New ppt.pptx

1. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 1 of 3
Crystal Structure
• In general, solids may be classified as crystalline,
amorphous or a combination of the two.
• Crystalline solids: Solids with regular, repetitive
arrangement of atoms in three dimensions.
• Amorphous solids: Arrangement of atoms in three
dimensions is not regular and not repetitive & the
pattern breaks at different planes.

2. 2
Crystalline solids Amorphous solids
The basic structural unit is
a crystal/grain.
The basic structural unit is a
molecule.
Number of crystals come
together to form a
crystalline solid.
Chains of molecules come
together to form an
amorphous solid.
Each crystal contains
number of repetitive blocks
called unit cells which are
neatly arranged in a
symmetrical order.
Chains of molecules are
random within the solid and
occur in no particular order.
They are irregular and lack
symmetry.

3. Density of crystalline solid
is generally high.
Density of amorphous solid
is generally low.
They have a stable
structure.
Their structure is unstable.
These melt at a definite
melting temperature.
These melt over a range of
temperature.
e.g. metals, alloys, NaCl
and many oxides
e.g. glass, polymer,
elastomer, etc.

4. • Space lattice : an infinite array
of points in three dimensions in
which every point has
surroundings or environment
identical to that of every point in
the array.
Lattice
points
Unit cell
Space lattice, unit cell & lattice points
Space lattice
Unit Cell: The smallest unit of the space lattice
(crystal) which exhibits full symmetry of the crystal is
called a unit cell.
• The crystal can be built up by repetition of the unit
cell in three dimensions.
Space lattice, unit cell & lattice points

5. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 5 of 3
• Crystallographic Axes:
The three axes OP, OQ & OR passing
through the three adjacent sides of the
unit cell are called crystallographic
axes.
• Interfacial angles:
The angles between the adjacent faces
of the unit cell are called interfacial
angles.
QOR = , POR =  & POQ = 
• Primitives or characteristic
intercepts:
The intercepts or edges ‘a’, ‘b’ & ‘c’ of
the unit cell along the crystallographic
axes which define the dimensions of
the unit cell are called primitives or
characteristic intercepts.
Representation of unit cell

6. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 6 of 3
Types of unit cells
SCC – POLINIUM
BCC- Cr, Mo, Ve, Na
FCC-Al, Ni, Cu, Ag, Au, Pb
HCP-Mg, Graphite

7. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 7 of 3
SCC –SIMPLE CUBIC CELL
Types of Unit Cell
Effective number of atoms per unit cell for SCC
• The contribution of each corner atom per unit cell is
1/8th(one-eighth)
• Therefore, the effective number of atoms per unit cell
may be calculated as follows:
• Contribution of 8 corner atoms/unit cell = 8 × 1/8 = 1
 Effective no. of atoms/unit cell of SCC = 1 atom

8. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 8 of 3

9. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 9 of 3
BCC Structure

10. Effective number of atoms per unit cell -
BCC
• The contribution of each corner atom per unit cell is
1/8th (one-eighth) and that of body centered atom is 1
(one).
• The effective number of atoms per unit cell may be
calculated as follows:
Contribution of 8 corner = 8 × 1/8 = 1 & 1 body
centered atoms/unit cell is = 1 × 1=1
Effective no. of atoms/unit cell of BCC 1+1= 2
atoms.
• Examples for BCC crystal structure are: Cr,
-Fe, -Fe, Mo, Ve, Na, Li, Ba, W, -brass, etc.

11. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 11 of 3
FCC Structure

12. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 12 of 3
Effective number of atoms per unit cell for FCC
The contribution of each face centered atom per unit cell
is ½ (half).
Effective number of atoms per unit cell may be
calculated as: Contribution of 8 corner atoms/unit cell =
8 × 1/8 = 1 and contribution of 6 face centered atom/unit
cell = 6 × 1/2 = 3.
Effective no. of atoms/unit cell of FCC is 1+3= 4
atoms
So FCC structure is more densely packed than BCC
structure.
Examples :-FCC crystal structure are: Al, Ni, Cu, Au, Ag,
Pb, Pt, -Fe, -Sn, -brass etc.

13. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 13 of 3
HCP Structure
Basic unit cell is a hexagonal prism. 12 corner atoms. At
the center of each hexagonal basal plane a face
centered atom is present, i.e., 2 face centered atoms. In
addition to these, there are 3 atoms lying inside at mid
height.

14. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 14 of 3
Effective number of atoms per unit cell for HCP
Contribution of 12 corner atoms/unit cell 12 × 1/6 = 2
Contribution of 2 face centered atom/unit cell
= 2 × 1/2 = 1
Contribution of 3 internal atoms/unit cell = 3 × 1 = 3
 Effective no. of atoms/unit cell of HCP is 6
atoms
• This indicates that HCP structure is more densely
packed than other structures.
• Examples:-Mg, Zn, Cd, Co, graphite, etc.

15. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 15 of 3
Derivation to find the height
of the HCP unit cell

16. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 16 of 3

17. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 17 of 3

18. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 18 of 3
18
Triangle AOD is right-angled at D.
AD = a/2,
 DAO = 30o.
0
0
2
cos30
2 2
. .,
cos30 3 3
2
a
AD
AO AO
a a
a
i e AO
 
  
Triangle AOQ right-angled at O
8
1.633
3
 
c a a
2 2 2
2
2
2
2
2 2
. .,
2 3
1 2
1
4 3 3
AQ AO OQ
c a
i e a
c
a a
 
 
 
 
   
   
   
  
   
   

19. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 19 of 3
COORDINATION NUMBER:
• The number of nearest and equidistant atoms with
respect to a given atom of the unit cell.
• Higher the coordination number, more closely
packed atoms.
Coordination number: SCC=6

20. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 20 of 3
In BCC unit cell the smallest distance
between atoms is = a√3 /2. Any
given atom in the unit cell has eight
neighbouring atoms spaced at this
distance, and therefore the
coordination number is 8.
Atom considered
Coordination number for FCC unit
cell:In FCC unit cell the smallest distance
between atoms is a√2 /2 . Any given
atom in the unit cell has twelve
neighbouring atoms spaced at this
distance, and therefore the coordination
number is 12
Coordination number for BCC unit cell:

21. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 21 of 3
Coordination number for HCP unit cell:
In HCP unit cell the smallest distance between atoms is
‘a’. Any given atom in the unit cell has twelve
neighbouring atoms spaced at this distance
(6 corner atoms + 3 atoms above + 3 atoms below),
and therefore the coordination number is12.
Considered atom

22. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 22 of 3
Atomic Packing Factor (APF) or efficiency():
It is defined as the percentage amount of volume
inside a unit cell that has been occupied by the
effective number of atoms of that unit cell.
It is given by the percentage ratio of total volume of
the effective number of atoms per unit cell to the
volume of the unit cell.

23. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 23 of 3
where, N = Effective number of atoms per unit cell,
r = Radius of the spherical atom
V = Volume of the unit cell,
The APF gives a measure of the closeness of packing
(or density of packing) of atoms inside the unit cell.
3
100
100
4
3
100
Volume of Effective number of atoms per unit cell
APF or
Volume of the unit cell
N Volume of a spherical atom
Volume of the unit cell
r
N
V


 

 
 
 
 
 

24. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 24 of 3
Atomic packing factor for BCC unit cell:
The relationship between ‘a’ and the atomic
radius ‘r’ can be found by considering the atomic arrangement on
one solid diagonal of a BCC unit cell as shown in fig.
Triangle XPY is right-angled at P.
a2 + a2 = (XY)2
2a2 = (XY)2
Triangle XYZ is right-angled at Y
with hypotenuse XZ = 4r
(XY)2 + (YZ)2 = (XZ)2
i.e., 2a2 + a2 = (4r)2 ; 3a2 = 16r2 
Effective number of atoms
NBCC = 2 per unit cell 4
3
r
a  a
a
r
r
r
r
X
Y
Z
P
a

25. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 25 of 3
3
3
3
3
3
3 3
4
3
100
4
3
100
4
2
3 8 3 3
100 100
3 64
4
3
68.02%
BCC
BCC
BCC
r
N
V
r
N
a
r
r
r
r





 
 
 
 
 
 
 
 
 
  
 
    
 
 
 


26. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 26 of 3
a
a
r
r
r
r
Relationship between ‘a’ and ‘r’
can be found by considering one
face of a FCC unit cell as shown
in fig. ; a2 + a2 = (4r)2
2a2 = 16r2
Effective number of atoms,
NFCC = 4 per unit cell
  4
2 2
2
r
a r
  
