An equilibrium problem for the Timoshenko-type plate containing a crack on the boundary of a rigid inclusion Текст научной статьи по специальности «Физика»

УДК 539.311

An Equilibrium Problem for the Timoshenko-type Plate Containing a Crack on the Boundary of a Rigid Inclusion

Nyurgun P. Lazarev*

North-Eastern Federal University, Belinsky, 58, Yakutsk, 677891; Institute of Hydrodynamics, SB RAS, Lavrentyeva, 15, Novosibirsk, 630090

Russia

Received 29.02.2012, received in revised form 10.06.2012, accepted 20.09.2012 An equilibrium problem for an elastic Timoshenko type plate containing a rigid inclusion is considered. On the interface between the elastic plate and the rigid inclusion, there is a vertical crack. It is assumed that at both crack faces, boundary conditions of inequality type are considered describing a mutual nonpenetration of the faces. A solvability of the problem is proved, and a complete system of boundary conditions is found. It is also shown that the problem is the limit one for a family of other problems posed for a wider domain and describing an equilibrium of elastic plates with a vertical crack as the rigidity parameter goes to infinity.

Keywords: crack, Timoshenko-type plate, rigid inclusion, energy functional, mutual non-penetration condition.

Introduction

The interest in studying mathematical models of bodies containing rigid inclusions is due to the wide application of composites. In [1-5], some problems of the theory of elasticity for bodies with cracks and rigid inclusions are considered. It is known that different problems in regard to bodies containing rigid inclusions may be successfully formulated and studied by using variational methods [6-13]. In particular, the theory of two-dimensional problems of the theory of elasticity with thin rigid inclusions and possible delamination is given in [6]. The three-dimensional case is considered in [7].

In studying the deformation of plates and shells, the Kirchhof-Love and Timoshenko-type models [14,15] are successfully used. A great number of papers [8-13] is devoted to investigating the models of the Kirchhof-Love plates containing rigid inclusions. In the [8-12] assumed that the rigid inclusion in a plate volumetric. In [13] the thin rigid inclusion is modeled by a plane two-dimensional curve.

In the present work, the equilibrium problem of elastic transversally isotropic Timoshenko-type plate containing a through crack on the boundary of a rigid inclusion is investigated. The unique solvability of the variational problem on the equilibrium of the plate with a crack is proved. The system of boundary-value conditions, which defines the equivalent boundary-value problem, is obtained from the variational formulation of the problem. For the equilibrium problem of the elastic Timoshenko-type plate with a crack, it is shown out that the equilibrium problem of the plate with a rigid inclusion is obtained as the rigidity parameter tends to the infinity at some fixed part of the plate.

* [email protected] © Siberian Federal University. All rights reserved

1. Formulation of the Problem

Consider the bounded domain Q c R2 with a smooth boundary r. Let the subdomain u be strictly inside Q, i.e., u n r = 0 and let its boundary S be sufficiently smooth. Let us consider that Sconsists of two parts 7 and S\y, moreover, d7 /7 (Fig. 1).

The thickness of the plate is considered to be constant and equal to 2h. We define a threedimensional Cartesian space {x1;x2,z} such that the set {QY} x {0} c R3 corresponds to the middle plane of the plate, where QY = Q\y. Here, the curve 7 defines a crack (a cut) in the plate. This means that the cylindrical surface of the through crack may be given by the relations x = (x1;x2) G 7, — h < z < h, where |z| is the distance to the middle plane. According to our arguments, the rigid inclusion is specified by the set u x [—h, h], i.e. the boundary of the rigid inclusion is given by the cylindrical surface S x [—h, h]. Elastic part of the plate corresponds to the domain Q\u.

