Junction problem for rigid and Timoshenko elastic inclusions in elastic ...

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Junction problem for rigid and Timoshenko elastic inclusions in elastic bodies

Mathematics and Mechanics of Solids 1–14 Ó The Author(s) 2015 Reprints and permissions: sagepub.co.uk/journalsPermissions.nav DOI: 10.1177/1081286515594655 mms.sagepub.com

AM Khludnev Lavrentyev Institute of Hydrodynamics of the Russian Academy of Sciences, Novosibirsk, Russian Federation L Faella Cassino University, Department of Mathematics, Cassino, Italy TS Popova North-Eastern Federal University, Department of Mathematics, Yakutsk, Russian Federation Received 14 May 2015; accepted 14 June 2015 Abstract This paper concerns an equilibrium problem for a two-dimensional elastic body with a thin Timoshenko elastic inclusion and a thin rigid inclusion. It is assumed that the inclusions have a joint point and we analyze a junction problem for these inclusions. The existence of solutions is proved and the different equivalent formulations of the problem are discussed. In particular, the junction conditions at the joint point are found. A delamination of the elastic inclusion is also assumed. In this case, the inequality-type boundary conditions are imposed at the crack faces to prevent a mutual penetration between the crack faces. We investigate the convergence to infinity and zero of a rigidity parameter of the elastic inclusion. It is proved that in the limit, we obtain a rigid inclusion and a zero rigidity inclusion (a crack).

Keywords Thin inclusion, rigid inclusion, non-linear boundary conditions, non-penetration, crack, variational inequality, junction conditions.

1. Introduction It is known that the mutual disposal of inclusions inside of composite materials is responsible for stresses and possible damage. In particular, one of the objectives in solid mechanics is using different inclusions to reach suitable deformations. The influence of material properties of inclusions with respect to stresses is also a great mathematical problem. We can distinguish between elastic and rigid inclusions and inclusions of zero rigidity (cracks). Rigid inclusions (called ‘stiffeners’ or ‘unticracks’), as well as elastic inclusions, are important objects in constructing composite materials. Cracks can also be seen as inclusions of zero rigidity [1–3]. Considering cracks, we have in mind high-level mathematical models Corresponding author: AM Khludnev, Lavrentyev Institute of Hydrodynamics of the Russian Academy of Sciences, Lavrentyev pr. 15, Novosibirsk 630090, Russian Federation. Email: [email protected]

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with unknown sets of a contact between the crack faces. On the other hand, it is possible to consider different thin elastic inclusions depending on the mathematical model accepted. In this paper we consider a Timoshenko thin inclusion which is considered to be more precise than the Euler–Bernoulli inclusion, since it takes into account a rotation angle of the inclusion. This means that the mechanical behavior of the elastic inclusion is modeled by the Timoshenko equilibrium equations, see the suitable results in [4, 5]. We should remark at this point that there are different approaches used to describe elastic inclusions (without delaminations) such as, in particular, those given in [6–8]. It is known that classical linear crack theory in elasticity is characterized by the linear boundary conditions imposed at the crack faces, see [9–11]. Such a linear approach allows the opposite crack faces to penetrate each other, which leads to inconsistency from the mechanical standpoint. In recent years a crack theory with non-penetration conditions has been under active investigation. This theory is characterized by the inequality-type boundary conditions at the crack faces and it leads to free boundary value problems. Khludnev and Kovtunenko [12] present results for crack models with the non-penetration conditions for a wide class of constitutive laws, see also [13–16]. The existence theorems and qualitative properties of the solutions for the equilibrium problems of elastic bodies with rigid inclusions can be found in [17–22], and the elastic behavior of bodies with cracks and rigid inclusions is analyzed in the book [23]. To describe suitably composite materials it is necessary to analyze high-level mathematical models for elastic bodies with thin elastic and rigid inclusions and cracks. The inclusions considered may be delaminated, and hence the crack approach with non-penetration conditions is to be used. In such a case, new types of non-linear boundary conditions appear which lead to free boundary value problems being correct from the mechanical standpoint. In the framework of similar models, many results have now been obtained [23]. This paper provides a rigorous analysis of a junction problem for elastic bodies with a thin rigid and delaminated thin Timoshenko inclusion. The free boundary approach is used to describe an equilibrium state of the structure with inequality-type boundary conditions at the crack faces. In particular, the junction conditions are found, an existence of the solutions is proved and the equivalent problem formulations are analyzed. We refer the reader to [24–31] where junction problems for different models are discussed. This paper is organized as follows. In Section 2 we discuss the problem formulation and find the junction conditions at a joint point. Section 3 is devoted to the analysis of a passage to a limit when a rigidity parameter of the elastic inclusion goes to infinity. A delaminated elastic inclusion is investigated in Section 4 and Section 5 is concerned with a passage to zero of the rigidity parameter of the elastic inclusion. In particular, we prove that at the limit a model of an elastic body with inclusion of zero rigidity which precisely corresponds to a crack is obtained.

