lecturenote 6-plastic deformation
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Mechanical Behavior of MaterialsTRANSCRIPT
Mechanical Behavior of Materials Lecture Note (6) 10/3/2010
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In order to understand why some materials bend easily while
others simply break one must think about the crystal structure,
and how a stress being applied to the material will influence
the atoms.
In all cases, one must think about how easy is it to get atoms to
move past one another. The easier this process is, the more
“bendable” or ductile a material is. The more difficult, the
more brittle that material is.
The motion of atoms past one another in a metal is known as slip.
The easier it is to induce slip in a material, the more ductile the material is. This is a
function of crystallography in the material.
There are certain preferred directions that atoms can move called slip directions, and
certain preferred planes of atoms that will moved when a stress is applied called slip
planes.
The combination of the slip directions and slip planes (multiplied together), give
much information about the mechanical response of a material called the slip
system. The larger the number of the slip system, the more ductile a material will
be.
D e f o r m a t i o n o f m a t e r i a l s o c c u r s
w h e n a l i n e d e f e c t ( d i s l o c a t i o n )
m o v e s ( s l i p ) t h r o u g h t h e m a t e r i a l .
Mechanical Behavior of Materials Lecture Note (6) 10/3/2010
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P l a s t i c d e f o r m a t i o n i s d u e t o t h e
m o t i o n o f a l a r g e n u m b e r o f
d i s l o c a t i o n s .
W h e n a s h e a r f o r c e i s a p p l i e d t o a
m a t e r i a l , t h e d i s l o c a t i o n s m o v e
R e a l m a t e r i a l s h a v e l o t s o f
d i s l o c a t i o n s , t h e r e f o r e t h e
s t r e n g t h o f t h e m a t e r i a l d e p e n d s
o n t h e f o r c e r e q u i r e d t o m a k e t h e
d i s l o c a t i o n m o v e , n o t t h e b o n d i n g
e n e r g y . ( U n l i k e i n e l a s t i c z o n e )
D i s l o c a t i o n s a l l o w d e f o r m a t i o n a t
m u c h l o w e r s t r e s s t h a n i n a
p e r f e c t c r y s t a l
If the top half of the crystal is slipping one plane at a time then only a small fraction of
the bonds are broken at any given time and this would require a much smaller force.
The propagation of one dislocation across the plane causes the top half of the crystal to
move (to slip) with respect to the bottom half but we do not have to break all the bonds
across the middle plane simultaneously (which would require a very large force).
T h e e a s e w i t h w h i c h d i s l o c a t i o n s
m o v e t h r o u g h a m e t a l c r y s t a l i s
Mechanical Behavior of Materials Lecture Note (6) 10/3/2010
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h i g h l y d e p e n d e n t u p o n t h e
p a r t i c u l a r c r y s t a l l o g r a p h i c p l a n e
a n d c r y s t a l l o g r a p h i c d i r e c t i o n
i n v o l v e d
S l i p i s e a s i e s t o n c l o s e p a c k e d
p l a n e s .
S l i p i s e a s i e s t i n t h e c l o s e p a c k e d
d i r e c t i o n .
G e n e r a l l y , o n e s e t o f
c r y s t a l l o g r a p h i c a l l y e q u i v a l e n t
s l i p s y s t e m s d o m i n a t e s t h e p l a s t i c
d e f o r m a t i o n o f a g i v e n m a t e r i a l .
H o w e v e r , o t h e r s l i p s y s t e m s
m i g h t o p e r a t e a t h i g h t e m p e r a t u r e
o r u n d e r h i g h a p p l i e d
s t r e s s .
Slip in close packed metals
Three slip directions in a close packed metal
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Q: What happens if the stress is applied, so
that it is not in line with one of the slip
directions?
A: The system will still relieve the stress by
slipping along the slip planes, although,
several planes may begin to move that are
closest in angle to the direction of the
stress.
Slip in FCC Materials
There are 4 different non-parallel slip
planes which each can move in 3
directions. Therefore there are 12 (4
x 3) ways that the close packed planes
in an FCC can move when a metal
deforms
What about in a HCP
Recall in a HCP metal, the stacking arrangement is
ABABA. With this arrangement, there are no new
slip planes formed as the layers are added the only
slip plane is in the close packing plane (an “A”
layer for instance. In the drawing, in the plane of
the paper)
Again, there are 3 slip directions in each plane, but
there is only 1 unique slip plane in a HCP metal.
Therefore, the total number of slip systems is (1
plane x 3 directions = 3)
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Slip in a body centered cubic (BCC)
Metal
Recall, BCC is not close packed, but because it is a dense
structure (not as dense as FCC or HCP) and the
arrangement of atoms, there are slip directions and planes
in a BCC as well.
But 6 slip unique planes
Therefore, there are 12 slip systems in a
BCC (2 directions * 6 planes).
BCC metals can contain up to 48 slip
systems (e.g. α-Fe). However, none of
the BCC slip planes are truly close-
packed – which means the slip systems
need additional energy (i.e. heat) to
operate. In other words, they are highly
temperature-dependent.
What does this mean?
FCC metals tend to be more ductile than HCP because there are more slip directions
HCP metals tend to be brittle
BCC metals are also ductile despite the different structure from FCC.