Denote by x = x(x) = (U, u) the vector of displacements of points of the mid-surface (x G Qy , U = (u1,u2) are the displacements in the plane {x1;x2}, and u are the displacement along the axis z. We denote the angles of rotation of a normal fiber by ^ = ^>(x) = (^1,^2) ( x G Qy ). Let us consider that the values x and ^ are infinitesimal by the classical theory of elasticity. In accordance with the direction of the outer (with respect to u) normal n = (n1; n2) to S, there is a positive face S+ and a negative face S- of the curve S. If the trace of a function v is chosen on the positive (from the side of the domain Q\u) face S+, we use the notation v+ = v| 3+ and if it is chosen on the negative face, then v- = v| 3- . The jump [v] of the function v on the curve S is found by the formula [v] = v| 3+ — v| 3-. Analogous notation is used for 7+ and y-.

Introduce the tensors describing the deformation of the plate

11 i)

£ij(^) = 2(^} (U) = 2(u,j +u,j )j i,j =12 (v,i = )'

The tensors of moments m(0) = {mij(0)} and stresses a(U) = {aij-(U)}, are expressed by the formulas (summation is performed over repeated indices)

mij (^) - aijrl£rl(^), Oij (U) - 3h aijrl £rl(U)? l - 1? 2?

where the nonzero components of elasticity tensor A = {aijrl} are expressed by the relations

aiiii D? aiijj D* aijij aijji D(1 *)/2, i = ^ j 1? 2?

where D and * are the constants: D is a cylindrical rigidity of the plate, * is the Poisson ratio, 0 < * < 1/2. The transverse forces in the Timoshenko-type model are specified by the expressions

qi(u,^) = J(u,i +^i), i = 1, 2, (1)

where J = 2k Gh, k2 is the shear coefficient, G is the shear modulus in areas perpendicular to the middle plane of the plate, and J, k2, G are the constants. Let BM(■, ■) be a bilinear form defined by the equality

Bm(£,n) = (mij(^),£ijW)M + J((u,i +^i), (v,i +^)}m + (<? ij(U),£ij(V)}m,

where (■, •)M is the scalar product in L2(M), for the subdomain M c Q, £ = (U, u, ^), n = (V, v, ^). The potential energy functional of the deformed plate occupying the region QY has the form

n(£) = 1 b^y(£,£) — </,£)Qy, £ = (U,u,^),

where / = (/1; /2, /3, ^1; ^2) G L2(Qy)5 is the vector specifying the external loads [15]. Introduce the Sobolev spaces

H 1,0(Q7) = {u G H 1(QY) u = 0 a.e. onrj , H = H 1l0(Q7)5

IlH'

Let £ G H. We write the relations for £ conditioned by the presence of rigid inclusion and

crack in the plate. It is known that deformations equal zero for a rigid body. With respect to

the considered rigid inclusion in the plate of the Timoshenko-type model, this means that in the domain u tangential strains eij(U), i, j = 1,2, bending strains eij(^), i, j = 1,2, and transverse strains u,i +^i, i =1,2 equal zero. Consequently, the restriction of the function £ to the domain u has the given structure (for example, see [16]):

U = p, ^ = d, ^ + Vu = 0 on u, p, d G R(u), (2)

where the space R(u) is defined as follows

R(u) = {p = (p1, P2) | p(x) = Bx + C, x G u},

B = ^ ^ , C = (c1, c2); 6, c1, c2 = const.

In component-wise notation, two last equalities in (2) take the form

u,i +^i = 0, i = 1, 2, on u,

^1 = d1 = 6x2 + c1 on u,

^2 = d2 = —6x1 + c2 on u.

This implies that u,12 = 6 and u,21 = —6. This means that 6 = 0 and the restriction of components of the function £ satisfies the relations

U = p, u = l, ^ + Vu = 0 on u, p G R(u), l G L(u), (3)

where

L(u) = {l | l(x) = ao + a1x1 + a2x2, ai G R, i = 0,1, 2, x = (x1, x2) G u}.

In turn, it is possible to define that equalities (3) follow from (2).

