Open access peer-reviewed chapter

# 2D Elastostatic Problems in Parabolic Coordinates

By Natela Zirakashvili

Submitted: November 3rd 2019Reviewed: January 9th 2020Published: February 12th 2020

DOI: 10.5772/intechopen.91057

## Abstract

In the present chapter, the boundary value problems are considered in a parabolic coordinate system. In terms of parabolic coordinates, the equilibrium equation system and Hooke’s law are written, and analytical (exact) solutions of 2D problems of elasticity are constructed in the homogeneous isotropic body bounded by coordinate lines of the parabolic coordinate system. Analytical solutions are obtained using the method of separation of variables. The solution is constructed using its general representation by two harmonic functions. Using the MATLAB software, numerical results and constructed graphs of the some boundary value problems are obtained.

### Keywords

• parabolic coordinates
• separation of variables
• elasticity
• boundary
• value problem
• harmonic function

## 1. Introduction

In order to solve boundary value and boundary-contact problems in the areas with curvilinear border, it is purposeful to examine such problems in the relevant curvilinear coordinate system. Namely, the problems for the regions bounded by a circle or its parts are considered in the polar coordinate system [1, 2, 3, 4], while the problems for the regions bounded by an ellipse or its parts or hyperbola are considered in the elliptic coordinate system [5, 6, 7, 8, 9, 10, 11, 12, 13], and the problems for the regions with parabolic boundaries are considered in the parabolic coordinate system [14, 15, 16]. The problems for the regions bounded by the circles with different centers and radiuses are considered in the bipolar coordinate system [17, 18, 19]. For that purpose, first the governing differential equations are expressed in terms of the relevant curvilinear coordinates. Then a number of important problems involving the relevant curvilinear coordinates are solved.

The chapter consists of five paragraphs.

Many problems are very easily cast in terms of parabolic coordinates. To this end, first the governing differential equations discussed in present chapter are expressed in terms of parabolic coordinates; then two concrete (test) problems involving parabolic coordinates are solved.

The second section, following the Introduction, gives the equilibrium equations and Hooke’s law written down in the parabolic coordinate system and the setting of boundary value problems in the parabolic coordinate system.

Section 3 considers the method used to solve internal and external boundary value problems of elasticity for a homogeneous isotropic body bounded by parabolic curves.

Section 4 solves the concrete problems, gains the numerical results, and constructs the relevant graphs.

Section 5 is a conclusion.

## 2. Problems statement

### 2.1 Equilibrium equations and Hooke’s law in parabolic coordinates

It is known that elastic equilibrium of an isotropic homogeneous elastic body free of volume forces is described by the following differential equation [20]:

where λ=/1+ν12ν, μ=E/21νare elastic Lamé constants; νis the Poisson’s ratio; Eis the modulus of elasticity; and Uis a displacement vector.

By projecting Eq. (1) onto the tangent lines of the curves of the parabolic coordinate system (see Appendix A), we obtain the system of equilibrium equations in the parabolic coordinates.

In the parabolic coordinate system, the equilibrium equations with respect to the function D,K,u,vand Hooke’s law can be written as [20, 21, 22]:

σηη=h01λu¯,ξ+λ+2μv¯,η+λ+μμh02ξu¯+ηv¯,σξξ=h01λ+2μu¯,ξ+λv¯,η+λ+μ+μh02ξu¯+ηv¯,τξη=μh01v,ξ+u,ηh02ξv¯+ηu¯,E3

where κ=41ν, u¯=hu/c2,v¯=hv/c2, h0=ξ2+η2, h=hξ=hη=cξ2+η2are Lamé coefficients (see Appendix A), u,vare the components of the displacement vector Ualong the tangents of η,ξcurved lines, and cis the scale factor (see Appendix A). And in the present paper, we take c=1, κ2/κμDis the divergence of the displacement vector, K/μis the rotor component of the displacement vector; σξξ,σηηand τξη=τηξare normal and tangential stresses; and sub-indexes ,ξand ,ηdenotes partial derivatives with relevant coordinates, for example, K,ξ=Kξ.

