NUMERICAL SOLUTION OF THE EQUILIBRIUM PROBLEM FOR A TWO-DIMENSIONAL ELASTIC BODY WITH A THIN SEMIRIGID INCLUSION Текст научной статьи по специальности «Математика»

Математические заметки СВФУ Январь—март, 2021. Том 28, № 1

UDC 517.9

NUMERICAL SOLUTION OF THE EQUILIBRIUM PROBLEM FOR A TWO-DIMENSIONAL ELASTIC BODY WITH A THIN SEMIRIGID INCLUSION T. S. Popova

Abstract. The equilibrium problem for a two-dimensional elastic body containing a thin semirigid inclusion is considered. The inclusion delaminates from the elastic matrix, forming a crack; therefore, the problem is posed in a nonsmooth domain with a cut. The mathematical model of the delaminated thin semirigid inclusion was developed on the assumption that the rigidity of the material differs in different directions. The problem statement is presented both in the form of a variational inequality and in the form of a boundary value problem. The boundary condition on the crack faces has a form of inequality and, as a result, the problem is non-linear. Consequently, the construction of an algorithm for the numerical solution of the problem requires the use of additional analytical methods. The methods of domain decomposition, the method of Lagrange multipliers, and the finite element method are used. An algorithm for the numerical solution of the problem is constructed and a computational example is provided.

DOI: 10.25587/SVFU.2021.12.25.005

Keywords: variational inequality, semirigid inclusion, thin inclusion, crack, non-penetration conditions, nonlinear boundary conditions, domain decomposition, Uzawa algorithm.

1. Introduction

The research is related to the equilibrium problem for the two-dimensional elastic body containing thin delaminated inclusion. The modeling of thin inclusions in elastic bodies received an increasing interest in view of the expansion of the use of fiber-composite materials and structures with reinforcement. One of the modern approaches in mathematical modeling of fibers is to apply the theory of elastic beams [1-5]. There is a wide range of works devoted to mathematical models of thin inclusions, exploiting the theory of elastic beams and focusing the interest on how to model the single thin inclusion delaminating from surrounding elastic media. The plane problems are investigated in which unknown functions are functions of displacements of the elastic body points, as well as normal and tangent displacements of inclusion points. Both the Euler-Bernoulli and Timoshenko models were used and a number of results were obtained in the field of solvability of problems, solutions properties, and numerical implementation (see, for example, [6-10]).

Delamination of a thin inclusion from an elastic matrix is modeled as a presence of crack in a body, and the inclusion is located on one of the face of the crack. The

© 2021 T. S. Popova

examples of non-physical mutual penetration of points on opposite sides of a crack in a case of classical boundary conditions on crack faces are well-known [11,12]. There are various approaches in crack modeling to overcome this difficulty [1315], in particular, setting the Signorini-type conditions [11,12,16-18]. One of the types of boundary conditions are the so-called non-penetration conditions, which exclude mutual penetration of points of opposite crack faces into each other. This Signorini-type condition has the form of an inequality and leads to the nonlinearity of the problem, a detailed description of these conditions can be found in [19,20]. In combination with a nonsmooth boundary, such a formulation leads to certain difficulties in its study, however, the method of variational inequalities gives satisfactory results in this field.

The mentioned mathematical models of delaminated thin inclusions constitute a system of problems related to each other by passage to the limit with respect to parameters that have a physical interpretation. For example, when the parameter characterizing the rigidity of the inclusion material tends to infinity, the limiting case corresponds to the problem of a thin rigid inclusion. The formulation and properties of solutions to this limiting problem can be found in [21-24]. On the contrary, as the rigidity parameter of a thin elastic inclusion tends to zero, the corresponding limit problem is a well-known problem of a crack in an elastic body (empty inclusion) [4]. Problems characterizing the relationship between thin rigid inclusions and volume rigid inclusions were also considered in [25]. We also mention the papers on the junction problems for different types of thin inclusions in an elastic body [26-29] and problems for viscoelastic body containing a thin inclusions [25, 30, 31].

At the same time, these models assume that the stiffness of the material of inclusions can have different values in different directions. As a result of the passage to the limit with respect to the rigidity parameter in one of the directions, with a fixed rigidity in the other direction, problems on the so-called semirigid inclusions are obtained. There are number of results for different cases of anisotropy and semirigid inclusions, for example [32, 33]. The present paper is devoted to the development of methods for the numerical implementation of the problem of a delaminated semirigid inclusion in two-dimensional elastic body. The problems of obtaining a model of a semirigid inclusion as a result of a limit passage in the model of an anisotropic elastic inclusion, the solvability of the problem for thin semirigid inclusion, as well as the study of problems of junction of semirigid inclusions of various types can be found in [4,27,32-35].

