APPLICATIONS OF POTENTIAL THEORY IN MECHANICS Selection of new results
V.I. Fabrikant
To my parents
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APPLICATIONS OF POTENTIAL THEORY IN MECHANICS Selection of new results
V.I. Fabrikant
To my parents
1
PREFACE
It is not easy to find something new in mathematics. It is difficult to find something new in something very old, like the Potential Theory, which was studied by the greatest scientists in the past centuries. And it is extremely difficult to make this find on an elementary level, with no mathematical apparatus involved which would be considered new even in the times of, say, Poisson. This is exactly what the author claims to have done in this book: a new and elementary method is described for solving mixed boundary value problems, and their applications in engineering. The method can solve non-axisymmetric problems as easily as axisymmetric ones, exactly and in closed form . It enables us to treat analytically non-classical domains. The major achievements of the method comprise the derivation of explicit and elementary expressions for the Green’s functions, related to a penny-shaped crack and a circular punch, development of the Saint-Venant type theory of contact and crack problems for general domains , and investigation of various interactions between cracks, punches, and external loadings. The method also provides, as a bonus, a tool for exact evaluation of various two-dimensional integrals involving distances between two or more points. It is believed that majority of these results are beyond the reach of existing methods. For over twenty years, since being a graduate student in Moscow, the author had been perplexed by an inconsistency between various solutions to the problems in Potential Theory and the way those solutions have been obtained, namely, that the solution was quite elementary, while the apparatus used was very complicated, involving various integral transforms or special functions expansions, which are beyond the comprehension of an ordinary engineer. The author’s search for a new method was based on the conviction that an elementary result should be obtainable by elementary means. The method has been found and is presented here in detail. One may just wonder why the method was not discovered at least a century ago.
2
3
The book is addressed to a wide audience ranging from engineers, involved in elastic stress analysis, to mathematical physicists and pure mathematicians. While an engineer can find in the book some elementary, ready to use formulae for solving various practical problems, a mathematical physicist might become interested in new applications of the mathematical apparatus presented, and a pure mathematician might interpret some of the results in terms of fractional calculus, investigate the group properties of the operators used, or, having noticed the fact that no attention is paid in the book to a rigorous foundation of the method, might wish to remedy the situation. Due to the mathematical analogy between mixed boundary value problems in elasticity and in other branches of engineering science, the book should be of interest to specialists in electromagnetics, acoustics, diffusion, fluid mechanics, etc. Though several such applications have been published by the author, the space considerations did not allow us to include them, but references are given at appropriate places in the book. The book is accessible to anyone with a background in university undergraduate calculus, but should be of interest to established scientists as well. Though the method is elementary, the transformations involved are sometimes very non-trivial and cumbersome, while the final result is usually very simple. The reader who is interested only in application of the general results to his/hers particular problems may skip the long derivations and use the final formulae which require little effort. The reader who wants to master the method in order to solve new problems has to repeat the derivations which are given in sufficient detail. The exercises are important in this regard. They vary from very simple to quite difficult. Some can be used as a subject for a graduate degree thesis. The book is based entirely on the author’s results, and this is why the work of other scientists is mentioned only when such a quotation is inevitable for some reason, like numerical data needed to verify the accuracy of approximate results, comparison with existing results, or pointing out some errors in publications. There are several books and review articles presenting an adequate account of the state-of-the-art in the field. Appropriate references are given for the reader’s convenience. The purpose of this book was neither to repeat nor to compete with them. The development of the method can by no means be considered completed, this is just a beginning. The results presented in the book may be compared to the tip of an iceberg, taking into consideration numerous applications which are still to come. The solution of fundamental problems in a simple form enables us to consider various more complicated problems which were not even attempted before. The method can be expanded to spherical, toroidal and other systems of coordinates, so that more complex geometries may benefit from it. The method proved useful in the generalized potential theory as well. Some of these results, though already published by the author, could not be included in the book due to severe restrictions on the book volume.
4
For the reader’s convenience, it was attempted to make each chapter (and section, wherever possible) self-contained. The reader can skip several sections and continue reading, without loosing the ability to understand material. On the other hand, this resulted in repetition of some definitions and descriptions. The author thinks that the additional convenience is worth several extra pages in the book. The author is grateful to Professor J.R. Barber from Michigan, Professor B. Noble from England, and Professor J.R. Rice from Harvard who agreed to read the manuscript and expressed their opinion. The book contains so much new material that some misprints and errors are inevitable, though every effort was made to eliminate them. The author would be grateful for every communication in this regard. All the readers’ comments are welcome. The address is V.I. Fabrikant prisoner # 167 932D Archambault jail Ste-Anne-des-Plaines Quebec, Canada J0N 1H0
CONTENTS
INTRODUCTION
7
CHAPTER 1. DESCRIPTION OF THE NEW METHOD 1.1 Integral representation for the reciprocal of the distance between two points 1.2 Properties of the L-operators 1.3 Some further integral representations 1.4 Internal mixed boundary value problem for a half-space 1.5 External mixed boundary value problem for a half-space 1.6 Some fundamental integrals CHAPTER 2. 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY General solution Point force solutions Internal mixed problem of type I External mixed problem of type I Integral representation for q 2/ R 3 Internal mixed problem of type II External mixed problem of type II Inverse crack problem in elasticity
15 22 25 31 43 63
CHAPTER 3. MIXED-MIXED BOUNDARY VALUE PROBLEMS 3.1 General formulation of the problem 3.2 Internal axisymmetric mixed-mixed problem 3.3 External axisymmetric mixed-mixed problem 3.4 Generalization for non-homogeneous half-space 3.5 Effect of a shearing force and a tilting moment on a bonded circular punch 3.6 Non-axisymmetric internal mixed-mixed problem 3.7 Non-axisymmetric external mixed-mixed problem Appendix A3.1 Appendix A3.2 Appendix A3.3
5
73 77 81 92 101 105 111 129
142 145 171 177 182 188 200 207 211 215
6
CHAPTER 4. APPLICATIONS IN FRACTURE MECHANICS 4.1 Flat crack under arbitrary normal loading 4.2 Point force loading of a penny-shaped crack 4.3 Concentrated load outside a circular crack 4.4 Plane crack under arbitrary shear loading 4.5 Penny-shaped crack under uniform pressure 4.6 Penny-shaped crack under uniform shear loading 4.7 Asymptotic behavior of stresses and displacements near the crack rim 4.8 Flat crack of general shape 4.9 General crack under uniform shear 4.10 Close interaction of pressurized coplanar circular cracks 4.11 Close interaction of coplanar circular cracks under shear loading Appendix A4.1 Appendix A4.2 Appendix A4.3 Appendix A4.4
220 227 229 231 243 248 252 259 273 275 292 307 311 312 317
CHAPTER 5. APPLICATION TO CONTACT PROBLEMS 5.1 Contact problem for a smooth punch. 324 5.2 Flat centrally loaded circular punch 328 5.3 Inclined circular punch on an elastic half-space 332 5.4 Flat punch of arbitrary planform under the action of a normal centrally applied force 336 5.5 Inclined flat punch of general shape 347 5.6 Curved punch of general planform 367 5.7 Flat flexible punch of general planform under shifting load 387 5.8 Reissner-Sagoci problem for general domains 393 5.9 Interaction between punches subjected to normal pressure 402 5.10 Interaction between flexible punches under shifting loading 412 5.11 Contact problem for a rough punch 422 Appendix A5.1 425 REFERENCES
439
SUBJECT INDEX
447
INTRODUCTION
A short survey of methods for solving mixed boundary value problems of potential theory is given, and some of their limitations are pointed out. The necessity for developing a new method is justified. A concise description of each chapter is given for the reader’s convenience. Various applications of potential theory in electrostatics, heat transfer, elasticity, diffusion, and other branches of engineering science are well known, and have attracted significant attention of scientists like Laplace, Poisson, Green, Beltrami, Kirchhoff, Lord Kelvin, Hobson, and others who made a significant contribution to the field during past centuries. The boundary value problems with mixed conditions are the most difficult to solve, and, at the same time, they are among the most important in various engineering applications. Recall the celebrated problem of an electrified circular disc. It is next to impossible even to name every scientist who solved the problem by one method or another. We can specify two categories of methods. The first one requires construction of the Green’s function, after which each particular solution can be presented in quadratures. The second category encompasses various integral transform methods. Let us briefly discuss both categories. Hobson (1899) has constructed the Green’s function for a circular disc and a spherical bowl using a method due to Sommerfeld. He has used toroidal coordinates τ, σ, φ (in our notation), which are related to the cartesian x , y , and z as follows: x =
a sinhτ cosφ , coshτ − cosσ
y =
Here a is the disc radius.
a sinhτ sinφ , coshτ − cosσ
z =
a sinσ . coshτ − cosσ
The potential function V at the point (τ0,σ0,φ0)
7
8
INTRODUCTION
external to the disc, which on the disc takes the values v (τ,φ), can be expressed (1 + coshτ) cos(σ /2) 0 1 ⌠⌠ V (τ0,σ0,φ0) = 2 π ⌡ ⌡ 2 aR [cosh2(α/2) − sin2(σ0/2)]1/2 S
+
z R3
tan-1
(1 − cosσ0)1/2
v (τ,φ) d S , (coshα + cosσ0)1/2
(0.1)
where R is the distance between the points (τ0,σ0,φ0) and (τ,σ,φ), and α is defined by coshα = coshτ0coshτ − sinhτ0sinhτ cos(φ−φ0).
The practical value of
expression (0.1) is quite limited, since there seems to be no way to evaluate the integral involved, even in the simplest case when v (τ,φ) is constant. Hobson had to use a very ingenious method in order to find the potential function for v =const and v =µ x , µ=const. On the other hand, it turns out that the integral in (0.1) is computable in elementary functions for any polynomial v (in Cartesian coordinates). This was one of the reasons that prompted the author to look for an alternative approach, which would be as general as (0.1), and, on the other hand, would allow elementary and straightforward computation of the integrals involved. Consideration of the mathematically equivalent problem of a circular punch pressed against an elastic half-space leads to the integral equation (Galin, 1953) 2π a
ω(ρ,φ) = H ⌠ ⌠ ⌡⌡ 0
σ(ρ0,φ0) R
ρ0dρ0dφ0.
(0.2)
0
Here a is the punch radius, H is an elastic constant, ω is displacement under the punch (a known function), σ stands for exerted by the punch (an unknown function), and R is the distance points (ρ,φ) and (ρ0,φ0). Leonov (1953) has obtained a closed solution of integral equation (0.2) by a very ingenious method. in our notation 2π a ω(ρ0,φ0) 1 ⌠⌠ σ(ρ,φ) = − ρ0dρ0dφ0 ∆ ⌡ R 4π2 H ⌡ 0 0
the normal the pressure between the form exact
His result reads
9
INTRODUCTION
2π a
2 ⌠ ⌠ R R − tan-1( ) + π ⌡ ⌡ η η 0
ω(ρ0,φ0) R3
ρ0dρ0dφ0.
(0.3)
0
Here η=[( a 2−ρ2)( a 2−ρ20)]1/2/ a and ∆ is the two-dimensional Laplace operator. One can observe the same handicap in (0.3): difficulty to evaluate the integrals directly, even in the simplest case when ω is constant. Again, it is clear that the integrals are computable in elementary functions for any prescribed polynomial displacement. This indicates a gap in our knowledge which needs to be filled. The integral transform method, involving dual integral equations, was originated, probably by Weber (1873) and Beltrami (1881), and continued by Busbridge (1938) and others. Significant achievements in the systematic application of the method to various problems belong to Sneddon. The reader is referred to the books by Sneddon (1951, 1966) and Ufliand (1967) for additional references. Some quite remarkable results were obtained by Ufliand (1977). Despite this success, it has always been the author’s conviction that our use of integral transforms generally indicates our inability to solve problems directly. To this end, two illustrative examples are presented. Here is how the problem of a circular disc, charged to a potential v 0=const, is solved by the dual integral equation method. It is necessary to find a harmonic function V , vanishing at boundary conditions on the plane z =0: V = v 0, for ρ≤ a ;
infinity,
∂V = 0, for ρ> a . ∂z
and
subjected
to
the
mixed
(0.4)
The solution is presented in the form ∞
dt V (ρ, z ) = ⌠ A ( t )e-tz J 0( t ρ) . t ⌡
(0.5)
0
Here J 0 is the Bessel function of zero order, and A ( t ) is the as yet unknown function which should be chosen to satisfy (0.4). Substitution of the boundary conditions (0.4) in (0.5) leads to the dual integral equations
10
INTRODUCTION
∞
⌠ A ( t ) J ( t ρ) d t = v , for 0≤ρ≤a ; 0 0 t ⌡ 0
∞
⌠ A ( t ) J ( t ρ)d t = 0, for ρ> a . 0 ⌡
(0.6)
0
By using the discontinuous Weber-Schafheitlin integrals, one can deduce that A(t) =
2 v sin( at ). π 0
(0.7)
The solution (0.5) can now be rewritten as ∞
2 ⌠ -tz dt V (ρ, z ) = v e sin( at ) J ( t ρ) , 0 t π 0⌡
for z ≥0,
(0.8)
0
and the charge density σ over the disc is given by v σ(ρ) =
∞ 0
⌠ J ( t ρ) sin( at ) d t . 0
π2 ⌡
(0.9)
0
Both integrals (0.8) and (0.9) are computable in elementary functions. example, the last integral yields v σ(ρ) =
0
π2( a 2 − ρ2)1/2
.
For
(0.10)
There seems to be a discrepancy between the simplicity of the final result and the apparatus used to obtain it. The general idea, that an elementary result should be obtained by elementary means, calls for a search for a new and elementary approach. The second example comes from consideration of the simplest case of a penny-shaped crack of radius a , subjected to axisymmetric pressure p (ρ). The corresponding potential function f can be found in (Kassir and Sih, 1975) as follows:
11
INTRODUCTION
∞
ds , f (ρ, z ) = ⌠ A ( s ) J 0(ρ s ) exp(− sz) s ⌡
(0.11)
0
where a
t
1 ⌠ rp(r) d r sin st d t ⌠ 2 A ( s) = − 2 1/2 . πµ ⌡ ⌡ (t − r ) 0
(0.12)
0
The user has to substitute the explicit expression for p in (0.12), and to evaluate two consecutive integrals. The result is to be substituted in (0.11), and an infinite integral with Bessel function is to be evaluated. It seems natural to try to spare one integration by substituting (0.12) in (0.11), changing the order of integration and evaluating the integral (Gradshtein and Ryzhik, 1963, formula 6.752.1): ∞
⌠ sinst J (ρs) exp(−sz) d s = sin-1 t , 0 s l (t) ⌡ 2
(0.13)
0
where the notation l (t) = 2
was introduced.
1 {[(ρ + t )2 + z 2]1/2 + [(ρ − t )2 + z 2]1/2}. 2
(0.14)
Formulae (0.11) and (0.12) can be combined to give
a t p ( ρ ) ρ dρ 0 0 0 1 ⌠ -1 t f (ρ, z ) = − sin dt ⌠ 2 . 2 1/2 πµ ⌡ l (t) ⌡ ( t − ρ0) 2
(0.15)
0
0
Note that expression (0.15) contains no trace of the integral transform, and thus gives us a hint that a direct and elementary solution is indeed possible. For example, in the case of a uniform loading, p =const., and the potential function (0.15) will take the form: a
p ⌠ t f (ρ, z ) = − t sin-1 dt. πµ ⌡ l (t) 0
2
(0.16)
12
INTRODUCTION
The integral in (0.16), though looking formidable due to (0.14), can be evaluated in elementary functions, namely, p 2 a (2 a + 2 z 2 − ρ2)sin-1( ) + l (ρ2 − l 21)1/2 − 2 z ( a 2 − l 21)1/2. 1 4πµ l
f (ρ, z ) = −
2
Here the abbreviations l
1
and l
2
(0.17) stand for l ( a ) and l ( a ) respectively, with l 1
2
2
defined by (0.14), and l (t) = 1
1 {[(ρ + t )2 + z 2]1/2 − [(ρ − t )2 + z 2]1/2}. 2
(0.18)
One can show that arbitrary polynomial loading in (0.15) will lead to an elementary expression for the potential function, and thus to an elementary complete solution. The situation can now be summarized. The Green’s function approach is the most general, the main impediment being the inability of direct derivation of results which were usually constructed due to some ingenious considerations. By contrast, the integral transform method allows a straightforward derivation of the results, but it is the least general, since each particular problem has to be solved from beginning to the end. The method is best suited to axisymmetric problems. In the general case, separate solutions have to be obtained for each harmonic. Non-axisymmetric problems involving various interactions (several arbitrarily located charged discs, interaction of punches and cracks, etc.) are extremely difficult to solve by the integral transform method. A new method has to be found which would be as general as the Green’s function method, and, at the same time, it has to be elementary and straightforward, with no integral transforms or special function expansions involved. When one problem has been solved by a complicated method, it is often possible to find another method to solve the same problem more simply. The new method would have been of little value if all it could do were to solve more easily the already solved problems. The main advantage of the new method presented here is its ability to solve non-axisymmetric problems as easily as axisymmetric ones, which in turn opens up new horizons, and allows us to solve some problems which were not even considered before, namely, the analytical treatment of nonclassical domains, and the solution to various interaction problems. The general description of the method is given in Chapter 1. It starts with a derivation of the basic integral representation for the reciprocal of the distance between two points, followed by several generalizations. A closed form exact solution is given to the non-axisymmetric mixed problem of potential theory
INTRODUCTION
for a half-space, with Dirichlet conditions prescribed inside a circle, and Neumann conditions given on the outside, and vice-versa. Some integrals, which are of fundamental value to the method, are evaluated in elementary functions. Chapter 2 is devoted to the mixed boundary value problems of an elastic half-space. The general solution is expressed in terms of three harmonic functions. A classification of internal and external mixed problems of type I and type II is introduced. The problems of the first type are characterized by mixed conditions with respect to normal parameters (the pressure and normal displacement), with the shear stress prescribed all over the boundary. These problems are solved for a non-homogeneous half-space, with the elasticity modulus assumed to be a power function of the depth. The case when the boundary conditions are mixed with respect to tangential displacements and stresses, with the normal stress being prescribed all over the boundary, is classified as the type II problem. Each type is considered separately. Exact closed form solution and various examples of punch and crack problems are presented. The problem is called mixed-mixed when the boundary conditions are mixed with respect to both normal and tangential parameters. This kind of problem is the most difficult to solve due to the coupling of the governing integral equations, which can no longer be solved separately. The axisymmetric and non-axisymmetric internal and external problems are considered in Chapter 3. The exact solution is given in terms of Fourier series expansions. A flat punch, bonded to a transversely isotropic elastic half-space and subjected to general loading is considered in detail. The interaction of exterior loading with the punch is also investigated. While the problems solved in Chapter 2 deal primarily with the stresses and displacements in the plane z =0, the complete solution to various crack problems is the subject of Chapter 4. The solution is called complete, when explicit expressions for the field of stresses and displacements is defined in the whole space. The cases of a penny-shaped crack under arbitrary normal and tangential loadings are considered separately. Explicit expressions are derived for all the Green’s functions involved. All the results are given in terms of elementary functions. These solutions enable us to solve more complicated problems of interaction of a penny-shaped crack and an exterior load. A set of non-singular governing integral equations is derived for the interaction between coplanar circular cracks, subjected to normal pressure and shear loading. An approximate analytical solution is presented for the case of a flat crack of general shape, subjected to a uniform pressure or a uniform shear. The solution is exact for an ellipse, and is expected to be satisfactory for a wide variety of shapes. A comparison is made with various numerical results, available in the literature, and a very good accuracy is established in many cases. The method is less accurate for domains with sharp angles and the aspect ratio far away from unity.
13
14
INTRODUCTION
A general solution in terms of one harmonic function is given in Chapter 5 to the problem of a smooth punch, penetrating a transversely isotropic elastic half-space. The main potential function and all the Green’s functions are expressed in terms of elementary functions for a circular punch of arbitrary profile. It is shown that the complete solution is also presentable in elementary functions for a general polynomial profile. The knowledge of a complete solution, combined with the reciprocal theorem, enables us to solve more complicated problems of interaction of exterior loadings with punches, and the interaction between punches. The general method is applied to the analytical treatment of nonclassical contact problems. Again, the solution is exact for an elliptical punch, and is expected to be satisfactory for a punch of arbitrary planform. The cases of a flat centrally and noncentrally loaded punch, and the case of a curved punch are considered in detail. An extensive comparison is made with the numerical results, available in the literature. As it was in the case of crack problems, the agreement is quite satisfactory, except for the domains with sharp angles and the aspect ratio far away from unity. An analytical solution is given to the problem of a non-smooth punch, subjected to normal and shear loading, with the Coulomb friction law assumed between the punch base and the elastic half-space. The best way to master a new method is through the exercises. Each chapter contains a certain number of them, the majority of exercises are supplied with an answer, a hint or a complete solution. The reader is encouraged to do them all. The transformations involved are elementary, though sometimes very non-trivial, and require some ingenuity.
CHAPTER 1 DESCRIPTION OF THE NEW METHOD
Some integral representations, which are of fundamental value to the method, are derived. An exact closed form solution is given to the mixed boundary value problem of potential theory for a half-space, with a circular line of division of boundary conditions.
1.1 Integral representation for the reciprocal of the distance between two points The author has decided to start with a derivation of a new integral representation for the reciprocal of the distance between two points located in the plane z =0 since this quantity is very important in potential theory. Here we repeat the derivation leading to such a representation, as it was given in (Fabrikant 1971e). Consider the expression 1 1 , 1+u = R ( ρ2 + ρ20 − 2ρρ0cos(φ − φ0))(1+u)/2 where u is a constant and -1< u <1. series will take the form ∞
1 = R 1+u ∞
=
Σ n=-∞
Σ n=-∞
in(φ-φ0)
e
2π
(1.1.1)
The standard expansion of (1.1.1) in Fourier
2π
ψ e-in dψ ⌠ ⌡ ( ρ2 + ρ2 − 2ρρ cosψ)(1+u)/2 0 0 0
in(φ-φ )
0 e 2πΓ[ n + (1 + u )/2] ρ n 1 + u 1 + u ρ2 F ( , n + , n +1; ). 2 2 Γ[(1 + u )/2] Γ( n + 1)ρ0 2πρ1+u ρ20 0
(1.1.2) 15
16
CHAPTER 1,
Description of the new method
Here F stands for the Gauss hypergeometric function. representation F(
By using another integral
1 + u 1 + u , n + 1; z ) , n + 2 2 1
2n+u 2 -(1+u)/2 2Γ( n + 1) ⌠ t (1 − t ) = dt, Γ[ n + (1 + u )/2] Γ[1 − (1 + u )/2] ⌡ 2 (1+u)/2 (1 − zt ) 0
expression (1.1.2) can be transformed into πu 1 2 1+u = π cos 2 R
∞
Σ n=-∞
in(φ-φ0)
min (ρ0,ρ)
e ⌠ (ρρ0)n ⌡ 0
x 2n+ud x
.
(1.1.3)
(ρ2 − x 2)(ρ2 − x 2)(1+u)/2 0
Summation in (1.1.3) finally gives 1 1 1+u = R ( ρ2 + ρ20 − 2ρρ cos(φ − φ ))(1+u)/2 0 0 x2 λ( , φ−φ ) x ud x 0 ρρ0
min (ρ0,ρ)
=
2 πu ⌠ cos π 2 ⌡ 0
.
(1.1.4)
(ρ2 − x 2)(ρ2 − x 2)(1+u)/2 0
Here the notation was introduced λ( k ,ψ) =
1 − k2 1 + k 2 − 2 k cosψ
.
(1.1.5)
After one cumbersome derivation is finished, we can always find a way to do it much simpler. Indeed, if we introduce a new variable η( x ) = [(ρ2 − x 2)(ρ20 − x 2)]1/2/ x , expression (1.1.4) may be rewritten as
(1.1.6)
17
Integral representation for the reciprocal distance
∞
1 2 π u ⌠ η-udη 1+u = π cos 2 2 2. R ⌡R + η
(1.1.7)
0
The integral in (1.1.7) can be evaluated by using formula (3.241.4) from (Gradshtein and Ryzhik, 1963), thus proving the identity. Note that parameter η will be used throughout the book also for the case when x >max(ρ,ρ0), and expression (1.1.6) in this case is interpreted as η( x ) = ( x 2 − ρ2)1/2( x 2 − ρ20)1/2/ x . One can deduce from (1.1.7) that in the particular case when u =0, the integral in (1.1.4) can be evaluated as indefinite, and we have a very important representation λ(
x2 , φ−φ0) d x ρρ0
(ρ2 − x 2)1/2(ρ20 − x 2)1/2
1 ⌠ = − tan-1 R ⌡ (ρ2 − x 2)1/2(ρ20 − x 2)1/2
xR
.
(1.1.8)
All the results above are related to the distance between two points in the plane z =0. We need to generalize them to represent 1 1 . 1+u = 2 2 [ρ + ρ0 − 2ρρ0cos(φ−φ0) + z 2](1+u)/2 R0
(1.1.9)
One can observe that representation (1.1.4) remains valid if we formally substitute ρ and ρ0 by arbitrary quantities l 1 and l 2. We need to choose them so that ρ2 + ρ20 − 2ρρ0cos(φ−φ0) + z 2 = l 21 + l 22 − 2 l 1 l 2cos(φ−φ0).
(1.1.10)
This leads to two equations l 21 + l 22 = ρ2 + ρ20 + z 2,
l 1 l 2 = ρρ0.
(1.1.11)
The solution will take the form l 1(ρ0,ρ, z ) =
1 {[(ρ + ρ0)2 + z 2]1/2 − [(ρ − ρ0)2 + z 2]1/2}, 2
(1.1.12)
18
CHAPTER 1,
l 2(ρ0,ρ, z ) =
Description of the new method
1 {[(ρ + ρ0)2 + z 2]1/2 + [(ρ − ρ0)2 + z 2]1/2}. 2
(1.1.13)
Hereafter the following abbreviations will be used: l 1( x ) ≡ l 1( x ,ρ, z ),
l 2( x ) ≡ l 2( x ,ρ, z ),
(1.1.14)
l 2 ≡ l 2( a ,ρ, z ).
