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Module 1
 Plastic deformation
 Plastic deformation of single crystals
 Plastic deformation by slip
 Plastic deformation by twinning.
 The happenings in terms of mechanics
1. No deformation / minute elastic deformation
2. Excessive elastic deformation
3. Yielding / Plastic deformation Failure
4. Fracture
 Are they desirable?
1. No deformation : Structural applications
2. Elastic deformation: Springs
3. Plastic deformation : Metal working applications,
Strengthening
4. Plastic deformation + Fracture : Machining, Testing
 If the solid body is loaded beyond the elastic limit, the
body will experience a permanent change in shape and
size, even if the load is removed.
 Plastic deformation of metals and alloys is generally
studied under two categories namely,
i. Plastic deformation of single crystals.
ii. Plastic deformation of polycrystalline materials
 Plastic deformation of single crystals involve the study of
one single crystal and observing how it behaves under
stress.
 A single crystal is nothing but a single grain and has no
grain boundaries.
 Plastic deformation in single crystals may take place by
i. Slip
ii. Twinning or
iii. a combination of both.
 Slip is the most common mode of plastic deformation
among crystals.
 When a single crystal in tension is stressed beyond its
elastic limit, a step appears such that the single crystal
divides into two blocks .
 When the tensile load is further increased, the blocks
become again divided and relative displacement takes place.
 Slip occurs due to the movement of dislocations through the
crystal as shown in below figure.
 To understand ‘how slip can lead to shape change?’;
we consider a square crystal deformed to a rhombus
(as Below).
 The movement of dislocations can be compared to the
movement of an caterpillar as its arches its back in
order to move forward.
 In twinning each plane of atoms move through a definite
distance and in the same direction.
 The extent of movement of each plane is proportional to its
distance from the twinning plane.
 When a shear stress is applied, the crystal will twin about
the twinning plane in such a way that the region to the left
of the twinning plane is not deformed whereas the region to
the right is deformed.
 The atomic arrangement on either side of the twinned plane
is in such a way that they are mirror reflections of each
other.
S.no Slip Twinning
1 Crystal slip is a line defect. Twinning is a surface defect grain boundary defect.
2 During slip, all atoms in a block move the same distance.
During twinning, the atoms in each successive plane in a block move
through different distances proportional to their distance from twinning
plane.
3
Slip is commonly observed in Body-centered Cubic (BCC) and
Face Centered Cubic (FCC) metals.
Twinning is commonly observed in Hexagonal Close Packing
(HCP) metals.
4 After the slip, the crystal axis remains the same. After twinning, the crystal axis is deformed.
5 The slipped crystal lattice has the same orientation. The twinned crystal lattice is the minor image of the original lattice.
6 The stress required for slip is comparatively low. The stress required for twinning is comparatively more.
7
The stress necessary to propagate slip is usually higher than
the stress required to start slip.
The stress necessary to propagate twinning is lesser than that required
starting it.
8 Slip can be seen as thin lines when viewed under microscope. Twinning can be seen as broad lines when viewed under microscope.
9
For slipping to occur, a threshold value of stress called critical
resolved shear stress is required.
For twinning to occur, no such threshold value of stress is required.
 Simple Cubic Structure (SC)
 Body Centered Cubic Structure (BCC)
 Face Centered Cubic Structure (FCC)
 Hexagonal Close-Packed Structure (HCP)
SIMPLE CUBIC STRUCTURE (SC)
Atoms are located at the corners of the cube only.
Rare due to low packing density (only Polyolefin (Po) has this structure)
Close-packed directions are cube edges.
Coordination # = 6
(# nearest neighbours)
 Atoms per unit cell
 𝑁𝐴𝑉 = 𝑁𝑐 + 𝑁𝐹 + 𝑁𝑖
=
8
8
+
0
2
+
0
1
= 1 + 0 + 0
= 1
Hari Prasad
BODY CENTERED CUBIC STRUCTURE (BCC)
 Atoms are located at the corners of the cube and one atom at the centre.
 ex: Cr, W, Tantalum, Molybdenum, Sodium etc
Coordination # = 8
2 atoms/unit cell: 1 center + 8 corners x 1/8
 Atoms per unit cell
 𝑁𝐴𝑉 = 𝑁𝑐 + 𝑁𝐹 + 𝑁𝑖
=
8
8
+
0
2
+
1
1
= 1 + 0 + 1
= 2
ATOMIC PACKING FACTOR:BCC
a
APF =
4
 ( 3 a/4 ) 3
3
2
atoms
unit cell atom
volume
a 3
unit cell
volume
Close-packed directions:
length = 4R = 3 a
• APF for a body-centered cubic structure = 0.68
2 a
3 a
FACE CENTERED CUBIC STRUCTURE (FCC)
Atoms are located at the corners of the cube and one atom at each face of the
cube.
Atoms touch each other along face diagonals.
