Upload
sollehah-suhaimi
View
221
Download
1
Embed Size (px)
Citation preview
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
1/36
1
4) Slip
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
2/36
Course Content
4) Slip
EXPLAIN concept of a slip, dislocations, twins, and their role inplastic deformation of a single crystal.
DEFINEcritical slip systems in BCC, FCC and HCP single crystals.
CALCULATEcritical resolved shear stress by using Schmidslaw
2
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
3/36
Course Content
4.1 Plastic deformation of metal single crystal
4.1.1 Slipbands and slip lines on the surface of metal crystals.
4.1.2 Plastic deformation in metal crystals by the slip mechanism
4.1.3 Slip systems
4.1.4 Critical resolved shear stress for metal single crystals.
4.1.5 SchmidsLaw
4.1.6 Twinning
4.2 Plastic deformation of polycrystalline metals
4.2.1 Effect of grain boundaries on strength of metals
4.2.2 Effect of plastic deformation on grain shape and dislocation arrangements
4.2.3 Effect of cold plastic deformation on increasing the strength of metals
3
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
4/36
4.1 Plastic deformation of metal single crystal
4.1.1 Slipbands and slip lines on the surface of metal crystals.
4
The slipbands are caused
by the slip or shear
deformation of metal
atoms on specific
crystallographic plane
called slip planes.
Figure 4.1
Slip bands
Consider the permanent deformation of a rod of a zinc single crystal by stressing itbeyond its elastic limit.
An examination after deformation shows that step markings appear on its surface,
which are called slipbands
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
5/36
4.1.1 Slipbands and slip lines on the surface of
metal crystals (cont.)
5
In single crystal of ductile FCC metals like copper andaluminum, slip occurs on multiple slip planes, and as a results
the slipbands pattern on the surface of these metals when
they deformed is more uniform (Figure 4.2)
A closer examination of the slipped surface of metals at high
magnification shows that slip has occurred on many slipplanes within the slipbands (Figure 4.3)
Figure 4.2: Slipbands pattern on
surface of copper single crystal after
0.9% deformation (Mag 100X)
Figure 4.3
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
6/36
4.1.1 Slipbands and slip lines on the surface of
metal crystals (cont.)
6
These fine steps arecalled slip lines
and usually about 50
to 500 atoms apart,
whereas slipbands
are commonlyseparated by about
10,000 atom
diameters.
Unfortunately, the
terms slipbandandslip line are often
used
interchangeably.
Figure 4.3
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
7/36
4.1.2 Plastic deformation in metal crystals
by the slip mechanism
7
Figure 4.4 shows a possible atomic model for the slippage of oneblock of atoms over another in a perfect metal crystal.
Calculations made from this model determine that the strength ofmetal crystals should be about 1000 to 10,000 times greater thantheir observed shear strength.
Thus, this mechanism for atomic slip in large real metal crystalsmust be incorrect.
Figure 4.4
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
8/36
4.1.2 Plastic deformation in metal crystals
by the slip mechanism Cont.)
8
In order for large metal crystals to deform at their observed
low shear strengths,
A high density of crystalline imperfections known as dislocation
must be present. These dislocations are created in large numbers (~106cm/cm3)
as the metal solidifies, and when the metal is deformed, many
more are created so that a highly deformed crystal may contain
as high as 1012
cm/cm3
of dislocations.
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
9/36
4.1.2 Plastic deformation in metal crystals by the
slip mechanism Cont.)
9
Figure 4.5 showsschematically how anedge dislocation canproduce a unit of slipunder a low shear
stress.
A relatively smallamount of stress isrequired for slip by this
process since only asmall group of atomsslips over each other atany instant.
Figure 4.5
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
10/36
4.1.2 Plastic deformation in metal crystals by the
slip mechanism Cont.)
10
Figure 4.5: Schematicillustration of how themotion of an edgedislocation produces aunit step of slip under alow shear stress.
(a) An edge dislocation(formed by an extrahalf plane of atoms.)
