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The Problem of the Crack in the “Generally”

Anisotropic Disk

N.I. Galidakis,

Supervisor Civil Engineer,

National Technical University of Athens,

Athens, Greece

March 1968

Dissertation for the scientiﬁc title of “Doctor of Civil Engineering” at the National

Technical University of Athens, submitted by N.I. Galidakis, Civil Engineer from Athens,

on the 10th of May 1967, accepted on the 14th of March 1968 and given by the University’s

doctoral committee the grade of “unanimously perfect”.1

1Translated and re-edited from the Greek published book into English L

A

T

E

X, by I.N. Galidakis. Final

typography partially veriﬁed courtesy of N.I. Rigos, Civil Engineer, graduate NTUA.

1

Presenter: Professor E. Panagiotounakos

Co-presenter: Professor P. Theoharis

2

To the Memory of My Father

3

Contents

Introduction 5

Appendix 8

Basic Notions and Theorems from the Theory of Functions of a Complex Variable 8

Expressions for the Previous Propositions when Cis the Unit Circle . . . . . . 10

A Plane Stress On the Anisotropic Disk - The Crack As a Container 11

A.1 General ..................................... 11

A.2 General Integral of (10). Form of stress functions - Boundary conditions . 16

A.3 The Muskhelishvili Transformation . . . . . . . . . . . . . . . . . . . . . . 22

B Solution of the “First Fundamental Problem” For the Three Basic

Charge Cases 25

B.1 Uncharged boundary - Given stresses very far away from the boundary . . 25

B.2 Continuous, uniformly distributed normal charge over section of the boundary 29

B.3 Continuous, uniformly distributed tangential charge over section of the

boundary .................................... 40

C Explanation of the Found Expressions - Conclusions 44

C.1 UnchargedBoundary.............................. 44

C.2 Normal Charge On the Boundary . . . . . . . . . . . . . . . . . . . . . . . 47

C.3 Tangential Charge On the Boundary . . . . . . . . . . . . . . . . . . . . . 56

References 61

4

Introduction

The usually unsurmountable mathematical diﬃculties which arise from trying to solve

even the simplest of problems, are well known to those who have involved themselves with

the subject of theoretical Elasticity.

The already known methods of approach to problems of Mathematical theory of Elas-

ticity, belong to two categories.

In the ﬁrst category belong the methods which concern themselves with the problem

of boundary values of functions on a given contour of the elastic medium, whereas in the

second category belong methods which derive from certain given general laws of “minima”,

which characterize the balanced state of the ideal elastic body.

In the current work the methods used are from the theory of Complex functions

belonging according to the method of study to the ﬁrst category.

The use of the Complex variable in the theory of plane Elasticity of an Isotropic body

was ﬁrst used by G.V. Kolossoﬀ in 1909 [11]. Despite this, a third of a century passed until

the theory being extended beyond the initial stage of the “typical” use of the complex

variable could enter into more substantial oﬀers in the area of theoretical Elasticity [17]

[21].

Today, the nature of the results of the plane theory of Elasticity cannot be considered

solely a conclusion of the work of Kolossof [11] or Filon [6], who, even though have found

certain complex expressions for stress, did not think to use them further.

First N.I. Muskhelisvili [17], by applying the above results on the “fundamental”

boundary problems by using a conformal mapping and the properties of the Cauchy inte-

gral, managed to give completeness and unity to the approximating methods to problems

relating to two dimensional Elasticity.

The reduction of the above problems into Fredholm integral equations of the second

type [14], [15], simpliﬁes generally the Mathematical description of the solution, the latter

succeeded by completely elementary methods in the case where the “nucleus” is a rational

function [17] [21].

The important contribution of the method is not so much the research by new means

of known or almost known problems, rather the push that was given to the study of new

problems, like for example, the cases of contact problems, by reducing them to the general

problem of “Linear Correlation” or otherwise to the Riemann-Hilbert problem [14] [15]

[17].

The extension of the theory of Muskhelisvili in the area of the two dimensional

anisotropic elastic body appears (completely independently and approximately the same

time period) in the West on one hand, and in Russia on the other, around 1950.

There are indications that the Russian works came ﬁrst, but they appear to have been

inaccessible not only for those outside Russia but also for many Russian Research Centers

[8].

A.E. Green and W. Zerna [7] give form to the Mathematical theory of Elasticity on

a more general basis so that it includes equally (at least in two dimensions) the isotropic

and the anisotropic medium. The generality of the appearance of the whole subject

however, fails to cover areas like those of investigating or even applying the austerely

general extracted solutions.

The research of general Anisotropy (covering 80 pages out of a total of 450) contains

5

the study of Complex Functional Potential and the speciﬁcation of such in very special

cases like that of the orthotropic half-space, the general similar charge of circular or ellip-

tical contours, the concentrated charge at inﬁnity plane, the inﬂuence of the uncharged

crack as well as a general introduction to the problem of contact (without friction) of two

elastic bodies.

A summary of the work of Russian researchers on the anisotropic elastic body was

published in 1963 by S.G. Lekhnitskii [12]. In the latter work the problems mentioned

in Green and Zerna are solved with additional material, such as the study of the half-

container having a parabolic limit, the expression of the equations for the Somigliana

problem (cylindrical body charged by external non-variable forces along the generational)

and extraction of conclusions from their investigation, the general problem of bending with

turning of beams of speciﬁc cross section, the turning of beams of double tenacity cross

section, as well as the presentation of the general equations of balance of an elastic body

having non-linear anisotropy.

S.G. Lemhniskii introduces essentially three independent Complex variables general-

izing Muskhelishvili’s method for anisotropic bodies having at least one level of symmetry

(elastic or charging) but he doesn’t give the general integral equations which describe the

solutions for problems of this kind, neither does he prove most of the conclusions from the

application of the above theory, referring usually to obscure because of non-circulation in

the West and because of language Russian Monographs.2

The present work refers to the problem of the crack in the generally anisotropic disk,

that is the elastic plane which has no elements of symmetry [13].

The crack considered as a simple Jordan curve, has the great advantage that it can

be mapped onto the unit circle by a rational function, succeeding thus in ﬁnding the

Complex Potential [7] [12] for cases of non-self-balanced charge.

The plane of the anisotropic disk (the regular Complex plane z) transforms by two

homoparallel transformations to the planes z1and z2which are considered “projections”

of a four-dimensional continuous bicylinder [3], “inside” which the functions of stress and

displacement are deﬁned.

When the homoparallel transformations are equal (K1=K2=K) or equivalently

if the planes z1and z2coincide, then the case reduces to “Pseudoanisotropy” and the

general two dimensional problem transforms itself properly to an isotropic problem [24].

The solution to the problem of the crack in the case of General Anisotropy is succeeded

by determination of the functions of Complex potential Φ1(z1) and Φ2(z2), which are

holomorphic functions on the planes z1and z2respectively.

By mapping conformally the above two planes into ζ1and ζ2respectively, such that

the limits of the crack image contour are mapped into the unit circles |ζ1|= 1 and |ζ2|= 1

and so that for all ziwe have |ζi| ≥ 1 we conclude that Φi(zi) will be holomorphic on the

planes ζifor all |ζi| ≥ 1, with the possible exception the points ζi=∞.

This property, which expresses mathematically the “regularity” of the intensity state

for the whole disk (with the possible exception of points on the contour), along with the

known values of the derivatives of the Φi(on the contour) are suﬃcient for the determi-

nation of the above functions by application of simple properties of Cauchy integrals.

In the current work, the aforementioned problem is solved almost elementarily, by

ﬁnding the functions of Complex potential in closed form for all the cases having a charge

2Some of them have been written in special Russian dialects, such as Georgian, etc.

6

upon the boundary of the crack.

Using appropriate limiting accessing methods we can ﬁnd from the general solution

the solutions to the problems referring to the half-space in the form of special cases, (since

for R→ ∞ the plane separates into the two half-planes) as well as extract conclusions

relating to the form and structure of the anisotropic medium.

It is mentioned lastly, that by this current work a new boundary (that of the crack)

appears, which along with the inﬁnite line of the half-plane contain the only known up to

now half-spaces upon which the two fundamental Problems of the Mathematical theory

of anisotropic Elasticity are solved for all cases of charging by closed form functions.

7

Figure 1: Jordan Curve

Mathematical Appendix:

Basic Notions and Theorems from the Theory of Functions of a

Complex Variable

1. By the term “Simple Jordan Curve” we mean a continuous normal curve which has

no double points and which can be thought of as a bijective image of a circular arc (Fig.

1a).

2. If the aforementioned curve is closed then we get the “Jordan curve” C, which has

the characteristic property of separating the plane into two profound domains G1and

G2, which, along with Ccover the entire Riemann sphere (Fig. 1b).

3. On the Jordan curve (Cab or Crespectively) we can deﬁne a positive direction,

therefore on its left we deﬁne G+and on its right G−(Fig. 1).

4. Let there be a function Φ(z) given and continuous on the “neighborhood” of C

(but not necessarily on C). We consider a point t∈c. We say that Φ(z) is continuous

left of t(or right of t) if the function tends to a limit, as ztends to talong any line which

lies left (or right) of C. The values of Φ(z) are called limiting values of the function and

are represented as Φ(t)∗or Φ−(t).

5. Let there be the function F(t), where t=x+yi a point of C. We say that

F(t) satisﬁes the H¨older condition on C, if for every pair t1and t2the inequality

|F(t1)−F(t2)| ≤ A|t1−t2|µ, holds, for Aand µpositive constants with 0 < µ ≤1.

6. The integral F(z) = 1

2πi Cff(t)

t−zdt (where f(t) is ﬁnite and integrable in the sense

of Riemann3on the plane z) taken over C, with zbeing a point of G+(or G−) is called

the Cauchy integral.

F(z) is a holomorphic function4on the entire plane, except at the points of C.

7. The function F(z) for z∈Chas meaning provided f(t) satisﬁes the H¨older

condition on the “neighborhood” of the point z.

3A function f(z) is called integrable in the usual sense or in the sense of Riemann, if it is deﬁned and

continuous on the domain G, and if the value of the curvilinear integral ∫γ2

γ1f(ζ)dζ is independent of the

curve joining the points γ1 and γ2.

4A single valued function f(z) of a complex variable z, whose derivative has a single value on every

point of the domain Gis called regular analytic or holomorphic function on G.

An equivalent deﬁnition can be gotten from the decomposition of f(z) into a series of powers of zin the

domain G.

