CC BY
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УДК 539.3
Cracks as limits of Eshelby inclusions
X. Markenscoff
Department of Mechanical and Aerospace Engineering, University of California, San Diego, La Jolla, CA, 92093, USA
As limiting behaviors of Eshelby ellipsoidal inclusions with transformation strain, crack solutions can be obtained both in statics and dynamics (for self-similarly expanding ones). Here is presented the detailed analysis of the static tension and shear cracks, as distributions of vertical centers of eigenstrains and centers of antisymmetric shear, respectively, inside the ellipse being flattened to a crack, so that the singular external field is obtained by the analysis, while the interior is zero. It is shown that a distribution of eigenstrains that produces a symmetric center of shear cannot produce a crack. A possible model for a Barenblatt type crack is proposed by the superposition of two elliptical inclusions by adjusting their small axis and strengths of eigenstrains so that the singularity cancels at the tip.
Keywords: Eshelby inclusions, transformation strain, cracks, Barenblatt crack, elasticity
DOI 10.24411/1683-805X-2018-16007
Трещины как пределы включений Эшелби
X. Markenscoff
Калифорнийский университет в Сан-Диего, Сан-Диего, Калифорния, 92093, США
В качестве предельного случая описания эллиптических включений Эшелби при деформации превращения могут быть получены решения задачи о трещине в статитической и динамической постановке (для самоподобно расширяющихся включений). В работе представлен подробный анализ статического растяжения и трещин сдвига как распределения вертикальных центров собственных деформаций и центров асимметричного сдвига, соответственно, внутри вырождающегося в трещину эллипса. Получено сингулярное внешнее поле при нулевом внутреннем поле. Показано, что распределение собственных деформаций, создающее симметричный центр сдвига, не приводит к зарождению трещины. Предложена модель трещины Баренблатта путем наложения двух эллиптических включений с выравниванием их малых осей и подбором собственных деформаций таким образом, что в вершине трещины исчезает сингулярность.
Ключевые слова: включения Эшелби, деформация превращения, трещины, трещина Баренблатта, упругость
1. Introduction
Mura [1] has presented a method for solving crack problems as Eshleby [2] ellipsoidal inclusion problems. Based on the constant stress Eshelby property Mura cancels the applied tractions on the faces of an ellipsoidal inclusion as the "vertical" axis of the ellipsoid tends to zero, so that the inclusion is flattened to a crack. The limit of the product of the eigenstrain times the vertical axis length, as the eigenstrain tends to infinity and the axes length tends to zero, is a finite quantity, the crack opening displacement.
Here is provided a complete analysis for the tension and shear Griffith cracks based on distributing centers of eigenstrain. In addition to the eigenstrains considered by Mura, are included eigenstrains so that all stresses vanish in the interior of the crack. The external field is obtained analytically, while Mura [1] only obtains the singularity
based on the energy-release rate known independently. We show that a crack cannot be produced as a symmetric center of shear with eigenstrains e*xx = -e*yy, because the internal stresses cannot be cancelled by an applied stress field in this case.
Analogously to the static, Markenscoff [3] has shown that the self-similarly expanding elliptical crack Burridge and Willis [4] can be obtained from the limit of the self-similarly expanding ellipsoidal inclusion with transformation strain [5-7], since the constant stress Eshelby property is valid also in self-similarly expanding Eshelby inclusions. It may be noted that the dynamic Eshelby fields in respective limits give both the static Eshelby inclusion (and hence the static cracks) and also Rayleigh waves as the axis expansion speed of the elliptically expanding crack tends to the Rayleigh wave speed.
© Markenscoff X., 2018
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2. The "flattened" elliptical cylinder with transformation strain as a crack
We consider that the crack will be a distribution of eigenstrains inside a flattened elliptical cylinder. The interior stresses of a flattened elliptical cylinder (a3 a2/a1 = = r|) with transformation strains e* are given by (also, [1])
a„ =
I
a,„, =
1 -v |
•£ —
n
2 + n
"^î+n2^ (1+n)2 "yy
- 2v-
1
1 + n
a „ =
1 -v |
(1+n)
£* -n(1 + 2n) £* -
2 xx /1 , „s2 yy z
(1+n)2
1 + n
, (1)
1 -v
1 * n * * -2v-exx -2v—!—£yv -2e*
1 + n xx 1 + n yy z
To obtain the crack |x| < a under tension = T, the interior stresses in the inclusion should be axx = a= 0,
a yy =-T •
xy
From Eqs. (1) the leading terms in the interior stresses as 0 are 1 -v
I
1 -v
4 = -2eXx -ne
yy
I
alyy = -neXx -n<y + O(n2).
