Open access peer-reviewed chapter

# 2D Elastostatic Problems in Parabolic Coordinates

Written By

Natela Zirakashvili

Submitted: 03 November 2019 Reviewed: 09 January 2020 Published: 12 February 2020

DOI: 10.5772/intechopen.91057

From the Edited Volume

## Solid State Physics - Metastable, Spintronics Materials and Mechanics of Deformable Bodies - Recent Progress

Edited by Subbarayan Sivasankaran, Pramoda Kumar Nayak and Ezgi Günay

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## 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]:

λ + 2 μ grad div U μ rot rot U = 0 E1

where λ = / 1 + ν 1 2 ν , μ = E / 2 1 ν are elastic Lamé constants; ν is the Poisson’s ratio; E is the modulus of elasticity; and U is 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 , v and Hooke’s law can be written as [20, 21, 22]:

a D , ξ K , η = 0 , c u ¯ , ξ + v ¯ , η = κ 2 / κμ h 0 2 D , b D , η + K , ξ = 0 , d v ¯ , ξ u ¯ , η = 1 / μ h 0 2 K . E2
σ ηη = h 0 1 λ u ¯ , ξ + λ + 2 μ v ¯ , η + λ + μ μ h 0 2 ξ u ¯ + η v ¯ , σ ξξ = h 0 1 λ + 2 μ u ¯ , ξ + λ v ¯ , η + λ + μ + μ h 0 2 ξ u ¯ + η v ¯ , τ ξη = μ h 0 1 v , ξ + u , η h 0 2 ξ v ¯ + η u ¯ , E3

where κ = 4 1 ν , u ¯ = hu / c 2 , v ¯ = h v / c 2 , h 0 = ξ 2 + η 2 , h = h ξ = h η = c ξ 2 + η 2 are Lamé coefficients (see Appendix A), u , v are the components of the displacement vector U along the tangents of η , ξ curved lines, and c is the scale factor (see Appendix A). And in the present paper, we take c = 1 , κ 2 / κμ D is the divergence of the displacement vector, K / μ is the rotor component of the displacement vector; σ ξξ , σ ηη and τ ξη = τ ηξ are normal and tangential stresses; and sub-indexes , ξ and , η denotes partial derivatives with relevant coordinates, for example, K , ξ = K ξ .

### 2.2 Boundary conditions

In the parabolic system of coordinates ξ , η < ξ < 0 η < , 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):

a D 1 = 0 < ξ < ξ 1 0 < η < η 1 , b D = ξ 1 < ξ < ξ 1 0 < η < η 1 , E4
a Ω 1 = 0 < ξ < ξ 1 η 1 < η < , b Ω = ξ 1 ξ < ξ 1 η 1 η < . E5

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

for ξ = ξ 1 : a σ ξξ = F 1 i η , τ ξη = F 2 i η or b u = G 1 i η , v = G 2 i η , E6
for η = η 1 : a σ ηη = Q 1 i ξ , τ ξη = Q 2 i ξ or b u = H 1 i ξ , v = H 2 i ξ , E7
for ξ = 0 : a v = 0 , σ ξξ = 0 or b u = 0 , τ ξη = 0 , E8
for η = 0 : a u = 0 , σ ηη = 0 , or b v = 0 , τ ξη = 0 , E9
for ξ 1 ± : σ ηη 0 , τ ξη 0 , u 0 , v 0 . E10
for η : σ ηη 0 , τ ξη 0 , u 0 , v 0 , E10a

where F i , Q i i = 1 2 with the first derivative and G i , H i with the first and second derivatives can be decomposed into the trigonometric absolute and uniform convergent Fourier series.

Boundary conditions on the linear parts ξ = 0 and η = 0 of 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 = 0 line (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 C 2 D (for D area shown in Figure 1b). The solution is presented by two harmonious φ 1 and φ 2 functions (see Appendix B). From formulas (B11)–(B13), after inserting α = η 1 and making simple transformations, we will obtain:

u ¯ = η φ 1 , η φ 2 , ξ + κ 1 φ ξ + η 1 2 η φ 1 , ξ + φ 2 , η κ 1 φ 2 η , v ¯ = η 1 2 η φ 1 , η φ 2 , ξ + κ 1 φ 1 η + η φ 1 , ξ + φ 2 , η κ 1 φ 2 ξ ; E11
D = κμ h 0 2 φ 1 , η φ 2 , ξ η φ 1 , ξ + φ 2 , η ξ , K = κμ h 0 2 φ 1 , η φ 2 , ξ ξ + φ 1 , ξ + φ 2 , η η ,

where

1 h 2 φ i , ξξ + φ i , 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)

φ i = n = 1 φ in , i = 1 , 2 , E14

where

φ 1 n = = A 1 n cosh cos , φ 2 n = = A 2 n sinh sin

or

φ 1 n = = A 1 n sinh sin , φ 2 n = = A 2 n cosh cos .