### 2.2 Boundary conditions

In the parabolic system of coordinates ξ,η<ξ<0η<, exact solutions of two-dimensional static boundary value problems of elasticity are constructed for homogeneous isotropic bodies occupying domains bounded by coordinate lines of the parabolic coordinate system (see Appendix A).

The elastic body occupies the following domain (see Figures 1 and 2):

aΩ1=0<ξ<ξ1η1<η<,bΩ=ξ1ξ<ξ1η1η<.E5

Boundary conditions that appear in the chapter have the following form:

forξ=ξ1:aσξξ=F1iη,τξη=F2iηorbu=G1iη,v=G2iη,E6
forη=η1:aσηη=Q1iξ,τξη=Q2iξorbu=H1iξ,v=H2iξ,E7
forξ=0:av=0,σξξ=0orbu=0,τξη=0,E8
forη=0:au=0,σηη=0,orbv=0,τξη=0,E9
forξ1±:σηη0,τξη0,u0,v0.E10
forη:σηη0,τξη0,u0,v0,E10a

where Fi,Qii=12with the first derivative and Gi,Hiwith the first and second derivatives can be decomposed into the trigonometric absolute and uniform convergent Fourier series.

Boundary conditions on the linear parts ξ=0and η=0of the consideration area enable us to continue the solutions continuously (symmetrically or anti-symmetrically) in the domain, that is, the mirror reflection of the consideration area in a relationship y=0line (see Figures 1b and 2b).

## 3. Solution of stated boundary value problems

In this section we will be considered internal and external problems for a homogeneous isotropic body bounded by parabolic curves.

### 3.1 Interior boundary value problems

Let us find the solution of problems (2), (3), (4a) (see Figure 1a), and (7)(10) in class C2D(for Darea shown in Figure 1b). The solution is presented by two harmonious φ1and φ2functions (see Appendix B). From formulas (B11)–(B13), after inserting α=η1and making simple transformations, we will obtain:

u¯=ηφ1,ηφ2,ξ+κ1φξ+η12ηφ1,ξ+φ2,ηκ1φ2η,v¯=η12ηφ1,ηφ2,ξ+κ1φ1η+ηφ1,ξ+φ2,ηκ1φ2ξ;E11
D=κμh02φ1,ηφ2,ξηφ1,ξ+φ2,ηξ,K=κμh02φ1,ηφ2,ξξ+φ1,ξ+φ2,ηη,

where

1h2φi,ξξ+φi,ηη=0,i=1,2.E12

The stress tensor components can be written as

h022μσηη=η12ηϕ1,ξξ+ϕ2,ξηκ2ϕ1,ηκ22ϕ2,ξη   +ηϕ1,ξηϕ2,ηη+κ22ϕ1,ξκ2ϕ2,ηξη12ηξ2+η2ϕ1,ηϕ2,ξηϕ1,ξ+ϕ2,ηξ,E13
h022μτξη=η12ηϕ1,ξηϕ2,ξξ+κ22ϕ1,ξκ2ϕ2,ηη   +ηϕ1,ξξ+ϕ2,ξηκ2ϕ1,ηκ22ϕ2,ξξη12ηξ2+η2ϕ1,ηϕ2,ξξ+ϕ1,ξ+ϕ2,ηη,
h022μσξξ=η12ηϕ1,ξξ+ϕ2,ξηκ42ϕ1,ηκ+22ϕ2,ξη    ηϕ1,ξηϕ2,ξξ+κ+22ϕ1,ξκ42ϕ2,ηξ+η12ηξ2+η2ϕ1,ηϕ2,ξηϕ1,ξ+ϕ2,ηξ.

From (12) by the separation of variables method, we obtain (see Appendix A)

φi=n=1φin,i=1,2,E14

where

φ1n==A1ncoshcos,φ2n==A2nsinhsin

or

φ1n==A1nsinhsin,φ2n==A2ncoshcos.

For n=0: φ10=A10+a02ξ++a03η+a04ξη, φ20=A20+b02ξ+b03η+b04ξη, where A10,a02,,b04are constant coefficients. When n=0and 0<ξ<ξ1, then the terms ξ,ηand ξηwill not be contained in φ10and φ20. If the foregoing solutions are presented in expressions of φ10and φ20, then it would be impossible on ξ=ξ1to satisfy the boundary conditions, and gradφi0=φi0,ξ+φi0,η/hi=12will not be bounded in the point M00.

