UDK 539.3:534.1	ISSN 1580-2949
Original scientific article/Izvirni znanstveni članek	MTAEC9, 44(4)219(2010)
A NEW SOLUTION OF THE HARMONIC FUNCTIONS IN THE THEORY OF ELASTICITY
NEKATERE ZNAČILNOSTI HARMONIČNIH FUNKCIJ V TEORIJI
O ELASTNIČNOSTI
Chygyryns'kyy Valeryy Victorovich1, Shevchenko Vladimir Grigorovich1, Ilija Mamuzic2, Belikov Sergey Borisevich1
1Zaporozhskyy National Technical University, Str. Zukovsky 64, Zaporizz'a, Ukraine 2University of Zagreb, Faculty of Mettalurgy Sisak, Str. A. N. heroja 3, Sisak, Croatia
mamuzic@simet.hr
Prejem rokopisa - received: 2008-11-21; sprejem za objavo - accepted for publication: 2009-08-17
A new approach to the solution of a plane problem of the theory of elasticity with the use of two harmonic functions with a Cauchy-Riemann analytical link is developed. The analysis of the harmonic functions shows that some allow a new approach to the solution of problems of the theory of elasticity. For the solution of linear differential equations a fundamental substitution is used, written in the general form ip(x,y) = y = C„ • exp 9, with 0 = 0(x,y) as a function of the strain centre.
The transformations are explained with the properties of harmonic functions, where the Cauchy-Riemann relations can be used. The considered variants extend the possibilities for solutions and, if necessary, to obtain suitable functions for predetermined tasks. The new method is universal and can be effectively used when the fields of stresses and strains are described with trigonometric expressions.
Key words: theory of elasticity, harmonic functions, Cauchy-Riemann expressions.
Razvit je bil nov način reševanja ravninskega problema teorije elastičnosti s Cauchy-Riemanovo analitično zvezo. Analiza harmoničnih funkcij pokaže, da nekatere omogočajo nov način za rešitve problemov iz teorije elastičnosti. Za rešitev linearnih diferencialnih enačb se uporablja temeljna substitucija, zapisana v splošni obliki z \p(x,y) = y = C„ • exp 9, z 9 = 9(x,y) kot funkcijo središča deformacije.
Transformacije smo razložili z lastnostmi harmoničnih funkcij, pri katerih je dovoljena uporaba Cauchy-Riemanovih povezav. Upoštevane variante razširjajo možnost rešitev in, če je potrebno, omogočijo, da dobimo rešitve za vnaprej načrtovano uporabo. Nova metoda je univerzalna in se lahko učinkovito uporabi, če so polja napetosti in deformacij opisana s trigonometričnimi odvisnostmi.
Ključne besede: teorija elestičnosti, harmonične funkcije, Cauchy-Riemanovi izrazi
where ^(x,y) is the coordinate function of the strain 1 INTRODUCTION AND FORMULATION OF	centre; A is the constant determining the elastic state of
THE TASK	a deformable medium; F is the coordinate function
characterizing the allocation of contact shearing The analysis of the peculiarities of the harmonic stresses; a is the slope angle of an element. functions shows that some of them allow new In place of equations (1) and (2), the biharmonic approaches to the solution of problems of the theory of equation (4) can be applied:
elasticity. Let us consider a plane problem of this theory.
d 4 (p d 4 (p d 4 (p
We have a set of equilibrium equations1.	V4 p = —-+2--—+—-	(4)
dx4 d2x-d2 y dy4
dox dtxy = dTx^ do^ =
dx + dy "0; dx + dy "0	(1) with p as a stress function.
The expression fulfils the boundary conditions (3) The equationof joint strains	t^ = ^p(x, y) - sin(AF)	(5)
V (o x +oy) = 0	(2) The stress difference in (3) is determined with
The stresses'boundary conditions	o -o = 2 - y) - cos(af)	(6)
o x - o y
:—---sin2a-txy • cos2a	(3)
2 SOLUTION OF THE TASK
Applying these expressions, the harmonic law of the	The fundamental substitution is often used during the
distribution of contact stresses is determined2, which solution of linear differential equations4, which can be
formally coincides with that in3:	written in the following general form
t n =-^(x, y) - sin( AF-2a)	^(x,y) = ^ = Co •exp 9	(7)
with 6 = 6(x,y) being the unknown coordinate function of the strain centre.