2. Setting up of the problem r  O: Let O  R2 be a bounded domain with a Lipschitz boundary G and g r a smooth curve, such that g e  O. Let g = g e[g r[ {(0, 0)} be a smooth curve such that Denote g e = (21,0) × {0} assuming that g an angle between g e and g r is zero at the point (0, 0), see Figure 1. Denote by n = (n1, n2) a unit normal . vector to g; t = (n2, 2n1), and put Og = O n g

Figure 1. Elastic body with inclusions g e ; gr .

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In what follows, the domain Og represents a region with an elastic material and g e and g r are thin elastic and rigid inclusions, incorporated in the elastic material, respectively. In particular, we consider g e as a Timoshenko beam. Let B = {bijkl}, i, j, k, l = 1,2, be a given elasticity tensor with the usual properties of symmetry and positive definiteness: bijkl = bjikl = bklij , i, j, k, l = 1, 2, bijkl 2 L‘ (O), bijkl jij jkl  c0 jjj2 ,

8jji = jij , c0 = const.0:

The summation convention over repeated indices is used and all functions with two lower indices are assumed to be symmetric in these indices. Introduce the space of infinitesimal rigid displacements R(g r ) = fr = (r1 , r2 ) j r(x1 , x2 ) = c(  x2 , x1 ) + (c1 , c2 ), (x1 , x2 ) 2 gr g,

where c, c1, c2 are arbitrary constants and denote W = f(u, v, w, u) 2 H01 (O)2 × H 1 (ge ) × H 1 (ge ) × H 1 (ge ) j v = un , w = ut on g e ; ujgr = (r1 , r2 ) 2 R(gr ); u(0) =  r2x (0)g:

The usual notation for H01 (O) and H1(g e) are used for the Sobolev spaces, un = un and ut = ut. We identify the functions defined on g e with functions of the variable x : x = x1, hx = dh/dx and (x1, x2) 2 g e. The last condition in the definition of W shows that the angle between g e and g r is fixed, i.e. there is no break between g e and g r at the point (0, 0). Now we are able to formulate an equilibrium problem for the elastic body Og with thin inclusions g e, g r. For the given external forces f = (f1, f2) 2 L2(O)2 acting on the body, we have to find the functions u = (u1, u2), v, w, u, s = {sij}, i, j = 1, 2, and r0 2 R(g r) such that  div s = f s  Be(u) = 0

in Og , in Og ,

ð1Þ ð2Þ

 vxx  ux = ½sn  on ge ,

ð3Þ

 wxx = ½st  on ge ,

ð4Þ

 uxx + vx + u = 0 (u, v, w, u) 2 W ; u = r0

on g e , on g r ,

vx + u = ux = wx = 0 for x =  1, Z  ½snr + (wx r1 )(0)  (ux r2x )(0) gr

ð5Þ ð6Þ ð7Þ

ð8Þ

+ ((vx + u)r2 )(0) = 0, 8r = (r1 , r2 ) 2 R(gr ):

Here [h] = h + 2 h2 is a jump of a function h on g, where h6 are the traces of h on the faces g 6, the signs 6 correspond to positive and negative directions of n, e(u) = {eij(u)} is the strain tensor, eij(u) = 1/2(ui,j + uj,i), i, j = 1,2, sn = (s1jnj, s2jnj), sn = sijnjni and st = sijnjt i. The function u = (u1, u2) describes a displacement field of the elastic body, the functions w, v fit to displacements of the inclusion g e along the x1-axis and x2-axis, respectively, and the function u describes a rotation angle of the inclusion g e. Observe that a part of the boundary conditions for the functions u, v, w is included in the condition (u, v, w, u) 2 W.