(1.1.15)
l 1 ≡ l 1( a ,ρ, z ), Note the limiting properties lim l 1( x ) = min( x ,ρ),
lim l 2( x ) = max( x ,ρ).
z→ 0
(1.1.16)
z→ 0
In view of the properties above, the representation (1.1.4) can be generalized 1 1 1+u = 2 2 R0 [ρ + ρ0 − 2ρρ0cos(φ−φ0) + z 2](1+u)/2 λ(
l1(ρ 0)
=
x2 , φ−φ0) x ud x ρρ0
2 πu ⌠ cos . π 2 ⌡ {[ l 21(ρ ) − x 2][ l 22(ρ ) − x 2]}(1+u)/2 0 0
(1.1.17)
0
Formula (1.1.17) simplifies when u =0 1 1 = 2 2 R0 [ρ + ρ0 − 2ρρ cos(φ−φ ) + z 2]1/2 0 0
l1(ρ 0)
λ(
x2 , φ−φ ) d x 0 ρρ 0
2 ⌠ = . π ⌡ {[ l 21(ρ ) − x 2][ l 22(ρ ) − x 2]}1/2 0 0
(1.1.18)
0
Again, one can notice that the integral in (1.1.18) may be evaluated as indefinite λ(
x2 , φ−φ ) d x 0 ρρ0
⌠ ⌡ {[ l 21(ρ0) − x 2][ l 22(ρ0) − x 2]}1/2
1 = − tan-1 R0
{[ l 21(ρ ) − x 2][ l 22(ρ ) − x 2]}1/2 0 0 xR 0
.
19
Integral representation for the reciprocal distance
(1.1.19) The last representation is very important and will be widely used throughout the book. Another series of useful formulae can be obtained from those above by a simple change of variables, namely, λ(
ρρ0 x2
, φ−φ0) d x
1 ⌠ tan-1 2 2 2 2 1/2 = R ⌡ {[ x − l 1(ρ0)][ x − l 2(ρ0)]} 0
{[ x 2 − l 21(ρ0)][ x 2 − l 22(ρ0)]}1/2 xR 0
, (1.1.20)
1 1 1+u = 2 2 R0 [ρ + ρ0 − 2ρρ0cos(φ−φ0) + z 2](1+u)/2 λ(
∞
=
2 πu cos π 2
l
2
ρρ0 x
2
, φ−φ0) x ud x
⌠ , ⌡ {[ x 2 − l 21(ρ )][ x 2 − l 22(ρ )]}(1+u)/2 0 0 (ρ )
(1.1.21)
0
1 1 = 2 2 R0 [ρ + ρ0 − 2ρρ0cos(φ−φ0) + z 2]1/2
∞
=
2 π
λ(
l
2
ρρ0 x2
λ(
ρρ0 x2
, φ−φ0) d x
⌠ , ⌡ {[ x 2 − l 21(ρ )][ x 2 − l 22(ρ )]}1/2 0 0 (ρ )
(1.1.22)
0
, φ−φ0) d x
⌠ ⌡ ( x 2 − ρ2)1/2( x 2 − ρ20)1/2
( x 2 − ρ2)1/2( x 2 − ρ20)1/2 1 . = tan-1 R xR
(1.1.23)
The representations above are useful for solving external mixed boundary value problems. Several modifications of (1.1.19) are available.
For example, we can write
20
CHAPTER 1,
λ(
x2 , φ−φ0) d x ρρ0
⌠ ⌡ (ρ2 − x 2)1/2[ρ20 − g 2( x )]1/2
1 = − tan-1 R0
Description of the new method
(ρ2 − x 2)1/2[ρ20 − g 2( x )]1/2 xR 0
. (1.1.24)
Here g ( x ) = x [1 + z 2/(ρ2 − x 2)]1/2.
(1.1.25)
It is important to notice that the function g ( x ) is inverse to l 1 for x 2<ρ2, and is inverse to l 2 for x 2>ρ2+ z 2.
Introduction of a new variable x = l 1( y ), y = g ( x )
transforms (1.1.24) into [ l 22( y ) − y 2]1/2
l 2( y )
1 , φ−φ d y ⌠ λ 0 ⌡ (ρ20 − y 2)1/2[ l 22( y ) − l 21( y )] ρρ0 1 = − tan-1 R0
(ρ20 − y 2)1/2[ l 22( y ) − y 2]1/2 yR 0
(1.1.26)
A particular case of (1.1.18), when z =0, reads 1 1 = 2 2 R [ρ + ρ0 − 2ρρ0cos(φ−φ0)]1/2 λ(
min (ρ0,ρ)
2 = ⌠ π⌡
x2 , φ−φ0) d x ρρ0
(ρ2 − x 2)1/2(ρ20 − x 2)1/2
0
.
(1.1.27)
The same result takes another form due to (1.1.22)
∞
1 2 = R π
λ(
ρρ0 x2
, φ−φ0) d x
⌠ . ⌡ ( x 2 − ρ2)1/2( x 2 − ρ20)1/2
max (ρ0,ρ)
(1.1.28)
21
Integral representation for the reciprocal distance
The integral representations of this section can be generalized for a sphere (see Fabrikant, 1987d), and yet another modification in toroidal coordinates is also possible. The book volume limitation does not allow us to go into detail. Exercise 1.1 Prove the identity (η is defined by 1.1.6) ρ2ρ20 − x 4 dη = − . dx x 3η
1.
2.
Prove the identity ( R is defined by the first line of 1.1.27) λ(
dη x2 xη , φ−φ ) = − 2 . 2 0 ρρ R + η dx 0
Hint : use the identity: x 2 + ρ2ρ20/ x 2 − 2ρρ cos(φ−φ ) = R 2 + η2, and the result 0
0
above. 3.
Prove the identity ρρ λ(
4.
0
x
2
, φ−φ ) = 0
dη xη 2 dx . R + η 2
Prove the identities ( l 22 − ρ2)1/2( l 22 − a 2)1/2 = z l ,
( a 2 − l 21)1/2(ρ2 − l 21)1/2 = z l ,
2
1
( a 2 − l 21)1/2( l 22 − a 2)1/2 = za , ( l 22 − ρ2)1/2(ρ2 − l 21)1/2 = z ρ. Reminder : l and l are understood as l ( a ,ρ, z ) and l ( a ,ρ, z ) respectively. 1
2
1
2
Hint :
use (1.1.11) 5.
Prove that g ( x ) is inverse to both l
1
and l , namely, prove that g ( l )= a , and 2
1
g ( l )= a . 2
6.
Prove the identities ∂l zl 1 1 = − 2 , ∂z l 2 − l 21
∂l
2
∂z
zl =
l 22
2
− l 21
,
22
∂l1 ∂ρ
CHAPTER 1,
=
a l 2 − ρl 1
=
l 22 − l 21
ρ( a 2 − l 21)
,
l 1( l 22 − l 21)
∂l2
ρl 2 − a l 1
=
∂ρ
l 22 − l 21
Description of the new method
=
ρ( l 22 − a 2) l 2( l 22 − l 21)
.
Hint : use the properties above. 7.
Evaluate the integral [ l 22( x ) − x 2]1/2
l ( x) x
1 ⌠ λ , φ − ψ . ⌡ (ρ20 − x 2)1/2 l 22( x ) − l 21( x ) l 2( x )ρ0 dx
1 Answer: − tan-1 R0
(ρ20 − x 2)1/2[ l 22( x ) − x 2]1/2 xR 0
.
Hint : use (1.1.26) 8.
Evaluate the integral ρρ ( x 2 − l 21( x ))1/2 dx 0 ⌠ λ 2 , φ−φ . 2 2 1/2 2 2 0 ⌡ ( x − ρ0) [ l 2( x ) − l 1( x )] l 2( x )
Answer:
1 tan-1 R 0
( x 2 − l 21( x ))1/2( x 2 − ρ20)1/2 xR
.
0
Hint : use (1.1.20) 9.
Establish the integral representation
min (ρ0,ρ)
2⌠ π⌡ 0
cos[κ(ρ2 − x 2)1/2(ρ20 − x 2)1/2/ x + (πν/2)] [(ρ2 − x 2)(ρ20 − x 2)](1+ν)/2
x2 e-κR λ , φ−φ0 x νd x = 1+ν. ρρ0 R
Note : try to use this integral representation for solving the Klein-Gordon equation.
1.2 Properties of the L-operators Introduce the integral L-operator as follows:
1.2 Properties of the L-operators
∞
Σ
1 L( k ) f (φ) = 2π
23
2π
-in φ
in φ k |n|e ⌠ e
⌡
n=-∞
∞
0
f (φ0)dφ0 =
Σk
|n|
in φ
f ne .
(1.2.1)
n=-∞
0
Here f n is the n -th Fourier coefficient of the function f .
In the case when k <1,
formula (1.2.1) can be rewritten as 2π
1⌠ L( k ) f (φ) = λ( k , φ−φ0) f (φ0) dφ0, 2π⌡
(1.2.2)
0
where λ(⋅,⋅) is defined by (1.1.5). The L-operator may also be called Poisson operator, since it was introduced by Poisson for solving the two-dimensional Dirichlet problem for a circle. The following properties of the L-operators are valid L( k 1)L( k 2) = L( k1k2 ),
limL( k ) f = f .
(1.2.3)
k→1
The proof is elementary, and left to the reader. These properties are widely used in various transformations throughout this book, and are essential to the new method. Exercise 1.2 1. Prove the identity L( k )λ( m ,φ)=λ( km ,φ), for k <1 and m <1. Hint : use (1.2.3) 2.
Evaluate the integral
2π
dφ ⌠ , for ρ>ρ0 and r > r 0. 2 2 ⌡ [ρ + ρ0 − 2ρρ0cos(φ−φ0)][ r2 + r20 − 2 rr0cos(φ−ψ)] 0
Answer:
3.
2π(ρ2 r 2 − ρ20 r 20) (ρ2 − ρ20)( r 2 − r 20)[ρ2 r 2 + ρ20 r 20 − 2ρρ0 rr0cos(φ0−ψ)]
Evaluate the integral
.
24
CHAPTER 1,
Description of the new method
2π
dφ ⌠ , for k <1 and k 1<1. 2 ⌡ [1 + k − 2 k cos(φ−φ0)]2[1 + k 21 − 2 k 1cos(φ−ψ)] 0
2k2 Answer: 1 + k 2 k 21 − 2 kk 1cos(φ0−ψ) (1 − k 2)3 2π
+
4.
1
k 21
1 + k 2 k 21 − 2 kk 1cos(φ0−ψ) 1 − k 21
1 − k 4 k 21
.
+
(1 − k 2)2
Evaluate the integral
2π
iφ
e dφ ⌠ , for k <1 and k 1<1. 2 ⌡ [1 + k − 2 k cos(φ−φ0)]2[1 + k 21 − 2 k 1cos(φ−ψ)] 0
iψ
k 31e Answer: [1 + k 2 k 21 − 2 kk 1cos(φ0−ψ)]2 1 − k 21 2π
iφ
ke +
5.
0
[2(1 + k 4 k 21) − kk 1(1 + k 2)e
] + k ei ψ(1 − 3 k 2)
- i (ψ-φ0)
1
.
(1 − k 2)3
Evaluate the integral
2π
2 iφ
e dφ ⌠ , for k <1 and k <1. 2 2 1 ⌡ [1 + k − 2 k cos(φ−φ0)] [1 + k 21 − 2 k 1cos(φ−ψ)]
0
2 iψ
k 21e Answer: [1 + k 2 k 21 − 2 kk cos(φ −ψ)]2 1 − k 21 1 0 2π
+
2 iφ
[e2 i ψ( k 4−3 k 2+1) − k 2e
- i (ψ-φ0)
2 kk 1e
0
2 iφ
] − k 2e
(1 − k 2)3
0
( k 2 + k 21−3 − 3 k 2 k 21)
.
25
1.3 Some further integral representations
1.3 Some further integral representations An integral representation for z / R 30 is presented here, for R 0=[ρ2 + ρ20 − 2ρρ0cos(φ−φ0) + z 2]1/2. ρ
I
1
1 1 d L( ) = ⌠ ρ ρ dρ ⌡ 0 0 0
Consider the integral:
0
xdx (ρ20 − x 2)1/2
[ x 2 − l 21( x )]1/2
l ( x) x
1 λ , φ−φ . 2 2 0 l ( x ) l 2( x ) − l 1( x ) 2
(1.3.1)
0
The integral (1.3.1) looks cumbersome, and the first impression is that one should be lucky just to express it in elliptic integrals. It will be shown below that the integral is quasi-elliptic, and is computable in elementary functions. By using the rule of differentiation ρ
d ⌠ dρ 0
ρ
0
f( x) dx f( x) + ρ ⌠ 2 2 1/2 = lim 0 x (ρ0 − x ) x→0 ⌡
⌡ 0
0
(ρ20
dx d f ( x ) 2 1/2 d x x , − x)
0
(1.3.2) the integral (1.3.1) can be rewritten as ρ
I
1
= ⌠
⌡
0
dx (ρ20 − x 2)1/2
2 2 1/2 l ( x) x d [ x − l 1( x )] 1 λ , φ−φ0. d x l 22( x ) − l 21( x ) l ( x )ρ 2 0
(1.3.3)
0
Introduce a new variable:
j =
(ρ20 − x 2)1/2 [ l 22( x ) − x 2]1/2 x
,
[ l 22( x ) − x 2]1/2 [ρ20 l 22( x ) − x 2 l 21( x )] dj j´ = = − . dx x 2[ l 22( x ) − l 21( x )] (ρ20 − x 2)1/2 Let us transform the expression for λ: l ( x) x
l 2( x )
l 2( x ) − ρ2ρ2/ l 2( x )
1 1 1 0 1 λ , φ−φ = λ , φ−φ = − 2 2 2 2 0 0 l 2( x )ρ0 ρρ0 l 1( x ) + ρ ρ0/ l 1( x ) − 2ρρ0cos(φ−φ0)
(1.3.4)
26
=
CHAPTER 1,
Description of the new method
− l 21( x ) + ρ2ρ20/ l 21( x ) l 21( x ) + ρ2ρ20/ l 21( x ) − 2ρρ cos(φ−φ ) + ρ2+ρ20+ z 2 − l 21( x )− l 22( x ) + x 2−ρ20 0 0
= −
x 2 l 21( x ) − ρ20 l 22( x ) x
2
( R 20
2
+ j)
= −
[ l 22( x ) − l 21( x )] (ρ20 − x 2)1/2 [ l 22( x )
− x]
2 1/2
j´ R 20
+ j2
(1.3.5)
The substitution of (1.3.4) and (1.3.5) in (1.3.3) allows us to continue the transformations: ρ
I
1
0
2 2 1/2 2 2 1/2 d [ x − l 1( x )] (ρ0 − x ) j´ dx [ l 22( x ) − x 2]1/2 R 20 + j 2
dx
= − ⌠
(ρ20 − x 2)1/2
⌡
0
ρ
= − ⌠
0
dx (ρ20 − x 2)1/2
⌡
2 2 1/2 d xz(ρ0 − x ) j´ d x l 22( x ) − x 2 R 20 + j 2
0
ρ
0
= −z ⌠
dx (ρ20 − x 2)1/2
⌡
2 2 3/2 d (ρ0 − x ) j ´ d x xj 2( R 20 + j 2)
0
ρ
r 0 (ρ20 − x 2) j ´ dj( x) ⌠ = −z 2 2 − xj ( R 0 + j 2) j 2( R 20 + j 2) ⌡ 0 0
ρ
(ρ20 − x 2) j ´ 1 1 j π z -1 = −z 2 2 + + tan = 2 2 3 2 R 30 R0 R0j R0 xj ( R 0 + j ) 0
(1.3.6)
0
Finally, (1.3.6) allows us to write the required representation: ρ
2 1 d z 3 = πρ L(ρ ) dρ ⌠ R0 0 0 0 ⌡ 0
0
xdx (ρ20 − x 2)1/2
[ x 2 − l 21( x )]1/2
l ( x) x
1 λ , φ−φ . 2 2 0 l 2( x ) − l 1( x ) l 2( x )
(1.3.7)
27
1.3 Some further integral representations
The second integral to consider is a
I2
[ l 22( x ) − x 2]1/2 ρ xdx 1 d ⌠ λ 2 , φ−φ0. = − L(ρ0) 2 2 1/2 2 2 ρ0 dρ0 ⌡ ( x − ρ0) l 2( x ) − l 1( x ) l 2( x ) ρ
(1.3.8)
0
Make use of the rule of differentiation a
a
d ⌠ F (ρ) dρ F(a)x dρ d F (ρ) = − + x⌠ 2 2 2 1/2 2 2 1/2 2 1/2 d x ⌡ (ρ − x ) a(a − x ) ⌡ (ρ − x ) dρ ρ x
x
a
1⌠ d F(a)a ρdρ = − 2 2 1/2 + x 2 2 1/2 dρ F (ρ). x( a − x ) ⌡ (ρ − x )
(1.3.9)
x
Expression (1.3.8) will take the form ( l 22 − a 2)1/2
ρρ
0 , φ−φ I2 = 2 2 1/2 2 2 λ 0 ( a − ρ0) ( l 2 − l 1) l 22 a
ρρ [ l 2( x ) − x 2]1/2 dx 0 ⌠ d 2 − λ 2 2 2 1/2 2 2 ⌡ ( x − ρ0) d x l 2( x ) − l 1( x ) l 2( x ) , φ−φ0. ρ
(1.3.10)
0
Introduce a new variable
h =
( x 2 − ρ20)1/2[ x 2 − l 21( x )]1/2
dh h´ = = dx
x
,
[ x 2 − l 21( x )]1/2[ x 2 l 22( x ) − ρ20 l 21( x )] x 2( x 2 − ρ20)1/2[ l 22( x ) − l 21( x )]
.
(1.3.11)
The expression for λ can be presented in the manner, similar to (1.3.5), namely,
28
CHAPTER 1,
ρρ0
λ 2 , φ−φ0 = l 2( x )
Description of the new method
[ l 22( x ) − l 21( x )]( x 2 − ρ20)1/2
h´
[ x 2 − l 21( x )]1/2
R 20 + h 2
.
(1.3.12)
Substitution of (1.3.11) and (1.3.12) in (1.3.10) yields ρρ
( l 22 − a 2)1/2
0 , φ−φ I2 = λ 0 ( a 2 − ρ20)1/2( l 22 − l 21) l 22 a
xz( x 2 − ρ20)1/2 h ´ dx ⌠ d − ⌡ ( x 2 − ρ20)1/2 d x [ x 2 − l 21( x )]( R 20 + h 2) ρ
0
( l 22 − a 2)1/2
ρρ
0 , φ−φ = 2 2 1/2 2 2 λ 0 ( a − ρ0) ( l 2 − l 1) l 22 a
( x 2 − ρ20)3/2 h ´ dx ⌠ d . − z 2 2 1/2 2 2 2 ⌡ ( x − ρ0) d x xh ( R 0 + h ) ρ
(1.3.13)
0
Integration by parts in (1.3.13) yields a
[ l 22( x ) − x 2]1/2 xdx 1 d ⌠ ρ , φ−φ L(ρ0) − λ 0 ρ0 dρ0 ⌡ ( x 2 − ρ20)1/2 l 22( x ) − l 21( x ) l 22( x ) ρ
0
z R 0 h = 3 + tan-1 . R 0 R0 h
(1.3.14)
Here h stands for h ( a ), as defined by the first expression of (1.3.11). In the limiting case, when a →∞, expression (1.3.14) gives yet another representation for z / R 30, namely,
29
1.3 Some further integral representations
∞
z R 30
[ l 22( x ) − x 2]1/2 xdx 2 d ⌠ ρ , φ−φ . L(ρ0) λ = − 0 πρ0 dρ0 ⌡ ( x 2 − ρ20)1/2 l 22( x ) − l 21( x ) l 22( x ) ρ
0
(1.3.15) The last expression is useful in external problems, while its equivalent (1.3.6) is needed in solving internal ones. Since the integral in (1.3.1) was evaluated earlier as indefinite, we may consider its generalization ρ
I3
1 1 d L( ) = ⌠ ρ0 ρ0 dρ0
0
⌡
[ x 2 − l 21( x )]1/2
xdx
l ( x) x 1 λ , φ−φ0. 2 2 l x ( ) l 2( x ) − l 1( x ) 2
(ρ20 − x 2)1/2
(1.3.16)
a
Make use of the rule of differentiation x
x
F(a)x d ⌠ F (ρ) dρ ⌠ 2 dρ 2 1/2 d F (ρ) = 2 2 1/2 + x d x ⌡ ( x 2 − ρ2)1/2 a(x − a ) ⌡ ( x − ρ ) dρ ρ a
a
x
1⌠ d F(a)a ρdρ = 2 2 1/2 + x 2 2 1/2 dρ F (ρ). x( x − a ) ⌡ (x − ρ )
(1.3.17)
a
Expression (1.3.16) will take the form
I3 = ρ
+ ⌠
⌡
( a 2 − l 21)1/2 (ρ20 − a 2)1/2[ l 22 − l 21] 0
dx (ρ20 − x 2)1/2
l 21
, φ−φ0 λ ρρ0
2 2 1/2 l ( x) x d [ x − l 1( x )] 1 λ , φ−φ0. d x l 22( x ) − l 21( x ) l 2( x )ρ0
(1.3.18)
a
By introducing the notation z R0 j(y) F ( y) = 3 + tan-1 , R 0 R 0 j(y)
(1.3.19)
30
CHAPTER 1,
Description of the new method
where j ( y ) is defined according to (1.3.4), expression (1.3.18) can be rewritten as
I
3
=
ρ20 − a 2 a
ρ
(ρ20 − y 2)3/2
0
dy d dF(a) + ⌠ 2 2 1/2 d y da ⌡ (ρ0 − y )
y
d F ( y ) . dy
(1.3.20)
a
Integration in (1.3.20) can be performed by parts, with a simple result F ( a ), which means establishment of another integral representation z R 0 j -1 + tan R 0 R 30 j ρ
1 1 d L( ) = ⌠ ρ0 ρ0 dρ0
0
⌡
[ x 2 − l 21( x )]1/2
xdx (ρ20 − x 2)1/2
l ( x) x
1 λ , φ−φ0. 2 2 l x ( ) l 2( x ) − l 1( x ) 2
(1.3.21)
a
1.
Exercise 1.3 Prove that h in (1.3.14) can be defined by any of the expressions z ( a 2 − ρ20)1/2 ( a 2 − ρ20)1/2( a 2 − l 21)1/2 h ≡ h(a) = = a ( l 22 − a 2)1/2
= 2.
l2
=
( a 2 − ρ20)1/2( l 22 − ρ2)1/2 l2
.
Prove that j in (1.3.4) can be defined in several equivalent ways: (ρ20 − a 2)1/2( l 22 − a 2)1/2 (ρ20 − a 2)1/2(ρ2 − l 21)1/2 j ≡ j(a) = = a l1
= 3.
[ l 22 − l 21(ρ0)]1/2[ l 22 − l 22(ρ0)]1/2
z (ρ20 − a 2)1/2 ( a 2 − l 21)1/2
Establish (1.3.15)
=
[ l 21(ρ0) − l 21]1/2[ l 22(ρ0) − l 21]1/2 l1
.
31
1.4 Internal mixed boundary value problem for a half-space
1.4 Internal mixed boundary value problem for a half-space The material in this section follows the paper (Fabrikant, 1986h). Introduce a set of cylindrical coordinates (ρ,φ, z ). Consider the problem of finding a potential function V , harmonic in the half-space z ≥0, vanishing at infinity, and subject to the boundary conditions on the plane z =0 V = v (ρ,φ),
for ρ≤ a ,
∂V = 0, ∂z
0≤φ<2π;
for ρ> a ,
0≤φ<2π.
(1.4.1)
The problem (1.4.1) can be interpreted as an electrostatic one of a charged disc, with a certain potential prescribed on its surface, or it can be interpreted as an elastic contact problem of a circular punch pressed against an elastic half-space; other interpretations are also possible. We call the problem internal because the non-zero conditions are prescribed inside the disc. The potential function V can be represented through the simple layer as follows: 2π a σ(ρ ,φ ) 0 0
V (ρ,φ, z ) = ⌠ ⌠ ⌡⌡ 0
R
0
0
2π ∞
ρ dρ dφ + ⌠ ⌠ 0 0 0 ⌡⌡ 0
σ(ρ ,φ ) 0
R
a
0
ρ dρ dφ . 0
0
Here R
0
= [ρ2 + ρ20 − 2ρρ cos(φ−φ ) + z 2]1/2, 0
and σ = −
0
0
(1.4.2)
0
1 ∂V . 2π ∂ z z=0
Substitution of (1.1.18) and (1.1.22) in (1.4.2) yields, after changing the order of integration l1
dx
V (ρ,φ, z ) = 4⌠ 2 2 1/2 ⌡ (ρ − x ) 0
∞
g(x)
ρ dρ
a
0 0 x σ(ρ ,φ) ⌠ 2 2 1/2 L ρρ 0 0 ⌡ [ρ0 − g ( x )] 2
g(x)
ρ dρ
ρρ
0 0 0 σ(ρ ,φ). ⌠ + 4⌠ 2 2 1/2 2 2 1/2 L 0 ⌡ ( x − ρ ) ⌡ [ g ( x ) − ρ0] x2 l
dx
(1.4.3)
a
2
Here the L-operator is defined by (1.2.1), g is given by (1.1.25), the abbreviations l and l are understood as l ( a ,ρ, z ) and l ( a ,ρ, z ) respectively; and 1
2
1
2
the following rule is used for changing the order of integration:
32
CHAPTER 1,
l1(ρ 0)
a
l1
⌠ dρ ⌠ d x = ⌠ d x ⌡ 0 ⌡ ⌡ 0 0
⌠ dρ , ⌡ 0
g(x)
0
∞
∞
a
Description of the new method
⌠ dρ ⌡ 0 a
∞
g(x)
⌠ d x = ⌠ d x ⌠ dρ . ⌡ ⌡ ⌡ 0
l (ρ0) 2
l
2
a
(1.4.4) The rule is illustrated by Fig. 1.4.1, where the domains of integration are shaded.
Fig.
1.4.1 Domains of integration.
Substitution of the boundary governing integral equation ρ
a
ρ dρ
condition
(1.4.1)
in
(1.4.3)
0 0 x σ(ρ ,φ) = v (ρ,φ). ⌠ 2 4⌠ 2 2 1/2 2 1/2 L ρρ 0 0 ⌡ (ρ − x ) ⌡ (ρ0 − x ) 0
dx
leads to
the
2
(1.4.5)
x
Expression (1.4.5) is now presented as a sequence of two Abel-type operators and one L-operator. We recall that the general Abel integral equation a
y ⌠ 2 F ( y ) 2d(1+u)/2 = f( x) ⌡ (y − x ) x
has the solution
(1.4.6)
33
1.4 Internal mixed boundary value problem for a half-space
a
2cos(π u /2) d ⌠ f( x) xdx . F ( r) = − d r ⌡ ( x 2 − r 2)(1-u)/2 π
(1.4.7)
r
Since the variables in the Abel operators of (1.4.5) are interwoven with those of the L-operator, we need to apply their combination, in order to invert (1.4.5). The first operator to be applied to both sides of (1.4.5) is t
ζ d ⌠ ρdρ ρ L L . 2 2 1/2 t dt ⌡ (t − ρ ) ζ
(1.4.8)
0
Here we introduced a dummy parameter ζ in order to make the parameter of the L-operator dimensionless, and also in order to claim it being less that unity almost everywhere in the interval of integration which is a precondition for usage of the properties (1.2.3). We call the parameter ζ ’dummy’ because it was introduced for formal reasons only; it will disappear in the final result, and has no real bearing on the transformations to follow. Of course, the introduction of dummy parameter does not rigorously validate our use of the properties (1.2.3). Such a validation is beyond scope of this book: the author is satisfied by the fact that the final result everywhere is proven to be correct. In order to make the intermediate transformations rigorous, one has to prove the theorem stating that one can use the properties (1.2.3) in the mathematical manipulations with an expression of the type L( k )M L( k )M L( k )..., where M are certain linear 1
1
operators, if the product k1k2k3 ...