2. Problem formulation

Let an elastic body in its natural undeformed state occupy a bounded domain Q C R2 with Lipschitz boundary T, and T = ID U ÎV, Tu n TN = 0. The body is fixed along the part of boundary Tu and subject to external loads on TN. By n we denote a unit outward normal vector to T. The shape of the thin inclusion in the domain Q is given by line 7 = (-1,1) x {0} C Q. We assume that 7 C S, where S is a curve that devides the domain Q onto the subdomains Q- and Q+ with

Lipschitz boundaries in such way, that meas(dO± n rD) > 0. By v and t we denote the unit vectors of the normal and tangent to £ respectively. Note that v = (0,1) and t = (1,0) on 7. A thin inclusion delaminates forming a crack; therefore, the problem is considered in a domain with a cut = SI \ 7 (Fig. 1). The domain defines the shape of a two-dimensional elastic body with a crack. For the cut edges, we introduce the notations: y± C dO±, where 7+ and 7- correspond to the upper and lower faces of the crack respectively, and v coincides with the outward normal to the boundary dO-. The value of the function £ on the corresponding face 7± of the crack are indicated by the superscript:

Fig. 1. Two-dimensional body with a thin inclusion.

Let the vector function u = (u1;u2) define a displacements of body, where ui corresponds to displacements along axis xi, i = 1,2. For the components of the strain tensor and stress tensor of the body, we introduce the following formulas:

1

i,j = 1, 2,

= Qijkl , i,j, l = 1 2>

where = the coefficients a.ijki(x), i,j,k,l = 1,2 are the components of the elasticity tensor A, satisfying the following conditions:

aijkl ajikl aklij >

ai jkl Çkl Çij > Co|£|2 VÇij = Çji,

where c0 is a positive constant. Throughout the text, summation is assumed over repeated indices.

We will consider the function w of displacements of the inclusion points along the x1 axis and the vector function x = (u, w). Since the inclusion is located on the lower edge of the crack, then the displacements of the inclusion coincide with the displacements of the body points on y-. The model of a thin semi-rigid inclusion in this study means that the functions of vertical displacements of body points on Y- have a predefined structure. Namely, the function u- of vertical displacements of body points on y- coincides with some element l(x1 ) of the following space

L(y) = {l | l(xi)

c0x1 + c1;

c0,c1 G

We assume that all functions defined on y are identified with functions of the variable

Xi.

The equilibrium problem for a two-dimensional elastic body containing delami-nated thin semirigid inclusion is formulating as follows. For the given functions f = (A, f2) we have to find u = (u^u2), o = {aij}, i, j = 1, 2, w and the element 1(x1) of the space L(y) such that the following equalities and conditions are fulfilled:

-&ij,j (u) = 0 in QY, i = 1,2, (1)

u1 = w, u2 = l on y-, (2)

-ESw,ii = [aT] on y- (3)

ui = 0 on rD, i = 1, 2, (4)

aj (u)nj = fi on IN, i =1, 2, (5)

w,1 = 0 for x1 = ±1, (6)

[uv] > 0, 0-+ < 0, = 0, [u„] = 0 on y, (7)

J[ovjids = 0 VÎ G l(y). (8)

Y

Here ov = (o1j-Vj, o2j-Vj), ov = aijVjVi, oT = (ov)t, i =1, 2; uv = uv; ov, oT are the normal and tangent components of vector oij-Vj respectively; E, S are the material constants of the inclusion: Young's modulus and cross- sectional area. The square brackets denote the jump of the function on the crack edges, i.e. [£] = —

Relations (1) are equilibrium equations for an elastic body. Equations (2), (3), (6), (8) describe the equilibrium of the semirigid inclusion with a free ends at the points x1 = ±1. The effect of the surrounding elastic media on the inclusion is taking into account in the right-hand side of the equation (3). According to equalities (2), vertical and horizontal displacements of the inclusion coincide with the normal and tangential displacements of the body on the lower edge y- of the crack. The second equality in (2) defines the structure of the functions of vertical displacements of the inclusion. Condition (4) specifies the fixing of the body on a part of the external boundary rD, external surface loads are defined on TN, see (5) . Relations (7) represent a standard system of boundary conditions describing the possible contact of the crack faces on y, including the condition of their mutual nonpenetration (the first of the presented relations). Condition (8) describes the presence of a thin semirigid inclusion y. This non-local relationship can be written in an equivalent form. For this, we use in (8) the decomposition of the vector aijVj into the normal and the tangent components, as well as the element I we write in the form c0x1 + c1. Then after substitution we have:

J[ov(u)j ds = 0; J[ov(u)jx1 ds = 0. (9)

Y Y

Relations (9) mean that the main vector of forces and moment acting on y are equal to zero.

