Construction of elasto-plastic boundaries using conservation laws Текст научной статьи по специальности «Математика»

UDC 539.374

Vestnik SibGAU Vol. 16, No. 2, P. 343-359

CONSTRUCTION OF ELASTO-PLASTIC BOUNDARIES USING CONSERVATION LAWS

S. I. Senashov*, E. V. Filyushina, O. V. Gomonova

Reshetnev Siberian State Aerospace University 31, Krasnoyarsky Rabochy Av., Krasnoyarsk, 660037, Russian Federation E-mail: sen@sibsau.ru

The solution of elasto-plastic problems is one of the most complicated and actual problems of solid mechanics. Traditionally, these problems are solved by the methods of complex analysis, calculus of variations or semi-inverse methods. Unfortunately, all these methods can be applied to a limited number of problems only.

In this paper, a technique of conservation laws is used. This technique allows constructing analytical formulas to determine the elasto-plastic boundary for a wide class ofproblems. As a result, the elasto-plastic boundaries were constructed for twisted straight rods with cross sections limited by piecewise smooth contour, for flexible consoles with constant cross-sections, as well as for anti-plane problems. Computer programs for construction of elasto-plastic boundaries for twisted straight rods were written using obtained technique.

In this work, the elasto-plastic boundary arising during the torsion of a straight beam of arbitrary cross section, which is limited by a piecewise smooth contour is constructed; and the elasto-plastic boundaries for the problems of a consol bending and anti-plane deformation are found. The plan of the paper is the following. In the first section the basic equations of elasticity and boundary problems are considered; in the second section the basic equations of the theory of ideal plasticity of von Mises are given; in the third section the conditions on the boundaries of the elastic and plastic domains are formulated. The fourth section is devoted to torsion of elastic prismatic rods; the fifth one describes elastic bending of bars; in the sixth section the plane problem of theory of elasticity is given. The seventh section covers an anti-plane problem of elasticity theory; in the eighth section, conservation laws for the equations of elasticity are constructed; in the ninth one, conservation laws of two-dimensional equations of plasticity are discussed. In the tenth section an elasto-plastic boundary of a twisted straight rod is found; in the eleventh one an elasto-plastic boundary in the bended console is given; and finally, in the twelfth section a method for the construction of elasto-plastic boundaries for large areas is described.

Keywords: conservation laws, elasto-plastic boundary, exact solutions, elasticity, plasticity, elasto-plasticity.

Вестник СибГАУ Т. 16, № 2. С. 343-359

ПОСТРОЕНИЕ УПРУГО-ПЛАСТИЧЕСКИХ ГРАНИЦ С ПОМОЩЬЮ ЗАКОНОВ СОХРАНЕНИЯ

С. И. Сенашов*, Е. В. Филюшина, О. В. Гомонова

Сибирский государственный аэрокосмический университет имени академика М. Ф. Решетнева Российская Федерация, 660037, г. Красноярск, просп. им. газ. «Красноярский рабочий», 31

Е-mail: sen@sibsau.ru

Решение упруго-пластических задач - одна из сложнейших и актуальных проблем механики деформируемого твердого тела. Традиционно эти задачи решаются или методами ТФКП, вариационного исчисления, или полуобратными методами. К сожалению, все эти методы могут быть применены лишь к ограниченному числу задач. В работе используется техника законов сохранения. Это позволяет построить аналитические формулы для нахождения упруго-пластической границы для широкого класса задач. В результате удалось построить упруго-пластические границы: для скручиваемых прямолинейных стержней, сечение которых ограничено кусочно-гладким контуром; для изгибаемых консолей постоянного сечения, а также для антиплоских задач. Разработанная методика позволила написать компьютерные программы для построения упруго-пластических границ для скручиваемых прямолинейных стержней. В предлагаемой работе построена упруго-пластическая граница, возникающая при кручении прямолинейного бруса произвольного сечения, которое ограничено кусочно-гладким контуром, а также упруго-пластическая граница в задачах об изгибе консоли и антиплоской деформации. В первом разделе статьи рассмотрены основные уравнения упругости и краевые задачи, во втором -даны основные уравнения теории идеальной пластичности Мизеса, в третьем - сформулированы условия на границах упругих и пластических областей, в четвертом - рассмотрено кручение призматических упругих стержней, в пятом - описан упругий изгиб брусьев, в шестом - рассмотрена плоская задача теории упругости,

Вестник СибГАУ. Том 16, № 2

в седьмом - описана антиплоская задача теории упругости, в восьмом - построены законы сохранения для уравнений упругости, в девятом - рассмотрены законы сохранения двумерных уравнений пластичности, в десятом - найдена упруго-пластическая граница в скручиваемом прямолинейном стержне, в одиннадцатом -найдена упруго-пластическая граница в изгибаемой консоли, в двенадцатом - предложена методика для построения упруго-пластических границ для областей больших размеров.

Ключевые слова: законы сохранения, упруго-пластическая граница, точное решение, упругость, пластичность, упруго-пластичность.

Introduction. Solution of elasto-plastic problems is one of the most complicated and actual problems of solid mechanics. It is determined by the fact that elasto-plastic boundary is not known in advance and should be defined during the solution of a problem. The elasto-plastic problems were considered by many well-known mechanicians. One can find a good review in works of B. D. Annin and G. P. Cherepanov [1], L. A. Galin [2; 3]. For the moment, a common approach for solving such problems has not been worked out yet. There are only a few single solutions for different special cases. As classical results one should consider an exact solution for the problem of elasto-plastic torsion of a rod with oval cross-section constructed by V. V. Sokolovsky, as well as solution of L. A. Galin for the problem of straining of a plane with circular hole.

An interesting theoretical result was obtained by B. D. Annin [1]. He proved the unique existence for the problem of elasto-plastic torsion of the rod with oval cross-section.

For solving of the elasto-plastic problem the methods of complex analysis, calculus of variations or semi-inverse methods were applied. In this paper, for construction of the elasto-plastic boundary the conservation laws were used. The conservation laws were applied in works [4-6] for the solving of the problem of 2-dimensional ideal plasticity; they allowed to obtain analytical solutions of Cauchy and Riemann problems. In following works of one of the co-authors of the present article the conservation laws were used for solving of some elasto-plastic problems [7; 8]. The obtained method allowed to write an algorithm (computer programs) for construction of elasto-plastic boundaries of twisted straight rods. For these programs the certificates of State registration are got [9; 10].