(b) A low stress causes ashift of atomic bonds tofree a new interleaved
plane.(c) Repetition of this
process causes thedislocation to moveacross the crystal
Figure 4.5
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
11/36
4.1.2 Plastic deformation in metal crystals by the
slip mechanism Cont.)
During shear, atoms do not slide over each other.
The slip occurs due to movement of dislocations.
Figure 4.511
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
12/36
4.1.2 Plastic deformation in metal crystals by the
slip mechanism Cont.)
12
Figure 4.5 (d): The ripple in the rug(carpet) analog.
A dislocation moves through a metal crystal
during plastic deformation in a manner
similar to a ripple that is pushed along acarpet lying on a floor.
In both cases a small amount of relative
movement is caused by the passage of the
dislocation or riple, and hence a relatively
low amount of energy is expended in this
process.
Figure 4.5
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
13/36
4.1.2 Plastic deformation in metal crystals by the
slip mechanism Cont.)
Dislocations in real crystals can beobserved in the transmission electronmicroscope in thin metal foils andappear as lines due to the atomicdisarray at the dislocations that interfere
with the transmission path of theelectron beam of the microscope.
Figure 4.6 shows a cellular wall patternof dislocations created by lightly
deforming an aluminum sample. The cells are relatively free from
dislocations but are separated by wallsof high dislocation density.
Dislocation cell structure in lightly
deformed AluminumFigure 4.6
Wall of high dislocation density
13
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
14/36
4.1.3 Slip systems
14
Dislocations produce atomic displacements on specific
crystallographic slip planes and in specific crystallographic slipdirections.
Slip occurs in densely or close packedplanes.
Since a lower stress for atomic displacement is required than for less
densely packed planes (Figure 4.7)
Figure 4.7
Close packed
plane
Non-close-packed
plane
If slip is restricted in close
planes due to local high stresses,
then less dense planes become
operative.
Less energyis required to move
atoms along denser planes from
one position to another if atoms
are closer together.
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
15/36
Slip systems are combination of slip planes andslip direction.
Each crystal has a number of characteristic slip systems.
In FCC crystal, slip takes place in{111}octahedral planes and in the directions.
4 (111) type planes and 3 [110] type directions.4 x 3 = 12 slip systems.
Table 6.3
4.1.3 Slip systems: FCC structure
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
16/36
16
Figure 4.8 (a) only 4 of the 8 {111} octahedral planes are considered slip planes, since plane opposite each other areconsidered the same plane.
(b) For each slip plane there are 3 slip direction, since opposite directions are considered as only one slipdirection
Note: slip direction only shown the upper four slip planes of the octahedral planes
Thus, there are 4 slip planes X 3 slip directions, giving total of 12 slip systems for the FCC crystal structure
4.1.3 Slip systems: FCC structure Cont)
Figure 4.8
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
17/36
17
BCC crystals are not close packed.
The slip predominantly occurs in {110} planes that has highest atomic density.
However, slip in BCC also occurs on {112} and {123} planes Since the slip planes in the BCC structure are not close-packed as in the case of the FCC structure,
higher shear stresses are necessary for slip in BCC than in FCC.
The slip direction in BCC metals is always of the type
Since there are six (110)-type slip planes of which each can slip in two directions,
there are 6 X 2 = 12 slip systems
4.1.3 Slip systems: BCC structure
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
18/36
18
The slip direction in BCC metals is always of the type
Since there are six (110)-type slip planes of which
each can slip in two directions,
there are 6 X 2 = 12 slip systems
4.1.3 Slip systems: BCC structure Cont.)
Figure 4.9
Figure 4.10
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
19/36
19
In the HCP structure, the basal plane (0001) is the closet-packed plane and is thecommon slip plane for HCP metals such as Zn, Cd and Mg that have high c/a ratios.
If HCP crystals have high c/a ratio, slip occurs alongbasal planes{0001}. However, for HCP metals such as Ti, Zr and Be that have low c/a ratios, slip also occurs
commonly on prism {1010} and pyramidal {1011} planes.