Provision. We are introducing the following assumptions:

1. ξ1is a sufficiently great positive number (see Appendix C).

2. The boundary conditions given on η=η1, i.e., stresses or displacements equal zero at interval ξ˜1<ξ<ξ1.

3. When stresses are given on η=η1, the main vector and main moment equal zero.

It is clear that

D=κσξξ+σηη/4,σξξ=4D/κσηη.

By ultimately opening expressions σηηand τξη(in details), we can demonstrate that at point M00, σηηand τξη(and naturally, σξξ, too) are determined, i.e., they are finite.

When at η=η1u¯and v¯are given, then it is expedient to take instead of them as their equivalent the following expressions:

1h02u¯η1+v¯ξ=η1φ1,ξ+φ2,ηκ1φ2,1h02u¯ξv¯η1=η1φ1,ηφ2,ξ+κ1φ1,E15

and if at η=η1h022μσηηand h022μσξηare given, then instead of them we have to take their equivalent following expressions:

12μσηηη1σξηξ=η1φ1,ξξ+φ2,ξηκ2φ1,ηκ22φ2,ξ,12μσηηξ+σξηη1=η1φ1,ξηφ2,ξξ+κ22φ1,ξκ2φ2,η.E16

Considering the homogeneous boundary conditions of the concrete problem, we will insert φ1and φ2functions selected from the (14) in the right sides of (15) or (16), and we will expand the left sides in the Fourier series. In both sides expressions which are with identical combinations of trigonometric functions will equate to each other and will receive the infinite system of linear algebraic equations to unknown coefficients A1nand A2nof harmonic functions, with its main matrix having a block-diagonal form. The dimension of each block is 2×2, and determinant is not equal to zero, but in infinite the determinant of block strives to the finite number different to zero.

It is very easy to establish the convergence of (11) and (13) functional series on the area D¯=ξ1ξξ10ηη1by construction of the corresponding uniform convergent numerical majorizing series. So we have the following:

Proposal 1. The functional series corresponding to (11) and (13) are absolute and uniform by convergent series on the area D¯=ξ1ξξ10ηη1.

### 3.2 Exterior boundary value problems

We have to find the solution of problems (2), (3), (5a) (see Figure 2a), (7), (8), (10), and (10′), which belongs to the class C2Ω(see region Ωon Figure 2b). The solution is constructed using its general representation by harmonic functions φ1, φ2(see Appendix B). From formulas (B11)–(B13), following inserting α=η1and simple transformations, we obtain the following expressions:

u¯=φ1,ξ+φ2,ηη1+φ1,ηφ2,ξξηη1κ1φ1+φ3,ηξκ1φ2φ3,ξη,v¯=φ1,ξ+φ2,ηξφ1,ηφ2,ξη1ηη1+κ1φ1+φ3,ηηκ1φ2φ3,ξξ,E17
D=κμh02φ1,ηφ2,ξηφ1,ξ+φ2,ηξ,K=κμh02φ1,ηφ2,ξξ+φ1,ξ+φ2,ηη,

where

1h2φi,ξξ+φi,ηη=0,i=1,2,3.E18

The stress tensor components can be written as:

h022μσηη=ϕ1,ξξ+ϕ2,ξηη1+ϕ1,ξηϕ2,ξξξηη0+κ2ϕ1,η+κ22ϕ2,ξϕ3,ξξη            +κ22ϕ1,ξκ2ϕ2,η+ϕ3,ξηξ+η2η12ξ2+η2ϕ1,ηϕ2,ξηϕ1,ξ+ϕ2,ηξ,
h022μτξη=ϕ1,ξξ+ϕ2,ξηξϕ1,ξηϕ2,ξξη`1ηη1κ2ϕ1,η+κ22ϕ2,ξϕ3,ξξξ            +κ22ϕ1,ξκ2ϕ2,η+ϕ3,ξηη +η2η12ξ2+η2ϕ1,ηϕ2,ξξϕ1,ξ+ϕ2,ηη ,E19
h022μσξξ=ϕ1,ξξ+ϕ2,ξηη1+ϕ1,ξηϕ2,ξξξηη0κ42ϕ1,η+κ+22ϕ2,ξϕ3,ξξη            κ+22ϕ1,ξκ22ϕ2,η+ϕ3,ξηξη2η12ξ2+η2ϕ1,ηϕ2,ξηϕ1,ξ+ϕ2,ηξ.