 --ex: Al, copper, gold, silver, calcium etc
Coordination # = 12
H
a
r
i
P
r
a
s
a
d
4 atoms/unit cell: 6 face x 1/2 + 8 corners x 1/8
 Atoms per unit cell
 𝑁𝐴𝑉 = 𝑁𝑐 + 𝑁𝐹 + 𝑁𝑖
=
8
8
+
6
2
+
0
1
= 1 + 3 + 0
= 4
ATOMIC PACKING FACTOR: FCC
H
a
r
i
P
r
a
s
a
d
• APF for a face-centered cubic structure = 0.74
maximum achievable APF
4
 ( 2 a/4 )3
3
4
atoms
unit cell
APF =
atom
volume
a3
unit cell
volume
Close-packed directions:
length = 4R = 2 a
Unit cell contains: 6 x
1/2 + 8 x 1/8
= 4 atoms/unit cell
• FCC Unit Cell
FCC STACKING SEQUENCE
• ABCABC... Stacking Sequence
• 2D Projection
BB
BA sites
B sites
C sites
C
B
C
B
C
B B
A
A
B
C
A B
+ +
FCC
=
Putting atoms in the B position in the II layer and in C positions in the III layer we get
a stacking sequence ABCABCABC….  The CCP(FCC) crystal
A
B
C
A
B
C
C
• Coordination # = 12
• APF = 0.74
HEXAGONAL CLOSE-PACKED STRUCTURE (HCP)
6 atoms/unit cell
ex: Mg, Ti, Zn
• c/a = 1.633
c
a
B sites
A sites Bottom layer
Middle layer
• Unit cell has an atom at each twelve corners of the hexagonal prism, one atom at the center
of the two hexagonal faces, three atoms at the body of the cell.
• 3D Projection • 2D Projection
A sites Top layer
APF FOR HCP C=1.633a
Number of atoms in HCP unit cell=
(12/6)+(2/2)+(3/1)=6atoms
Vol.of HCP unit cell=
area of the hexagonal face X height of the hexagonal
Area of the hexagonal face=area of each triangle X6
a
h
a
𝒃𝒉
Area of triangle =
𝟐
=
𝒂𝒉
𝟐
𝟏 𝒂 𝟑
= 𝒂.
𝟐 𝟐
Area of hexagon = 𝟔.
𝒂 𝟐
𝟑
𝟒
Volume of HCP= 𝟔.
𝒂 𝟑𝟐 𝟐
𝒂 𝟑
𝟒 𝟒
. 𝐂 = 𝟔. . 𝟏.𝟔𝟑𝟑𝐚
APF= 6∗
𝟒𝝅𝒓 𝟑
𝟑
𝟑 𝟒
/( ∗ 𝟔 ∗ 𝟏. 𝟔𝟑𝟑 ∗ 𝐚𝟑)
a=2r
APF =0.74
Deformation

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Deformation

  • 2.  Plastic deformation  Plastic deformation of single crystals  Plastic deformation by slip  Plastic deformation by twinning.
  • 3.  The happenings in terms of mechanics 1. No deformation / minute elastic deformation 2. Excessive elastic deformation 3. Yielding / Plastic deformation Failure 4. Fracture  Are they desirable? 1. No deformation : Structural applications 2. Elastic deformation: Springs 3. Plastic deformation : Metal working applications, Strengthening 4. Plastic deformation + Fracture : Machining, Testing
  • 4.  If the solid body is loaded beyond the elastic limit, the body will experience a permanent change in shape and size, even if the load is removed.  Plastic deformation of metals and alloys is generally studied under two categories namely, i. Plastic deformation of single crystals. ii. Plastic deformation of polycrystalline materials
  • 5.  Plastic deformation of single crystals involve the study of one single crystal and observing how it behaves under stress.  A single crystal is nothing but a single grain and has no grain boundaries.  Plastic deformation in single crystals may take place by i. Slip ii. Twinning or iii. a combination of both.
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  • 7.  Slip is the most common mode of plastic deformation among crystals.  When a single crystal in tension is stressed beyond its elastic limit, a step appears such that the single crystal divides into two blocks .  When the tensile load is further increased, the blocks become again divided and relative displacement takes place.  Slip occurs due to the movement of dislocations through the crystal as shown in below figure.
  • 8.  To understand ‘how slip can lead to shape change?’; we consider a square crystal deformed to a rhombus (as Below).
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  • 12.  The movement of dislocations can be compared to the movement of an caterpillar as its arches its back in order to move forward.
  • 13.  In twinning each plane of atoms move through a definite distance and in the same direction.  The extent of movement of each plane is proportional to its distance from the twinning plane.  When a shear stress is applied, the crystal will twin about the twinning plane in such a way that the region to the left of the twinning plane is not deformed whereas the region to the right is deformed.  The atomic arrangement on either side of the twinned plane is in such a way that they are mirror reflections of each other.