In all cases the slip direction remains
4.1.3 Slip systems: HCP structure
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
20/36
20
4.1.3 Slip systems: HCP structure cont.)
Figure 4.11
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
21/36
Critical resolved shear stress, cis the stress required to cause slip
in pure metal single crystal.
Stress required to cause slip plane in metal single crystal depends on
Crystal Structure
Atomic bonding characteristics
Temperature during deformation
Orientation of the active slip planes relative to shear stress
Slip begins within the crystal when the shear stress on the slip planein the slip direction reaches a required level of critical resolved
shear stress c
This value is equivalent to yield stress.
4.1.4 Critical resolved shear stress for metal
single crystals.
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
22/36
4.1.4 Critical resolved shear stress for metal single crystals cont.)
22
Table 6.4 lists values for the critical resolves shear stresses of some pure-metalsingle crystals at room temperature.
HCP metal (Zn, Cd, Mg) have low critical resolved shear stresses ranging from0.18 to 0.77 MPa.
HCP metal (Ti) has a very high cof 13.7 MPa
It is believed that some covalent bonding mixed with metallic bonding is partlyresponsible for this high value of
c
Pure FCC metals such as Ag and Cu have lowc
values of 0.48 and 0.65MPa, because of their multiple slip systems
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
23/36
4.1.5 hmidsLaw
The relationship between uniaxial stress action on a single cylinder
of pure metal single crystal and resulting resolved shear stress producedon a slip system is give by;
r=Shear Force
Shear Area
CosCosA
F
CosA
CosF
A
Fr
./
.
001
CosCos ..
1A
Fr
r
Slip
direction
Normal to
Slip plane
0
A
F
A1=Area ofSlip plane
Figure 4.12
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
24/36
4.1.5 hmidsLaw cont.)
24
Consider a uniaxial tensile stress acting on a metal cylinder (Fig. 4.12)
Let A0be the area normal to the axial force F, and A1 the area of the slip
plane or shear area on which the resolved shear force Fr is acting.
Figure 4.12
We can orient the slip
plane and slip direction
by defining the angles
and
is the angle between
the uniaxial force F and
the normal to the slip
plane are A1
is the angle between
the axial force and the
slip direction
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
25/36
4.1.5 hmidsLaw cont.)
25
In order for dislocations to move in the slip system, a sufficient resolved
shear stress acting in the slip direction must be produced by the applied
axial force.
Figure 4.12
The resolved shear stress is;
The resolved shear force Fr isrelated to the axial force F by;
The area of the slip plane (sheararea);
Then:
is called Schmidslaw
=
( )=
=
=
=
=
=
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
26/36
Example 4.1
26
Calculate the resolved shear stress on the (111) [011] slip
system of a unit cell in an FCC nickel single crystal if a stress
of 13.7 Mpa is applied in the [001] direction of a unit cell
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
27/36
4.1.6 Twinning In twinning, a part of atomic lattice is deformed and forms mirror image of
the undeformed lattice next to it (Fig. 4.13).
The crystallographic plane of symmetry between the undeformed and deformedparts of the metal lattice is called the twinning plane.
Twinning, like slip, occurs in a specific direction called the twinning direction
Figure 4.13: Schematic diagram
of the twinning process in an FCC lattice
However, in slipthe atoms on one
side of the slip plane all moveequal distances (Fig 4.5),
Whereas in twinningthe atomsmove distances proportional totheir distance from the twinning
plane (Fig. 4.13)
Figure 4.5
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
28/36
4.1.6 Twinning Cont.)
28
Fig 4.14 illustrates the basic difference between slip and twinning on
the surface of a metal after deformation.
(a) slipsleaves a series of steps (lines)
(b) twinningleaves small but well-defined regions of the crystal
deformed
Figure 4.14
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
29/36
4.1.6 Twinning Cont.)
29
Twinning only involves a small fraction of the total volume of the metalcrystal, and so the amount of overall deformation that can be produced by twinning
is small.
However, the important role of twinning in deformation is that thelattice orientation changes that are caused by twinning may place new
slip systems into favorable orientation with respect to the shear stress andthus enable additional slip to occur
Of the three major metallic unit-cell structures (BCC, FCC and HCP),twinning is most important for the HCP structure because of its small
number of slip systems.
