1. Plastic deformation and
Strengthening Mechanism
Dr.Rahul Sen, Associate Professor, Department of Mechanical Engineering, Poornima College of Engineering, Jaipur
2. Plastic deformation
• Material remains intact
• Original crystal structure is not destroyed
• Crystal distortion is extremely localized
• Possible mechanisms:
– Translational glide (slipping)
– Twin glide (twinning)
Dr.Rahul Sen, Associate Professor, Department of Mechanical Engineering,
Poornima College of Engineering, Jaipur
3. Translational glide
• The principle mode of plastic deformation
• Slip planes: preferred planes with greatest interplanar
distance, e.g., (111) in fcc crystals
• Slip directions: with lowest resistance, e.g., closed packed
direction
• Slip lines: intersection of a slip plane with a free surface
• Slip band: many parallel slip lines very closely spaced
together
Slip plane
4. Existence of defects
• Theoretical yield strength predicted for perfect
crystals is much greater than the measured strength.
• The large discrepancy puzzled many scientists until
Orowan, Polanyi, and Taylor (1934).
• The existence of defects (specifically, dislocations)
explains the discrepancy.
Dr.Rahul Sen, Associate Professor,
Department of Mechanical
5. Defects
• Point defects: vacancies, interstitial atoms,
substitional atoms, etc.
• Line defects: dislocations (edge, screw, mixed)
– Most important for plastic deformation
• Surface defects: grain boundaries, phase boundaries,
free surfaces, etc.
6. Edge dislocations
• Burgers vector: characterizes the “strength” of dislocations
• Edge dislocations: b ⊥ dislocation line
D.R. Askeland and P.P. Phule, The Science and Engineering of Materials, Brooks/Cole (2003).
7. Screw dislocations
• Burgers vector b // dislocation line
D.R. Askeland and P.P. Phule, The Science and Engineering of Materials, Brooks/Cole (2003).
Dr.Rahul Sen, Associate Professor,
Department of Mechanical
9. Observation of dislocations
• Transmission Electron microscopy (TEM): diffraction
images of dislocations appear as dark lines.
M.F. Ashby and D.R.H. Jones, Engineering Materials 1, 2nd
ed. (2002)
Dr.Rahul Sen, Associate Professor,
Department of Mechanical
10. Glide of an edge dislocation
• Break one bond at a time, much easier than
breaking all the bonds along the slip plane
simultaneously, and thus lower yield stress.
D.R. Askeland and P.P. Phule, The Science and Engineering of Materials, Brooks/Cole (2003).
11. Motion of dislocations
William D. Callister, Jr., Materials Science and Engineering, An Introduction, John Wiley & Sons, Inc.
(2003)
12. Force acting on dislocations
• Applied shear stress (τ) exerts a force on a dislocation
• Motion of dislocation is resisted by a frictional force (f, per
unit length)
• Work done by the shear stress (Wτ) equals the work done
by the frictional force (Wf).
( ) bllW ×= 21ττ
( ) 21 lflWf ×=
bfWW f ττ =⇒=
M.F. Ashby and D.R.H. Jones, Engineering Materials 1, 2nd
ed. (2002)
13. Lattice friction stress
• Theoretical shear strength:
• Lattice friction stress for dislocation motion:
• Lattice friction stress is much less than the theoretical shear
strength
• Dislocation motion most likely occurs on closed packed planes
(large a, interplanar spacing) in closed packed directions (small
b, in-plane atomic spacing).
π
τ
2
max
G
=
−==
b
a
G
b
f
f
π
τ
2
exp
Dr.Rahul Sen, Associate Professor,
Department of Mechanical
14. Interactions of dislocations
• Two dislocations may repel or attract each other,
depending on their directions.
Repulsion Attraction
15. Line tension of a dislocation
• Atoms near the core of a dislocation have a higher energy
due to distortion.
• Dislocation line tends to shorten to minimize energy, as if
it had a line tension.
• Line tension = strain energy per unit length
T
T
2
2
1
GbT ≈
Dr.Rahul Sen, Associate Professor,
Department of Mechanical
16. Dislocation bowing
• Dislocations may be pinned by solutes, interstitials, and
precipitates
• Pinned dislocations can bow when subjected to shear
stress, analogous to the bowing of a string.
τbL
T T
θ
R
θ/2θ/2
bLT τ
θ
=
2
sin2
τ2
Gb
R =
2
2
1
GbT ≈
R
L
17. Dislocation multiplication
• Some dislocations form during the process of crystallization.
• More dislocations are created during plastic deformation.
• Frank-Read Sources: a dislocation breeding mechanism.
Dr.Rahul Sen, Associate Professor,
Department of Mechanical
18. Frank-Read sources in Si
Dash, Dislocation and Mechanical Properties of Crystals, Wiley (1957).
19. Strengthening mechanisms
• Pure metals have low resistance to dislocation motion,
thus low yield strength.
• Increase the resistance by strengthening:
– Solution strengthening
– Precipitate strengthening
– Work hardening
Dr.Rahul Sen, Associate Professor,
Department of Mechanical
20. Solution strengthening
• Add impurities to form solid solution (alloy)
• Example: add Zn in Cu to form brass, strength
increased by up to 10 times.
Cu Cu Cu Cu Cu Cu
Cu Cu Cu
Cu Cu Cu Cu
Zn Zn
Bigger Zn atoms make the
slip plane “rougher”, thus
increase the resistance to
dislocation motion.
Dr.Rahul Sen, Associate Professor,
Department of Mechanical
21. Precipitate strengthening
• Precipitates (small particles) can promote strengthening
by impeding dislocation motion.
Dislocation bowing and looping.
Critical condition at semicircular
configuration:
TbL 2=τ
L
Gb
bL
T
≈=
2
τ
M.F. Ashby and D.R.H. Jones, Engineering Materials 1, 2nd
ed. (2002)
22. Work-hardening
• Dislocations interact and obstruct each other.
• Accounts for higher strength of cold rolled steels.
σ
ε
σYU
σYL
Strain hardening
×
σUTS
εf
Dr.Rahul Sen, Associate Professor,
Department of Mechanical
23. Polycrystalline materials
• Different crystal orientations in different grains.
• Crystal structure is disturbed at grain boundaries.
D.R. Askeland and P.P. Phule, The Science and Engineering of Materials, Brooks/Cole (2003).Dr.Rahul Sen, Associate Professor,
Department of Mechanical
24. Plastic deformation in polycrystals
• Slip in each grain is constrained
• Dislocations pile up at grain boundaries
• Gross yield-strength is higher than single crystals
(Taylor factor)
• Strength depends on grain size (Hall-Petch).
YY τσ 3=
2/1
0
−
+= KdY σσ