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Relations (1) and (2) are the equilibrium equations for the elastic body and Hooke’s law and equations (3)–(5) are the Timoshenko equilibrium equations for the elastic inclusion g e. According to the condition (u, v, w, u) 2 W, the vertical (along the axis x2) and tangential (along the axis x1) displacements of the elastic body coincide with the inclusion displacements at g e. The right-hand sides of equations (3) and (4), [sn], and [st], describe the forces acting on g e from the surrounding elastic media. The non-local boundary condition (8) guarantees an equilibrium state for the rigid inclusion g r, i.e. the principal vector of forces and the principal moment of forces acting on g r are equal to zero. This can easily be checked by taking into account a structure of functions r involved in identity (8). Indeed, denoting (sn)6 by (s1, s2)6 on g 6 r , the condition (8) can be rewritten in the following form: Z Z 1 ½s  = wx (0), ½s2  = (vx + u)(0), ð9Þ gr

gr

Z

(½s2 x1  ½s1 x2 ) =  ux (0):

ð10Þ

gr

We can also write junction conditions included in the definition of W, w(0) = r01 (0), v(0) = r02 (0), u(0) =  r02x (0),

ð11Þ

and consider equations (9)–(11) as a complete system of junction conditions at the joint point x = 0. Consequently, the non-local condition (8) can be seen as a part of the junction conditions. It is clear that equation (9) provide zero forces applied to g r and equation (10) provides a zero moment acting on g r. Conditions (11) are kinematical ones. Remark. A zero angle between g e and g r at the point x = 0 is considered for simplicity. For a general case, denote by n = (n1, n2) a unit normal vector to g r, see Figure 2. Let s = (n2, 2n1) be a tangent vector on g r, and instead of the last condition in equation (11) we require that u(0) = 

d (rn)(0): ds

ð12Þ

Condition (12) should be included in the definition of the space W, and the identity (8) in this case should be changed by Z d  ½snr + (wx r1 )(0)  (ux (rn))(0) ds gr

+ ((vx + u)r2 )(0) = 0, 8r = (r1 , r2 ) 2 R(gr ):

We should note that the problem (1)–(8) admits a variational formulation; indeed, consider the energy functional

Figure 2. Non-zero angle between ge ; gr .

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p1 (u, v, w, u) =

1 2

Z Og

+

1 2

s(u)e(u) 

Z fu Og

Z

fw2x + u2x + (vx + u)2 g:

ge

Here s(u) = s is defined by equation (2), i.e. s(u) = Be(u), and, for simplicity, we write s(u)e(u) = sij(u)eij(u) and fu = fiui. Consider the following minimization problem. Find (u, v, w, u) 2 W such that p1 (u, v, w, u) = inf p1 : W

ð13Þ

This problem has a unique solution satisfying the identity (u, v, w, u) 2 W , Z Z Z  s(u)e( u)  f u + f(vx + u)(vx + u) Og

ð14Þ

ð15Þ

ge

Og

 x g = 0,  x + ux u + wx w

 2 W:  , u) 8 (u, v, w

To prove that problem (13) has a solution, it is sufficient to establish a coercivity of the functional p1 on the space W, since its weak lower semicontinuity is obvious. We omit this proof since the proof is similar to that given in [4]. In what follows we show an equivalence between the equations (1)–(8) and equations (14) and (15) for smooth solutions. Theorem 1. The problem formulations (1)–(8) and (14) and (15) are equivalent, provided that the solutions are smooth.  2 W and multiply equations (1) and (3)–(5) by  , u) Proof. Let equations (1)–(8) be fulfilled, take (u, v, w  respectively. Integrating over Og and g e, respectively, we get  , u, u, v, w Z Z    (uxx  vx  u)ug (  div s  f ) u+ f(vxx + ux )v  wxx w Og