2
2
3
is less than unity, thus allowing for any
particular k to be greater than unity. willing and able to prove the theorem.
We hope that some readers might be
The result of application of (1.4.8) to both sides of (1.4.5) is a
2π⌠
ρ0dρ0
2 2 1/2 ⌡ (ρ0 − t ) t
t
t ζ d ⌠ ρdρ ρ v (ρ,φ). L σ(ρ0,φ) = L 2 2 1/2 L ζ ρ t d t 0 ⌡ (t − ρ ) 0
The second operator to be applied to both sides of (1.4.9) is a
y d⌠ tdt ζ L L , 2 2 1/2 ζ d y⌡ ( t − y ) t y
(1.4.9)
34
CHAPTER 1,
Description of the new method
with the result a
t
ζ2 d ⌠ ρdρ 1 y d⌠ tdt ρ σ( y ,φ) = − 2 L L 2 2 2 1/2 2 2 1/2 L ζ v (ρ,φ). ζ d y d t t πy ⌡ (t − y ) ⌡ (t − ρ ) y
0
(1.4.10) The rules of differentiation of integrands and the properties of the L-operators allow us to rewrite (1.4.10) a
1 Φ( a , y ,φ) ⌠ 2 d t 2 1/2 d Φ( t , y ,φ). σ( y ,φ) = 2 2 2 1/2 − π (a − y ) ⌡ (t − y ) dt
(1.4.11)
y
Here t
1 ⌠ ρdρ d ρ y Φ( t , y ,φ) = ρL v (ρ,φ). 2 2 1/2 t ⌡ ( t − ρ ) dρ t 2
(1.4.12)
0
Note that the dummy parameter ζ has disappeared from the final solution, and the combined parameter of the L-operator is less than unity, as it should be. In the future we shall no longer use the dummy parameter explicitly, assuming that the use of the properties (1.2.3) is justified. Using integration by parts and the fact that λ( k ,ψ) satisfies the two-dimensional Laplace equation in polar coordinates, the following identity can be established t
d ρdρ ρy ∆ v (ρ,φ), Φ( t , y ,φ) = ⌠ 2 2 1/2 L 2 dt t ⌡ (t − ρ )
(1.4.13)
0
where ∆ is the two-dimensional Laplace operator in polar coordinates. Substitution of (1.4.13) in (1.4.12) leads to another form of solution, namely, a
t
1 Φ( a , y ,φ) ⌠ 2 d t 2 1/2 ⌠ 2 ρdρ 2 1/2 Lρ2y ∆ v (ρ,φ), σ( y ,φ) = 2 2 2 1/2 − t π (a − y ) ⌡ (t − y ) ⌡ (t − ρ ) y
0
(1.4.14) Interchange of the order of integration in (1.4.14) and integration with respect to t (see (1.1.23)) yields
35
1.4 Internal mixed boundary value problem for a half-space
σ( y ,φ) =
1 Φ( a , y ,φ) π2 ( a 2 − y 2)1/2
2π a
∆ v (ρ,ψ) ρdρdψ ( a 2 − ρ2)1/2( a 2 − y 2)1/2 1⌠ ⌠ -1 − tan . 2π⌡ ⌡ a [ρ2 + y 2 − 2ρy cos(φ−ψ)]1/2 [ρ2 + y 2 − 2ρy cos(φ−ψ)]1/2 0
0
(1.4.15) The solution obtained here consists of two parts: the first part is singular at the boundary while the second one vanishes at the boundary. In various applications it is required that the solution be nonsingular at the boundary. The necessary and sufficient condition then is Φ( a , a ,φ)=0. In elastic contact problems this condition defines the radius of the contact domain. Notice also that in the case when v is a two-dimensional harmonic function, the non-trivial solution is singular. Now it is of interest to express the potential V in the half-space directly through its value v prescribed inside the disc ρ= a . Substitution of (1.4.10) in (1.4.3) yields, after subsequent integration g(x)
l1
rdr 2 dx x2 d ⌠ V (ρ,φ, z ) = ⌠ 2 L L( r ) v ( r ,φ). 2 1/2 2 2 π ⌡ (ρ − x ) ρg ( x ) d g ( x )⌡ [ g ( x ) − r2]1/2 0
0
(1.4.16) Here the following property of the Abel operators was used a
a
⌠ 2 d r 2 1/2 d ⌠ 2tf ( t ) d2 t1/2 = − π f ( y ). 2 ⌡ (r − y ) dr⌡ (t − r ) y
(1.4.17)
r
Introduction of a new variable t = g ( x ), x = l 1( t ), transforms (1.4.16) into a t d l 1( t ) ρ0dρ0 l 21( t ) 2⌠ d ⌠ V (ρ,φ, z ) = L L(ρ0) v (ρ0,φ). π⌡ [ρ2 − l 21( t )]1/2 ρ t 2 d t ⌡ ( t 2 − ρ20)1/2 0
0
By changing the order of integration in (1.4.18), according to the rule
(1.4.18)
36
CHAPTER 1,
a
r
Description of the new method
a
a
⌠ F ( r)d r d ⌠ ρ2 f (ρ) d2 ρ1/2 = −⌠ f (ρ)dρ d ⌠ 2F ( r) rd2r 1/2, dr ⌡ (r − ρ ) dρ ⌡ ( r − ρ ) ⌡ ⌡ 0
0
(1.4.19)
ρ
0
the following expression can be obtained a
t d l 1( t ) ρ 2⌠ d ⌠ ρ V (ρ,φ, z ) = − v (ρ ,φ)dρ0. L ( ) L π⌡ 0 dρ ⌡ ( t 2 − ρ20)1/2[ρ2 − l 21( t )]1/2 l 22( t ) 0 0ρ 0 a
0
(1.4.20) The integral in curly brackets can be evaluated using (1.3.14), with the result 2π a
R0 h z ⌠ ⌠ V (ρ,φ, z ) = 2 + tan-1 3 v (ρ0,φ0) ρ0dρ0dφ0. R 0 R 0 π ⌡ ⌡ h 0 1
(1.4.21)
0
Here R = [ρ2 + ρ20 − 2ρρ cos(φ−φ ) + z 2]1/2, 0 0 0
h = ( a 2 − l 21)1/2( a 2 − ρ20)1/2/ a .
(1.4.22) V Formulae (1.4.18) and (1.4.21) define the potential function in the half-space z ≥0, expressed directly through its value v prescribed inside the disc ρ= a , z =0. Expression (1.4.18) is useful when an explicit evaluation of the integrals is possible, while expression (1.4.21) is more convenient for numerical integration. Note that in the limiting case, when z =0, equation (1.4.21) transforms into a known result, namely, V (ρ,φ,0) = v (ρ,φ),
V (ρ,φ,0) =
(ρ2 − a 2)1/2 π2
2π a
for ρ≤ a ; and v (ρ ,φ )ρ dρ dφ
0 0 0 0 0 ⌠⌠ , for ρ> a . 2 2 1/2 2 2 ⌡ ⌡ ( a −ρ0) [ρ + ρ0 − 2ρρ0cos(φ−φ0)] 0 0
The solution of the first mixed boundary value problem is completed. results are given by formulae (1.4.10) and (1.4.18).
The main
Consider now another internal problem, characterized by the following mixed conditions on the boundary z =0:
37
1.4 Internal mixed boundary value problem for a half-space
∂V = − 2πσ(ρ,φ), ∂z
for ρ≤ a , and 0≤φ<2π; for ρ> a , and 0≤φ<2π.
V = 0,
(1.4.23)
The problem (1.4.23) can be interpreted as an electrostatic one of a charged disc ρ≤ a inside an infinite grounded diaphragm ρ> a . Mathematically similar problem arises in the consideration of a penny-shaped crack subjected to an arbitrary pressure σ. Substitution of (1.4.23) in (1.4.3) leads to the integral equation, for ρ> a , a ρ dρ 0 0 d x x 2 σ(ρ ,φ) ⌠ 2 ⌠ L 2 1/2 2 2 1/2 0 ρρ0 ⌡ (ρ − x ) ⌡ (ρ0 − x ) a
0
x
∞
ρ dρ
x
ρρ
dx 0 0 0 σ(ρ ,φ) = 0. ⌠ 2 + ⌠ 2 2 1/2 2 1/2 L 0 x2 ⌡ ( x − ρ ) ⌡ ( x − ρ0) ρ
(1.4.24)
a
Notice that σ in the first term of (1.4.24) is known from (1.4.23), while σ in the second term is yet to be determined. By using the integral representations (1.1.27) and (1.1.28), equation (1.4.24) can be rewritten as ∞
ρ dρ
x
ρρ
0 0 0 σ(ρ ,φ) ⌠ ⌠ 2 2 2 1/2 2 1/2 L 0 x2 ⌡ ( x − ρ ) ⌡ ( x − ρ0)
dx
ρ
a
∞
a
ρ dρ
ρρ
0 0 0 σ(ρ ,φ). ⌠ = −⌠ 2 L 0 ⌡ ( x − ρ2)1/2 ⌡ ( x 2 − ρ20)1/2 x 2
dx
ρ
0
Operation on both sides of (1.4.25) by ∞
ρdρ d ⌠ 1 L( t ) L d t ⌡ (ρ2 − t 2)1/2 ρ t
leads to
(1.4.25)
38
CHAPTER 1,
t
ρ0dρ0
⌠ 2 2 1/2 ⌡ ( t − ρ0)
a
ρ
a
ρ dρ
Description of the new method
ρ
0 0 0 0 σ(ρ ,φ). L σ(ρ0,φ) = −⌠ 2 L 2 1/2 0 t t ⌡ ( t − ρ0)
(1.4.26)
0
The next operator to apply is ρ
tdt 1 d ⌠ L L( t ), 2 ρ dρ ⌡ (ρ − t 2)1/2 a
and the final result takes the form 2 ⌠ σ(ρ,φ) = − 2 π(ρ − a 2)1/2 ⌡
a(a2
ρ2 − ρ20
0
= −
2π a 2 (a
1 π (ρ − a ) 2
2
2 1/2
− ρ20)1/2ρ0dρ0
ρ
0 L σ(ρ ,φ) 0 ρ
− ρ20)1/2σ(ρ0,φ0) ρ0dρ0dφ0
⌠⌠ ⌡ ⌡ ρ2 + ρ20 − 2ρρ0cos(φ−φ0)
0
.
(1.4.27)
0
Formula (1.4.27) defines the value of σ outside the circle ρ= a directly through its value inside. Now σ is known all over the plane z =0, and substitution of (1.4.27) in the second term of (1.4.3) allows us to express the potential function V directly through the prescribed value of σ. The first integration yields l1
dx
V (ρ,φ, z ) = 4⌠
2 2 1/2 ⌡ (ρ − x ) 0
∞
a
ρ0dρ0
x σ(ρ ,φ) ⌠ 2 2 1/2 L ρρ 0 0 ⌡ [ρ0 − g ( x )] 2
g(x)
ρ0dρ0 ρρ dx 0 σ(ρ ,φ). ⌠ L − 4⌠ 2 0 ⌡ ( x − ρ2)1/2 ⌡ [ g 2( x ) − ρ20]1/2 x 2 l 2
a
(1.4.28)
0
Here the following integral was employed ρ
ydy π ⌠ 2 , 2 1/2 2 2 1/2 2 2 = 2 2 1/2 2 2(ρ − r ) ( a − r 2)1/2 ⌡ (ρ − y ) ( y − a ) ( y − r ) a
for r < a .
39
1.4 Internal mixed boundary value problem for a half-space
(1.4.29) The first term in (1.4.29) can be transformed by using (1.1.22), in the following manner: l1
ρ dρ
a
0 0 x σ(ρ ,φ) ⌠ 2 2 1/2 L ρρ 0 0 ⌡ [ρ0 − g ( x )]
dx
⌠ 2 2 1/2 ⌡ (ρ − x )
2
g(x)
0
∞
a
= ⌠ ρ0dρ0 ⌡ 0
l2
l
2
ρρ
dx ⌠ 0 σ(ρ ,φ) 2 2 1/2 2 2 1/2 L 0 ⌡ ( x − ρ ) [ g ( x ) − ρ0] x2 (ρ ) 0
g(x)
ρ dρ
ρρ
0 0 0 σ(ρ φ) ⌠ ⌠ 2 2 1/2 2 2 1/2 L 0 x2 ⌡ ( x − ρ ) ⌡ [ g ( x ) − ρ0]
=
dx
0
l (0) 2 ∞
a
ρ dρ
ρρ
0 0 0 σ(ρ φ). ⌠ 2 + ⌠ 2 2 1/2 2 1/2 L 0 ⌡ ( x − ρ ) ⌡ [ g ( x ) − ρ0] x2 l
dx
(1.4.30)
0
2
Substitution of (1.4.30) in (1.4.29) yields l2
g(x)
ρ dρ
ρρ
0 0 0 σ(ρ φ). ⌠ V (ρ,φ, z ) = 4 ⌠ 2 2 1/2 2 2 1/2 L 0 x2 ⌡ ( x − ρ ) ⌡ [ g ( x ) − ρ0]
dx
(1.4.31)
0
l (0) 2
Introduction of a new variable t = g ( x ), x = l 2( t ), transforms (1.4.31) into a
t
d l 2( t )
ρ dρ
ρρ
0 0 0 σ(ρ ,φ). ⌠ 2 V (ρ,φ, z ) = 4⌠ 2 2 1/2 2 1/2L 2 ⌡ [ l 2( t ) − ρ ] ⌡ ( t − ρ0) l 2( t ) 0 0
(1.4.32)
0
An interchange of the order of integration in (1.4.32), and integration with respect to t (see 1.1.23), yields
40
CHAPTER 1,
Description of the new method
2π a
V (ρ,φ, z ) =
h 2⌠ ⌠ 1 tan-1 σ(ρ0,φ0) ρ0dρ0dφ0. R π⌡ ⌡ R 0 0 0
(1.4.33)
0
Formulae (1.4.31−33) give three equivalent representations of the potential function V , the first two being more convenient for explicit evaluation of the integrals involved, while the third one has some advantages for numerical integration. Two examples are considered below. Example 1. Let the potential prescribed inside the disc be v (ρ,φ)= v nρncosn φ, v n=const.
The solution due to (1.4.16) is l
2vn
1 x 2nd x Γ( n + 1) V (ρ,φ, z ) = cosn φ ⌠ 2 2 1/2 √π ρn Γ( n + 1) ⌡ (ρ − x ) 0 2
2 2 1/2 l2 − a2 2Γ( n + 1) ( l 2 − a ) 1 1 3 2 . = v nρ cosn φ 1 − F −n, ; ; 2 1 l 2 2 2 l 2 2 √πΓ( n + ) n
(1.4.34)
2
The hypergeometric function in (1.4.34) can be expressed in elementary functions (Bateman and Erdelyi, 1955) F(
1 1 3 (1 − ζ)n+1/2 dn ζn-1/2 sin-1√ζ. − n , ; ; ζ) = 2 2 2 Γ( n + 1) dζn √1 − ζ
Example 2. Let the charge distribution be prescribed σ(ρ,φ)=σnρncosn φ, σn=const. The solution is given by (1.4.32)
(1.4.35) in
the
form
b
x 2n+2d x Γ( n + 1) V (ρ,φ, z ) = 2√π σnρncosn φ ⌠ 2 2 n+1 3 ⌡ (x + z ) Γ( n + ) 0 2 = √π
Γ( n + 1) b 2n+3 3 5 b2 σnρncosn φ 2n+2 F ( n + 1, n + ; n + ; − 2 ), 5 2 2 z z Γ( n + ) 2
(1.4.36)
where b =( a 2 − l 21)1/2, and the hypergeometric function can be expressed in elementary (Bateman and Erdelyi, 1955)
41
1.4 Internal mixed boundary value problem for a half-space
2 n + 3 dn 1 1 1 + √ζ 5 3 . F ( n +1, n + ; n + ; ζ) = − ln 1 Γ( n + 1) dζn ζ 2√ζ 1 − √ζ 2 2
(1.4.37) Bibliographical note. The degree of effectiveness of the new approach can be established by comparison with the results reported in the literature. Formulae, similar to (1.4.21), can be found in the works of Lord Kelvin. Heine, Hobson and others. As was mentioned before, their practical use was quite limited. The equivalent expressions, like (1.4.18), allow us to evaluate the integrals involved in a straightforward and elementary manner. This was demonstrated by two examples above. The general solution has now become a working tool, available to anyone with even an undergraduate background in mathematics. Copson (1947) was very close to the discovery of the new method. established the identity
He
2π
cosn (φ − ψ) dφ ⌠ ⌡ [ρ2 + ρ20 − 2ρρ cos(φ−φ )]1/2 0 0 0
=
4cosn (φ − ψ) 0 (ρρ0)n
min (ρ0,ρ)
⌠ ⌡ 0
x 2nd x . (ρ2 − x 2)1/2(ρ20 − x 2)1/2
(1.4.38)
Copson has obtained a solution, similar to (1.4.10), in the form of a Fourier expansion. It remains unclear why he did not perform the summation in (1.4.38) leading to (1.1.27) which, in a way, is the foundation of the new method. Galin (1953) obtained the nonsingular part (corresponding to the second term in (1.4.15) of the general solution of equation (1.4.5). Leonov (1953) obtained the closed form complete solution (0.3). Both solutions had the same limitation as the previous classical solutions: difficulty in practical use. Another type of solution of equation (1.4.5) can be expressed through the z -derivative of (1.4.21) for z =0, due to the relationship σ = − The result is
1 ∂V . 2π ∂ z z=0
(1.4.39)
42
CHAPTER 1,
Description of the new method
2π a
η 1 ⌠⌠ 1 σ(ρ,φ) = tan-1( ) ω(ρ0,φ0) ρ0dρ0dφ0 3 −∆ R R 2π 0⌡ ⌡ 0
−
ω(ρ ,φ ) 1 − tt 0 0 ⌠⌠ ρ ρ φ d d ⌡ ⌡ (1 − t )(1 − t ) ( a 2 − ρ20)1/2 0 0 0 2π a
a ( a 2 − ρ2)3/2
0
0
x ρ0dρ0 ρρ0 dx 1 ⌠ ω(ρ ,φ) ⌠ 2 = 2 −∆ 2 2 1/2 2 1/2 L 2 0 x π ( x − ρ ) ⌡ ( x − ρ0) ⌡ 0 ρ a
−
ρ0dρ0 ρρ0 ω(ρ ,φ ), ⌠ 2 L ( 2 1/2 0 0 a2 ⌡ ( a − ρ0) a
a ( a 2 − ρ2)3/2
(1.4.40)
0
where t =
ρρ0 a
2
i(φ-φ0)
e
,
and the overbar everywhere (1.4.40) corresponds to the developed apparatus can be coaxial (Fabrikant, 1987e) and
1.
(1.4.41) indicates the complex conjugate value. Formula solution by Mossakovskii et al. (1985). The used for solving the problem of several charged arbitrarily located (Fabrikant, 1988a) circular discs.
Exercise 1.4 A circular conducting disc is kept at the potential v 0.
Find the potential
function V . Answer: V (ρ, z ) =
2 2 a v sin-1( l 1/ρ) = v sin-1( ). π 0 π 0 l2
Hint : use formula (1.4.18) 2. Subject to the conditions of the previous problem, find the charge distribution σ by using formulae (1.4.10) and (1.4.39). Prove that in both cases the result is the same. v0 Answer: σ = 2 2 . π ( a − ρ2)1/2
1.5 External mixed boundary value problem for a half-space
3.
Solve problems 1 and 2 for v = v 1ρcosφ, v 1= const.
Answer: V (ρ,φ, z ) =
σ (ρ,φ) = 4.
2 a a √1 − ( a / l 2)2, v 1ρcosφsin-1( ) − π l2 l2 2 v 1ρcosφ π2( a 2 − ρ2)1/2
.
A uniform charge density σ=σ0=const is prescribed over a circular disc of
radius a .
Find the potential function.
a Answer: V (ρ, z ) = 4σ0( a 2 − l 21)1/2 − z sin-1( ). l 2
5.
Solve the previous problem for the case where σ=σ1ρcosφ, σ1=const.
Answer: V (ρ,φ, z ) = 6.
a a2 8 3 3 z sin-1( ). σ ρcosφ ( a 2 − l 21)1/2 − 2 − l 3 1 2 2 2l2 2
The potential function is given by the expression
a a2 8 3 2 2 1/23 z sin-1( ). V (ρ,φ, z ) = σ ρcosφ ( a − l 1) − 2 − l 3 1 2 2 2 l 2 2 Find the charge distribution on the plane z =0. Answer: σ = σ ρcosφ, for ρ≤ a ; σ=0, for ρ> a . 1 Hint : use (1.4.39). 7. Solve the mixed boundary value problem of potential theory for a sphere. Hint : see (Fabrikant, 1987i).
1.5 External mixed boundary value problem for a half-space The material in this section follows the paper (Fabrikant, 1986f). The problem is called external when non-zero boundary conditions are prescribed outside the disc. As in the previous section, we consider two types of problem. Problem 1. It is necessary to find a function, harmonic in the half-space z ≥0, vanishing at infinity, and subject to the mixed boundary conditions on the plane z =0, namely,
43
44
CHAPTER 1,
∂V = 0, ∂ z z=0
for ρ< a , 0≤φ<2π;
V = v (ρ,φ),
for ρ≥ a , 0≤φ<2π.
Description of the new method
(1.5.1)
The problem (1.5.1) can be interpreted as an electrostatic one of a charged diaphragm, or as an external elastic contact problem. The potential V is presented through a simple layer distribution (1.4.3). Substitution of the boundary conditions (1.5.1) in (1.4.3) leads to the governing integral equation ∞
ρ dρ
x
ρρ
0 0 0 σ(ρ ,φ) = v (ρ,φ). ⌠ 4⌠ 2 L 0 ⌡ ( x − ρ2)1/2 ⌡ ( x 2 − ρ20)1/2 x 2
dx
ρ
(1.5.2)
a
Its solution is obtained in exactly the same manner as that of (1.4.5), and is ρ ∞ ρ dρ 0 0 d ⌠ 1 1 d ⌠ xdx 2 1 σ(ρ,φ) = − 2 L L( x ) 2 2 1/2 2 2 1/2 L ρ v (ρ0,φ). ρ ρ d d x 0 πρ ⌡ (ρ − x ) ⌡ (ρ0 − x ) a
x
(1.5.3) The rules of differentiation (1.3.2) and (1.3.9) allow us to rewrite (1.5.3) as follows ρ
dx ∂ 1 χ( a ,ρ,φ) ⌠ σ(ρ,φ) = − 2 + χ ( x , ρ , φ ) , π (ρ2 − a 2)1/2 (ρ2 − x 2)1/2 ∂ x ⌡
(1.5.4)
a
where dρ
∞
χ( x ,ρ,φ) = x⌠
0
2 2 1/2 ⌡ (ρ0 − x ) x
∂ x2 L v (ρ0φ). ∂ρ ρρ 0 0
The following transformation can now be performed: ∞ dρ 0 ∂ ∂ ⌠ χ( x ,ρ,φ) = x (L v )´ 2 ∂x ∂ x ⌡ (ρ0 − x 2)1/2 x
(1.5.5)
45
1.5 External mixed boundary value problem for a half-space
∞
dρ0
= ⌠ 2 2 1/2 [(L v )´ + ρ0(L v )´´ − 2(L´ρ0 v )´] ⌡ (ρ0 − x ) x
∞
ρ dρ
0 0 Lv ´´ + 1 v ´ − L´´ + 1 L´v . = ⌠ 2 2 1/2 ρ0 ρ0 ⌡ (ρ0 − x )
(1.5.6)
x
Here, for the sake of brevity, the primes (´) indicate the partial derivatives with respect to ρ0, L stands for L( x 2/ρρ0), v ≡ v (ρ0,φ), and the following identity was used ρ0 ∂ x2 ∂ x2 L = − 2 L . ∂ x ρρ0 x ∂ρ0 ρρ0 Since L
1 ∂2 v 1 ∂2L = v, ρ20 ∂φ2 ρ20 ∂φ2
its addition to and subtraction from (1.5.6) yields ∞ ρ0dρ0 ∂ L∆ v − (∆L) v , ⌠ χ( x ,ρ,φ) = 2 2 1/2 ∂x ( ρ − x ) ⌡ 0
(1.5.7)
x
where ∆ is the two-dimensional Laplace operator in polar coordinates. is a harmonic function, ∆L=0, and (1.5.7) simplifies to ∞ ρ0dρ0 ∂ ⌠ χ( x ,ρ,φ) = 2 2 1/2 L∆ v . ∂x ⌡ (ρ0 − x ) x
Substitution of (1.5.8) in (1.5.4) yields
Since λ
(1.5.8)
46
CHAPTER 1,
Description of the new method
ρ ∞ ρ0dρ0 x 1 χ( a ,ρ,φ) d x 2 ∆ v (ρ ,φ). ⌠ ⌠ σ(ρ,φ) = − 2 2 L + 2 2 1/2 2 2 1/2 0 π (ρ − a 2)1/2 ⌡ (ρ − x ) ⌡ (ρ0 − x ) ρρ0 a
x
It should be noticed that the first term in while the second term vanishes at the edge a harmonic function, the second term in represented by the first term only. Further possible in (1.5.9), after changing the order result is σ(ρ,φ) = −
(1.5.9) (1.5.9) becomes singular when ρ→ a , of the disc. In the case of v being (1.5.9) vanishes, and the solution is integration with respect to x becomes of integration and using (1.1.8). The
1 χ( a ,ρ,φ) π2 (ρ2 − a 2)1/2
2π ∞ ∆ v (ρ0,φ0) ρ0dρ0dφ0 (ρ2 − a 2)1/2(ρ20 − a 2)1/2 1⌠⌠ -1 + tan . 2π⌡ ⌡ [ρ2 + ρ20 − 2ρρ cos(φ−φ )]1/2 a [ρ2 + ρ20 − 2ρρ cos(φ−φ )]1/2 0
0
a
0
0
0
(1.5.10) Solutions, like (1.5.3) and (1.5.9), are appropriate for use when an exact evaluation of the integrals is possible, while the solution in the form (1.5.10) has some advantages when numerical integration is to be employed. Now we can express the potential function V directly through its boundary value v . Since σ=0 inside the circle ρ= a , the potential function (1.4.3) takes the form ∞
g(x)
ρ dρ
ρρ
0 0 0 σ(ρ ,φ). ⌠ V (ρ,φ, z ) = 4⌠ 2 2 1/2 2 2 1/2 L 0 ⌡ ( x − ρ ) ⌡ [ g ( x ) − ρ0] x2 l
dx
2
a
(1.5.11) Substitution of (1.5.3) in (1.5.11) yields, after the first integration, ∞ ∞ ρ dρ 2 0 0 2⌠ dx ρ g ( x ) ∂ 1 ⌠ V (ρ,φ, z ) = − L L v (ρ0,φ). 2 2 1/2 2 2 2 1/2 π⌡ ( x − ρ ) x ∂ g ( x ) ⌡ [ρ0 − g ( x )] ρ 0 l
2
g(x)
(1.5.12) Here the properties of the L-operators (1.2.3) were used, along with the following identity, valid for the Abel-type operators
47
1.5 External mixed boundary value problem for a half-space
ρ
x
⌠ 2 d x 2 1/2 d ⌠ f2( t ) t d2 t1/2 = π f (ρ). 2 ⌡ (ρ − x ) d x ⌡ ( x − t ) a
(1.5.13)
a
Introduction of a new variable y = g ( x ), x = l ( y ), in (1.5.12), allows us to rewrite 2
(1.5.12) ∞ ∞ dl ( y) ρ dρ l 21( y ) 2 0 0 2⌠ d ⌠ 1 V (ρ,φ, z ) = − L L v (ρ ,φ). π⌡ [ l 22( y ) − ρ2]1/2 ρ d y ⌡ (ρ20 − y 2)1/2 ρ 0 0 a
y
(1.5.14) Interchange of the order of integration in (1.5.14) yields ∞
V (ρ,φ, z ) = −
2⌠ π⌡ a
ρ
ydl ( y) 0 l 21( y ) 1 d 2 v (ρ ,φ)dρ . Lρ dρ ⌠ 2 2 1/2 2 2 1/2 L 0 ρ 0 (ρ0 − y ) [ l 2( y ) − ρ ] 0 ⌡ 0 a (1.5.15)
Here the general formula was used ∞
∞
∞
x
⌠ F (ρ)dρ d ⌠ 2xf ( x )d2x 1/2 = −⌠ f ( x ) d x d ⌠ ρ2F (ρ) 2dρ1/2 . dx ⌡ (x − ρ ) dρ⌡ ( x − ρ ) ⌡ ⌡ ρ
a
a
(1.5.16)
a
The integral in curly brackets of (1.5.15) can be evaluated, according to (1.3.21), with the result 2π ∞ R j 1 ⌠⌠ z 0 -1 + tan V (ρ,φ, z ) = 2 v (ρ ,φ )ρ dρ dφ . 0 0 0 0 0 R 0 π ⌡ ⌡ R 30 j 0
(1.5.17)
a
Here R
0
is defined by (1.4.22), and
j( x) =
(ρ20 − x 2)1/2[ l 22( x ) − x 2]1/2 x
.