Let us examine the harmonic functions AF and 6. The analytical link between them is admitted by the Cauchy-Riemann expressions4,5
6 x =±AFy 6 y =±AFx	(8)
After the derivation of equation (5), consideration of equation (7) and substitution in the equilibrium equations we obtain do x
-d^T+Co ■ 6y ■ exp 6 • sin(AF) +
+Co ■ AFy ■ exp 6■ cos(A6)=0
do y
-dy-+Co ■ 6x ■ exp 6 ■ sin(AF) +
+Co ■ AFx ■ exp6■ cos(AF)= 0
with 6y, AFx as the partial derivatives of the appropriate functions of the coordinates y and x. Passing from one variable to the other with the help of (8), we obtain, after integration and simplifications, the normal and shearing stresses
ox = Co ■ exp 6 ■ cos(AF) + f(y) + C ^ = Co ■ exp 6 ■ cos(AF)+f(x)+C	(9)
y
^ ^ = C o
■ exp 6 ■ sin( AF)
Substituting f(y) = f(x) = 0 in (9), we obtain the relation (6) that fulfils the boundary conditions for equation (3).
Considering (9), the sum of the stresses is
o x + o y = 2C
and the equation of joint strains (2) is automatically fulfilled. It is interesting that during the evaluation of the Laplacian for each value Co ■ exp 6 ■ cos(AF) and the substitution (8) the identity 0 ^ 0 is obtained. Using this peculiarity, the sum of stresses can be expressed as a product of the functions
o'=ox + oy = n ■ C ■ exp 6 ■ cos(AF) (10)
with n as the number that defines the influences of hydrostatic pressure on the medium of the stressed state in the strain zone.
By substituting (10) in (2) we obtain
V'(ox + oy) = V' [n ■ Co ■ exp 6■ cos(AF)_ = =n ■ C o ■ exp 6{[6 ^^ +(6 x - AFy )(6 x + AFy)+ +6yy +(6y - AFx)(6y + AFx)]■ cos(AF)-
- 26AFx + 26yAFy + AFxx + AFyy ]■ sin(AF)} (11)
It is clear from the analysis of the differential equation (11), that it turns to identity under the condition of
6 x =±AFy 6 y =±AFx
This is the relation (8), which was introduced as an assumption during the solution of the equilibrium equations. Differentiating further, we obtain
=±AF,„
^xy =±AFyy
=±AF^ =±AF,
The last relations convert equation (11) into identity. The last expressions show that the indicated functions are harmonic, i.e.
:+ 6 yy = 0

AFyy = 0
It is remarkable that the operators of the trigonometrical functions are equal to zero. This peculiarity shows that the function (10) also fulfils the biharmonic equation (4). Considering equation (10), the Laplace of the equation has the form
V=V' [co ■ exp 6 ■ cos(AF)] =
=Co ■exp6
'6xx +(62 - AFy'
+ 6 yy +(6 2 - AFx
cos( AF )-
(2 ■ 6 xAFx + 26 yAFy +AF + AF,„
sin(AF)
=0
+ 62 - AFy + 6yy +62 - AFx
Let us introduce the symbolisms L(x, y) = L =(
M(x, y) = M = 2■ 6x • AFx + 26y • AFy + AFxx + AF^^
Then, with consideration of the symbolisms, the accurate form of the Laplace equation is obtained
a(x, y) = a = L■ cos(AF)- M ■ sin(AF)= 0
If the factors in the trigonometrical functions are equal to zero, also the operators L = M = 0. For a more integrated analysis let us write a Laplacian for the function a(x,y)
d4 ^ d4 ^ d4 (p
V4 p =
d2
dx4 d2 ^
d2x■d2y dy4 (L■ cos(AF)- M ■ sin(AF
= V4a = d 2 2
= (Lxx - L ■ AF^; - 2Mx • AFx - M ■ AFxx + L.^ - L ■ AFy; -
-2My • AFy - M ■ AFyy) ■ cos( AF )-(2Lx • AFX +
+L ■ AFXX + Mxx - M ■ AFx + 2Ly • AFy + L ■ AFyy +
+Myy -My ■ AFy2) ■ sin(AF)
It is expected that the partial derivatives from zero functions are equal to zero, thus, V4 a ^ 0. Let us write the partial derivatives of separate operators and track the mechanism of the turning into identity of the harmonic functions
L. = (6xxx + 6yyx)+(26x 6xx -2AF.AFyx + +(26y 6yx -2AFx AFxx)
Following (8), we have
Ö =-AF 6„ = AF^
?xx =-AFyx
^^ =-AFyy
6 yy = AF^
6 yx = AFxx
6_ =-AF 6.....= AF„

= AFxxx
(12)
Let us substitute (12) in the operator Lx.
d (0xx + 0yy) + [20x 0xx - 2(eX )(0xx) ]+
L =
dx
'20 y 0 yx - 2(0 y )(e ^x )]- 0
We have obtained the identity, "quod erat demonstrandum".