Z

ge

(½sn v + ½st  w) = 0:

ge

Hence, Z

(s(u)e( u)  f  u) +

Og

x g  + ux u

Z

Z

½snu +

g

(½sn v + ½st  w) 

Z

 + wx w x f(vx + u)(vx + u)

ge

(vx + u)vj01

ð16Þ 

 j01 wx w



 01 = 0: ux uj

ge

 2 W , the boundary conditions (7)  , u) We have that ½snu = ½sn un + ½st ut on g e. Accounting for (u, v, w and the identity (8), then the identity (15) follows from equation (16). Conversely, let equations (14) and (15) be fulfilled. We take test functions of the form  = (c, 0, 0, 0), c 2 C0‘ (Og )2 . It gives the equilibrium equation (1). Next, from equation (15) it  , u) ( u, v, w follows that

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Z

½sn u+

g

Z

 + (vx + u)ug    uxx u f(vxx + ux )v  wxx w ð17Þ

ge

 j01 + wx w

 01 + ux uj

+ (vx + u)vj01

 2 W:  , u) = 0; 8(u, v, w

 = 0 at x = 21,0, the following relation is derived:  = v = u Choosing here w Z Z Z  ½sn  un  ½st ut  ½snu ge



Z

ge

gr

ð18Þ

  (vx + u)ug  = 0:  + uxx u f(vxx + ux )v + wxx w

ge

 2 W , from equation (18), equations (3)–(5) follow.  , u) Consequently, by the condition that (u, v, w From (17) we obtain that Z  j01  ½sn u + wx w ð19Þ gr  01 + (vx + u)vj01 = 0, 8(u, v, w  2 W:  , u) + ux uj

In particular, the identity (19) provides vx + u = ux = wx = 0 as x =  1

and thus from equation (19) we derive that Z   )(0) + (ux u)(0)þ  ½sn u + (wx w gr

 2 W,  , u) + ((vx + u)v)(0) = 0, 8(u, v, w

which coincides with equation (8). Hence, the equivalence of equations (1)–(8) and equations (14) and (15) is established. Theorem 1 is proved.

3. Rigidity tends to infinity In practical situations a solution of the problem (1)–(8) depends on the rigidity parameter of the elastic inclusion. In the model (1)–(8), this parameter is taken to be equal to one. In this section we introduce a parameter d . 0 to the model and analyze its passage to infinity. The parameter d characterizes a rigidity of the elastic inclusion g e. To this end, the energy functional and the minimization problems are considered with d . 0: Z Z Z 1 d pd (u, v, w, u) = s(u)e(u)  fu + f(vx + u)2 + w2x + u2x g: 2 2 Og

Og

ge

There exists a unique solution of the following problem. Find (ud , vd , wd , ud ) 2 W such that pd (ud , vd , wd , ud ) = inf pd : W

The solution of this minimization problem satisfies the identity (ud , vd , wd , ud ) 2 W ,

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ð20Þ

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Z

d

s(u )e( u) 

Og

Z

Z f u + d

 f(vdx + ud )(vx + u) ð21Þ

ge

Og

 x g = 0, 8(u, v, w   , u)  x + ud u + wdx w

2 W:

Our aim is to pass to the limit in equations (20) and (21) as d! + N. From equation (21) it follows, with a 2 R, that Z Z Z d d d s(u )e(u )  fu 6a ((wd )2 + (vd )2 ) Og

Z

+d

ge

Og

f(vdx

d 2

+u )

ð22Þ

+ (wdx )2 + (udx )2 g = 0:

ge

For small a . 0, due to vd = udn , wd = udt on g e and the imbedding theorems, the following relation holds: Z Z 1 s(ud )e(ud )  a ((wd )2 + (vd )2 )  0: 2 ge

Og

Consequently, from equation (22) and Lemma 1, we obtain, as d  a, that  2  2   c0 ud H 1 (O)2 + a(vd , wd , ud )H 1 (g )3  c1 ud 1, Og , c0 .0: e