Now we at the position to provide a variational formulation of the problem. Let us consider the following Sobolev space

Hd = {x = (u, w) € H)2 x H 1(7) | u = 0 on ID}.

The set of admissible displacements is defined in the following form

Ko = {x € HrD | [uv] > 0, u- = w, u- = l on y, l € L(y)}.

Boundary value problem (1)-(8) can be formulated in variational form as the problem of minimizing of the energy functional

1 f 1

- / CTij{u)£ij{u)dx + -

n(X) = ^ J Vij{u)eij{u) dx + iJ ESw2! ds - J ftut ds

over the set K0, i.e. we have to find the element x € K0, providing a minimum to the functional n:

n(x) = _inf n(x).

X6K 0

The problem has a unique solution that satisfies the variational inequality [4]

X € Ko : J <Jij(u)£ij(u-u) dx +J ESwii(w,i —w,i)ds > J fi(u — ui)ds Vx € Ko.

Oy Y r„

3. Approximate problem

To construct an algorithm for the numerical implementation in a domain with a cut the domain decomposition method is applied. We will consider linear equilibrium problems in the domains O- and O+ with further search for the saddle point of the corresponding Lagrangian. It means that in this section we consider the problems of calculating the vector functions x = (u-,w) in O- and u+ in O+, where u± = (u±,u±) are functions defined in subdomains O±.

Let £ = (—2,2) x {0} satisfy the conditions of section 2, and Yg = £ \ 7. We define the following functional spaces

V- = {x = (u-,w) € H 1(O-)2 x H 1(y) | u- = w on y; u- = u- = 0on dO- nTD}, V+ = {u+ € H 1(O+)2 | u+ = u+ = 0 on dO+ n rD}

with the norms

llxlly- = / <ij(u-)£ij(u-)dx + / ESw21 ds; ||u+||V+ = J (u+)£ij(u+)dx,

O- Y O+

and the subspace

V- = {x € V- | u- = l on Y, l € L(y)}. Consider a convex set of the following form

K1 = {(x, u+) € V- x V + | (u+ — u-) = 0, (u+ — u-) = 0 on Yg; u+ — u- > 0 on 7}.

For the junctionals

IIi(x) = ^ J tTtj{u )elj(u )dx + i J ESw^ds- J fu ds,

n- y rNndn-

II2(u+) = — J (Jij{u+)eij(u+) dx — J fu+ds

n+ rwndn+

we will solve the following minimization problem

i+nf (ni(x)+ n2(u+)). Let us introduce the notation A = (Ac, Av, AT) and for every p > 0 define the

sets

u- = {x e V- | HxHv- < p}, u+ = {u+ e v + | ||u+||y + < p}, Ap = {Ac e L2(y) | 0 < Ac < p on7}, A; = {Av e L2(lg) |-p < Av < p on Yg}, Ap = {AT e L2(7g) | -p < AT < p on Yg}. Consider on a set W = U- x U+ x Ap x A^ x A; the Lagrange function of the form

u+)

L(x,u+,A) = ni(x)+ n2(u+)^ Ac(u- - u+) ds

+ y Av(u2 — u+) ds + y AT(u- — u+) ds.

Yg Yg

Denote ^p = (^p , , ) and for the considered form of the Lagrangian L we formulate the following problem of finding a saddle point. Find the vector (xp,up+, Mp) e W, such that the following inequalities are fulfilled

L(xp,up+,A) < L(xp,up+ ,Mp) < L(^,u+,Mp) V(^,u+,A) e W, (10)

where xp = (up-,wp), £ = (u-,w). The sets U±, Ap, A;, A; are closed, convex and bounded in the corresponding reflexive spaces. The Lagrange functional L is convex and lower semicontinuous on Up x Up , concave and upper semicontinuous on Ap x A; x A;. Then the problem (10) has a solution for every fixed p > 0.