The same approaches take place during the evaluation of the operator Lxx. Let us write it as
Lxx = (0xxxx - 0yyxx ) + (20xx 0xx - 2 AFyx AFyx ) + +(20x 0xxx -2AFyAFyxx)-(2AFxxAFxx -20yx 0yx)--(2AFxAFXXX -20y 0yxx)
Substituting (12) into the expression for Lxx and factoring out a flexion on x from the first brackets, after conversion the identity 0 ^ 0 is obtained. Thus, the operators
Lxx = Lyy = Mx = Myy = Lx = Ly = M^ = My = L = M = 0
demonstrate that the function (10) fulfils the biharmonic equation (4) and it can be used for the evaluation of the components of the stress tensor. It is necessary to ensure that the field of stresses and the stress function are described, as a matter of fact, with identical expressions (9) and (10) linked analytically6 with
^ =
d2 <p
dy'
o„ =
d2 p
dx2
r„. =
d2 p dydx
dx dy
dx dy
By analogy with (9) and with integration and simplifications we obtain the stress tensor components
o x = C „ • exp 0 • cos(Ae)+o + f{y )+C oy =-Co •exp0•cos(Ae) + o+f{x)+C
r ^ = C o • exp 0 • sin(Ae)	(14)
With: 0x =±A0 , 0y
Such schemes of transformations are explained with the properties of harmonic functions where the Cauchy-Riemann relations can be applied. The considered variants allow us to extend the possibility of solutions and, if necessary, to obtain suitable functions for the development of a predetermined result.
Let us return to the expressions for the stress tensor components and consider the equilibrium equations in the components of the stress deviator. Let us introduce the symbols
o x= o x - o - f (y) - C o y= o y - o - f(x) - C	(13)
with o being the mean stress.
Considering (13), the equilibrium equation (1), can be rewritten in the form6
d'o:+d:!xL=0 ^TxL+^=0
It follows from expressions (14) that their deviator part for the normal stresses Co • exp 0• cos(AF) coincides with the shifting part o in (10). Considering (13), (6) is fulfilled and the boundary conditions (3) are satisfied.
The outcome (14) can be generalized. The analytical link of the functions with the opposite signs is obtained in relations (8) and different signs of an index in an exponential curve result can be obtained. Therefore, the index of an exponential curve in a solution will be not unique. The exponential function can be written in the form of a sum with the use of the hyperbolic cosine or sine in the general form
ox =[c 1 • ch(0)±C2 • sh(0)] cos(AF)+o + f(y)+C oy =-[c 1 •ch(0)±C2 •sh(0)] cos(AF)+o + f(x)+C rxy =+[c 1 • ch(0)±C2 • sh(0)] sin(AF) (15)
In these expressions it is assumed that the arguments of the trigonometric and exponential functions can be represented in the form of a series of harmonic functions interlinked with the Cauchy-Riemann relations.
3 COMPARISON TO OTHER SOLUTIONS
The solutions of a plane problem with the help of a trigonometric series are often used. For example, the following combination of functions is often met3:
p = sin(a • x) •[c 1 • exp(a • y)+C 2 • exp(-a • y)] (16)
Let us ascertain whether the Cauchy-Riemann relation exists between the arguments of trigonometric and exponential functions
AF = a • x AFy = 0
)=±a •y ^ =±a
AFx = a 0x =x 0
The obtained relations take place for the functions
= + AF = 0=0
=±AF =±a= a
The peculiarity of these solutions is that they are common and do not contradict known partial solutions. The arguments AF and 0 are harmonic functions of the coordinates x and y. They can be rather complicated and cannot be determined from the linear dependences for one coordinate.
Let us analyze the possibilities of the solution (14). The elementary variant of a harmonic function of two variables is AF = A• x • y. Applying the relations (8), it is written as
0 = + 2• A(x2 -y2)
Thus,
0 x = +AFy = +A • x = A^ x ey =±AFx =±A •y = A •y Each function fulfils the Laplace equations.
4	CONCLUSION
A new approach to the solution of a plane problem of the theory of elasticity based on the use of two harmonic functions with the Cauchy-Riemann analytical link is developed. The new method is universal and can be effectively used when the fields of stresses and strains are described with trigonometric expressions.
5	REFERENCES
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