - high kinetic
energy - high temperature. So no definite arrangement / no structure.
• when the temperature brought down - some atoms may group together -
not permanent - but frequently break up and regroup
• Life of the group is based on temperature and the size of the group
MECHANISM OF CRYSTALLISATION
High temperature High K.E High atomic mobility Iife of the atoms group
JOE JEBA RAJAN, AST PROF - MECH, RIET
68. UNDERCOOLING
• embryos (more like solid) is formed - freezing point
• @ this stage both liquid and solid states exist with same K.E But different P.E.
(latent heat)
• Solid have lowest P.E result in dense atomic group -latent heat (heat of fusion or
heat of solidification).
• Super cooling is necessary to create a connection between liquid and solid atomic
groups.
HOMOGENOUS NUCLEATION
• nucleus grows further by adding more and more atoms to it-stable solid regions -
nuclei- latent heat of fusion - heat released -so more new nuclei formed
throughout solid
HETEROGENEOUS NUCLEATION
• by the presence of some impurity atoms-(grain refiner)-imperfection or a grain
boundary
• much low undercooling is enough
• Grain refiners - fine grained grains Dendritic Growth growth continues in three
dimensions-usually along the axes of the crystal-tree like structure – dendrite
• primary arms, secondary, tertiary etc. arms -continue to grow and thicken Grains
MECHANISM OF CRYSTALLISATION
JOE JEBA RAJAN, AST PROF - MECH, RIET
69. GRAIN SHAPE
• Nature of the container will also affect the shape and number of grains.
• 1st zone is near to the mould walls fine grains
• 2nd zone long columnar grains
• 3rd zone towards the centre of mould having coarse, equi-axed grains.
• Mould is narrow, or longer- the equi-axed zone may be absent.
• Metallic mould - produce a fine grained structure
• Sand mould - coarse grained structure.
• Thin sections are subjected to relatively larger rate of cooling.
• Equi-axed grain - same dimensions along the three axes directions.
• Other commonly observed grain shapes are columnar, plate-like or dendritic
MECHANISM OF CRYSTALLISATION
JOE JEBA RAJAN, AST PROF - MECH, RIET
70. GRAIN SIZE
✅ Rate of cooling at the freezing point affects crystal size
✅ Slow cooling - less nuclei formation - crystal size will be large
✅ Rapid cooling - vice versa - crystal size will be fine
✅ Grain size in materials is also controlled by the thermo mechanical treatments
MECHANISM OF CRYSTALLISATION
JOE JEBA RAJAN, AST PROF - MECH, RIET
71. ORIENTATION OF GRAINS
• Usually, the grains have random orientations.
• With metallurgical techniques oriented along specified direction.
• Grains of iron oriented along the [100] direction a higher magnetic permeability than any
other orientation.
MECHANISM OF CRYSTALLISATION
JOE JEBA RAJAN, AST PROF - MECH, RIET
72. GRAIN BOUNDARY
• In polycrystalline material - disturbed region lattice of a few atomic diameters wide
• In some areas of Boundary
atoms so closer create region of compression
• And in some other areas
atoms in far apart create region of tension
Low-angle grain boundary (tilt boundary)
• misalignment is of the order of a few degrees (upto 10°)
• High-angle grain boundary will have more mismatch
MECHANISM OF CRYSTALLISATION
JOE JEBA RAJAN, AST PROF - MECH, RIET
73. GRAIN SIZE VS YIELD STRENGTH OFA MATERIAL
σ0 yield strength (the stress needed to cause plastic deformation)
D average diameter of grains,
σi and k constants for the material.
• Equation is valid for grains as small as about 50nm.
• Equation finds application in design of components made of metallic materials
Hall-Petch Equation
JOE JEBA RAJAN, AST PROF - MECH, RIET