In present paper the elasto-plastic boundary for the problem of torsion of a straight beam of arbitrary cross-section, which is limited by a piecewise smooth contour is constructed; and the elasto-plastic boundaries for the problems of a consol bending and anti-plane deformation are found. For convenience, the article is divided into sections. In the first section the basic equations of elasticity and boundary problems are considered; in the second section the basic equations of the theory of ideal plasticity of von Mises are given; in the third section the conditions on the boundaries of the elastic and plastic domains are formulated. The fourth section is devoted to torsion of elastic prismatic rods; the fifth one describes elastic bending of bars; in the sixth section the plane problem of theory of elasticity is given. The seventh section covers an anti-plane problem of elasticity theory; in the eighth section, conservation laws for the equations of elasticity are constructed; in the ninth one, conservation laws of two-dimensional equations of plasticity are discussed. In the

tenth section an elasto-plastic boundary of a twisted straight rod is found; in the eleventh one an elasto-plastic boundary in the bended console is given; and finally, in the twelfth section a method for the construction of elasto-plastic boundaries for large areas is described.

1. The Basic Equations of Elasticity and Boundary Problems. Let's consider steady-state equations of linear isotropic elasticity.

The equilibrium equations look like:

+ X = 0,

cctx + CXxy + Cxxz

Cx Cy Cz

Cx „, Ост y Cx ,„

+ y i yz

ox Cy Cz

+ Y = 0,

(1)

drxz dzyz 9ct z —— +—— +—- + Z = 0,

dx dy dz

here ctx, cty, ctz, x, xxz, xyz are components of a stress tensor, X, Y, Z are components of an external force affected to a unit of volume. The components of a stress tensor related to components of a strain tensor by means of Hook's law:

ctx =Хе + 2|аех, ст = Xe + 2|isv, ст z =Хе + 2це

у Z

Xxy = 2|8xy , Xxz 2|exz, Xyz = 2|eyz •

(2)

^xy xz

xz> yz

-yz'

du dv dw du dv

Here e=--1---1--, e x =—, e =—,

dx dy dz dx dy

dw ^ du dv ^ du dw ^ du e z =—, 2e = — +—, 2e = — +—, 2e yz =— + dz dy dx dz dx dz

dw

+--, and X, u are constants of Lame, u, v, w are com-

dy' ' "

ponents of a vector of deformations, ex, e , ez, e^, exz, eyz

are components of a strain tensor.

Taking into account (2), equations of theory of elasticity can be written using displacements:

de

(X + |u)--b|uAu = 0,

dx

Cte

(X + |)— + |Av = 0,

Cs

(X + |)— + |Aw = 0,

here A is the Laplace operator.

On account of (2) the components of a stress tensor are in accord with the compatibility equations along with the equilibrium equations (1). The compatibility equations here are written in case of absence of external forces:

5 2e

5 2e

(1 + v)AaI +— = 0, (1 + v)Aay + — = 0;

5x

5 2e

5y 2 5 2e

(1 + v) Actz + — = 0, (1 + v) Ax xy +-—= 0, (4)

5z

dxdy

5 2e

5 2e

(1 + v)AXxz *HXdZ - (1 + V)AX"Z * dydz | = ст„ +ct„ + ctz, v is a Poisson's ratio.

= 0,

Problems for elasticity equations are usually pose either using displacements (in this case one have to solve equations (3)) or using stresses (in that case one solves equations (1)-(3)).

If a problem is written using stresses, one should add boundary conditions to equations (1), (3):

X = CTxl + Xxym + Xxzn,

Y =Xxyl + CT ym +X yzn, Z =X xzl +X yzm + CT zn,

(5)

here l, m, n are direction cosines of an external normal line to the boundary surface at point under study, X, Y, Z are components of a vector of superficial forces affected to a unit of area.

If the problem is written in displacements then on a boundary S these displacements are specified:

u /_ = u, v /_ = v, w /_ = w,

(6)

*x -3eJ +(ay -3eJ +(az -ioÏ +

+ 2x2 + 2x2 + 2x2 = 2k2

xy xz yz

(7)

here k is a yield point under simple shear.

In the case of plane deformation the plasticity law (7) can be reduce to form:

(X -CTy)2 + 4x2y = 4k2.

(8)

In plastic domain, the components of deviator of the strain tensor relate to the components of tensor of a strain rate with correlations

y3 e = s; =Авх, ay -13 e = Sy =Aey

CTz - к e = Sz = Aez, x„ = AeX

>3 e=s

(9)

X yz Aeyz, Xxz Aexz ,

here A is a nonnegative function obtained from (7):

au1 cX

ey =■

5U2

du_

~dz

du1 du2

2e =-+-

5y 5x

5u1 5u3

2exz =-+-

5z 5x

(10)

du2 du3

2e =-+-,

dz dy

here u1, u2, u3 are components of the vector of strain rate.

3. Conditions on the Boundaries of the Elastic and Plastic domains. Determination of a boundary separating elastic and plastic domains is one of the most difficult problems of the solid mechanics. The boundary is not known in advance and is defined during the elasto-plastic problem solving. In some cases a shape of the boundary can be guessed by general considerations.

Assume that an elastic state of a medium continuously changes over to a yield state. In this case close to elasto-plastic boundary and on each side of it the Hook's law is applies. This fact leads to the continuity of all components of the stress tensor and strain tensor, on the elasto-plastic boundary.

4. Torsion of Elastic Prismatic Rods. Let's consider an elastic prismatic rod with a cross-section of a variable form. It's lateral surface is free from efforts, face planes have forces equivalent to rotational moment M.

Let the coordinate origin is placed in an arbitrary point of the face plane and axis z is parallel to generatrix of the rod. The boundary conditions (3) will look like:

here u, v, w are certain functions on S.

Remark. There are others problems in the theory of elasticity, but they are not adduced in this article.

2. The Basic Equations of the Theory of Plasticity of Mises. For steady-state equations of the theory of plasticity of Mises one should add the plasticity law of Mises to the equilibrium equations (1). This law looks like:

CTxl + Xxym =0,

Xxyl + CTym = ^

X xzl +xyzm =0,

(11)

and on the face planes of the rod (z = 0, z = l)

\\%xzdxdy = ^ jjx yzdxdy = ^

Q Q

jjCTzdxdy = 0, jj xct zdxdy = 0, (12)

Q Q

jj ya zdxdy = 0,

Q

jjj(xxyz - yxxz) dxdy = M, (13)

Q

here q is the area of a cross-section. As is the convention in the theory of torsion:

a x =ay =Xxy = a (14)

and the remaining components of the stress tensor are in accord with equilibrium equations (1) which are the following form taking into account (14):

5x

5x yz

xz = 0, —y± = 0,

5x

5xxz 5x

yz

5z

5ctz

(15)

- + —= 0.

dy 5z

ex =

ez =

Compatibility equations (4) will look like

• = 0,

5 2 a,

d2a,

5 2 a,

5 2 a,

dx2

dy2

(1 + v)AXxZ +

dz 2

dxdy

d2az

dxdy

= 0,

(16)

(1 + v)ATyZ +

d 2a,

• = 0.

dxxz dx

yz

dx dy

= 0,

AXxz = 0, Ax yz = 0.