(2)
yy
In order for ayy = -T, we set eyy = e ^ n ^ 0, and
ne = T(1 - v)/Moreover, for axx = 0, from Eq. (2) there should exist eigenstrain eXx = -ne^/2. Such eigenstrains were not considered by Mura, but they need to be considered to obtain all stress and displacement fields correctly. We give below the fields for a vertical center of eigenstrain
C = e,eXx =eXy =eL =0, and M4na2) = q obtain-
a^0
ed by means of the Airy stress function
U (r, 0) = -|q/(n(K + 1))(2log r + cos (20)), from which we obtain
3
ux (x, y) =
q
2n(K +1)
(5 -K)4 - 4 ^
Uy (x, y) = (K-1)4 + 4^
r r
ux (X, 0) = -
q(K-1) 1,
2n(K +1) x'
(x, 0) = q S(x)sgn y,
a xx( x, y) = ayy(x, y) =
axy (X, y) =
2|q
n(K + 1) 2|q
n(K + 1)
4|q
2 4
>4 + 8 xr
4 6
r r
-3 -2+12 £L - ^^
r r r
(3)
n(K + 1)
xy - 4 xy
4 4 6
r r
axx (X, 0) = a yy (X, 0) =
2|q 1 n(K + 1) x2
a xy ( x, 0) = 0.
3. The stress field of a crack in tension as a distribution of vertical centers of eigenstrain
The crack stress field is obtained by distributing in the interior of the ellipse (crack) vertical centers of eigenstrain e = e with density proportional to the thickness X2 of
the ellipse, so that q(x) = q0 a2/a1 ^aj2 - x2 so that from Eqs. (3) we have
° XX (X,°) = yy (X,0) =
2i a2 q 0 4a12 ds =
2I 02 K+1 a1
q0
1 -
ijx2 -
-H (| x| -a1)
a1
(4)
where K denotes the Kolosov constant and the integral in the sense of principal value was evaluated according to Kaya and Erdogan [8]. Requiring that on |x| < a1, ayy (X, 0) = = -T, in the limit as a2 ^ 0 and e we set
2|ia2q0/(a1(K +1)) = T, superpose a = T and obtain the total stresses
a xx ( x, 0) = T
-1 +
\jx2 -
-H (| x| - a1)
a1
x
ayy(x, 0) = T i22
■\jx - a1
a xy ( x, 0) = 0.
H (| x| - 01),
(5)
4. The stress field of a crack in shear as a distribution of antisymmetric centers of eigenstrain
For the shear crack centers of eigenstrain exy = e, exx = = e*yy =0 (antisymmetric centers of shear) are distributed similarly with the stress fields derived from the stress function U(r, 0) = 2|qsin(20)/(TC(K+1)) we have the fields
ux( X y) =
Uy (^ y) =
q
n(K + 1)
q
(k + 3)4 - 4y
n(K + 1)
(k + 3)4 - 4 4
Ux (X, 0) = qS(x)sgn y, Uy (x, 0) =
q(K-1) 1, n(K + 1) x'
(6)
aXX (X, y) =
ayy (X, y) = axy (X, y) =
8|q
n(K +1) 8|q
n(K +1)
4|q
n(K + 1) 8|q „,
xy + ^^
3 4 6
r r
^y - 4 xy
4 4 6 r r
2 4
+ 8X-
r r6
a xx( 0) =-78'( x)s8n y.
K + 1
a xy ( x, 0) = 4' ayy ( x, 0) = 0.
n(K +1) x2
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By distributing the centers of shear eigenstrain inside the ellipse and integrating as before we obtain the stresses 1, |x| < flj , Ixl
°x, (x, 0) = -
xx (x, 0) = -
4M-^0 a2
K+1 a1 8|iq0 a2
l-
aT < |x| <
K+1 aj^ja2 -x2 ctyy (x, 0) = 0 and the displacements
l~2 2 ■\jx - aT
x
sgn y, x < aT
(7)
ux (x, 0+ ) - ux (x, 0 ) =
2q0a2
2
x , x < aj,
uv (x, 0)=qq^K-j| x'|x| <aj
(8)
ai K+J [x - yj x2 - aj2 sgn x, aT < x
where we set 4|a/(K + j)(a2l aj)q0 = 5, and we superpose oxy = S to obtain the total stresses for the Griffith shear crack. We may note that uy (x, 0) = x for |x| < aj means that the shear crack rotates.
