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Again, detailed calculations are
complicated and must be done numerically in most cases. For practical usage,
however, we will find simple approximations by using the
energy formula already derived; this
will be good enough for most cases. |
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First, we have to see that for the movement of a dislocation
on its glide plane, we only need to consider the shear
stress on this plane. This is so, because only force components
lying in the glide plane of the dislocation
can have any effect on dislocation motion in the glide plan. The normal
components of the the stress in the glide plane system act perpendicular to the
glide plane and thus will not contribute to the dislocation movement. |
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Both shear stress components in the glide plane act on the
dislocation. Important, however, is only their combined effect in the direction
of the Burgers vector, which is called the resolved shear stress τres. |
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However, while the resolved shear stress points into the
direction of the Burgers vector, the direction of the
force component acting on, i.e. moving the dislocation, is always perpendicular to the line direction! This
is so because the force component in the line direction does not do anything -
a dislocation cannot move in its own direction. or, if you like that better: If
it would - nothing happens! The whole situation is outlined below |
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Under the influence of the force
F acting on the dislocation
and which we want to calculate, the dislocation moves and work W is
done given by W = Force · distance. Lets look at
the ultimate work that can be done by moving one dislocation. |
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If the dislocation moves in
total through the crystal on a glide plane with the area
A, the upper half of the crystal moves by b
relative to the lower half which is the distance on which work has been done.
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This only happens if a shear force acts on the crystal, and
this force obviously does some work W. This
work is done bit by bit by moving the dislocation through the crystal, so we
must identify the force that does work with the force F acting on
the dislocation. |
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The acting shear stress in this case is then τ = Force F/area A. and
force F is that component of the external force that is contained
or "resolved" in the glide plane of the dislocation as discussed
above. |
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For the total work W
done by moving half of the crystal a distance equal to the Burgers vector
b we obtain |
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With A · τ =
Force; b = Burgesvector = distance. |
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We just as well can divide
W into incremental steps dW, the incremental work
done on an incremental area that consists of an incremental piece dl of the dislocation moving an incremental
distance ds, as shown below |
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The relation between the incremental work dW to
the total work W then is just the ratio between the incremental
area to the total area; we have |
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Putting everything together, we
obtain |
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dW |
= |
A · τ ·
b · |
dl · ds
A |
= |
τ · b · dl ·
ds |
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An incremental piece of work dW can always be
expressed as a force times an incremental distance ds; i.e.
dW = F · ds. The force
F acting on the incremental length dl of
dislocation then obviously is F = τ
· b · dl. |
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If we now redefine the force on a dislocation slightly and
refer it to the (incremental) unit length
dl, i.e. we take F* =
F/dl, we obtain a very simple formula for the magnitude of the force (it
is not a vector!) acting on a unit length of a dislocation: |
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This is easy - but
beware of the sign of the force! You must get all the
signs right (Burgers vector, line vector, τ) to get the correct sign of the force! We also will
drop the "*" in what follows, because as with all other
properties of dislocations, it is automatically per unit length if not
otherwise specified. |
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The important part is τ. It
is the component of the shear strain in the glide plane in the direction of
b. This is normally not a known quantity but must be
calculated, e.g. by a coordinate transformation of a given external stress
tensor to a coordinate system that contains the glide plane as one of its
coordinate planes. |
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Again, we must realize that the Force
F as defined above is always
perpendicular to the dislocation line; even if τ = constant everywhere on the glide plane. |
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This is somewhat counterintuitive, but always imagine the
limiting case of a pure edge and screw
dislocation: The same external τ must
exert a force on a screw dislocation that is perpendicular to the force on an
edge dislocation to achieve the same deformation (think about that; looking at
the pictures helps!) |
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This calls for a little exercise |
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In reality,
dislocations can rarely move in total because they are usually firmly anchored
somewhere. For a straight edge dislocation anchored at two points (e.g. at
immobile dislocation knots) responding to a constant τ, we have the following situation. |
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The forces resulting from the resolved shear stress (red
arrows) will "draw out" the dislocation into a strongly curved
dislocation (on the right). A mechanical equilibrium will be established as
soon as the force pulling back the dislocation (its own line tension) exactly
cancels the external force. |
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The middle picture shows an intermediate stage where the
dislocation is still moving. |
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It is possible to write down the
force on a dislocation as a tensor equation which automatically takes care of
the components - but this gets complicated: |
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First we need to express the force as a vector with components
in the glide plane and perpendicular to it. We define F =
Force on a dislocation = (FN,
FG) |
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With FG = component in the
glide plane, FN = component vertical to the glide
plane. Only FG is of interest, it is given
by |
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Note that scalars, vectors and tensors are combined to form
ultimately a vector. The colors of the brackets code the respective property as
outlined in the margin. |
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The consequences of this equation and
the quantities used are illustrated below. Also note that you have many ways to
confuse signs! |
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Using the formulas derived so far, we
can find an important quantity, the shear stress
necessary to maintain a certain radius of curvature for a
dislocation. |
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If we look at the illustration
above, we see that for a certain stress, the force will draw the
dislocation into a curved line, but for some configuration there will be a
balance of force, because the line tension of the dislocation pulls back.
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We can calculate the
balance of power by looking at an incremental piece of dislocation with a
radius of curvature R. The acting force F is
balanced by the line tension T |
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Let's assume we increase the radius
R of an incremental curved piece by dl. The acting
force needed for this is F = τ
· b · dl and we have dl = R ·
dΘ. The picture below shows a very large
dΘ for clarity. |
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The line tension T = Gb2 is
"pulling back", but only a small component
TF is directly opposing F.
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The component TF is given by
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T |
= T · sin(dΘ/2)
» dΘ/2
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for small dΘ |
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Since there are two components we have the balance of
power |
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T · dΘ =
Gb2· dΘ = τ
· b · dl = τ
· b ·R · dΘ |
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The equilibrium radius
R0 obtained for a shear stress τ0 is thus |
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This will have important consequences becuase the equation
states that a dislocation will move "forever" if τ > Gb/Rmin with
Rmin denoting some minimal radius of curvature that
cannot be decreased anymore. |
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From looking at force balance, we now
can answer the questions posed
before for a dislocation network: |
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The sum of the line tensions at a knot must be zero, too (or
at least very small), otherwise the knot and the dislocations with it will
move. We thus expect that 3-knots will always show angles of
(approximately) 120o. |
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Knots with more than three dislocations will,
as a rule, split into 3-knots, since otherwise there can be no easy
balance of line tensions. In real cases, however, you must also consider the
geometry of the anchor points (are they fixed, can they move?), the change of
line energy with the character of the dislocation and the new total length of
the dislocations. |
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© H. Föll