8

8. If we deﬁne the positive direction on Cin a way such that G+is the ﬁnite domain

and G−is the inﬁnite domain of the plane which Cencloses, then:

a) If f(z) is a holomorphic function on G+and continuous on G+∪C, we have

1

2πi C

f(t)

t−zdt =f(z), for z∈G+

1

2πi C

f(t)

t−zdt = 0, for z∈G−

b) If f(z) is a holomorphic function on G−and continuous on G−∪C, we have

1

2πi C

f(t)

t−zdt =−f(z) + f(∞), for z∈G−

1

2πi C

f(t)

t−zdt =f(∞), for z∈G+

9. According to the above follow the two basic propositions of the theory of limiting

values of functions:

a) A suﬃcient and necessary condition for a given f(t) on Cto be the limiting

value of a holomorphic function on G+is

1

2πi C

f(t)

t−zdt = 0, for all z∈G−

b) A suﬃcient and necessary condition for a given f(t) on Cto be the limiting

value of a holomorphic function on G−is

1

2πi C

f(t)

t−zdt =α, for all z∈G+

9

Expressions for the Previous Propositions when Cis the Unit Cir-

cle

1. Let Γ be the unit circle |ζ|= 1 and σits points, where σ=eiθ, for 0 ≤θ≤2π.

We call Σ+and Σ−the interior and exterior of Γ, or the same, for the domains where

|ζ|<1 and |ζ|>1 respectively. Let also Φ(ζ) be a function deﬁned in Σ+(or Σ−).

We deﬁne the function

Φ∗(ζ) = Φ 1

ζ= Φ 1

ζ

From the deﬁnition it follows that if Φ(ζ) is holomorphic on Σ+(or Σ−), Φ∗(ζ) will

be holomorphic on Σ+(or Σ−) and conversely.

If we assume that Φ(ζ) deﬁned on Σ+has the limiting value Φ(σ) for ζ→σthen

Φ∗(ζ) has the limiting value Φ−

∗(σ) deﬁned on Σ−and such that

Φ−

∗(σ) = Φ+(σ)

or, Φ+

∗(σ) = Φ−(σ)

2. According to the above, conditions 1.9 a),b) can be expressed as follows:

a) A suﬃcient and necessary condition for the function f(σ) (continuous on the

circle Γ) to be the limiting value of a holomorphic function inside of Γ is:

1

2πi Γ

f(σ)

σ−ζdσ =α, for all |ζ|<1 and where α=f(0)

b) A suﬃcient and necessary condition for the function f(σ), which is continuous

on the circle Γ to be the limiting value of a holomorphic function outside of Γ is:

1

2πi Γ

f(σ)

σ−ζdσ = 0, for all |ζ|>1

10

A Plane Stress On the Anisotropic Disk - The Crack

As a Container

A.1 General

1. As is well known the static intensity state at any point of the continuous disk is

determined completely if the stress components are given on any two linear elements ∆s1

and ∆s2passing through the aforementioned point.

If the linear elements are parallel to the axes xand yrespectively, we say that the

stress tensor Ton the given point has components the linear magnitudes σx,σy,τxy and

τyx. (Fig. 2).

If we “project” the above tensor against the linear element ∆s(passing through the

initial point), which is signed with respect to direction as ⃗n, we ﬁnd by using the Cauchy

relations the two components Xnand Ynof Taccording to ⃗n as follows5: (Fig. 2)6

Xn=σxcos(n, x) + τxy cos(n, y)

Yn=τyx cos(n, x) + σycos(n, y)(1)

We accept the magnitudes of σx,σy,τxy and τyx as continuous functions of the

coordinates of the plane xand y, hence we also accept the existence of their ﬁrst (at

least) partial derivatives.

2. We consider now one element Fof the balanced plane Disk. The corresponding

components of all acting forces on it must be zero7, in the xand yaxes, Hence,

Γ

Xnds = 0, and Γ

Ynds = 0

(where Γ is the contour of F) or because of (1) and by applying Green’s Theorem:

Γ

[σxcos(n, x) + τxy cos(n, y)] ds =

F∂σx

∂x +∂τxy

∂y dF = 0

Γ

[τyx cos(n, x) + σycos(n, y)] ds =

F∂τyx

∂x +∂σy

∂y dF = 0

5Translator’s Note: The book the dissertation was published in has the equations Xn=σxσυν (n, x)+

τxyσυν (n, y) and Yn=τyx συν(n, x) +σyσυν(n, y) in Greek, for which the operator συ ν translates as cos,

which is of course a one-argument function. It appears that the author has nto be the unit normal to the

linear element ∆sand Tto be a two-tensor. The translator thinks that the author means Xn=T(ˆn, ˆx)

and Yn=T(ˆn, ˆy). In other words, treating T(ˆn, ∗) as a vector, Xnand Ynare the “pro jected” components

in the directions of the axes xand y. In that case, since Thas a matrix expression in the standard basis

given by the σ’s and τ’s, we see that this is consistent with the interpretation which has συν(n, ∗) = P∗(n),

or the “projection” in the direction of ∗, namely that the vector nwhen expressed in the standard basis

is (συν(n, x), συ ν(n, y)). Furthermore, if nis assumed to be a unit vector, then the component of nin

the xdirection is precisely the cosine of the angle between them. So this also matches the interpretation

which has συν(n, ∗) = cos(n, ∗) in the case that nis a unit, in which case the more austere notation

cos ⟨ˆn, ˆ

∗⟩ could have been used.

6Stresses are linear magnitudes and hence Fig. 2 depicts vectors having components equal to the

stresses.

7In the above as well as in what follows, we accept that there are no forces due to Mass.

11

Figure 2: Component Stresses on the Linear Element ∆s

and because the container Fis random the following relationships must hold:

∂σx

∂x +∂τxy

∂y = 0

∂τyx

∂x +∂σy

∂y = 0

(2)

Equations (2) in combination with the resultant from the vanishing of torque in the

elementary rectangle τxy =τyx comprise the balance system, known as Cauchy system,

which holds for any continuous body irrespectively of its elastic behavior.

3. From the geometry of the deformation [9] [17] [21] we have8

ϵx=∂u

∂x ,ϵy=∂v

∂y ,γxy =∂v

∂x +∂u

∂y (3)

where u(x, y) and v(x, y) are continuous functions (since we preclude breaking of the

body), which implies the existence of their ﬁrst (at least) partial derivatives.

Relations (3) as is known connect the three characteristics of deformation with the two

components of the vector of displacement. Consequently the ﬁrst cannot be independent

between themselves.

Accepting the existence of derivatives of uand vup to third order, we can eliminate

these functions, so we get the following relationship:

∂2γxy

∂x∂y =∂2ϵx

∂y2+∂2ϵy

∂x2(4)

8It is not important that we usually consider γxy =1

2(∂v

∂x +∂ u

∂y )to make calculations for the expres-

sion of stress deformation easier.

12

This equation is called relation of compromise of B. De Saint-Venant (or relation

of continuity) and is the necessary and suﬃcient condition so that the disk retains its

consistency even after its deformation.

4. After the above we form the following relations which connect the characteristics

of deformation with the components of the stress tensor, in the general case where there

are no elements of elastic symmetry in the homogeneous anisotropic elastic body:

ϵx=α11σx+α12 σy+α13τxy

ϵy=α21σx+α22 σy+α23τxy

γxy =α31σx+α32 σy+α33τxy

or

ϵx

ϵy

γxy

=

α11 α12 α13

α21 α22 α23

α31 α32 α33

σx

σy

τxy

(5)

The relations (5) express the “Generalized Hook’s Law” and can also be written

conversely:

σx=A11ϵx+A12 ϵy+A13γxy

σy=A21ϵx+A22 ϵy+A23γxy

τxy =A31ϵx+A32 ϵy+A33γxy

or

σx

σy

τxy

=

A11 A12 A13

A21 A22 A23

A31 A32 A33

ϵx

ϵy

γxy

(6)

where as usual, (αij )(Akl) = I(identity matrix).

If we call VGreen’s (hypothetically in the beginning) inserted function of elastic

Potential [9] [13] [22], and we assume that the deformation of the absolutely elastic body

takes place adiabatically or isothermally9, then we know10 that Vindeed exists, and we

will have (Castigliano):

σx=∂V

∂ϵx

,σy=∂V

∂ϵy

,τxy =∂V

∂γyx

Vis continuous and well-deﬁned, which implies:

∂σx

∂ϵy

=∂σy

∂ϵx

,∂σx

∂γxy

=∂τxy

∂ϵx

,∂σy

∂γxy

=∂τxy

∂ϵy

or because of (6): A12 =A21,A13 =A31 and A23 =A32.

Easily from the above we conclude also that α12 =α21,α13 =α31 and α23 =α32

5. Going back to the equations of balance (2) we ﬁnd that the ﬁrst relation ∂ σy

∂x =

−∂τxy

∂y expresses the necessary and suﬃcient condition for the existence of a function

X(x, y) such that:

∂X

∂y =σx, and ∂X

∂x =−τxy (7)

Similarly, from the second equation of (2) we conclude that there exists a function

Ψ(x, y) such that:

9With generally diﬀerent constants of elasticity.

10Sir W. Thomson. On the Thermoelastic Properties of Mater. Math. and phys. papers 1. 1882.

13

∂Ψ

∂x =σx, and ∂Ψ

∂y =−τxy

From the second relations of the above we conclude ∂X

∂x =∂Ψ

∂y , hence the following

must hold:

X=∂Φ

∂y , and Ψ = ∂Φ

∂x (8)

and ﬁnally because of (7) and the subsequent equations,

σx=∂2Φ

∂y2,σy=∂2Φ

∂x2,τxy =−∂2Φ

∂x∂y (9)

If we substitute equations (9) into (5) and then the resultants in the condition of com-

promise and we perform the necessary diﬀerentiations, we ﬁnd, that the wanted function

Φ will satisfy the following diﬀerential equation:

α22

∂4Ψ

∂x4−2α23

∂4Ψ

∂x3∂ y +(2α12 +α33)∂4Ψ

∂x2∂y2

−2α13

∂4Ψ

∂x∂y3+α11

∂4Ψ

∂y4= 0

(10)

If we represent the fourth order diﬀerential operator with D:

α22

∂4

∂x4−2α23

∂4

∂x3∂ y +(2α12 +α33)∂4

∂x2∂y2

−2α13

∂4

∂x∂y3+α11

∂4

∂y4

we can decompose the above into a product of four ﬁrst order linear diﬀerential operators

of the form:

Di=∂

∂y −ρi∂

∂x , where ρi(i= 1,2,3,4) are the roots of the algebraic equation:

α11ρ4−2α13 ρ3+ (2α12 +α33)ρ2−2α23 ρ+α22 = 0 (11)

We will prove that (11) has always imaginary roots11.

