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The Problem of the Crack in the "Generally" Anisotropic Disk

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Recently translated father's 1968 Ph.D. Thesis, appears to be related to Hilbert's Problem #21, in the context of Theory of Elasticity using Complex Variables. The Hilbert kernel is stressed on various points and a complete solution for its elastic profile is given up to compromise. The crack is described using analytic functions.
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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 tc. 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|t1t2|µ, holds, for Aand µpositive constants with 0 < µ 1.
6. The integral F(z) = 1
2πi Cff(t)
tzdt (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 zChas 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)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 Gis 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)
tzdt =f(z), for zG+
1
2πi C
f(t)
tzdt = 0, for zG
b) If f(z) is a holomorphic function on Gand continuous on GC, we have
1
2πi C
f(t)
tzdt =f(z) + f(), for zG
1
2πi C
f(t)
tzdt =f(), for zG+
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)
tzdt = 0, for all zG
b) A sufficient and necessary condition for a given f(t) on Cto be the limiting
value of a holomorphic function on Gis
1
2πi C
f(t)
tzdt =α, for all zG+
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=e, for 0 θ2π.
We call Σ+and Σthe 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 ζσthen
Φ(ζ) has the limiting value Φ
(σ) defined on Σand 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)
σζ=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)
σζ= 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=Tn, ˆy). In other words, treating Tn, ) 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
∂x42α23
4Ψ
∂x3∂ y +(2α12 +α33)4Ψ
∂x2y2
2α13
4Ψ
∂x∂y3+α11
4Ψ
∂y4= 0
(10)
If we represent the fourth order differential operator with D:
α22
4
∂x42α23
4
∂x3∂ y +(2α12 +α33)4
∂x2y2
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:
α11ρ42α13 ρ3+ (2α12 +α33)ρ22α23 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
2Φ2Φ
∂x2y22α23
2Φ2Φ
∂x2x∂y 2α13
2Φ2Φ
∂y2∂x∂y
(13)
If we search for the constants λisuch that for Φ
∂y λiΦ
∂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
α11ρ42α13 ρ3+ (2α12 +α33)ρ22α23 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) = 2{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= 2{κ1Φ
1(z1) + κ2Φ
2(z2)}+u0
v= 2{λ1Φ
1(z1) + λ2Φ
2(z2)}+v0, with
κi=α11ρ2
iα13ρi+α12
λi=α12ρiα23 +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) + 1y.
Similarly for z2,L2(x+yi) = x+ρ2y= (x+γ2y) + 2y.
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
αnzn
1+
n=0
βnzn
2(where αn, βncomplex constants)
Going back to equations (15) we conclude that
Φ′′
1(z1) =
n=0
αnzn
1and Φ′′
2(z2)
n=0
βnzn
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 zz′′.
b) For continuous, uniformly distributed, tangential charge qon a section of the
contour (Fig. 7b). Xn=q,Yn= 0 for zbetween zz′′.
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=ethe 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
zz
2i,z2=z+z
2+ρ2
zz
2i
and by eliminating zwe find the relation
(ρ1ρ2)ω(ζ0) = (ρ2i)ω1(ζ0)(ρ1i)ω2(ζ0), where ζ0=e(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 zz1and zz2(the
relation ωω1ω2is 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
i1̸= 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
ζ1ln(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
ζ1and Ψ0
2(ζ2) = Φ0
2R
2ζ2+1
ζ2are 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=ethe 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
σζand we integrate on the unit
circle.