(18)

(19)

d T dx v

dy

dxx

dx dy

= 0.

(20)

_d_

dx

dxx

dx ^

yz

dy dx It follows from (19) and (20)

= 0.

dxxz dx

yz

dy dx

= C,

(21)

(22)

here C is an arbitrary constant.

A system (18), (19) may be replaced be equations (18) and (22).

As from (12) one can get

T du dw

xxz =M T- + -T"

^ dz dx

then

dxxz dx

yz

dv dw x y, = u|--1--

yz n dz dy

d T du dv

dy

It is known that

dx

1T du

^ dz dy dx

(23)

(24)

2 ^dy dx of the vector rot (u, v, w). We'll obtain dx

dy

dx dz

(25)

C = -2u-—^ = -2u9.

dz

(26)

We obtained finally that the problem of torsion of elastic prismatic rod comes to integration the following equations

dxxz dx

yz

dy dx

= 0,

dx dx

yz

dy dx

= -2^9,

dydz

From the equations (16) one can get

ctz = Azy + Bzx + Dx + Ey + Fz + H, (17)

here A, B, D, E, F, H are arbitrary constants.

By substituting (17) into (13) on gets that ctz = 0 in all alternate cross-sections of the rod. Therefore equations (15), (16) are reduced to the following:

with boundary conditions

xxzl + xyzm =°-

(27)

(28)

Let's transform the equations (18), (19). For this purpose we'll derive the equation (18) on x and subtract from it the first equation (19):

Now we'll derive the equation (18) on y and subtract from it the second equation (19).

As the equations (27) come to Poisson's equation then it is the base of numerous examples of solving the problem of torsion of elastic prismatic rods.

This fact leads to analogies which permit to reduce the problem (27), (28) to others mechanical problems which solution is described by the same equations. Here are some of them: membrane analogy, some fluid-flow analogies, electrodynamic analogy. One can introduce a term torsional hardness C = M / 8.

It is considered, the bigger a torsional hardness the better a rod resists to the torsion.

It was shown that among all the prismatic rods with the same area of lateral face, the biggest torsional hardness belongs to a rod with a circular cross-section.

Moreover, it is proved that among all the prismatic rods with multiply connected cross-section of the defined area and the defined total area of holes, the rod with ring-shaped cross-section which is bounded by two concentric circles has the highest torsional hardness. These and other problems of the theory of torsion of elastic bodies one can find in [11].

Saint-Venant noted an interesting fact: the maximum tangential stress as a rule is achieved upon the lateral face of a rod in the points the closest to a center of gravity of a cross-section.

5. Elastic Bending of Bars. Let's consider a prismatic rod bending by two equal and opposite moments M in one of the principal plane (fig. 1).

The coordinate origin is in the centre of gravity of a cross-section, the plane xz is in the main plane of bending. One gets the following elementary solution of the equations (1) in case of absence of body forces:

Ex ~R ,

a y = a x = x xy = x xz = x yz = 0

(29)

is the third component

here R is a radius of curvature of the bended rod; E is Young's modulus of stretch and compression.

Let's consider a common case of bending of a console with the constant cross-section, which is under the action of a force P applied to an end and which is parallel to one of a main axes of the cross-section (fig. 2).

Let's suppose that in console case, stresses allocate in a distance z from the fixed end in the same way as (29):

a z =

P (l - z)j l

(30)

here —L is the angle of a torsion per unit length of fiber

dz

of the rod. This angle is called twist and is denoted 8. From (25) and (22) one gets

Let's suppose now that in every point of the cross-sections tangential stresses xxz and x yz affected and the

other components of the stress tensor ctx, ct

equal to zero.

y xy

are

У

' X

Fig. 1. Bending of a Bar

Fig. 2. Bending of a Console

By such suppositions, in case of absence of volume forces one gets from the equations (1)

dxx

dz

= 0,

dx

yz

dz

= 0,

dxxz dx

yz

dx dy

Px l

= -—. (31)

It is follows from (31) that tangent stresses do not depend on z, and they are the same for every cross-section. The compatibility equations come to following:

Ax = —. P . , xz l (1 + v)

Ax yz =0.

(32)

One gets in the same way as in the previous paragraph:

_d_f dx yz dxx dy

dx dy

P

l (1 + v )

d Г dx„ dx Vz ^

(33)

0.

dx ^ dy dx It is obtained from the formulas (33)

dxxz dx

yz

Py

dy dx

l (1 + v )

+ C,

here C is a constant. It is possible to show that C = 0 [11]. Then equations of the bending of console look like:

dxxz + dx yz

dx dy

dxxz dxxz

Px,

"T,

Py

(34)

dy dx l (1 + v)

One should add a boundary condition to these equations which is the following on the frontier of the contour

Xxzl +xyzm = 0 .

6. A Plane Problem of Elasticity Theory. In this section the equations of a plane problem of elasticity theory in displacements are given and some boundary problems are posed.

For a plane problem the following conditions are valid:

= u (x,y), v = v(x,y), w = 0.

Then from (3) one gets

F = (X + |)

F2 = (X + |)

d2u d2v

2„\ f

dx2 dxdy

2

d2u d2v

2

dxdy dy2

+ 1

2

du du dx2 dy2

2

2

d2 v d2 v dx2 dy2

= 0,

= 0.

(35)

The boundary conditions (5) look like

xl + X xym =

/. u du dv I du ( +|)|"dx + "dy /J + |äX

l +

du ôv | — + || — + — I m = X, dx J

(36)

i du dv ) x y + ct ym = ul--1--1/ +

xy y ^dy dx)

J du dv | dv —

_(x+u)bx+dy J+udy Jm=7 •

7. An Anti-plane Problem of Elasticity Theory.

Let's consider equations (1), (2) when u = v = 0,

w = ra(x, y). This case corresponds to so-called antiplane elastic state.

Equations (1) come to

dxxz dx

yz

dx dy

+ Z = 0, X = Y = 0,

dx

d_

dy

dy

-pg = 0,

X xz X yz = 0.

(37)

dy

Definition. Conserved current for the system F1 = 0, F2 = 0, is a vector (A, B) which is

d-A+d-B = nF +n 2 F2, dx dy

(38)

here ni are some differentiation operators. It is assumed that both of them are not equal to zero simultaneously.

Let's find conservation laws for equations from the sections 4-7.