0

Hence, uniformly for d  d0,    d 2 u  1 2 + (vd , wd , ud )2 1 3  c: H (O) H (g )

ð23Þ

On the other hand, the relation (22) implies for d  d0, Z d f(vdx + ud )2 + (wdx )2 + (udx )2 g  c:

ð24Þ

e

0

ge

From equations (23) and (24), we can assume that, as d! + N, weakly in H01 (O)2 ,

ð25Þ

weakly in H 1 (ge ), vx + u = 0 in ge ,

ð26Þ

ud ! u vd ! v

wd ! w weakly in H 1 (ge ), wx = 0 in g e ,

ð27Þ

ud ! u weakly in H 1 (g e ), ux = 0 in g e :

ð28Þ

In particular, v(x) = c0 + c1 x, w(x) = b0 , u(x) =  c1 , x 2 (  1, 0); b0 , c0 , c1  const:

Denote rd = ud jgr , rd 2 R(g r ); we can assume that rd ! r0 almost everywhere on g r ,

and thus

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ð29Þ

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r0 (x1 , x2 ) = b(  x2 , x1 ) + (c1 , c2 ), (x1 , x2 ) 2 g r ; b, c1 , c2  const:

From equation (29) we derive that ujge = (w, v) = r1 ; r1 (x1 , x2 ) = c1 (  x2 , x1 ) + (b0 , c0 ), (x1 , x2 ) 2 g e :

Meanwhile (u, v, w, u) 2 W, i.e. u(0) =  r02x (0) and r1(0) = r0(0); thus (b0, c0) = (c1, c2),c1 = b. Consider the space Wr = fu 2 H01 (O)2 jujg 2 R(g)g,

where R(g) = fr = (r1 , r2 ) j r(x1 , x2 ) = d(  x2 , x1 ) + (d 1 , d 2 ), (x1 , x2 ) 2 gg, d, d 1 , d 2  const:

Consequently, ujg = r0 2 R(g) and r0(x1, x2) = b(2x2, x1) + (c1, c2), (x1, x2) 2 g.   , u). We take  u 2 Wr : Then (u, u2 jge , u1 jge ,  (u2 jge )x ) 2 W : Denote (u, u2 jge , u1 jge ,  (u2 jge )x ) by (u, v, w  in equation (21) as a test function and pass to the limit as d!N. In the limit, we  , u) Substitute ( u, v, w obtain that Z Z u 2 Wr , s(u)e(u)  f u = 0; 8 u 2 Wr , ð30Þ Og

Og

where u is taken from equation (25). Problem (30) admits an equivalent differential formulation: find displacements u = (u1, u2), a stress tensor s = {sij},i, j = 1, 2, and r0 2 R(g) such that  div s = f , s  Be(u) = 0

in Og ,

u = 0 on G; u = r0 on g, Z ½snr = 0, 8r 2 R(g):

ð31Þ ð32Þ ð33Þ

g

Thus we have proved the following statement. Theorem 2. The solutions of the problem (20) and (21) converge to the solution of (30) in the sense (25)– (28) as d!N. The model (31)–(33), or (30) describes an equilibrium state for the elastic body with the rigid inclusion g. This means that in the limit we have only rigid inclusion g with the displacement field r0. The identity (33) provides the equilibrium conditions for the rigid inclusion g. At the end of this section we formulate the statement used in the proof of Theorem 2. Lemma 1. There exists a constant c . 0 such that Z fv2 + w2 + w2x + u2x + (vx + u)2 g ge

 ck(v, w, u)k2H 1 (ge )3 8(v, w, u) 2 H 1 (ge )3 :

The proof of Lemma 1 is given in [4].