If we consider18 that:
1
2πi γ
σdσ
σζ= 0 and 1
2πi γ
σ(σζ)=1
ζ
and
1
2πi γ
Ψ0
1(σ)
σζ=Ψ0
1(ζ) and 1
2πi γ
Ψ0
2(σ)
σζ=Ψ0
2(ζ)
and also that:
1
2πi γ
Ψ0
1(σ)
σζ=1
2πi γ
Ψ0
2(σ)
σζ= 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(ρ1ρ2)ρ2σ
y+τ
xy 1
ζ1
Ψ0
2(ζ2) = R
2(ρ1ρ2)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+ 21
ρ1ρ2ρ2σ
y+τ
xy 1
ζ2
11+1
ρ1ρ2ρ1σ
y+τ
xy 1
ζ2
21
σx=σ
x+ 2ρ2
1
ρ1ρ2ρ2σ
y+τ
xy 1
ζ2
11+ρ2
2
ρ1ρ2ρ1σ
y+τ
xy 1
ζ2
21
τxy =τ
xy + 2ρ1
ρ1ρ2ρ2σ
y+τ
xy 1
ζ2
11+ρ2
ρ1ρ2ρ1σ
y+τ
xy 1
ζ2
21
(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→∞ 2{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:
(λ1ρ1κ1)α0+ (λ2ρ2κ2)β0+ (λ1ρ1κ1)α0+ (λ2ρ2κ2)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),:
(ρ1κ1λ1) = α11ρ3
1α13ρ2
1+α23 α22
ρ1
Going back to (11) we find the following relation:
α22
ρ1
=α11ρ3
12α13ρ2
1+ (2α12 +α33)ρ12α23
which we substitute in its previous. We therefore get:
(ρ1κ1λ1) = 2α11ρ3
13α13ρ2
1+ (2α12 +α33)ρ1α23
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+α2α1α2) ln(σ)
ρ1f0
1(σ) + ρ2f0
2(σ) + ρ1f0
1(σ) + ρ2f0
2(σ) = ρ1α1+ρ2β1ρ1α1ρ2β1ln(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 zz′′ 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+ρ2β1ρ1α1ρ2β12πi
Y0=α1+β1α1β12πi (33a)
Thinking similarly for the deformation functions (16), we must have, because uand
vare single-valued functions, the following relations:
κ1α1+κ2β1κ1α1κ2β1= 0
λ1α1+λ2β1λ1α1λ2β1= 0 (33b)
30
The first of (33b) can be written:
0 =α1α11ρ2
1α13ρ1+α12 +β1α11ρ2
2α13ρ2+α12
α1α11ρ2
1α13ρ1+α12 β1α11ρ2
2α13ρ2+α12
or:
α11 ρ2
1α1+ρ2
2β1ρ2
1α1ρ2
2β1α13 ρ1α1+ρ2β1ρ1α1ρ2β1
+α12 α1+β1α1β1= 0
and because of (33a)
ρ2
1α1+ρ2
2β1ρ2
1α1ρ2
2β1=α13X0+α12 Y0
α112πi (33c)
The second of (33b) can be written:
α1α12ρ1α23 +α22
ρ1+β1α12ρ2α23 +α22
ρ2
+α1α12ρ1α23 +α22
ρ1+β1α12ρ2α23 +α22
ρ2= 0
and using similar to the previous case transformations, it takes the following form.
1
ρ1
α1+1
ρ2
β11
ρ1
α11
ρ2
β1=α23Y0+α12 X0
a222πi
Concluding, we write the entire linear system which can be solved for the constants
α1and β1
α1+β1α1β1=Y0
2πi
ρ1α1+ρ2β1ρ1α1ρ2β1=X0
2πi
ρ2
1α1+ρ2
2β1ρ2
1α1ρ2
2β1=α13X0+α12 Y0
α112πi
1
ρ1
α1+1
ρ2
β11
ρ1
α11
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/(σζ) and integrating on the
unit circle γwe get21:
1
2πi γ
f0
1(σ)
σζ+1
2πi γ
f0
2(σ)
σζ+1
2πi γ
f0
1(σ)
σζ+1
2πi γ
f0
2(σ)
σζ
=1
2πi γ
s
0Ynds
σζY0
(2πi)2γ
ln(s)
σζ
ρ1
2πi γ
f0
1(σ)
σζ+ρ2
2πi γ
f0
2(σ)
σζ+ρ1
2πi γ
f0
1(σ)
σζ+ρ2
2πi γ
f0
2(σ)
σζ= 0
We also take into account that22
1
2πi γ
f0
1(σ)
σζ= 0, 1
2πi γ
f0
2(σ)
σζ= 0
1
2πi γ
f0
1(σ)
σζ=f0
1(ζ), 1
2πi γ
f0
2(σ)
σζ=f0
2(ζ)
The above system takes the form
f0
1(ζ) + f0
2(ζ) = 1
2πi γYnds
σζ+Y0
(2πi)2γ
ln(σ)
σζ
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
σζ=1
2πi σ2
σ1
p·z
σζ+1
2πi σ1
σ2
pz′′
σζ
Calculating separately each integral we have:
σ2
σ1
pz
σζ=pR
2σ2
σ1σ+1
σ
σζ
=pR
2σ2
σ1
(σζ) + ζ+1
σ
σζ
=pR
2(σ2σ1) + ζln σ2ζ
σ1ζ+1
ζσ2ζ
σ1ζln σ2
σ1
=pR
2(σ2σ1) + ζ+1
ζln σ2ζ
σ1ζln σ2
σ1
21For the writing of the system we made use of the equations α1+β1α1β1=Y0
2πi and X0= 0.
