Mumbai University Mechanical engineering SEM III Material Technology module 1.3 Deformation: Definition, elastic and plastic deformation, Mechanism of deformation and its significance in design and shaping, Critical Resolved shear stress, Deformation in single crystal and polycrystalline materials, Slip systems and deformability of FCC, BCC and HCP lattice systems.

1. Module 1

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.

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).

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.

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

20. Hari Prasad

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

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