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4. Delaminated elastic inclusion In this section we assume that the elastic part g e of the inclusion g is delaminated. This means that we have a crack located between g e and the elastic body. To fix a situation, the delamination is assumed to be at the positive side of g e. In this case, the displacements v,w of the inclusion should coincide with the displacements of the elastic body at g  e : In our model, the inequality-type boundary conditions are considered at the crack faces to prevent a mutual penetration between the faces. First of all, introduce a set of admissible displacements K = f(u, v, w, u) 2 HG1 (Oge )2 × H 1 (ge ) × H 1 (ge ) × H 1 (ge ) j ½un  0,  v = u n , w = ut on g e ; ujgr = (r1 , r 2 ) 2 R(g r ), u(0) =  r 2x (0)g,

e and where Oge = O n g HG1 (Oge ) = ff 2 H 1 (Oge ) j f = 0 on Gg:

Notice that the inequality [u]n  0 included in the definition of K provides a mutual non-penetration between the crack faces g 6 e . An equilibrium problem for the elastic body with a delaminated inclusion g e and the rigid inclusion g r can be formulated as follows. We have to find functions u = (u1, u2),v, w, u, s = {sij}, i, j = 1,2, and r0 2 R(g r) such that  div s = f s  Be(u) = 0

ð34Þ

in Og ,

ð35Þ

in Og ,

 vxx  ux = ½sn  on ge ,

ð36Þ

 wxx = ½st  on ge ,

ð37Þ

 uxx + vx + u = 0

ð38Þ

on g e ,

(u, v, w, u) 2 K; u = r0

ð39Þ

on gr ,

+ + s+ n  0, s t = 0, sn ½un  = 0

on ge ,

ð40Þ

vx + u = ux = wx = 0 for x =  1, Z  ½snr + (wx r1 )(0)  (ux r2x )(0)

ð41Þ

ð42Þ

gr

+ ((vx + u)r2 )(0) = 0; 8r = (r1 , r2 ) 2 R(gr ):

We remark that the problem (34)–(42) admits a variational formulation. Indeed, denote by (u, v, w, u) a solution of the following minimization problem. Find (u, v, w, u) 2 K such that p1 (u, v, w, u) = inf p1 : K

ð43Þ

This problem has a unique solution satisfying the variational inequality Z Og

s(u)e( u  u) 

Z Og

(u, v, w, u) 2 K, Z   vx  u) f (u  u) + f(vx + u)(vx + u

ð45Þ

ge

 x  ux )g  0, + wx ( wx  wx ) + ux (u

ð44Þ

 2 K:  , u) 8 (u, v, w

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Theorem 3. Problem formulations (34)–(42) and equations (44) and (45) are equivalent provided that the solutions are smooth.  2 K and multiply equations (34)–(38) by  , u) Proof. Let equations (34)–(42) hold. We choose (u, v, w   u, respectively. Integrating over Og and g e we obtain that    w and u u  u, v  v, w Z Z (  divs  f )( u  u)  f(vxx + ux )(v  v) + wxx ( w  w) ge

Og

  u)g  + (uxx  vx  u)(u

Z

(½sn (v  v) + ½st ( w  w)) = 0:

ge

Hence, integrating by parts gives Z Z ðs(u)e( u  u)  f (u  u)) þ ½sn(u  u) g

Og

Z +

ð46Þ

  vx  u) + wx (  x  ux )) ((vx + u)(vx + u wx  wx ) + ux (u

ge



Z

(½sn (v  v) + ½st ( w  w))

ge

  u)j01 = 0: w  w)j01  ux (u (vx + u)(v  v)j01  wx (

First note that Z

½sn( u  u) =

g

Z

½sn ( un  un ) +

ge

Z

½st (ut  ut ) +

ge

Z

½sn(u  u):

gr

We see that by taking into account the boundary conditions in equation (41) and the second conditions from equation (40) we can derive the variational inequality (45) from equation (46), provided that, in equation (46), the following inequality holds: Z Z + sn (½ un   ½un ) + ½sn(u  u)  (vx + u)(v  v)(0) ð47Þ ge gr   u)(0)  wx ( w  w)(0)  0: ux (u