22See Appendix 2.2.b) and 1.8.b).i.
32
pσ1
σ2
z′′
σζ=pz′′ ln σ2ζ
σ1ζconsequently
1
2πi γYnds
σζ=pR
4πi (σ2σ1)+(zz′′) ln σ2ζ
σ1ζ1
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)
σζand differentiating with respect to ζ
(ζ) = 1
2πi γ
ln(σ)d1
σζ=1
2πi ln(σ)
σζγ
+1
2πi γ
σ(σζ)
but it is
1
2πi γ
σ(σζ)=1
ζand 1
2πi ln(σ)
σζσ2=ei(θ+2π)
σ1=e
=1
σ1ζ
as such we conclude that [17]
(ζ) = 1
σ1ζ1
ζ
therefore,
Ω(ζ) = ln(σ1ζ)ln(ζ) = ln σ1ζ
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(σ2σ1) + (zz′′) ln σ2ζ
σ1ζR
2ζln σ2
σ1
+Y0
2πi ln σ1ζ
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
ρ1ρ2p
2πi R
2ζ1
ln σ2
σ1+ (z1z′′) ln σ2ζ1
σ1ζ1+Y0
2πi ln σ1ζ1
ζ1
f0
2(ζ2) = ρ1
ρ1ρ2p
2πi R
2ζ2
ln σ2
σ1+ (z2z′′) ln σ2ζ2
σ1ζ2+Y0
2πi ln σ1ζ2
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
ρ1ρ2×
×p
2πi R
2ζ1
ln σ2
σ1+ (z1z′′) ln σ2ζ1
σ1ζ1+Y0
2πi ln σ1ζ1
ζ1
+α1ln(ζ1)
Φ
2(z2) = f2(ζ2) = ρ1
ρ1ρ2×
×p
2πi R
2ζ2
ln σ2
σ1+ (z2z′′) ln σ2ζ2
σ1ζ2+Y0
2πi ln σ1ζ2
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
ρ1ρ2p
2πi A1Y0σ1ζ1
2πi R
2(σ1ζ1)(ζ2
11)+α1ζ1
R
2(ζ2
11)
Φ′′
2(z1) = ρ1
ρ1ρ2p
2πi A2Y0σ1ζ2
2πi R
2(σ1ζ2)(ζ2
21)+β1ζ2
R
2(ζ2
21)
(40)
with24
A1=1
ζ2
11ln σ2
σ1+ ln σ2ζ1
σ1ζ1+(z1z′′)(σ2σ1)ζ2
1
(σ1ζ1)(σ2ζ1)R
2(ζ2
11)
A2=1
ζ2
21ln σ2
σ1+ ln σ2ζ2
σ1ζ2+(z1z′′)(σ2σ1)ζ2
2
(σ1ζ2)(σ2ζ2)R
2(ζ2
21)
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 σ2ζi
σ1ζi= 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
ρ1ρ2·pR
2ζ1
Φ
2(z2) = ρ1
ρ1ρ2·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(zz′′) = Y0and σ2σ1=σ0or
pR
2(σ1σ2)σ1σ2
σ1σ2=pR
2(σ1σ2)σ2
01
σ2
0
=Y0(42)
If we set:
p(σ1σ2) = Aand pln σ2
σ1=B
we will have:
lim
σ2σ1
B
A=ln(σ2)ln(σ1)
(σ2σ1)=1
σ0
or B=A
σ0
and also
A·R
2·σ2
01
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
01·1
ζ
or
lim
σ2σ1p
2πi ·R
2·1
ζln σ2
σ1=Y0
2πi ·σ0
σ2
01·1
ζ(43)
Similarly we find, that if (42) holds,
lim
σ2σ1
(zz0)p
2πi ln σ2ζ
σ1ζ= (zz0)1
2πi ·Y0
σ0ζ·σ2
0
R
2(σ2
01) (44)
With the substitution of (43) and (44) in (39), we find finally:
35
f1(ζ1) = ρ2Y0
(ρ1ρ2)2πi σ0
(σ2
01)ζ1
+z1z0
R
2(σ0ζ1)+ ln σ0ζ1
ζ1+α1ln(ζ1)
f2(ζ2) = ρ1Y0
(ρ1ρ2)2πi σ0
(σ2
01)ζ2
+z2z0
R
2(σ0ζ2)+ ln σ0ζ2
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
ρ1ρ2·Y0
2πi ln σ0ζ1
ζ1+α1ln(ζ1)
f2(ζ2) = ρ1
ρ1ρ2·Y0
2πi ln σ0ζ2
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
R0fi(ζi), for i= 1,2
However, because for R0, the “finiteness” of zassumes |ζi|→∞, we conclude
that:
lim ln σ0ζi
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
11) =α1
1
R
2(ζ1ζ1
1)
But for: |ζi| → ∞,ζi1
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