For completeness we present the fields of a horizontal center of eigenstrain and of the center of dilatation. Horizontal center of expansion:
M4
U =-
n(K + l)
ux (x y) =■
[-2 log r + cos (29)],
q
2n(K + j)
Uy (x, y) = q ux (x,0) =
(K-j)4 + 4 ^
2n(K + j) q(K + 3) j
(5 -K)4 - 4y
Uy (x, 0) =
2n(K + j) x
q(3 -K)
xx (x, y) = xy ( x y) =
CTyy (x, y) =
2(k+1) 2Mq
8( x)sgn y,
n(K + 1)
4^q
n(K + 1) 2^q
1 2 4
1 „x nx
- + 4 -4 - 8 -6
r r r
xy + 4 xy_
3 4 4 6
r r
n(K + 1)
2 4
l^T + 8 ^T
4 6
r r
xx (x 0) = -
CTxy (x, 0) = -
yy (x, 0) =
6|jq 1 n(K +1) x2
2|Xq 8, K+1 2|oq 1
8'( x)sgn y,
n(K +1) x2' Center of dilatation:
U =-(KqT-log r, n(K + 1)
ux (x, y) =
Uy ^ y) = ux (x 0) =
2q x n(K + l) r2 '
2q y
n(K + l) r2 '
2q J n(K +1) x ' 2q
Uy(x, 0) = —i4-8(x)sgny
xx(x, y) =
k+1 4\xq
xy ( x y) =-
n(K + l)
2Mq xy
2
T - 2 x_
2 2 4
r r
n(K + l) r4
yy (x, y) =
4^q
xx (x 0) = -
n(K + l) 4|oq l
2
--L + 2
r 2 + 2 r4
n(K +1) x2^
2|Xq 8'( k+1 4^q l
xy (x, 0) = x)sgn У,
CTyy (x, 0) =-
n(K + 1) x2'
We also note that two dimensional inclusions with eigenstrain were solved by the Airy stress functions in Ref. [9].
5. Symmetric centers of shear do not produce a crack
A symmetric center of shear consists of eigenstrains
e = —e = e e = 0
°xx °yy e °xy
As Eqs. (1) indicate as e ^^ so that eq ^ const, ctxx ^ ^ in the interior of the crack. It is not possible to cancel these stresses by remote tractions as to produce a cavity for the crack. However, for a self-similarly expanding inclusion with transformation strain containing symmetric centers of shear, dynamic fields would be emitted outside and the shear stress would be singular at the tips. The stress ctxx would be symmetric in x, while CTxy would be antisymmetric.
6. Concluding remarks for the Barenblatt crack
A detailed analysis shows how a crack can be obtained as the limit of an Eshelby ellipsoidal (in 2D elliptical) inclusion as the eigenstrain tends to infinitity, the small axis length tends to zero, while their product to a length that gives the crack opening displacement. This is due to the Eshelby constant stress property so that the constant applied traction on the crack faces can be cancelled. This property is true also in the dynamic generalization of the Eshelby inclusion problem where the Burridge and Willis [4] self-similarly expanding elliptical crack is obtained, and at the corresponding axis speed Rayleigh waves, as M waves emitted by the expanding ellipsoid.
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MarkenscoffX. / Физическая мезомеханика 21 6 (2018) 45-48
Cl\ = Cl2
а\ > a2
Fig. 1. Superposition of two elliptical inclusions to produce a crack with vanishing stress at the tip
A question may arise whether a Barenblatt type crack can be produced by distribution of eigenstrains [10]. We may consider the superposed fields of two elliptical inclusions with axes lengths (aT, bT), (a2, b2 ) as in the Fig. 1, so that all stresses in the interior of the inner ellipsoid vanish as to produce a crack. Only in the case aT = a2 we can adjust the eigenstrain densities qT and q2 that the singularity cancels at that point. It would be interesting to obtain the fields and compare with the Barenblatt crack [10].
Acknowledgments
Support of NSF grant CMMI No. 1745960 is acknowledged.
References
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Received November 08, 20l8; revised November 08, 2018, accepted November 15, 2018
Сведения об авторе
Xanthippi Markenscoff, Distinguished Professor, University of California, USA, [email protected]
