SJTU多尺度材料模拟与计算
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Dislocation and Stacking Fault
Name:Wu lingling(user023) Student number:016050910054
1 Calculations of Lattice constant and volume modulus
Using molecular dynamics,we can simulate crystals in edge dislocation,screw dislocations and stacking fault, also we can calculate the dislocation strain energy and dislocations. Comparing the method of molecular dynamics calculation values and theoretical, we can analysis its error.Through this experiment, deepen para fault, fault, and the understanding of molecular dynamics simulation.
For edge dislocation, strain for per unit length:
20ln
4(1)e e
Gb R
E r πn =−
For a screw dislocation, strain for per unit length:
20ln
4s
e
Gb R E r π
= Molecular dynamics is dislocation of strain energy method:
()/MD dislocated ref E E E L
=−
In actual crystal structure, the closed normal stacking sequence may be damaged and staggered,
which named the stacking fault.Cambium mistake almost do not produce lattice distortion, but it undermines the integrity of the crystal and the normal cyclical, anomalous diffraction effect in the electronic, allowing the energy of the crystal increased, this part of the increased energy is called the stacking fault energy.
The mathod using Molecular dynamics to calculation approach stacking fault:
SFE =
tot ref
E E S
γ−
2 Results and Analysis
2.1 helical dislocation
-91512.1172811518-(-91519.9264975819)7.80921643s
E ev =
e
l =4.45685A A
,so
57.80921643 1.5326.0/7A 3A 95s s e
s
E ev
E ev l ===A A When calculating Cu,0 3.639A a =A
,
0/2b a =,43.55G GPa =,we can get the result:
20
ln 0.6595(/A)
4s e
Gb R
E ev r π==A
2.2 stacking fault
-91512.1172811518-(-91519.9264975819)7.80921643s
E ev =
e
l =5.09537A A
When calculating Cu,0 3.639A a =A
，0/2b a =，43.55G GPa =, we can get the
result:
20
ln 0.6595(/A)
4s e
Gb R
E ev r π==A
2.3 stacking fault
E=
2
44.0461825440018(mJ/m )
For edge dislocation,when chocing x1=[6.81205 126.341 109.114] A and x2=[6.87178 126.333 141.831] A ，we can get 141.831109.11432.717A d −==A
，so
2102 1.30(10/)81s b J m m n πn −+•=−，2
2
21=39.7347(/)81s b mJ m d
m n γπn +••=− For helical dislocation,when chocing x1=[127.087 125.957 2.27685] A ，x2=[137.353 124.044 1.02476] A ，we can get 137.353127.08710.266A d −==A
，so
210
230.585(10/)81s b J m m n πn −−•=−，222-31=56.9842(/)81s b mJ m d m n γπn ••=−.
3 Summary form