[next] [prev] [prev-tail] [tail] [up]

3.4  Dislocation in an anisotropic continuum

Steeds [72] developed a treatment to derive the dislocation energy within the anisotropic elasticity. According to this treatment, the dislocation is considered straight and extended to infinity along the z-axis. This assumption simplifies the problem to a plane strain problem where no quantity depends on the z-coordinate, so

∂ ---= 0. ∂z
(3.17)

The Burgers vector of the dislocation is b. Displacements are given by the functions ux, uy and uz. The displacements ux and uy correspond to the edge component of the considered dislocation whereas uz corresponds to the screw component. The strain components are

εxx = ∂ux ∂x--,εxz = 1 2-∂uz -∂x-,
εzz = 0,εxy = 1- 2( ) ∂ux- + ∂uy- ∂y ∂x, (3.18)
εyy = ∂uy ---- ∂y,εyz = 1 -- 2∂uz ---- ∂y.
The previous equations relate the six strain components to the three components of the lattice displacement and this implies that relations exist between the various εij. These relations are known as the compatibility equations  [45]:
∂2εxx- ∂y2 + ∂2εyy- ∂x2 = 2∂2εxy- ∂x∂y, (3.19a)
∂εyz- ∂x -∂εxz- ∂y = 0. (3.19b)

Hooke’s law for an anisotropic material is written as:

( ) ( ) ( ) εxx s11 s12 s13 s15 s15 s16 σxx || εyy || || s21 s22 s23 s24 s25 s26|| || σyy|| | εzz | | s31 s32 s33 s34 s35 s36| | σzz| || || = || || || || . | 2εyz| | s41 s42 s43 s44 s45 s46| | σyz| ( 2εxz) ( s51 s52 s53 s54 s55 s56) ( σxz) 2εxy s61 s62 s63 s64 s65 s66 σxy
(3.20)

The infinite straight dislocation lies along the z-axis and therefore εzz = 0. This last relation yields an additional condition:

0 = εzz = s21σxx + s22σzz + s23σyy + s24σyz + s25σxy + s26σxz,
σzz = (- s21σxx - s23σyy - s24σyz - s25σxy - s26σxz) ∕s22. (3.21)
Steeds [72] expressed the stresses as a function of the Airy functions, F and ϕ
σxx = ∂2F ---2 ∂y,σxz = -∂ϕ --- ∂y,
σxy = -∂2F--- ∂x∂y,σyz = -∂-ϕ ∂x,
σyy = 2 ∂-F- ∂x2. (3.22)
Substituting the expressions (3.21) and (3.22) into the compatibility equations (3.19) yields
(s11 + s22) ∂4F ---2--2 ∂x ∂y + s12∂4F ---4 ∂y + s21∂4F ---4 ∂x+
s63 s33( ) --∂4F-- -∂4F--- -∂4F--- --∂4ϕ-- --∂4ϕ-- s31∂x3 ∂y + s32 ∂x∂y3 - s36 ∂x2∂y2 - s34∂x2 ∂y1 + s35∂x ∂y3 + +
s66 4 --∂-F-- ∂x2 ∂y2 + s15 4 ∂-ϕ- ∂y4 + s25 4 -∂-ϕ--- ∂x2∂y2 + s64 4 --∂-ϕ-- ∂x2 ∂y1 =
s13 --- s33( ∂4F ∂4F ∂4F ∂3ϕ ∂4 ϕ) s31---2--2 + s32---4 - s36-----3-- s34------2 + s35--4 ∂x ∂y ∂y ∂x ∂y ∂x ∂y ∂y
s23 s 33( ) ∂4F- --∂4F-- -∂4F--- ∂3-ϕ --∂4ϕ-- s31 ∂x4 + s32∂x2 ∂y2 - s36∂x3 ∂y - s34∂x3 + s35∂x2∂y2 +
s16 4 -∂-F--- ∂x ∂y3 + s26 4 -∂-F--- ∂x3 ∂y + s61 4 -∂--F-- ∂x3 ∂y + s62 4 -∂-F-- ∂x ∂y+
s14 3 -∂--ϕ-- ∂x ∂y2 + s24 3 ∂-ϕ- ∂x3 + s65 4 -∂-ϕ--- ∂x∂y3, (3.23a)
s11 4 --∂-F-- ∂x2 ∂y2 + s12 4 ∂-F- ∂y4 + s21 4 ∂-F- ∂x4 + s22 2 ∂-F- ∂y2+
s -63 s33( ∂4F ∂4F ∂4F ∂3ϕ ∂4 ϕ ) s31-------+ s32 -------- s36------- - s34------- + s35------- ∂x3 ∂y ∂x1∂y3 ∂x2 ∂y2 ∂x2 ∂y1 ∂x ∂y3 +
s66 ∂4ϕ ---2--2 ∂x ∂y + s15∂4ϕ ---4 ∂y + s25 ∂4ϕ --2---2 ∂x ∂y + s64 ∂3ϕ ---2--- ∂x ∂y =
s13 s33( ) --∂4F-- ∂4F- -∂3F--- --∂3ϕ-- ∂4-ϕ s31∂x2 ∂y2 + s32 ∂y4 - s36∂x ∂y3 - s34∂x ∂y2 + s35∂y4 +
s16 4 -∂-F--- ∂x ∂y3 + s23 s33( 4 4 4 3 4 ) s31 ∂-F-+ s32--∂-F-- - s36-∂--F--- s34∂--ϕ + s35--∂-ϕ--+ ∂x4 ∂x2 ∂y2 ∂x3 ∂y ∂x3 ∂x2∂y2 +
s26 ∂4F ---3--- ∂x ∂y + s61 ∂4F ---3--- ∂x ∂y + s62 ∂4F -----3- ∂x ∂y + s14 ∂3ϕ -----2- ∂x ∂y + s24∂3 ϕ ---3 ∂x + s65 ∂3ϕ -----3- ∂x ∂y. (3.23b)

