This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Chapter 7: Deformation and Strengthening Mechanisms OUTLINE 7.1 Deformation Mechanisms
‘ 7.1.1 Relation between Dislocations and Plastic Deformation 7.1.2Dislocan'on Generation 
7.1.3 Dislocation Motion a) Basic Concepts " b) Characteristics of Dislocations
‘ c) Slip Systems . ' d) Slip in Single vs Polycrystalline Materials 7.1.4 Slip in SingleCrystals vs Slip in Polycrystals
7.2 Strengthening Mechanisms 7.2.1 Strain hardening
7.2.2 Solid solution strengthening
7.2.3 Precipitation strengthening
7.2.4 Grain size strengthening _____________——————————— 7.1 Deformation Mechanisms
7.1.1 Relation between dislocations and plastic defamation We have discussed in the past lectures, the macroscopic plastic ﬂow of materials.
However, at the atomic level, the macroscopic ﬂow occurs by the motion of dislocations. ‘
Therefore the following question arises: How are dislocations related with plastic ﬂow?
In order to answer this question, let us consider the strain produced by the motion of a single dislocation. Suppose we have a crystal in the shape of a cube, where the edge
length is L (Figure 1473). Suppose now the dislocation is allowed to move across the
crystal (Figure l47b). The plastic shear strain associated with passage of the dislocation can be expressed as re
L! where b is the Burgers vector. Let us now imagine that the dislocation is not able to move
 completely across the crystal. What is now the plastic strain? Well, it has to be less than
before, but proportional to the distance moved. This strain can be expressed as FLA
L L 125 so that, when x =1. =>y =%. In other words, the overall shear strain is proportional to the fraction of plane area a dislocation has traversed. Figure 147: a) Applied shear stress in a cube crystal of edge length L. b) The passage of a
dislocation across the entire crystal produces a displacement equal to the Burgers vector of the dislocation. Now, less us imagine that instead of one dislocation, we have N dislocations, all moving
an average distance 3 (Figure 148). In this case, the resulting strain can be written as E b Nib
Y'N'II'Z' L2 Rearranging gives, =N5EbL= NLfb
7 Lz'L L3 where NL is the total length of dislocation line and L3 is the volume. Thus, NyL’ = p (dislocation density). As a consequence, we can write Y=pb7c In other words, the shear strain is the product number of dislocations that have moved per
unit volume, the Burgers vector and the average distance each dislocation has moved. 126 Figure 148: Plastic deformation of a crystal by the passage of N dislocations. Another important point which arises from observing deformed crystals is that an
examination of the slip steps wouldreveal that several thousand dislocations have passed
through the crystal on a given slip plane. However, if we would examine the same crystal
before deformation, we would ﬁnd only a few dislocations on any slip plane. Thus, where
did the extra dislocations come from? Well, somewhere in the sample, there are
dislocation sources, which are able to generate dislocations during plastic ﬂow at low
stresses. This led us to the next section in our discussion. 7.7.2 Dzs‘ location generation In considering dislocation generation, we need ﬁrst to understand the effective force
acting on a dislocation. Let us think like this: an applied shear stress 1 causes dislocation
motion, which leads to plastic deformation of the crystal, which means that the stress
does’Work in order to lower the potential energy of the system. On this basis, we can
write for the force acting on the dislocation 4 _ d (work)
' d(position) .