Summary of Slip in Metal Crystals
HCP FCC BCC
Close Packed? Yes Yes No
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Packing Efficiency 74% 74% 68%
Stacking Sequence of Close Packed
Planes ABABABAB ABCABC n/a
No. of unique, non-parallel slip
planes 1 4 6
No. slip directions in each plane 3 3 2
No. of total slip systems 3 12 12
Ductile or Brittle? Brittle Ductile Ductile
Atoms per cell 6 4 2
Examples: Mg, Zn
Al, Fe
(austenite),
Au, Ag, C
Fe (ferrite),
W, Nb
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Slip geometry: the critical resolved shear stress
Schmidt’s Law, Slip in Single Crystals
In order for a dislocation to move in its slip
system, a shear force acting in the slip
direction must be produced by the applied
force.
It is observed experimentally that slip occurs
when the shear stress acting in the slip direction
on the slip plane reaches some critical value.
This critical shear stress is related to the stress
required to move dislocations across the slip
plane.
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The tensile yield stress of a material is the applied stress required to start plastic
deformation of the material under a tensile load. We want to relate the tensile stress
applied to a sample to the shear stress that acts along the slip direction. This can be done
as follows. Consider applying a tensile stress along the long axis of a cylindrical single
crystal sample with cross-sectional area A:
The applied force along the tensile axis is
The area of the slip plane is
Where: is the angle between the tensile axis and the slip plane normal.
The component of the axial force F that lies parallel to the slip direction is
cosF
The resolved shear stress on the slip plane parallel to the slip direction is therefore given
by:
coscos cos
cos
r
F F
A A
It is found that the value of τr at which slip occurs in a given material with specified
dislocation density and purity is a constant, known as the critical resolved shear stress
τc. This is Schmid's Law.
The quantity, “cos υ cos λ”, is called the Schmid factor. The tensile stress at which the
crystal starts to slip is known as the yield stress σy, and corresponds to the quantity F/A
in the above equation. Symbolically, therefore, Schmid's Law can be written:
cos cosc y
In a given crystal, there may be many available slip systems. As the tensile load is
increased, the resolved shear stress on each system increases until eventually τc is reached
on one system. The crystal begins to plastically deform by slip on this system, known as
the primary slip system. The stress required to cause slip on the primary slip system is
cos
A
F A
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the yield stress of the single crystal. As the load is increased further, τc may be reached
on other slip systems; these then begin to operate.
From Schmid's Law, it is apparent that the primary slip system will be the system with
the greatest Schmid factor. It is possible to calculate the values of cos υ cos λ for every
slip system and subsequently determine which slip system operates first.
Maximum value of (cosυ cosλ) corresponds to:
Slip will occur first in slip systems oriented close to this angle (υ = λ = 45o)
with respect to the applied stress
Schmid postulated that:
• Initial yield stress varies from sample to sample depending on, among
several factors, the position of the crystal lattice relative to the loading
axis.
• It is the shear stress resolved along the slip direction on the slip plane
that initiates plastic deformation.
• Yield will begin on a slip system when the shear stress on this system
first reaches a critical value (critical resolved shear stress, crss),
independent of the tensile stress or any other normal stress on the lattice
plane.
The Dot Product is a vector operation which will return a scalar value (single number),
which for unit vectors is equal to the cosine of the angle between the two input vectors
(for non-unit vectors, it is equal to the length of each multiplied by the cosine, as shown
in the equation below). We can represent the Dot Product equation with the ● symbol.
* *cosA B A B
Mechanical Behavior of Materials Lecture Note (6) 10/3/2010
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The cosine of the angle between the vectors is:
2 2 2 2 2 2
cos
cos*
x x y y z z
x y z x y z
A B AB
A B A B A BA B
AB A A A B B B
Slip in a Single Crystal
Each step (shear band) result from the generation
of a large number of dislocations and their
propagation in the slip system with maximum
resolved shear stress.
Shear bands
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Plastic Deformation of Polycrystalline Materials
Grain orientations with respect to applied stress are
random.
The dislocation motion occurs along the slip systems
with favorable orientation (i.e. that with highest resolved
shear stress).
Larger plastic deformation corresponds to elongation of
grains along direction of applied stress.
Slip directions vary from crystal to crystal ⇒ Some
grains are unfavorably oriented with respect to the
applied stress (i.e. cosυ cosλ low)
Even those grains for which cosυ cosλ is high may
be limited in deformation by adjacent grains which
cannot deform so easily
Dislocations cannot easily cross grain boundaries
because of changes in direction of slip plane and
disorder at grain boundary
As a result, polycrystalline metals are stronger
than single crystals (the exception is the perfect
single crystal without any defects, as in whiskers
Whiskers are microscopic single-crystal metal fibers, thinner than a
human hair.
This incandescent lamp uses a single crystal of silica carbide for a filament
rather than a tungsten wire. Another material, hafnium carbide, has also been tried in
experiments.
The very thin "whisker" filament is stronger than tungsten, and has a high resistance that
changes little during the lamp's operation. A major difficulty lies in growing the single
crystals long enough to be useful.