The elastic potential is determined by the following relation:

2V= (σxσyτxy)

α11 α12 α13

α21 α22 α23

α31 α32 α33

σx

σy

τxy

where (αij ) is the Matrix of coeﬃcients of equations (5), hence:

2V=α11σ2

x+α22σ2

y+α33τ2

xy + 2α12σxσy+ 2α13 σxτxy + 2α23σyτxy (12)

11The proof of Green-Zerna [7] does not refer to the above roots ρand is considerably more complex,

whereas that of S.G. Lekhnitskii [12], based on the lemma of arbitrariness of stresses, lacks elegance.

14

Expression (12) is a homogeneous quadratic form of the variables σx,σyand τxy,

which however, must be “positively deﬁned” [1], given that V > 0 always.

Relation (12) by virtue of (9) can be written:

2V=α11 ∂2Φ

∂y22

+α22 ∂2Φ

∂x22

+α33 ∂2Φ

∂x∂ y 2

+2α12

∂2Φ∂2Φ

∂x2∂y2−2α23

∂2Φ∂2Φ

∂x2∂x∂y −2α13

∂2Φ∂2Φ

∂y2∂x∂y

(13)

If we search for the constants λisuch that for ∂Φ

∂y −λi∂Φ

∂x we have V= 0, we ﬁnd

that the λisatisfy (11). Therefore since (13) has imaginary roots, the same must hold

for (11).

The roots ρ1,ρ2,ρ1and ρ2are called complex parameters and depend on the constants

of elasticity of the disk.

15

A.2 General Integral of (10). Form of stress functions - Boundary

conditions

1. Let ρ1,ρ2,ρ1and ρ2be the roots of the algebraic equation

α11ρ4−2α13 ρ3+ (2α12 +α33)ρ2−2α23 ρ+α22 = 0 (14)

We also assume that ρ1̸=ρ2(also ρ1̸=ρ2)12.

The general solution of (10) therefore is [12] [10] the following:

Φ = Φ1(z1)+Φ2(z2) + Φ3(z1)+Φ4(z2)

where

z1=x+ρ1y,z1=x+ρ1y

z2=x+ρ2y,z2=x+ρ2y

Because Φ must be a real function, it follows:

Φ3= Φ1and Φ4= Φ2

Hence the used general solution can be written as:

Φ = Φ1(z1)+Φ2(z2) + Φ1(z1) + Φ2(z2) = 2ℜ{Φ1(z1)+Φ2(z2)}

hence:

σx=ρ2

1Φ′′

1(z1) + ρ2

2Φ′′

2(z2) + ρ2

1Φ′′

1(z1) + ρ2

2Φ′′

2(z2)

σy= Φ′′

1(z1)+Φ′′

2(z2) + Φ′′

1(z1) + Φ′′

2(z2)

τxy =−ρ1Φ′′

1(z1)−ρ2Φ′′

2(z2)−ρ1Φ′′

1(z1)−ρ2Φ′′

2(z2)

(15)

If we substitute equations (15) into (5) and integrate, after performing all steps we

will have:

u= 2ℜ{κ1Φ′

1(z1) + κ2Φ′

2(z2)}+u0

v= 2ℜ{λ1Φ′

1(z1) + λ2Φ′

2(z2)}+v0, with

κi=α11ρ2

i−α13ρi+α12

λi=α12ρi−α23 +α22

ρi

, (i= 1,2)

(16)

According to the previous then, we see that the functions of stress and displacement

are not deﬁned on the initial plane z, but on a new, four-dimensional region W(bi-cylinder

[3]), which has as projections the planes z1and z2.

12The problem of the plane tensile situation in the case where ρ1=ρ2is solved by reducing it to an

“iconic” isotropic case, based on geometrical transformations [24].

16

Figure 3: Generally Anisotropic Disk

The common domain of the functions, belongs both to Wand to the planes z1and

z2, and is the intersection of the aforementioned planes.

The plane z1, results from the homoparallel transformation L1such that L1(x+yi) =

x+ρ1y, or the same, L1(x+yi) = x+ρ1y= (x+γ1y) + iδ1y.

Similarly for z2,L2(x+yi) = x+ρ2y= (x+γ2y) + iδ2y.

Consequently, the intersection of the planes z1and z2is the real axis.

The fact that the crack contour is found on Was well as on the intersection of the

new complex planes, is the reason the fundamental problems 1 and 2 [17] [21], which are

related to the aforementioned contour, admit a closed form solution.

2. In the given problem, the region which deﬁnes the elastic anisotropic disk is the

inﬁnite plane, excluding the points of a linear segment of length 2R.

We choose the coordinate axes as in Fig. 3 and assume known the constants of

elasticity of the plane with respect to the directions xand y, as well as those which

depend on the ordered tuple xy.

We transform the plane zinto the planes z1and z2according to the homoparallel

transformations L1and L2such that L1(z) = z+ρ1yand L2(z) = z+ρ2y.

Fig. 3 therefore transforms into Fig. 4 and Fig. 5 respectively13.

The contour C, because of the special nature of its curve remains intact with respect

to the transformations.

As far as the stress functions are concerned, deﬁned on the bi-cylinder (which projects

onto the planes z1and z2), for any charge on the contour they must be holomorphic

everywhere (including ∞) with the possible exception of a countable number (of points)

which project against the part of the axis |x| ≤ R[17] [21] [7] [19].

Every holomorphic function of such properties is expressed as follows:

13If the complex parameters are pure imaginary then the points ρ1and ρ2will lie on the yaxis,

preserving thus the main directions also on the planes z1and z2(Orthotropy case).

17

Figure 4: Plane z1

Figure 5: Plane z2

18

Figure 6: Contour Con plane z

F=

∞

n=0

αnz−n

1+

∞

n=0

βnz−n

2(where αn, βncomplex constants)

Going back to equations (15) we conclude that

Φ′′

1(z1) =

∞

n=0

αnz−n

1and Φ′′

2(z2)

∞

n=0

βnz−n

2

and by integrating:

Φ′

1=α0z1+α1ln(z1)+Φ0

1

Φ′

2=β0z2+β1ln(z2)+Φ0

2

(17)

where Φ0

1and Φ0

2are holomorphic functions everywhere on the planes z1and z2(including

∞) with the possible exception of a countable number of points on the boundary of C.

3. If we consider a contour Con the complex plane z(Fig. 6) and we deﬁne as usual

the positive direction for its traversal to be that direction which leaves the interior to the

left, we determine the vector ⃗n such that the system of n,shas the same orientation as

that of xand y.

If we accept that there is a charge on the contour (in this case on the crack), the stress

must result as a limiting value of the functions which describe the stresses (9).

The components Xn,Ynof that stress on the contour however are given by (1) as

follows:

Xn=σxcos(n, x) + τxy cos(n, y)

Yn=τxy cos(n, x) + σycos(n, y)

19

Because we have14

cos(n, x) = dy

ds and cos(n, y) = −dx

ds

we ﬁnd:

Xn=∂2Φ

∂y2

dy

ds +∂2Φ

∂x∂y

dx

ds =d

ds ∂Φ

∂y

Yn=−∂2Φ

∂x∂y

dy

ds −∂2Φ

∂x2

dx

ds =d

ds ∂Φ

∂x

and ﬁnally

∂Φ

∂y =s

O

Xnds and ∂Φ

∂x =−s

O

Ynds (18)

with the point Ochosen arbitrarily along the contour C.

Relations (18), which are the equations which describe the boundary conditions in the

case of the crack, can also be written as follows:

ρ1Φ′

1(z) + ρ2Φ′

2(z) + ρ1Φ′

1(z) + ρ2Φ′

2(z) = s

O

Xnds

Φ′

1(z)+Φ′

2(z) + Φ′

1(z) + Φ′

2(z) = −s

O

Ynds

(19)

where zis a real variable such that 0 ≤ |z| ≤ Rand ds =−dz because of cos(n, x)=0

and cos(n, y) = 1.

The components Xn,Ynare found as follows for the three basic cases of charge:

a) For continuous, uniformly distributed, normal charge on a section of the contour

(Fig. 7a). Xn= 0, Yn=−pfor zbetween z′z′′.

b) For continuous, uniformly distributed, tangential charge qon a section of the

contour (Fig. 7b). Xn=q,Yn= 0 for zbetween z′z′′.

c) For given stresses σ∞

x,σ∞

yand τ∞

xy where Xn=Yn= 0.

4. Examining the structure of the functions Φ′

1(z1), Φ′

2(z2) in equations (17) we

observe:

a) The stresses resulting from substitution into equations (15) are exhausted at

inﬁnity.

b) If the stresses at inﬁnity are non-zero, then at least one of the complex coeﬃ-

cients α0,β0must be non-zero.

Otherwise (σ∞

x=σ∞

y=τ∞

xy = 0), we must have α0=β0= 0.

c) Based on the boundary relations (19) and for a full traversal of the contour, the

integrals Xnds and Ynds are incremented (for every full traversal) by −X0and +Y0,

where X0,Y0are the components of the total stress charge in the direction of the axes x

and y. Consequently, these functions are multi-valued, for the case of a non-self-balanced

system with external forces.

14Translator’s Note: See footnote 5 on page 11.

20

Figure 7: Cases a) and b)

Therefore, in order for equations (19) to hold, the corresponding left-hand-side func-

tions must also be multi-valued functions.

Thus for X0+Y0i̸= 0, we conclude: α1, β1̸= 0 whereas for X0+Y0i= 0, α1=β1= 0.

21

A.3 The Muskhelishvili Transformation

1. The search for a function z=ω(ζ), which maps the complex plane zonto the plane

ζand such that the contour Cof a normal closed curve is mapped onto the perimeter

of the unit circle |ζ|= 1, despite the known Theorem of Existence by Riemann [3] [2], is

obstructed in most cases by such diﬃculties which allow ω(ζ) to be given in closed form

only in very few cases.

The above problem which we face during the solution of problems of the Mathematical

Theory of Elasticity of an Isotropic Body is multiply compounded in the case of a generally

anisotropic body.

In fact, in the second case one looks for three functions ω(ζ), ω1(ζ1) and ω2(ζ2), which

map the outlines C(on the zplane) and the transformed C1(on the z1plane) and C2

(on the z2plane), onto the unit circles |ζ|= 1, |ζ1|= 1 and |ζ2|= 1, on the planes ζ,ζ1

and ζ2respectively.

In order to ﬁnally have closed-form expressions for the functions of stress - dis-

placement, not only the functions ω(ζ), ω1(ζ1) should be rational15[17], but also for

all ζ=ζ1=ζ2=eiθ the found z1=ω1(ζ) and z2=ω2(ζ) must be “projections” of one

and the same z.