dgF,  g2 , = d(position) 1'} where PA is the applied force acting on the slip plane. So, let us suppose we again have a
cube of length L. Thus, we can write for FA FA =1'L2 and for the displacement d we write 127 d=[£)b,whichmeansthat ifx=L=>d=b L
Thus,now
, 2,31
F=ﬂit L Ldeerb
" dx dx ﬁ=rbL=>€4=rb Thus, the Force/unit length acting on the dislocation is merely the product of the applied
stress and the Burgers vector. The forces act in a direction perpendicular to the
dislocation line or parallel to the direction parallel to the slip. So, now let us imagine we
have a dislocation segment AB of length l and pinned at points AB (Figure 149a). Figure 149: a) Pinned dislocation of length L at points A and B. b) Bowing of the
dislocation due to the applied shear stress. As we apply a shear stress to the dislocation, a line tension develops in the dislocation,
which will start counteracting the bowing caused by the applied stress. Hence, there are
two competing forces, namely the line tension = 2Tsin9 , and the applied force = 1M. At
equilibrium, we have then tbt’ =2T sin6. Therefore, when 6 = 90° the force due to line
tension is maximum. Thus, at 6 = 90° we obtainvb! = 2T. Since T = sz where G is the
shear modulus we can write 26b1=H_£§
bl ‘ e t: 128 On this basis, let us see what happens during dislocation generation (Figure 150). As the
segment bows out, the dislocation line essentially pivots around the pinning points A and
B. Successive stages of this bowing process are shown in Figure 150. When the bowing
has reached sufﬁcient proportions so that the two sides swing around and come into
contact on the backside of the pinning points we see an interesting phenomenon. The
portions of the dislocation line that come into contact have the same Burgers vector but
are of opposite sign, and consequently annihilate one another. When these two portions
annihilate, there remains a dislocation loop plus a dislocation segment identical to the
initial segment AB. The source continues to generate new dislocations which expand and move away as far as 1') 26b 3 This type of source is called FrankRead source. Figure 150: Mechanism of dislocation generation by a FrankRead source. 7.1.3 Dislocation motion Now that we understand how dislocations can be generated, we should discuss one of the
most important aspects of plastic deformation, which is the motion of dislocations. We
will try to answer some basic questions of dislocation motion, such as 1) How do
dislocations move?, 2) In which planes do dislocations move?, and 3) In which direction do dislocations move? 129 Let us think first about a human being. How do we humans move? We move by placing
the two legs forward in an alternative manner (Fig. 151) What about a worm? The worm moves by fanning a hump near its anterior end 'and
propagating the hump towards the posterior end. When the hump reaches the end, the entire worm has moved (Fig. 152). Figure 151: Motion of a human being. Figure 152: Motion of a worm. 7 What about the motion of dislocations? In fact, the motion of a dislocation is very similar
to a worm. The worm and its motion correspond to the extrahalf plane of atoms in the dislocation model of plastic deformation (Figure 153). D E ::> § . . a E  .::>g E
i a i a E . c i 0
Figure 153: Motion of a dislocation. 130 Now we understand why the stresses required to move dislocations are much less than the
stresses required to shear planes. This is because during dislocation motion, only a few
bonds need to be broken and reformed. Therefore, materials with directional bonds, such
as covalent materials exhibit a higher lattice resistance to dislocation motion. Hence,
dislocations move across the crystal by shifting the bonds. The process by which plastic
deformation is produced is termed “slip”, and the plane and direction in which dislocation
moves are the slip plane and the slip direction. Secondly, we should understand in
which planes and directions do dislocations move? Thus, we should realize that
dislocations do not move with the same degree of case on all planes of atoms. Why?
Because dislocations will choose a plane and a direction that will minimize the atomic
distortion, in order to spend the least amount of energy. The combination of a slip plane
and a slip direction is deﬁned as the slip system. Therefore the following question arises:
What will be the most likely slip systems in a particular material? The ﬁrst aspect we
should have in mind about slip systems is that: a) Slip planes must contain the Burgers vector and the dislocation line. b) Slip planes have the most densely atomic packing.
c) Slip directions correspond to directions that are the most closely packed with atoms. On this basis, let us now discuss the most likely slip systems in the most common crystal
structures FCC In the fcc crystal structure, the {111} type planes are the most closed packed.
Observation of these planes show that the most closedpacked directions are the <110>
type directions. Since in each single (111) plane there are 3 <110> directions, this means
that in the fcc crystal structure there are 12 slip systems. This is shown schematically as. 4 {111} planes x
; <110> directions for each plane
12 The most likely Burgers vector in fcc crystal structures is then b = g[l 10] . BCC In body centered cubic structures, slip occurs in the closepacked direction [111].