Solutions for the previous equations with cylindrical symmetry have the following form  [72]:

F = Bg(x + py) ,ϕ = Cf(x + py) , (3.24)
where g and f are functions of a linear combination of the coordinates x and y. Substituting (3.24) into (3.23) and eliminating B and C yields a sextic equation for the parameter p:
p6( 2 ) S11S55 - S15-
2p5(S11S45 - S15(S14 + S56) + S16S55) +
p4(S S + S (2S + S ) - (S + S )2 - 2S (S + S ) + 4S S ) 11 44 55 12 66 14 56 15 25 46 16 45-
2p3(S45(2S12 + S66) - (S14 + S56)(S25 + S46) - S15S24 + S16S44 + S26S55 ) +
p2(S44(2S12 + S66) - 2S24 (S14 + S56) + S22S55 - (S25 + S46)2 + 4S26S45 )-
2p(S22S45 - S24(S25 + S46) + S26S44) +
S22S44 - S242 = 0 (3.25)
where Slm = slm -((s3ls3m)∕s33). Roots of this equation occur as pairs of complex conjugates p1 and p1*, p 2 and p2*, and p 3 and p3*. Therefore the most general solution for the functions F and ϕ are
F = ∑ n=13 * * (Bngn (ζn) + B ngn (ζn)) , (3.26a)
ϕ = ∑ n=13(C f (ζ ) + C *f (ζ*)) n n n n n n , (3.26b)

where n = 1, 2, 3, ζn = x + pny, and gn and fn denote, respectively, the functions g and f belonging to the root pn. In order to satisfy the physical expectations that the stress field decays as 1/r, it is required that

∂fn- ∂ζn = 2 ∂-gn- ∂ζ2n = 1- ζ. (3.27)
Substituting (3.26) and (3.27) into (3.22) causes the stress components to become
σxx = ∑ n=13( ) p2nBn p*n2B *n ----- + ---*-- ζn ζn,
σxy = -∑ n=13( ) pnBn- p*nB-*n ζn + ζ* n,
σyy = ∑ n=13( *) Bn-+ Bn- ζn ζ*n,
σxz = ∑ n=13( p C p*C *) -n--n+ -n*-n ζn ζn,
σyz = -∑ n=13( ) Cn Cn* ---+ -*- ζn ζn, (3.28)
The displacement components are obtained combining (3.18) and (3.26):
ux = ∫ εxxdx =
∑ n=13 {[( ) ] S11p2n - S16pn + S12 Bn + (S15pn - S14)Cn lnζn
[( *2 * ) * * *] *} S11p n - S16pn + S12 Bn + (S15pn - S14)C n ln ζn, (3.29a)
uy = ∫ εyydz =
∑ n=13 {[( ) ] S12p2n - S26pn + S22 Bn + (S25pn - S24)Cn lnζn
[( *2 * ) * * *] *} S12p n - S26pn + S22 Bn + (S25pn - S24)C n ln ζn, (3.29b)
uz = ∫ εxzdx =
∑ n=13 {[( ) ] S15p2 - S56pn + S25 Bn + (S55pn - S45)Cn lnζn n
(( ) ) } S15p *2n - S56pn + S25 Bn*+ (S55p*n - S45) C*n ln ζ*n. (3.29c)

In order to evaluate the quantities Bn and Cn and their complex conjugates, six relationships are required.

