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The
theory of
elasticity is quite difficult just for simple homogeneous media (no
crystal), and even more difficult for crystals with dislocations - because the
dislocation core cannot be treated with the linear approximations always used
when the math gets tough. |
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Moreover, relatively simple analytical solutions for e.g. the
elastic energy stored in the displacement field of a dislocation, are only
obtained for an infinite crystal, but then
often lead to infinities. |
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As an example, the energy of one dislocation in an otherwise perfect infinite
crystal comes out to be infinite! |
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This looks not very promising.
However, for practical purposes, very
simple relations can be obtained in good approximations. |
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This is especially true for the energy per unit length, the line energy of a dislocation, and for the forces
between dislocations, or between dislocations and other defects. |
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A very good introduction into the
elasticity theory as applied to dislocation is given in the text book Introduction to
Dislocations of D.
Hull and D. J.
Bacon. We will essentially follow the presentation in this book.
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The atoms in a crystal containing a
dislocation are displaced from their perfect lattice sites, and the resulting
distortion produces a
displacement field in the crystal
around the dislocation. |
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If there is a displacement field, we automatically have a
stress field and a strain field, too. Try not to mix up
displacement, stress and
strain! |
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If we look at the picture of the
edge dislocation,
we see that the region above the inserted half-plane is in
compression - the
distance between the atoms is smaller then in equilibrium; the region below the
half-plane is in tension.
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The dislocation is therefore a source of
internal stress in the crystal. In all
regions of the crystal except right at the dislocation core, the stress is
small enough to be treated by conventional linear elasticity theory. Moreover, it is generally
sufficient to use isotropic theory,
simplifying things even more. |
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If we know what is called the elastic field, i.e. the
relative displacement of all atoms, we can
calculate the force that a dislocation exerts on other dislocations, or, more
generally, any interaction with elastic fields from other defects or from
external forces. We also can then calculate the energy contained in the elastic
field produced by a dislocation. |
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The first element of elasticity
theory is to define the displacement
field u(x,y,z), where
u is a vector that defines the displacement of atoms or,
since we essentially consider a continuum, the displacement of any point
P in a strained body from its original (unstrained) position to the
position P' in the strained state. |
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Displacement of P to P'
by displacement vector u |
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The displacement vector u(x, y, z) is
then given by |
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u(x, y, z) |
= |
ux(x, y, z)
uy(x, y, z)
uz(x, y, z)] |
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The components ux ,
uy , uz represent projections
of u on the x, y,
z axes, as shown above. |
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The vector field u, however,
contains not only uninteresting rigid body translations, but at some point
(x,y,z) all the summed up displacements from the other parts of
the body. |
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If, for example, a long rod is
just elongated along the x-axis, the resulting
u field, if we neglect the contraction, would be |
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ux |
= |
const · x |
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uy |
= |
0 |
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uz |
= |
0 |
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But we are only interested in the local deformation, i.e. the deformation that acts on
a volume element dV after it
has been displaced some amount defined by the environment. In other words, we
only are interested in the changes of the shape of a volume element that was a perfect cube in
the undisplaced state. In the example above, all volume element cubes would deform into a
rectangular block. |
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We thus resort to the local strain ε, defined by the nine components of the strain
tensor acting on an elementary cube. That this is true for small strains you
can prove for yourself in the next exercise. |
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Applied to our case, the nine components of the
strain tensor are directly given in terms of the first
derivatives of the displacement components.
If you are not sure about
this, activate the link. |
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We obtain the normal
strain as the diagonal elements of the strain tensor. |
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εxx |
= |
dux
dx |
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εyy |
= |
duy
dy |
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εzz |
= |
duz
dz |
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The shear
strains are contained in the rest of the tensor: |
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εyz |
= |
εzy |
= |
½ · |
( |
duy
dz |
+ |
duz
dy |
) |
εzx |
= |
εxz |
= |
½ · |
( |
duz
dx |
+ |
dux
dz |
) |
εxy |
= |
εyx |
= |
½ · |
( |
dux
dy |
+ |
duy
dx |
) |
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Within our basic assumption of linear
theory, the magnitude of these components is << 1. The normal
strains simply represent the fractional change in length of elements parallel
to the x, y, and z axes respectively.
