Integral representation discontinuous solution of the problem of bending of anisotropic plates Текст научной статьи по специальности «Математика»

UDK 539.3

Levada V. S., Ph.D., Khizhnyak V. K., Ph.D., Levitskaya T. I., Ph.D.

Zaporozhye National Technical University

INTEGRAL REPRESENTATION DISCONTINUOUS SOLUTION OF THE PROBLEM OF BENDING OF ANISOTROPIC PLATES

Leaning on the ratios connecting deflection derivatives as the generalized function, with usual derivatives, the differential equation which right part contains the generalizedfunctions having jumps of a deflection, tilt angles, the moments and generalized shear forces are resived. The solution of the equation is received in the form of convolution of the fundamental decision with the right part. From the found representation the boundary integrated equations (BIE) for the solution of the problem can be received. These BIE can be solved by method of boundary elements.

Key words: bending, an anisotropic plate, defects, discontinuous solution, generalizedfunction, boundary value problem.

Introduction

The structures of many machines have plate elements. In these plates can be formed cracks. In addition they may contain thin inserts of other materials. The research of stress-strain state of such plates is an important problem. At the same time, the solution of the corresponding boundary value problems causing serious mathematical difficulties. To solve these problems G.Y. Popov proposed a generalized method of integral transforms [1]. This method was developed in the work of G.A. Morar [2]. S. Crouch proposed a method of discontinuous shifts, alternatively boundary element method (BEM) [3]. The corresponding boundary elements for anisotropic media were obtained in [4, 5]. There used the relationship between ordinary and generalized derivatives of regular generalized functions. This technique is used in this work.

If the boundary of the extended straight portion includes a rigid or articulated fixing, then can be used as G(x, y, n) Green's function obtained in [7, 8].

Formulating and solving problems

Consider the following problem:

Ai

5 4w

dx

4 + 4Di6

+ 2(Di2 + 2D66) x

d 4 w

2a 2 + 4D26

dx dy

d4 w dx3dy

44

d w d w

—3+d22T-T=q( x yx (i)

dxdy'

dy 4

where w(x,y) - the deflection at the point (x,y);

(x, y) e B <z R2, B - limited area, l0 - piecewise smooth

boundary of B , k = AiBi (i = 1, k) - smooth curves lying in b . These curves can be closed, they can coincide ends.

The end of one arc may be an interior point of the other. On the line l0 are given two boundary conditions. Also, two

conditions are given by on the lines li.

Equation (1) describes the bending of anisotropic plate with rigidity anisotropy D„, Di2 , Di6, D26, D22, D66.

Curves li simulate cracks or thin inclusions or reinforcements.

k

If the curve l c B \ y li, then are set there the following

i=0

functions

M = -

n

( ^2 ~,2 ~,2 A

_ d w ^ d w „ ^ d w

D11 ~T + D12 TT + 2D16 d-dy dx dy dxdy

nX -

(2 D12 ^ + D.

d2 w

2... A

'12 - 2 + D22 ^ 2 + 2D26

ax2

dy 2

d 2 w dxdy

n2y -

- 2

( 2 2 d w ^ d w

D

!6-rT + D26~T~2 + 2D66

ax2

dy2

cj2 A

a w

dxdy

nxny ;

Htn =

~j2 -¡2 .„a w _ a w

(D11^T + D12^2 + 2D16 dx dy

d 2w

- D1

a 2w

dxdy 12 dx2

-D

^^ - 2D

22 - 2 2 26

dy2

a2 w

dxdy

)nxny +

( 2 2 d w ^ d w

D

16 "TT + D26^~T + 2D66

ax2

dy2

cj2 A

a w

dxdy

(n2 - nl).

x

+

© Levada V. S., Khizhnyak V. K., Levitskaya T. I., 2015

N = -

n

(

d w

d3w

d3w

Dii —T + D TTT" + D + 2D66) —T +

dx dx dy dxdy

+ D.

A (

d w

26 3

dv3

d3w

5 3w

D„ "dxr + d_2 + 2D66) ^+

+ 3D.

d3w

+ D2

cj3 A

d w

26 dxdy2 22 dv3

ny ; Qn = Nn +

dHn 5s

dw dw dw dw dw

dw

■ = — nx +--ny • — =--ny +--nx

5n dx dy dx Sx dy '

There Mn - bending moment, Htn - twisting moment,

dw

Nn - shear force, gn - generalized shear force, — -

dw

normal angle of inclination, — - tangent angle of

5 4w

5 4w

by replacing the

Next is obtained from .

dy dx4

variable x on y .