3
3
3 3
4
4
3 16 2 2
100 100
3 64
4
2
74.04%
r
r
r
r


 
 
 
    
 
 
 
 

3
3
3
4
3
100
4
3
100
FCC
FCC
FCC
r
N
V
r
N
a



 
 
 
 
 
 
 
 
Atomic packing factor for FCC unit cell

27. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 27 of 3
a
a
h
Q
P
R
Area of HCP basal
plane
Atomic packing factor for
HCP unit cell:
Volume of the hexagonal unit cell,
V = Area of the hexagonal plane ×
height. Plane can be divided into
six equilateral triangles as shown
in fig.
Then, area of the hexagon
= 6 × area of triangle PQR

28. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 28 of 3
3
2
a
3
2
a 2
3
4
a
Altitude of the triangle PQR, h = tan 60o × a/2 =
Area of triangle PQR = ½ × a × h = ½ × a × =
2
3
4
a
2
3 3
2
a
Area of the hexagonal basal plane = 6 × =
Volume of the hexagonal unit cell, V = Area of the hexagonal basal
plane × height, Height of HCP unit cell,
8
1.633
3
c a a
  In the exam derivation is must (ie C=1.633a)

29. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 29 of 3
2
3 3
2
a 8
3
a 3
3 2a
 V = × =
It may be observed that, a = 2r
3
3
3
3
3
3 3
4
3
100
4
3
100
3 2
4
6
3 24 1
100 100
3
3 2 (2 ) 24 2
74.05%
HCP
HCP
HCP
r
N
V
r
N
a
r
r
r r





 
 
 
 
 
 
 
 
 
 
 
    



30. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 30 of 3
Crystallographic Planes and Directions –
Miller Indices:
A set of integers(whole numbers) used to represent
crystallographic planes and directions is known as Miller
indices.
A directional line which is used to indicate a direction of
slip with in the crystal.
A crystallographic direction is basically a vector between
two points in the crystal. Any direction can be defined by
following a simple procedure.

31. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 31 of 3
31
[111]
Notation of crystallographic directions:
Directions are enclosed in square brackets. Negative
directions are represented by putting a bar above the
appropriate integer, e.g. is the opposite direction
to [111]. The choice of negative directions is arbitrary
but it is essential to be consistent.

32. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 32 of 3
Step 3: Since a=b=c, the intercepts
will be: ½, 1 & 0. Multiplying
throughout by 2 and enclosing
within square brackets we get,
[120] to be the direction indices of
the given vector.
Intercept on X axis Intercept on Y axis Intercept on Z axis
a/2 b 0
Find the Miller indices for the vector shown in the
unit cell where, a=b=c.
Step 1: The given vector is passing through the origin of the
coordinate system.
Step 2: Take the intercepts of the vector on the X, Y & Z axes.

33. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 33 of 3
Few examples of Crystallographic directions

34. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 34 of 3
Crystallographic Planes:
•The planes along which the atoms are arranged is
known as crystallographic planes.
Notation of crystallographic planes:
Planes are enclosed in round brackets (parenthesis).
Negative directions are represented by putting a bar
above the appropriate integer, e.g. is the plane
opposite to the plane (111).
(111)

35. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 35 of 3
Miller indices for crystallographic plane:
Find the Miller indices for the plane shown in fig.1. where,
a=b=c.
Fig.1
Intercep
t on X
axis
Intercep
t on Y
axis
Intercept
on Z
axis
 -b c/2
 -1 1/2
(OOO)

36. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 36 of 3
Miller indices for crystallographic plane:
Step 1: Since plane passes through the origin, shift it to
the adjacent unit cell as shown in fig.(b).
Step 2: Find the intercepts of the plane with the X, Y &
Z axes:
Step 3: Take the reciprocals of the intercepts we get: 0,
-1 & 2
Step 4: Enclose the indices in round brackets
(parenthesis). So Miller Indices is of the plane is .
(012)

37. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 37 of 3
Miller indices for Crystallographic planes
Try This: (001), (010), (121), (246), (426), (-4 -6 8)

38. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 38 of 3
Examples
1. A crystal makes intercepts 3, 2 and 1 on the
crystal axes x, y, z. Find out its miller indices. Show
the crystal plane.
Intercepts of the plane x=3, y=2 and z=1; Reciprocals of
the intercepts 1/3, 1/2 & 1/1 ; Convert into smallest
possible integers by maintaining the same ratio ; 1/3 X6,
1/2 X6 & 1/1 x6 that is =2 3 6
Enclose in a bracket
(2 3 6) is the Miller
Indices.
3
2
1
(2 3 6)
x
z
y
o

39. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 39 of 3
Conventionally, a direction in
analytical geometry is expressed by
a vector normal to the plane under
consideration
For example, the miller indices for a
plane perpendicular to X axis is (100)
and the direction indices of a vector
normal to it is [100]
Relationship between crystallographic
planes & directions:

40. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 40 of 3
Crystal imperfections / Defects:
Types of defects:
The main types of crystal defects (imperfections) are:
1.Point Imperfections (or Zero Dimensional Defects)
2.Line or Linear Imperfections (or One Dimensional
Defects)
3.Surface or Plane Imperfections (or Two
Dimensional Defects)
4.Volume Imperfections (or Three Dimensional
Defects)

41. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 41 of 3
Point Imperfections include:
Vacancy Defect –
lattice points at which atoms are absent.
Interstitial Defect – presence of extra
atom in between regular lattice points.
Substitutional defect:
If a foreign atom substitutes a
parent atom in the regular lattice
structure

42. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 42 of 3
Miss
ing
Cati
on
Cations Anions
Missi
ng
Anio
n
Schotty’s Defect in an ionic crystal
Frenkel’s Defect:
•Formed when cat ion displaces
from its regular location to an
interstitial location.
• Usually, the cations being the
smaller sized ions, are the ones
which are displaced to the
interstitial position.
Disp
lace
d
Cati
on
Cations Anions
Initial
position
of
displace
d Cation
Frenkel’s Defect in an ionic
crystal
Ionic defects
Schottky’s Defect:
• Occurs in ionic crystals.
• Pair of ionic vacancy (cation &
anion) in an ionic crystal.

43. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 43 of 3
Linear Imperfections or Line Defects or
Dislocations:
The Line/Linear defects in crystalline solids that result
from lattice distortion centered about a line is called
dislocation.
Dislocations are of two types:
Edge Dislocation
Screw Dislocation

44. Edge Dislocation:
An edge dislocation in its cross section is essentially a
localized distortion of the crystal lattice due to the
presence of an extra half plane of atoms. Dislocations
occur when an extra incomplete plane is inserted. The
dislocation line is at the end of the plane.

45. Dislocation movement in slip plane

46. Fig.: Screw
dislocation.
Localized distortion of the crystal lattice such that the
atomic planes are bent into a helical surface about a
central distortion line. If we go around the centre of
distortion so produced, we move parallel to the line of
distortion by one inter atomic distance. Similar to
screw moving forward or backward by a distance
equal to its pitch when turned through 360o. This
centre line of distortion is ‘Screw Dislocation Line’.
Screw dislocation

47. Burger’s Circuit & Burger’s Vector:
The direction and magnitude of permanent deformation
(by slip) in any dislocation is given by the Burger’s
Vector.

48. Figure shows the Burger’s circuit A-B-C-D-E drawn
around a region of edge dislocation. Trying to maintain
equal number of horizontal and vertical interatomic
movements on the opposite arms of the Burger’s circuit,
results in the non-closure of the circuit. This indicates
that the region of the crystal where the circuit is drawn
possesses a dislocation. An extra inter atomic
movement E-A, directed from E to A is required to
complete the circuit. The length of EA gives the
magnitude of Burger’s Vector and the direction from E to
A gives the direction of the Burger’s Vector. It can be
clearly noted in figure 2.26(b) that the Burger’s Vector is
perpendicular to the edge dislocation line.

49. Figure 2.26(c) shows the
Burger’s circuit A-B-C-D-E-F
drawn around a region of
screw dislocation. The circuit
is incomplete in between C &
D. It can be clearly noted that
for the Burger’s Circuit to
close, a Burger’s Vector from
C to D must be drawn. Also it
can be noted that the Burger’s
Vector is parallel to the screw
dislocation line.