Besides, it can be proved that there exists a constant G such that for every p > G the saddle point (xp,up+,^p), xp = (up-,wp) satisfies the equations and inequalities

J ffij (up-)£ij(u-) dx + | ESwpLw7j1 ds + J ^pu2 ds

n- Y Y

+ /MPu- ds + j ^ ds = j fu- ds =(u-,W) e K-, (11)

Yg Yg ndn-

J CTjj (up+ )eij- (u+) dx — y ^pu+ ds — J ^pu+ ds — J ^pu+ ds

Yg

y ds Vu+ G V +, (12) y Ac(up- — up+) ds < J Mp(up- — up+) ds VAc G Ap,

0 + Y Yg Yg

J AV(0r — u?+)ds < J № — u?+)ds VAV G Ap,

YY

J AT(u?- — up+ ) ds < y Mp(u?- — up+) ds VAT G Ap.

YY

The proof can be carried out similarly to the proof of Theorem 2 in paper [10]. Hence, the solution to the saddle point problem on the set W coincides with the solution to the problem on the entire Vs- x V + x Ap x Ap x Ap. In addition, following the method outlined in [23], one can prove that the solution (xp, up+) converges strongly to (x, u+) in V- x V + as p ^ œ.

4. Auxiliary problem

We consider the equilibrium problem of a two-dimensional elastic body occupying the domain w G R2, on the boundary dw of which a thin semirigid inclusion is located: y C dw, wherein dw = yd U yn, y c yn, Y n YD = 0. The shape of the inclusion and its position on the coordinate plane coincide with that specified in section 2, i.e. y = (—1,1) x {0}.

We formulate the problem as follows. For a given function g = (gi,g2) G L2(yn)2 we have to find the functions u = (u?,u2), Oj, i,j = 1, 2, w and the element Z = c0x1 + c1 G L(y), such that the following relations are fulfilled:

—ojj(u) = 0 in w, i = 1, 2,

ESw,?? = oT on y, u =0 on yd, i = 1, 2, oj (u)vj = gî on yn, i = 1, 2, u? = w, u2 = Z on y, w,? =0 for x = ±1,

1

y Ov(u)[=y g2Î VZ = c°xi + 51 G L(y).

13)

14)

15)

16)

17)

18)

YY

Then we consider the space

H = {x = (u, w) G H ?(w)2 x H ?(y ) | ui = w on y ; u = 0 on yd }

and take the set of admissible displacements in the form

k = {x = (u, w) e h | U2 e l(y) on y}. Then the problem (13)-(19) is equivalent to the following variational equation

x e K : J aij (u)£ij (u) dx + J ESwi1w7j1 ds = J giui ds Vx = (u, w) e K. (20)

w Y Yn

The inner product in a space H can be taken by the formula

(x,x)=y O j (u)£ij (u) dx + / ESw^w^ds. (21)

w y

Thus the following problem can be considered in this space. For a given function g we have to find xg = (ug, wg) e H such that

(xg, x) = y giu ds Vx e h, (22)

YN

Problem (22) is equivalent to the following:

—CTij,j (ug) = 0 in w, i = 1, 2, ug = 0 on yD, i =1, 2, o-ij (ug)vj = gi on yn \ J, i = 1, 2, 022(ug)= g2 on Y,

g

ul = w on y, ESwgu = 012 (ug) — gi on y, wgL = 0 for x = ±1.

Considering the problem

x e k : (xg — x, x) = 0 Vx e k, (23)

one can conclude that the problem (20) can be written in the form (23). This means that the element x is the orthogonal projection of the element xg onto the set K with respect to the inner product (20), thus, x is uniquely determined by xg. The set of all elements xg — x is the orthogonal complement to K, i.e.

h = k © kxg = x + x*, x e k, x* e k^.

Thus, it is necessary to construct an algorithm for finding the element x e K, if xg e H is known.

First, we define sufficient conditions for the element x* e H to belong to K Let x* = (u*,w*) be an arbitrary element of Kwhere u* = (u*,u2). Then

(x*,x) = 0 Vx e K. (24)

Take u e C0°(w)2 and substitute in (24) an element of the form x = (u, 0). Obtain

(x*,X) = J Oj(u*)ejj(U) dx = 0.

u

This implies

—Ojj(u*) = 0, in w, i = 1, 2. Moreover, by the definition of space H

u* = 0 on yd .