5 4 w

For „ o have the four options representations.

dx dy

The first option.

44

d w Id w

d3w

dx3dy I ôx3dy I lôx2dy

hX0)8(/o)+

(

dx

(d 2" 1nx1)«('o) +£ ({£ I^Co)

{dxdy J

(wny°)5(/o))+]r

dx'

d3w dx2dy

^(l ) +

inclination. The tangent vector t is chosen such that the

three vectors n , t , k form a right-handed vectors. Let us introduce the notation:

n(0) - unit normal vector to the line l0 outward region B ; n(i) = (nX!),n^) - arbitrarily chosen unit normal vector to l,, i = \J;

t(0) - unit tangent vector to l0; T(i) = (TXi), t^)) -unit tangent vector to li;

n« = cos(n(i), x), nf = cos(n(i),y), t« = -n« , Tp = nf i = 0,1

Denote the Daw(x, y) - derivative of w(x,y) e D'(R2); {Daw(x,y)} - ordinary derivative

w(x,y); (x,y) e R2\\Jlt .

i=0

dx

d 2w dxdy

4°S(l )

A 5 2 ( + —

dx

dw

nx°S(/; )

+ |3 (wjnfSdi ))

dx

The second option.

^4 I ^4

d w I d w

d3w

dx3dy I dx3dy I I dx2dy I

^fSdo)-

dx

{H lnx">8<'o) +p' ({f k>8('o) ) +

dx dy i=i

d3w dx2dy

Sx

d 2w dxdy

4°S(l )

A 52 + —-

dx

dw dx

n(;Mii ) 1+

Using the relationship between Daw(x, y) and

{Da w( x, y)}, we get:

d4w I d4w I I d3w I (o),., .

" + Jnxo)S(/o) +

dx4 I dx4

dx■

+ ^ ({£ }nx°)§(lo) ^+-1x3 (wn x0)^(/o))+

d3w dx3

dw

nfsd- )+

Sx

d 2w dx2

n^S(li )

dx2

dx

«S(li )]+* ([wjnfSdi ))

dx'

dx 2dy

(w]nxi)S(li ))

The third option.

44

d w I d w

d3w

dx3dy Idx3dy I Idx2dy

hxo)s(io) +

dx

d3

^ Inyo)S(lo) (iiw

Sx2}y o dxdy [{dx

^xo)s(lo)

dx2dy

(wnfSdo))

d3w dx2dy

n(xi)S(li ) +

nx -

5

+

d

+

5

+

5

+

3

d

d

+

d

+

i=i

2

d

+

n

dx

d2 w dx2

n«5(lj )

A 52 + -

dxdy

dw dx

n«5(li )| +

dx

dx'

dw

2r«x0)S(lo) +^TT ^K^o)

dy

■'y

dx 2dy

The fourth option.

d 4w

dx3dy I 5x3dy

d w | + I 53w I (o)

K 8(/o)+

dx'

dx 2dy

(wny%(/o))+i

d3w

Sx

d 2 w

d7

nïH )

cx 2

dxdy

dw dy

№(/, ) +

«y°s(/j )

dy

'fSdo)"

d | I dw

Sx2dy

(w«x0)S(/o))+I

dxdy I [ dx ^ (r

^x0)8(/o)

d3w

Sx3

K°s(/i ) +

d3

+ aX2dy № —

The third option.

d 4 w

5 4 w

d3w

2a 2 I ^ 2^ 2 I ' U"TTf"x S('o)

dx dy I dx dy J I dxdy

|«xo)s(/o) +

dy

d 2w dx2

'fSd,- )

A 32 + -

dxdy

dw dx

"xWi ) | +

dx

I—

[Sxdy J

fs^-

dxdy

40)S(/o)

d3

\wPx -vi;

dx dy Four options for

d4w

5 4w

„ „ 3 are obtained from . 3. the

dxdy dx dy

permutation of letters x and y .

d 4w

For Q-yZy2 , there are six options representations. The first option.

