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Stress Fields and Energies of Dislocation
Stress Field Around Dislocations
• Dislocations are defects; hence, they introduce stresses and strains in the surrounding lattice of a material.
• The mathematical treatment of these stresses and strains can be substantially simplified if the medium is considered to be isotropic and continuous.
• Under conditions of isotropy, a dislocation is completely described by the line and Burgers vectors.
• With this in mind, and considering the simplest situation, dislocations are assumed to be straight, infinitely long lines.
• Figure 14-1 shows a hollow cylinder sectioned along the longitudinal direction. This is an idealization of the strains around an edge dislocation.
Figure 14-1. Simple model for edge dislocation.
• The deformation fields can be obtained by cutting a slit longitudinally along a thick-walled cylinder and displacing the surface by b perpendicular to the dislocation line.
Edge Dislocation
Figure 14-2b. Deformation of a circle containing an edge dislocation. The unstrained circle is shown by a dashed line. The solid line represents the circle after the dislocation has been introduced.
• The cylinder, with external radius R, was longitudinally and transversally displaced by the Burgers vector b, which is perpendicular to the cylinder axis in the representation of an edge dislocation.
• An internal hole with radius ro is made through the center.
• This is done to simplify the mathematical treatment.
• In a continuous medium, the stresses on the center would build up and become infinite in the absence of a hole; in real dislocations the crystalline lattice is periodic, and this does not occur.
• In mechanics terminology, this is called singularity. A “singularity” is a spike, or a single event. For instance, the Kilimanjaro is a singularity in the African plans.
• Therefore, we “drill out” the central core, which is a way of reconciling the continuous-medium hypothesis with the periodic nature of the structure.
• To analyze the stresses around a dislocation, we use the formal theory of elasticity.
• For that, one has to use the relationships between stresses and strains (constitutive relationships), the equilibrium equations, the compatibility equations, and the boundary conditions.
• Hence, the problem is somewhat elaborate.
22
22 )3(
yx
yxbyox
222
22
)(
)(
yx
yxbyoy
222
)(yx
vyv o
yxz
(14.1)
(14.2)
(14.3)
Stress Field Due to Edge Dislocations
where
)1(2 vG
o
(14.4)
222
22
)(
)(
yx
yxbxoxy
0 yzxy
(14.5)
(14.6)
z
yxy
xyx
edge
bygivenisndislocatioedgeforfieldstressThe
00
0
0
:
(14.7)
• The largest normal stress is along the x-axis. – This is compressive--- above slip plane.
– tensile---------- below slip plane.
xy shear stress is maximum in the slip plane, i.e. when y=0
• The stress field can also be written in Polar Coordinates, and this is given as:
r
bor
sin
rb
orr
cos
(14.8)
(14.9)
x
Figure 14-2a. Simple model for screw dislocation.
• The deformation field can be obtained by cutting a slit longitudinally along a thick-walled cylinder and displacing a surface by b parallel to the dislocation line.
Screw Dislocation
Stress Field Due to Screw Dislocations
• This has complete cylindrical symmetry
• The non zero components are:
• In Cartesian coordinate, the stress field matrix is given as:
222 yx
yGbxz
222
yxxGb
yz
(14.10)
(14.11)
0
00
00
yzxz
yz
xz
screw
(14.12)
• There are no extra half plane of atoms.
Therefore, there are no compressive or tensile normal stresses.
• The stress field of the screw dislocation can also be expressed in Polar-coordinate system as:
rGb
2 (14.13)
Strain Energy
• The elastic deformation energy of a dislocation can be found by integrating the elastic deformation energy over the whole volume of the deformed crystal. The deformation energy is given for
(a) Edge Dislocation
1 1 cos21
21 2rr
rr oro o r
drbbdrU
orr
vGb
U 12
ln)1(4
(14.14)
(14.15)
(b) Screw Dislocation
• Note that:
for both edge and screw dislocations
• If we add the core energy (ro ~ b), the total Energy will be given by:
o
rr z r
rGbbdrU
o
12
ln42
11
(14.16)
brGb
Ut1
2
ln4
(14.18)
2GbU (14.17)
• For an annealed crystal: r1~ 10-5cm, b ~ 2 x10-8 cm
Therefore,
• Strain energy of dislocation ~ 8eV for each atom plane threaded by the dislocation.
• Core energy ~0.5eV per atom plane
• Free energy of crystal increases by introducing a dislocation.
2ln 1
br
2
2GbU (14.19)
Forces on Dislocations
• When a sufficiently high stress is applied to a crystal:
Dislocation move
Produce plastic deformation
Slip (glide) Climb (high Temperatures)
• When dislocations move it responds as though it experiences a force equal to the work done divided by the distance it moves
• The force is regarded as a glide force if no climb is involved.
b
dl
ds
Figure 14-3. Force acting on a dislocation line.
• The crystal planes above & below the slip plane will be displaced relative to each other by b
• Average shear displacement =
where, A is the area of the slip plane
• The external force on the area is
Therefore, work done when the elements of slip occur is:
bAdlds
A
bAdsdl
AdW
(14-20)
(14-21)
• The glide force F on a unit length of dislocation is defined as the work done when unit length of dislocation moves unit distance.
Therefore,
• Shear stress in the glide plane resolved in the direction of b
bF
bdAdW
dldsdW
F
(14.22)
(14.23)
Line Tension:
• In addition to the force due to an externally applied stress, a dislocation has a line tension, T which is defined as the energy per unit length = force tending to straighten the line
• The is analogous to the surface tension of a soap bubble or a liquid.
• Consider the curved dislocation. The line tension will produce forces tending to straighten the line & so reduce the total energy of the line.
2GbUel (Energy)
Figure 14-4. Forces on a curved dislocation line.
• The direction of the net force is perpendicular to the dislocation and towards the center of curvature
• For small , F ~ 2T
• But
dislocation segment Radius of curvature
Rds 1
*2
sin
sin2TF (14.24)
• The line will only remain curved if there is a shear stress which produces a force on the dislocation line in the opposite sense. [recall equation 14.23]
• equation 14.25 & 14.26 gives
RTds
Rds
TF
1*
2*2
(14.25)
bdsF (14.26)
RTds
bds
Recall
Stress required to bend a dislocation to a radius R
bRT
2GbT
RGb
bRGb
2
(14.27)
(14.28)
• A more general form of eqn 14.23 is given as
where, t is the dislocation line vector
• Expanding equation 14.29 gives:
Gt
or
btF
)(
bG
321
321
GGG
ttt
kji
F
(14.29)
(14.30)
• Note eqns. 14.26, 14.29 & 14.31 are the same
• A particular direct applicant of these is in the understanding of the Frank-Read dislocation multiplication source
kGtGtjGtGtiGtGtF ))()( 122131132332
kGtGtjGtGtiGtGtF ))()( 211213313223 (14.31)
• Forces on dislocations can be due to other dislocations, precipitates, point defects, thermal gradients, second-phases, etc.
3132121111 bbbG
3232221212 bbbG
3332321313 bbbG
3
2
1
333231
232221
131211
b
b
b
bG
(14.32)
(14.33)