For n = 0 : φ 10 = A 10 + a 02 ξ + + a 03 η + a 04 ξη , φ 20 = A 20 + b 02 ξ + b 03 η + b 04 ξη , where A 10 , a 02 , , b 04 are constant coefficients. When n = 0 and 0 < ξ < ξ 1 , then the terms ξ , η and ξη will not be contained in φ 10 and φ 20 . If the foregoing solutions are presented in expressions of φ 10 and φ 20 , then it would be impossible on ξ = ξ 1 to satisfy the boundary conditions, and grad φ i 0 = φ i 0 , ξ + φ i 0 , η / h i = 1 2 will not be bounded in the point M 0 0 .

Provision. We are introducing the following assumptions:

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

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

3. When stresses are given on η = η 1 , the main vector and main moment equal zero.

It is clear that

D = κ σ ξξ + σ ηη / 4 , σ ξξ = 4 D / κ σ ηη .

By ultimately opening expressions σ ηη and τ ξη (in details), we can demonstrate that at point M 0 0 , σ ηη and τ ξη (and naturally, σ ξξ , too) are determined, i.e., they are finite.

When at η = η 1 u ¯ and v ¯ are given, then it is expedient to take instead of them as their equivalent the following expressions:

1 h 0 2 u ¯ η 1 + v ¯ ξ = η 1 φ 1 , ξ + φ 2 , η κ 1 φ 2 , 1 h 0 2 u ¯ ξ v ¯ η 1 = η 1 φ 1 , η φ 2 , ξ + κ 1 φ 1 , E15

and if at η = η 1 h 0 2 2 μ σ ηη and h 0 2 2 μ σ ξη are given, then instead of them we have to take their equivalent following expressions:

1 2 μ σ ηη η 1 σ ξη ξ = η 1 φ 1 , ξξ + φ 2 , ξη κ 2 φ 1 , η κ 2 2 φ 2 , ξ , 1 2 μ σ ηη ξ + σ ξη η 1 = η 1 φ 1 , ξη φ 2 , ξξ + κ 2 2 φ 1 , ξ κ 2 φ 2 , η . E16

Considering the homogeneous boundary conditions of the concrete problem, we will insert φ 1 and φ 2 functions 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 A 1 n and A 2 n of 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 ¯ = ξ 1 ξ ξ 1 0 η η 1 by 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 ¯ = ξ 1 ξ ξ 1 0 η η 1 .

### 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 C 2 Ω (see region Ω on Figure 2b). The solution is constructed using its general representation by harmonic functions φ 1 , φ 2 (see Appendix B). From formulas (B11)–(B13), following inserting α = η 1 and simple transformations, we obtain the following expressions:

u ¯ = φ 1 , ξ + φ 2 , η η 1 + φ 1 , η φ 2 , ξ ξ η η 1 κ 1 φ 1 + φ 3 , η ξ κ 1 φ 2 φ 3 , ξ η , v ¯ = φ 1 , ξ + φ 2 , η ξ φ 1 , η φ 2 , ξ η 1 η η 1 + κ 1 φ 1 + φ 3 , η η κ 1 φ 2 φ 3 , ξ ξ , E17
D = κμ h 0 2 φ 1 , η φ 2 , ξ η φ 1 , ξ + φ 2 , η ξ , K = κμ h 0 2 φ 1 , η φ 2 , ξ ξ + φ 1 , ξ + φ 2 , η η ,

where

1 h 2 φ i , ξξ + φ i , 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 = η 1 , then we take φ 3 = 0 , and when h 0 2 2 μ σ ηη and h 0 2 2 μ σ ξη is given for η = η 1 , then φ 3 = κ 2 2 φ 2 .