dx 2dy y

(yo>5(/o))+i

d3w

dxdy

n^/j ) +

dx

d2 w dxdy

^(/j )

A d 2 ( +

dxdy

dw dy

«x°s(/j )

53

+ ax2dy w —

The fourth option.

d 4 w

d 4w

ax25y2 I ax2cy2

d 3w

dx2dy y

■«yo)s(/o) +

5 4w

5 4 w

d3w

ny0)8(/o)"

dx2dy2 Icx2dy2 J ' i5x2dy | y o

(fa 2

dy

2 y

yo)s(/o)-

53

dxdy (

d2 ifdw]

dy2 H dx J (

nx0)S(/o)

d3w dx 2dy

_d_

dy

d

d2 w dx2

4°8(/i )

A 52 +

dy

dw dx

)+

40S(/j )

2 Iwi'x

dxdy

The second option.

dy

d 2w I

I

dxdy J

d2 (I5w

■«xo)s(/o) ^N^I«yo)s(/o) | +

53

dxdy (

dxdy ^ | dx r

d3w

dx 2dy

yps/ )+

d_

dy

d 2w dxdy

«*s(/, )

A 52 +

dxdy

dw dx

«y)S(/I. )| +

d3

2^ -1 x y I

dxdy The fifth option.

d4 w

d4 w

d3w

dx2dy2 I 3x2dy2

dxdy

2 |«x0)S(/o) +

d 4 w

d 4 w

d3w

2a 2 I ^ 2^ 2 I ' U"TTf"x S('o)

dx dy I dx dy J I dxdy

ko)s(/o)+

2

d

d

+

3

3

d

d

i=1

5

5

+

+

5

+

3

5

2

5

5

d

+

3

d

+

5

+

5

3

+

+

+

2

2

I =1

+

+

+

dx

w

yo)s(i0)N^ k0^) 1+

n u(i _____ u(i

I dxdy ' y dvd, I I ^ I y

d3

dxdy

r

-(wnX°)8(l0))+]T

dxdy ^ | ox r d3w

dxdy

^)+

d_

dx

d2 w dxdy

ny°s(ii)

A d 2 + -

dxdy

dw dx

ny°s(ii) |+

d3

dxdy The sixth option.

d4w _[ d4w ] [ d3w

ax^dy2iaxvf+teyp "('0)

K^« +

dy

i—}'

[dxdy J

xo)8(i0)-

,2 r

Sx5y

^xo)s(i0)

+-()8(i!)) (())+

"dy

dx2

+ (T5(i)S(ii ))+^ (^Sfli))

dy'

dxdy

dx3

dx 2dy

where

(i 7i)s(ii ))+£ (/«sdi))

dy3

(r^Sdi ))^Td3r ($8(4))

I(i) _ D

■1 _Dn

d3w ox3

dxdy

n« + D2

d 3w

ly3

n^) +

+ 3D

16

+3D2,

d3w ox 2dy

d3w dxdy2

nl!) + D

l6

n(i) + D

ny + d26

d 3w dx3

d3w

^y3

ni:) +

nx) +

(2)

d3

2 _ wny -C0)

ox dy

5r

+Z

i _1

d3w

Sx2dy

^(i.-) +

dy

d 2w dxdy

m^o,)

A 5 2 r

dxdy

dw dy

№(it)

+D

12

+ 2D6,

d3w ox 2dy

ni + D12

d3w dxdy 2

np+

d3w dxdy 2

n() + 2D6

d3w dx2dy

(i)

dx dy

There [g(x, y)] -jump function g (x, y) when passing through a curve i in the selected direction of the normal; S(ii) - delta function concentrated on the curve it. If p(x, y) e Cm (R2) - the basic function, then

(DVx,y)8(i), cp(x,y)) _ _ (-l)x,y)8(i), Dap(x,y)) _ _ (-1)1 x,y) Dap(x,y)di.