If u¯and v¯are given for η=η1, then we take φ3=0, and when h022μσηηand h022μσξηis given for η=η1, then φ3=κ22φ2.

From (18), by the separation of variables method, we obtain

φi=n=1φin,i=1,2,3,E20

where

φ1n=B1nesin,φ2n=B2necos,φ3n=κ22nB2nesin

or

φ1n=B1necos,φ2n=B2nesin,φ3n=κ22nB2necos.

When n=0, then φ10=A10+a02ξ+a03η+a04ξη,φ20=A20+b02ξ+b03η+b04ξη,where A10,a02,,b04are constants. From limited of functions φi0i=12in ηand satisfying boundary condition for ξ=ξ1, it implies that a02=0, b02=0, a03=0, b03=0,a04=0, b04=0. Therefore, φ10=0, φ20=A20or φ10=A10, φ20=0.

Provision. As in the previous subsection we make the following assumptions:

• ξ1is a sufficiently large positive number (see Appendix C).

• At η=η1given boundary conditions, i.e., displacements or stresses on interval ξ˜1<ξ<ξ1, will equal zero.

• When stresses are given on η=η1, the main vector and main moment will equal zero.

When u¯and v¯are given at η=η1, then instead of them, it is expedient to take the following expressions as their equivalent:

1h02κ1u¯ξv¯η1=φ1,1h02κ1u¯η1+v¯ξ=φ2,E21

and if at η=η1h022μσηηand h022μσξηare given, then instead of them we have to take the following expressions as their equivalent:

12μσηηη1σξηξ=κ2φ1,η,12μσηηξ+σξηη1=κ22φ1,ξφ2,η.E22

Just like that in the previous subsection, considering the homogeneous boundary conditions of the concrete problem, we will insert φ1and φ2functions selected from (20) in Eq. (21) or (22), and we will expand the left sides in the Fourier series. Both sides of the expressions, which show the identical combinations of trigonometric functions, will equate to each other and will receive the infinite system of linear algebraic equations to unknown coefficients A1nand A2nof harmonic functions, with its main matrix having a block-diagonal form. The dimension of each block is 2×2, and the determinant does not equate to zero, but in the infinity, the determinant of block tends to the finite number different from zero.

As in the previous subsection, we received the following:

Proposition 2. The functional series corresponding to (17) and (19) are absolute and a uniformly convergent series on region Ω¯=ξ1ξξ1η1η<.

## 4. Test problems

In this section we will be obtained numerical results of internal and external problems for a homogeneous isotropic body bounded by parabolic curves when normal stress distribution is applied to the parabolic border.

### 4.1 Internal problem

We will set and solve the concrete internal boundary value problem in stresses. Let us find the solution of equilibrium equation system (2) of the homogeneous isotropic body in the area Ω1=0<ξ<ξ10<η<η1(see Figure 1a), which satisfies boundary conditions (7a), (8a), (9a), and (10).

From (14), (8a), and (9a)

φi=n=1φin,i=1,2,E23

where φ1n==A1nsinhsin,φ2n==A2ncoshcos.

By inserting (23) in (11) and (13), we will receive the following expressions for the displacements:

u¯=n=1nηξcoshA1n+A2n+κ1ξsinhA1nsin+nη12sinhA1n+A2nκ1ηcoshA2ncos,v¯=n=1nη12coshA1n+A2n+κ1ηsinhA1nsin+nηξsinhA1n+A2nκ1ξcoshA2ncos,E24

but for the stresses the following:

h022μσηη=n=1n2η12sinhA1n+A2n+coshκ2A1nκ22A2nsin+n2ηξcoshA1n+A2n+sinhκ22A1nκ2A2ncosη12η2ξ2+η2coshA1n+A2nsinsinhA1n+A2ncos,
h022μτξη=n=1n2η12coshA1n+A2n+sinhκ22A1nκ2A2ncosn2ηξsinhA1n+A2n+coshκ2A1nκ22A2nsinη12η2ξ2+η2coshA1n+A2nsin+sinhA1n+A2ncos,E25
h022μσξξ=n=1n2η12sinhA1n+A2n+coshκ42A1nκ+22A2nsinn2ηξcoshA1n+A2n+sinhκ+22A1nκ42A2ncos+η12η2ξ2+η2coshA1n+A2nsinsinhA1n+A2ncos.

We have to solve problem (2), (7a), (8a), and (9a) when Q1ξ=Pand Q2ξ=0, i.e., at η=η1boundary the normal load 12μσηη=Ph02is given, but tangent stress is equal to zero. From (16), and (23), we obtain the following equations:

n=1n2η1sinhnη1A1n+A2nncoshnη1κ2A1nκ22A2nsin=Pη1ξ2+η12,n=1n2η1coshnη1A1n+A2n+nsinhnη1κ22A1nκ2A2ncos=ξ2+η12.

From here an infinite system of the linear algebraic equations with unknown A1nand A2ncoefficients is obtained:

n2η1sinhnη1nκ2coshnη1A1n+n2η1sinhnη1+nκ22coshnη1A2n=F˜1n,n2η1coshnη1+nκ22sinhnη1A1n+n2η1coshnη1nκ2sinhnη1A2n=F˜2n,n=1,2,E26

where F˜1nand F˜2nare the coefficients of expansion into the Fourier series f1ξ=n=1F˜1nsinand f2ξ=n=1F˜2ncos, respectively, f1ξ=Pη1ξ2+η12and f2ξ=ξ2+η12functions.

As seen, the main matrix of system (26) has a block-diagonal form, dimension of each block is 2×2. Thus, two equations with two A1nand A2nunknown values will be solved. After solving this system, we find A1nand A2ncoefficients, and in putting them into formulas (24) and (25), we get displacements and stresses at any points of the body.

Numerical values of displacements and stresses are obtained at the points of the finite size region bounded by curved lines η=η1and ξ=ξ1(see Figure 1a), and relevant 3D graphs are drafted. The numerical results are obtained for the following data: ν=0.3, E=2×106kg/cm2, P=10kg/cm2, 0.1η13, ξ1=2π, ξ1=4π, and ξ1=6π. Numerical calculations and the visual presentation are made by MATLAB software.

Figures 3 and 4 show the distribution of stresses and displacements in the region bounded by curved lines η=η1and ξ=ξ1kξ1(see Figure 1a), when (7a), (8a), and (9a) boundary conditions are valid and normal stress is applied to the parabolic boundary. Following conditions (8a) and (9a), at points of the linear parts ξ=0and η=0of consideration area σξξ0η, σηηξ0stresses and uξ0, v0ηdisplacements equal zero which is seen in Figures 3 and 4.

### 4.2 External problem

We will set and solve the concrete external boundary value problem in stresses. Let us find the solution of equilibrium equation system (2) of the homogeneous isotropic body in the region Ω1=0<ξ<ξ1η1<η<, which satisfies the following boundary conditions: (7a), (8a), (10), and (10′).

From (20) and (8a)

φi=n=1φin,i=1,2,3,E27

where φ1n=B1nesin,φ2n=B2necos,φ3n=κ22nB2nesin.

By inserting (27) in (17) and (19), we will obtain the following expressions for displacements:

u¯=n=0neB1nB2nη1cos+B1nB2nξsinηη1eκ1B1nκ2B2nξsinκ2eB2nηcos,v¯=n=1neB1nB2nξcos+B1nB2nη1sinηη1+eκ1B1nκ2B2nηsinκ2eB2nξcos,E28

and for the stresses, we obtain the following formula:

h022μσηη=n=1n2eB1nB2nη1sin+B1nB2nξcosηη1neκ2B1nηsinκ22B1n+B2nξcosη2η12ξ2+η2neB1nB2nηsin+ξcos,
h022μτξη=n=1n2eB1nB2nξsinB1nB2nη1cosηη1neκ2B1nξsinκ22B1n+B2nηcos,η2η12ξ2+η2neB1nB2nξsin+ηcos,E29
h022μσξξ=n=1n2eB1nB2nη1sin+B1nB2nξcosηη1+neκ42B1n+2B2nηsin+κ+22B1nξcos+η2η12ξ2+η2neB1nB2nηsin+ξcos.