The abbreviation j in (1.5.17) stands for j ( a ). expression (1.5.17) simplifies to
(1.5.18) In the particular case, when z =0,
48
CHAPTER 1,
Description of the new method
2π ∞ v (ρ0,φ0) ρ0dρ0dφ0 1 2 2 1/2⌠ ⌠ V (ρ,φ,0) = 2 ( a − ρ ) , ⌡ ⌡ (ρ20 − a 2)1/2[ρ2 + ρ20 − 2ρρ0cos(φ−φ0)] π 0
V (ρ,φ,0) = v (ρ,φ),
a
for ρ< a ;
for ρ≥ a .
(1.5.19)
The general solution is completed. The charge density σ is given by the two equivalent expressions (1.5.3) and (1.5.10), while the potential is in the two forms (1.5.14) and (1.5.17), the first one being more convenient for exact evaluation of the integrals involved, while the second is better suited for numerical integration. Problem 2. Consider the problem of finding a harmonic function, vanishing at infinity, and subject to the mixed conditions on the plane z =0 V = 0,
for ρ≤ a ,
∂V = − 2πσ(ρ,φ), ∂z
0≤φ<2π;
for ρ> a ,
0≤φ<2π.
(1.5.20)
The problem may be interpreted as an electrostatic one of a charged infinite diaphragm, with a grounded disc inside, or as an external crack problem in elasticity. Substitution of the boundary conditions (1.5.20) in (1.4.3) leads to the governing integral equation ρ
a
ρ dρ
0 0 x σ(ρ ,φ) ⌠ 2 ⌠ 2 2 1/2 2 1/2 L ρρ 0 0 ⌡ (ρ − x ) ⌡ (ρ0 − x )
dx
0
2
x
∞
= − ⌠
x
dx
2 2 1/2 ⌡ (x − ρ ) a
ρ dρ
ρρ
0 0 0 σ(ρ ,φ). ⌠ 2 2 1/2 L 0 x2 ⌡ ( x − ρ0)
(1.5.21)
a
One should notice that σ in the second term of (1.5.21) is known from the boundary condition (1.5.20), while the value of σ in the first term is yet to be determined. The right hand side of (1.5.21) can be transformed, by using (1.1.26), ρ
a
ρ dρ
0 0 x σ(ρ ,φ) ⌠ 2 ⌠ 2 2 1/2 2 1/2 L ρρ 0 0 ( ρ − ) ( ρ − ) x x ⌡ ⌡ 0 0
dx
x
2
49
1.5 External mixed boundary value problem for a half-space
ρ
ρ dρ
∞
0 0 x σ(ρ ,φ), ⌠ = − ⌠ 2 2 1/2 2 2 1/2 L ρρ 0 0 ⌡ (ρ − x ) ⌡ (ρ0 − x )
dx
0
2
a
with an immediate result a
ρ0dρ0
⌠ 2 2 1/2 ⌡ (ρ0 − x )
x
∞ ρ0dρ0 x x ⌠ σ(ρ ,φ) = − L L σ(ρ0,φ). 2 2 1/2 0 ρ0 ρ0 ⌡ (ρ0 − x )
(1.5.22)
a
Application of the operator a
L(ρ)
xdx d ⌠ 1 L dρ ⌡ ( x 2 − ρ2)1/2 x ρ
to both sides of (1.5.22) gives, after necessary transformations ∞
σ(ρ,φ) = −
π( a
(ρ20 − a 2)1/2
2 ⌠ − ρ2)1/2 ⌡
2
ρ20
a
− ρ
2
ρ L( ) σ(ρ ,φ) ρ dρ , 0 0 0 ρ
for ρ< a ,
0
(1.5.23) or, interpreting the L-operator, we obtain 2π ∞
1 ⌠⌠ σ(ρ,φ) = − 2 2 π ( a − ρ2)1/2 ⌡ ⌡ 0
(ρ20 − a 2)1/2σ(ρ ,φ ) ρ dρ dφ 0
ρ + 2
a
ρ20
0
0
0
− 2ρρ0cos(φ−φ0)
0
.
(1.5.24)
Now the value of σ is known all over the plane z =0, and (1.4.3) can be used in order to express the potential V directly through the prescribed σ. Substitution of (1.5.23) in (1.4.3) yields, after the first integration l1
∞
ρ dρ
0 0 x σ(ρ ,φ) ⌠ 2 V (ρ,φ, z ) = − 4⌠ 2 2 1/2 2 1/2 L ρρ 0 0 ⌡ (ρ − x ) ⌡ [ρ0 − g ( x )] 0
dx
a
2
50
CHAPTER 1,
∞
ρ dρ
g(x)
Description of the new method
ρρ
0 0 0 σ(ρ ,φ). ⌠ + 4⌠ 2 2 1/2 2 2 1/2 L 0 ⌡ ( x − ρ ) ⌡ [ g ( x ) − ρ0] x2 l
dx
(1.5.25)
a
2
The second term in (1.5.25) is equivalent to the second term in (1.4.2), which, in turn, can be represented by using (1.1.18), as l (ρ )
∞
1 0 dx x 2 σ(ρ ,φ) ρ dρ . 4⌠ ⌠ L 2 2 1/2 2 2 1/2 0 0 0 ρρ0 ⌡ ⌡ (ρ − x ) [ρ0 − g ( x )] 0 a The following scheme of changing the order of integration is enacted l1(ρ 0)
∞
l
∞
l1
1
(∞)
⌠ dρ ⌠ d x = ⌠ d x ⌠ dρ + ⌠ d x ⌡ ⌡ 0 ⌡ ⌡ 0 ⌡ l 0 a
a
0
1
∞
⌠ dρ , ⌡ 0
g(x)
and the second term in (1.5.25) can be rewritten as l1
ρ dρ
∞
0 0 x σ(ρ ,φ) ⌠ 2 4⌠ 2 2 1/2 2 1/2 L ρρ 0 0 ⌡ (ρ − x ) ⌡ [ρ0 − g ( x )]
dx
2
a
0
l
1
(∞)
dx ⌠ + 4 2 ⌡ (ρ − x 2)1/2 l 1
ρ dρ
∞
0 0 x σ(ρ ,φ). ⌠ 2 2 1/2 L ρρ 0 0 ⌡ [ρ0 − g ( x )] 2
(1.5.26)
g(x)
Substitution of (1.5.26) in (1.5.25) gives, by virtue of l 1(∞)=ρ, ρ
dx V (ρ,φ, z ) = 4⌠ 2 ⌡ (ρ − x 2)1/2 l 1
∞
ρ dρ
0 0 x σ(ρ ,φ). ⌠ 2 2 1/2 L ρρ 0 0 ⌡ [ρ0 − g ( x )] 2
(1.5.27)
g(x)
Interchange of the order of integration in (1.5.27), and integration with respect to x , according to (1.1.24), results in
51
1.5 External mixed boundary value problem for a half-space
2π ∞
V (ρ,φ, z ) =
2⌠ ⌠ π⌡ ⌡ 0
1 j tan-1( ) σ(ρ0,φ0) ρ0dρ0dφ0, R0 R0
a
(1.5.28)
where R 0 is defined by (1.4.22), and j stands for j ( a ), as defined by (1.3.4). The second problem is now solved. Expression (1.5.24) defines the charge density σ inside a circle directly in terms of its values outside. The potential V is given by two equivalent expressions (1.5.27) and (1.5.28), the first one to be used for exact evaluation of the integrals, while the second has some advantages in the case of numerical integration. Some specific examples are considered below. Example 1. Consider boundary conditions at z =0 V = v 0/ρn, ∂V = 0, ∂z
an
external
mixed
problem
with
the
following
for ρ≥ a , 0≤φ<2π; for ρ< a , 0≤φ<2π.
(1.5.29)
The conditions (1.5.29) correspond to those of Problem 1. The solution is given by (1.5.14) and (1.5.3). Substitution of (1.5.29) in (1.5.14) yields, after the integration 2 v 0 Γ[( n + 1)/2] ∞ dx ⌠ V (ρ,φ, z ) = , 2 Γ( n /2) ⌡ ( x − ρ2)1/2 g n( x ) √π l
(1.5.30)
2
where g ( x ) is defined by (1.1.25), and the following integral was employed (Gradshtein and Ryzhik, 1963) ∞
√π Γ( n /2) ⌠ n 2 dρ 2 1/2 = . 2Γ[( n + 1)/2] x n ⌡ ρ (ρ − x )
(1.5.31)
x
The integral in (1.5.30) can be evaluated in terms of elementary functions for any integer n , but the procedure is slightly different for even and odd values of n . For example, for even n =2 k , the problem reduces to the evaluation of the integral
52
CHAPTER 1,
∞
Description of the new method
⌠ (x − ρ ) dx , ⌡ x 2k( x 2 − ρ2 − z2)k l 2
2 k-1/2
2
which can be evaluated by introduction of a new variable t = x /( x 2 − ρ2)1/2. final result is 2 v 0 Γ[( n + 1)/2]
V (ρ,φ, z ) =
Γ( n /2) z
Am
k
Σ
2m-1 2 m − 1 [1 − Q 0 ] m=1
n
( Q m-1 − Q m-1) − ( Q m-1 − Q m-1), 2 3 4 1 − m 1 m=2 B
k
+ 2 B 1ln Q +
√π
Σ
The
m
(1.5.32)
where =
1 dm-1 (η − 1)k-1 , ( m − 1)! dηm-1 ( r 2 − η)k
for η=0, and r 2=1+ρ2/ z 2;
B k-m+1 =
1 dm-1 ( t 2 − 1)k-1 , ( m − 1)! d t m-1 t 2k( r 2 + t )k
for t =√1+ρ2/ z 2;
A
k-m+1
Q0 =
Q1 =
Q3 =
( l 22 − ρ2)1/2 l2
,
l 2[(ρ2 + z 2)1/2 + ( l 22 − a 2)1/2]
Q =
z [(ρ2 + z 2)1/2 + ( l 22 − a 2)1/2] l 21 z [(ρ2 + z 2)1/2 + z ] ρ2
,
a [(ρ2 + z 2)1/2 + z ]
, Q4 =
, Q2 =
z [(ρ2 + z 2)1/2 − ( l 22 − a 2)1/2]
z [(ρ2 + z 2)1/2 − z ] ρ2
l 21 .
,
(1.5.33)
For the case of an odd n =2 k +1, the integration can be performed by using the substitution t =( x 2−ρ2− z 2)1/2, and the final result is v 0 Γ[( n + 1)/2] V (ρ,φ, z ) = Γ( n /2) √π
k
Σ
Cm
2 2 m-1/2 m=1 (2 m − 1)( a − l 1)
k+1
+
Σ m=1
D m E m ,
(1.5.34)
53
1.5 External mixed boundary value problem for a half-space
where Cm =
1 dk-m ( t + z 2)k , k-m 2 2 k+1 ( k − m )! d t ( t + ρ + z )
for t =0;
Dm =
1 dk+1-m ( t + z 2)k , ( k + 1 − m )! d t k+1-m tk
for t =−(ρ2+ z 2);
Em =
√t (−1)m dm-1 1 , tan-1 2 m-1 ( m − 1)! d t √ t ( a − l 21)1/2
for t =ρ2+ z 2. (1.5.35)
Substitution of (1.5.29) in (1.5.3) yields, after integration σ(ρ,φ) =
v 0Γ[( n +1)/2] π3/2Γ( n /2)
1
n 2 2 1/2 a ( ρ − a )
n (ρ2 − a 2)1/2 1 a 2 1 3 − − +1, ; ; 1 F n , 2 2 2 ρn+2 ρ2
(1.5.36) and the Gauss hypergeometric function can be expressed in elementary functions (Bateman and Erdelyi, 1955), namely, for even n =2 k , k =1,2,3 ..., F k + 1,
1 3 1 dk k-1/2 1 + √ t , ; ; t = t ln 2 2 2k! dtk 1 − √t
and for odd n =2 k +1, k =0,1,2, ..., F k +
3 1 3 , ; ; t = 2 2 2
dk t k+1/2 . k 3 2√ t Γ( k + ) d t √1 − t 2 √π
(1.5.37) The charge density distribution, evaluated due to (1.5.36), is non-negative for n =1, and changes sign when n ≥2, its negative maximum increases with n , while the total charge stays at zero. Example 2. Consider the boundary conditions at z =0 in φ V = ( v /ρn) e , n
∂V = 0, ∂z where
v
n
is
for ρ≥ a , 0≤φ<2π;
for ρ< a , 0≤φ<2π,
constant.
The
solution
(1.5.38) is
given
by
Substitution of (1.5.38) in (1.5.14) yields, after integration
(1.5.3)
and
(1.5.14).
54
CHAPTER 1,
Description of the new method
∞
V (ρ,φ, z ) =
2Γ( n + 1/2) n in φ ⌠ dx ρe . 2n 2 ⌡ x ( x − ρ2)1/2 √πΓ( n ) l
(1.5.39)
2
The final integration gives V (ρ,φ, z ) =
2vn √πρ
n
n
in φ
e
Σ
(−1)k-1Γ( n + 1/2) (1 − Q 2k-1 0 ), (2 k − 1)Γ( k )Γ( n + 1 − k )
(1.5.40)
k=1
where Q 0 is defined by (1.5.33).
Substituting (1.5.38) in (1.5.3), we get, after
integration,
σ(ρ,φ) =
Γ( n + 1/2)
v nein φ
π3/2Γ( n )
ρn(ρ2 − a 2)1/2
.
(1.5.41)
Evidently, expression (1.5.41) can also be obtained by differentiation of (1.5.40) with respect to z for z =0. Example 3. Consider a case related to Problem 2, with the boundary conditions V = 0,
for ρ≤ a ,
σ0 ∂V = −2π n ∂z ρ
0≤φ<2π;
for ρ> a ,
0≤φ<2π;
The solution is given by (1.5.23) and (1.5.27). (1.5.27) yields, after integration using (1.5.31),
(1.5.42) Substitution of (1.5.42) in
ρ
V (ρ,φ, z ) = 2√πσ0
Γ[( n − 1)/2] ⌠ dx , 2 Γ( n /2) ⌡ (ρ − x 2)1/2 g n-1( x ) l
(1.5.43)
1
where g ( x ) is defined by (1.1.25). The technique used in the previous example can be employed here for further integration. The final result depends on the value of n being even or odd. For even n =2 k , k =1,2,3, ..., the potential is
55
1.5 External mixed boundary value problem for a half-space
V (ρ,φ, z ) = 2√πσ0
Σ(2m
1 −
a
2m-1
B m+1
Σ mz
+
( a 2 − l 21)1/2
− 1) z 2m-1
m=1
k-1
Am
k-1
+
Γ[( n − 1)/2] 2 B 1ln Q Γ( n /2)
m m m m m [ Q 1 − Q 2 − ( Q 3 − Q 4 )].
(1.5.44)
m=1
Here Q , Q , Q , Q , and Q are defined by (1.5.33), and 1
A
B
k-m
=
k-m+1
2
3
4
1 dm-1 ( t − z 2)k-1 ( m − 1)! d t m-1 (ρ2 + z 2 − t )k =
for t =0;
1 dm-1 ( t 2 − z 2)k-1 , ( m − 1)! d t m-1 t 2k-2[(ρ2 + z 2)1/2 − t ]k
for t =−(ρ2 + z 2)1/2. (1.5.45)
For odd n =2 k +1, the result is σ Γ[( n − 1)/2] V (ρ,φ, z ) = 2√π
0
Γ( n /2) z n-1
Gm ( l 22 − a 2)1/2 2m-1 − H L . 2m − 1 m m a m=1 k
Σ
(1.5.46) Here G k-m+1 =
1 dm-1 (1 + t )k-1 , Γ( m ) d t m-1 (ξ + t )k
H k-m+1 =
1 dm-1 (1 + t )k-1 , for t =−(ρ2+ z 2)/ z 2; m-1 k Γ( m ) d t t
Lm =
for t =0, ξ=(ρ2 + z 2)/ z 2;
(−1)m dm-1 1 √t tan-1 ( l 22 − a 2)1/2, Γ( m ) d t m-1 √ t a
for t =(ρ2+ z 2)/ z 2. (1.5.47)
Example 4. Consider the boundary conditions on the plane z =0: V = 0,
for ρ≤ a ,
0≤φ<2π;
56
CHAPTER 1,
∂V = − 2π(σ /ρn)ein φ, n ∂z
for ρ> a ,
Description of the new method
0≤φ<2π;
(1.5.48)
Substitution of (1.5.48) in (1.5.27) yields, after integration, σ Γ( n − 1/2) V (ρ,φ, z ) = 2√π
n
− z
n
Γ( n )ρn
ein φ( l 22 − a 2)1/2 − z
(−1)kΓ( n ) [1 − (1 − l 21/ a 2)k-3/2], Γ( k ) Γ( n − k +1)(2 k − 3)
Σ
(1.5.49)
k=2
and on the plane z =0 σ Γ( n − 1/2) V (ρ,φ,0) = 2√π
n
Γ( n )ρn
ein φℜ(ρ2 − a 2)1/2.
The symbol ℜ indicates the real part sign. according to (1.5.23), σ Γ( n − 1/2) σ(ρ,φ) =
n
+
n
√π Γ( n )ρ
n
ein φℜ1 −
The charge density is defined,
a ( a − ρ2)1/2 2
(−1)kΓ( n ) [1 − (1 − ρ2/ a 2)k-3/2]. Γ( k ) Γ( n − k +1)(2 k − 3)
Σ
k=2
(1.5.50)
A more general case of boundary conditions, namely, V = 0,
for ρ≤ a ,
∂V = − 2π(σ /ρj)ein φ, jn ∂z
0≤φ<2π; for ρ> a ,
0≤φ<2π;
(1.5.51)
can also be considered, by using the same technique as in the previous examples, and the final result can always be expressed in elementary functions. The form of the result will be different for ( j + n ) even, and for ( j + n ) odd. As an example, the following expression can be obtained by substituting (1.5.51) in (1.5.23), for the case when j + n =2 k
57
1.5 External mixed boundary value problem for a half-space
σjn
a 1 − σ(ρ,φ) = j e ℜ1 − 2 ρ ( a − ρ2)1/2 in φ
k
Γ( m − 3/2) ρ2m-2 , 2√πΓ( m ) a
Σ m=2
(1.5.52)
and for odd j + n =2 k +1 2σjn
-1 ρ a σ(ρ,φ) = e ℜ sin (a ) − 2 j 2 1/2 1 − πρ (a − ρ ) in φ
k
√πΓ( m − 1) ρ2m-2 , 4Γ( m + 1/2) a
Σ m=2
(1.5.53) Expressions (1.5.52) and (1.5.53) represent general formulae which cover all the particular cases considered in Examples 3 and 4. The examples above have demonstrated the simplicity of the method. The generation of the solution is reduced to a straightforward and elementary procedure.
1.
Exercise 1.5 The following boundary conditions are prescribed at z =0 V = v 0/ρ,
for ρ≥ a , 0≤φ<2π;
∂V = 0, for ρ< a , 0≤φ<2π. ∂z Find the potential function and the charge distribution. 2v0 2 2 1/2 -1(ρ + z ) sin , Answer: V (ρ,φ, z ) = l2 π(ρ2 + z 2)1/2 σ(ρ,φ) =
v0a
π2ρ2(ρ2 − a 2)1/2 The total charge is equal v .
.
0
2.
The following boundary conditions are prescribed at z =0 V = v 0/ρ2,
for ρ≥ a , 0≤φ<2π;
∂V = 0, for ρ< a , 0≤φ<2π. ∂z Find the potential function and the charge distribution.
58
CHAPTER 1,
v0
1 − ρ2 + z 2
Answer: V (ρ,φ, z ) =
+
z ln (ρ + z 2)1/2
3.
( l 22 − ρ2)1/2 l2
l 2[(ρ2 + z 2)1/2 + ( l 22 − a 2)1/2] a [(ρ + z )
2
σ(ρ,φ) =
Description of the new method
2
2 1/2
+ z]
,
v0
1 1 ρ + (ρ2 − a 2)1/2 − ln . ρ a 2πρ2(ρ2 − a 2)1/2
The following boundary conditions are prescribed at z =0 V = v 0/ρ3,
for ρ≥ a , 0≤φ<2π;
∂V = 0, for ρ< a , 0≤φ<2π. ∂z Find the potential function and the charge distribution. 2 4v0 l 21( a 2 − l 21)1/2 z Answer: V (ρ,φ, z ) = 2 − (ρ + z 2)2 ( a 2 − l 21)1/2 2a2 +
ρ2 − 2 z 2
(ρ2 + z 2)1/2 sin . l2 -1
2(ρ2 + z 2)1/2 Note that the potential V (0,0,0)=4 v 0/(3π a 3).
σ(ρ,φ) = 4.
at
the
2 v 0(2 a 2 − ρ2) π2 a ρ4(ρ2 − a 2)1/2
coordinate
origin
.
The following boundary conditions are prescribed at z =0 V = v 0/ρ4,
for ρ≥ a , 0≤φ<2π;
∂V = 0, for ρ< a , 0≤φ<2π. ∂z Find the potential function and the charge distribution.
is
finite,
namely,
59
1.5 External mixed boundary value problem for a half-space
( l 22 − ρ2)1/2 ρ2 − z2 1 − Answer: V (ρ,φ, z ) = l2 2(ρ2 + z 2)2 ρ2 + z 2 3v0
( l 22 − ρ2)1/2 3 l 22 z 1 1 − + ( l 22 − a 2)1/2 − z − 2 2 2 3 l2 2(ρ + z )a
−
z (2 z 2 − 3ρ2) 2(ρ + z ) 2
2 3/2
l [(ρ2 + z 2)1/2 + ( l 22 − a 2)1/2] ln
2
a [(ρ2 + z 2)1/2 + z ]
,
The last expression simplifies at z =0: v
( a 2 − ρ2)1/2 ρ2( a 2 − ρ2)1/2 , V (ρ,φ,0) = 4 1 − − a 2a3 ρ 0
and V (ρ,φ,0) = v /ρ4, 0
for ρ> a .
for ρ≤ a ;
Note that the potential at the coordinate
origin is finite, namely, V (0,0,0)=3 v /(8 a 4). 0
3 a 2 − ρ2 3 ρ + (ρ2 − a 2)1/2 ln σ(ρ,φ) = − . ρ a 8πρ4 a 2(ρ2 − a 2)1/2 3v
5.
0
The following boundary conditions are prescribed at z =0 iφ V = ( v 1/ρ) e ,
for ρ≥ a , 0≤φ<2π;
∂V = 0, for ρ< a , 0≤φ<2π. ∂z Find the potential function. v ( a 2 − l 21)1/2 1 iφ e 1 − Answer: V (ρ,φ, z ) = ρ a 6.
The following boundary conditions are prescribed at z =0 iφ V = ( v 2/ρ2) e2 ,
for ρ≥ a , 0≤φ<2π;
60
CHAPTER 1,
Description of the new method
∂V = 0, for ρ< a , 0≤φ<2π. ∂z Find the potential function. 3v ( a 2 − l 21)1/2 3 ( a 2 − l 21)1/2 2 1 2 iφ . 1 − − Answer: V (ρ,φ, z ) = 1 − 2 e 3 a a 2ρ
7.
The following boundary conditions are prescribed at z =0 V = ( v /ρ3) e3 i φ, 3
for ρ≥ a , 0≤φ<2π;
∂V = 0, for ρ< a , 0≤φ<2π. ∂z Find the potential function. 15 v ( a 2 − l 21)1/2 3 1 e3 i φ 1 − Answer: V (ρ,φ, z ) = 2 a 4ρ3
( a 2 − l 21)1/2 3 ( a 2 − l 21)1/2 5 1 1 + 1 − . 1 − − 3 10 a a
8.
Let the following boundary conditions be prescribed at z =0: for ρ≤ a ,
V = 0,
0≤φ<2π;
∂V = − 2πσ /ρ2, σ =const, for ρ> a , 0≤φ<2π. 0 0 ∂z Find the potential function and the charge distribution. 2πσ l [(ρ2 + z 2)1/2 + ( l 22 − a 2)1/2] 0 2 , Answer: V (ρ,φ, z ) = 2 2 1/2 ln (ρ + z ) a [(ρ2 + z 2)1/2 + z ] σ
σ
a , σ(ρ,φ) = 2 ℜ1 − 2 ( a − ρ2)1/2 ρ 0
9.