ρ1ρ2·Y0
2πi ln ζ1i
ζ1+i
Φ
2(z2) = ρ1
ρ1ρ2·Y0
2πi ln ζ2i
ζ2+i(49)
Differentiating equations (49) with respect to z1and z2respectively, we find after
performing the calculations:
Φ′′
1(z1) = ρ2
ρ1ρ2·Y0
2πz1·2ζ1
ζ2
11
Φ′′
2(z2) = ρ1
ρ1ρ2·Y0
2πz2·2ζ2
ζ2
21
(50)
In order to have zifinite while R→ ∞ we conclude that
lim
R→∞
ζi=±i
Therefore,
lim
R→∞
Φ′′
1(z1) = ρ2Y0
2πi(ρ1ρ2)·1
z1
lim
R→∞
Φ′′
2(z2) = ρ1Y0
2πi(ρ1ρ2)·1
z2
for (z)>0 (51a)
37
Figure 12: Charge between the points zand z′′
lim
R→∞
Φ′′
1(z1) = ρ2Y0
2πi(ρ1ρ2)·1
z1
lim
R→∞
Φ′′
2(z2) = ρ1Y0
2πi(ρ1ρ2)·1
z2
for (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(zz′′)
ρ1ρ2·p
2πi ln(R) and setting
Y0=p(zz′′) we get
f1(ζ1) = ρ2
ρ1ρ2
A1+ (κ+α1) ln(ζ1) (52)
where26
A1=R
2·1
ζ1
ln σ2
σ1+ (z1z′′) ln σ2ζ1
σ1ζ1+ (zz′′) ln(R)(σ1ζ1)
κ=ρ2
ρ1ρ2·p
2πi (zz′′ )
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
11) ln σ2
σ1=1
ζ2
11ln σ2
σ1
and because for R→ ∞,ζi→ −iand σ2σ1, 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= (z1z′′) ln σ2ζ1
σ1ζ
and for R→ ∞ (we accept the variable ζin the region of i) we will have:
σ2=x2x2
21x2i1x2
2
2,x2=z′′
R0
σ1=x1x2
11x1i1x2
1
2,x1=z
R0
ζ1=xx21xi1x2
2,x=z1
R0
which finally gives:
lim
R→∞
(z1z′′) ln x2x
x1x·1ix+x2
2
1ix+x1
2
= (z1z′′) ln x2x
x1x
= (z1z′′) ln z′′ z1
zz1
Similarly we find:
lim
R→∞
(zz′′) ln(R)(σ1ζ1) = (zz′′ ) ln(zz1)
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
ρ1ρ2
p
2πi {(z1z′′ ) ln(z1z′′ )(z1z) ln(z1z)}
Φ
2(z2) = ρ1
ρ1ρ2
p
2πi {(z2z′′ ) ln(z2z′′ )(z2z) ln(z2z)}
(53)
for (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+β2α2β2) ln(σ)
ρ1Q0
1(σ) + ρ2Q0
2(σ) + ρ1Q0
1(σ) + ρ2Q0
2(σ) = Xnds =(ρ1α2+ρ2β2ρ1α2ρ2β2) 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
σζX0
(2πi)2γ
ln(σ)
σζ(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+β2α2β2= 0
ρ1α2+ρ2β2ρ1α2ρ2β2=X0
2πi
ρ2
1α2+ρ2
2β2ρ2
1α2ρ2
2β2=α13
α11 ·X0
2πi
1
ρ1
α2+1
ρ2
β21
ρ1
α21
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 (σ2σ1) + q
2πi (zz′′ ) ln σ2ζ
σ1ζ
+qR
2πi ·1
ζln σ2
σ1+X0
2πi ln σ1ζ
ζ
and finally: (in correspondence to the general solution (39) and (40))
Ψ
1(z1) = Q1(ζ1) = 1
ρ1ρ2×
×q
2πi R
2ζ1
ln σ2
σ1+ (z1z′′) ln σ2ζ1
σ1ζ1+X0
2πi ln σ1ζ1
ζ1
+α2ln(ζ1)
Ψ
2(z2) = Q2(ζ2) = 1
ρ1ρ2×
×q
2πi R
2ζ2
ln σ2
σ1+ (z2z′′) ln σ2ζ2
σ1ζ2+X0
2πi ln σ1ζ2
ζ2
+β2ln(ζ2)
(56)
By differentiating equations (56) we get
Ψ′′
1(z1) = 1
ρ1ρ2q
2πi A1X0σ1ζ1
2πi R
2(σ1ζ1)(ζ2
11)+α2ζ1
R
2(ζ2
11)
Ψ′′
2(z1) = 1
ρ1ρ2q
2πi A2X0σ1ζ2
2πi R
2(σ1ζ2)(ζ2
21)+β2ζ2
R
2(ζ2
21)
(57)
with27
27Translator’s Note: See footnote 24 on page 34.