Derive the nonpenetration conditions of the crack faces. Since the dependence of horizontal displacements U(z) on z for the Timoshenko-type model is expressed by the following formulas [15]:

ui(z) = ui + z^i, i = 1, 2, |z| ^ h, U(z) = (u1(z), u2(z)),

after scalar product of the jump of the displacement vector [x(z)] = ([u1 (z)], [u2(z)], [u]) by the normal to the vertical crack surface with coordinates (n1,n2,0), we have

([u1(z)], [u2(z)], [u]) ■ (n1, n2, 0) = [U(z)] ■ n = [U] ■ n + z[^] ■ n ^ 0 |z| ^ h, on 7.

Hereinafter, by ” ■ ” denote the operation of scalar product. Substitution of z = h and z = —h into this inequality gives the condition of the mutual non-penetration.

[U] ■ n ^ h| [^] ■ n| on 7.

Note that the vertical displacements u are absent in this inequality. It is due to the fact that the generatrix of the cylindrical surface, which bounds the rigid inclusion, are perpendicular to the mid-surface of the plate. Next, taking into consideration the fact that the traces of functions U and ^ on y- are defined by (3), from the last inequality, we derive

(U — p) ■ n ^ h (^ + Vl) ■ n on 7 +, (4)

where U = p, u = l on u, p G R(u), l G L(u).

The equilibrium problem of the plate with a crack on the boundary of the rigid inclusion may be formulated in the form of the minimization problem

inf n(£), (5)

S w

where K = j £ = (U, u, ^) G H £ satisfies (3) , (4)j is the set of admissible functions. Note that the inclusion £ G H assumes that the homogeneous boundary-value conditions hold:

u = 0, ^ = U = 0 = (0, 0) on r. (6)

By virtue of the equivalence of (2) and (3), for the functions £ G K, the following equalities

hold: eij(U) = 0, eij (^) = 0, ^i + u,i = 0 a.e. in u, i, j = 1, 2. Hence, the energy functional may

be represented in the form

n(£) = 1 Bn\*(£, £) — </, £)^, £ = (U, u, ¿) G K.

One can easily show that the set Ku is convex and closed in the Hilbert space H. Due to the estimate

bOy (£,n) < c1|

where the constant c1 > 0 is independent of £ G H and n G H, the symmetric bilinear form of

Bqy (£, n) is continuous with respect to H. The coercivity of the functional n(£) follows from

the inequality

B^(£,£) > c21£|2, £ G H (7)

where the constant c2 > 0 is independent of £ (see [17]). The above properties of energy functional n(£), form Bqy(-, ■), and set Ku allow one to state on the unique solvability of problem (5)

(see [18]).

In what follows, by £ = (U,u, ^) we denote the solution of problem (5). Let the functions p0 G R(u) and l0 G L(u) be defined by equalities p0 = U and l0 = u on u, where U and u are the components of the solution £.

Symmetry and continuity of the bilinear form Bqy (■, ■) and the properties of the set Ku provide (see [18]) the equivalence of problem (5) to the variational inequality

£ G K, Bn\j(£, n — £) > (/,n — £)oy for any n =(V,v,^) G K. (8)

Compare two inequalities obtained by substituting of the test functions n = £ + n and n = £ — V where n = (V,v,V’) G C“(Q\u)5. As a result, we get an equality, which may be written as follows: < ) < ) < ) <

(mij ,£ij (V0) n\_ + ( Gij ,£ij (V)) n\_ + ( q^ (v,i ^ ^ ^ = (/,n)Q\^, (9)

where mij = mij(^), aij = aij(U), qi = qi(u, ^), i, j = 1, 2; U, u, ^ are the components of the solution £. From (9), with allowance for the independence of v1, V2, v, V’1, V2, we have

<aij ,£ij (V))Q\— = </i,vi)Q\^ for any V G

<qi, (v,i ))q\^ = </3,v)Q\^ forany v G <mij,£ij(V))n\_ + <qi,'iV%\— = <Mi,V^i^ forany V G co0^^2.