σ(0,0) = −
0
2a2
Let the following boundary conditions be prescribed at z =0: V = 0,
for ρ≤ a ,
0≤φ<2π;
∂V = − 2πσ /ρ3, σ =const, 0 0 ∂z
for ρ> a ,
0≤φ<2π.
.
61
1.5 External mixed boundary value problem for a half-space
Find the potential function and the charge distribution. 4σ0 ( l 22 − a 2)1/2 z (ρ2 + z 2)1/2 -1 Answer: V (ρ,φ, z ) = 2 − 2 sin , a l2 ρ + z2 (ρ + z 2)1/2 σ(ρ,φ) = 10.
2σ0
1 3 1 sin-1(ρ) − 2 2 1/2 , for ρ< a ; σ(0,0) = −2σ0/(3π a ). a ρ (a − ρ ) πρ 2
Let the following boundary conditions be prescribed at z =0: for ρ≤ a ,
V = 0,
0≤φ<2π;
∂V = − 2πσ0/ρ4, σ0=const, for ρ> a , 0≤φ<2π. ∂z Find the potential function and the charge distribution. πσ0 l 22 2 z 2 2 1/2 2 2 1/2 Answer: V (ρ,φ, z ) = ( a − l 1) − 3 z + 2 ( l 2 − a ) a 2(ρ2 + z 2)2 a
ρ2 − 2 z 2 (ρ2 + z 2)1/2
ln
l 2[(ρ2 + z 2)1/2 + ( l 22 − a 2)1/2] a [(ρ2 + z 2)1/2 + z ]
σ0
2 a 2 − ρ2 σ(ρ,φ) = 4 ℜ1 − , ρ 2 a ( a 2 − ρ2)1/2 11.
,
σ(0,0) = −
Consider the boundary conditions on the plane z =0: V = 0,
for ρ≤ a ,
0≤φ<2π;
∂V iφ = − 2π(σ1/ρ)e , for ρ> a , 0≤φ<2π. ∂z Find the potential function and the charge distribution. iφ
Answer: V (ρ,φ, z ) = 2π(σ1/ρ) e [( l 22 − a 2)1/2 − z ], iφ
σ(ρ,φ) = (σ1/ρ)e ℜ[1 − a /( a 2 − ρ2)1/2]. 12.
Consider the boundary conditions on the plane z =0: V = 0,
for ρ≤ a ,
0≤φ<2π;
σ0 8a4
.
62
CHAPTER 1,
Description of the new method
∂V = − 2π(σ /ρ2)e2 i φ, for ρ> a , 0≤φ<2π. 2 ∂z Find the potential function and the charge distribution.
Answer: V (ρ,φ, z ) = π(σ /ρ2) e2 i φ[( l 22 − a 2)1/2 − 2 z + z ( a 2 − l 21)1/2/ a ], 2
σ(ρ,φ) = (σ /ρ2)ei φℜ1 −
2
13.
2 a 2 − ρ2 . 2 a ( a 2 − ρ2)1/2
Consider the boundary conditions on the plane z =0: for ρ≤ a ,
V = 0,
0≤φ<2π;
∂V = − 2π(σ /ρ3)e3 i φ, for ρ> a , 0≤φ<2π. 3 ∂z Find the potential function and the charge distribution. 3πσ 3 8 3 i φ 2 Answer: V (ρ,φ, z ) = ( l 2 − a 2)1/2 − z 3 e 3 4ρ ( a 2 − l 21)1/2 3 z 1 2 2 1/2 , + 2 ( a − l 1) − z a 3 a σ σ(ρ,φ) = 14.
e3 i φℜ1 −
3
ρ3
3a 3( a 2 − ρ2)1/2 ( a 2 − ρ2)3/2 − . + 4a 8( a 2 − ρ2)1/2 8a3
Prove that the total charge Q
T
in Problem 2 (1.5.20) can be expressed
directly in terms of the given charge density σ as 2π ∞
Q
T
=
2⌠ ⌠ a σ(ρ,φ) cos-1( ) ρdρdφ. π⌡ ⌡ ρ 0
a
Hint : integrate (1.5.23). 15.
Solve the problem above in the case when σ=σ /ρn. 0
2σ √π Γ[( n − 1)/2] Answer: Q
T
=
0
( n − 2) Γ( n /2) a n-2
.
63
1.6 Some fundamental integrals
1.6 Some fundamental integrals The integrals are called fundamental because of their primary importance to the new method, and also because almost all the integral representations, derived earlier, are just particular cases of the fundamental ones to be evaluated here. Consider three points in the system of cylindrical coordinates, namely, M (ρ,φ, z ), M 0(ρ0,φ0, z 0), and N ( r ,ψ,0). The following notation is introduced: l 1(t) =
1 {[(ρ + t )2 + z 2]1/2 − [(ρ − t )2 + z 2]1/2} , 2
(1.6.1)
l 2(t) =
1 {[(ρ + t )2 + z 2]1/2 + [(ρ − t )2 + z 2]1/2} , 2
(1.6.2)
l 10(t) =
1 {[(ρ0 + t )2 + z 20]1/2 − [(ρ0 − t )2 + z 20]1/2} , 2
(1.6.3)
l 20(t) =
1 {[(ρ0 + t )2 + z 20]1/2 + [(ρ0 − t )2 + z 20]1/2}. 2
(1.6.4)
According to the earlier convention, l 10 stands as an abbreviation for l 10( a ), etc.;
R (⋅,⋅) denotes the distance between two points. Consider the following integral: 2π a
h
z 1 0 rd rdψ , I1 = ⌠ ⌠ 3 tan-1 ⌡ ⌡ R ( M,N) R ( N,M0) R ( N,M0) 0
(1.6.5)
h 0 = [ a 2 − l 10]1/2 [ a 2 − r 2]1/2/ a
(1.6.6)
0
where
Make use of the integral representation (1.1.23) a h0 d l 20( x ) ρ0 r 1 -1 tan = ⌠ , ψ−φ0, 2 2 1/2 2 2 1/2 λ 2 R ( N,M ) R ( N,M ) l x x r l x [ ( ) − ρ ] ( − ) ( ) ⌡ 20 0 20 0 0 r
(1.6.7)
where λ(⋅,⋅) is defined by (1.1.5). The
substitution
of
(1.6.7)
in
(1.6.5)
yields,
after
changing
the
order
of
64
CHAPTER 1,
Description of the new method
integration: d l 20( x ) ρ0 r rdr z ⌠ ⌠ ⌠ I1 = dψ , ψ−φ0 3 . 2 2 1/2 2 2 1/2 λ 2 l 20( x ) R ( M,N) ⌡ [ l 20( x ) − ρ0] (x − r ) ⌡ ⌡ 0 2π
a
x
0
0
(1.6.8) By substituting the integral representation (1.3.7), z R 3( M,N )
≡
z [ρ2 + r 2 − 2 r ρ cos(φ − ψ) + z 2]3/2
r l 1( t ) t [ t 2 − l 21( t )]1/2 tdt 2 1 d ⌠ = L( ) λ , φ−ψ, 2 2 1/2 2 2 πr r dr ⌡ (r − t ) l 2( t ) − l 1( t ) l 2( t )
(1.6.9)
0
in (1.6.8), the following result can be obtained, after integration with respect to ψ: a
I 1 = 4⌠
x
d l 20( x )
dr
⌠ 2 2 1/2 2 1/2 ⌡ [ l 20( x ) − ρ0] ⌡ ( x − r ) 0
0
r
d tdt ⌠ 2 d r ⌡ ( r − t 2)1/2 0
l 1( t ) t ρ0 [ t 2 − l 21( t )]1/2 × λ 2 , φ−φ0. l 20( x ) l 2( t ) l 22 − l 21
(1.6.10)
Here the following property of the L-operators was used: L( k ) λ( k 1,⋅) = λ( k k 1,⋅) , for k , k 1 <1
(1.6.11)
The well known property of the Abel-type operators, namely, x
r
dr d ⌠ f(t) tdt π ⌠ 2 2 1/2 d r 2 2 1/2 = 2 f ( x ) , ⌡ (x − r ) ⌡ (r − t ) 0
0
allows us to simplify (1.6.10) significantly:
(1.6.12)
65
1.6 Some fundamental integrals
[ x 2 − l 21( x )]1/2 [ x 2 − l 210( x )]1/2
a
l ( x) l ( x)
1 10 λ , φ−φ d x . I 1 = 2π⌠ 2 2 2 2 0 l x l x ( ) ( ) ⌡ l 2( x ) − l 1( x ) l 20( x ) − l 10( x ) 2 20 0
(1.6.13) It is noteworthy that the integrand in (1.6.13) is symmetric with respect to the points M and M while it did not look so in the original expression (1.6.5). 0
The integrand in (1.6.13) is a perfect differential, so that the integral can be evaluated as indefinite: [ x 2 − l 21( x )]1/2 [ x 2 − l 210( x )]1/2
l ( x) l ( x)
1 10 ⌠ , φ−φ d x λ 0 ⌡ l 22( x ) − l 21( x ) l 220( x ) − l 210( x ) l 2( x ) l 20( x ) 1 = 2R
Θ ( x) -1
tan
1
R
1
1
1 + 2R
Θ ( x) -1
tan 2
2
R
,
(1.6.14)
2
where R R
1
= [ρ2 + ρ20 − 2ρρ cos(φ − φ ) + ( z − z )2]1/2 ,
2
= [ρ2 + ρ20 − 2ρρ cos(φ − φ ) + ( z + z )2]1/2 ,
0
0
0
0
Θ ( x ) = θ( x ) + zz /θ( x ) , 1
0
0
Θ ( x ) = θ( x ) − zz /θ( x ) ,
0
2
0
θ( x ) = [ x 2 − l 21( x )]1/2 [ x 2 − l 210( x )]1/2/ x .
(1.6.15)
Notice that when z =0, θ( x ) transforms into h ( x ) as it is defined by (1.3.11), 0
and the integral (1.6.14) reduces to the one considered in Exercise 1.1.8. Correctness of the integral in (1.6.14) can be verified by differentiation. The algebra involved is not trivial. Here we present some intermediate transformations: θ( x ) [ l 22( x ) l 220( x ) − l 21( x ) l 210( x )] ∂ θ( x ) = , ∂x x [ l 22( x ) − l 21( x )] [ l 220( x ) − l 210( x )] l ( x) l ( x)
λ , φ−φ = 0 l ( x ) l ( x ) 1 2
10 20
l 22( x ) l 220( x ) − l 21( x ) l 210( x ) 2x
2
1 1 + 2 2 2 R 1 + Θ1( x ) R 2 + Θ22( x )
(1.6.16)
66
CHAPTER 1,
Description of the new method
(1.6.17) Formula (1.6.14) allows us to evaluate the integral (1.6.5): 2π a
h
1 z 0 rd rdψ ⌠⌠ tan-1 3 R ( N,M0) ⌡ ⌡ R ( M,N) R ( N,M0) 0 0
| zz | Θ Θ | zz | 0 0 | z | 1 -1 1 1 -1 2 π π , = π tan + tan + − 2 2 zz z R 1 zz0 R 2 R 1 R 2 0
where the contractions Θ
1
(1.6.18) stand for Θ ( a ) and Θ ( a ) respectively. Note
and Θ
2
1
an important particular case when z 0=0.
2
Formula (1.6.18) in this case transforms
into: 2π a
1
z
⌠⌠ tan ⌡ ⌡ R 3( M,N) R ( N,N0) 0
-1
( a 2 − r 2)1/2 ( a 2 − ρ20)1/2 a R ( N,N0)
0
=
rd rdψ
2π h , for ρ < a , tan-1 0 R ( M,N 0) R ( M,N ) 0
and the integral vanishes when ρ0≥ a .
(1.6.19)
Here the point N 0 has the cylindrical
coordinates (ρ0,φ0,0), and h is defined by (1.4.22).
The second fundamental integral to be considered is: 2π a
z
0 I2 = ⌠ ⌠ 3 ⌡ ⌡ R ( N,M0) 0 0
where h 0 is defined by (1.6.6).
R ( N,M 0) h0
-1
+ tan
h0
rd rdψ , R ( N,M 0) R ( M,N )
(1.6.20) Make use of the integral representation for the
reciprocal distance r l 1( x ) x [ l 22( x ) − x 2]1/2 dx 1 2 ⌠ = λ , φ−ψ, 2 2 1/2 2 2 π ⌡ (r − x ) R ( M,N ) l 2( x ) − l 1( x ) l 2( x ) r 0
which is a variation of (1.1.26).
By substituting (1.6.21)
(1.6.21) in (1.6.20), the
67
1.6 Some fundamental integrals
following expression results, after changing the order of integration: 2π
I2
a 2 a [ l 2( x ) − x 2]1/2 rdr 2 ⌠ dx ⌠ = dψ ⌠ 2 2 2 2 1/2 π ⌡ ⌡ l 2( x ) − l 1( x ) ⌡ (r − x ) 0
x
0
l ( x) x
R ( N,M )
z
1 0 × λ , φ − ψ 3 l 2( x ) r R ( N,M0)
0
h
h -1
+ tan
0
,
0
R ( N,M ) 0
(1.6.22)
We use the integral representation (1.3.14): z
R ( N,M )
3 R ( N,M ) 0 0
0
h
h -1
+ tan
0
0
R ( N,M ) 0
a l (t) [ l 220( t ) − t 2]1/2 tdt 10 L( r ) d ⌠ = − , φ −ψ. λ 0 r d r ⌡ ( t 2 − r 2)1/2 l 220( t ) − l 210( t ) t l ( t ) 20 r
(1.6.23) The substitution of (1.6.23) in (1.6.22) yields, after integration with respect to ψ: a
I 2 = −4 ⌠
[ l 22( x ) − x 2]1/2
2 2 ⌡ l 2( x ) − l 1( x )
a
dx ⌠
0
dr
2 2 1/2 ⌡ (r − x ) x
[ l 220( t ) − t 2]1/2
a
d tdt ⌠ 2 d r ⌡ ( t − r 2)1/2 r
l ( x) l ( t) x
1 10 × 2 λ , φ−φ . 2 0 l 20( t ) − l 10( t ) l 2( x ) l 20( t ) t
(1.6.24)
We recall another well known property of the Abel operators: a
a
dr d ⌠ f(t)tdt π ⌠ 2 2 1/2 d r 2 2 1/2 = − 2 f ( x ). ⌡ (r − x ) ⌡ (t − r ) x
r
Application of (1.6.25) to (1.6.24) yields: a
[ l 22( x ) − x 2]1/2 [ l 220( x ) − x 2]1/2
l ( x) l ( x)
1 10 λ , φ−φ d x . I 2 = 2π⌠ 2 2 2 2 0 ⌡ l 2( x ) − l 1( x ) l 20( x ) − l 10( x ) l 2( x ) l 20( x ) 0
(1.6.25)
68
CHAPTER 1,
Description of the new method
(1.6.26) Note certain similarity between (1.6.26) and (1.6.13). The integrand in (1.6.26) is a perfect differential, and can be evaluated in elementary functions: [ l 22( x ) − x 2]1/2 [ l 220( x ) − x 2]1/2
l ( x) l ( x)
1 10 ⌠ λ , φ − φ0 d x 2 2 2 2 ⌡ l 2( x ) − l 1( x ) l 20( x ) − l 10( x ) l 2( x ) l 20( x ) 1 tan-1 = − 2R1
Ξ1( x ) R1
1 − tan-1 2R2
Ξ2( x ) R2
,
(1.6.27)
where R 1 and R 2 are defined by (1.6.15), and Ξ1( x ) = ξ( x ) + zz0/ξ( x ) ,
Ξ2( x ) = ξ( x ) − zz0/ξ( x ) ,
ξ( x ) = [ l 22( x ) − x 2]1/2 [ l 220( x ) − x 2]1/2/ x
(1.6.28)
Again, one can notice that in the case when z 0=0, ξ( x ) transforms into j ( x ) as it is defined in (1.3.4), and the integral (1.6.27) coincides with (1.1.26). As before, correctness of the integration can be verified by differentiation, using (1.6.17) and the property: ξ( x ) [ l 22( x ) l 220( x ) − l 21( x ) l 210( x )] ∂ ξ( x ) = − , ∂x x [ l 22( x ) − l 21( x )] [ l 220( x ) − l 210( x )]
(1.6.29)
Finally, the integral (1.6.20) can be evaluated as follows: 2π a
z
0 ⌠⌠ 3 ⌡ ⌡ R ( N,M0) 0 0
R ( N,M 0) h0
-1
+ tan
h0
r d r dψ R ( N,M 0) R ( M,N)
| z 0| 1 Ξ1 Ξ2 π 1 π -1 -1 . = π − tan + − tan z0 R 1 2 R1 R2 2 R 2
(1.6.30)
According to our convention, Ξ1 and Ξ2 denote Ξ1( a ) and Ξ2( a ) respectively. Consider a particular case, when z 0=0, and ρ0> a .
Due to the relationship
69
1.6 Some fundamental integrals
az 0 (a
2
−
l 210)1/2
→ (ρ20 − a 2)1/2 , for z 0→0 ,
the integral (1.6.30) will take the form: 2π a
(ρ2 − a 2)1/2
r d r dψ ⌠⌠ 0 2 2 1/2 ⌡ ⌡ ( a − r ) R ( M,N) R 2( N0, N) 0 0
2π π − tan-1 = R0 2
( l 22 − a 2)1/2(ρ20 − a 2)1/2
.
(1.6.31)
Here R 0= R ( M,N 0)=[ρ2 + ρ20 − 2ρρ0cos(φ−φ0) + z 2]1/2.
The integration in (1.6.30)
aR 0
for z 0=0 and ρ0< a yields π / R 0. 2
The case of z =0 can be considered in a
similar manner. The integrals evaluated above may be called internal because the domain of integration was the interior of a disc. We can also evaluate relevant external integrals. For example, consider the integral 2π ∞
j
z 1 0 I3 = ⌠ ⌠ r d r dψ, tan-1 3 R ( N,M 0) ⌡ ⌡ R ( M,N) R ( N,M0) 0
(1.6.32)
j 0 = ( r 2 − a 2)1/2( l 220 − a 2)1/2/ a .
(1.6.33)
a
where
The integral representations due to (1.1.26) and (1.3.15), namely, r j0 l 210( x ) [ l 220( x ) − x 2]1/2 d x 1 tan-1 = ⌠ λ , φ0−ψ, 2 2 1/2 2 2 R ( N,M 0) R ( N,M 0) r ρ ⌡ ( r − x ) [ l 20( x ) − l 10( x )] 0 a
(1.6.34) ∞ [ l 22( t ) − t 2]1/2 tdt ρ z 2 d ⌠ L( r ) , φ−ψ, = − λ 2 3 2 2 1/2 2 2 π r d r R ( M,N ) l 2( t ) − l 1( t ) l 2( t ) ⌡ (t − r ) r
(1.6.35) can be substituted in (1.6.32) yielding
70
CHAPTER 1,
∞
I
3
= −4⌠
⌡
a ∞
= 2π⌠ ⌡ a
[ l 220( x ) − x 2]1/2 l 220( x ) − l 210( x )
∞
[ l 22( t ) − t 2]1/2
dr
dx ⌠
Description of the new method
2 2 1/2 ⌡ (t − r ) x
[ l 22( x ) − x 2]1/2 [ l 220( x ) − x 2]1/2
l 2 ( x )ρ
10 λ , φ−φ0 2 2 2 l 2( t ) − l 1( t ) l 2( t )ρ 0
l ( x) l ( x)
1 10 λ , φ−φ0 d x . l 22( x ) − l 21( x ) l 220( x ) − l 210( x ) l 2( x ) l 20( x )
(1.6.36)
This is the integral which was already evaluated in (1.6.27), so we may write the final result 2π ∞
j
z 1 0 ⌠⌠ r d r dψ tan-1 3 R ( N,M ) ⌡ ⌡ R ( M,N) R ( N,M0) 0 0 a
Ξ Ξ | zz0| | zz0| | z | 1 -1 1 π 1 -1 2 π , = π tan − + tan + z R R zz R R zz 2 2 1 1 0 2 2 0
(1.6.37) Comparison with the relevant internal integral (1.6.18) indicates similarity, except for substitution of Θ by Ξ. The second external integral is 2π ∞
z
0 I4 = ⌠ ⌠ 3 ⌡ ⌡ R ( N,M0) 0
R ( N,M 0) j0
a
where j 0 is defined by (1.6.33).
-1
+ tan
j0
rd rdψ , R ( N,M 0) R ( M,N )
(1.6.38) Make use of the integral representations (see
Exercise 1.1.8 and (1.3.21)) ∞ l 1( x ) r [ x 2 − l 21( x )]1/2 dx 1 2 ⌠ = λ , φ − ψ, R ( M,N) π ⌡ ( x 2 − r 2)1/2 l 22( x ) − l 21( x ) l 2( x ) x r
(1.6.39) z0
R 3( N,M 0)
R ( N,M 0) j0
-1
+ tan
j0
R ( N,M 0)
71
1.6 Some fundamental integrals
r l ( x) x [ x 2 − l 210( x )]1/2 xdx 10 1 1 d ⌠ λ L( ) , ψ−φ0. = r r d r ⌡ ( r 2 − x 2)1/2 l 220( x ) − l 210( x ) l ( x ) 20 a
(1.6.40) Substitution of (1.6.39) and (1.6.40) in (1.6.38) leads to ∞
[ x 2 − l 21( x )]1/2 [ x 2 − l 210( x )]1/2
l ( x) l ( x)
1 10 λ I = 2π⌠ 2 , φ−φ d x . 2 2 2 4 0 l x l x ( ) ( ) − − l x l x l x l x ( ) ( ) ( ) ( ) ⌡ 2 1 20 10 2 20 a
(1.6.41) This integral was evaluated in (1.6.14), and the final result is 2π ∞
z
0 ⌠⌠ 3 ⌡ ⌡ R ( N,M0) 0
R ( N,M )
j
0
j
a
rd rdψ R ( N,M ) R ( M,N ) 0
-1
+ tan
0
0
|z | 1 Θ Θ π 1 π 0 1 2 -1 -1 . = π − tan + − tan 2 2 z R R R R 0
1
1
2
2
(1.6.42)
One can notice the same similarity between the internal (1.6.30) and the external (1.6.42) integrals. The similarity goes further. By using the property ( l 22( x ) − x 2)( x 2 − l 21( x )) = x 2 z 2, we deduce, that for zz >0, 0
Ξ ( x ) = ξ( x ) + θ( x ),
Ξ ( x ) = ξ( x ) − θ( x ),
Θ ( x ) = θ( x ) + ξ( x ),
Θ ( x ) = θ( x ) − ξ( x ).
1
2
1
This means that Ξ =Θ 1
1
2
and Ξ =−Θ 2
2
(1.6.43) In the case when zz <0, the
for zz >0. 0
0
relationships change, namely, Ξ =−Θ and Ξ =Θ . 1
1
2
2
Exercise 1.6 Introduce
the
following
points:
M (ρ,φ, z ),
M (ρ ,φ , z ), 0
0
0
0
N ( r ,ψ,0),
N (ρ ,φ ,0), 0
0
0
72
CHAPTER 1,
Description of the new method
P (ρ,φ,0); as before, R (⋅,⋅) stands for the distance between two points. 1.
Evaluate the integral 2π a
z
R ( N,M 0)
0 ⌠⌠ 3 ⌡ ⌡ R ( N,M0) 0 0
h0
2π π − tan-1 Answer: R ( M 0, P ) 2
-1
+ tan
h0
rd rdψ , R ( N,M 0)R ( P , N )
( l 220 − a 2)1/2(ρ2 − a 2)1/2 aR ( M 0, P )
for ρ> a .
.
Hint : use (1.6.30) for z =0. 2.
Evaluate the integral above for ρ< a .
Answer: π2/ R ( M 0, P ). 3.
Evaluate the integral 2π ∞
z 1 ⌠⌠ tan-1 3 R N,N ( ) ⌡ ⌡ R ( M,N) 0 0
( r 2 − a 2)1/2(ρ20 − a 2)1/2
a
a R ( N,N0)
(ρ20 − a 2)1/2( l 22 − a 2)1/2 2π -1 . Answer: tan R ( M,N0) a R ( M,N0) 4.
Evaluate the integral 2π ∞
( a 2 − ρ2)1/2
r d r dψ 0 ⌠⌠ 2 2 1/2 ⌡ ⌡ ( r − a ) R ( M,N) R 2( N0, N) 0 a
2π π − tan-1 Answer: R ( M,N 0) 2
( a 2 − l 21)1/2( a 2 − ρ20)1/2 a R ( M,N 0)
.
rd rdψ.
CHAPTER 2 MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
Elastic half-space has been proven to be a useful mathematical model for consideration of various contact and crack problems in finite bodies, provided that the domain of contact or the crack size is much smaller than the characteristic dimension of the body. A general solution in terms of three harmonic functions is presented for the case of transverse isotropy. Exact closed form solutions are given to the mixed problems of the first and second type, with various applications considered. The material in this Chapter is based on the papers (Fabrikant, 1970, 1971b, 1971c, 1985b, 1986g).
2.1 General solution Consider a transversely isotropic elastic body which is characterized by five elastic constants A ik defining the following stress-strain relationships: σx = A 11
∂ux ∂x
+ ( A 11 − 2 A 66)
σy = ( A 11 − 2 A 66)
σz = A 13
τxy = A 66(
∂ux ∂x ∂ux ∂y
∂ux ∂x
+ A 13
+
∂uy
∂uy ∂x
+ A 11
∂y ) ,
∂uy ∂y ∂uy ∂y
+ A 33
+ A 13
∂w , ∂z
+ A 13
∂w , ∂z
∂w , ∂z
τyz = A 44(
73
∂uy ∂z
+
∂w ) , ∂y
74
CHAPTER 2.
τzx = A 44(
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
∂ux ∂w ). + ∂x ∂z
(2.1.1)
The equilibrium equations are: ∂σx ∂x ∂τzx ∂x
+
+
∂τxy ∂y ∂τyz ∂y
+
+
∂τzx ∂z ∂σz ∂z
∂τxy
= 0 ,
∂x
+
∂σy ∂y
+
∂τyz ∂z
= 0 ,
= 0 .