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

φ i = n = 1 φ in , i = 1 , 2 , 3 , E20

where

φ 1 n = B 1 n e sin , φ 2 n = B 2 n e cos , φ 3 n = κ 2 2 n B 2 n e sin

or

φ 1 n = B 1 n e cos , φ 2 n = B 2 n e sin , φ 3 n = κ 2 2 n B 2 n e cos .

When n = 0 , then φ 10 = A 10 + a 02 ξ + a 03 η + a 04 ξη , φ 20 = A 20 + b 02 ξ + b 03 η + b 04 ξη , where A 10 , a 02 , , b 04 are constants. From limited of functions φ i 0 i = 1 2 in η and satisfying boundary condition for ξ = ξ 1 , it implies that a 02 = 0 , b 02 = 0 , a 03 = 0 , b 03 = 0 , a 04 = 0 , b 04 = 0 . Therefore, φ 10 = 0 , φ 20 = A 20 or φ 10 = A 10 , φ 20 = 0 .

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

• ξ 1 is a sufficiently large positive number (see Appendix C).

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

• When stresses are given on η = η 1 , the main vector and main moment will equal zero.

When u ¯ and v ¯ are given at η = η 1 , then instead of them, it is expedient to take the following expressions as their equivalent:

1 h 0 2 κ 1 u ¯ ξ v ¯ η 1 = φ 1 , 1 h 0 2 κ 1 u ¯ η 1 + v ¯ ξ = φ 2 , E21

and if at η = η 1 h 0 2 2 μ σ ηη and h 0 2 2 μ σ ξη are given, then instead of them we have to take the following expressions as their equivalent:

1 2 μ σ ηη η 1 σ ξη ξ = κ 2 φ 1 , η , 1 2 μ σ ηη ξ + σ ξη η 1 = κ 2 2 φ 1 , ξ φ 2 , η . E22

Just like that in the previous subsection, considering the homogeneous boundary conditions of the concrete problem, we will insert φ 1 and φ 2 functions 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 A 1 n and A 2 n of 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 ¯ = ξ 1 ξ ξ 1 η 1 η < .

## 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 Ω 1 = 0 < ξ < ξ 1 0 < η < η 1 (see Figure 1a), which satisfies boundary conditions (7a), (8a), (9a), and (10).

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

φ i = n = 1 φ in , i = 1 , 2 , E23

where φ 1 n = = A 1 n sinh sin , φ 2 n = = A 2 n cosh cos .

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

u ¯ = n = 1 nηξ cosh A 1 n + A 2 n + κ 1 ξ sinh A 1 n sin + n η 1 2 sinh A 1 n + A 2 n κ 1 η cosh A 2 n cos , v ¯ = n = 1 n η 1 2 cosh A 1 n + A 2 n + κ 1 η sinh A 1 n sin + nηξ sinh A 1 n + A 2 n κ 1 ξ cosh A 2 n cos , E24

but for the stresses the following:

h 0 2 2 μ σ ηη = n = 1 n 2 η 1 2 sinh A 1 n + A 2 n + cosh κ 2 A 1 n κ 2 2 A 2 n sin + n 2 ηξ cosh A 1 n + A 2 n + sinh κ 2 2 A 1 n κ 2 A 2 n cos η 1 2 η 2 ξ 2 + η 2 cosh A 1 n + A 2 n sin sinh A 1 n + A 2 n cos ,
h 0 2 2 μ τ ξη = n = 1 n 2 η 1 2 cosh A 1 n + A 2 n + sinh κ 2 2 A 1 n κ 2 A 2 n cos n 2 ηξ sinh A 1 n + A 2 n + cosh κ 2 A 1 n κ 2 2 A 2 n sin η 1 2 η 2 ξ 2 + η 2 cosh A 1 n + A 2 n sin + sinh A 1 n + A 2 n cos , E25
h 0 2 2 μ σ ξξ = n = 1 n 2 η 1 2 sinh A 1 n + A 2 n + cosh κ 4 2 A 1 n κ + 2 2 A 2 n sin n 2 ηξ cosh A 1 n + A 2 n + sinh κ + 2 2 A 1 n κ 4 2 A 2 n cos + η 1 2 η 2 ξ 2 + η 2 cosh A 1 n + A 2 n sin sinh A 1 n + A 2 n cos .