Next, we will obtain the numerical results of the following example.

We have to solve problem (2), (7a), and (8a), when Q1ξ=Pand Q2ξ=0, i.e., at η=η1boundary the normal load 12μσηη=Ph02is given, but tangent stress is equal to zero. From (22) and (27), we obtain the following equations:

n=1nenη1κ2B1nsin=Pη1ξ2+η12,n=1nenη1κ22B1n+B2ncos=ξ2+η12.

Consequently, we obtain the infinite system of the linear algebraic equations with unknown B1nand B2ncoefficients:

n=1nenη1κ2B1nsin=n=1P˜1nsin,n=1nenη1κ22B1n+B2ncos=n=1P˜2ncos,i.e.,

nenη1κ2B1n=P˜1n,nenη1κ22B1n+B2n=P˜2n,n=1,2,.E30

Hence,

B1n=2κnenη1P˜1n,B2n=enη1nP˜2n+κ2κP˜1n,

where P˜1nand P˜2nare the coefficients of expansion into the Fourier series of functions f1ξ=Pη1ξ2+η12and f2ξ=ξ2+η12, respectively (f1ξ, according to sinuses, and f2ξ, according to cosines).

As it can be seen, the main matrix of system (30) has a block-diagonal form, and the dimension of each block is 2×2. Thus, two equations with two B1nand B2nunknown values will be solved. After solving this system, we find the values of B1nand B2ncoefficients and put them into formulas (28) and (29) to get displacements and stresses at any points of the body.

Numerical results are obtained for some characteristic points of the body, in particular, M10η1, M2ξ1η1points (see. Figure 2a), for the following data: ν=0.3, E=2106kg/cm2, P=10kg/cm2, 0.01η13, ξ1=2π, ξ1=4π, and ξ1=6π.

The above-presented graphs (see Figures 5 and 6) show how displacements and stresses change at some characteristic points of body, namely, at points M1j0η1jand M2jξ1η1jj=128, when 0.01η13(see Figure 7).

From the presented results, we obtain the following:

• At points M1j0η1j, maxut<maxun,vt=vn=0.

• At points M2jξ1η1j, maxσξξt>maxσξξn,maxut>maxun,maxvt<maxvn.

• When ξ1, then displacements and stresses tend to zero, that is, the boundary conditions (10) are satisfied.

• When η1, then displacements and stresses tend to zero, that is, the boundary conditions (10′) are satisfied.

• When η10(in this case there is a crack), then (a) at points M1j0η1jtangential stresses and normal displacements tend to , but other components equal to zero. It can be seen from the boundary conditions (8a) (b) at points M2jξ1η1jthat all components of the displacements and stresses tend to .

Here superscript tand ndenote the tangential and normal displacement or the stress, respectively.

## 5. Conclusion

The main results of this chapter can be formulated as follows:

• The equilibrium equations and Hooke’s law are written in terms of parabolic coordinates.

• The solution of the equilibrium equations is obtained by the method of separation of variables. The solution is constructed using its general representation by harmonic functions.

• In parabolic coordinates, analytical solutions of 2D static boundary value problems for the elasticity are constructed for homogeneous isotropic finite and infinite bodies occupying domains bounded by coordinate lines of parabolic coordinate system.

• Two concrete internal and external boundary value problems in stresses are set and solved.

The bodies bounded by the parabola are common in practice, for example, in building, mechanical engineering, biology, medicine, etc., the study of the deformed state of such bodies is topical, and consequently, in my opinion, setting the problems considered in the chapter and the method of their solution is interesting in a practical view.

In orthogonal parabolic coordinate system ξ,η(<ξ<,0η<, see Figure A1) [23, 24]; we have

hξ=hη=h=cξ2+η2,x=cξ2η2/2,y=cξη,

where hξ,hηare Lame's coefficients of the system of parabolic coordinates, cis a scale coefficient, x,yare the Cartesian coordinates.