(2.1.2)
Substitution of (2.1.1) in (2.1.2) yields:
A 11
A 66
∂2 u x ∂x2 ∂2 u y ∂x2
+ A 66
+ A 11
∂2 u x ∂y2 ∂2 u y ∂y2
+ A 44
+ A 44
∂2 u x ∂ z2 ∂2 u y ∂ z2
+ ( A 11 − A 66)
+ ( A 11 − A 66)
∂2 u y ∂x∂y ∂2 u x ∂x∂y
+ ( A 13
∂2 w + A 44) = 0 , ∂x∂z
+ ( A 13
∂2 w + A 44) = 0 , ∂y∂z
∂2 u x ∂2 u y ∂2 w ∂2 w ∂2 w = 0 . A 44 + A 33 2 + ( A 44 + A 13) + 2 + 2 ∂ ∂ ∂ ∂ x z y z ∂x ∂y ∂z (2.1.3) Introduce complex tangential displacements u = u x+i u y, and u = u x−i u y. This will allow us to reduce the number of equations in (2.1.3) by one, and to rewrite these equations in a more compact manner, namely, 1 ∂2 u 1 ∂w ( A 11 + A 66)∆ u + A 44 + ( A 11 − A 66)Λ2 u + ( A 13 + A 44)Λ = 0, 2 ∂z 2 2 ∂z A 44∆ w + A 33
∂2 w ∂ 1 2 + 2( A 13+ A 44) ∂ z (Λu + Λ u ) = 0. ∂z
(2.1.4)
Here the following differential operators were used: ∆ =
∂2 ∂2 + , ∂x2 ∂y2
Λ =
∂ ∂ + i , ∂x ∂y
and the overbar everywhere indicates the complex conjugate value.
(2.1.5) Note also
75
2.1 General solution
that ∆=ΛΛ.
One can verify that equations (2.1.4) can be satisfied by
u = Λ( F 1+ F 2 + i F 3),
w = m1
∂F1 ∂z
+ m2
∂F2
(2.1.6)
∂z
where all three functions F k satisfy the equation (Elliott, 1948): ∂2 F k
∆ F k + γ2k
∂ z2
for k = 1,2,3,
= 0,
(2.1.7)
and the values of m k and γk are related by the following expressions (Elliott, 1948): A 44 + m k( A 13+ A 44) A 11
=
m k A 33 m k A 44 + A 13 + A 44
= γ2k,
for k =1,2;
γ3 = A 44/ A 661/2.
(2.1.8)
Introducing the notation z k= z /γk, for k =1,2,3, we may call function F k= F ( x,y,zk ) harmonic. Note the property m 1 m 2=1, which seems to have escaped the attention of previous researchers, and which will help us to simplify various expressions to follow. The other elastic constants which will be used throughout the book are: G 1 = β + γ 1γ 2 H ,
H =
(γ1 + γ2) A 11 2 2π( A 11 A 33 − A 13)
,
G 2 = β − γ 1γ 2 H , α =
( A 11 A 33)1/2 − A 13 A 11(γ1 + γ2)
,
β =
γ3 2π A 44
. (2.1.9)
Introduce the following inplane stress components: σ1 = σx + σy ,
σ2 = σx − σy + 2iτxy ,
τz = τzx + iτyz. (2.1.10)
This will simplify expressions (2.1.1), namely σ1 = ( A 11 − A 66)(Λu + Λ u ) + 2 A 13
∂w , ∂z
σ2 = 2 A 66Λ u ,
76
CHAPTER 2.
σz =
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
1 ∂w , A (Λu + Λ u ) + A 33 2 13 ∂z
τz = A 44
∂u
∂z
+ Λ w .
(2.1.11) We have now only four components of stress, instead of six, as it was in (2.1.1). The substitution of (2.1.6) in (2.1.11) yields: σ1 = 2 A 66
∂2 2 2 2 2 2 {[γ1 − (1 + m 1)γ3] F 1 + [γ2 − (1 + m 2)γ3] F 2} , ∂z
σ2 = 2 A 66Λ2( F 1 + F 2 + i F 3) , σz = A 44
∂2 2 2 2 [(1 + m 1)γ1 F 1 + (1 + m 2)γ2 F 2] ∂z
= − A 44 ∆[(1 + m 1) F 1 + (1 + m 2) F 2] , τz = A 44Λ
∂ [(1 + m 1) F 1 + (1 + m 2) F 2 + i F 3] . ∂z
(2.1.12)
Here we used the fact that each F k satisfies equation (2.1.7), and the relation: A 11γ2k− A 13 m k= A 44(1+ m k), (for k =1,2) which is an immediate consequence of (2.1.8). Expressions (2.1.6) and (2.1.12) give a general solution, expressed in terms of three harmonic functions F k. It is very attractive to express each function F k through just one harmonic function as follows: F k( x,y,z) = c k F ( x,y,zk ) ,
(2.1.13)
where z k= z /γk, and c k is an as yet unknown complex constant. As we shall see further, this is possible indeed. All the results obtained in the book are valid for isotropic solids, provided that we take γ1 = γ2 = γ3 = 1, β =
1 + ν , πE
G1 =
H =
1 − ν2 , πE
(2 − ν)(1 + ν) , πE
α =
1 − 2ν , 2(1 − ν)
G2 =
ν(1 + ν) , πE
where E is the elastic modulus, and ν is Poisson coefficient.
(2.1.14)
77
2.2 Point force solutions
1.
Exercise 2.1 Establish the equivalence of (2.1.3) and (2.1.4)
2. Prove that the solution (2.1.6) satisfies (2.1.3), provided that the condition (2.1.7) is met. 3.
Prove the identity m 1 m 2=1.
4.
Prove the identity A 11γ2k− A 13 m k= A 44(1+ m k).
2.2 Point force solutions The field of stresses and displacements due to a concentrated load is important for the integral equation formulation of various mixed boundary value problems. Two cases are considered here: an arbitrary point load in a transversely isotropic elastic space, and the action of an arbitrary concentrated force on the boundary of a similar half-space. Though these problems have been solved by many authors, we follow here the results given in (Fabrikant, 1970). The main reason for this is the simplification of the elastic coefficients, which seems to have escaped the attention of other authors. Here is an example: one of the coefficients in (Chen, 1966) reads A 13 + A 44
A m1
33 2 2 A 11 A 44(γ1 − γ2) γ21
− A 13.
This expression, after simplification, reduces to 1/(1− m 2). Let a point force, with components T x, T y, and P in Cartesian coordinates be applied at the point N 0 inside a transversely isotropic elastic space. We may assume, without loss of generality, that the polar cylindrical coordinates of N 0 are (ρ0,φ0,0). We need to find the field of stresses and displacements at the point M (ρ,φ, z ). Introduce the complex tangential force T = T x+i T y. The general solution can be expressed through the three potential functions: F1 =
1 1 γ m (Λχ + Λχ ) + P ln( R + z ), 1 1 1 1 4π A 44( m 1 − m 2) 2 1 2
F2 = −
1 1 γ m (Λχ + Λχ ) + P ln( R + z ), 2 2 2 2 4π A 44( m 1 − m 2) 2 2 1
78
CHAPTER 2.
F3 = i
γ3 8π A 44
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
(Λχ3 − Λχ3). (2.2.1)
Here the notation was introduced χk( z ) = χ( z k), R k = [ρ2 + ρ20 − 2ρρ0cos(φ−φ0) + z 2k]1/2, for k =1,2,3; χ( z ) = T [ z ln( R 0 + z ) − R 0].
(2.2.2)
The displacements are defined by (2.1.6) as follows:
u =
−
+
1 1 T q2T γ1 m 2− + 2 4π A 44( m 1 − m 2) 2 R R 1( R 1 + z 1) 1
z z T q2T 1 − P 1 − 2 γ2 m 1− + R 2 2 R2 R 2( R 2 + z 2)2 q R 1
γ3
q2T T + , 2 8π A 44 R 3 R 3( R 3 + z 3)
(2.2.3)
z1 z2 m1 m2 1T T 1 . w = − − + + + P R 2 γ 2 R 2 4π A 44( m 1 − m 2) 2q γ1 R 1 q R1
(2.2.4) Here q = ρei
φ
iφ
− ρ0e 0.
The stress field is defined by (2.1.12). σz and τz. Here they are:
(2.2.5) We shall need only the expressions for
γ1 γ2 1 1 σz = − (T q + Tq) + 4π2 ( m 1 − 1) R 31 ( m 2 − 1) R 32
m1 z1 m2 z2 , + P + ( m 1 − 1) R 31 ( m 2 − 1) R 32
79
2.2 Point force solutions
(2.2.6) z1
z3
T + − 3 3 3 8π ( m 1 − 1) R 1 ( m 2 − 1) R 2 R 3
τz =
−
z2
2R 3 + z3 2R 2 + z2 2R 1 + z1 Tq 2 + + 8π ( m 1 − 1) R 31( R 1 + z 1)2 ( m 2 − 1) R 32( R 2 + z 2)2 R 33( R 3 + z 3)2 m1 m2 Pq + − 3 . 4π γ1( m 1 − 1) R 31 γ2( m 2 − 1) R 2
(2.2.7)
Consider a transversely isotropic elastic half-space z ≥0. Let a concentrated force, with components T x, T y, and P , be applied at the point N 0(ρ0,φ0,0). We need to find the field of stresses and displacements in the half-space. The potential functions are defined by
F1 =
F2 =
H γ1
1 γ (Λχ + Λχ ) + P ln( R + z ), 1 1 1 1 m 1 − 1 2 2 H γ2
1 γ (Λχ + Λχ ) + P ln( R + z ), 2 2 2 2 m 2 − 1 2 1 γ3
F3 = i
4π A 44
(Λχ3 − Λχ3).
(2.2.8)
Substitution of (2.2.8) in (2.1.6) yields
u =
γ3 4π A 44
q2T T + R 3 R 3( R 3 + z 3)2
+
1 T q2T Pq + + γ1 − m2 − 1 2 R2 R 2( R 2 + z 2) R 2( R 2 + z 2)2
+
1 T q2T Pq + + , γ2 − m1 − 1 2 R1 R 1( R 1 + z 1) R 1( R 1 + z 1)2
H γ2
H γ1
(2.2.9)
80
CHAPTER 2.
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
γ2 m 1 γ1 m 2 1 + w = H (T q + Tq) 2 ( m 2 − 1) R 2( R 2 + z 2) ( m 1 − 1) R 1( R 1 + z1)
m1 m2 . + P + ( m 2 − 1) R 2 ( m 1 − 1) R 1
(2.2.10)
We shall need expressions for the following stress components: σz =
1 1 1 1 γ1γ2( T q + T q ) + Pz − 3 + 3, 2π(γ1 − γ2) 2 R 1 R 2
γ2
Tz
T q 2(2 R + z )
γ
Tz
T q 2(2 R + z )
1 1 1 2 2 1 − − 2 − τz = 3 3 2 3 3 2 4π(γ1 − γ2) R 1 4 ( ) π γ − γ R 1( R 1 + z 1) R 2( R 2 + z 2) 1 2 R2
T q 2(2 R 3 + z 3) 1 Tz 3 Pq + − 1 + 1 . + 3 − 2 4π R 33 2 ( ) π γ − γ R 3( R 3 + z 3) R 31 R 32 1 2
(2.2.11)
Expressions (2.2.9) and (2.2.10) simplify for the case when z =0 u =
1 1 P T Tq2 + G2 , G1 3 − Hα 2 2 R R q
T P w = H α ℜ + H . R q
(2.2.12)
(2.2.13)
Here H , α, G 1, and G 2 are defined by (2.1.9), and R = [ρ2 + ρ20 − 2ρρ0cos(φ−φ0)]1/2.
(2.2.14)
Expressions (2.2.12) and (2.2.13) will be used for the integral equation formulation of various mixed boundary value problems in an elastic half-space. The following classification of mixed boundary value problems may be suggested. The problem is called internal mixed when the normal/tangential displacements are prescribed inside a finite domain, while the relevant tractions are prescribed on the rest of the half-space boundary. In the case when the displacements are given outside a finite domain, the problem is called external.
81
2.3 Internal mixed problem of type I
We can specify two types of internal problems. The internal problem of type I : the normal displacements are prescribed inside a finite domain S , the normal traction is given outside the domain S , while the tangential tractions are known all over the plane z =0. The internal problem of type II : the tangential displacements are prescribed inside S , and the shear tractions are given outside, while the normal traction is known all over the plane z =0. The external problems of types I and II are defined in the same way as the internal ones above, with an interchange of the terms traction and displacement. These are the four types of problems which will be considered in this chapter. We shall call a problem mixed-mixed when the boundary conditions are mixed with respect to both normal and tangential components. These problems will be considered in the next chapter.
1.
Exercise 2.2 Establish (2.2.3) and (2.2.4).
2.
Verify (2.2.6) and (2.2.7).
3. Derive expressions for σ1 and σ2 in both cases of concentrated load, considered in section 2.2. 4.
Derive the equivalent solutions for an isotropic body.
5. Consider the half-space.
case of
arbitrary
point
loading
applied inside
an elastic
2.3 Internal mixed problem of type I. Consider a transversely isotropic elastic half-space z ≥0. Introduce a set of polar cylindrical coordinates (ρ,φ, z ). Let the following boundary conditions be prescribed on the plane z =0: w = w (ρ,φ),
ρ≤ a , 0≤φ<2π,
σ = σ(ρ,φ),
ρ> a , 0≤φ<2π,
τ = τ(ρ,φ),
0≤ρ<∞, 0≤φ<2π,
(2.3.1) Here σ stands for the normal loading, and τ is the complex shear loading, namely, τ=τzx+iτyz. The governing integral equation can be written by using (2.2.13), namely,
82
CHAPTER 2.
2π a
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
σ(ρ ,φ ) ρ dρ dφ
0 0 0 0 0 = f (ρ,φ). H⌠ ⌠ ⌡ ⌡ [ρ2 + ρ20 − 2ρρ0cos(φ−φ0)]1/2 0
(2.3.2)
0
We use the same notation σ for the unknown normal loading inside the circle ρ≤ a , as well as for the prescribed function σ outside the circle. This should not create any confusion since the argument (ρ0,φ0) provides a clear distinction. Function f is known from the conditions (2.3.1), and is 2π ∞
σ(ρ0,φ0) ρ0dρ0dφ0
f (ρ,φ) = w (ρ,φ) − ⌠ ⌠
⌡ ⌡ [ρ2 + ρ20 − 2ρρ0cos(φ−φ0)]1/2 0 a
2π ∞
− H αℜ⌠ ⌠
τ(ρ0,φ0) ρ0dρ0dφ0
⌡⌡ 0
iφ
ρe
0
i φ0
.
(2.3.3)
− ρ0e
As soon as equation (2.3.2) is solved, and the value of σ inside the circle becomes known, the tangential displacements in the plane z =0 can be defined by (2.2.12): 2π ∞
τ(ρ0,φ0) ρ0dρ0dφ0 1 ⌠⌠ G1 2 ⌡ ⌡ [ρ2 + ρ20 − 2ρρ0cos(φ−φ0)]1/2
u =
0
2π ∞
+
1 ⌠⌠ G 2 2⌡ ⌡ 0
0
i φ0
[ρei φ − ρ0e
]2τ(ρ0,φ0) ρ0dρ0dφ0
[ρ2 + ρ20 − 2ρρ0cos(φ−φ0)]3/2
0
2π ∞
− H α⌠ ⌠
⌡⌡
0
Integral equation (2.3.2) was solved in section 1.4. here a more general case: 2π a
H⌠ ⌠
σ(ρ0,φ0) ρ0dρ0dφ0
⌡ ⌡ [ρ2 + ρ20 − 2ρρ0cos(φ−φ0)](1+κ)/2 0
0
σ(ρ0,φ0) ρ0dρ0dφ0 -i φ
ρe
-i φ0
.
− ρ0e
(2.3.4) It seems useful to consider
= f (ρ,φ),
(2.3.5)
0
where −1<κ<1. This type of equation arises in the problems of nonhomogeneous elastic half-space, with the modulus of elasticity E being a power function of z , namely, E = E 0 z κ. Of course, in the nonhomogeneous case H is no longer defined
83
2.3 Internal mixed problem of type I
by (2.1.9). The reader is referred to the paper by Rostovtsev (1964) for details. Rostovtsev (1964) obtained an exact solution of (2.3.5) in Fourier series. Here we present a closed form solution. By using the integral representation (1.1.4), integral equation (2.3.5) can be rewritten as ρ
πκ ⌠ 4 H cos 2 ⌡
κ
a
x dx (ρ2 − x 2)(1+κ)/2
0
ρ dρ
0 0 x ⌠ 2 2 (1+κ)/2 L ρρ σ(ρ0,φ) = f (ρ,φ). 0 ⌡ (ρ0 − x ) 2
(2.3.6)
x
Integral equation (2.3.6) represents a sequence of two Abel operators and one L-operator. The solution procedure is similar to that of (1.4.5). The first operator to be applied to both sides of (2.3.6) is t
ρdρ 1 d ⌠ L 2 2 (1-κ)/2 L(ρ). t t d ⌡ (t − ρ )
(2.3.7)
0
The result of application of (2.3.7) to both sides of (2.3.6) is a
2π Ht
κ⌠
ρ0dρ0
t
2 2 (1+κ)/2 ⌡ (ρ0 − t )
t
ρdρ t 1 d ⌠ L σ(ρ0,φ) = L L(ρ) f (ρ,φ). (2.3.8) 2 ρ0 t d t ⌡ ( t − ρ2)(1-κ)/2 0
The second operator to be applied to both sides of (2.3.8) is a
d⌠ t 1-κd t 1 L( y ) L , κ)/2 2 2 (1d y⌡ ( t − y ) t y
with the result cos(πκ/2) d⌠ σ( y ,φ) = − L( y ) 2 d y⌡ π Hy y
a
(t2
t 1-κd t − y 2)(1-κ)/2
84
CHAPTER 2.
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
t
ρdρ 1 d ⌠ ×L 2 2 2 (1-κ)/2 L(ρ) f (ρ,φ). t d t ⌡ (t − ρ ) 0
(2.3.9) The rules of differentiation of integrands and the properties of the L-operators allow us to rewrite (2.3.9) in the form a
cos(πκ/2) Φ( a , y ,φ) ⌠ 2 d t2 (1-κ)/2 d Φ( t , y ,φ). σ( y ,φ) = 2 2 2 (1-κ)/2 − dt πH (a − y ) ⌡ (t − y )
(2.3.10)
y
Here t
Φ( t , y ,φ) =
1 t 1+κ
1+κ ⌠ 2 ρ d2ρ(1-κ)/2 d ρ1-κLρ2y f (ρ,φ). dρ t ⌡ (t − ρ )
(2.3.11)
0
Yet another form of solution can be found in (Fabrikant, 1971e). The problem solved above has two major applications: contact problems of a smooth punch pressed against an elastic half-space, and that of an external circular crack in an infinite elastic body. Let us consider both cases in more detail. Example 1. The smooth punch problem. In elastic contact problems, we have σ=0, for ρ> a , and τ=0 all over the plane z =0, so that function f = w (see 2.3.9). It becomes possible to compute the resultant force P and the tilting moments M x and M y directly in terms of the prescribed displacement w . Since 2π a
P = ⌠ ⌠ σ(ρ,φ) ρdρdφ, ⌡⌡ 0
(2.3.12)
0
substitution of (2.3.9) in (2.3.12) yields directly the resultant force 2π a
cos(πκ/2) ⌠ w (ρ,φ) ρdρdφ ⌠ P = . ⌡ ⌡ ( a 2 − ρ2)(1-κ)/2 π2 H 0
(2.3.13)
0
For computation of the tilting moments M x and M y, it is convenient to introduce the complex parameter
85
2.3 Internal mixed problem of type I
2π a
M = M x + i M y = −i ⌠ ⌠
⌡⌡
0
σ(ρ,φ) ei φ ρ2dρdφ.
(2.3.14)
0
By using (2.3.9), the final expression for the tilting moment is found to be φ w (ρ,φ) ei ρ2dρdφ ⌠ ⌠ . M = −i 2 π H (1 + κ) ⌡ ⌡ ( a 2 − ρ2)(1-κ)/2 0 2π a
2cos(πκ/2)
(2.3.15)
0
Expressions (2.3.14) Rostovtsev (1964).
and
(2.3.15)
are
in
agreement
with
similar
results
of
By reviewing the derivation of expression (2.3.6), one may find that it is valid for evaluating the normal displacements outside the contact region, if the upper limit of integration ρ is replaced by a . Substitution of (2.3.9) into the modified form of (2.3.6) results in a x ρ0dρ0 ρ 2cos(πκ/2) ⌠ dx d ⌠ 0w(ρ ,φ), L w (ρ,φ) = 2 2 (1+κ)/2 d x 2 2 (1-κ)/2 π ρ 0 ⌡ (ρ − x ) ⌡ ( x − ρ0) 0
0
for ρ> a . (2.3.16) Performing differentiation of the integrand, and then integrating by parts, we obtain 2π
a w (ρ0,φ0) ρ0dρ0dφ0 1 πκ 2 2 (1-κ)/2⌠ ⌠ w (ρ,φ) = 2cos( )(ρ − a ) , 2 ⌡ ⌡ ( a 2−ρ20)(1-κ)/2[ρ2 + ρ20 − 2ρρ0cos(φ−φ0)] π 0
0
for ρ> a . (2.3.17) Here the following identities were employed (Bateman and Erde´ lyi, 1955) d (1+κ)/2 1+κ 1+κ 3+κ 1+κ -(1-κ)/2 ζ F , ; ; ζ = ζ (1 − ζ)-(1+κ)/2, dζ 2 2 2 2 x
x
d ⌠ df(t) f(t)t dt = f (0) x κ + x ⌠ 2 2 (1-κ)/2. d x ⌡ ( x 2 − t 2)(1-κ)/2 ⌡ (x − t ) 0
(2.3.18)
0
All the quantities of interest, namely, the pressure exerted by the punch σ, the resultant force P , the tilting moment M , and the normal displacement outside the
86
CHAPTER 2.
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
punch, can be expressed directly through the prescribed normal displacement w by formulae (2.3.9), (2.3.13), (2.3.15), and (2.3.17) respectively. Example 2.
External crack in non-homogeneous elasticity.
κ
Consider a
non-homogeneous elastic space with modulus of elasticity E = E 0| z | , E 0=const, |κ|<1. This space is weakened by a circular external crack ρ≥ a . An arbitrary pressure σ(ρ,φ) is applied to both faces of the crack in opposite directions. The problem is to find the normal stress in the crack neck, the normal displacements of the crack faces, the stress intensity factor, and the work required to open up the crack. Due to the symmetry of the problem, it may be reduced to the mixed boundary value problem of a half-space, subject to the boundary conditions at z =0: w = 0,
τ = 0,
for 0≤ρ≤ a ,
0≤φ<2π;
σ = σ(ρ,φ),
τ = 0,
for a <ρ<∞,
0≤φ<2π.
(2.3.19)
The governing integral equation takes the form (2.3.5), with the known function 2π ∞
σ(ρ0,φ0) ρ0dρ0dφ0
f (ρ,φ) = − H ⌠ ⌠
.
(2.3.20)
⌡ ⌡ [ρ2 + ρ20 − 2ρρ0cos(φ−φ0)](1+κ)/2
0
a
Its solution can be found in exactly the same manner as that of (1.5.21), and is
σ(ρ,φ) = −
cos(πκ/2) π (a 2
2
− ρ)
2 (1-κ)/2
2π ∞
⌠⌠ ⌡⌡ 0 a
(ρ20 − a 2)(1-κ)/2σ(ρ0,φ0) ρ0dρ0dφ0 ρ2 + ρ20 − 2ρρ0cos(φ−φ0)
.
(2.3.21)
Expression (2.3.21) gives the normal stress in the crack neck through the pressure applied to the crack faces. Note that (1.5.24) may be considered as a particular case of (2.3.21), when κ=0. The normal displacement of the crack faces can be evaluated as a superposition of the displacement caused by the applied pressure, and the displacement due to the normal stress in the crack neck. By using a procedure analogous to the one described in section 1.5, we can obtain the expression ∞ x ρ0dρ0 ρρ x κd x πκ ⌠ 0σ(ρ ,φ) ⌠ w (ρ,φ) = 4 H cos L 2 2 (1+κ)/2 2 2 (1+κ)/2 2 x2 0 ⌡ ( x − ρ0) ρ⌡ ( x − ρ ) a
87
2.3 Internal mixed problem of type I
κ
ρ0dρ0 x2 ⌠ ⌠ + 2 2 (1+κ)/2 2 2 (1+κ)/2 L ρρ σ(ρ0,φ), for ρ> a . 0 ⌡ (ρ − x ) ⌡ (ρ0 − x ) a
a
x dx
0
(2.3.22)
x
Substitution of (2.3.21) in (2.3.22) leads, after simplification, to ρ
∞
x κd x πκ ⌠ ⌠ w (ρ,φ) = 4 H cos 2 ⌡ (ρ2 − x 2)(1+κ)/2 ⌡ a
x
ρ0dρ0 (ρ20 − x 2)(1+κ)/2
x2 L σ(ρ0,φ), ρρ0
for ρ> a . (2.3.23) The normal displacements of the crack faces are now defined in terms of the applied pressure. Introduce the stress intensity factor as k 1(φ) = lim[( a − ρ)(1-κ)/2σ(ρ,φ)].
(2.3.24)
ρ→ a
Substitution of (2.3.21) in (2.3.24) gives
k 1(φ) =
2cos(πκ/2) π(2 a )(1-κ)/2
∞
ρ dρ
0 0 a ⌠ 2 2 (1+κ)/2 L ρ σ(ρ0,φ). 0 ⌡ (ρ0 − a )
a
Introduce the stress intensity function : 2cos(πκ/2)
K 1(ρ,φ) =
π(2ρ)(1-κ)/2
∞
ρ dρ
0 0 ρ ⌠ 2 2 (1+κ)/2 L ρ σ(ρ0,φ). 0 ⌡ (ρ0 − ρ )
(2.3.25)
ρ
It is obvious that the limiting case of the stress intensity function, when ρ→ a , is the stress intensity factor. By using the property of the L-operators (1.2.3) we may rewrite (2.3.23) as ρ
∞
κ
x dx πκ ⌠ x w (ρ,φ) = 4 H cos L ⌠ κ)/2 2 2 (1+ 2 ⌡ (ρ − x ) ρ ⌡ a
x
ρ0dρ0 (ρ20
− x)
2 (1+κ)/2
x L σ(ρ0,φ) ρ0
88
CHAPTER 2.
ρ
= 2
πH ⌠
(3-κ)/2
⌡
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
x (1+κ)/2d x x 2 (1+κ)/2 L ρ K 1( x ,φ). (ρ − x ) 2
(2.3.26)
a
The energy W may be defined by the integral 2π ∞
W = ⌠⌠
σ(ρ,φ) w (ρ,φ)ρ dρdφ.