Expressing the above, we have:

z1=z+z

2+ρ1

z−z

2i,z2=z+z

2+ρ2

z−z

2i

and by eliminating zwe ﬁnd the relation

(ρ1−ρ2)ω(ζ0) = (ρ2−i)ω1(ζ0)−(ρ1−i)ω2(ζ0), where ζ0=eiθ (20)

which must hold (along with ω1,ω2,ωbeing rational) so that we can have a solution by

functions in closed form.

2. Following the previous, we enter the investigation of ﬁnding a mapping of the

planes z,z1and z2onto ζ,ζ1and ζ2such that the outlines C,C1and C2are mapped

respectively onto the unit circles |ζ|= 1, |ζ1|= 1 and |ζ2|= 1.

The mapping function has the same form in all three cases because of the invariance

of the initial contour Cunder the homoparallel transformations z→z1and z→z2(the

relation ω≡ω1≡ω2is also suﬃcient for the validity of (20)), and there is:

z=R

2ζ+1

ζ, consequently

z1=R

2ζ1+1

ζ1

z2=R

2ζ2+1

ζ2, with

|ζ| ≥ 1, |ζ1| ≥ 1, |ζ2| ≥ 1

(21)

The coincidence of the points z,z1and z2on the boundary Cimplies the same for

ζ,ζ1and ζ2on the unit circles (and conversely), a conclusion which allows us to refer to

one unit circle(c) the one found on the “intersection” of the planes ζ1and ζ2(Fig. 8).

15Muskhelishvili Theorem for the case of an isotropic elastic body.

22

Figure 8: Intersection of planes z1,z2and ζ1,ζ2

In order for transformation (21) to be invertible it must have a non-vanishing ﬁrst

derivative therefore it must have ζ2

i−1̸= 0 or ζi̸=±1, a result which was expected

because of the existence of a cusp16 in the above positions. The transformations (21)

substituted into relations (17) give:

Φ′

1(z1) = f1(z1) =α0

R

2ζ1+1

ζ1+α1ln R

2ζ1+1

ζ1+ Φ0

1R

2ζ1+1

ζ1

=α0

R

2ζ1+α1ln(ζ1) + f0

1(ζ1)

Φ′

2(z2) = f2(z2) =β0

R

2ζ2+1

ζ2+β1ln R

2ζ2+1

ζ2+ Φ0

2R

2ζ2+1

ζ2

=β0

R

2ζ2+β1ln(ζ2) + f0

2(ζ2)

(22)

where f0

1(ζ1) and f0

2(ζ2) are holomorphic functions on the planes ζ1and ζ2for |ζ1|>1

and |ζ2|>1 respectively.

One can conclude easily that f0

1(ζ1) (and f0

2(ζ2) respectively) can be expressed via

the following relation.

16Translator’s Note: “Cusp” in modern terminology would be the equivalent of a Branch Point.

23

f0

1(ζ1)=Φ0

1R

2ζ1+1

ζ1+α0

R

2

1

ζ1

+α1ln(R)

+α1ln ζ1+1

ζ1−ln(ζ1), for |ζ1|>1

as a sum of holomorphic functions therefore it (f0

1(ζ1)) is also holomorphic on the entire

plane ζ1(including ∞) with the exception of the points inside the unit circle.

24

B Solution of the “First Fundamental Problem” For

the Three Basic Charge Cases

B.1 Uncharged boundary - Given stresses very far away from the

boundary

17

1. All the previous take the following form:

Xn=Yn= 0. All σ∞

x,σ∞

y,τ∞

xy are known.

Therefore according to section A.2.3 for α1=β1= 0 equations (22) take the following

form:

Φ′

1(z1) = f1(ζ1) = α0

R

2ζ1+1

ζ1+ Ψ0

1(ζ1)

Φ′

2(z2) = f2(ζ2) = β0

R

2ζ2+1

ζ2+ Ψ0

1(ζ2)

(23)

where Ψ0

1(ζ1) = Φ0

1R

2ζ1+1

ζ1and Ψ0

2(ζ2) = Φ0

2R

2ζ2+1

ζ2are holomorphic func-

tions on the entire planes ζ1and ζ2respectively, for |ζ1| ≥ 1, |ζ2| ≥ 1.

The boundary conditions (19) on the unit circle, given that for σ=eiθ the relation

σ=1

σholds, are written:

α0

R

2σ+1

σ+ Ψ0

1(σ) + β0

R

2σ+1

σ+ Ψ0

2(σ)

+α0

R

21

σ+σ+ Ψ0

1(σ) + β0

R

21

σ+σ+ Ψ0

2(σ) = 0

ρ1α0

R

2σ+1

σ+ρ1Ψ0

1(σ) + ρ2β0

R

2σ+1

σ+ρ2Ψ0

2(σ)

+ρ1α0

R

21

σ+σ+ρ1Ψ0

1(σ) + ρ2β0

R

21

σ+σ+ρ2Ψ0

2(σ) = 0

or after performing the calculations:

Ψ0

1(σ) + Ψ0

2(σ) + Ψ0

1(σ) + Ψ0

2(σ) =

−R

2σ+1

σα0+β0+α0+β0

ρ1Ψ0

1(σ) + ρ2Ψ0

2(σ) + ρ1Ψ0

1(σ) + ρ2Ψ0

2(σ) =

−R

2σ+1

σρ1α0+ρ2β0+ρ1α0+ρ2β0

(24)

17The solution for the ﬁrst charge case is drafted roughly in general terms and using a diﬀerent method

in [7] page 366. See also [13], where the Russian Method is presented, but the solution of problems with

sectional charge of the contour is avoided. The solution 1 in the current work concerns on one hand the

unity of charge cases, on the other the extraction of conclusions in the case of Gen. Anisotropy.

25

We now multiply both sides of equations (24) by 1

2πi

dσ

σ−ζand we integrate on the unit

circle.

If we consider18 that:

1

2πi γ

σdσ

σ−ζ= 0 and 1

2πi γ

dσ

σ(σ−ζ)=−1

ζ

and

1

2πi γ

Ψ0

1(σ)

σ−ζdσ =−Ψ0

1(ζ) and 1

2πi γ

Ψ0

2(σ)

σ−ζdσ =−Ψ0

2(ζ)

and also that:

1

2πi γ

Ψ0

1(σ)

σ−ζdσ =1

2πi γ

Ψ0

2(σ)

σ−ζdσ = 0

We ﬁnally ﬁnd

Ψ0

1(ζ)+Ψ0

2(ζ) = −R

2α0+β0+α0+β01

ζ

ρ1Ψ0

1(ζ) + ρ2Ψ0

2(ζ) = −R

2ρ1α0+ρ2β0+ρ1α0+ρ2β01

ζ

(25)

We observe that in equations (23) that for z→ ∞ (in which case z1,z2,ζ1and ζ2

also go to inﬁnity) Φ′

1(z1) can be expressed as:

α0R

2ζ1+1

ζ1

or the same:

Φ′

1(z1) = O(α0z1)

The same holds for Φ′

2(z2) which is expressed by β0z2or the same, Φ′

2(z2) = O(β0z2)19

Therefore diﬀerentiating with respect to z1and to z2respectively, we have:

lim

z1→∞ Φ′′

1(z1) = α0and lim

z1→∞ Φ′′

2(z2) = β0(26)

We substitute the above results into (15) and we get: (given that we know the stress

at inﬁnity)

α0+β0+α0+β0=σ∞

y

ρ1α0+ρ2β0+ρ1α0+ρ2β0=−τ∞

xy

ρ2

1α0+ρ2

2β0+ρ2

1α0+ρ2

2β0=σ∞

x

(27)

18See Mathemat. Appendix in the following order: 1.8.a2, 1.8.b1, 2.2.b, 1.8.b1, 1.8.b1. It is also known

that Ψ0

1(∞)=Ψ0

2(∞) = 0.

19The functions Ψ0

1(ζ1) and Ψ0

1(ζ1), being holomorphic for |ζ1|>1, |ζ2|>1, can be expanded as usual

into series of the form A0+A1

ζi+A2

ζ2

i

+···, hence for ζi→ ∞, we have lim

ζi→∞

Ψ0

i(ζi) = A0.

26

Using the ﬁrst two equations from the above in system (25) we have

Ψ0

1(ζ) + Ψ0

2(ζ) = −Rσ∞

y

2

1

ζ

ρ1Ψ0

1(ζ) + ρ2Ψ0

2(ζ) = Rτ∞

xy

2

1

ζ

and ﬁnally (considering the corresponding functions’ domains of deﬁnition)

Ψ0

1(ζ1) = R

2(ρ1−ρ2)ρ2σ∞

y+τ∞

xy 1

ζ1

Ψ0

2(ζ2) = −R

2(ρ1−ρ2)ρ1σ∞

y+τ∞

xy 1

ζ2

(28)

The ﬁnding of functions Ψ0

1(ζ1) and Ψ0

2(ζ2) solves the problem of determining the

stress on the disk, without needing to investigate system (27) further, the solution of

which would be necessary for the determination of the constants α0,β0,α0and β0.

Indeed if we use relations (23) which if we diﬀerentiate with respect to z1and z2

respectively, we substitute in relations (15), we ﬁnd:

σy=σ∞

y+ 2ℜ−1

ρ1−ρ2ρ2σ∞

y+τ∞

xy 1

ζ2

1−1+1

ρ1−ρ2ρ1σ∞

y+τ∞

xy 1

ζ2

2−1

σx=σ∞

x+ 2ℜ−ρ2

1

ρ1−ρ2ρ2σ∞

y+τ∞

xy 1

ζ2

1−1+ρ2

2

ρ1−ρ2ρ1σ∞

y+τ∞

xy 1

ζ2

2−1

τxy =τ∞

xy + 2ℜρ1

ρ1−ρ2ρ2σ∞

y+τ∞

xy 1

ζ2

1−1+−ρ2

ρ1−ρ2ρ1σ∞

y+τ∞

xy 1

ζ2

2−1

(29)

Equations (29) give the solution to the problem.

2. Concluding, referring to the linear system (27), it wouldn’t be pointless to add the

following:

It is obvious that the equations are insuﬃcient for the determination of the constants

α0and β0.

It is true that the structure of the functions of stress requires only the sums α0+β0+

α0+β0,ρ1α0+ρ2β0+ρ1α0+ρ2β0,ρ2

1α0+ρ2

2β0+ρ2

1α0+ρ2

2β0, which are determined

(27), but this is not true for the calculation of displacements (16), where the separate

determination of the constants α0and β0is required.

In this case, we use the condition of vanishing of torque at inﬁnity [7] [12] [17], which

is expressed by the relation:

lim e∞=1

2∂v

∂x −∂u

∂y = 0

and because of (16)

lim

z→∞ 2ℜ{λ1Φ′′

1(z1) + λ2Φ′′

2(z2)−ρ1κ1Φ′′

1(z1)−ρ2κ2Φ′′

2(z2)}= 0

27

because lim

z1→∞ Φ′′

1(z1) = α0and lim

z2→∞ Φ′′

2(z2) = β0, the above equation is written:

(λ1−ρ1κ1)α0+ (λ2−ρ2κ2)β0+ (λ1−ρ1κ1)α0+ (λ2−ρ2κ2)β0= 0 (30)

Equation (30) is the missing fourth equation of the system (27).