1. The equations describing the elastic torsion (27) in convenient denotation look as follows

du dv

— + — = ux + vy = 0, dx dy

(39)

= uy - vx

and compatibility conditions for deformations come to equation

dyXxz -dxXyz =

Let the elastic body be affected by only its dead weigh, then if the axis Oz is up-directed one receives z = -pg , here p is a constant density.

Finally equations describing elastic state on condition that the deformation is anti-plane look as follow

dx _ dx v

du dv dy fyx here u =xxz; v = x yz; a = -2|6. Let

A =a1u + P1v + Y1, B = a 2u + p2v + y2, (40)

where a1, pl, yl are arbitrary functions of x, y.

From (38) with respect to (40) and (39) one obtains

a1xu +a1ux +p1xv + pVx +y1x +a2yu + a2uy +

+P2v + P2vy + y2 = a1 (( + vy ) + a2 ( - vx - a).

Here and further, subscript signifies a corresponding variable derivative. One can get hence

a1 =P2, P1 = -a2, yx +y2y = -a2a, a1, +a2y =0, px +py = 0. Or after simple conversion

a2 =-p\ p2 = a1,

Tx + y2y = -a2a, ax -ßl, = 0,

Px +ay = 0.

Therefore, a conserved current look as follows

(41)

8. Conservation Laws for the Equations of Elasticity. Conservation laws are the fundamental laws of nature, they were determined in the beginning of the XXth century. A concept of the conservation laws appeared later after researches of E. Noether and her followings. The wide application of these laws to solving and investigations of some differential equations is relative to the last quarter of the XXth century. But significance and usefulness of this concept is not properly understood by majority of researchers even nowadays.

In this work the simplest definition of the conservation laws is given. For more details see [11] and cited literature there.

Let's Fj = 0, F2 = 0 is a system of two differential equations for two sought functions u = u (x, y).

A = a1u +ß1v + y1, B = -ß1M +a1v + y2,

(42)

here (a1, p1 ) is a solution of the Cauchy-Riemann system; y1, y2 are determined from the equation (41).

With respect to (42) the conservation law may be defined in the following form

J (a1u + p1v + y1 ) dy - (-p1« + a1v + y2 ) dx = 0, (43)

where Г is an arbitrary piecewise smooth closed contour.

Let (x0, y0 ) a point inside the domain bounded by r. One can choose

(х - xo ) + (y - Уо )

Р1 =-

У - Уо

(44)

(х - хо ) +(У - Уо )

2 '

Let Г1 is a circle (х - х0 )2 + (у - у0 )2 = R2 (fig. 3).

Fig. 3. A Circle Г1

It's not complicated to indicate that integral (44)

J^o/m +pV + Y1 )dy -(-p1« + aV + y2 }dx =

r

= - J (M +p1v + y1 jdy-(-p1« + a1v + y2)dx.

(45)

J (u + p1v + у1 ) dy - (-Р1« + a1v + у2 ) dx =

Г1

2л

= J

cos 0 sin 0 1 .

-u--v + у IR cos 0 +

R R 1

. sin 0 cos 0 2|„ ■ n

+ |-u +-v + y2 | R sin 0

R R

d 0 =

= J [u + y1R cos 0 + у2R sin e]d0 =

0

2л 2л

= J ud 0+ R J (cos 0 + y2 sin 0)d 0.

Let this time

a = -

У - Уо

(х - хо ) +(у - Уо )

Р1 =

(х - хо )2 +(y - Уо )2

■ = a

= Р2.

(47)

Let's calculate the circulation integral on r in this case. In polar coordinates x - x0 = R cos 8,

y - y0 = R sin 8 :

J=J

sin 0 cos 0 1

■u +--v + у

R

R

R cos 0 +

cos 0 sin 0 2

-u +--v + у

R

R

R sin 0|> d 0 =

= J[v + R (y1 cos e + y2sin e)]d9.

0

On conditions that R ^ 0 one obtains from the last formula and (46):

v(хо,Уо) = J (a2u + Р2v + у1 ) dy --(-Р2 u +a2v + у2 )).

(48)

Let's calculate the circulation integral on ri using polar coordinates x - x0 = R cos 8, y - y0 = R sin 8 :

Expressions (47) and (48) allow to calculate values u, v at any internal point of the domain enclosed by r if the values u, v on the contour are known. But u = x xz, v = x yz therefore these two values are not known on r, it known only the expression xxzl + x yzm = 0. Hence

formulas (47), (48) don't allow to calculate the values of the stress tensor inside the domain and then don't allow to solve the problem of the rod torsion. But as one will see later these formulas allow to resolve the elasto-plastic problem which is more complicated.

Remark. The similar formulas can be obtained easily for the equations of anti-plane theory of elasticity, which is given in the section 7.

Equations describing the torsion of a bar have a form (34). Let us assume in these equations

u = x „, v = x „„,

-P7i =-1. -pyi(1

+ v )

= ю2

then

uх + vy =®1. ^ -vy =-2.

(49)

In the last expression R tends to zero (R ^ 0), and using mean-value one gets

| (a1u + p1 v + y1) dy - (-p1u + a1 v + y2 ) dx = 2nu (x0, y0 ).

Now from (45)

u(x0, y0) =

= "21-j (a1u + p1v + y1) dy - (-p1u + a1v + y2 ) dx. (46)

The conserved current for this system will be found in a form

A = a1u +p1v + y1, B = a2u + p2v + y2,

here a', p!, y' are fuctions of x,y only.

Analogously to the previous clause, one can get

Ax + By = a1xu + a1ux + p1xv + p1vx + ylx + + a2yu +a2uy + p^yv + p2 +y2y =

(5о)

г о

+

= a (ux +vy ) + a2 ( -vx -®2).

One can obtain from here

a1 =p2, p1 = -a2, yx +y2 =-a1ra1 -a2ra2,

ax +a 2 = 0, px +p2 = 0.

After not complicated calculations

a2 =-p1, p2 = a1,

yx +y2y = -aS -a 2ra2, ax -py = 0, px +ay =0.

(52)

d51 + d52 = 0 dx dy

Making substitution of (54) into (55) one can get

, d2a, d2 y1 , d2a,

(X + 2|)—21 + —dL (x + |) +1—21 = 0,

dx 2

dxdy

2

dy2

2

dx2 dxdy dy2

(51) or

„ , d2p1 d28w„ , d2p1 „ dx2 dxdy dy2

x ry r^ y

Hence, the conserved current looks like A = au + pv + y1, B = -pu +av + y2, here (a, p) is an arbitrary solution of Cauchy-Riemann

equations; y1 , y2 are determined from the equation (52).