⌡⌡ 0
(2.3.27)
a
Substitution of (2.3.26) in (2.3.27) gives ρ
∞
2π
x (1+κ)/2 d x x 2 2 (1+κ)/2 L ρ K 1( x ,φ) ⌡ (ρ − x )
W = 2(3-κ)/2π H ⌠ dφ ⌠ σ(ρ,φ)ρ dρ ⌠
⌡
⌡
0
a
a
∞
2π
∞
σ(ρ,φ)ρ dρ x π H ⌠ dφ ⌠ x (1+κ)/2d x ⌠ 2 2 (1+κ)/2 L ρ K 1( x ,φ) ⌡ ⌡ ⌡ (ρ − x ) 0
(3-κ)/2
= 2
a
∞
2π
= 2
x
∞
ρ dρ x π H ⌠ dφ ⌠ K 1( x ,φ) x (1+κ)/2d x ⌠ 2 2 (1+κ)/2 L ρ σ(ρ,φ). ⌡ ⌡ ⌡ (ρ − x ) 0
(3-κ)/2
a
x
Here the interchange of the order of integration was used twice. comparison of the last expression with (2.3.25) yields the final result 2π ∞
2 πH ⌠⌠ [ K 1(ρ,φ)]2ρ dρ dφ. cos(πκ/2) ⌡ ⌡ 1-κ 2
W =
Now
0
(2.3.28)
a
Expression (2.3.28) interprets the stress intensity function squared as being proportional to the energy per unit area required to open up the crack. In the case of axial symmetry, formula (2.3.28) simplifies to ∞
22-κπ3 H ⌠ W = [ K (ρ)]2ρ dρ, cos(πκ/2) ⌡ 1 a
89
2.3 Internal mixed problem of type I
with ∞
2cos(πκ/2)
K 1(ρ) =
σ(ρ ,φ )ρ dρ
⌠ 2 0 0 20 (1+κ)/20 . ⌡ (ρ0 − ρ )
π(2ρ)(1-κ)/2
ρ
All the results obtained become valid for a transversely isotropic space provided that κ=0, and H is defined by (2.1.9). When H is defined by (2.1.14), we have the results for isotropic body, namely, 2π ∞
1 − ν2 ⌠ ⌠ W = 2π E ⌡⌡ 0
[ K 1(ρ,φ)]2ρ dρ dφ,
(2.3.29)
a
with ∞
x dx ρ σ( x ,φ). 2 1/2 L x (x − ρ )
√2 ⌠ K 1(ρ,φ) = π√ρ ⌡
2
ρ
(2.3.30)
The case of axial symmetry simplifies (2.3.29) and (2.3.30) as follows: ∞
1 − ν2 ⌠ W = 4π E ⌡ 2
[ K 1(ρ)]2ρ dρ,
(2.3.31)
a
with ∞
K 1(ρ) =
√2 ⌠ π√ρ ⌡ ρ
σ( x ) x d x . ( x 2 − ρ2)1/2
(2.3.32)
Exercise 2.3. 1. The normal displacements under a flat circular punch are given by w (ρ,φ)= w 0+θρcosφ, with w 0=const, and θ=const. Find the traction distribution σ exerted by the punch. cos(πκ/2) w 0 + (2θρcosφ)/(1 + κ) Answer: σ(ρ,φ) = . ( a 2 − ρ2)(1-κ)/2 π2 H 2. In the problem above find the relationships between the applied force P , the tilting moment M , and the punch settlement w 0 and the inclination angle θ.
90
CHAPTER 2.
Answer: P =
2 w 0 a 1+κcos(πκ/2) π H (1 + κ)
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
M =
,
4θ a 3+κcos(πκ/2) . π H (1 + κ)(3 + κ)
3. The normal displacements under a paraboloidal punch are w (ρ,φ)= w 0− c ρ2, with w 0=const, and c =const. Find the traction distribution σ and the radius of contact a . Solution : utilization of (2.3.9) yields σ(ρ) =
cos(πκ/2) 2 c [ a 2(1 − κ) − 2ρ2] w + 0 . π2 H ( a 2 − ρ2)(1-κ)/2 (1 + κ)2
The radius of contact a is found from the condition σ( a )=0. a =[(1 + κ) w 0/(2 c )]1/2, and the final expression for the traction is σ(ρ) =
2 w 0cos(πκ/2) π a H (1 + κ) 2 2
The result is
( a 2 − ρ2)(1+κ)/2.
4. In the problem above find the relationship between the punch settlement w 0 and the applied force P . 4 a 1+κcos(πκ/2) Answer: P = . π H (1 + κ)(3 + κ) 5. Consider an external circular crack ρ> a . Find the traction distribution σ in the crack neck due to the action of a pair of equal concentrated forces P , applied normally to the crack faces in opposite directions at the point ( b ,ψ). P πκ b 2 − a 2(1-κ)/2 1 Answer: σ(ρ,φ) = − 2 cos 2 . 2 2 2 2 a − ρ π ρ + b − 2 b ρcos(φ−ψ) 6. Consider an external circular crack ρ> a . Let a uniform pressure σ0 be applied in opposite directions to the annulus b ≤ρ≤ c , ( b > a ), the rest of the crack faces being traction free. Find the stress distribution σ in the crack neck and the stress intensity factor k 1. Answer: σ(ρ) = −
2σ0cos(πκ/2)
c
⌠
π( a 2 − ρ2)(1-κ)/2 ⌡ b
(ρ20 − a 2)(1-κ)/2ρ0 dρ0 ρ20 − ρ2
, for ρ< a .
In general, the last integral can be computed in terms of hypergeometric functions. In the particular case of an isotropic body (κ=0), the integral is computable in elementary functions:
91
2.3 Internal mixed problem of type I
σ(ρ) = −
2 ( c 2 − a 2)1/2 − ( b 2 − a 2)1/2 σ0 π ( a 2 − ρ2)1/2
2 c 2 − a 2 1/2 − a 21/2 -1b + tan . − tan-1 2 a − ρ2 a 2 − ρ2
The stress intensity factor is 2σ0cos(πκ/2)[( c 2 − a 2)(1-κ)/2 − ( b 2 − a 2)(1-κ)/2] k1 = − . π(2 a )(1-κ)/2(1 − κ) 7.
In the preceding problem find the crack opening displacement w . min(ρ,c)
⌠ 4cos(πκ/2) Answer: w (ρ) = σ0 ⌡ 1 − κ a min(ρ,b)
−
⌠ ⌡ a
8.
( c 2 − x 2)(1-κ)/2 κ x dx (ρ2 − x 2)(1+κ)/2
( b 2 − x 2)(1-κ)/2 κ x x d , for ρ> a . (ρ2 − x 2)(1+κ)/2
By using formula (2.3.9), prove the identity 2π a
⌠ ⌠ σ(ρ,φ)ρ1+|n| ein φ dρ dφ ⌡⌡ 0 0
2π a
(1 + κ)Γ(1 + | n |) πκ ⌠ ⌠ f (ρ,φ)ρ1+|n| ein φ dρ dφ = cos . 2 ⌡⌡ ( a 2 − ρ2)(1-κ)/2 2π2Γ[| n | + (1 + κ)/2] 0 0
Note : in the particular cases n =0 and n =−1, the last identity transforms into (2.3.13) and (2.3.15) respectively. 9.
Define the stress intensity function K 1(ρ,φ) in terms of the displacement w . ρ
cos(πκ/2) 1 d ⌠ x dx Answer: K 1(ρ,φ) = (1-κ)/2 2 (1+κ)/2 L 2 2 (1-κ)/2 L( x ) w ( x ,φ). d ρ ρ 2 π Hρ ⌡ (ρ − x ) a
Hint : perform the inversion of (2.3.26). 10.
Prove the identity
92
CHAPTER 2.
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
d ρ( x 2 − a 2)(1-κ)/2 π(1 − κ) f ( a ) ⌠ 2 f x x . ( ) d = lim 2 (1-κ)/2 ρ 2cos(πκ/2) d ρ→ a (ρ − x ) ⌡ a Hint : use the substitution t =(ρ2− x 2)/( x 2− a 2). 11. Use the identity above to express the stress intensity factor in terms of the displacement w . w (ρ,φ) 1 − κ Answer: k 1(φ) = 2-κ lim . ρ → a (ρ − a )(1-κ)/2 2 πH Hint : compute the limit ρ→ a of the result in Exercise 9.
2.4 External mixed problem of type I. The problem is characterized by the following mixed boundary conditions on the plane z =0: w = w (ρ,φ),
ρ> a , 0≤φ<2π,
σ = σ(ρ,φ),
ρ≤ a , 0≤φ<2π,
τ = τ(ρ,φ),
0≤ρ<∞, 0≤φ<2π,
Here the same notation is used as in the previous section. integral equation can be written by using (2.2.13), namely, 2π ∞
σ(ρ0,φ0) ρ0dρ0dφ0
H⌠ ⌠
⌡ ⌡ [ρ2 + ρ20 − 2ρρ0cos(φ−φ0)]1/2 0
= f (ρ,φ).
(2.4.1) The governing
(2.4.2)
a
We use again the same notation σ for the unknown normal loading inside the circle ρ≤ a , as well as for the prescribed function σ outside the circle. Function f is known from the second condition (2.4.1), and is 2π a
σ(ρ0,φ0) ρ0dρ0dφ0
f (ρ,φ) = w (ρ,φ) − ⌠ ⌠
⌡ ⌡ [ρ2 + ρ20 − 2ρρ0cos(φ−φ0)]1/2 0 0
2π ∞
− H αℜ⌠ ⌠
⌡⌡ 0 0
τ(ρ0,φ0) ρ0dρ0dφ0 iφ
ρe
i φ0
− ρ0e
.
(2.4.3)
93
2.4 External mixed problem of type I
As soon as equation (2.4.2) is solved, and the value of σ inside the circle becomes known, the tangential displacements in the plane z =0 can be defined by (2.3.4). Integral equation (2.4.2) was solved in section 1.5. We consider again a more general case: 2π ∞
σ(ρ0,φ0) ρ0dρ0dφ0
H⌠ ⌠
⌡ ⌡ [ρ2 + ρ20 − 2ρρ0cos(φ−φ0)](1+κ)/2 0
= f (ρ,φ),
(2.4.4)
a
where −1<κ<1.
The new method allows us to present a closed form solution.
By using the integral representation (1.1.21), for z =0, the integral equation (2.4.4) can be rewritten as ∞
πκ ⌠ 4 H cos 2 ⌡ ρ
x κd x
x
( x 2 − ρ2)(1+κ)/2
ρ dρ
ρρ
0 0 0σ(ρ ,φ) = f (ρ,φ). ⌠ 2 2 (1+κ)/2 L x2 0 ⌡ ( x − ρ0)
(2.4.5)
a
Integral equation (2.4.5) represents a sequence of two Abel operators and one The solution procedure is similar to that of (1.5.2). The first L-operator. operator to be applied to both sides of (2.4.5) is ∞
d ⌠ ρdρ 1 L( t ) L . 2 2 (1κ)/2 d t ⌡ (ρ − t ) ρ
(2.4.6)
t
The result of application of (2.4.6) to both sides of (2.4.5) is ρ0dρ0
t
−2π Ht κ ⌠
2 2 (1+κ)/2 ⌡ ( t − ρ0)
ρ0 L σ(ρ0,φ) t
a
∞
d ⌠ ρdρ 1 = L( t ) L f (ρ,φ). d t ⌡ (ρ2 − t 2)(1-κ)/2 ρ t
The second operator to be applied to both sides of (2.4.7) is
(2.4.7)
94
CHAPTER 2.
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
y
1 d⌠ t 1-κd t L L( t ) , y d y⌡ ( y 2 − t 2)(1-κ)/2 a
with the result y
cos(πκ/2) 1 d ⌠ t 1-κd t σ( y ,φ) = − L y d y ⌡ ( y 2 − t 2)(1-κ)/2 π2 Hy a
∞
d ⌠ ρdρ 1 ×L( t 2) L f (ρ,φ). 2 2 (1κ)/2 d t ⌡ (ρ − t ) ρ
(2.4.8)
t
The rules of differentiation of integrands and the properties of the L-operators allow us to rewrite (2.4.8) in the form y
cos(πκ/2) Φ( a , y ,φ) ⌠ 2 d t2 (1-κ)/2 d Φ( t , y ,φ). σ( y ,φ) = − 2 2 2 (1-κ)/2 − dt ( y − a ) πH ⌡ (y − t )
(2.4.9)
a
Here ∞
Φ( t , y ,φ) = t
1-κ
2 ⌠ 2 dρ2 (1-κ)/2 d L t f (ρ,φ). dρ ρ y ⌡ (ρ − t )
(2.4.10)
t
We can consider again two major applications: contact problems of a smooth punch pressed against an elastic half-space, and that of a penny-shaped crack in an infinite elastic body. Let us consider both cases in more detail. Example 1. The smooth punch problem. In elastic contact problems we have σ=0, for ρ< a , and τ=0 all over the plane z =0, so that function f = w . It becomes possible to compute the resulting force P and the tilting moments M x and M y directly in terms of the prescribed displacement w . Since 2π ∞
P = ⌠⌠
⌡⌡
0
σ(ρ,φ) ρdρdφ,
a
substitution of (2.4.8) in (2.4.11) yields directly the resultant force
(2.4.11)
95
2.4 External mixed problem of type I
∞ 2-κ cos(πκ/2) y t t d ⌠ ⌠ 2 dρ2 (1-κ)/2 d ⌠ w(ρ,φ) dφ. P = lim− 2 2 2 (1-κ)/2 dρ ⌡ πH y→∞ ⌡ (y − t ) ⌡ (ρ − t ) 0 a t 2π
(2.4.12) The tilting moment can be found in a similar manner. We can also express the normal displacement inside the circle ρ≤ a directly in terms of the prescribed displacement w outside the circle. We substitute (2.4.8) in (2.4.5) keeping in mind that, for ρ≤ a , the lower limit of integration of the first integral will be a instead of ρ. By using the properties of Abel operators and the L-operators, the following expression can be obtained ∞
∞
2 πκ ⌠ dx d ⌠ w (ρ,φ) = − cos π 2 ⌡ ( x 2 − ρ2)(1+κ)/2 d x ⌡ a
x
ρ0dρ0 (ρ20
− x)
2 (1-κ)/2
ρ L w (ρ0,φ). ρ0
(2.4.13) Carrying out the differentiation of the integrand, interchanging the order of integration, and then integrating with respect to x yields ∞
2cos(πκ/2) ⌠ w (ρ,φ) = − π(1 + κ) ⌡ a
ρ20 − a 2
(1+κ)/2 ρ20 − ρ2
ρ2 − a 2 1 + κ 1 + κ 3 + κ 0 d L ρ w(ρ ,φ) dρ . ×F , ; ; 2 2 0 2 2 2 0 ρ0 − ρ dρ0 ρ0 Integration by parts and the differential properties of the Gauss hypergeometric functions (2.3.18) allow us to simplify the last expression, namely, ∞ ρ0dρ0 2 πκ 2 2 (1-κ)/2⌠ ρ w (ρ,φ) = − cos (a − ρ ) 2 2 (1-κ)/2 2 2 L ρ w (ρ0,φ) π 2 (ρ0 − ρ ) 0 ⌡ (ρ0 − a ) a
∞ w (ρ0,φ0)ρ0 dρ0 dφ0 1 πκ 2 2 (1-κ)/2⌠ ⌠ = − 2 cos ( a −ρ ) . 2 ⌡ ⌡ (ρ20−a 2)(1-κ)/2[ρ2 + ρ20 − 2ρρ0cos(φ−φ0)] π 2π
0
a
(2.4.14) Expression (2.4.14) gives the normal displacements inside a circle ρ≤ a directly in terms of the prescribed displacement outside the circle. We note a certain similarity between (2.3.17) and (2.4.14).
96
CHAPTER 2.
Example 2.
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
Penny-shaped crack in non-homogeneous elasticity. κ
Consider
a non-homogeneous elastic space with modulus of elasticity E = E 0| z | , E 0=const, |κ|<1. This space is weakened by a penny-shaped crack ρ≤ a . The crack is opened by arbitrary pressure σ(ρ,φ). The problem is to find the normal stress on the plane z =0 outside the crack, the crack opening displacement, the stress intensity factor, and the work required to open up the crack. Due to the symmetry of the problem, it may be reduced to the mixed boundary value problem of a half-space, subject to the boundary conditions at z =0: τ = 0,
w = 0, σ = σ(ρ,φ),
τ = 0,
for a <ρ<∞, 0≤φ<2π; for 0≤ρ≤ a , 0≤φ<2π.
(2.4.15)
The governing integral equation takes the form (2.4.2), with the known function 2π a
σ(ρ ,φ ) ρ dρ dφ
0 0 0 0 0 f (ρ,φ) = − H ⌠ ⌠ . ⌡ ⌡ [ρ2 + ρ20 − 2ρρ0cos(φ−φ0)](1+κ)/2 0
(2.4.16)
0
Its solution can be found in exactly the same manner as that of (1.4.25), and is
σ(ρ,φ) = −
cos(πκ/2)
2π a
⌠⌠
π2(ρ2 − a 2)(1-κ)/2 ⌡ ⌡ 0 0
( a 2 − ρ20)(1-κ)/2σ(ρ0,φ0) ρ0dρ0dφ0 ρ2 + ρ20 − 2ρρ0cos(φ−φ0)
.
(2.4.17)
Again, one should notice the similarity between (2.3.21) and (2.4.17). Expression (2.4.17) gives the normal stress in the plane z =0 outside the crack in terms of the pressure applied to the crack faces. Note that (1.4.27) may be considered as a particular case of (2.4.17), when κ=0. The crack opening displacement can be evaluated as a superposition of the displacement caused by the applied pressure, and the displacement due to the normal stress (2.4.17) outside the crack. By using a procedure analogous to the one described in section 1.4, we obtain the expression x ∞ ρ0dρ0 ρρ x κd x πκ ⌠ 0σ(ρ ,φ) ⌠ w (ρ,φ) = 4 H cos L 2 2 (1+κ)/2 2 2 (1+κ)/2 2 x2 0 ⌡ (x − ρ ) ⌡ ( x − ρ0)
a
a
97
2.4 External mixed problem of type I
ρ
x κd x
ρ0dρ0 x 2 σ(ρ ,φ), for ρ< a . ⌠ ⌠ + L 2 2 (1+κ)/2 2 2 (1+κ)/2 ρρ0 0 ⌡ (ρ − x ) ⌡ (ρ0 − x ) a
(2.4.18)
x
0
Substitution of (2.4.17) in (2.4.18) leads, after simplification, to x κd x
ρ0dρ0 ρρ πκ ⌠ 0σ(ρ ,φ), ⌠ w (ρ,φ) = 4 H cos L 2 ⌡ ( x 2 − ρ2)(1+κ)/2 ⌡ ( x 2 − ρ20)(1+κ)/2 x 2 0 a
x
ρ
0
for ρ< a . (2.4.19) The crack opening displacements are now defined in terms of the applied pressure. Introduce the stress intensity factor as k 1(φ) = lim[(ρ − a )(1-κ)/2σ(ρ,φ)].
(2.4.20)
ρ→ a
Substitution of (2.4.17) in (2.4.20) gives
k 1(φ) =
2cos(πκ/2) π(2 a )(1-κ)/2
ρ dρ
a
ρ
0 0 0 ⌠ 2 2 (1+κ)/2 L a σ(ρ0,φ). ⌡ ( a − ρ0)
(2.4.21)
0
Introduce the stress intensity function :
K 1(ρ,φ) =
2cos(πκ/2) π(2ρ)(1-κ)/2
ρ
ρ dρ
ρ
0 0 0σ(ρ ,φ). ⌠ 2 L 2 (1+κ)/2 ρ 0 ⌡ (ρ − ρ0)
(2.4.22)
0
One can see that the limiting case of the stress intensity function, when ρ→ a , is the stress intensity factor. By using the property of the L-operators (1.2.3) we may rewrite (2.4.19) as x ρ0dρ0 ρ x κd x πκ ⌠ ρ ⌠ 0σ(ρ ,φ) w (ρ,φ) = 4 H cos L L 2 ⌡ ( x 2 − ρ2)(1+κ)/2 x ⌡ ( x 2 − ρ20)(1+κ)/2 x 0 a
ρ
0
98
CHAPTER 2.
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
a
x (1+κ)/2d x ρ 2 2 (1+κ)/2 L x K 1( x ,φ). ⌡ (x − ρ )
πH ⌠
(3-κ)/2
= 2
(2.4.23)
ρ
The energy W may be defined by the integral 2π a
W = ⌠ ⌠ σ(ρ,φ) w (ρ,φ)ρ dρdφ.
(2.4.24)
⌡⌡ 0 0
Substitution of (2.4.23) in (2.4.24) gives 2π
a
a
x (1+κ)/2 d x ρ K ( x ,φ) 2 2 (1+κ)/2 L x 1 ⌡ (x − ρ )
π H ⌠ dφ ⌠ σ(ρ,φ)ρ dρ ⌠
(3-κ)/2
W = 2
⌡
⌡
0
ρ
0
2π
a
x
σ(ρ,φ)ρ dρ ρ K ( x ,φ) π H ⌠ dφ ⌠ x (1+κ)/2 d x ⌠ 2 2 (1+κ)/2 L x 1 ⌡ (x − ρ ) ⌡ ⌡ 0
(3-κ)/2
= 2
0
2π
a
π H ⌠ dφ ⌠ K 1( x ,φ) x
(3-κ)/2
= 2
0
⌡ 0
⌡ 0
x
(1+κ)/2
ρ dρ ρ σ(ρ,φ). 2 (1+κ)/2 L x ⌡ (x − ρ )
dx ⌠
2
0
Here the interchange of the order of integration was used twice. comparison of the last expression with (2.4.22) yields the final result 2π a
2 πH ⌠⌠ [ K (ρ,φ)]2ρ dρ dφ. cos(πκ/2) ⌡ ⌡ 1 1-κ 2
W =
Now
0
(2.4.25)
0
Expression (2.4.25) interprets the stress intensity function squared as being proportional to the energy per unit area required to open up the crack. In the case of axial symmetry, formula (2.4.25) simplifies to
99
2.4 External mixed problem of type I
a
22-κπ3 H ⌠ W = [ K (ρ)]2ρ dρ, cos(πκ/2) ⌡ 1 0
with K 1(ρ) =
ρ
2cos(πκ/2)
σ(ρ0,φ0)ρ0 dρ0
⌠ 2 2 (1+κ)/2 . ⌡ (ρ − ρ0)
π(2ρ)(1-κ)/2
0
All the results obtained are valid for a transversely isotropic space provided that κ=0, and H is defined by (2.1.9).
1.
Exercise 2.4 Find the equivalent of (2.4.25) for an isotropic body. 2π a
1 − ν2 ⌠ ⌠ Answer: W = 2π [ K 1(ρ,φ)]2ρ dρ dφ, E ⌡⌡ 0 0
with ρ
√2 ⌠ K 1(ρ,φ) = π√ρ ⌡
x dx x σ( x ,φ). 2 1/2 L ρ (ρ − x ) 2
0
2.
Solve the problem above for the case of axial symmetry.
1 − ν2 ⌠ Answer: W = 4π E ⌡ 2
a
[ K 1(ρ)]2ρ dρ,
0
with ρ
√2 ⌠ σ( x ) x d x . K 1(ρ) = π√ρ ⌡ (ρ2 − x 2)1/2 0
Note : these results were obtained by Sneddon (1965). 3. A penny-shaped crack in a non-homogeneous elastic space is opened by the pressure σ(ρ,φ)=σ0+σ1ρcosφ, with σ0=const and σ1=const. Find the normal stress in the plane z =0 outside the crack. Answer: σ(ρ,φ) = −
2cos(πκ/2) a 3-κ 3-κ 3-κ 5-κ a 2 ; ; ( ) σ F , 2 2 π(3 − κ) ρ 0 2 ρ
100
CHAPTER 2.
+
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
2 a 2 F 3-κ, 5-κ; 7-κ; (a )2 σ ρcosφ, for ρ> a . 2 ρ 1 2 5 − κ ρ 2
The result can be expressed in elementary functions for a homogeneous body: 2 a a − sin-1( )σ0 π(ρ2 − a 2)1/2 ρ
σ(ρ,φ) = −
3ρ2 − a 2 ρ -1 a + − 3 sin ( ) σ ρ cos φ . a ρ 1 3ρ(ρ2 − a 2)1/2
In the problem above find the crack opening displacements w . 2 4cos(πκ/2) 2 − ρ2)(1-κ)/2 σ0 + σ1ρcosφ. Answer: w = 2 (a 3 − κ (1 − κ) 4.
5. Consider a penny-shaped crack ρ≤ a . Find the stress distribution σ in the plane z =0 outside the crack due to the action of a pair of equal concentrated forces P , applied normally to the crack faces in opposite directions at the point ( b ,ψ), b < a . πκ a 2 − b 2(1-κ)/2 P 1 Answer: σ(ρ,φ) = − 2 cos 2 . 2 2 2 2 π ρ − a ρ + b − 2 b ρcos(φ−ψ) 6.
Prove the identity for a penny-shaped crack: 2π ∞
2π a
⌠ ⌠ σ(ρ,φ)ρ ⌡⌡ 0
|n|+1
in φ
e
φ dρ dφ = − ⌠ ⌠ σ(ρ,φ)ρ|n|+1 ein dρ dφ.
0
⌡⌡
0
a
Note : the identity states that the normal stress in the plane z =0 is in equilibrium, which is not the case for an external crack. 7.
Express the stress intensity function K 1(ρ,φ) in terms of the displacement w . a
cos(πκ/2) d ⌠ x dx 1 Answer: K 1(ρ,φ) = − (1-κ)/2 2 (1+κ)/2L(ρ) 2 2 (1-κ)/2L x w ( x ,φ). ρ d 2 π Hρ ⌡ (x − ρ ) ρ
Hint : perform the inversion of (2.4.23). 8.
Prove the identity
2.5 Integral representation for q 2/ R 3
101
d ( a 2 − x 2)(1-κ)/2 π(1 − κ) f ( a ) ⌠ f x x . ( ) d = − lim 2 2 (1κ)/2 2cos(πκ/2) dρ ⌡ ( x − ρ ) ρ→ a ρ a
Hint : use the substitution t =( x 2−ρ2)/( a 2− x 2). 9. Use the identity above to express the stress intensity factor in terms of the displacement w . 1 − κ w (ρ,φ) Answer: k 1(φ) = 2-κ lim . 2 π H ρ→a ( a − ρ)(1-κ)/2 Hint : compute the limit ρ→ a of the result in Exercise 7. 10. Consider a transversely isotropic elastic space weakened by a penny-shaped crack ρ≤ a in the plane z =0, subjected to arbitrary pressure σ(ρ,φ). Find the complex tangential displacements u in the plane z =0. 2π a
( a 2 − ρ20)1/2
Answer: u = − H α ⌠ ⌠
σ(ρ0,φ0) ρ0dρ0dφ0
⌡ ⌡ ( a 2 − ρρ e-i(φ-φ0) )1/2 0 0 0
2π a
u = −
-i φ
ρe
-i φ0
− ρ0e
, for ρ≤ a ;
( a 2 − ρ20)1/2
2 Hα ⌠ ⌠ π ⌡ ⌡ ( a 2 − ρρ e-i(φ-φ0) )1/2 0 0 0
-i(φ-φ0) 1/2
( a 2 − ρρ0e
×tan-1
(ρ2 − a 2)1/2
)
σ(ρ0,φ0) ρ0dρ0dφ0 -i φ
ρe Hint : see (Fabrikant, 1987a) for details.