41
A1=1
ζ2
11ln σ2
σ1+ ln σ2ζ1
σ1ζ1+(z1z′′)(σ2σ1)ζ2
1
(σ1ζ1)(σ2ζ1)R
2(ζ2
11)
A2=1
ζ2
21ln σ2
σ1+ ln σ2ζ2
σ1ζ2+(z1z′′)(σ2σ1)ζ2
2
(σ1ζ2)(σ2ζ2)R
2(ζ2
21)
for a total charge of the boundary we find:
Ψ′′
1(z1) = +q
ρ1ρ2·1
ζ2
11Ψ′′
2(z2) = q
ρ1ρ2·1
ζ2
21(58)
The relations (56) and (57) solve completely the problem for the case of a constant
tangential distributed charge between the points zand 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
ρ1ρ2·X0
2πi ln σ0ζ1
ζ1+α2ln(ζ1)
Q2(ζ1) = 1
ρ1ρ2·X0
2πi ln σ0ζ2
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
ρ1ρ2·X0
2πiz1
Ψ′′
2(z2) = 1
ρ1ρ2·X0
2πiz2
for (z)>0 (61a)
Ψ′′
1(z1) = 1
ρ1ρ2·X0
2πiz1
Ψ′′
2(z2) = +1
ρ1ρ2·X0
2πiz2
for (z)<0 (61b)
d) Constant tangential charge normally distributed at a section of the boundary
y= 0 and between the points zand z′′ (from equations (53)):
42
Ψ
1(z1) = +1
ρ1ρ2·q
2πi {(z1z′′ ) ln(z1z′′)(z1z) ln(z1z)}for (z)>0
Ψ
2(z2) = 1
ρ1ρ2·q
2πi {(z2z′′ ) ln(z2z′′)(z2z) ln(z2z)}for (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+ 2σ
y·1
σ21where σ=ethe 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 + 2τxy
1
σ21= 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=e , therefore:
σx=σ
x+ 2ρ1ρ2(ρ2ρ1)
ρ1ρ2
σ
y1
σ21
or
σx=σ
x1
sin(θ)ρ1ρ2σ
y+ (ρ1+ρ2)τ
xy e
i
if in the relation above we substitute the value of the complex parameters.
ρ1=γ1+1where δ1>0
ρ2=γ2+2where 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+ [(γ1γ2δ1δ2)(γ1δ2γ2δ1) 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) = γ1γ2δ1δ2
γ1δ2+γ2δ1
or θ0= arctan γ1γ2δ1δ2
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)|=|πθ0|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 zand z′′ onto σ1and σ2(Fig. 17).