For the current (52) a conservation law can be written as follows

J (au + pv + y1) dy - (-pu + av + y2 ) dx = 0.

r

Acting much as the previous clause one can obtain finally

u( x0. y0) = 2-J(u +pV + y1 )dy -(-p1u +a1v + y2 )dx,

v(x0.y0) = "2!- J (a2u + p2v + y1) dy - (-p2u + a2v + y2 ) dx.

Remarks of the previous item are also correct for this problem.

Conservation laws of the plane theory of elasticity. Let's find some conservation laws for equations describing 2-dimensional resilience (35) as it is in the work [12]. The conserved current is looked for in a form

A =a1 (X + 2l) ux + p1 Iuy +y Ix + (X + 2l) vy , B = a2ux + p2uy + y2vx + 82vy ,

here ai, pi, yi, 8i, some functions of x, y. From the relation

d-A +d-B = a1F1 +y1F2. (53)

dx dy

From (53) one can obtain

a2 =-p1 +(X + |)УJ, p2 =a1|, y 2 =-§1 +(X + |)a1, (54)

82 =(X + 2|)y1,

(x+2|))a1=0, V ' dx dy

^ + 0p2 = 0, ^ + dy2 = 0, (55)

dx dy dx dy

d 281

I^1 + + + (X + 21))^. = 0.

dx2 'dxdyy ' dy2

It means that (a1, y1) and (p1,81) are arbitrary solutions of the equations (35) which are coupled by correlations (55). This fact permits to construct an infinite system of conservation laws on a base of the exact solutions of equations of elasticity.

9. Conservation Laws of Two-dimensional Equations of Plasticity. Let us consider the following equations of two-dimensional plane theory of plasticity which can be obtained with ease from the equation of the section 2:

2k i06 cos 26+—sin 29! = 0, (56)

dx V dx dy J

2k i— sin 26-—cos 26^ = 0. (57)

dy \dx dy J

Here c is a hydrostatic pressure, 6 = (1, x) --4, (1, x)

is an angle between the main direction of the stress tensor and the axis Ox.

Let's find conservation laws of a system (86), (57) in the form C = C (ct, 6) for which equality

dC dD „ — +— = 0

dx dy

or owing to Green's formula

( Ddx - Cdy = 0 (58)

r

is correct on account of system, i. e. relation

dC dCT dC dD dCT dD d6 „ --+-+--+--= 0 (59)

dCT dx d6dx dCT dy d6 dy

has to be performed for all its solutions in a domain bounded by the smooth contour r.

Let's determine a system of plasticity in a normal matrix form [5]:

d 281

f0CT> f cos 26 2k 1 f OctI

ox sin 26 sin 26 dy

d6 1 cos 26 06

VOx J V 2k sin 26 sin 26 J Vdy J

= 0.

(60)

dC dC

Multiplying this system by the vector | —, — I one

can get the following equation

dC de dC de ( dC cos2e dC - +---1--+ -

1

дст

дст дх д0 дх ^дст sin20 50 2ksin20) dy дС 2k 5C cos 20^ 50 л

+--I—= о.

дст sin 20 50 sin 20 ) dy

(61)

Comparing the equations (60) and (61) it is possible to obtain two exppressions for the functions C and D: dD dC cos28 dC 1

дст дст sin 20 50 2k sin 20 dD =-_д£ 2k дС cos20 д0 = дст sin 20 д0 sin 20 .

(62)

дС „, I ^D sin 20 + дС

-+ 2k I —<

д0 1 JCT дст

5D I ' 3D cos 20 - дС

--2k I

д0 1 "дст

(64)

This fact permits to use all the proprieties of this system during the conservation laws construction.

Thus, the linearization of the plasticity system is achieved without the requirement of the non being zero to Jacobian.

Further using the substitution

§ = CT + 2k8, -q = CT-2k8,

the system (63) comes to equations:

dD-dC

| = о,

dD dC n

-+-ctg8 = 0.

d^ d^

If insert new independent functions 9, y 9 = D- tg8C, y = D + ctg8C, it is possible to obtain the system

|-^tg0(y-9) = 0,

(65)

(66)

(67)

Finally, by setting p = 9 cos e one can come to equation:

P

-4 = 0.

And this is the well-known telegraph equation.

(68)

Thereby the construction of conservation laws for the plasticity equations comes to solving of the linear systems for which a lot of methods of resolution of equations and boundary problems are developed.

10. Elasto-plastic Boundary of a Twisted Straight Rod. Let's consider an elasto-plastic torsion of a straight rod which cross-section is bounded by a convex contour r.

If the twisting moment is rather significant, a plastic domain P forms in the rod. This domain arises on the external contour r. Suppose that the plastic domain is covered completely by the contour. In this case in the cross-section two domains appear, a plastic one P and an elastic one F. L is a boundary of these domains (fig. 4).

dC dD

Lets express the components —,— of the linear

d8 d8

system (62) in an explicit form:

(63)

It is possible to remark that by substitution C = -y (ct, 8), D = x (ct, 8), the system (63) coincides with a linearized plasticity system

y8 - 2k (-yCT cos 28 + xCT sin 28) = 0, x8 -2k (yCT sin28 + xCT cos 28) = 0.

Fig. 4. Cross-section of the Twisted Rod

There are a lot of works devoted to solution of the problem of the stressed state of an elasto-plastic rod, but most of them are based on some assumptions concerning the form of boundary L which is not known in advance. A novel method of the determination of unknown boundary is proposed by B. D. Annin [1]. This method is based on contact transformations and it permits to define the boundary between elastic and plastic domains in the rods with oval cross-section. This problem one can find in [1] and in the bibliography cited there.

In the present work the stress state is defined in all internal points of the rod by means of conservation laws, and formulas for analytical calculations of these stresses are proposed in the case of a piecewise-smooth directed boundary of the cross-section. The conservation laws is used for a long time in a fruitful way for solving of various mathematical and mechanical problems. A summary of results and solved problems in different domains of mechanics can be found in [4; 5; 7].

Problem Definition

Let xxz, xyz are single non-zero components of the

stress tensor. In the elastic domain they satisfy the equilibrium equation

3x„ dx y

дх ду

=о

and the equations x „ = G0

ду

дх

-y

x yZ = G0

ду

ду

+ х

(69)

(7о)

Here function 8y(x, y) determines a deplanation

warpingof the cross-section, 8 isa constant, G isa Young modulus by shear.

Let's introduce the stress function 9 as following

O9 dm

x xz =—, x yz (71)

0y dx

then to determine of 9 in the elastic domain one can get the equation

d 2 9 d2rn —7 + —7 = a, doc2 dy2

(72)

here a = -2G6 is non-zero constant.

In the plastic domain the components xxz, x yz along

with the equilibrium equation satisfy the plasticity condition

Xxz +X2yz = 1

(73)

Here, to simplify the further calculations, the plasticity constant equals to one.