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  • 16. S.no Slip Twinning 1 Crystal slip is a line defect. Twinning is a surface defect grain boundary defect. 2 During slip, all atoms in a block move the same distance. During twinning, the atoms in each successive plane in a block move through different distances proportional to their distance from twinning plane. 3 Slip is commonly observed in Body-centered Cubic (BCC) and Face Centered Cubic (FCC) metals. Twinning is commonly observed in Hexagonal Close Packing (HCP) metals. 4 After the slip, the crystal axis remains the same. After twinning, the crystal axis is deformed. 5 The slipped crystal lattice has the same orientation. The twinned crystal lattice is the minor image of the original lattice. 6 The stress required for slip is comparatively low. The stress required for twinning is comparatively more. 7 The stress necessary to propagate slip is usually higher than the stress required to start slip. The stress necessary to propagate twinning is lesser than that required starting it. 8 Slip can be seen as thin lines when viewed under microscope. Twinning can be seen as broad lines when viewed under microscope. 9 For slipping to occur, a threshold value of stress called critical resolved shear stress is required. For twinning to occur, no such threshold value of stress is required.
  • 17.  Simple Cubic Structure (SC)  Body Centered Cubic Structure (BCC)  Face Centered Cubic Structure (FCC)  Hexagonal Close-Packed Structure (HCP)
  • 18. SIMPLE CUBIC STRUCTURE (SC) Atoms are located at the corners of the cube only. Rare due to low packing density (only Polyolefin (Po) has this structure) Close-packed directions are cube edges. Coordination # = 6 (# nearest neighbours)
  • 19.  Atoms per unit cell  𝑁𝐴𝑉 = 𝑁𝑐 + 𝑁𝐹 + 𝑁𝑖 = 8 8 + 0 2 + 0 1 = 1 + 0 + 0 = 1
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  • 22. BODY CENTERED CUBIC STRUCTURE (BCC)  Atoms are located at the corners of the cube and one atom at the centre.  ex: Cr, W, Tantalum, Molybdenum, Sodium etc Coordination # = 8 2 atoms/unit cell: 1 center + 8 corners x 1/8
  • 23.  Atoms per unit cell  𝑁𝐴𝑉 = 𝑁𝑐 + 𝑁𝐹 + 𝑁𝑖 = 8 8 + 0 2 + 1 1 = 1 + 0 + 1 = 2
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  • 25. ATOMIC PACKING FACTOR:BCC a APF = 4  ( 3 a/4 ) 3 3 2 atoms unit cell atom volume a 3 unit cell volume Close-packed directions: length = 4R = 3 a • APF for a body-centered cubic structure = 0.68 2 a 3 a
  • 26. FACE CENTERED CUBIC STRUCTURE (FCC) Atoms are located at the corners of the cube and one atom at each face of the cube. Atoms touch each other along face diagonals.  --ex: Al, copper, gold, silver, calcium etc Coordination # = 12 H a r i P r a s a d 4 atoms/unit cell: 6 face x 1/2 + 8 corners x 1/8
  • 27.  Atoms per unit cell  𝑁𝐴𝑉 = 𝑁𝑐 + 𝑁𝐹 + 𝑁𝑖 = 8 8 + 6 2 + 0 1 = 1 + 3 + 0 = 4
  • 28. ATOMIC PACKING FACTOR: FCC H a r i P r a s a d • APF for a face-centered cubic structure = 0.74 maximum achievable APF 4  ( 2 a/4 )3 3 4 atoms unit cell APF = atom volume a3 unit cell volume Close-packed directions: length = 4R = 2 a Unit cell contains: 6 x 1/2 + 8 x 1/8 = 4 atoms/unit cell
  • 29. • FCC Unit Cell FCC STACKING SEQUENCE • ABCABC... Stacking Sequence • 2D Projection BB BA sites B sites C sites C B C B C B B A A B C
  • 30. A B + + FCC = Putting atoms in the B position in the II layer and in C positions in the III layer we get a stacking sequence ABCABCABC….  The CCP(FCC) crystal A B C A B C C
  • 31. • Coordination # = 12 • APF = 0.74 HEXAGONAL CLOSE-PACKED STRUCTURE (HCP) 6 atoms/unit cell ex: Mg, Ti, Zn • c/a = 1.633 c a B sites A sites Bottom layer Middle layer • Unit cell has an atom at each twelve corners of the hexagonal prism, one atom at the center of the two hexagonal faces, three atoms at the body of the cell. • 3D Projection • 2D Projection A sites Top layer
  • 32. APF FOR HCP C=1.633a Number of atoms in HCP unit cell= (12/6)+(2/2)+(3/1)=6atoms Vol.of HCP unit cell= area of the hexagonal face X height of the hexagonal Area of the hexagonal face=area of each triangle X6 a h a 𝒃𝒉 Area of triangle = 𝟐 = 𝒂𝒉 𝟐 𝟏 𝒂 𝟑 = 𝒂. 𝟐 𝟐 Area of hexagon = 𝟔. 𝒂 𝟐 𝟑 𝟒 Volume of HCP= 𝟔. 𝒂 𝟑𝟐 𝟐 𝒂 𝟑 𝟒 𝟒 . 𝐂 = 𝟔. . 𝟏.𝟔𝟑𝟑𝐚 APF= 6∗ 𝟒𝝅𝒓 𝟑 𝟑 𝟑 𝟒 /( ∗ 𝟔 ∗ 𝟏. 𝟔𝟑𝟑 ∗ 𝐚𝟑) a=2r APF =0.74