Now let x = (u, w) e K, and substituting this element into (24), we have (x*,x) = J Oj(u*)ejj(u) dx + J ESw*1w,ids = 0.

u Y

We apply Green's formula, then we get

J oj-(u*)vju ds — J ESw*11w ds + ESw*1w|-1 = 0.

9u Y

Assume that w* = 0 on y, this implies

CTjj (u*)vj = 0 on yN \ Y, i =1, 2.

y 022(u*)ids = 0 vi e l(y).

Y

Therefore, taking into account the obtained equations, it follows

ESw *11 = o12(u*) on y, w *1 = 0 for x1 = ±1. Taking into account the geometry of the problem, we write

Oj Vj = {011V1 + O12V2, O21V1 + O22V2} = {012, 022} = {o>, Ov}, Then the element x* e K^ satisfies the following conditions:

, j (u*) = 0 in w, i = 1, 2, (25)

u* = 0 on yd, (26)

u1 = w*, w *11 = c12(u*) on y, (27)

CTjj (u*)vj = 0 on yn \ YS i = 1, 2, (28)

ESw*1 = 0 for xi = ±1, (29)

y ^22 (u*)I ds = 0 VZ G L(y) (30)

Y

and conversely, the element satisfying these equations, is an element of KThe last condition (30) can be written in an equivalent form:

J o22(u*)x? ds = 0, J o22(u*) ds = 0.

To each vector a = (a?,a2) G R2 we associate an element Za = a?x? + a2 of the space L(y) and consider the following problem. Find an element xa G H, Xa = (ua, wa) such that the condition uSi = uf — Za take place on y and the following equation is hold

y Ojj (ua)£jj (u) dx + J ESw^w,!ds = 0 VX G H°, (31)

w y

where H° = {x G H ?(w)2 x H ?(y ) | u? = w on y ; u? = u2 = 0 on yd , u2 = 0 on y}. For each a G R2 problem (31) has a unique solution and is equivalent to next boundary value problem:

—Oj'j(ua) = 0 in w, i = 1, 2,

ui = 0 on yd, i = 1, 2, u; = wa, ESw?? = o?2(ua) on Y,

Ojj (ua)vj = 0 on yn \ Y, i = 1, 2,

w;? =0 for x = ±1,

a g

u2 = u2 — a?x? — a2 on y. Consider the following functions

h?(a) = J o22(ua)x?, h2(a) = J O22(ua),

YY

where xa = (ua,wa) is a solution of (31). It can be proved that there is a unique vector a G R2 such that h?(a) = 0, h2(a) = 0. In this case xa G KIndeed, the element u; depends linearly on a, so xa can be represented as

Xa = a?X(?' + a2X(2) + X(0).

Here x(0) is a solution of (31) for a = (0, 0), and functions x(m) = (u(m), w(m)), m = 1,2 are solutions of the following problems: find x(m) G H, such that u2m) = Z(m) on y and the following relation is fulfilled

y Ojj (u(m))£jj (u) dx + y ESw(m)w,?ds = 0 VX G H°, (32)

W J

where

Z(?) = —x?, Z(2) = —1.

Problem (32) is equivalent to the following boundary value problem:

-Gy ,j(u(m))=0in w, i = 1, 2,

ESw (1"1) = ai2(u(m)) on y,

u(m) = 0 on yd, i = 1, 2,

Gy (u(m))vj = 0 on yn \ Y

(1 ) (1 ) (1 ) (1 ) u1 — w( ), u2 = 1( ) on y,

w (1) =0 for x = ±1.

The functions h1 and h2 also linearly depend on a, so we can write

h1(a) = a1 J ^22(u(1))x1 ds + a2 J g22(u(2) )x1 ds + J g22(u(0) )x1 ds,

Y Y Y

h2(a) = a1 J g22(u(1) ) ds + a2 J G22(u(2)) ds + J G22(u(0) ) ds.

Y Y Y

There is a unique vector a G R2 such that (a) = 0, k = 1,2. To prove this statement calculate the following expressions:

(x(m),x(1)) = -J g 22 (u(m) )x1 ds, m = 1, 2,

Y

(x(m),x(2)) = -J g 22 (u(m) ) ds, m = 1, 2,

Y

(X(0),X(1)) = -/G22(u(0) )x1 ds, (X(0),X(2)) = -J G22(u(0) ) ds.