We prove below that the determinant of the coeﬃcients of the unknowns is (for ρ1̸=

ρ2) always non-zero.

If we examine the coeﬃcient of α0in (30), it can be written because of (16),:

(ρ1κ1−λ1) = α11ρ3

1−α13ρ2

1+α23 −α22

ρ1

Going back to (11) we ﬁnd the following relation:

−α22

ρ1

=α11ρ3

1−2α13ρ2

1+ (2α12 +α33)ρ1−2α23

which we substitute in its previous. We therefore get:

(ρ1κ1−λ1) = 2α11ρ3

1−3α13ρ2

1+ (2α12 +α33)ρ1−α23

If we substitute this equation (as well as in those which result by exchanging ρ1with

ρ2) into (30), using also the ﬁrst three equations of system (27), we transform (30) as

follows:

2α11ρ3

1α0+ 2α11ρ3

2β0+ 2α11ρ3

1α0+ 2α11ρ3

2β0= 3α13σ∞

x+ (2α12 +α33)τ∞

xy +α23σ∞

y

or after simplifying

ρ3

1α0+ρ3

2β0+ρ3

1α0+ρ3

2β0=3α13

2α11

σ∞

x+2α12 +α33

2α11

τ∞

xy +α23

2α11

σ∞

y(31)

The determinant of the coeﬃcients of the unknowns of system (27) and of (31) can

be written as follows:

D=

1 1 1 1

ρ1ρ2ρ1ρ2

ρ2

1ρ2

2ρ2

1ρ2

2

ρ3

1ρ3

2ρ3

1ρ3

2

it is therefore non-zero, being of type Vandermonde, for ρ1̸=ρ2.

28

Figure 9: Charge on plane zand on planes z1and z2

B.2 Continuous, uniformly distributed normal charge over sec-

tion of the boundary

1. The image of the charge on the plane z, as well as that on the planes z1and z2are

shown on Figs. (9a) and (9b) respectively.

The conformal mapping of Fig. 9b) using the transformations zi=R

2ζi+1

ζi,

(i= 1,2) gives the image of Fig. 10.

Additionally we have:

Yn=−p,Xn= 0, ds =−dz

According to the above and after we set α0=β0= 0 (vanishing stress at inﬁnity)20,

the boundary conditions (19) transferred onto the unit circle take the following form:

α1ln(σ) + f0

1(σ) + β1ln(σ) + f0

2(σ)−α1ln(σ) + f0

1(σ)−β1ln(σ) + f0

2(σ)

=−ps

0

dz

ρ1α1ln(σ) + ρ1f0

1(σ) + ρ2β1ln(σ) + ρ2f0

2(σ)

+ρ1α1ln(σ) + ρ1f0

1(σ)−ρ2β1ln(σ) + ρ2f0

2(σ) = 0

or after simplifying:

f0

1(σ) + f0

2(σ) + f0

1(σ) + f0

2(σ) = −ps

0

dz −(α1+α2−α1−α2) ln(σ)

ρ1f0

1(σ) + ρ2f0

2(σ) + ρ1f0

1(σ) + ρ2f0

2(σ) = −ρ1α1+ρ2β1−ρ1α1−ρ2β1ln(σ)

(32)

20See investigation of the function forms of Complex potential in A.2.4

29

Figure 10: Fig. 9b) under the transformations zi=R

2ζi+1

ζi, (i= 1,2)

2. Before integrating equations (32) it is required on one hand the investigation of the

integral ps

0dz and a search for necessary conditions for the determination of constants

α1and β1on the other.

Because stress p(as an external charge) acts only on section z′z′′ we assume that on

this section of the contour pdz takes the form −pz, whereas on section z′′M z′(Fig. 9)

it is constant and equal to −pz′′. [17]

As far as the characterization of the constants α1and α2we think as follows:

Integrals Ynds and Xnds for every traversal of the contour acquire increases −X0

and Y0respectively, where X0and Y0are the components with respect to the axes xand

yof the total applied charge.

They are therefore multi-valued functions.

Because the function ln(σ) is also multi-valued (being the inverse of a periodic function

with period 2πi) the following equations must hold, in order to ensure the single-valued-

ness of f0

1and f0

2:

X0=−ρ1α1+ρ2β1−ρ1α1−ρ2β12πi

Y0=α1+β1−α1−β12πi (33a)

Thinking similarly for the deformation functions (16), we must have, because uand

vare single-valued functions, the following relations:

κ1α1+κ2β1−κ1α1−κ2β1= 0

λ1α1+λ2β1−λ1α1−λ2β1= 0 (33b)

30

The ﬁrst of (33b) can be written:

0 =α1α11ρ2

1−α13ρ1+α12 +β1α11ρ2

2−α13ρ2+α12

−α1α11ρ2

1−α13ρ1+α12 −β1α11ρ2

2−α13ρ2+α12

or:

α11 ρ2

1α1+ρ2

2β1−ρ2

1α1−ρ2

2β1−α13 ρ1α1+ρ2β1−ρ1α1−ρ2β1

+α12 α1+β1−α1−β1= 0

and because of (33a)

ρ2

1α1+ρ2

2β1−ρ2

1α1−ρ2

2β1=−α13X0+α12 Y0

α112πi (33c)

The second of (33b) can be written:

α1α12ρ1−α23 +α22

ρ1+β1α12ρ2−α23 +α22

ρ2

+α1α12ρ1−α23 +α22

ρ1+β1α12ρ2−α23 +α22

ρ2= 0

and using similar to the previous case transformations, it takes the following form.

1

ρ1

α1+1

ρ2

β1−1

ρ1

α1−1

ρ2

β1=α23Y0+α12 X0

a222πi

Concluding, we write the entire linear system which can be solved for the constants

α1and β1

α1+β1−α1−β1=Y0

2πi

ρ1α1+ρ2β1−ρ1α1−ρ2β1=−X0

2πi

ρ2

1α1+ρ2

2β1−ρ2

1α1−ρ2

2β1=−α13X0+α12 Y0

α112πi

1

ρ1

α1+1

ρ2

β1−1

ρ1

α1−1

ρ2

β1=α12X0+α23 Y0

a222πi

(34)

System (34) always has a solution because the determinant of the coeﬃcients of the

unknowns is non-zero. Indeed we have:

1111

ρ1ρ2ρ1ρ2

ρ2

1ρ2

2ρ2

1ρ2

2

1

ρ1

1

ρ2

1

ρ1

1

ρ2

=−1

ρ1ρ2ρ1ρ2

1 1 1 1

ρ1ρ2ρ1ρ2

ρ2

1ρ2

2ρ2

1ρ2

2

ρ3

1ρ3

2ρ3

1ρ3

2

31

And this as of type Vandermonde for ρ1̸=ρ2never vanishes.

3. Coming back to the system of limit values of the functions f1(ζ1) and f2(ζ2) (32),

we can proceed with its integration (according to Cauchy), as follows:

Multiplying both sides of the equations by 1/(2πi)dσ/(σ−ζ) and integrating on the

unit circle γwe get21:

1

2πi γ

f0

1(σ)

σ−ζdσ +1

2πi γ

f0

2(σ)

σ−ζdσ +1

2πi γ

f0

1(σ)

σ−ζdσ +1

2πi γ

f0

2(σ)

σ−ζdσ

=1

2πi γ

−s

0Ynds

σ−ζdσ −Y0

(2πi)2γ

ln(s)

σ−ζdσ

ρ1

2πi γ

f0

1(σ)

σ−ζdσ +ρ2

2πi γ

f0

2(σ)

σ−ζdσ +ρ1

2πi γ

f0

1(σ)

σ−ζdσ +ρ2

2πi γ

f0

2(σ)

σ−ζdσ = 0

We also take into account that22

1

2πi γ

f0

1(σ)

σ−ζdσ = 0, 1

2πi γ

f0

2(σ)

σ−ζdσ = 0

1

2πi γ

f0

1(σ)

σ−ζdσ =−f0

1(ζ), 1

2πi γ

f0

2(σ)

σ−ζdσ =−f0

2(ζ)

The above system takes the form

f0

1(ζ) + f0

2(ζ) = 1

2πi γYnds

σ−ζdσ +Y0

(2πi)2γ

ln(σ)

σ−ζdσ

ρ1f0

1(ζ) + ρ2f0

2(ζ) = 0

(35)

The ﬁrst integral of the right hand side of the ﬁrst equation of system (35), in view

of the remark in B.2.2, can be written:

1

2πi γYnds

σ−ζdσ =1

2πi σ2

σ1

p·z

σ−ζdσ +1

2πi σ1

σ2

pz′′

σ−ζdσ

Calculating separately each integral we have:

σ2

σ1

pz

σ−ζdσ =pR

2σ2

σ1σ+1

σ

σ−ζdσ

=pR

2σ2

σ1

(σ−ζ) + ζ+1

σ

σ−ζdσ

=pR

2(σ2−σ1) + ζln σ2−ζ

σ1−ζ+1

ζσ2−ζ

σ1−ζ−ln σ2

σ1

=pR

2(σ2−σ1) + ζ+1

ζln σ2−ζ

σ1−ζ−ln σ2

σ1

21For the writing of the system we made use of the equations α1+β1−α1−β1=Y0

2πi and X0= 0.

22See Appendix 2.2.b) and 1.8.b).i.