The coordinate axes are parabolas

y2=2cξ02xcξ02/2,ξ0=const,y2=2cη02x+cη02/2,η0=const.

Laplace’s equation Δf=0, where f=fξη, in the parabolic coordinates has the form

f,ξξ+f,ηη/c2ξ2+η2=0.

We have to find solution of the equation in following form

f=XξEη,

and then by separation of variables, we will receive

1c2ξ2+η2X"X+E'E=0.

From here

X"+mX=0,E"mE=0,

where mis any constant, their solutions are [25]

X=C1cos+C2sin,E=C3e+C4e=C3cosh+C4sinh.

So

fξη=C3e+C4eC1cos+C2sinorfξη=C3cosh+C4sinhC1cos+C2sin,

We solve the system of partial differential equations (2).

We have introduced φ1harmonic function, and if we take

then Eqs. (2a) and (2b) will be satisfied identically, while Eqs. (2c) and (2d) will receive the following form:

au¯,ξ+v¯,η=κ2φ1,ηηφ1,ξξ,bv¯,ξu¯,η=κφ1,ηξ+φ1,ξη,EB2
au¯,ξ+v¯,η=κ2φ1,ηηφ1,ξξ,bv¯κφ1η,ξ=u¯+κφ1ξ,η.EB3

From equation (B3b) imply that exists such type harmonic function φ, for which fulfill the following

u¯=φ,ξκφ1ξ,v¯=φ,η+κφ1η.EB4

Considering (B4), from Equation (B3a), the following will be obtained:

h2Δφ=φ,ξξ+φ,ηη=κφ1+κφ1,ξξκφ1κφ1,ηη+κ2φ1,ηηφ1ξξ=2φ1,ξξφ1,ηη.EB5

General solution of the system (B2) can be written in the form u¯=ψ1,v¯=ψ2, where

ψ1,ξ+ψ2,η=0,ψ2,ξψ1,η=0.

The full solution of equation system (B2) is written in the following form:

u¯=φ,ξκφ1ξ+ψ1,v¯=φ,η+κφ1η+ψ2,EB6

where φis the partial solution of the (B5).

If we take κ=const, then

φ=ξ2η22φ1,

and (B6) formula will receive the following form:

u¯=ξ2η22φ1,ξκ1φ1ξ+ψ1,v¯=ξ2η22φ1,η+κ1φ1η+ψ2.

From here

u¯=ξ2η22φ1,ξ+ξηφ1,ηξηφ1,ηκ1φ1ξ+ψ1,v¯=ξ2η22φ1,ηξηφ1,ξ+ξηφ1,ξ+κ1φ1η+ψ2.

Without losing the generality, the expression in brackets can be taken as zero, because we already have in u¯and v¯of the solutions Laplacian (we mean ψ1and ψ2). Therefore, the solutions of system (2) are given in the following form:

ah02D=κμφ1,ηηφ1,ξξ,bh02K=κμφ1,ηξ+φ1,ξη,cu¯=ξηφ1,ηκ1φ1ξ+ψ1,dv¯=ξηφ1,ξ+κ1φ1η+ψ2.EB7

Now we have to write down three versions of ψ1and ψ2function representation. In the first version

ψ1=φ¯1,η+φ˜1,η+φ2,η,ψ2=φ¯1,ξ+φ˜1,ξ+φ2,ξ,EB8

φ¯1,φ˜1,φ2are harmonic functions; in addition, φ¯1,φ˜1are selected so that at η=α, where α=η1or α=η2, the following equations will be satisfied:

ξηφ1,ηκ1φ1ξ+φ¯1,η+φ˜1,η=0,ξηφ1,ξ+κ1φ1ξ+φ¯1,ξ+φ˜1,ξ=0,

In the second version

ψ1=αξ2ηα22φ1,ξ+ξηφ1,η+ξ2η22φ2,ξ+ξηφ2,η,ψ2=αξηφ1,ξξ2ηα22φ1,η+ξ2η22φ2,ηξηφ2,ξ,EB9

where φ2is the harmonic function.