However the most closedpacked planes are several, and of course, they are not as packed
as in the fcc structure. The most closedpacked planes are the {110}, {112} and {123}
type planes. Each of these planes contain <111> slip directions. Hence, for the bcc structure the slip systems are 131 6 {110}.p1anes x 2 <111> directions =12 slip systems
12 {112} planes x l <lll> direction =12 slip systems
24 {123} planes x l <111> direction = 24 slip systems Thus, there are a total of 48 slip systems in ECG and the most likely Burgers vector is
b =3[111].
2 HCP The Jaasal plane is a closedpacked plane, and the close—packed directions are the
<1120 > type directions. Although this behavior is anticipated, the hop metals are not
ideally packed (which requires a ratio c/a = 1.633) and thus some metals slip more easily on the {101 0} planes. So for hcp metals the most likely slip systems are l{0001}plane"l 3 < 1120 > directions = 3
1{1010}plane* 3 < 1120 > directions = 3
6{10T l}planes *1 < 1120 .> direction = 6 Therefore, there are 12 slip systems in the hop structure. 7.1.4 Slip in singlecrystals vs slip in polycrystals When we discussed the stressstrain curve, we have always thought of it as the test used
for a polycrystalline material. However, in some applications, singlecrystals might be of
interest and thus, the stressstrain behavior of these materials should be known. Let us
then look in more detail into a stressstrain test of a single crystal (Figure 154). In this
Figure we can identify three stages: In stage I, called the “easy glide stage” dislocations
arerather mobile and thus they produce a signiﬁcant strain. The transition from stages I to
II is due to the fact that multiple slip occurs. This means that dislocations begin to
intersect in certain planes, which leads to strain hardening. In stage III, the stress is so
high that dislocations can unblock from intersections or other obstacles and thus the strain
hardening decreases. The same thing happens in each grain of a polycrystalline material
but now, each crystal is oriented differently with respect to the tensile axis, and thus, the
materials behaves differently overall. 132 Figure 154: Stressstrain behavior of a singlecrystal. 7.2 Strengthening mechanisms 7.2.1 Strain hardening On the basis of our previous discussions, we now know that the stressstrain curve is
related to the number of dislocations moved and the average distance moved by the
dislocations. However, the following question still arises: Why is there an increase in
su'ess in the stressstrain curve, despite the fact that dislocations are moving? Well, in
fact, there is something else other than just dislocations moving. It is called “strain
hardening”. This phenomenon is related to an increase in strength caused by an increase
in dislocation density and consequent interaction between dislocations. We will see that
strain hardening is very important ﬁom a technological perspective. As an example, A1 alloys, which are relatively soﬁ, are usually coldrolled before they are put into service. In order to understand the mechanism of strain hardening we should consider that when
dislocations are present in the material, atoms around a dislocation are displaced ﬁom
their normal lattice sites. As a result, all dislocations can be considered to have a stress
ﬁeld or strain ﬁeld surrounding themselves. The elastic energy stored inthis strain field
represents what is known as the elastic energy of the dislocation. For screw dislocations, the shear strain is given by 133 where b is the elastic displacement and 2m is the distance over which diSplacement
occurs. In addition, since, 1' = Gr , we can write Gb r=—
270'
which is a shear pure elastic distortion. Thus, the elastic energy per unit length of dislocation can be given by Ewe ’11 2 1 1 z
=  21td =— =—
Ln, 5267 n 2'7 267
Wecanthenwrite
E r1 b )’
M: '_ __
L L2 21tr andr
r1 Gb2 erz
= l_ 2 d = l—
5241t’r? nr r J"041170".