If we want to find σyon the boundary we will have:
σy= 2{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
σ21ln σ2
σ1+ + ln σ2σ
σ1σ+(zz′′)(σ2σ1)σ2
(σ1σ)(σ2σ)R
2(σ21)
Y0σ1σ
2πi R
2(σ1σ)(σ21) +(α1+β1)σ
R
2(σ21)
p
2πi σ2
σ21ln σ2
σ1+ ln σ2σ
σ1σ+(zz′′)(σ2σ1)σ2
(σ1σ)(σ2σ)R
2(σ21)
+Y0σ2
2πi R
2(σ1σ)(σ21) (α1+β1)σ
R
2(σ21)
The above equation after simplifications and based on the relation α1+β1α1β1=
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 σ2σ
σ1σln σ2σ
σ1σ (64)
Setting,
σ2σ=r2e2σ2σ=r2e2
σ1σ=r1e1σ1σ=r1e1
We find
ln σ2σ
σ1σln σ2σ
σ1σ= ln r2
r1+i(ϕ2ϕ1)ln r2
r1+i(ϕ2ϕ1)
= 2i(ϕ2ϕ1)
and equation (64) takes the following final form:
σy=p
2πi ln σ2
σ12i(ϕ2ϕ1)(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
ϕ2ϕ1=ϕ=ω
2=1
2iln σ2
σ1
The above result substituted in (65) gives:
σy=p
2πi ln σ2
σ1ln 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=πω
2= (ϕ1π)+(πϕ2) = ϕ1ϕ2
hence
ϕ2ϕ1=ω
2π=1
2iln σ2
σ1π
The above result substituted in (5) gives
σy=p
2πi ln σ2
σ1ln 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= 2ρ2
1f
1(σ) + ρ2
2f
2(σ)
ρ2
1f
1(σ) + ρ2
2f
2(σ) = ρ1ρ2p
2πi ×
×
ln σ2
σ1
σ21+ ln σ2σ
σ1σ+(zz′′)(σ2σ1)σ2
(σ1σ)(σ2σ)R
2(σ21)
+Y0σ1σρ1ρ2
2πi R
2(σ1σ)(σ21) +(ρ2
1α1+ρ2
2β1)σ
R
2(σ21)
ρ2
1f
1(σ) + ρ2
2f
2(σ) = ρ1ρ2p
2πi ×
×
σln σ2
σ1
σ21+ ln σ2σ
σ1σ+(zz′′)(σ2σ1)σ2
(σ1σ)(σ2σ)R
2(σ21)
Y0σ2ρ1ρ2
2πi R
2(σ1σ)(σ21) (ρ2
1α1+ρ2
2β1)σ
R
2(σ21)
50
and by adding we get28
σx=p
2πi ρ1ρ2+σ2ρ1ρ2
σ21ln σ2
σ1ρ1ρ2ln σ2σ
σ1σ+ρ1ρ2ln σ2σ
σ1σ
α12Y0σ
α11πiR(σ21) +Y0σ(ρ1ρ2σ1ρ1ρ2σ)
πiR(σ1σ)(σ21)
p(ρ1ρ2ρ1ρ2) (zz′′)(σ2σ1)σ2
2πi(σ1σ)(σ2σ)R
2(σ21)
(66)
On the above expression (66) we perform the following transformations and substitu-
tions:
ρ1=γ1+1where δ1>0
ρ2=γ2+2where δ2>0
σ1=e1
σ2=e2θ2> θ1
σ=eθ̸=π
2±π
2
Y0=pR (cos(θ1)cos(θ2))
We also examine the expression
U=ρ1ρ2ln σ2σ
σ1σ+ρ1ρ2ln σ2σ
σ1σ
We set
σ2σ=r2e2hence σ2σ=r2e2
σ1σ=r1e1hence σ1σ=r1e1
from which we conclude:
U= (ρ1ρ2+ρ1ρ2) ln r2
r1(ρ1ρ2+ρ1ρ2)i(ϕ2ϕ1)
or
U=2i(γ1δ2+γ2δ1) ln r2
r12i(γ1δ1δ1δ2)i(ϕ2ϕ1)
With respect to the cases C.2.1 a) b) we have,
(θθ2)(θθ1)>0ϕ2ϕ1=θ2θ1
2
(θθ2)(θθ1)<0ϕ2ϕ1=θ2θ1
2π
28We also use the third equation from (34).
51
therefore
U1=U2+ 2πi(γ1γ2δ1δ2) (67)
Consequently, the found σxwill have a jump discontinuity at the points zand z′′ of
magnitude (γ1γ2δ1δ2)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= [θ2θ1] cot(θ) + 2 lnr2
r1
+[cos(θ)cos(θ2)] [sin(θθ1)sin(θθ2) + sin(θ1θ2)]
sin(θ) [1 cos(θθ1)] [1 cos(θθ2)]
A2=(γ1γ2δ1δ2) [1 cos(θθ1)] (γ1δ2+γ2δ1) sin(θθ1)
1cos(θθ1)α12
α11
or after further calculations and simplifications the following final form
σx= (γ1δ2+γ2δ1)(θ2θ1) cos(θ) + sin(θ2)sin(θ1)
sin(θ)+ 2 ln r2
r1
+γ1γ2δ1δ2α12
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(θθ2
2)
sin(θθ1
2)
),