Substituting the expressions obtained in (1), we obtain

- + 2(D1? +

d 4w

d 4 w

L(w) _ D11 ^ + 4D16 d^T^D12

4

\ O w

+ 66) _ 2~ 2 + 4D26 - - 3 + d22~T4 dx dy dxdy dy

44

d w ^ d w + D2

■■q( x, y) + ZZ (■« 8(ii) + 2° 8(ii))+

i_0 dxy

■ (i) _ D ■ 2 D11

d 2 w cx 2

n<? + 2D

16

d 2w dxdy

nx(i) +

+ D1

+D

12

d 2w cx 2

d 2w

n(i) + D

ny d26

d2 w

ly2

nf +

nl° + 2D,

66

d 2 w dxdy

( )

J(i) _ D

3 22

d 2w

"dy2

n(i) + D

ny +D16

d 2w dx 2

n(') +

+ 2D.

26

d 2w oxdy

n(i) + D

ny d26

+ D1

d 2w dx 2

+ 2D66

■(i) _ D

ox

nl!) + D

16

+ D

16

dw dx

ni + D12

dw

dy

d 2w

~dyT

d2 w dxdy

dw

dy _

(i)

np+

(i)

nl!) +

d

+

+

3

i_1

+

3

d

+

5

+

+

+

3

5

+

l5° - D.

22

dw ¥

n(i) + D

ny + d26

dw dx

ny) +

+D2,

dw

"dy

n{i) + D.

dw dx

,(0

I(i) - 2D I6 _2D16

dw

cx

ni!) + 2D.

+ 2D6,

dw dy

ni!) + 2D6

26

dw

cx

dw (i)

n« +

I« - Dn[w]n« + D16[w^ 4° - D22[w^ + D26[w]n«; I« -3Di6[w]n« + D12[w]) + 2D66[w] I« - 3D26 [w]« + D12 [w] + 2D66 [w].

r(0

I

(i)

Multiply 14

(n«)2 + (nff - 1.

After transformations we get

L(w) - ?(x,y) + ]T(-[, ]s(/i)-

i6°( i - 0, k ) on

dx2

dw 5n,

(^(n«)2 +A2(n?))2 +

+ 2D16nx,)ny,))5(/, )--

dy2

dw

(D12(n«)2

+ D22(ny )2 + 2D66npny)?>(li)-

- 2-

SrSy

dw

(^(n«)2 + D26(nff +

+ 2^« n(,,))S(l, ))-

cx 2

dw

^«ny:* -D^nf +

+ D16(n(,,))2 -D^n«)2)s(l,))-

v16

il

dy2

dw St«

-D^nf +

+ D2fi(n^'))2 -D26(nx°)2)s(l,))-

ôx5y

dw St«

D16n«n^) -D26n()ny) +

+ D66(ny))2 -D66(nx°)2)s(l,))-

(Mn, k ))-dT(ô ( ] ))

dn(i)

d

dx' d3

+-^1(11^« + D,6n(i)

11" x M6" y

) ))+

+ (fe^} + D-J? ) )) +

2d. (D^n« + Dun(;) + 2D66ny))))+

dx dy

d3

+—T (,w]3D26n(y} + Dun(:) + 2D66n^ ) )) (3) dxdy

The solution of equation (3) is obtained as the convolution of the fundamental solution of the operator

L(w) with the right part (3).

The fundamental solution r( x, y) of the operator L(w) is a solution of equation

(i)

(i)

L(r( x, y))-S( x)S( y).

(4)

This solution was obtained in [6] and has two options. The first option corresponds to a case when equation

DuV) + 4D^3 + 2( D12 + 2D66)^2 +

+4D26|a + D22 _ 0 has four different roots:

M-1,2 _a1 ±''P1 , ^3,4 _a2 ± ''P2, ai e R , Pi > 0 (i _ 1,2).

The second option is obtained when the reduced equation has multiple roots

M-1,2 _a + iP, M-3,4 _a- iP, ae R , P> 0.

The fundamental solution for the first variant of the roots has the form:

r(x,y) _-1-jp2(a2 +a2 +P2 -P2 -

- 2a1a 2)

r1 ln r1 + 2P1x(y + a1x)61

+ P1 (a2 +a2 +P12-P2 -2a1a2)

r2 ln r2 +

+ 2P2x(y + a2x)92]+ 2P1P2 (a2 -a^ x

r1 01 - 2P1x( y + a1x) ln r1 -

-r2 62 + 2P2x(y +a2x)lnr2]}, where

(5)

r, - (x2(a2 +P2) + 2a,xy + y2)

2^2,

5

+

,-0

2

d

+

2

d

2

d

+

x

2

d

r = ( x2 (a 2-P2) + 2a.xy + y2) 6. = arctg

2)>2.