By introducing the stress function in this equation one can get

+ M2 = 1.

dx J ^dy

(74)

Boundary Conditions. Let the lateral surface be free

from stresses. It means that — = 0 on the contour r. Here

dl

l = ((1, l2 ) is a tangent vector to the contour r. It follows that 9 = const among the contour. As r is a simply connected contour then 9 = 0 on it.

Finally one gets the following problem. It is necessary to resolve the following equation in the domain bounded by the curve L:

d 2 9 d29 —t + —rr = a.

dx2 dy2

(75)

In the domain bounded by curves L and r, i. e. in the plastic domain, the function 9 satisfies the equation

^T+(V|2=1.

dxJ ^ dy

(76)

ô9 d<p d9

— = 0 or —11 + —12 = 0 ,

dl dx dy

(78)

Fi = ux + vy - a =

u 2 + v2 = 1.

(80)

Owing to the denotation the following equality occurs: F2 = uy - vx = 0. (81)

Definition. A vector (A, B) is a conserved current for the system of the equations (79), (81) if there is the following correlation

5 xA + dyB = AjFj +A2 F2 =0. (82)

Here Aj, A 2 are some linear differential operators. It means that for functions A and B the conservation is correct law for all solutions of the system (79), (81):

5 xA + 5yB = 0. (83)

The conservation law (83) owing to the equations (79), (77) look like

Ax + AUx + AVx + By + Bu«y + BvVy = 0

or taking into account ux = a - vy and uy = vx,

Ax + A«a + AVx + By + BuVx + BvVy = 0.

From the last expression follows that functions A and B satisfy the equations

Ax + Aua + By = 0, Bv - A, = 0, Av + Bu = 0.

(84)

(85)

(84), (85) are Cauchy-Riemann equations.

Let's consider a domain D with the boundary r on condition that plastic domain P comprises completely the elastic domain F. Let r be a smooth directed contour, i. e. continuously differentiable without singular points. It follows from the conservation law

||(d xA + d yB ) dxdy = 0.

(86)

From (86), using Green's formula one can obtain

Ady - Bdx = 0.

(87)

Our objective is to find a domain F belonging with its boundary to the domain D where inequality u2 + v2 < 1 applies.

Let A = au + pv, B = av - pu + y then

The following conditions apply on the contour r for the function 9

9 = 0, (77)

Ax =a xu +ßxv + ßvx

By = a yv + avy-ßyu-ßuy +yy.

(88) (89)

(90)

on the frontier L the function 9 is continued.

It is necessary to find 9 in elastic and plastic domains and to determine a frontier L.

Let's introduce the denotation9x = u , 9y = v . Then

equations (75), (76) come to

(79)

According to the conservation law (83) one can get the equation

Ax + By =axu +aux +pxv + pvx +

+ ayv + avy -pyu-puy +yy = 0, which contains conditions on functions a, p and y.

ax -py = 0, ■px +ay =0,

aa + yy =0.

Let's consider two solutions of the system (90) The first one is

(x - x0 ) +^.y - y0 )

D

Р1 =

У - Уо

(х - хо ) +(У - Уо )

(92)

у1у =

then

(х - хо ) +(У - Уо ) У - Уо

у1 = -a • arctg-

Respectively, the second one is

a =_y-y_

°2 = / \2 / \2 : (x - xo ) + (y - yo )

p =_x - xo_

(x - xo )2 +(y - yo )2

Y = „ y- yo

(93)

(94)

2 У / ч2 / \ 2

(х - хо ) + (У - Уо )

then

у 2 =- 2 •ln ((х_хо )2 +(y - Уо )2 ).

Let's note the equation (87) for the functions A and B

(j Ady - Bdx = (( (au + pv) dy - (av - pu + y) dx =

Г

I l2 Л

-a-2- + Р

V l1

vdy -

I l Л

a ^-p l

(

l2

-a-2- + Р

h

V 2 )

dy -

a ^-p

I l Л

-a—

vdy -

V '2 )

I l1 ^

a —

udx - (( ydx =

Г

dx - ( ydx =

дф — i

дх

(95)

dx -

V '2 )

- (j^dx + (pi — dy + — dx I =

Г Г V^ дх )

= (J audy - (av + у) dx = о.

Let's decompose the boundary r into parts, i. e. r =

= r1 + r2 + r3 + r4; r3 is a circle (x - x0)2 + (y - y0)2 = R2

(fig. 5).

Fig. 5. Boundary r

In this case

Ady - Bdx = audy - (av + y) dx =

= (( audy - (av + y) dx + ( audy -

r r2

-(av + y) dx + ((audy -(av + y) dx + (96)

T3

+ (J audy - (av + y) dx = o.

r4

Obviously, (Jaudy-(av + y)dx + (Jaudy - (av +

r2 T4

+ y)dx = o. Taking into account this condition the equation (93) looks like

(J audy - (av + y) dx = - J audy - (av + y) dx. (97)

r3 r

Let's calculate an integral (J, r1 is a circle of the

r1

radius R.

Let

a = a1 =

Р = Р1 =-

(x - хо ) +(У - Уо )

У - Уо (x - хо )2 +(y - Уо )2 У - Уо

(98)

у = у1 = -a • arctg-

Introduce the polar coordinate system i x - xo = R cos 9

|y - yo = R sin 9,

(99)

then

dx = R sin фd ф dy = R cos фd ф'

cos ф _ sin ф

a =-, p =--, у = -аф.

R

R

As a result with R ^ 0 one obtains

(j audy - (av + y) dx = -u (x0, y0 ).

Analogically using a = a2, p = p2, y = y2

(j audy - (av + y) dx = -u (x0, y0 ).

Finally one can get from (82)

(j a1udy -(a1v + y1) dx = -u (x0, y0 ),

T3

(( a2udy - (a2 v + y2 ) dx = -u (x0, y0 ).

(100)

(101)

(Ю2)

(Ю3) (Ю4)

Let's determine the curve r3 in parametric form:

x = f (t), y = 9(t), 0 < t < T, (105)

f'(t), 9' (t) are derivatives of the functions f (t) and 9(t) respectively.

Hen functions u (x0, y0), v(x0, y0) from (98), (102) are found from the formulas

(

1 t

u(x0,y0)= -J •

(f (t )-X0 ))(f'(t ))2 +(cp'(t ))2 f) x0 ) + (p(t )-y0 )

+ af'(t )arctg

p(t )- y0 f (t )- x0

dt;

( ) 1T ((p(t )-yp ))(f'(t ))2 +(p'(t ))2

- 01 4(f (t)-x0)2 +(p(t)-y0)2

+ af'(t )ln ((f (t )-x0 )2 +(p(t )-y0 )2 ) dt.