In the third version

ψ1=α2ξ2η22φ1,ξ+ξηφ1,η+ξ2η22φ2,ξ+ξηφ2,η,ψ2=α2ξηφ1,ξξ2η22φ1,η+ξ2η22φ2,ηξηφ2,ξ.EB10

Inserting (B8) in (B7c and d), we will get

au¯=ξηφ1,ηκ1φ1ξ+φ¯1,η+φ˜1,η+φ2,η,bv¯=ξηφξ+κ1φ1ξ+φ¯1,ξ+φ˜1,ξ+φ2,ξ.EB11

Inserting (B9) in (B7c and d), we will have

au¯=αξ2ηα22φ1,ξ+ξηφ1,ηξηφ1,ηκ1φ1ξ+ξ2η22φ2,ξ+ξηφ2,η,bv¯=αξηφ1,ξξ2ηα22φ1,η+ξηφ1,ξ+κ1φ1η+ξ2η22φ2,ηξηφ2,ξ.EB12

Inserting (B10) in (B7c and d), we will get

au¯=α2ξ2η22φ1,ξ+ξηφ1,ηξηφ1,ηκ1φ1ξ+ξ2η22φ2,ξ+ξηφ2,η,bv¯=α2ξηφ1,ξξ2η22φ1,η+ξηφ1,ξ+κ1φ1η+ξ2η22φ2,ηξηφ2,ξ.EB13

After the boundary value problem with relevant boundary conditions on ξ=ξ1=ξ11is solved, the following condition is examined: F11/F10<ε.

εis a sufficiently small positive number given in advance (ε=0,0010,0001).

F11=0η1σξξ+σηη+τξηhdηξ=ξ1,F10=0η1σξξ+σηη+τξηhdηξ=gξ˜1.

gnumber will be selected so that on boundary η=η1, point Mgξ˜1η1should correspond to the highest value of expressionσηηgξ˜1η12+τξηgξ˜1η12(when stresses are given) or to the highest value of expression u¯gξ˜1η12+v¯gξ˜1η12(when displacements are given).

If condition F11/F10<εis not valid forξ1=ξ11, the same problem will be solved at the beginning, but ξ1=ξ12will be used instead of ξ1=ξ11. In addition, ξ12>ξ11. Then, if condition F12/F10<εis not still valid, we will continue with the boundary problem, where ξ1=ξ13; besides, ξ13>ξ12>ξ11, and we will examine condition F13/F10<ε. The process will be over at the kth stage, if condition F1k/F10<εis valid.

Finding such ξ1=ξ1k, for which F1k/F10<ε.

Distance lbetween surfaces ξ=ξ1and ξ=ξ˜1, which gives the guarantee for condition F1k/F10<εto be valid in the parabolic coordinate system, will be taken along the axis of the parabola , and the following expression will be obtained:

ξ1=l/c+ξ˜12.

By relying on the known solutions of the relevant plain problems of elasticity, it is purposeful to admit that l/c=4,5,6,,which allows finding ξ1from the relevant equation. Let us note that when l/c=4, we will denote value ξ1by ξ11, when l/c=5; by ξ12, when l/c=6; by ξ13, etc. If after selecting ξ1=ξ1k, inequality F1k/F10<εis valid; in order to check the righteousness of the selection, it is necessary to once again make sure that, together with condition F1k/F10<ε, condition ε>F1k/F10>F1k+1/F10>F1k+2/F10>is valid, too.

## Notations

x,y

Cartesian coordinates

ξ,η

parabolic coordinates

E and v

modulus of elasticity and Poisson’s ratio

λ, μ

elastic Lamé constants

Uuv

displacement vector

σξξ,σηη, τξη=τηξ

normal and tangential stresses

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Natela Zirakashvili (February 12th 2020). 2D Elastostatic Problems in Parabolic Coordinates, Solid State Physics - Metastable, Spintronics Materials and Mechanics of Deformable Bodies - Recent Progress, Subbarayan Sivasankaran, Pramoda Kumar Nayak and Ezgi Günay, IntechOpen, DOI: 10.5772/intechopen.91057. Available from:

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