( y aA + —

V xPi Pi y

Q = P4 + P2 + a4 + a2 + 6afa2 - 2pfp2 +

+ 2Pj2aj2 + 2Pj2a2 - 4pi2a1a2 + ipfaf +

+ 2p2a2 - 4p2a1a2 - 4aj3a2 - 4a1 ai. For the second variant of the roots

r( x. y) =-1-(x2 (a2 +P2) + 2axy + y2 )x

8np3D11 V '

x ln(x2(a2 +P2) + 2axy + y2 ) . (6)

Thus.

r( x. y) * D(ax. y)(^( x. y)S(/)) = = J D(x, y)(r( x - Ç. y - n)ds(Ç.n =

= (-1)1a J D^n) (r(x -Ç. y - n)MÇ. .

i

Denoting. G(x. y. Ç. n) = r(x - Ç. y - n). we get Denoting. G(x. y. Ç. n) = r(x - Ç. y - n). we get r( x. y) * D^y-M x. y)S(/)) =

= (-1)laljD(Ç.n)(G(x.y.Ç. nMÇ.n)ds(Ç.n . (7)

Thus. the solution of equation (3) has the form:

W ( x. y) = JJ G( x. y. Ç. n)q(Ç. n)dÇdn +

k

+ 1

i=o

-J

(

dw

dn(l)

J[[„(i) ].n)G(x. y.Ç. n)ds(Ç.n) -(D11(nÇI'))2 + D12(«ni))2 +

(Ç.n)

+ 2D66»Ç0»n°)

dw

ds.

(Ç.n) '

-J

dn(I)

(Ç.n)

d 2G( x. y. Ç. n)

dÇ2

(D12(«ÇI))2 + D22(«ni))2 +

+ 2D66«Ç°«n0)

5 2G( x. y. Ç. n)

ds

(Ç.n) '

- 2 J

dw

(Ç.n)

5n2

(D16(nÇI))2 + D26(nnI))2 +

+ 2D66«ÇI)«nI))

ds(

(Ç.n) '

-J

dw

(Ç.n)

d G(x. y. Ç. n) 5Ç5n

((Dn -Dïi^n® +

+ D16((«nI))2 - («Ç0)2))

dw

2.. d G(x.y.Ç.n)

-J

dT(I)

dÇ2

((Du - Du)«0«0 +

ds

(Ç.n) '

(Ç.n)

+ D26((«nI))2 -(«ÇI))2))

2„ d G(x. y. Ç. n)

- 2J

dw

(Ç.n)

5n2

((D\6 - Di6)nÇI)«f +

ds

(Ç.n) '

+D66((«nI))2 - («Ç0)2))

2„ d G(x. y. Ç. n)

-Jk l

dG( x. y. Ç. n) Ç.n) d«(I)(Ç. n)

5Ç5n

ds(Ç.n) +

ds(

(Ç.n) '

J[h

+ .[ | dG(x.y.Ç.n) dS(Çn)-

-J[w](Ç.n)( Dn«ÇI) + D^«0)

d3G( x. y. Ç. n) dÇ3

ds

(Ç.n) '

-J[w](Ç.n)(D22«nI) + Di6«ÇI))

d3G( x. y. Ç. n)

dn3

ds

(Ç.n) '

-JHç.T,)(3A6«Ç° + ( Ai + 2D66)«nI))

xd 3g( X. y.Ç.n)

x (Ç.n)

dÇ dn t1

J[w](Ç.n)(3D:

■26«n') +

+ (D + 2D )n(D) &G(x.y.Ç.n) ds ^

+ (D12 + 2D66)nÇ ) 2 dS(Ç.n)

oÇon

(8)

G(x,y, n) can be considered as a deflection at a point

(£,n) when the unit load concentrated at a point (x, y).

