Lecture 20: The mechanism of plastic deformation

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Lecture 20 The mechanism of plastic deformation

• PHYS 430/603 material
• Laszlo Takacs
• UMBC Department of Physics

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• Original defect-free crystal
• Elastic deformation bonds are compressed and
  stretched, but not broken.
• Plastic deformation by the introduction of
  dislocations.
• Dislocations organize into parallel lines,
  forming low-angle grain boundaries.
• Recrystallization via annealing defect free
  sample with new shape is obtained.

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• Engineering stress-strain curve - the most
  important source of information on the plastic
  properties of a material - for impure iron and
  for copper.
• Notice the Lüders region for iron and the more
  substantial strain hardening of copper.

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Plastic deformation

• Macroscopically it is always shear. Compression
  and tension do not result in plastic deformation
  directly.
• Microscopic mechanisms
• Slip mediated by dislocations is the most
  important mechanism by far.
• Mechanical twinning is active only if slip is not
  possible.
• Creep by thermally activated diffusion over time.

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Surface roughening of an AlMg(0.8) alloy due
toplastic deformation in tensionImage by
confocal optical microscopy.Eric Moore, Ph.D.
Thesis, UMBC 2006
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Twinning results in plastic deformation. It is a
cooperative phenomenon, layers of atoms must move
simultaneously. Consequently, it requires large
local stress.
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Shear stress required for slip

• The theoretical shear stress is estimated
  assuming that an entire layer of atoms moves
  simultaneously. Assume that the potential felt by
  atoms is
• For very small displacement, well within the
  elastic limit
• From here the theoretical shear strength can be
  expressed. It is typically three orders of
  magnitude too large, yet a useful reference

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Peierls stress the stress required for slip by
the movement of a dislocation.
If b d and
Slip is easiest for planes with low Miller
indeces, that is for small b and large d.
Peach-Koehler equation
The displacement of a dislocation relates to work
done by the external stress, thus we can
introduce the concept of force on a
dislocation. This is not a force in the sense of
Newtons second law dislocations do not have a
mass and they are not accelerated by a force. But
equilibrium of forces on a dislocation remains a
valid concept.
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• In fcc, d is the largest for the
• 1 1 1 planes, the dominant slip system is 1 1
  1lt1 1 0gt.
• Notice that we need to give the slip plane and
  the slip direction within the plane.
• In fact, this is the only active slip system, but
  there are 12 different combinations 4 different
  1 1 1 planes and 3 different slip directions/
  edge directions for each. This many slip systems
  can easily mediate any shape change. (The minimum
  number of slip systems required is 5.)

Notice that the actual geometry of dislocations
depends on the crystal structure - it is not like
the square grid used for illustration.
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Slip lines in AlThe orientation of the
individual crystallite determines which slip
system(s) are active.
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A dislocation may break into two parallel partial
dislocations separated by a band of stacking
fault. The magnitude of dislocation and stacking
fault energies determines whether this indeed
happens and how far the two partials separate
from each other.
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• A full dislocation moves atoms from one position
  to and equivalent one. In fcc, this can split
  easily into two partials according to the
  equation (Shockley partial dislocation)
• a/2 1 -1 0 ? a/6 2 -1 -1 a/6 1 -2 1

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• At the theoretical 1.63 c/a ration, the 0 0 0
  1lt1 1 -2 0gt system is identical to the slip
  system of fcc. But it operates in only one plane
  (the basal plane) and three possible directions.
  Not sufficient for an arbitrary shape change that
  would require at least five slip systems.

If c/a lt 1.63, other slip systems become
available, the material is ductile (Ti). For c/a
gt 1.63, the material is either brittle, or the
deformation takes place in part by twinning.
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• In the bcc structure, there is the
• lt1 1 1gt direction in which atoms are in direct
  contact it defines the slip direction. But slip
  can take place almost equally easily along three
  different slip planes
• 1 1 0lt111gt is the best. With many active slip
  systems, bcc metals are quite ductile, although
  not as much as fcc, as the slip planes are not
  close packed.
• In more complicated lattice structures - larger
  b/d - slip is more difficult. Complex compounds,
  including intermetallic phases, are usually
  brittle. This is an important consideration
  technologically.

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Relationship between applied stress and stress
on a slip plane

• Although we usually put the sample under tension
  or compression, slip takes place due to shear
  along the slip system that reaches the critical
  shear stress first. If tension ? is applied, the
  resulting shear stress is
• ? ? cos(?) cos(?)
• Here cos(?) comes from taking the component of F
  in the direction of the slip and cos(?) describes
  the fact that this force acts on the larger cross
  section A/ cos(?).
• The sheer stress is the largest when the slip
  direction is in the plane of F and N. In this
  case ? ? 90, ? ? sin(?) cos(?). This
  expression is the largest when ? 45. As a
  result, slip takes pace at 45 in isotropic
  materials.

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Preferred slip systems for some metals

• Notes
• The easiest slip for Ti is not along (0 0 0 1)
• Impurities usually increase the critical shear
  stress.

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• Except for a core about as wide as a single line
  of atoms, a dislocation can be represented with
  its elastic stress field.
• This is useful to calculate the energy of the
  dislocation. The result depends on the type of
  dislocation, but it is of the order of
• E(d) 1/2 G b2
• per unit length.