The physical meaning of the shear strains is shown in the following
illustration |
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A small area element ABCD in the
xy plane has been strained to the shape A B' C' D'
without change of area. The angle between the sides AB and
AD, initially parallel to x and y,
respectively, has decreased by 2εxy. By rotating, but not deforming, the
element as shown on the right-hand side, it is seen that the element has
undergone a simple shear. The simple shear strain often used in engineering
practice is 2εxy, as
indicated. |
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The volume V of a small volume
element is changed by strain to |
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V + DV |
= |
V · (1 + εxx ) · (1
+ εyy) · (1 + εzz) |
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The fractional change in volume D, known as the dilatation, is therefore |
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Δ |
= |
ΔV
V |
= εxx + eyy + εzz |
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Note that D is
independent of the orientation of the axes x, y, z |
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Simple elasticity theory links the strain
experienced in a volume element to the forces acting on this element. (Difficult elasticity theory links the strain
experienced in any volume element to the forces acting on the macroscopic
body!). The forces act on the surface of the element and are expressed as
stress, i.e. as force per area. Stress is
propagated in a solid because each volume element acts on its neighbors. |
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A complete description of the stresses acting
therefore requires not only specification of the magnitude and direction of the
force but also of the orientation of the surface, for as the orientation
changes so, in general, does the force. |
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Consequently, nine components must be defined to
specify the state of stress. This is shown in the illustration below. |
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We have a volume element, here a little
cube, with the components of the stress shown as vectors |
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Since on any surface an arbitrary force vector can be applied,
we decompose it into 3 vectors at right angles to each other. Since we
want to keep the volume element at rest (no translation and no rotation), the
sum of all forces and moments must be zero, which leaves us with 6
independent components. |
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The stress vectors on the other 3 sides are exactly the
opposites of the vectors shown |
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A picture of the components of the strain tensor would look
exactly like this, too, of course. |
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The component sij, where i and j can be
x, y, or z, is defined as the force
per unit area exerted in the + i direction on a face with outward normal
in the + j direction by the material outside upon the material inside. For a face with outward normal in the
j direction, i.e. the bottom and back faces in the figure above,
sij is the force per unit area
exerted in the i direction. For example, syz acts in the positive y
direction on the top face and the negative y direction on the
bottom face. |
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The six components with i unequal to
j are the shear stresses. It is
customary to abbreviate shear stresses with t. In dislocation studies
t without an index then often represents the shear
stress acting on the slip plane in the
slip direction of a crystal. |
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As mentioned above, by considering moments of
forces taken about x, y, and z axes
placed through the centre of the cube, it can be shown that rotational
equilibrium of the element, i.e. net momentum = 0, requires |
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τzy |
= |
τzy |
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τzx |
= |
τxz |
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τxy |
= |
τyx |
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It therefore does not matter in which order the
subscripts are written. |
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The three remaining components sxx, syy, szz are the normal components of the stress. From the definition
given above, a positive normal stress
results in tension, and a negative one in compression. We can define an
effective pressure acting on a volume
element by |
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For some problems, it is more
convenient to use cylindrical polar
coordinates (r, q, z).
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This is shown below; the proper
volume element of cylindrical polar coordinates, is essentially a
"piece of cake". |
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Is it a piece of cake indeed? Well -
no! |
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The picture above is straight form
the really good book of
Introduction to Dislocations of D. Hull and D. J. Bacon, but it is a little bit wrong.
But since it was only used a basic illustration, it did not produce faulty
reasoning or equations. |
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The picture below shows it right -
think about it a bit yourself. (Hint: Imagine a situation, where you apply an
uniaxial stress and try to keep your volume element in place) |
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The stresses are defined as shown
above; the stress σz, q, e.g., is the stress in
z-direction on the q plane. The
second subscript j thus denotes the plane or face of the "slice of
cake" volume element that is perpendicular to the axis denote by the
sunscript - as in the cartesian coordinate system above. The yellow plane or face thus is the q plane, green corresponds to the r
plane and pink denotes the z plane |
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In the simplest approximation (which
is almost always good enough) the relation between
stress and strain is taken to be linear, as in most "material
laws" (take, e.g. "Ohm's law", or the relation between
electrical field and polarization expressed by the dielectric constant); it is
called "Hooke's
law". |
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Each strain component is linearly proportional to each stress
component; in full generality for anisotropic media we have, e.g. |
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ε11 |
= |
a11σ11 +
a22σ22 +
a33σ33 +
a12σ12 +
a13σ13 +
a23σ23 |
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For symmetry reasons, not all aij are
independent; but even in the worst case (i.e. triclinic lattice) only 21
independent components remain. |
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For isotropic solids,
however, only two independent
aij remain and Hooke's law can be written
as |
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σxx |
= |
2G · εxx + l · (εxx + eyy + εzz) |
σyy |
= |
2G · εyy + l · (εxx + eyy + εzz) |
σzz |
= |
2G · εzz + l · (εxx + eyy + εzz) |
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The two remaining material parameters l and G are known as Lamé
constants, but G is more commonly known as the
shear modulus. |
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It is customary to use different
elastic moduli, too. But for isotropic cubic crystals there are always only
two independent constants; if you have
more, some may expressed by the other ones. |
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Most
frequently used, and most useful are Young's modulus, Y , Poisson's ratio, n, and the bulk
modulus, K. These moduli
refer to simple deformation experiments: |
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Under uniaxial, normal loading in the longitudinal direction,
Young´s modulus
Y equals the ratio of longitudinal stress to longitudinal
strain, and
Poisson's ratio ν equals the negative ratio of lateral strain to
longitudinal strain. The
bulk modulus K is
defined to be p/DV
(p = pressure, DV =
volume change). Since only two material parameters are required in Hooke's law,
these constants are interrelated by the following equations |
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Typical values of Y and
n for metallic and ceramic solids are in the
ranges Y = (40-600) GNm2 and n = (0.2 0.45), respectively. |
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A material under strain contains
elastic energy - it is just the sum of the energy it takes to move atoms off
their equilibrium position at the bottom of the potential well from the binding
potential. Since energy is the sum over all displacements time the force needed
for the displacement, we have: |
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Elastic strain energy
Eel per unit volume = one-half the product of stress
times strain for each component. The factor 1/2 comes from counting
twice by taking each component. Thus, for an element of volume
dV, the elastic strain energy is |
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dEel |
= ½ · |
S
i = x,y,z |
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S
j = x,y,z |
σij ·
εij |
· dV |
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For polar or cylindrical coordinates we would get a similar
formula. |
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We do not actually have to calculate
the energy with this formula (be grateful), but you must remember: If we have
the stress field, we can calculate the strain field. If have both, we can
calculate the energy. |
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And if we have the energy we have (almost) everything!
Minimizing the energy gives the equilibrium configuration; gradients of the
energy with respect to coordinates give forces, and so on. |
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© H. Föll