Denote Mn(G(x,y,n%,n) the bending moment

corresponding deflection G(£,n). Similarly, denote the remaining bending characteristics. After the transformation in (8) we obtain

W (x, y) = JJ G( x, y, n)d|dn +

B

X

B

k

+ 1

i-0

(

i[[n(i) L^ x, У, Ç n)ds( li

dw(Ç, n)

(Ç,n) '

-i

Mn(i)(G(x n))(^,n)ds(Ç,n) -

J(Ç,n)

Hn(i) (G(x У,Ç,n))(E,n)

(Ç,n) dG(x, y, Ç, n)

i

+ J! (Ç, n)] x,y, Ç n)

ds(ç n) +

dr(0(Ç, n)

ds(ç,n) +

-J[w(Ç, n)] (G( x, y, Ç, n))(ç

n)

(9)

Let 4^ «i2), b, (bi1, bi2) (i -1,k ), T1 j tJ 2) -corner points l0 .

dw(Ç, n)

i

dr(i)

Htn(i)(G(x, y,Ç n))ds(ç,n) -

- w^X n(S2) + 0)Htn(i) (G(x, y, Ç(S2), n(S2 )) --w(Ç(s1),n(s1) + 0) H (i) (G(x,y, Ç(S1), n(s1)) -

-J w(Ç(s), n(s) +

dHfm (G(x, y, Ç(s), n(s))

S1

ds

-ds -

-w(Ç(S2 ), n(S2 ) - 0)Htn(i) (G(x, y, Ç(S2 ), n(S2 )) + +w(Ç(s1 ), n(s1) - 0) H , ) (G( x, y, Ç(S1), n(s1 )) +

S2

+ J w(Ç(s), n(s) - 0)

S1

dH t (¿) (G(x,y, Ç(s),n(s))

---:-ds -

ds

-[w(bn, b^)]] )(G( x, y, bn, bt 2)) -- [w(ai1, «i2 )]tJO (G(x У, «1, «i2 )) -

- J [w(Ç, n)]

H (i)(G(x, y, Ç, n))

AiBi

ds(Ç, n)

ds

(Ç,n).

AiBi ^(Ç,,)

- [Htn(i) w(bi1, bi2 )G(x y, bi1, bi2 ) -

- [Htn(i) w(«i1, ai2)]p( x У, «1, ai 2 ) -

-J

dHtn(i)(Ç, n)

ds

G( x, y, Ç, n)ds(ç,

n),

i _ 1, k.

Similar transformations could be done for the curve i0 at the points (t, tj 2), j _ 1, m.

Given the expression for 0n(i), we get

r (x, y) _ JJ G(x, y, n)q(|, n)d|dn +

k

+1

i-1

J

J[(i)fe n)]]G( ^ y, ç n)ds,

li

dw(Ç, n)

(Ç,n) '

dn

(i)

Mn(i)(G( x, y,Ç n))(ç,n) ds(ç,

n)

(Ç,n)

(Ç-nJG^ A

h

A(Ç,n) "

(Ç,n)

+ J[w(Ç, n)]en(i) (G(x,y,Ç, n))ds(Ç,n) + k

+ [w(ai1, ai2 )]Htn(i) (G(x, У, ai1, ai2 )) --[w(bi1, b, 2)]Htn(i) (G(x, y, bi1, bu )) + + [Htn(i) (bi1, bi 2 )]G(X, У, bi1, bi 2 ) -

-[Htn(i) («h , «i 2 )]g(x, У, ai1, ai2 )) +

+ JQn(0)G(x, У,ç, n)ds(Ç,n) -

-J

dw(Ç, n) dn

M (00) (G(x, y,Ç, n))ds(Ç ,) +

J Mn(0)(Ç, n)

dG( x, y, Ç, n)

dn(0)

ds(Ç,n) +

J w(Ç, n)Q n(0)(G( x, У, ç, n))ds(Ç,

n)

m ,

+ L(w(tj1, tj2 )(Htn(01) (G(x, y, tj1, tj2 )) -

j-1

-(Htn(02) (G(^У, tj1,tj2)) -

-G(X, y, tj1, t] 2 )(Htn(01) (tj1, t] 2 ) - Htn(02) (tj1, t] 2 ))) ,

there n(01) and n(02) - normal direction at the point Tj (tj1, tj2), corresponding to the beginning and end of a bypass l0 from Tj in Tj ( j -1,m ). If the curve lt - AiBi has no points in common with lk ( k ^ i ), then as a result of continuous jumps on the li outintegrated terms at the

+

B

AB

0

0

0

0

AB

points Ai and Bi equals zero. From four integrals at

usually unknown jumps are contained in two. If models the crack, the unknowns are the jumps of deflection and the normal angle of inclination. If simulates the thin insert rigidly engaged with the plate, the unknown jumps moment and generalized shear force. Knowing the two boundary conditions on and two conditions on (), to find the unknown functions can be obtained the system of boundary integral equations possibly strongly singular.