The first set of boundary conditions is derived from the force equilibrium state of the media, i.e., the condition of zero net force on the dislocation  [72], expressed as

∫ ∫S ∑ jσijnjdS = 0, (3.30)
for an arbitrary cylindrical surface S enclosing the dislocation line. nj denotes components of the outer normal to the integration surface S. For an infinite straight dislocation along the z-axis, the previous equation becomes
∫∫S ∑ jσijnjdS = 0, (j = 1,2 ) . (3.31)
It is assumed that the cylinder S has a height L with a circular base C on the xy-plane. Based on these assumptions, the previous equation yields for i = x
0 = ∫∫S(σxxnx + σxyny) dS = L∫ Cd( ) ∂F ---- ∂y = L[ ] ∂F ---- ∂y, (3.32)
and when substituting the expression for F from (3.26) and using (3.27) one gets  [72]
[ ] ∑3 * * * (Bnpn ln ζn + Bnp nlnζn) n=1C = 0. (3.33)
The start and the end point of the closed integration loop C differ by the argument Δθn = 2π. Since the logarithm of a complex argument can be written as
ln ζn = ln rn + iθn, (3.34a)
ln ζn* = ln r n*- iθ n, (3.34b)

equation (3.33) becomes

∑ n=13[Bnpn (lnrn + i2π - lnr ) + B *np*n (ln r*n - i2π - ln r*n)] = 0. (3.35)
Simplifying, one gets
∑ n=13 * * (Bnpn - Bnpn ) = 0. (3.36)
Analogously, another two conditions are obtained from equation (3.33) for i = y
∑ n=13(Bn - B*) n = 0, (3.37)
and for i = z
∑ n=13(Cn - C*n) = 0. (3.38)
The second set of boundary equations is provided by the displacement relations. The integral of the displacement acquisitions along the Burgers circuit encircling the dislocation line must be equal to the Burgers vector b:
∮ du = b. (3.39)
Substituting equations (3.29) for the displacements components into the last relation yields a set of three equations of the form bi∕2πi = QniBn - Qni*B n*. More explicitly the last three equations become
bx = 2πi∑ n=13[ ( 2 * *2) * * ] S11 Bnp n - B npn + S15 (Cnpn - C npn), (3.40a)
by = 2πi∑ n=13[ ( 2 * *2) ( 3 * *3) 2S16 Bnp n - B npn - S11 Bnp n - B npn
+ (S + S )(C p - C *p*) - S (C p2 - C *p*2)] 56 14 n n n n 15 n n n n, (3.40b)
bz = 2πi∑ n=13[ ( ) ] S15 Bnp2n - B *np*n2 + S55 (Cnpn - C *np*n). (3.40c)

The stress field of the dislocation is fully described by equations (3.28) where constants pn are the roots of equation (3.25) and constants Bn are the solution of the system of linear equations (3.36), (3.37), (3.38) and (3.40).

The calculation of the dislocation energy is performed by substituting the stress components in equation (3.15) along the cut surface described by the coordinate r such that x = r sin φ and y = r cos φ. The components of the outer normal n are nx = cos φ, nz = 0 and ny = sin φ. The inner and outer cut-off radii are rc and R, respectively. Then one obtains

dEd ---- dy = 1 -- 2 ∫ rcR ∑ n=13[ (bxBnpn + byCn - bzBn )pn sinφ + (- bxBnpn - byCn + bzBn )cosφ ----------------------------------------------------------------+ cos φ + pnsin φ
+ ] (bxB*np*n-+-byC-*n --bzB*n)-p*nsinφ-+--(--bxB-*np*n---byCn*+-bzB-*n)cosφ-- cosφ + p* sin φ ndr- r =
= 1- 2 ∑ n=13[Bn (- bxpn + bz) - Cnby + B * (- bxp* + bz) - C*by] n n n ln R- rc =
= K ln R -- rc, (3.41)
where K is the so called pre-logarithmic coefficient of the dislocation energy. K depends on the elastic constants, the Burgers vector b and the particular direction of the dislocation line with respect to the crystallographic axes.

[next] [prev] [prev-tail] [front] [up]