(106)

The solutions (102) and (103) were used respectively to obtain these relations.

Let's calculate now a value of the expression

2 2 u 2 + v2

Let the component ct z of the stress tensor is distributed along the consol as in the case of pure bending

p (i - z )x

CTz =--1-—.

z l

Let components of the stress tensor are

CT x = CT y =CTxy =0.

Then the residual components of the stress tensor satisfy the equations

5Xxz =5Xyz = n 5XxZ , 5xyz = px

dz dz

■ = 0.

5x dx l

(108)

Usually the equations (108) replace by two compatibility equations

Ax = — y ,

xz l (1 + v )

Ax yz = 0,

(107)

in a point (x0, y0). The points where (107) is greater than

or equal to one belong to the plastic domain, the points where the expression (107) is less than one belong to the elastic domain.

On the base of the formulas (103), (104) the programs were developed; they permit to construct plastic and elastic domains in a twisted rod with indicated accuracy.

The solutions obtained using the programs coincident rather well with the known solutions.

In this journal one can find some examples of calculation of elasto-plastic boundaries for some cross-section of the rolling section. These results belong to A. V. Kondrin and to the authors of the article. The article [13] gives examples of the calculation of elastic - plastic rods borders for rolling profile.

11. Elasto-plastic Boundary in the Bended Consol. Let's consider a consol with the permanent cross-section bounded by the contour r. The consol is under the concentrated force P on a free end in parallels to principal axes (fig. 6).

here A is the Laplacian, v is a Poisson's ratio. This system is usually resolved by semi-inverse Saint-Venant method.

Let's rewrite the system (108) in terms of the vector of deformations (u; v; w). A boundary problem will be posed and resolved using conservation laws.

Using the formulas (2) which connect components of the stress tensor and strain tensor one can get

ct x = Xe + 2u— = 0, ct „ = Xe + 2u— = 0,

dx dy

X + 2 dw p (l - z ) ct z = Xe + 2u— =----- = ct,

z dz l

( du dv ^ (du dw ^ ( )

xxy +x J = °, xxz = J = x1,

( dv dw

x ^ = l = x2'

here X, u are Lamé coeffisients, e = — + — + —, x,, x2

dx dy dz

are sought functions of x, y.

From the first, the second and the third equation one can get

du dv

— = A1 xz + B1 x, — = A2 xz + B2 x, dx dy

dw

— = A3 xz + B3 x, dz 3 3

(1Ю)

here constants Ai, Bi can be evaluated with ease per

X, p, l.

From the equations (110) one can get

A xz 2

w = ^2--+ B3 xz + ra(x, y ),

A z3 B Z3

u = —3---3--+ (x1 / u - ®x )z + U (x, y),

ct 2

v = (x2/y ) + V (x, y) here ra,U,V are sought functions. From the relation x = 0 one gets

dy +lx = (T1 y / U-®^^ )Z + Uy +(x2x / U-raxy ) + Vx = a

Relations

du

dx = "(dl1x / U-®xx )Z + Ux = A1xz + B1x,

Cv

cy

= (( / УУ )z + Uy = A2xz + B2 x

are substituted in the previous equation; the result is

Px

М(хх + A1 x) + ^(ro yy + A2 x) = -

l

Suppose that the lateral surface of the beam if free from the stresses. It means that

Xxzn2 -Xyzn1 = ° (111)

here (n1, n2 ) are an external normal line to the contour r.

Suppose also that the plastic flow begins from the external side of the lateral surface of the beam. In this case the plasticity condition of Von Mises looks like

(112)

x2 +x2 = k 2

xz yz ,

®xx +-yy = ax.

(113)

under the following conditions on T:

I k A1 x2 '

n1k ——

-y =-

(n2k - A2 хУ )

(114)

Remark. There are two domains in the cross-section of the beam, a plastic one and an elastic one. If the following condition satisfies in a point of the cross-section

xL +< a

Then the point falls into the elastic domain. The other points including the boundary of the contour T belong to the plastic domain. The relation of a form

d xA + dyB = о,

(115)

is called a conservation law of (115) for the equation (113) if the equation (115) is correct for all solutions of the equation (113). Let the conserved current looks like

A =a( x y ) +p(x, y)y +Y(x, y ) B = a1 (x,y) +p1 (y) +y! (y).

One can obtain from (113) and (115)

a(a-ayy ) + ax +pc°xy +px roy +Yx +

УУ J x x > xy rx y ' x + a1-xy +ay-x +p1-yy +РУ-y +уУ о.

(116)

The relation (116) is correct for all solutions of the equation (113) therefore it is follows from (116)

ax-py = o, Px +Oy = o, aa + yx +yly = o. (117)

The conservation law (115) can be written using the Green's formula:

J-(arax +Proy + y) dy +(-prax +aray + y1 ) dx = o.

Let's consider two solutions of the equations (117). The first one is the following

a1 =-

У - Уо

x хо_ p1 =__

(x - хо )2 + (y - Уо )2 (x - хо )2 +(y - Уо )2

I

у x = о, a1 xa = -уУ, у1 = -ax • arctg

IУ - Уо ^

The second one looks like

a2 =■

У - Уо

(x - хо ) + (у - Уо )

Р2 =

(x - хо ) + (у - Уо )

here k is a constant of plasticity. Solving the system (111), (112) one can get

TXz = ±nk, X yz = +n2k.

Choosing the upper sign in these relations, one can pose the following problem.

It is necessary to resolve the equation

y y = a a2 xa = -y x.

Using the conservation law and applying it to the contour represented on the fig. 6 one can obtain on the analogy with the previous section

j(arax +p® y + y) dy -(-prax +ara y +y1) dx =

= - j (a®x +p® y +y) dy -

(x - x„ )2 +(y - y0 ) = R2

-(-p®x +ara y +y1) dx.

Let's calculate the second integral for the first and the second solutions of the equation (117). As a result the formulas for finding of rax (x0, y0), ray (x0, y0) are

obtained.

M-

M-

BecmHUK CuôrÂY. TOM 16, № 2

These formulas permit to find a stress state in any point (0, y0 ) . It means that it is possible to determine for

every point of the domain its belonging to either elastic or plastic zone.

The expounded method allow to construct a boundary between an elastic and a plastic domains with any prescribed accuracy using the computer. Preliminary calculations confirm this conclusion.

12. Elasto-plastic Boundaries for Large Areas. In this section only domains with smooth convex boundaries are regarded.

Let's consider a domain bounded by the contour r (fig. 7).

__—__ r

(CT-k cos26) l + k sin26m = X, k sin 26l + (ct + k cos 26) m = Y.