Taking into account that the representations (mij-,£ij (V))n\— = (mij-, (V,j ))n\_ and

(aij, £ij (V)}n\_ = (aij, (V,j ))n\_ take place, three previous equalities imply that the follow-

ing equilibrium equations hold in the sense of distributions:

aij,j = —/i, i = 1, 2, in Q\u, (10)

qi,i = —/3 in Q\u, (11)

mij,j — qi = —Mi, i = 1, 2, in Q\u. (12)

2. Differential Statement of the Problem

Below, we get the equivalent differential statement of problem (5). Namely, by the variational inequality (8), using an appropriate choice of test functions, we derive a complete set of boundary-value conditions on the curve 7. In order to find the relations on the inner boundary 7 from inequality (8), we use Green’s formulas. Since the derivatives of functions in the space H 1(QY) have generally no traces, we assume a sufficient smoothness £ in this paragraph.

For the domain Q\u, Green’s formula is valid (see [19], [20])

f aij £ij(V) = — / CTjjvi — /aijnjvi, V =(v1,v2) G H1>0(Q7)2, (13)

Jq\— Jq\— J3

where n = (n1,n2) is the normal to S. Similarly, for the arbitraries V G H 1,0(QY)2 and v G H 1,0(QY), Green’s formulas take place

(mij, £ij(V))n\_ = — <mij,j, ^^n\_ — / mijnjV\ (14)

(Vu, Vv)n\— = — Au,v>n\— — I dnv, (15)

<^ Vv)q\— = —<(^,i), v)q\— — /(^ ■ n)v (16)

Let the function n = (V, v, V) G K_ such that n G Hq (Q)5. Substitution of the test functions of

0(

the form n = £ ± n into the variational inequality (8) yields the following integral identity valid for all n G HQ (Q)5:

(mij,£ij(V))^ + (aij,£ij(V))^ + (q^ (v,i +?/>^^ = (/,n)ny. (17)

Next, integrating by parts, with allowance for (3) in (17), we derive

— ^Ojj nj pi + ^mj nj l,i — J^(^ ■ n + )l = J (F ■ p + /3l — Mi l,i),

(18)

where F = (/1, /2). Taking into consideration the independence of p and l in (18), we have

— Jaijnjpi = J F ■ p for any p G R(u), (19)

^mijnjl,i — J^(^ ■ n + )l = J (/3l — Mil,i) for any l G L(u). (20)

Comparing two inequalities obtained by substituting the test functions n = 0 and n = 2£ into

(8), we derive the identity

du N

— j +aijnjui — j +mijnjft — Jj ■ n + — )u = j (F ■ po + /3lo — Milo,i). (21)

Here S+ denotes that the values of functions on S are taken on the positive face. Consider the function n = £±n, where n = (V, V, V) such that V = 0, V = 0 in QY, the function v G H 1,0(QY), v = 0 in u. By construction, it is obvious that n G K_. Note that the values of function v are equal to zero on the curve S\7 and arbitrary on 7+. Substitute the function n into (8) and transform the obtained relation by using formulas (15), (16). As a result, we get

f du

J +(^ ' n + dn)’ ^ 0.

'7

Hence, with allowance for the arbitrariness of values v on the boundary 7+, we find

^ ■ n + =0 on 7+. (22)

dn

The integral over the boundary in (13) may be represented as follows (see [20]):

J aij nj vi = J (a„(V ■ n) + aTiVTi), (23)

where ann and aT = (aT 1, aT2) are normal and tangential components of the vector an = {aijnj}; here the following relations are valid:

aijnj — anni + aT^ an — aijnjni, T — ( n2,n1),

V = (v1, v2), vi = (V ■ n)ni + VTi, i =1, 2, Vt = (Vt 1,Vt2).