=93: Iﬁﬂ=£mm=ibtlni
47: ’or L 41: 7;, Therefore, the energy/unit length of a screw dislocation is proportional to the square of its
Burger vector. As a result of the presence of stress ﬁelds, dislocations are able to interact
and exert forces on one another. Some examples of dislocation/dislocation interaction can be seen in Figure 155. I No Interaction I a
lRepulsion I l Attraction
T Figure 155: Dislocations that are very far apart do not interact. In addition, dislocations
of the same sign repel each other, whereas dislocations of different sign attract each other. 134 In general, one determine out the increase in yield strength due to strain hardening by
equating the dislocation density. This is given by a = 020in—: The increase in dislocation density is usually obtained by deformation. A common
method is to use coldrolling. In this case, the heavy deformation will increase
signiﬁcantly the density of dislocations. The amount of cold—work can be calculated
according to the following expression. Ad
% coldwork = [19—— * 100 0 where A0 is the initial crosssectional area and Ad is the ﬁnal crosssectional area. 7.2.2 Solid Solution Strengthening Another technique to strengthen and harden materials is to add impurity atoms that go
into substitutional or interstitial sites. In a ﬁrst stage, impurity atoms diffuse to the region
below the dislocation core, which is in a state of tensile stress, in order to reduce the
lattice strain energy (Figure 156). Subsequently, a very strong binding energy develops
between the dislocation core and the impurity atoms. Thus, upon the application of an
external stress, the dislocation want to move to release the stress, but the solute atoms will inhibit 1t. .‘dl'lml a.
A": n anh‘cnﬁi‘ A A
1' ‘7 un'gzgtygpr ‘D’ a aware .: a: a: as, a i . ~31}; 3;} " .
1.9.9.. a¢iﬂn Compressnon
wee" "cow
an n Al an A an (A II
V" " "3*." i
n n A n an an n A. \‘o'e‘e‘e' " "0'7
'\ n n II A. A; n 4'
VW$$WWV ‘\ n n n ‘I a' “Nwwv Figure 156: State of stresses around a dislocation. 135 7.2.3 Precipitation Strengthening In this case, the alloy is designed in such a way, that precipitates will form in the matrix.
Thus, as an applied stress is imposed on the material, dislocations will move. However,
due to the presence of precipitates, dislocations motion will be inhibited and an increase
in yield strength will occur. This mechanism is depicted in Figure 157, whichshows the
bowing of dislocations around the precipitates. Clearly, for larger separation distances the
curvature of the bowing will be pronounced, and thus dislocations can more easily move.
In general, the increase in yield strength can be expressed as 26b 1:— " t where t’ is the closest distance between the precipitates. (a) ' (b) (c) (a) Figure 157: Bowing of a dislocation around precipitates. 7.2.3 Grain Size Strengthening In our previous discussions about grain sizes, we have said that grain growth can occur
and deteriorate several of the mechanical properties of the materials. The increase in grain
size with temperature and time can be determined according to the expression d’ —d: = kotexp(Q/RT) where d is the instantaneous grain size, do is the initial grain size, k0 a constant related to
grain boundary mobility, t is the time, Q is the activation energy, R is the molar gas constant and T is the temperature. 136 Before discussing in more detail why the grain size affects the mechanical properties, let
us ﬁrst understand the effect of grain boundaries on dislocation motion. In general, the grain boundary acts as a barrier to dislocation motion for two reasons a) Since the two grains have different orientations, a dislocation passing from
grain A into grain B will have to change its direction of motion.
b) The misorientation at the grain boundary leads to a discontinuity in the slip planes. Based on these arguments, what is then the role of grain Size? There are two aspects that
should be considered, namely 1) a decrease in grain size will increase the amount of grain
boundary area, Which will more frequently impede dislocation motion, and 2) an increase
in the number of dislocations in a pileup is expected for an increase in grain size, This will facilitate propagation of slip into the next grain (Figure 158). As a result of these
mechanisms the increase in yield strength with grain size can be expressed as 0'" = 00 + It’d7g
where 00 is average yield strength of a single grain, lg, is related to the effectiveness of
the grain boundary in raising the yield strength, and d is the grain size diameter. Figure 158: Slip transfer across adjacent grains. 137 ...
View
Full Document
 Spring '08
 MEYERS

Click to edit the document details