We have to solve problem (2), (7a), (8a), and (9a) when Q 1 ξ = P and Q 2 ξ = 0 , i.e., at η = η 1 boundary the normal load 1 2 μ σ ηη = P h 0 2 is given, but tangent stress is equal to zero. From (16), and (23), we obtain the following equations:

n = 1 n 2 η 1 sinh n η 1 A 1 n + A 2 n n cosh n η 1 κ 2 A 1 n κ 2 2 A 2 n sin = P η 1 ξ 2 + η 1 2 , n = 1 n 2 η 1 cosh n η 1 A 1 n + A 2 n + n sinh n η 1 κ 2 2 A 1 n κ 2 A 2 n cos = ξ 2 + η 1 2 .

From here an infinite system of the linear algebraic equations with unknown A 1 n and A 2 n coefficients is obtained:

n 2 η 1 sinh n η 1 n κ 2 cosh n η 1 A 1 n + n 2 η 1 sinh n η 1 + n κ 2 2 cosh n η 1 A 2 n = F ˜ 1 n , n 2 η 1 cosh n η 1 + n κ 2 2 sinh n η 1 A 1 n + n 2 η 1 cosh n η 1 n κ 2 sinh n η 1 A 2 n = F ˜ 2 n , n = 1 , 2 , E26

where F ˜ 1 n and F ˜ 2 n are the coefficients of expansion into the Fourier series f 1 ξ = n = 1 F ˜ 1 n sin and f 2 ξ = n = 1 F ˜ 2 n cos , respectively, f 1 ξ = P η 1 ξ 2 + η 1 2 and f 2 ξ = ξ 2 + η 1 2 functions.

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 A 1 n and A 2 n unknown values will be solved. After solving this system, we find A 1 n and A 2 n coefficients, 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 = η 1 and ξ = ξ 1 (see Figure 1a), and relevant 3D graphs are drafted. The numerical results are obtained for the following data: ν = 0.3 , E = 2 × 10 6 kg / cm 2 , P = 10 kg / cm 2 , 0.1 η 1 3 , ξ 1 = 2 π , ξ 1 = 4 π , and ξ 1 = 6 π . 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 = η 1 and ξ = ξ 1 k ξ 1 (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 = 0 and η = 0 of consideration area σ ξξ 0 η , σ ηη ξ 0 stresses and u ξ 0 , v 0 η displacements 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 Ω 1 = 0 < ξ < ξ 1 η 1 < η < , which satisfies the following boundary conditions: (7a), (8a), (10), and (10′).

From (20) and (8a)

φ i = n = 1 φ in , i = 1 , 2 , 3 , E27

where φ 1 n = B 1 n e sin , φ 2 n = B 2 n e cos , φ 3 n = κ 2 2 n B 2 n e sin .

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

u ¯ = n = 0 ne B 1 n B 2 n η 1 cos + B 1 n B 2 n ξ sin η η 1 e κ 1 B 1 n κ 2 B 2 n ξ sin κ 2 e B 2 n η cos , v ¯ = n = 1 ne B 1 n B 2 n ξ cos + B 1 n B 2 n η 1 sin η η 1 + e κ 1 B 1 n κ 2 B 2 n η sin κ 2 e B 2 n ξ cos , E28

and for the stresses, we obtain the following formula:

h 0 2 2 μ σ ηη = n = 1 n 2 e B 1 n B 2 n η 1 sin + B 1 n B 2 n ξ cos η η 1 ne κ 2 B 1 n η sin κ 2 2 B 1 n + B 2 n ξ cos η 2 η 1 2 ξ 2 + η 2 ne B 1 n B 2 n η sin + ξ cos ,
h 0 2 2 μ τ ξη = n = 1 n 2 e B 1 n B 2 n ξ sin B 1 n B 2 n η 1 cos η η 1 ne κ 2 B 1 n ξ sin κ 2 2 B 1 n + B 2 n η cos , η 2 η 1 2 ξ 2 + η 2 ne B 1 n B 2 n ξ sin + η cos , E29
h 0 2 2 μ σ ξξ = n = 1 n 2 e B 1 n B 2 n η 1 sin + B 1 n B 2 n ξ cos η η 1 + ne κ 4 2 B 1 n + 2 B 2 n η sin + κ + 2 2 B 1 n ξ cos + η 2 η 1 2 ξ 2 + η 2 ne B 1 n B 2 n η sin + ξ cos .