32

pσ1

σ2

z′′

σ−ζdσ =−pz′′ ln σ2−ζ

σ1−ζconsequently

1

2πi γYnds

σ−ζdσ =pR

4πi (σ2−σ1)+(z−z′′) ln σ2−ζ

σ1−ζ−1

ζln σ2

σ1 (36)

The second integral can be found as follows. If we represent it by Ω(ζ), we will have:

Ω(ζ) = 1

2πi γ

ln(σ)

σ−ζdσ and diﬀerentiating with respect to ζ

Ω′(ζ) = −1

2πi γ

ln(σ)d1

σ−ζ=−1

2πi ln(σ)

σ−ζγ

+1

2πi γ

dσ

σ(σ−ζ)

but it is

1

2πi γ

dσ

σ(σ−ζ)=−1

ζand 1

2πi ln(σ)

σ−ζσ2=ei(θ+2π)

σ1=eiθ

=1

σ1−ζ

as such we conclude that [17]

Ω′(ζ) = −1

σ1−ζ−1

ζ

therefore,

Ω(ζ) = ln(σ1−ζ)−ln(ζ) = ln σ1−ζ

ζ+C(37)

By substituting equations (36) and (37) in system (35), the latter transforms into:

f0

1(ζ) + f0

2(ζ) = p

2πi R

2(σ2−σ1) + (z−z′′) ln σ2−ζ

σ1−ζ−R

2ζln σ2

σ1

+Y0

2πi ln σ1−ζ

ζ

ρ1f0

1(ζ) + ρ2f0

2(ζ) = 0

(38)

By solving (38) with respect to the unknown functions f0

1(ζ) and f0

1(ζ) (after we

consider their domains of deﬁnition), we ﬁnd23.

f0

1(ζ1) = −ρ2

ρ1−ρ2p

2πi −R

2ζ1

ln σ2

σ1+ (z1−z′′) ln σ2−ζ1

σ1−ζ1+Y0

2πi ln σ1−ζ1

ζ1

f0

2(ζ2) = ρ1

ρ1−ρ2p

2πi −R

2ζ2

ln σ2

σ1+ (z2−z′′) ln σ2−ζ2

σ1−ζ2+Y0

2πi ln σ1−ζ2

ζ2

23In the ﬁnal expressions of Complex Potential constants are ignored as not inﬂuencing the tensile

situation.

33

therefore:

Φ′

1(z1) = f1(ζ1) = −ρ2

ρ1−ρ2×

×p

2πi −R

2ζ1

ln σ2

σ1+ (z1−z′′) ln σ2−ζ1

σ1−ζ1+Y0

2πi ln σ1−ζ1

ζ1

+α1ln(ζ1)

Φ′

2(z2) = f2(ζ2) = ρ1

ρ1−ρ2×

×p

2πi −R

2ζ2

ln σ2

σ1+ (z2−z′′) ln σ2−ζ2

σ1−ζ2+Y0

2πi ln σ1−ζ2

ζ2

+β1ln(ζ2)

(39)

The above found functions Φ1(z1) and Φ2(z2) solve completely the initially posed

problem.

The calculation of stresses can now be performed easily by diﬀerentiating with respect

to z1and z2respectively the functions f1(ζ1) and f2(ζ2) as follows:

Φ′′

1(z1) = −ρ2

ρ1−ρ2p

2πi A1−Y0σ1ζ1

2πi R

2(σ1−ζ1)(ζ2

1−1)+α1ζ1

R

2(ζ2

1−1)

Φ′′

2(z1) = ρ1

ρ1−ρ2p

2πi A2−Y0σ1ζ2

2πi R

2(σ1−ζ2)(ζ2

2−1)+β1ζ2

R

2(ζ2

2−1)

(40)

with24

A1=1

ζ2

1−1ln σ2

σ1+ ln σ2−ζ1

σ1−ζ1+(z1−z′′)(σ2−σ1)ζ2

1

(σ1−ζ1)(σ2−ζ1)R

2(ζ2

1−1)

A2=1

ζ2

2−1ln σ2

σ1+ ln σ2−ζ2

σ1−ζ2+(z1−z′′)(σ2−σ1)ζ2

2

(σ1−ζ2)(σ2−ζ2)R

2(ζ2

2−1)

From equations (40), based on relations (15) and (16), we ﬁnd the stresses and the

displacements.

For a total charge on the boundary relations (39) are simpliﬁed as follows:

We have then

ln σ2

σ1= 2πi, ln σ2−ζi

σ1−ζi= 0, Y0= 0

so25,

24Translator’s Note: The terms A1and A2are given separately for easier reading.

25Equations (41) are found in a diﬀerent form also in [7] because of the various constants of elasticity

and parameters used.

34

Φ′

1(z1) = ρ2

ρ1−ρ2·pR

2ζ1

Φ′

2(z2) = −ρ1

ρ1−ρ2·pR

2ζ2

(41)

4. In each case using appropriate limiting access we can, by using equations (39), solve

the problem of the crack with concentrated charge (moving to the limit with respect to

the charge) or the problems of the half-space with concentrated or distributed charges

(moving to the limit with respect to the boundary).

a) Concentrated normal charge against the boundary of the crack at the point Z0.

In this case is: p(z′−z′′) = Y0and σ2→σ1=σ0or

pR

2(σ1−σ2)−σ1−σ2

σ1σ2=pR

2(σ1−σ2)σ2

0−1

σ2

0

=Y0(42)

If we set:

p(σ1−σ2) = Aand pln σ2

σ1=B

we will have:

lim

σ2→σ1

B

−A=ln(σ2)−ln(σ1)

−(σ2−σ1)=−1

σ0

or B=A

σ0

and also

A·R

2·σ2

0−1

σ2

0

=Y0

Hence:

−p

2πi ·R

2·1

ζln σ2

σ1=−B

2πi ·R

2·1

ζ=−A

2πi ·R

2σ0·1

ζ=−Y0

2πi ·σ2

0

σ2

0−1·1

ζ

or

lim

σ2→σ1−p

2πi ·R

2·1

ζln σ2

σ1=−Y0

2πi ·σ0

σ2

0−1·1

ζ(43)

Similarly we ﬁnd, that if (42) holds,

lim

σ2→σ1

(z−z0)p

2πi ln σ2−ζ

σ1−ζ= (z−z0)1

2πi ·Y0

σ0−ζ·σ2

0

R

2(σ2

0−1) (44)

With the substitution of (43) and (44) in (39), we ﬁnd ﬁnally:

35

f1(ζ1) = −ρ2Y0

(ρ1−ρ2)2πi −σ0

(σ2

0−1)ζ1

+z1−z0

R

2(σ0−ζ1)+ ln σ0−ζ1

ζ1+α1ln(ζ1)

f2(ζ2) = ρ1Y0

(ρ1−ρ2)2πi −σ0

(σ2

0−1)ζ2

+z2−z0

R

2(σ0−ζ2)+ ln σ0−ζ2

ζ2+β1ln(ζ2)

(45)

If we set zi=R

2ζi+1

ζi, for i= 1,2 in the equations (45) and perform the calcula-

tions, we ﬁnd (ignoring constants):

f1(ζ1) = −ρ2

ρ1−ρ2·Y0

2πi ln σ0−ζ1

ζ1+α1ln(ζ1)

f2(ζ2) = ρ1

ρ1−ρ2·Y0

2πi ln σ0−ζ2

ζ2+β1ln(ζ2)

(46)

b) Concentrated charge (normal) at the point z0= 0 of the inﬁnite disk.

In this case we are looking for

lim

R→0fi(ζi), for i= 1,2

However, because for R→0, the “ﬁniteness” of zassumes |ζi|→∞, we conclude

that:

lim ln σ0−ζi

ζi=πi

Therefore, as long as we ignore the constants for the functions of Complex Potential,

we have initially:

Φ′

1(z1) = α1ln(ζ1)

and

Φ′′

1(z1) = α1

ζ1

R

2(ζ2

1−1) =α1

1

R

2(ζ1−ζ−1

1)

But for: |ζi| → ∞,ζi−1

ζi=O(ζi) = ζi+1

ζi, (i= 1,2).

Hence,

Φ′′

1=α1

z1

Φ′′

2=β1

z1

(47)

Therefore the functions of Complex Potential, can be written, by integrating (47) as:

36

Figure 11: Two successive charges

Φ′

1(z1) = α1ln(z1)

Φ′

2(z2) = β1ln(z1)(48)

where α1,β1are determined as usual from the solution of the system (34).

c) Concentrated normal charge at a point z0= 0 of the half-space.

We consider the situation (as two successive charges) as in (Fig. 11),

and by applying (46) and because σ0=i,σ′

0=−i, we ﬁnd:

Φ′

1(z1) = −ρ2

ρ1−ρ2·Y0

2πi ln ζ1−i

ζ1+i

Φ′

2(z2) = ρ1

ρ1−ρ2·Y0

2πi ln ζ2−i

ζ2+i(49)

Diﬀerentiating equations (49) with respect to z1and z2respectively, we ﬁnd after

performing the calculations:

Φ′′

1(z1) = −ρ2

ρ1−ρ2·Y0

2πz1·2ζ1

ζ2

1−1

Φ′′

2(z2) = ρ1

ρ1−ρ2·Y0

2πz2·2ζ2

ζ2

2−1

(50)

In order to have ziﬁnite while R→ ∞ we conclude that

lim

R→∞

ζi=±i

Therefore,

lim

R→∞

Φ′′

1(z1) = −ρ2Y0

2πi(ρ1−ρ2)·1

z1

lim

R→∞

Φ′′

2(z2) = ρ1Y0

2πi(ρ1−ρ2)·1

z2

for ℑ(z)>0 (51a)

37

Figure 12: Charge between the points z′and z′′

lim

R→∞

Φ′′

1(z1) = −ρ2Y0

2πi(ρ1−ρ2)·1

z1

lim

R→∞

Φ′′

2(z2) = ρ1Y0

2πi(ρ1−ρ2)·1

z2

for ℑ(z)<0 (51b)

d) Continuous, uniformly distributed normal charge pon the vicinity of the half-

space y= 0 and between the points z′,z′′ (Fig. 12).

Adding to the ﬁrst of equations (39) the constant ρ2(z′−z′′)

ρ1−ρ2·p

2πi ln(R) and setting

Y0=p(z′−z′′) we get

f1(ζ1) = −ρ2

ρ1−ρ2

A1+ (κ+α1) ln(ζ1) (52)

where26

A1=−R

2·1

ζ1

ln σ2

σ1+ (z1−z′′) ln σ2−ζ1

σ1−ζ1+ (z′−z′′) ln(R)(σ1−ζ1)

κ=ρ2

ρ1−ρ2·p

2πi (z′−z′′ )

Because R→ ∞, we can set α1+κ= 0 considering the charge on (Fig. 12) to be

balanced with an equal and opposite force applied at inﬁnity and thus not aﬀecting the

tensile situation on the considered domain.

The ﬁrst term of A1of (52), diﬀerentiated with respect to z1gives:

A′(z1) = R

2·1

ζ2

1·ζ2

1

R

2(ζ2

1−1) ln σ2

σ1=1

ζ2

1−1ln σ2

σ1

and because for R→ ∞,ζi→ −iand σ2→σ1, we conclude: lim

R→∞

A′

z1= 0. Therefore A

equals a constant, which can be ignored.

26Translator’s Note: See footnote 24 on page 34.

