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Prof. Yo-Sep Min Fusion Technology of Chemical & Materials Engineering Lecture 11, Chapter 7 - 1
ISSUES TO ADDRESS...
• Why are the number of dislocations present
greatest in metals?
• How are strength and dislocation motion related?
• Why does heating alter strength and other properties?
Chapter 7:
Dislocations & Strengthening
Mechanisms
Prof. Yo-Sep Min Fusion Technology of Chemical & Materials Engineering Lecture 11, Chapter 7 - 2
Dislocations & Materials Classes
• Covalent Ceramics
(Si, diamond): Dislocation
motion difficult
- directional (angular) bonding
• Ionic Ceramics (NaCl):
Motion difficult
- need to avoid nearest
neighbors of like sign (- and +)
+ + + +
+ + +
+ + + +
- - -
- - - -
- - -
• Metals (Cu, Al):
Motion easiest
- non-directional bonding
- close-packed directions
for slip electron cloud
ion cores
+
+
+
+
+ + + + + + +
+ + + + + +
+ + + + + + +
Prof. Yo-Sep Min Fusion Technology of Chemical & Materials Engineering Lecture 11, Chapter 7 - 3
Dislocation Motion
Dislocation motion & plastic deformation
• Metals - plastic deformation occurs by slip – an edge
dislocation (extra half-plane of atoms) slides over adjacent
plane half-planes of atoms.
• If dislocations can't move, plastic deformation doesn't occur!
Prof. Yo-Sep Min Fusion Technology of Chemical & Materials Engineering Lecture 11, Chapter 7 - 4
Dislocation Motion
• For an edge dislocation, the motion of the dislocation line is parallel to the applied shear stress.
Edge dislocation
Screw dislocation
• For a screw dislocation, the motion of the dislocation line is perpendicular to the stress direction.
Prof. Yo-Sep Min Fusion Technology of Chemical & Materials Engineering Lecture 11, Chapter 7 - 5
Dislocation Density
• Dislocation density: total dislocation length per unit volume
or number of dislocations that intersect a unit area
Ex) carefully solidified metal: ~ 103/mm2
heavily deformed metal: 109~1010/mm2
typical ceramics: 102 ~ 104/mm2
Si wafer: 0.1 ~ 1/mm2
Prof. Yo-Sep Min Fusion Technology of Chemical & Materials Engineering Lecture 11, Chapter 7 - 6
Lattice Strains Around Dislocations
• In an edge dislocation, compressive, tensile, and shear lattice strains are imposed on the neighboring atoms.
• For a screw dislocation, lattice strains are pure shear only.
Prof. Yo-Sep Min Fusion Technology of Chemical & Materials Engineering Lecture 11, Chapter 7 - 7
Strain Field Interaction
Prof. Yo-Sep Min Fusion Technology of Chemical & Materials Engineering Lecture 11, Chapter 7 - 8
Slip system: combination of slip plane and slip direction
– Slip plane - plane on which easiest slippage occurs
• Highest planar densities (and large interplanar spacings)
– Slip directions - directions of movement
• Highest linear densities
Slip Systems
Prof. Yo-Sep Min Fusion Technology of Chemical & Materials Engineering Lecture 11, Chapter 7 - 9
Slip Systems
• FCC Slip occurs on {111} planes (close-packed) in <110> directions (close-packed)
Number of slip systems in FCC: 12
Prof. Yo-Sep Min Fusion Technology of Chemical & Materials Engineering Lecture 11, Chapter 7 - 10
Slip Systems
Relatively ductile
Relatively brittle
Prof. Yo-Sep Min Fusion Technology of Chemical & Materials Engineering Lecture 11, Chapter 7 - 11
Single Crystal Slip
Slip in a Zn single crystal
Prof. Yo-Sep Min Fusion Technology of Chemical & Materials Engineering Lecture 11, Chapter 7 - 12
Stress and Dislocation Motion
• Resolved shear stress, tR
– results from applied tensile stresses
slip plane
normal, ns
Resolved shear stress: tR = F s /A s
AS
tR
tR
FS
Relation between s and tR
tR = FS /AS
F cos l A / cos f
l F
FS
f nS
AS A
Applied tensile stress: = F/A s
F A
F
flst coscosR
Prof. Yo-Sep Min Fusion Technology of Chemical & Materials Engineering Lecture 11, Chapter 7 - 13
• Condition for dislocation motion: CRSS ttR
• Ease of dislocation motion depends
on crystallographic orientation.
10-4 to 10-2 GPa
typically
flst coscosR
Critical Resolved Shear Stress
tR maximum at l = f = 45º
tR = 0
l = 90°
s
tR = s /2 l = 45° f = 45°
s
tR = 0
f = 90°
s
Prof. Yo-Sep Min Fusion Technology of Chemical & Materials Engineering Lecture 11, Chapter 7 - 14
Ex: Deformation of single crystal
So the applied stress of 45 MPa will not cause the crystal to yield.
t s cosl cosf
s 45 MPa
l = 35°
f = 60° tcrss = 20.7 MPa
a) Will the single crystal yield?
b) If not, what stress is needed?
s = 45 MPa
MPa 7.20 MPa 4.18
)41.0( MPa) 45(
)60)(cos35cos( MPa) 45(
crss tt
t
Prof. Yo-Sep Min Fusion Technology of Chemical & Materials Engineering Lecture 11, Chapter 7 - 15
Ex: Deformation of single crystal
What stress is necessary (i.e., what is the
yield stress, sy)?
)41.0(cos cos MPa 7.20 yycrss sflst
MPa 0.5541.0
MPa 0.72
coscoscrss
y fl
ts
MPa 5.50y ss
So for deformation to occur the applied stress must
be greater than or equal to the yield stress
Prof. Yo-Sep Min Fusion Technology of Chemical & Materials Engineering Lecture 11, Chapter 7 - 16
Slip Motion in Polycrystals s
300 mm
• Polycrystals stronger than
single crystals – grain
boundaries are barriers
to dislocation motion.
• Slip planes & directions
(l, f) change from one
grain to another.
• tR will vary from one
grain to another.
• The grain with the
largest tR yields first.
• Other (less favorably
oriented) grains
yield later.
Deformed Cu
polycrystals
Prof. Yo-Sep Min Fusion Technology of Chemical & Materials Engineering Lecture 11, Chapter 7 - 17
• Can be induced by deformation in a polycrystalline metal
- Before
235 mm
- after
- anisotropic
since deformation affects grain
orientation and shape.
deformation direction
Anisotropy in sy
- isotropic
since grains are equiaxed &
randomly oriented.
Prof. Yo-Sep Min Fusion Technology of Chemical & Materials Engineering Lecture 11, Chapter 7 - 18
Deformation by Twinning
• In addition to slip, Plastic deformation in some metallic
materials can occur by the formation of mechanical twins or
twinning.
Prof. Yo-Sep Min Fusion Technology of Chemical & Materials Engineering Lecture 11, Chapter 7 - 19
ANNOUNCEMENTS
Reading: pp. 211 ~ 225