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

We have to solve problem (2), (7a), and (8a), when Q 1 ξ = P and Q 2 ξ = 0 , i.e., at η = η 1 boundary the normal load 1 2 μ σ ηη = P h 0 2 is given, but tangent stress is equal to zero. From (22) and (27), we obtain the following equations:

n = 1 ne n η 1 κ 2 B 1 n sin = P η 1 ξ 2 + η 1 2 , n = 1 ne n η 1 κ 2 2 B 1 n + B 2 n cos = ξ 2 + η 1 2 .

Consequently, we obtain the infinite system of the linear algebraic equations with unknown B 1 n and B 2 n coefficients:

n = 1 ne n η 1 κ 2 B 1 n sin = n = 1 P ˜ 1 n sin , n = 1 ne n η 1 κ 2 2 B 1 n + B 2 n cos = n = 1 P ˜ 2 n cos , i.e.,

ne n η 1 κ 2 B 1 n = P ˜ 1 n , ne n η 1 κ 2 2 B 1 n + B 2 n = P ˜ 2 n , n = 1 , 2 , . E30

Hence,

B 1 n = 2 κn e n η 1 P ˜ 1 n , B 2 n = e n η 1 n P ˜ 2 n + κ 2 κ P ˜ 1 n ,

where P ˜ 1 n and P ˜ 2 n are the coefficients of expansion into the Fourier series of functions f 1 ξ = P η 1 ξ 2 + η 1 2 and f 2 ξ = ξ 2 + η 1 2 , respectively ( f 1 ξ , according to sinuses, and f 2 ξ , 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 B 1 n and B 2 n unknown values will be solved. After solving this system, we find the values of B 1 n and B 2 n coefficients 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, M 1 0 η 1 , M 2 ξ 1 η 1 points (see. Figure 2a), for the following data: ν = 0.3 , E = 2 10 6 kg / cm 2 , P = 10 kg / cm 2 , 0.01 η 1 3 , ξ 1 = 2 π , ξ 1 = 4 π , and ξ 1 = 6 π .

The above-presented graphs (see Figures 5 and 6) show how displacements and stresses change at some characteristic points of body, namely, at points M 1 j 0 η 1 j and M 2 j ξ 1 η 1 j j = 1 2 8 , when 0.01 η 1 3 (see Figure 7).

From the presented results, we obtain the following:

• At points M 1 j 0 η 1 j , max u t < max u n , v t = v n = 0 .

• At points M 2 j ξ 1 η 1 j , max σ ξξ t > max σ ξξ n , max u t > max u n , max v t < max v n .

• When ξ 1 , then displacements and stresses tend to zero, that is, the boundary conditions (10) are satisfied.

• When η 1 , then displacements and stresses tend to zero, that is, the boundary conditions (10′) are satisfied.

• When η 1 0 (in this case there is a crack), then (a) at points M 1 j 0 η 1 j tangential stresses and normal displacements tend to , but other components equal to zero. It can be seen from the boundary conditions (8a) (b) at points M 2 j ξ 1 η 1 j that all components of the displacements and stresses tend to .

Here superscript t and n denote 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 < , 0 η < , see Figure A1) [23, 24]; we have

h ξ = h η = h = c ξ 2 + η 2 , x = c ξ 2 η 2 / 2 , y = cξη ,

where h ξ , h η are Lame's coefficients of the system of parabolic coordinates, c is a scale coefficient, x , y are the Cartesian coordinates.

The coordinate axes are parabolas

y 2 = 2 c ξ 0 2 x c ξ 0 2 / 2 , ξ 0 = const , y 2 = 2 c η 0 2 x + c η 0 2 / 2 , η 0 = const .

Laplace’s equation Δ f = 0 , where f = f ξ η , in the parabolic coordinates has the form

f , ξξ + f , ηη / c 2 ξ 2 + η 2 = 0 .

We have to find solution of the equation in following form

f = X ξ E η ,

and then by separation of variables, we will receive

1 c 2 ξ 2 + η 2 X " X + E ' E = 0 .

From here

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

where m is any constant, their solutions are [25]

X = C 1 cos + C 2 sin , E = C 3 e + C 4 e = C 3 cosh + C 4 sinh .