These conditions can be written as

ct = X', 6= Y . (120)

Thus, one gets the Cauchy problem on the boundary r in the plastic domain. Solving this problem using formulas of the section 9, one obtains two families of characteristic curves (fig. 8).

L J~~ 1

Fig. 7. Domain with the Contour r

Let boundary conditions (5) applies on r. They will look like

ct xl + Xxym = x, V + ct ym = Y=

(118)

Moreover, suppose that the loadings (118) bring the entire boundary and the nearby points to the plastic state. Then it is possible to introduce variables ct, 6 by the following way

ctx = ct - k cos 26, cty = ct + k cos 26, t^y = k sin 26. (119) In this case conditions (118) look like

Fig. 8. Characteristic Curves

For these curves one can construct an envelope curve L. This line is the sought elasto-plastic boundary. It is enough to solve an elastic problem inside the domain to resolve completely elasto-plastic problem.

It is possible to set Cauchy problem for the system of

plasticity (56), (57).

Let on the contour curve r denominated as SP there are starting data:

isp

ct,

el sp

ef

(121)

Let describe a characteristic curve PR : ^0 = const from the point P and a characteristic curve RS : ^0 = const from the point S. Then a solution of Cauchy problem is determined in a curvilinear triangle ASPR (fig. 9).

ct

>

Fig. 9. Cauchy Problem

Thus it is necessary to find coordinates of a cross point R of the characteristic curves. If the coordinates of the point R (xR, yR) and values §0, r0 of are known

then it is possible to find values of functions c, 0.

One can calculate an integral over the closed contour SPR. Using the correlation (58) in the Stokes theorem for a plane [11], one can conclude that this integral is equal to zero:

j Ddx - Cdy =j + j + j = 0.

SP PR:r|=r|0 RS-^0

SPR

1. For the coordinate xR one can get:

J Ddx - Cdy = J | D - C-d-- |dx.

SPR

SPR

Integrals J and J are integrated by parts:

PR:^^

RS:—^

J| D - С^ ]dx = J (D + Cctg0)dx =

PR V dx ) PR

PR ,|x=xR

= Jydx = x у|x=xP - J x^ d—

d—

J I D - СсСу 'dx = J (D - Cctg0) dx =

RS

dx,

J фdx = хфх

RS

x _

=xR

J x dл.

RS RS ^

Assuming 9 = 1, y = 0 one can get the boundary conditions for the system (65) in the form:

9 RS = 1 y| PR = 0 (122)

Under such assumption the final expression for the coordinate xR looks like:

j Ddx - Cdy = j (Ddx - Cdy ) + x y|^=J + x 9|x=^ = 0,

SPR SP

xR = j (Ddx - Cdy) + xS . (123)

SP

2. Similarly fort the coordinate yR :

j Ddx - Cdy = j I D — - C |dy.

SPR SPR V dy )

Integrals j and j are integrated by parts:

i( D% - С > -J

PR V UX ) PR

D - С^У-dx

dy dx

D + Cctg0 , г у

dy = J

J У

PJR -ctg0

Уу

RS

ctg0

\\D% - Ф =J

У=yR

PR -ctg0

d I У

y=yP PR

d—V ctg0 D - Ctg0

dy = dy = d I,

dy =

= J^dy = ^

RJS tg0 tg0

=xS

-J y-^Ы d л.

RS M tg0j

x=xR RS

Assuming 9 = tge, y = o one can get the boundary conditions for the system (65) in the form:

фи = tg ^ у PR = о'.

2

(124)

Under such assumption the final expression for the coordinate yR looks like:

J Ddx - СсУ :

SPR

J (Ddx - Cdy) -

ху

ctg0

У=yR

y=yP

Уф tg0

y=yS

= о, (125)

y=yR

yR = J (Ddx - ссУ) + yS .

It remains to resolve the problems (65), (122) and (65), (124). These problems can be redused to the equation (66). Namely, taking into account that fuctions 9, y are related with function p in the following way

ф = -

P

У =-

2 dp

cos 8 sin 8

one can obtain the problems

d2p P

д—5л 4

|=|о 2

d|

(126)

= о (127)

Л=Ло

and

C2P P

д—5л 4

- 4 = о^ P|,

I — = sin^^, dp

2 d—

о. (128)

Л=Ло

A general solution of the first problem (128) (for the coordinate xR ) is the following function:

P = P1 (—л) = 1о (\/(|- —))cos

По -—о

1 л

1 J Iо ((о)(л - x))

sin LJk dx.

ло

moreover

dPL = Icos Л^о i d— 2 2

((- — )(Л-Ло ))

4 J I1 (о)(Л-Х))

ло V

Л - Ло

|-|о

,^sin I-1 dx.

f |-|о 2

The solution of the second problem (128) (for the coordinate yR ) is the function:

p = P2 (I, л) = 1о (>/(—-—о )(л - Ло ) ) 1 л

+2 J 1о ((оХл-x))

sin

cos -—— d x .

ло

RS

when

SP

SP

PR

PR

Ф2 = 1sln По 2

Ii (¡¡Чо)(Л"Ло))

= _sin^^Ii ¡/¡-^)(Л-Ло+

П-то cosтЧо

П-По

1} I1 ((¡R))

По ' ъ Ъо

cos-—— d т.

In all solutions the function I0 is the Bessel function of the first kind of an imaginary argument I0 (0) = 1,

i0 (0) = 0.

The functions 9, y can be found using formulas

(126). From the relation (66) one can obtain the components of conservation laws:

^ ytg6+mctg6 . 2 2n

D = ^—-r b =ysin26 + 9cos26,

С = -

tge+ctge

у-ф tge+ctge

= ¡/-ф)1п e cos e.

Substituting the obtained С and D in (123) and (124) one can get the coordinates of the point R. Thus Cauchy problem for the system of plasticity (56), (57) with starting data (121) is resolved completely.

Conclusion. The small range of problems considered in this article, concerning the construction of elasto-plastic boundaries reveals good prospects of the application of conservation laws for the the boundary problems solving. By now, the authors have solved some other problems of solid mechanics and prepare them to publish.

More results in the study of equations of elasticity and plasticity can be found in articles [14—17].

Acknowledgements. Research is supported by Ministry of Education and Science of Russian Federation, the project Б-18о-14 and the FSP "Researches and Elaboration on the Priority Directions of Development of a Scientific and Technological Complex of Russia for 2о14-2о2о", the project No. 14.574.21.оо82.

Благодарности. Работа поддержана Министерством образования и науки РФ проект № Б 18о-14 и ФЦП FSP «Исследования и разработки по приоритетным направлениям развития научного и технологического комплекса России на 2о14-2о2о», проект № 14.574.21.оо82.

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© Senashov S. I., Filyushina E. V., Gomonova O. V., 2о15