Conclusions

Obtained an integral representation for the deflection of the anisotropic plate containing defects (curves on which the discontinuities of the first kind: deflections, tilt angles, moments or generalized shear forces). The resulting representation allows us to reduce the boundary value problem of the bending to a system of integral equations.

Literature

1. Попов Г. Я. Концентрация упругих напряжений возле штампов, разрезов, тонких включений и подкреплений / Попов Г. Я. - М. : Наука, 1982. - 344 с.

2. Морарь Г. А. Метод разрывных решений в механике деформируемых тел / Морарь Г. А. - Кишинев : Штин-ца, 1990. - 130 с.

3. Крауч С., Старфилд А. Методы граничных элементов в механике твердого тела. - М. : Мир, 1987. - 326 с.

4. Левада В. С. О разрывных решениях в теории анизотропных пластин / В. С. Левада, В. К. Хижняк // Прикладная механика, 1997, т. 33. - № 8. - С. 89-91.

5. Левада В. С. О концевом граничном элементе трещины в плоской задаче теории упругости для анизотропных сред (случай разных корней) / В. С. Левада, П. В. Цоко-тун // Новi матерiали i технологи в металургй та маши-нобудуванш. - 2004. - № 2. - С. 99-102.

6. Левада В. С. Построение фундаментального решения для задачи изгиба анизотропной пластины / Левада В. С. -Запорожье, 1996. - 8 с. - Рукопись. Деп. в ГНТБ Украины № 476. - Ук 96.

7. Левада В. С. Изгиб полубесконечной анизотропной пластины с жестко закрепленным краем, находящейся под действием сосредоточенной загрузки / В. С. Левада,

B. К. Хижняк, Т. И. Левицкая // Новi матерiали i технологи в металургй та машинобудуванш. - 2011. - № 2. -

C. 117-119.

8. Левада В. С. Изгиб полубесконечной анизотропной пластины с шарнирно-опертым краем, находящейся под действием сосредоточенной загрузки / В. С. Левада, В. К. Хижняк, Т. И. Левицкая // Методи розв'язання прикладних задач механжи деформiвного твердого тша, 2012. - Вип. 13. - С. 254-259.

Одержано 28.07.2015

Левада В. С., Хижняк В.К., Левицька XI Интегральне подання розривного розв'язка задачi згину ашзотропноТ пластини

Опираючись на стввгдношення, що зв 'язують пох1дт прогину, якузагальненог функци, 3i звичайними похгдними, одержали диференщальне ргвняння, у правш частинг якого м1стяться узагальненг функцП, що мають стрибки прогину, кутгв нахилу, моментгв i узагальнених перергзних сил. Розв 'язокргвняння отримано у виглядг згортки фундаментального розв 'язку iз правою частиною. 3i знайденого подання можуть бути отриманi граничнi ттегральт рiвняння (Г1Р) для розв 'язання поставленог задачi. Ц Г1Р можуть виршуватися методом граничних елементiв.

Ключовi слова: згин, атзотропна пластина, дефекти, розривний розв 'язок, узагальнена функщя, крайова задача.

Левада В.С., Хижняк В.К., Левицкая Т.И. Интегральное представление разрывного решения задачи изгиба анизотропной пластины

Опираясь на соотношения, связывающие производные прогиба, как обобщенной функции, с обычными производными, получили дифференциальное уравнение, в правой части которого содержатся обобщенные функции, имеющие скачки прогиба, углов наклона, моментов и обобщенных перерезывающих сил. Решение уравнения получено в виде свертки фундаментального решения с правой частью. Из найденного представления могут быть получены граничные интегральные уравнения (ГИУ) для решения поставленной задачи. Эти ГИУ могут решаться методом граничных элементов.

Ключевые слова: изгиб, анизотропная пластина, дефекты, разрывное решение, обобщенная функция, краевая задача.