YY

Thus, the equations

h1(a) = 0, h2(a)=0

are equivalent to equations

a1(x(1) ,X(1)) + a2(x(2) ,X(1)) + (x(0), X(1)) = 0,

a1(x(1) ,X(2)) + a2(x(2) ,X(2)) + (x(0), X(2)) = 0. Defining the matrix B = {bmn}, with bmn = (x(m) , X(n) ); m,n = 1,2, and a vector C = (c1 ,c2), where cm = (x(0) , X(m)), m = 1, 2, it follows that the solving of the equations h1(a) = 0 and h2(a) = 0 is equivalent to solving the matrix equation

aB + C = 0. (33)

Let a = -CB-1 be a solution of (33) and a = (a1, a2) correspond to xa = (ua, wa), Xa = a1x(1) + a2x(2) + x(0), Xa G KThen the solution to the original problem of the equilibrium of an elastic body with thin inclusion on the boundary can be found from the relation

X = xg - xa.

5. Iterative algorithm and numerical example

Taking into account the results of the previous section, we construct an algorithm for the numerical solution of the problem for a delaminated semirigid inclusion. Let p be chosen and satisfying the condition p > G, we also assume that PAp , PAp, PAp are the operators of projection onto sets Ap, Ap, Ap in L2(y), L2(Yg) and L2(Yg) respectively.

And now we compose an iterative algorithm for the numerical solution of the original problem based on the Uzawa algorithm in the following form. Denote

V- = {x e V- | u- = 0 on Y},

thus, if the vector function x = (u-, w) is an element of the space V-, then u- = 0 on dO- n rD and u- = 0 on (dO- 0 rD) U y. Then the iterative algorithm for solving the original problem will look as follows.

1. Define auxiliary functions

X(m) G V-, x(m) = (u(m),w(m)), 4m) = 1(m)

on y,

such that

y CTj (u(m))£ij (U) dx + J ESw(1m)W,ids = 0 VX G V

m = 1, 2, 1(1) = -xi, 1(2) = -1.

2. Calculate the matrix B = {bmn}m,n=1-

3. Iteration k = 0. Setting initial values G Ap, G Ap, G Ap.

4. For k > 0 find the solution u+'k of the following problem

J o-j (u+'k)ejj (u) dx — y Mc,kU2 ds — J u2 ds — J u1 ds

O + Y Yg Yg

= y /jUj dx Vu G V +.

rwndO+

5. Find the solution xg'k = (ug'k) G V- of the problem

y Oij (ug'k)£ij(u) dx + J ESwg1kw,i + J U2 ds

O- Y Y

+ J U2 ds + y MT,kUi ds = y /jUj ds VX G V

Yg Yg ndO-

6. Define another auxiliary function x(°''k = (u(°''k, w(°''k) G Vsuch that u2°),k = u2'k on y and the following equality is fulfilled

y Ojj(u(°)'k)£jj(U) dx + y ESw(°)'kw, ids = 0 VX G V-.

n-

7. Calculate the components of the vector Ck = (ck, ck).

8. Calculate the components of the vector Ak = (A, ) by the formula Ak = -C k B-1.

9. Find the element = Akx(1) + Akx(2) + X(0)'k.

10. Find an intermediate solution xk = Xg'k - XA , Xk = (u-'k, wk).

11. For next iterations find

Mc'k+1 = Pap(Mc'k + 9(u-k - u+'k)), Mv'k+1 = PAp(Mv'k + 9(u-k - u+'k)), MT'k+1 = PAp (MT'k + 9(u-'k - u+'k)).

12. Stop if criterion is fulfilled and define xk, u+'k as the solution of the original problem. Otherwise, the transition with the next iteration k + 1 to step 4.

To illustrate the described method for solving the problem of a thin semirigid inclusion in an elastic body, consider the following example.

Take the square (-2,2) x (-2, 2) as a domain Q and the elastic body is assumed homogeneous, isotropic, fixed on the sides rD = {(±2)} x (-2, 2) and the surface forces f1 and f2 are specified at the lower and upper boundaries of the Q, respectively. Plane-stress assumption is considered in this analysis. The thin semirigid inclusion Y = (-1,1) x {0} delaminates, forming a crack. Decomposition gives the subdomains Q- = (-1,1) x (-1,0) and Q+ = (-1,1) x (0,1). Under these conditions, on (-2, -1) x {0} and (1, 2) x {0}, the glue conditions are satisfied for the functions of displacement of points of the body, while the system of boundary conditions for a crack with possible contact of the faces and equilibrium equations for a thin semirigid inclusion are hold on (-1, 1) x {0}.