38

The second term of A1of (52) can be written:

B= (z1−z′′) ln σ2−ζ1

σ1−ζ

and for R→ ∞ (we accept the variable ζin the region of −i) we will have:

σ2=x2−x2

2−1∼x2−i1−x2

2

2,x2=z′′

R→0

σ1=x1−x2

1−1∼x1−i1−x2

1

2,x1=z′

R→0

ζ1=x−x2−1∼x−i1−x2

2,x=z1

R→0

which ﬁnally gives:

lim

R→∞

(z1−z′′) ln x2−x

x1−x·1−ix+x2

2

1−ix+x1

2

= (z1−z′′) ln x2−x

x1−x

= (z1−z′′) ln z′′ −z1

z′−z1

Similarly we ﬁnd:

lim

R→∞

(z′−z′′) ln(R)(σ1−ζ1) = (z′−z′′ ) ln(z′−z1)

Summarizing the above results and taking into account that in the distribution of

stress the functions of Complex Potential are invariant under addition or subtraction of

constants, we have:

Φ′

1(z1) = −ρ2

ρ1−ρ2

p

2πi {(z1−z′′ ) ln(z1−z′′ )−(z1−z′) ln(z1−z′)}

Φ′

2(z2) = ρ1

ρ1−ρ2

p

2πi {(z2−z′′ ) ln(z2−z′′ )−(z2−z′) ln(z2−z′)}

(53)

for ℑ(z)<0.

Equations (53) solve the posed problem, much more generally than [23], showing for

one more time the importance of the general solution (39).

39

Figure 13: Distorted boundary

B.3 Continuous, uniformly distributed tangential charge over sec-

tion of the boundary

1. In this case, we work exactly as in the problem of normal charge.

We have: Xn= +q,Yn= 0, and ds =−dz

Functions (22) already take the following form:

Ψ′

1(z1) = Q1(ζ1) = α2ln(ζ1) + Q0

1(ζ1)

Ψ′

2(z2) = Q1(ζ2) = β2ln(ζ2) + Q0

2(ζ2)

while the boundary conditions on the unit circle γcan be written:

Q0

1(σ) + Q0

2(σ) + Q0

1(σ) + Q0

2(σ) = −(α2+β2−α2−β2) ln(σ)

ρ1Q0

1(σ) + ρ2Q0

2(σ) + ρ1Q0

1(σ) + ρ2Q0

2(σ) = Xnds =−(ρ1α2+ρ2β2−ρ1α2−ρ2β2) ln(σ)

By integrating according to Cauchy on the unit circle, we get the corresponding to

(35) equations:

Q0

1(ζ) + Q0

2(ζ) = 0

ρ1Q0

1(ζ) + ρ2Q0

2(ζ) = −1

2πi γXnds

σ−ζdσ −X0

(2πi)2γ

ln(σ)

σ−ζdσ (54)

where the constants α2,β2are calculated by solving the corresponding to (34) general

system.

For Y0= 0 the system takes the form:

40

α2+β2−α2−β2= 0

ρ1α2+ρ2β2−ρ1α2−ρ2β2=−X0

2πi

ρ2

1α2+ρ2

2β2−ρ2

1α2−ρ2

2β2=−α13

α11 ·X0

2πi

1

ρ1

α2+1

ρ2

β2−1

ρ1

α2−1

ρ2

β2=−α12

α22 ·X0

2πi

(55)

After the above and in full correspondence to the solution of the functional system

(38), we ﬁnd:

Q0

1(ζ) + Q0

2(ζ) = 0

ρ1Q0

1(ζ) + ρ2Q0

2(ζ) = + qR

4πi (σ2−σ1) + q

2πi (z−z′′ ) ln σ2−ζ

σ1−ζ

+qR

2πi ·1

ζln σ2

σ1+X0

2πi ln σ1−ζ

ζ

and ﬁnally: (in correspondence to the general solution (39) and (40))

Ψ′

1(z1) = Q1(ζ1) = 1

ρ1−ρ2×

×q

2πi −R

2ζ1

ln σ2

σ1+ (z1−z′′) ln σ2−ζ1

σ1−ζ1+X0

2πi ln σ1−ζ1

ζ1

+α2ln(ζ1)

Ψ′

2(z2) = Q2(ζ2) = −1

ρ1−ρ2×

×q

2πi −R

2ζ2

ln σ2

σ1+ (z2−z′′) ln σ2−ζ2

σ1−ζ2+X0

2πi ln σ1−ζ2

ζ2

+β2ln(ζ2)

(56)

By diﬀerentiating equations (56) we get

Ψ′′

1(z1) = 1

ρ1−ρ2q

2πi A1−X0σ1ζ1

2πi R

2(σ1−ζ1)(ζ2

1−1)+α2ζ1

R

2(ζ2

1−1)

Ψ′′

2(z1) = −1

ρ1−ρ2q

2πi A2−X0σ1ζ2

2πi R

2(σ1−ζ2)(ζ2

2−1)+β2ζ2

R

2(ζ2

2−1)

(57)

with27

27Translator’s Note: See footnote 24 on page 34.

41

A1=1

ζ2

1−1ln σ2

σ1+ ln σ2−ζ1

σ1−ζ1+(z1−z′′)(σ2−σ1)ζ2

1

(σ1−ζ1)(σ2−ζ1)R

2(ζ2

1−1)

A2=1

ζ2

2−1ln σ2

σ1+ ln σ2−ζ2

σ1−ζ2+(z1−z′′)(σ2−σ1)ζ2

2

(σ1−ζ2)(σ2−ζ2)R

2(ζ2

2−1)

for a total charge of the boundary we ﬁnd:

Ψ′′

1(z1) = +q

ρ1−ρ2·1

ζ2

1−1Ψ′′

2(z2) = −q

ρ1−ρ2·1

ζ2

2−1(58)

The relations (56) and (57) solve completely the problem for the case of a constant

tangential distributed charge between the points z′and z′′. We can now apply these,

using the same methods of limiting approach as in section B, to solve the corresponding

problems in B.2.4. a), b), c) and d).

2. a) Concentrated tangential charge on the boundary at the point z0(from (46)):

Q1(ζ1) = +1

ρ1−ρ2·X0

2πi ln σ0−ζ1

ζ1+α2ln(ζ1)

Q2(ζ1) = −1

ρ1−ρ2·X0

2πi ln σ0−ζ2

ζ2+β2ln(ζ2)

(59)

b) Concentrated charge (horizontal) at the point z0= 0 of the inﬁnite disk (from

(47)):

Ψ′′

1(z1) = α2

z1

, or Ψ′

1(z1) = α2ln(z1)

Ψ′′

2(z2) = β2

z2

, or Ψ′

2(z2) = β2ln(z2)

(60)

c) Charge (horizontal) concentrated at one point z0= 0 of the boundary y= 0 of

the half-space (from (51a) and (51b)):

Ψ′′

1(z1) = +1

ρ1−ρ2·X0

2πiz1

Ψ′′

2(z2) = −1

ρ1−ρ2·X0

2πiz2

for ℑ(z)>0 (61a)

Ψ′′

1(z1) = −1

ρ1−ρ2·X0

2πiz1

Ψ′′

2(z2) = +1

ρ1−ρ2·X0

2πiz2

for ℑ(z)<0 (61b)

d) Constant tangential charge normally distributed at a section of the boundary

y= 0 and between the points z′and z′′ (from equations (53)):

42

Ψ′

1(z1) = +1

ρ1−ρ2·q

2πi {(z1−z′′ ) ln(z1−z′′)−(z1−z′) ln(z1−z′)}for ℑ(z)>0

Ψ′

2(z2) = −1

ρ1−ρ2·q

2πi {(z2−z′′ ) ln(z2−z′′)−(z2−z′) ln(z2−z′)}for ℑ(z)<0

(62)

43

C Explanation of the Found Expressions - Conclu-

sions

The general formulas (28), (39) and (56) found in section B solve completely the

initially posed problem.

Via relations (16) we can also ﬁnd the displacements on any point on the disk by

determining based on initial conditions the undetermined constants u0and v0.

In what follows we examine the operational details of the found expressions by applying

them to the stress relations on the boundary.

The transition from the complex variable to the real variable can be performed easily

via the known transformations, omitting mostly the operations on the usual trigonometric

functions.

C.1 Uncharged Boundary

1. Relations (29) give the stresses on the disk for uniformly distributed stresses at

inﬁnity, that is for given σ∞

x,σ∞

yand τ∞

xy .

It can be proved initially that existence of a crack oriented parallel to the applied

stresses does not aﬀect the tensile situation of the disk (as in the case of isotropy).

Indeed for σ∞

x=σ∞

y= 0, it is: σx=σ∞

xand σy=τxy = 0.

2. If we look for the stresses on the boundary we will have:

a) σy=σ∞

y+ 2ℜσ∞

y·1

σ2−1where σ=eiθ the common value of the variables

ζ1and ζ2(θ̸=π

2±π

2) or

σy=σ∞

y+ 2σ∞

y· ℜ cos(θ)−isin(θ)

sin(θ)·1

2i= 0

similarly

τxy =τ∞

xy + 2ℜτxy

1

σ2−1= 0 (σ̸=±1)

The above results were expected and were found simply for the veriﬁcation of the

correctness of expressions (29).

b) To ﬁnd σxon the boundary we set ζ1=ζ2=σ=eiθ , therefore:

σx=σ∞

x+ 2ℜρ1ρ2(ρ2−ρ1)

ρ1−ρ2

σ∞

y1

σ2−1

or

σx=σ∞

x−1

sin(θ)ρ1ρ2σ∞

y+ (ρ1+ρ2)τ∞

xy e−iθ

i

if in the relation above we substitute the value of the complex parameters.

ρ1=γ1+iδ1where δ1>0

ρ2=γ2+iδ2where δ2>0

44

Figure 14: Diagram σxbecause of σ∞

y>0, for upper section of the crack

we get after the calculations.

σx=σ∞

x+ [(γ1γ2−δ1δ2)−(γ1δ2−γ2δ1) cot(θ)] σ∞

y

+ [(γ1+γ2)−(δ1+δ2) cot(θ)] τ∞

xy

(63)

We can easily plot the graph of equation (63) for the two separate cases: the vanishing

of τxy and of σy.

Taking into account the relations |γ1|< δ1and |γ2|< δ2(they hold since we are in

the elliptic region) as well as (γ1δ2+γ2δ1)>0 and σy>0 we draw on (Fig. 14) the

function σx(θ) for the upper section of the crack.

The vanishing point of σxcan be found from the relation:

tan(θ) = γ1γ2−δ1δ2

γ1δ2+γ2δ1

or θ0= arctan γ1γ2−δ1δ2

γ1δ2+γ2δ1

therefore:

θ0=arg(ρ1) + arg(ρ2) , for arg(ρ1ρ2)< π,

arg(ρ1) + arg(ρ2)−π, for arg(ρ1ρ2)> π.

We ﬁnd the diagram of σxfor the lower section from (Fig. 14) by reﬂecting with

respect to the axes xand y.