According to equation (42), Z

  u)(0) ½sn( u  u)  (vx + u)(v  v)(0)  ux (u

gr

wx ( w  w)(0) = 0,

and due to boundary conditions (40), the validity of equation (47) is clear. Hence from equation (46) the variational inequality (45) follows. Conversely, let equations (44) and (45) be fulfilled. First, it is easy to derive the equilibrium equation (34) from equations (44) and (45); however, we omit the details here. Note that it is sufficient to check equation (42). All the rest of the equations and boundary conditions  = (u, v, w, u)6(~u, ~v, w ~  , u) ~ , u) can be verified as those given in [4]. We substitute the test functions (u, v, w

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~ 2 K, ½~u = 0, ~un = ~v, ~ut = w ~ =  r2x , as ~ , u) ~ on g e ; ~ujgr = r 2 R(g r ), u in equation (45), where (~u, ~v, w x = 0, giving Z Z Z ~ x g = 0: ~ + wx w ~ x + ux u s(u)e(~ u)  f~ u+ f(vx + u)(~vx + u) Og

ge

Og

Hence, integrating here by parts, we obtain Z ~ ~  (uxx  vx  u)ug f(vxx + ux )~v  wxx w ge



Z

ð48Þ ~ j01 + ½sn~ u + (vx + u)~vj01 + wx w

~ x j01 = 0: + ux u

g

~ and consequently it coincides ~ , u) The identity (48) holds for all of the above-mentioned functions (~u, ~v, w with equation (17). We next repeat the arguments following equation (17), and from equation (48), equation (42) follows. The proof of Theorem 3 is now complete. Note that junction conditions at the point x = 0 for the problem (34)–(41) coincide with equations (9)–(11).

5. Rigidity goes to zero In this section we consider the delaminated elastic inclusion g e with the rigidity parameter d . 0 and justify a passage to the limit as the parameter goes to zero. In this case, instead of equations (44) and (45), for any fixed d . 0, we have the following variational inequality: (ud , vd , wd , ud ) 2 K, Z Z Z  s(ud )e( u  ud )  f (u  ud ) + d f(vdx + ud )(vx + u Og

ð50Þ

ge

Og

ð49Þ

 x  udx )g  0, 8 (u, v, w  2 K:  , u) wx  wdx ) + udx (u vdx  ud ) + wdx (

Our aim is to analyze a passage to the limit in the problem of equations (49) and (50) as d! 0. From equations (49) and (50) we know that Z Z Z s(ud )e(ud )  fud + d f(vdx + ud )2 + (wdx )2 + (udx )2 g = 0: ð51Þ Og

Og

ge

Hence, we have an estimate uniform in d,  d 2 u  1

HG (Oge )2

 c:

ð52Þ

On the other hand, the relation (51) implies for all d that Z d f(vdx + ud )2 + (wdx )2 + (udx )2 g  c:

ð53Þ

ge

By equation (52), we obtain that Z Z Z Z 2 d 2 d 2 d 2 (v ) = (un )  c, (w ) = (ud t )  c: ge

ge

ge

ge

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ð54Þ

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Mathematics and Mechanics of Solids

Hence, in view of equation (53) and Lemma 1, the following estimate holds for d  d0:  2  2  2 dvd H 1 (g ) + dwd H 1 (g ) + dud H 1 (g )  c: e

e

e

We can assume that, as d! 0, ud ! u

weakly in HG1 (Oge )2 , ujgr = r0 2 R(gr ), pffiffiffi d dv ! ~v weakly in H 1 (ge ), pffiffiffi d ~ weakly in H 1 (g e ), dw ! w pffiffiffi d ~ weakly in H 1 (g e ): du ! u

ð55Þ ð56Þ ð57Þ ð58Þ

By equations (55)–(58), a passage to the limit in equations (49) and (50) is possible. Define a set of admissible displacements for a limit problem Kr = fu 2 HG1 (Oge )2 j ½un   0 on ge ; ujgr 2 R(gr )g:   2 Kr , ujgr = (r1 , r2 ), such that u We choose u u u n,  t are smooth at g e , ( n )x (0) = r2x (0), and define the     2 K with u  =  (un jge )x on g e, and a substitution of this  = ut on g e. Then (u, v, w  , u) functions v =  un , w test function in equation (50) implies that Z Z s(ud )e( u)  f (u  ud ) Og