However, even with the assistance of twinning, HCP metals like Zn andMg are still less ductile than the BCC and FCC metals that have moreslips systems.
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
30/36
4.2 Plastic deformation of polycrystalline metals4.2.1 Effect of grain boundaries on strength of metals
30
Almost all engineering alloys are polycrystalline
Single-crystal metals and alloys are used mainly for researchpurposes and only in a few cases for engineering applications.
Grain boundaries strengthen metals and alloys by acting as barriersto dislocation movement except at high temperature At high temperature, grain boundaries become regions of weakness
For most applications where strength is important, a fine grain sizeis desirable, and so most metals are fabricated with a fine grain size.
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
31/36
4.2.1 Effect of grain boundaries on strength of metals
cont.)
31
Next chapter will be discussed ASTM grain size number and a method todetermine the average grain diameter of a metal using metallographic techniques. These parameter allow us to make a relative comparison of grain density and
therefore grain boundary density in metals.
Accordingly, for two components made of the same alloy, the component that hasa larger ASTM grain size number or a smaller average grain diameter is stronger.
The relationship between strength and grain size is of great importance to engineers
The well known Hall-Petch equation, is an empirical (based on experimentalmeasurement and not on theory) equation that relates the yield strength of a metal ytoits average diameterd as follows:
Where 0and kare constants related to the materials of interest
A similar effect exists between hardness (Vickers microhardness test) and grain size.
It is important to note that the Hall-Petch equation does not apply to (1) extremely coarse or extremely fine grain sizes, and (2) metals used at elevated temperatures
= 0 + ()1/2
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
32/36
32
Fig. 4.15; Slip lines change directions
at grain boundaries.
Thus, each grain has its own set ofdislocations on its own preferred slipplanes,
Which have different orientationsfrom those of neighboring grains
4.2.1 Effect of grain boundaries on strength of metals
cont.)
Figure 4.15
As the number of grains increases and grain diameter becomes smaller,
Dislocations within each grain can travel a smaller distance before they
encounter the grain boundary, at which point their movement is
terminated (dislocation pileup)
It is for this reason that fine-grained materials posses a higher strength
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
33/36
4.2.2 Effect of plastic deformation on grain shape
33
Fig 4.16 shows the microstructures of copper sheet that was cold-rolled to reduction of (a) 30% and (b) 50%
Upon plastic deformation the grains are sheared relative to each other
by the generation, movement and rearrangement of dislocations
Note that with increased cold rolling the grains are more elongated inthe rolling direction as a consequence of dislocation movements.
Figure 4.16
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
34/36
4.2.3 Effect of cold plastic deformation on increasing
the strength of metals
34
The dislocation density increases with increased cold deformation.
The exact mechanism by which the dislocation density is increased bycold working is not completely understood.
New dislocations are created by the cold deformation and must interactwith those already existing.
As the dislocation density increases with deformation, it becomes moreand more difficult for the dislocations to move through the existingforest of dislocations, and thus the metal work or strain hardens withincreased cold deformation.
When ductile metals such as copper, aluminum and iron that havebeen annealed are cold-worked at room temperature, they strain-hardenbecause of the dislocation interaction as described before.
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
35/36
35
Fig. 4.17 shows how cold-working at room
temperature increases the tensile strengthof unalloyed copper from anount 30 ksi(200 MPa) to 45 ksi (320 MPa) with 30 %cold work.
Associated with the increase in tensile
strength, however, is a decrease inelongation (ductility)
With 30% cold work, the elongationdecreases from about 52 to 10 % elongation.
4.2.3 Effect of cold plastic deformation on increasing
the strength of metals cont.)
Figure 4.17
Cold working or strain hardening is one of the most important methodsfor strengthening some metals. For example, pure copper and aluminum can be strengthened significantly only by
this method. Thus, cold-drawn unalloyed copper wire can be produced with different strengths
(within certain limitations) by varying the amount of strain hardening
7/24/2019 EBT211 - Lecture 6 - Slip.pdf
36/36
36