The representation of the form (23) is valid for the integral over the boundary in (14) as well. .

Substitute the test functions of the form n = £ ± n, where n = (V, v, V) G H such that V = 0,

V = 0 in Qy, V = 0 in u, and V ■ n = 0 on 7+ into the variational inequality (8). By using

Green’s formulas, the transformation of the obtained integrals yields

J" aij njV' ----- ^ (an(V ■ n) + aTi'Vri) ---------- ^ aTi'Vri ------ 0.

Hence, by virtue of the arbitrariness of the values V on 7+, we infer that aT - 0 on 7 +. By using analogous calculations and choosing the test function of special form (considering V, which satisfy V ■ n - 0), we can get that mT - 0 on 7+. Now let n = £ + n, where n = (V, v, V) G H such that V = 0 in QY, V1 = 0, V = 0 in u, V ■ n - h|V ■ n|, V ■ n > 0 on 7+. It is easy to note that n G K_. Substitute it into (8). By using Green’s formulas, we find

i ( — a„(V ■ n) ± hm„(V ■ n)) > 0.

The choice of functions VV , VV allows one to obtain the following relation from the last inequality:

—ha„ > |m„ | on 7+. (24)

From identities (18) and (21), we have

— aijnj(ui — poi) — mijnj (^ + lo,i) — J^ (^ • n + )(u — lo) = 0. (25)

From (25), taking into account the equalities aT - mT - 0 on 7+, (3), (22) and (24), we get

a„(U — po) • n + m„(^ + Vl„) • n - 0 on 7 + .

As a result, from the variational inequality (8), we derive equilibrium equations (10)-(12), identities (19), (20), and the following boundary-value conditions on 7 +:

du

^ • n + t— - 0, aT - 0, mT - 0, —han ^ |mn|,

dn (26)

a„(U — po) • n + m„(^ + Vl„) • n - 0.

We can also establish the converse; the solution to boundary-value problem (10)-(12), (19), (20) with conditions (6), (26) is the solution to the variational problem (8) as well.

Thus, the following statement is valid.

Theorem 1. The smooth function £ - (U, u, ^) is the solution to the variational problem (5) if and only if it is the solution to the boundary-value problem consisting of equilibrium equations

(10)-(12), relations (19), (20) and conditions (6), (26).

3. Passage to the Limit

Problem (5) turns out to be considered as limit one for the family of problems describing the equilibrium of plates, each of which occupies the domain . The family is characterized by the parameter A > 0 and the limit case corresponds to A - 0. In other words, for each A > 0, the

subdomain u corresponds to the elastic inclusion in the plate, and in the limit, each point x G u

has the displacement (p0(x), l0(x)), where p0 G R(u) and l0 G L(u).

Let the tensor of elastic moduluses AA - {aj} and coefficient JA depend on A as follows:

A __ j aijrl in ; jA _________________ j J in (27)

°ijrl [ A-1aijr; in u ’ [ A-1 J in u. ( )

For i j - 1, 2, define the functions

mAj(^) = aAjr;£W(^), aAj(U) = 3aA.wh-2£W(U), qA(u, ^) = JA(u,i +^i).

Let the bilinear form BA be defined by formula

(£,n) = (mj (^),£ij (V))qy + jA((u,i + ^ (v,i +Vi))^Y + (aj (U ),£ij (V ))^Y .

Formulate the equilibrium problem of the elastic plate containing a crack in the variational form

inf nA(£), (28)

where nA(£) = 2BAy(£,£) — </,£)Qy and

K - {£ G H | [U] • n > h[^j • n a.e. on 7 }.