The stress tensor components for elastic body are calculated by the formulas:

gu(«) = (2^ + A)en(u) + Ae22, g 22 (u) = Aen + (2^ + A)e22(u),

G12 (u) = G21 (u) = 2^12 (u),

where A = y^H, n = 2(i+v) are Lamé coefficients and the physical constants of the material are taken equal Em = 21 x 107 Pa, v = 0.28. Parameter characterizing the stiffness of a thin inclusion is equal to ES = 21 x 106 Pam2. The algorithm parameters are taken equal 9 = 107, p = 1010, the criterion for stopping the algorithm is chosen in the form

llxk|

V -

The spaces V , V + are approximated by finite element spaces consisting of piecewise smooth functions. Consider the plane-stress problem with f1 = —0.001^x1, f2 =

u

0.00018 0.00016 0.00014 0.00012 0.0001 8x10"5 6x10"5 4x10"5 2xl0-5

- 'jump vGrt-txt' —

1........-

Fig. 2. Normal displacements jump [uv] on S = (-2, 2) x {0}.

0.0002 0.00018 0.00016 0.00014 0.00012 0.0001 8x10"5 6xl0"5 4x10"5 2x10"5 0

— 'disp_hor_l.txt'- \ 'disp_hor_2.b<t' -

- /

-1.5 -1 -0.5 0

Fig. 4. Horizontal displacements on S = (-2, 2) x {0}.

8xl0"5 6xl0"5 4x10"5 2x10"5 0

-2xl0"5 -4x10"5 -6x10"5 -8xl0"5 -0.0001 -0.00012

Fig. 3. Normal stresses on S = (-2, 2) x {0}.

misp vert l.txt' -

; / \ -

—I............;........x .......-1..........

-2 -1.5 -1 -0.5

Fig. 5. Vertical displacements U2 on S = (-2, 2) x {0}.

0.001^xi. The subdomains O- and 0+ are divided into 3708 and 3654 triangles with 1951 and 1924 vertices, respectively.

As can be seen from the Fig. 2, for a given type of loading the crack has partial closure in the interval (—1, —0.4) and on the rest of the y the normal displacement jump has a positive value, which means crack opening. Fig. 3 shows a graph of the normal stress function at E points. In accordance with the conditions (7), normal stresses are zero at the points of crack opening, while at the points of crack closure they are negative. At the same time, at the points at the ends of the crack, there are singularities both at point xi = 1, at which the crack is opened, and at the point xi = —1 when the crack is closed. The Fig. 4 and 5 show graphs of the functions of horizontal and vertical displacements on E+ and E-. The vertical displacements function is linear which satisfies the semirigid inclusion model. Horizontal and vertical displacements coincide at the points of crack closure. Also, according to the glue conditions, displacements coincide at the points on E \ y.

6. Conclusions

The presence of a thin semirigid inclusion on one of the crack faces necessitates the use of additional analytical methods to develop a method for calculating the solution of a linear problem in the corresponding subdomain. The presented results of a numerical experiment correspond the properties of solutions to the original nonlinear problem of a delaminated semirigid inclusion in a two-dimensional elastic body.

In some cases of application of the finite element method to solve problems of cracks with classical boundary conditions such as equalities on the crack faces, the effect of penetration of points of opposite crack faces can be observed. The implementation of the developed algorithm in this work allows avoiding the non-physical result of the numerical solution of such problems. The non-penetration condition is fulfilled along the entire length of the crack. At the points of crack opening, the normal displacement jumps are positive and normal stresses on the positive crack face have zero values. Zero jumps and non-positive values of normal stresses were obtained at the points of crack closure. In addition, the domain decomposition method using the Lagrangian leads to the fulfillment of the glue conditions the boundary points of two subdomains at points outside the crack. As a result of calculating the displacement functions of the points of a thin semirigid inclusion on the basis of the properties of the orthogonal complement of the set, solutions are obtained that satisfy the stated conditions of the problem. Thus, the functions of horizontal displacements satisfy the equations of elasticity, while the graph of vertical displacements is linear.

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Submitted January 14, 2021 Revised February 26, 2021 Accepted February 26, 2021

Tatiana S. Popova

Ammosov North-Eastern Federal University, 48 Kulakovsky Street, Yakutsk 677000, Russia [email protected]