So

f ξ η = C 3 e + C 4 e C 1 cos + C 2 sin or f ξ η = C 3 cosh + C 4 sinh C 1 cos + C 2 sin ,

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

We have introduced φ 1 harmonic function, and if we take

a D = κμ h 0 2 φ 1 , η η φ 1 , ξ ξ , b K = κμ h 0 2 φ 1 , η ξ + φ 1 , ξ η , EB1

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

a u ¯ , ξ + v ¯ , η = κ 2 φ 1 , η η φ 1 , ξ ξ , b v ¯ , ξ u ¯ , η = κ φ 1 , η ξ + φ 1 , ξ η , EB2
a u ¯ , ξ + v ¯ , η = κ 2 φ 1 , η η φ 1 , ξ ξ , b v ¯ κφ 1 η , ξ = u ¯ + κφ 1 ξ , η . EB3

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

u ¯ = φ , ξ κφ 1 ξ , v ¯ = φ , η + κφ 1 η . EB4

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

h 2 Δ φ = φ , ξξ + φ , ηη = κφ 1 + κφ 1 , ξ ξ κφ 1 κφ 1 , η η + κ 2 φ 1 , η η φ 1 ξ ξ = 2 φ 1 , ξ ξ φ 1 , η η . EB5

General solution of the system (B2) can be written in the form u ¯ = ψ 1 , v ¯ = ψ 2 , where

ψ 1 , ξ + ψ 2 , η = 0 , ψ 2 , ξ ψ 1 , η = 0 .

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

u ¯ = φ , ξ κφ 1 ξ + ψ 1 , v ¯ = φ , η + κφ 1 η + ψ 2 , EB6

where φ is the partial solution of the (B5).

If we take κ = const , then

φ = ξ 2 η 2 2 φ 1 ,

and (B6) formula will receive the following form:

u ¯ = ξ 2 η 2 2 φ 1 , ξ κ 1 φ 1 ξ + ψ 1 , v ¯ = ξ 2 η 2 2 φ 1 , η + κ 1 φ 1 η + ψ 2 .

From here

u ¯ = ξ 2 η 2 2 φ 1 , ξ + ξηφ 1 , η ξηφ 1 , η κ 1 φ 1 ξ + ψ 1 , v ¯ = ξ 2 η 2 2 φ 1 , η ξηφ 1 , ξ + ξηφ 1 , ξ + κ 1 φ 1 η + ψ 2 .

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 ψ 1 and ψ 2 ). Therefore, the solutions of system (2) are given in the following form:

a h 0 2 D = κμ φ 1 , η η φ 1 , ξ ξ , b h 0 2 K = κμ φ 1 , η ξ + φ 1 , ξ η , c u ¯ = ξηφ 1 , η κ 1 φ 1 ξ + ψ 1 , d v ¯ = ξηφ 1 , ξ + κ 1 φ 1 η + ψ 2 . EB7

Now we have to write down three versions of ψ 1 and ψ 2 function representation. In the first version

ψ 1 = φ ¯ 1 , η + φ ˜ 1 , η + φ 2 , η , ψ 2 = φ ¯ 1 , ξ + φ ˜ 1 , ξ + φ 2 , ξ , EB8

φ ¯ 1 , φ ˜ 1 , φ 2 are harmonic functions; in addition, φ ¯ 1 , φ ˜ 1 are selected so that at η = α , where α = η 1 or α = η 2 , the following equations will be satisfied:

ξηφ 1 , η κ 1 φ 1 ξ + φ ¯ 1 , η + φ ˜ 1 , η = 0 , ξηφ 1 , ξ + κ 1 φ 1 ξ + φ ¯ 1 , ξ + φ ˜ 1 , ξ = 0 ,

In the second version

ψ 1 = α ξ 2 η α 2 2 φ 1 , ξ + ξηφ 1 , η + ξ 2 η 2 2 φ 2 , ξ + ξηφ 2 , η , ψ 2 = α ξηφ 1 , ξ ξ 2 η α 2 2 φ 1 , η + ξ 2 η 2 2 φ 2 , η ξηφ 2 , ξ , EB9

where φ 2 is the harmonic function.