The diagram of σxbecause of τ∞

xy for the lower section is found by reﬂecting (Fig. 15)

with respect to the xand yaxes.

c) From equation (63) as well as from studying the diagrams (Fig. 14) and (Fig.

15) we conclude the following:

c1) In the case where τ∞

xy = 0, σxbecomes unbounded in absolute value close to

the edges of the crack. This phenomenon appears only in the case of general anisotropy

45

Figure 15: Diagram σxbecause τ∞

xy >0, for upper section of the crack

and happens because the material tends to slide towards the direction x, despite the fact

that it is stressed only in the direction y.

c2) If we have orthotropic material or balance then σxis constant and compressive

on the entire boundary and equal to −δ1δ2σ∞

yin the ﬁrst case, and equal to −σ∞

yin the

second case.

c3) In the case of general anisotropy the magnitudes of |γ1|and |γ2|are much

smaller than δ1and δ2, which means that the angle |π−(θ1+θ2)|=|π−θ0|is very

small and the vanishing point of σxis found very close to the edges of the crack.

c4) If σ∞

y= 0 then σxin all three cases, general anisotropy, orthotropy and

isotropy, becomes unbounded in magnitude the closer we are to the edges of the crack.

The only diﬀerence between the ﬁrst case and the other two is that in general anisotropy

an additional constant σ0

xappears on the entire boundary which equals (γ1+γ2)τ∞

xy .

The vanishing point of σxin the above case is found at Rγ1+γ2

√(γ1+γ2)2+(δ1+δ2)2from the

middle, which is quite small, but non-zero nevertheless for γ1+γ2̸= 0 or the same for

α13 ̸= 0.

We can also add that the unboundedness of the stress at the end points happens

because of the existence of an ideal cusp, a thing which cannot happen in reality on the

one hand because of the nature of the material and on the other because of the immediate

lamination which will happen in cases of such great stress on the aforementioned points.

46

Figure 16: Normal charge on the disk

C.2 Normal Charge On the Boundary

1. Relations (40) give as is known the functions Φ′′

1(z1) and Φ′′

2(z2) based on which

we can calculate the stresses on any point of the disk for a charge shown on (Fig. 16).

The mapping of the boundary of the crack on the circumference of the unit circle

maps the points z′and z′′ onto σ1and σ2(Fig. 17).

If we want to ﬁnd σyon the boundary we will have:

σy= 2ℜ{Φ′′

1(σ)+Φ′′

2(σ)}= Φ′′

1(σ)+Φ′′

2(σ) + Φ′′

1(σ) + Φ′′

2(σ)

= Φ′′

1(σ)+Φ′′

2(σ) + Φ′′

11

σ+ Φ′′

21

σ

given that for σon the unit circle the relation σσ = 1 holds, we have:

σy=p

2πi 1

σ2−1ln σ2

σ1+ + ln σ2−σ

σ1−σ+(z−z′′)(σ2−σ1)σ2

(σ1−σ)(σ2−σ)R

2(σ2−1)

−Y0σ1σ

2πi R

2(σ1−σ)(σ2−1) +(α1+β1)σ

R

2(σ2−1)

−p

2πi σ2

σ2−1ln σ2

σ1+ ln σ2−σ

σ1−σ+(z−z′′)(σ2−σ1)σ2

(σ1−σ)(σ2−σ)R

2(σ2−1)

+Y0σ2

2πi R

2(σ1−σ)(σ2−1) −(α1+β1)σ

R

2(σ2−1)

The above equation after simpliﬁcations and based on the relation α1+β1−α1−β1=

Y0

2πi (see system (34)) results in the following:

47

Figure 17: Correspondence between points z′,z′′ and σ1,σ2

σy=−p

2πi ln σ2

σ1+p

2πi ln σ2−σ

σ1−σ−ln σ2−σ

σ1−σ (64)

Setting,

σ2−σ=r2eiϕ2σ2−σ=r2e−iϕ2

σ1−σ=r1eiϕ1σ1−σ=r1e−iϕ1

We ﬁnd

ln σ2−σ

σ1−σ−ln σ2−σ

σ1−σ= ln r2

r1+i(ϕ2−ϕ1)−ln r2

r1+i(ϕ2−ϕ1)

= 2i(ϕ2−ϕ1)

and equation (64) takes the following ﬁnal form:

σy=−p

2πi ln σ2

σ1−2i(ϕ2−ϕ1)(65)

We examine below the following two cases:

a) The point σis found outside the arc deﬁned by σ1σ2(outside the charged area),

in which case we have the geometric image (Fig. 18).

One can easily deduce that in this case:

48

Figure 18: Unit circle with σnot in σ1σ2

ϕ2−ϕ1=ϕ=ω

2=1

2iln σ2

σ1

The above result substituted in (65) gives:

σy=−p

2πi ln σ2

σ1−ln σ2

σ1= 0

b) The point zis found inside the charge area hence σlies between σ1σ2. In this

case we will have the geometric image (Fig. 19), from which we conclude the following:

ω′=π−ω

2= (ϕ1−π)+(π−ϕ2) = ϕ1−ϕ2

hence

ϕ2−ϕ1=ω

2−π=1

2iln σ2

σ1−π

The above result substituted in (5) gives

σy=−p

2πi ln σ2

σ1−ln σ2

σ1+ 2πi=−p

The results of cases a) and b) were expected. The calculations were performed so

that the operational details of the found functions could be shown. These functions being

continuous on the entire planes ζ1and ζ2(except at the points ±1) give limiting functions

on the boundary which satisfy the initially posed conditions.

49

Figure 19: Unit circle with σin σ1σ2

2. The calculation of τxy on the boundary is simpler, because it can be concluded

immediately from the second of the limiting conditions (32) or by simple calculations,

that τxy = 0 on the entire boundary. (Exceptions are the usual points σ=±1).

3. The calculation of σxon the boundary is a little involved because of the length of

the used equations. Indeed, we have:

σx= 2ℜρ2

1f′

1(σ) + ρ2

2f′

2(σ)

ρ2

1f′

1(σ) + ρ2

2f′

2(σ) = −ρ1ρ2p

2πi ×

×

ln σ2

σ1

σ2−1+ ln σ2−σ

σ1−σ+(z−z′′)(σ2−σ1)σ2

(σ1−σ)(σ2−σ)R

2(σ2−1)

+Y0σ1σρ1ρ2

2πi R

2(σ1−σ)(σ2−1) +(ρ2

1α1+ρ2

2β1)σ

R

2(σ2−1)

ρ2

1f′

1(σ) + ρ2

2f′

2(σ) = −ρ1ρ2p

2πi ×

×

σln σ2

σ1

σ2−1+ ln σ2−σ

σ1−σ+(z−z′′)(σ2−σ1)σ2

(σ1−σ)(σ2−σ)R

2(σ2−1)

−Y0σ2ρ1ρ2

2πi R

2(σ1−σ)(σ2−1) −(ρ2

1α1+ρ2

2β1)σ

R

2(σ2−1)

50

and by adding we get28

σx=p

2πi −ρ1ρ2+σ2ρ1ρ2

σ2−1ln σ2

σ1−ρ1ρ2ln σ2−σ

σ1−σ+ρ1ρ2ln σ2−σ

σ1−σ

−α12Y0σ

α11πiR(σ2−1) +Y0σ(ρ1ρ2σ1−ρ1ρ2σ)

πiR(σ1−σ)(σ2−1)

−p(ρ1ρ2−ρ1ρ2) (z−z′′)(σ2−σ1)σ2

2πi(σ1−σ)(σ2−σ)R

2(σ2−1)

(66)

On the above expression (66) we perform the following transformations and substitu-

tions:

ρ1=γ1+iδ1where δ1>0

ρ2=γ2+iδ2where δ2>0

σ1=eiθ1

σ2=eiθ2θ2> θ1

σ=eiθ θ̸=π

2±π

2

Y0=pR (cos(θ1)−cos(θ2))

We also examine the expression

U=−ρ1ρ2ln σ2−σ

σ1−σ+ρ1ρ2ln σ2−σ

σ1−σ

We set

σ2−σ=r2eiϕ2hence σ2−σ=r2e−iϕ2

σ1−σ=r1eiϕ1hence σ1−σ=r1e−iϕ1

from which we conclude:

U= (−ρ1ρ2+ρ1ρ2) ln r2

r1−(−ρ1ρ2+ρ1ρ2)i(ϕ2−ϕ1)

or

U=−2i(γ1δ2+γ2δ1) ln r2

r1−2i(γ1δ1−δ1δ2)i(ϕ2−ϕ1)

With respect to the cases C.2.1 a) b) we have,

(θ−θ2)(θ−θ1)>0ϕ2−ϕ1=θ2−θ1

2

(θ−θ2)(θ−θ1)<0ϕ2−ϕ1=θ2−θ1

2−π

28We also use the third equation from (34).

51

therefore

U1=U2+ 2πi(γ1γ2−δ1δ2) (67)

Consequently, the found σxwill have a jump discontinuity at the points z′and z′′ of

magnitude (γ1γ2−δ1δ2)p.

According to the previous substitutions and transformations and after certain calcu-

lations expression (66) takes the following form.

σx=−p

2π(γ1δ2+γ2δ1)A1+cos(θ1)−cos(θ2)

sin(θ)A2

where29

A1= [θ2−θ1] cot(θ) + 2 lnr2

r1

+[cos(θ)−cos(θ2)] [sin(θ−θ1)−sin(θ−θ2) + sin(θ1−θ2)]

sin(θ) [1 −cos(θ−θ1)] [1 −cos(θ−θ2)]

A2=(γ1γ2−δ1δ2) [1 −cos(θ−θ1)] −(γ1δ2+γ2δ1) sin(θ−θ1)

1−cos(θ−θ1)−α12

α11

or after further calculations and simpliﬁcations the following ﬁnal form

σx= (γ1δ2+γ2δ1)(θ2−θ1) cos(θ) + sin(θ2)−sin(θ1)

sin(θ)+ 2 ln r2

r1

+γ1γ2−δ1δ2−α12

α11 cos(θ1)−cos(θ2)

sin(θ)

(68)

Expression (68) gives the stress on the boundary as a function of the angle θ, which

is connected as usual to the abscissa from the middle of the section based on the relation

x=Rcos(θ).

We also observe that the expression (68) consists of two terms. The ﬁrst stands only

in the case of general anisotropy (γ1·γ2̸= 0) and its magnitude becomes unbounded

at the points which have abscissas equal to Rcos(0), Rcos(θ1), Rcos(θ2) and Rcos(π),

because for θtending to 0 or π, sin(θ) vanishes, while for θtending to θ1or θ2(given

that ln r2

r1= ln

sin(θ−θ2

2)

sin(θ−θ1

2)

),