Z

Og d

d

s(u )e(u ) + d

 f(vdx + ud )2  (vdx + ud )(vx + u)

ge

Og

+ (wdx )2

Z

 x g:  x + (udx )2  udx u  wdx w

Taking the lower limit as d! 0 in both parts of this inequality, we derive that Z Z s(u)e(u  u)  f (u  u)  0: Og

ð59Þ

Og

The inequality (59) holds for all functions u 2 Kr , ujgr = (r1 , r2 ), such that un , ut are quite smooth at g  e and ( u u 2 Kr . To prove this statement it is sufficient n )x (0) = r2x (0). We state that it will be valid for all  to check that for any given u 2 Kr we can choose a sequence un from Kr converging to u in HG1 (Oge )2 , n unt are smooth on g  un u 2 Kr be any fixed function such such that  unn ,  e and ( n )x (0) = r2x (0). Indeed, let  that  u = r on g r , r 2 R(g r ). We divide the domain Og into two subdomains O1 and O2, as shown in Figure 3. Consider the restriction ujO2 2 H 1 (O2 )2 and extend this function to O as a function from H01 (O)2 . Denote this extension by v, note that v = r on g r and set ~u = u  v. It is clear that ½~un   0 on g e 2 n ‘ and ~ u = 0 in O2, and thus ~un = 0, ~ut = 0 at g  e . Next we choose a sequence v 2 C0 (O) such that

Figure 3. Division of domain O into two subdomains.

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Khludnev et al.

13

vn = r on gr ; vn ! v strongly in H01 (O)2 :

In this case ~ u + vn ! u strongly in HG1 (Oge )2 : n On the other hand, u~ + vn 2 Kr and u~n + vnn , ~ut + vnt are smooth functions at g  u 2x (0). e , (~ n + vn )x (0) = r Hence, we have proved the statement. Hence, the limit function u from equation (55) satisfies the variational inequality:

Z

u 2 Kr , Z s(u)e( u  u)  f (u  u)  0, 8u 2 Kr :

Og

ð60Þ ð61Þ

Og

Thus we have proved that the limit problem for (49) and (50) as d! 0 coincides with the problem describing the equilibrium of the elastic body with the crack g e and the rigid inclusion g r. Therefore, we have proved the following theorem. Theorem 4. The solutions of the problem of equations (49) and (50) converge to the solution ofequations (60) and (61), in the sense (55)–(58), as d! 0. The problem (60) and (61) has an equivalent differential formulation for smooth solutions. Find the functions u = (u1, u2), s = {sij}, i, j = 1, 2, defined in Og, and r0 2 R(g r) such that  div s = f s  Be(u) = 0 u=0

in Og , in Og ,

on G; u = r0 on g r ,

6 ½un   0, s6 n  0, ½sn  = 0, s t = 0, sn ½un  = 0 on g e , Z ½snr = 0, 8r 2 R(g r ):

ð62Þ ð63Þ ð64Þ ð65Þ ð66Þ

gr

We do not provide a proof of the equivalence of the problems of equations (60) and (61) and equations (62)–(66). The model (62)–(66) takes into account a mutual non-penetration between the crack faces g6 e , see the first inequality in (65). We refer the reader to [12, 23] and many other papers where different crack problems with non-penetration conditions are analyzed.

6. Conclusion In this paper, we have analyzed a junction problem for a thin Timoshenko inclusion and a thin rigid inclusion located inside of an elastic body. The thin inclusions have a joint point and no break was assumed at this point. In the frame of the free boundary approach, both the differential and variational problem formulations were analyzed and a set of junction conditions was found. In particular, we proved that the differential setting of the problem is characterized by a non-local boundary condition describing an equilibrium state for the rigid inclusion. This boundary condition provides, in fact, a part of the junction conditions. We investigated passages to zero and infinity of a rigidity parameter of the Timoshenko inclusion and the limit models were analyzed. Conflict of interest None declared.

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Funding This work was supported by Gruppo Nazionale per l’Analisi Matematica, la Probabilita` e le loro Applicazioni of the Istituto Nazionale di Alta Matematica (grant number U 2014/000851).

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