In the third version

ψ 1 = α 2 ξ 2 η 2 2 φ 1 , ξ + ξηφ 1 , η + ξ 2 η 2 2 φ 2 , ξ + ξηφ 2 , η , ψ 2 = α 2 ξηφ 1 , ξ ξ 2 η 2 2 φ 1 , η + ξ 2 η 2 2 φ 2 , η ξηφ 2 , ξ . EB10

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

a u ¯ = ξηφ 1 , η κ 1 φ 1 ξ + φ ¯ 1 , η + φ ˜ 1 , η + φ 2 , η , b v ¯ = ξηφξ + κ 1 φ 1 ξ + φ ¯ 1 , ξ + φ ˜ 1 , ξ + φ 2 , ξ . EB11

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

a u ¯ = α ξ 2 η α 2 2 φ 1 , ξ + ξηφ 1 , η ξηφ 1 , η κ 1 φ 1 ξ + ξ 2 η 2 2 φ 2 , ξ + ξηφ 2 , η , b v ¯ = α ξηφ 1 , ξ ξ 2 η α 2 2 φ 1 , η + ξηφ 1 , ξ + κ 1 φ 1 η + ξ 2 η 2 2 φ 2 , η ξηφ 2 , ξ . EB12

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

a u ¯ = α 2 ξ 2 η 2 2 φ 1 , ξ + ξηφ 1 , η ξηφ 1 , η κ 1 φ 1 ξ + ξ 2 η 2 2 φ 2 , ξ + ξηφ 2 , η , b v ¯ = α 2 ξηφ 1 , ξ ξ 2 η 2 2 φ 1 , η + ξηφ 1 , ξ + κ 1 φ 1 η + ξ 2 η 2 2 φ 2 , η ξηφ 2 , ξ . EB13

After the boundary value problem with relevant boundary conditions on ξ = ξ 1 = ξ 11 is solved, the following condition is examined: F 11 / F 10 < ε .

ε is a sufficiently small positive number given in advance ( ε = 0 , 001 0 , 0001 ).

F 11 = 0 η 1 σ ξξ + σ ηη + τ ξη hdη ξ = ξ 1 , F 10 = 0 η 1 σ ξξ + σ ηη + τ ξη hdη ξ = g ξ ˜ 1 .

g number will be selected so that on boundary η = η 1 , point M g ξ ˜ 1 η 1 should correspond to the highest value of expression σ ηη g ξ ˜ 1 η 1 2 + τ ξη g ξ ˜ 1 η 1 2 (when stresses are given) or to the highest value of expression u ¯ g ξ ˜ 1 η 1 2 + v ¯ g ξ ˜ 1 η 1 2 (when displacements are given).

If condition F 11 / F 10 < ε is not valid for ξ 1 = ξ 11 , the same problem will be solved at the beginning, but ξ 1 = ξ 12 will be used instead of ξ 1 = ξ 11 . In addition, ξ 12 > ξ 11 . Then, if condition F 12 / F 10 < ε is not still valid, we will continue with the boundary problem, where ξ 1 = ξ 13 ; besides, ξ 13 > ξ 12 > ξ 11 , and we will examine condition F 13 / F 10 < ε . The process will be over at the k th stage, if condition F 1 k / F 10 < ε is valid.

Finding such ξ 1 = ξ 1 k , for which F 1 k / F 10 < ε .

Distance l between surfaces ξ = ξ 1 and ξ = ξ ˜ 1 , which gives the guarantee for condition F 1 k / F 10 < ε to be valid in the parabolic coordinate system, will be taken along the axis of the parabola , and the following expression will be obtained:

ξ 1 = l / c + ξ ˜ 1 2 .

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 ξ 1 from the relevant equation. Let us note that when l / c = 4 , we will denote value ξ 1 by ξ 11 , when l / c = 5 ; by ξ 12 , when l / c = 6 ; by ξ 13 , etc. If after selecting ξ 1 = ξ 1 k , inequality F 1 k / F 10 < ε is valid; in order to check the righteousness of the selection, it is necessary to once again make sure that, together with condition F 1 k / F 10 < ε , condition ε > F 1 k / F 10 > F 1 k + 1 / F 10 > F 1 k + 2 / F 10 > 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

U u v

displacement vector

σ ξξ , σ ηη , τ ξη = τ ηξ

normal and tangential stresses

## References

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Written By

Natela Zirakashvili

Submitted: 03 November 2019 Reviewed: 09 January 2020 Published: 12 February 2020