Here [g] - g+ — g- the jump of the function g on 7. Problem (28) has the unique solution £A for each A G (0, A0), which satisfies the variational inequality (see [17])

£A G K, BAy(£a, n — £A) > </,n — £A)Qy for any n G K. (29)

Here, if the solution £A is sufficiently smooth, then one can show that it is also the solution of

the boundary-value problem (see [17])

aj j = —fi, i - 1, 2, a.e. in Q7,

qAi = —/3 a.e. in ,

mAj j — qA = —^i, i - 1, 2, a.e. in Q7,

UA - ^A - 0, uA - 0 on r,

[UA] • n > h|[^A] • n|, [aA] = [mA] = 0, aA = mA = 0 on 7,

duA

—-----+ ^A • n - 0, aA[UA] • n + m^[^A] • n - 0, — haA > |mA| on 7.

Here aA - aA(UA), mA - mA(^A), qA - qA(uA,^A), the values aA, aAn, and mA, mAn are

the components of vectors {aAnj}, {mAnj} respectively (for them, the formulas, which are analogous to those in (23), are valid). Further, we substantiate the passage to the limit as A ^ 0

in (29). It turns out that the limit problem completely coincides with (8). The comparison of

two inequalities obtained by substituting n - 2£A, n = 0 into (29) yields

<mj ,£ij (^A))qy + <aj ,£ij (U A))qy + <qA (u A, i +(^i)A))nY = ^ £A)^y .

Hence, by using (7), we get two following estimates, which are uniform in A:

II£AII < C3,

A(£a,£a) < c4. (30)

Choosing subsequences if necessary, one can consider that as A ^ 0

£a ^ £* weakly in H. (31)

Then, by (30), we have

£ij(U*)=0, ^* + Vu* =0, £ij (^*) = 0 in u, ij = 1, 2.

According to the arguments coming about from deriving (2) and (3), it follows from the last equalities that there exist functions p*, l* such that

U* = p*, u* = l*, ^* + Vu* = 0 on u, p* G R(u), l* G L(u).

Weak convergence £A ^ £* in H implies strong convergence £A ^ £* in L2(7)5. Choosing subsequences if necessary, one can consider that £A ^ £* a.e. on 7 as A ^ 0. Passing to the limit, as A ^ 0, in the relation

[UA] • n > h|[^A] • n| a.e. on 7+,

as A ^ 0 we find

(U* - p*) • n > h|(^>* + V/*) • n| a.e. on y+.

This means that the limit function £* belongs to the set K.. Fix the arbitrary function n G Kw. It is obvious that one can substitute it into (29) as a test one. We obtain

With allowance for weak convergence (31) and relations (27), it is possible to pass to the lower limit in (32) as A ^ 0 that gives

Due to the uniqueness of the solution to problem (5), we get that £* - £. Thus, for the family of problems (28) describing the equilibrium of elastic plates, as A ^ 0, we have the equilibrium problem (5) of the plate with the rigid inclusion as a limit one.

This work was partly supported by the grant no. 4402 of the Ministry of Education and Science of the Russian Federation and the grant no. 12-01-31076-mol-a by RFBR.

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(32)

£* G K, BQ\^(£*,n - £*) > (/,n - £*)qt for all n =(V,v, ^) G K

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Задача о равновесии пластины Тимошенко, содержащей трещину на границе жесткого включения

Нюргун П. Лазарев

Исследуется нелинейная задача о равновесии пластины, содержащей жесткое включение. Предполагается, что пластина имеет вертикальную трещину вдоль некоторой части поверхности, ограничивающей жесткое включение. Деформирование пластины описывается моделью Тимошенко. На кривой, задающей трещину, налагаются нелинейные условия вида неравенств, описывающие взаимное непроникание противоположных берегов трещины. В работе установлена однозначная разрешимость задачи о равновесии пластины. Получены соотношения, описывающие контакт противоположных берегов трещины. Показано, что задача является предельной для семейства задач, моделирующих равновесие упругих пластин при стремлении параметра жесткости к бесконечности в той области, которая соответствует жесткому включению.

Ключевые слова: трещина, пластина Тимошенко, жесткое включение, функционал энергии, условие непроникания.