-i φ0
− ρ0e
, for ρ> a .
2
2.5 Integral representation for q / R
3
While it was sufficient to know the integral representation for 1/ R in order to solve the mixed boundary value problems of type I, this is no longer the case for the problems of type II. The need to know the integral representation for q 2/ R 3 is quite obvious from (2.2.12). We recall that q is defined by (2.2.5), and R is given by (2.2.14). The original derivation was made by the author many years ago. It was very long and cumbersome. Here we present only the idea used, and the final result.
102
CHAPTER 2.
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
Since q 2/ R 3= q /( q R ),
(2.5.1)
we may use the following expansion: i φ0
ρei
φ
− ρ0e
ρe-i
φ
− ρ0e
ρei
φ
− ρ0e
ρe-i
φ
-i φ0
i φ0 -i φ0
− ρ0e
2 i φ0
= e
∞ ρ 2 ρ k -ik(φ-φ0) ρ i(φ-φ0) − e 1 − ρ ρ e , for ρ<ρ0; ρ0 0 0 k=0
Σ
ρ0 2 ∞ ρ0 k ik(φ-φ ) ρ e 0 − 0e-i(φ-φ0) , for ρ>ρ ; = e 1 − 0 ρ ρ ρ k=0
Σ
2 iφ
Now we have to substitute (2.5.2) and (1.1.27) in (2.5.1). for ρ<ρ0, 2 i φ0 q2 1 − ρ 2 = e ρ0 R3
ρ k e-ik(φ-φ0) − ρ ei(φ-φ0) ρ0 ρ0
Σ
∞
∞
(2.5.2) This procedure yields,
k=0
ρ ( x 2/ρρ0)|n|d x 2 in(φ-φ0) ⌠ e 2 2 1/2 2 2 1/2 . π ⌡ (ρ − x ) (ρ0 − x )
Σ
×
n=-∞
0
(2.5.3) A very tedious procedure follows: grouping together the terms belonging to each harmonic. The derivation does not end there: we shall also need to use the identities: n
min (ρ0,ρ)
⌠ ⌡ 0
min (ρ0,ρ)
=⌠
⌡
0
(ρ20
− x) 2
Σρ
2(n-k) 2k
x
− ρ2n x 2
k=0
(ρ2 − x 2)1/2(ρ20 − x 2)1/2
dx
x 2n[(2 n + 1)ρ20 − (2 n + 2) x 2] (ρ2 − x 2)1/2(ρ20 − x 2)1/2
dx, (2.5.4)
and
2.5 Integral representation for q 2/ R 3
103
min (ρ0,ρ)
(2 n − 1)ρ2 − 2 n x 2 2n-2 dx 2 2 1/2 2 2 1/2 x (ρ − x ) (ρ0 − x )
ρ20 ⌠
⌡
0
min (ρ0,ρ)
2 n ρ2 − (2 n + 1) x 2 x 2nd x , for n =1,2,3, ... (ρ2 − x 2)1/2(ρ20 − x 2)1/2
= ⌠
⌡ 0
(2.5.5)
The first identity may be proven by the mathematical induction method, the second one may be established by using the property min (ρ0,ρ)
⌠ ⌡
d[ x 2n-1(ρ2 − x 2)1/2(ρ20 − x 2)1/2] = 0.
0
A similar procedure is required for the case ρ>ρ0. min (ρ0,ρ)
∞ 2 i φ in(φ-φ0) q 2 e e ⌠ = π ρ2 (ρρ0)n ⌡ R3 n=0 0
Σ
2
∞
+
Σ
e
n=0
−
2 i φ0
-in(φ-φ0)
e ρρ0
(2 n + 1)ρ2 − (2 n + 2) x 2 2n x dx (ρ2 − x 2)1/2(ρ20 − x 2)1/2
min (ρ0,ρ)
e ⌠ 2 ρ0 (ρρ0)n ⌡
i (φ+φ ) 0
The final result is
0
(2 n + 1)ρ20 − (2 n + 2) x 2 (ρ − x ) 2
2 1/2
(ρ20
− x)
2 1/2
x 2n d x
min (ρ0,ρ)
⌠ ⌡
0
x2 dx . (ρ2 − x 2)1/2(ρ20 − x 2)1/2
(2.5.6)
Expression (2.5.6), though looking cumbersome, will prove very useful for solving internal mixed boundary value problem of the type II. We need yet another integral representation which is useful in external problems. The procedure is as tedious as the one described above, with the final result ∞ ∞ (2 n + 1) x 2 − (2 n + 2)ρ20 i( φ φ ) q 2 2 e i φ (e 0 ρρ0)n ⌠ 3 = π 2n+2 2 2 1/2 2 2 1/2 d x − ρ − ρ R ) (x ⌡ x (x 0) n=0 max (ρ0,ρ) 2
Σ
104
CHAPTER 2.
∞
+
Σ
2 i φ0
e
n=0
∞
(e
-i(φ-φ0)
⌡
ρρ0 ⌠
− e
ρρ0)n ⌠
(2 n + 1) x 2 − (2 n + 2)ρ2 dx x 2n+2( x 2 − ρ2)1/2( x 2 − ρ20)1/2
max (ρ0,ρ) ∞
i (φ+φ ) 0
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
⌡
dx . x 2( x 2 − ρ2)1/2( x 2 − ρ20)1/2
max (ρ0,ρ)
(2.5.7)
Here the following identities were used: n
( x 2 − ρ2)
∞
⌠ ⌡
2k 2(n-k) ρ0
− ρ2 x 2n
k=0
x 2n+2( x 2 − ρ2)1/2( x 2 − ρ20)1/2
max (ρ0,ρ)
∞
=
Σx
(2 n + 1) x 2 − (2 n + 2)ρ2
⌠ ⌡
ρ2n 0
dx
max (ρ0,ρ)
x 2n+2( x 2 − ρ2)1/2( x 2 − ρ20)1/2
dx,
and ∞
2 n ρ20 − (2 n − 1) x 2
⌠ ⌡
x 2n( x 2 − ρ2)1/2( x 2 − ρ20)1/2
dx
max (ρ0,ρ)
∞
= ρ ⌠ 2
⌡
(2 n + 1)ρ20 − 2 n x 2 x 2n+2( x 2 − ρ2)1/2( x 2 − ρ20)1/2
dx.
max (ρ0,ρ)
These identities can be derived from (2.5.4) and (2.5.5) by a formal substitution of x by ρρ0/ x . As usual, when the final result is achieved, one can find an easier way to do it. We shall discuss further some generalizations of the integral representations derived here (see section 2.7).
1.
Exercise 2.5 Prove the identities (2.5.4) and (2.5.5).
105
2.6 Internal mixed problem of type II
2.
Establish the representation (2.5.6).
3.
Establish the representation (2.5.7).
2.6 Internal mixed problem of type II The material in this section follows essentially the paper (Fabrikant, 1971c). Consider a transversely isotropic elastic half-space z ≥0. Let the normal traction σ be prescribed all over the plane z =0. An arbitrary tangential displacement u = u x+i u y is specified inside a circle ρ= a , while the complex shear loading τ is known outside the circle. The problem is to find the shear traction inside the circle. The mathematical formulation of the boundary conditions is σ = σ(ρ,φ),
for 0≤ρ<∞,
0≤φ<2π;
τ = τ(ρ,φ),
for a <ρ<∞,
0≤φ<2π;
u = u (ρ,φ),
for 0≤ρ≤ a ,
0≤φ<2π;
(2.6.1)
The governing integral equation can be written due to (2.2.12): 2π a
1 ⌠⌠ G 2 1⌡ ⌡ 0
τ(ρ0,φ0) ρ0dρ0dφ0 R
2π a
1 + G2⌠ ⌠ 2 ⌡⌡ 0
0
q τ(ρ0,φ0) ρ0dρ0dφ0 qR
= χ(ρ,φ).
(2.6.2)
0
Function χ is known from the boundary conditions (2.6.1): 2π ∞
χ(ρ,φ) = u + H α⌠ ⌠
⌡⌡ 0 0
2π ∞
1 − G1⌠ ⌠ 2 ⌡⌡ 0
σ(ρ0,φ0) ρ0dρ0dφ0
τ(ρ0,φ0) ρ0dρ0dφ0 R
-i φ0
ρe-i φ − ρ0e
2π ∞
1 − G ⌠⌠ 2 2⌡ ⌡ 0
a
q τ(ρ0,φ0) ρ0dρ0dφ0 qR
.
a
(2.6.3) Though a closed form exact solution of (2.6.2) is possible, we present first its exact solution in Fourier series. Assume validity of the expansions: ∞
τ(ρ,φ) =
Στ (ρ) n
n=-∞
∞
in φ
e ,
χ(ρ,φ) =
Σχ (ρ) n
n=-∞
ein φ.
(2.6.4)
106
CHAPTER 2.
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
Substitution of (2.5.6) and (2.6.4) in (2.6.2) leads to an infinite set of integral equations 2G1 ρ
a
x 2n+2d x
τn+1(ρ0) dρ0
⌠ ⌠ ρn+1 ⌡ (ρ2 − x 2)1/2 ⌡ ρn0(ρ20 − x 2)1/2 0
+
2G2 ρ
⌠
ρn+1 ⌡
x
x 2n-2d x
a
(2 n − 1)ρ2 − 2 n x 2
⌠
(ρ2 − x 2)1/2 ⌡
0
2 2 1/2 ρn-2 0 (ρ0 − x )
x
τ-n+1(ρ0) dρ0 = χn+1(ρ), (2.6.5)
2G1 ρ
a
x 2n-2d x
τ-n+1(ρ0) dρ0
⌠ ⌠ 2 2 1/2 ρn-1 ⌡ (ρ2 − x 2)1/2 ⌡ ρn-2 0 (ρ0 − x ) 0
+
2G2 ρ
⌠
x
a
x 2n-2d x
⌠
(2 n − 1)ρ20 − 2 n x 2
ρn-1 ⌡ (ρ2 − x 2)1/2 ⌡ 0
ρn0(ρ20 − x 2)1/2
x
τn+1(ρ0) dρ0 = χ-n+1(ρ),
Equations (2.6.5) and (2.6.6) are valid for n =1,2,3, ... . symmetry, n =0, and the integral equation takes the form
(2.6.6) In the case of axial
ρ a G 1τ1(ρ0) − G 2τ1(ρ0) x 2d x 2 ⌠ ⌠ dρ0 = χ1(ρ). ρ ⌡ (ρ2 − x 2)1/2 ⌡ (ρ20 − x 2)1/2 0
(2.6.7)
x
Its solution is elementary, namely, a
x
2 d ⌠ dx d ⌠ τ1(ρ) = − 2 2 2 dρ 2 2 1/2 d x x ⌡ π ( G 1 − G 2) ⌡ (x − ρ ) ρ
0
G 1χ1(ρ0) + G 2χ1(ρ0) ( x 2 − ρ20)1/2
dρ0.
(2.6.8) The general solution of the system (2.6.5) and (2.6.6) can be presented in the form a
f
(t) dt
G
-n+1 ρn-1 2 C + D τ-n+1(ρ) = ρ ⌠ + , n ( a 2 − ρ2)1/2 G 1 n ⌡ ( t 2 − ρ2)1/2 n-1
ρ
107
2.6 Internal mixed problem of type II
a
a
f n+1( t ) d t
τ (ρ) = ρ ⌠ n+1 ⌡ ( t 2 − ρ2)1/2 n-1
ρ
2n + n+1⌠ y 2n-1d y ⌠ ρ ⌡ ⌡ ρ
a
f n+1( t ) d t ( t 2 − y 2)1/2
y
a
ρn-1 2n ⌠ y 2n-1 d y + C n 2 + . ( a − ρ2)1/2 ρn+1 ⌡ ( a 2 − y 2)1/2 ρ
Here f k are the as yet unknown complex functions, and C n and D n
(2.6.9) are the as
yet unknown constants. By substitution of (2.6.9) in (2.6.5) and (2.6.6), we obtain, after interchanging the order of integration and integration with respect to ρ0, πG1 ρn+1
ρ
a
ρ ⌠ t 2n-1 f ( t ) d t + C ρa 2n-1 − ⌠ (ρ2 − t 2)1/2 t 2n-1 f ( t ) d t n+1 n n+1 ⌡ ⌡ 0
0
+
πG2
ρ
⌠ (ρ2 − t 2)1/2 t 2n-1 f ( t ) d t = χ (ρ), -n+1 n+1
ρn+1 ⌡
(2.6.10)
0
πG1 ρ
n-1
ρ
x 2n-2d x
⌠ ⌡ (ρ2 − x 2)1/2 0
−
πG2
ρ
1 1 Γ( ) Γ( n − ) 2 2 2n-2 ⌠ f (t) dt + Dn ρ 2Γ( n ) ⌡ -n+1 a
x
x 2n-2d x
⌠
a
⌠ f ( t ) d t = χ (ρ). n+1 -n+1
ρn-1 ⌡ (ρ2 − x 2)1/2 ⌡ 0
(2.6.11)
x
Expression (2.6.10) can be simplified if we define C n as a
Cn = −a
-2n+1
⌠ t 2n-1 f ( t ) d t . n+1 ⌡ 0
Application of the operator
(2.6.12)
108
CHAPTER 2.
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
r
r
-2n+2
ρndρ d ⌠ 2 d r ⌡ ( r − ρ2)1/2 0
to both sides of (2.6.11) yields a
rχ (ρ)ρndρ -n+1
a
π ⌠ d ⌠ G f ( t ) d t − G 2⌠ f n+1( t ) d t + G 1 D n = r -2n+2 . r ⌡ ( r 2 − ρ2)1/2 d 2 1 ⌡ -n+1 ⌡ 2
r
0
r
(2.6.13) Since expression (2.6.13) should stay valid in the limiting case of r → a , this defines D n as follows:
Dn =
aχ (ρ)ρndρ -n+1
2 d ⌠ 2n-2 d a 2 2 1/2. π G1a ⌡ (a − ρ )
(2.6.14)
2
0
Both constants are now defined. (2.6.13) lead to the system
Inversion of (2.6.10) and differentiation of
− G 1 f n+1( r ) + G 2 f -n+1( r ) = ψn+1( r ), − G 1 f -n+1( r ) + G 2 f n+1( r ) = ψ-n+1( r ). (2.6.15) Here the notations were introduced r
2 d ⌠ ψn+1( r ) = 2 2n-1 dr ⌡ πr
d[ρn+1χn+1(ρ)] ( r 2 − ρ2)1/2
,
0
r ρnχ (ρ) -n+1
dρ 2 d -2n+2 d ⌠ ψ-n+1( r ) = 2 r 2 2 1/2 . d r d r π ⌡ (r − ρ ) 0
(2.6.16) Solution of the system (2.6.15) yields
f -n+1( r ) = −
G 1ψ-n+1( r ) + G 2ψn+1( r ) G 21 − G 22
,
109
2.6 Internal mixed problem of type II
f n+1( r ) = −
G 1ψn+1( r ) + G 2ψ-n+1( r ) G 21 − G 22
.
(2.6.17)
The general solution is now completed. It is given by formulae (2.6.9), with the constants C and D defined by (2.6.12) and (2.6.14), and functions f defined by (2.6.17) and (2.6.16). Example. Consider a transversely isotropic elastic space weakened by an external circular crack ρ≥ a in the plane z =0. Two identically oriented equal concentrated forces P are applied normally to the crack faces at the points (ρ0,φ0,0±), ρ0> a . Let us find the tangential stress in the crack neck. Due to the antisymmetry of the applied load, the problem can be reduced to the one of a half-space, with the tangential displacements and the normal stress vanishing in the crack neck. The mathematical formulation of the boundary conditions is σ = P δ(ρ−ρ0)δ(φ−φ0)/ρ
for 0≤ρ<∞,
τ = 0,
for a <ρ<∞,
0≤φ<2π;
u = 0,
for 0≤ρ≤ a ,
0≤φ<2π;
0≤φ<2π;
(2.6.18)
Here δ(⋅) is the Dirac delta-function. The governing integral equation corresponds to (2.6.2), with the right hand side χ, defined by (2.6.3), i.e. χ(ρ,φ) =
PH α
= −
-i φ0
ρe-i φ − ρ0e
PH α
-i φ0
ρ0e
∞
Σ(e
-i(φ-φ0)
n=0
ρ n ). ρ0
The general solution, presented above, yields the following results i φ0
PH αe χ-n+1(ρ) = − ρ0
f -n+1 = f n+1 = C n =0,
i φ0 ρ n-1
(e
ρ0
) ,
Dn = −
χn+1(ρ) = 0, 2 PH αΓ( n )
-i φ0 n
π3/2 G 1Γ( n − 12)(ρ0e
)
(2.6.19) Substitution of (2.6.19) in (2.6.17) and (2.6.9) leads to the solution
110
CHAPTER 2.
2 PH α τ(ρ,φ) = − 3/2 π G1
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
∞
ρne-in φ 2 2 1/2. -i φ0 n+1 1 (a − ρ ) ) Γ( n + 2) n=0 (ρ0e
Σ
Γ( n + 1)
The summation can be performed, according to the scheme ∞
ΣΓ(n
Γ( n + 1)
n=0
1
ζ = π n
-1/2
+ 2)
π-1/2 ζ 1/2 1 F (1, 1; ; ζ) = 1 + sin-1√ζ. − ζ 2 1 − ζ 1
Here we have used the well known property of the hypergeometric functions (Bateman and Erde´ lyi, 1955). Now the final result will take the form τ(ρ,φ) = −
2 PH α
-i φ0
π2 G 1ρ0e
( a 2 − ρ2)1/2
1 b 1/2 1 + sin-1√ b , − 1 − b 1 b
-i(φ-φ )
where b =(ρ/ρ0)e 0 . In the case of isotropy, the last formula coincides with the result of Ufliand (1967). Exercise 2.6 1. Consider a transversely isotropic elastic half-space z ≥0. The tangential displacement u = u 0=const is prescribed inside a circle ρ= a . The shear traction outside the circle is equal to zero, and the normal pressure vanishes all over the plane z =0. Find the shear traction inside the circle. 2u0 Answer: τ(ρ) = 2 . π G 1( a 2 − ρ2)1/2 2.
In the problem above find the normal displacement w in the plane z =0. -i φ
Answer: w =
w =
4ℜ ( u 0 e ) H α a , π G 1ρ
for ρ> a ;
-i φ 4ℜ( u 0e ) H α[ a − ( a 2 − ρ2)1/2]
π G 1ρ
,
for ρ≤ a ;
3. Subject to the conditions of the first problem, displacement outside the circle ρ= a . G a (ρ2 − a 2)1/2 2 i φ 2 a Answer: u = u 0sin-1( ) + u 0 2 e . π ρ G1 ρ2
find
the
tangential
111
2.7 External mixed problem of type II
2.7 External mixed problem of type II Consider a transversely isotropic elastic half-space z ≥0. Let the normal traction σ be prescribed all over the plane z =0. An arbitrary tangential displacement u = u x+i u y is specified outside a circle ρ= a , while the complex shear loading τ is known inside the circle. The problem is to find the shear traction outside the circle, the tangential displacement inside, and the normal displacement at the plane z =0. The mathematical statement of the boundary conditions is σ = σ(ρ,φ),
for
0≤ρ<∞,
0≤φ<2π;
τ = τ(ρ,φ),
for
0≤ρ< a ,
0≤φ<2π;
u = u (ρ,φ),
for
a ≤ρ<∞,
0≤φ<2π.
(2.7.1)
The governing integral equation can be written due to (2.2.12): 2π ∞
1 G1⌠ ⌠ 2 ⌡⌡ 0
τ(ρ0,φ0) ρ0dρ0dφ0 R
a
2π ∞
1 G 2⌠ ⌠ + 2 ⌡⌡ 0
q τ(ρ0,φ0) ρ0dρ0dφ0 qR
= χ(ρ,φ).
(2.7.2)
a
Function χ is known due to the boundary conditions (2.7.1): 2π ∞
χ(ρ,φ) = u + H α⌠ ⌠ ⌡⌡ 0 0
2π a
−
1 G ⌠⌠ 2 1⌡ ⌡ 0 0
σ(ρ0,φ0) ρ0dρ0dφ0
τ(ρ0,φ0) ρ0dρ0dφ0 R
-i φ0
ρe-i φ − ρ0e
2π a
−
1 G⌠⌠ 2 2⌡ ⌡ 0
q τ(ρ0,φ0) ρ0dρ0dφ0 qR
.
0
(2.7.3) Once equation (2.7.2) has been solved, the normal displacement is found from 2π ∞ 2π ∞ τ(ρ0,φ0) ρ0dρ0dφ0 σ(ρ0,φ0) ρ0dρ0dφ0 ⌠ ⌠ ⌠ ⌠ w (ρ,φ) = H α ℜ H + . i φ0 R ⌡ ⌡ iφ ⌡ ⌡ ρe − ρ0e 0 0 0 0
We present the exact solution of (2.7.2) in Fourier series. the expansions:
(2.7.4) Assume validity of
112
CHAPTER 2.
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
∞
τ(ρ,φ) =
∞
Στ (ρ)
in φ
χ(ρ,φ) =
e ,
n
n=-∞
Σχ (ρ)
in φ
e .
n
(2.7.5)
n=-∞
Substitution of the integral representation (2.5.7) and (2.7.5) in (2.7.2) leads to an infinite set of coupled integral equations ∞
2 G 1ρn+1⌠
⌡x
2n+2
ρ x
×⌠
(x − ∞
2 G 1ρ ⌠ n-1
ρ20)1/2
×⌠ a
ρ20)1/2
xτ
(ρ0)ρn0 dρ0
⌠ ⌡ x 2n-2( x 2 − ρ2)1/2 ⌡ ( x 2 − ρ20)1/2
τn+1(ρ0)ρn0 dρ0 ( x 2 − ρ20)1/2
⌡
(x − 2
∞
-n+1
⌡x
2n+2
ρ
( x − ρ2)1/2 2
for n =0,1,2, ...
∞
+ 2 G 2ρ ⌠ n-1
(2 n −1) x 2 − 2 n ρ2
⌡ x 2n( x 2 − ρ2)1/2
(2.7.6)
dx
ρ
a
= χ-n+1(ρ),
dx
+ 2 G 2ρn+1 ⌠
τ-n+1(ρ0)ρn0 dρ0 = χn+1(ρ),
dx
ρ x
2 1/2
a
2
a
⌠ ⌡
(x − ρ ) 2
2 n x 2 − (2 n + 1)ρ20
⌡
τn+1(ρ0)ρn+2 dρ0 0
x
dx
for n =1,2,3, ...
(2.7.7)
Here ρ
χn+1(ρ) = u n+1(ρ) + 2π H αρ
-n-1
⌠ σn(ρ0) ⌡
a
ρn+1 0 dρ0
0
a
×⌠
⌡ x
2 x 2nd x − n+1 ⌠ 2 2 1/2 ρ ⌡ (ρ − x ) 0
G 1 x 2τn+1(ρ0) + G 2[2 n ρ2 − (2 n + 1) x 2]τ-n+1(ρ0) ρn0(ρ20 − x 2)1/2
dρ0,
(2.7.8)
and ∞
χ-n+1(ρ) = u -n+1(ρ) + 2π H αρ
n-1
⌠ σ (ρ ) ⌡ -n 0
ρ
a
ρ-n+1 0 dρ0
2 x 2n-2d x − n-1 ⌠ 2 2 1/2 ρ ⌡ (ρ − x ) 0
113
2.7 External mixed problem of type II
a
×⌠
G 1ρ20τ-n+1(ρ0) + G 2[(2 n − 1)ρ20 − 2 n x 2]τn+1(ρ0) ρn0(ρ20 − x 2)1/2
⌡ x
dρ0,
(2.7.9)
The case of axial symmetry corresponds to n =0, and we have only one equation (2.7.6) to solve. The general solution of the system (2.7.6) and (2.7.7) may be presented in the form ρ f (t) dt Dn n+1 1 ⌠ τn+1(ρ) = n+1 2 2 1/2 + 2 2 1/2 , for n =0,1,2, ... ⌡ (ρ − t ) ρ (ρ − a ) a
ρ
y
1 ydy d 2n ⌠ τ-n+1(ρ) = n+1 ⌠ 2 2 1/2 d y y ρ ⌡ (ρ − y ) ⌡ a
f -n+1( t ) t 2n+1
dt
a
ρ
2n 1 y 2nd y ⌠ , for n =1,2,3, ... + + C n n+1 2 ρ (ρ − a 2)1/2 ρn+1 a 2n+1 ⌡ (ρ2 − y 2)1/2 a
Here functions f are to be determined, and C complex constants. We used the same notation that this would not produce any confusion, and clearly that, for example, C n of this section previous one. Substitution of (2.7.10) in simplification, ∞
(2.7.10) and D are the as yet unknown as in the previous section hoping that the reader would understand is not equal to C n from the equations (2.7.6) yields, after
x
dx G ⌠ f ( t ) dt + D 2n+2 2 n x ( x − ρ2)1/2 1 ⌡ n+1
πρn+1 ⌠
⌡
ρ
a
x
− G 2 ⌠ f -n+1( t ) d t + C n = χn+1(ρ), ⌡ a
(2.7.11)
114
CHAPTER 2.
∞
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
x
⌠ πρ G 1 ⌠ 2 2 1/2 ⌡ ( − ρ ) x ⌡ xdx
n-1
ρ
f -n+1( t ) d t t 2n+1
+
Cn
a 2n+1
a
∞
+ πρ G 2 ⌠ ⌡ n-1
( x 2 − ρ2)1/2 x 2n+1
ρ
f n+1( x ) d x = χ-n+1(ρ).
(2.7.12)
The following identity was used during the procedure of integration by parts −
d ( x 2 − ρ2)1/2 (2 n − 1) x 2 − 2 n ρ2 = . dx x 2n x 2n+1( x 2 − ρ2)1/2
Looking at the first integral in (2.7.12), one may conclude that it converges only if the term in square brackets tends to zero when x →∞. This condition defines C n as follows: ∞
Cn = −a
⌠ f ( t ) t -2n-1d t . ⌡ -n+1
2n+1
(2.7.13)
a
By dividing both sides of (2.7.11) by ρn(ρ2− r 2)1/2, integrating with respect to ρ from r to ∞, differentiating with respect to r , and multiplying the result by r 2n+2/π, we obtain r
r
G 1 ⌠ f n+1( t ) d t + D n − G 2 ⌠ f -n+1( t ) d t + C n ⌡ ⌡ a
a
∞
2 d ⌠ = − 2 r 2n+2 dr ⌡ π
χn+1(ρ) dρ ρn(ρ2 − r 2)1/2
.
(2.7.14)
r
Proceeding to the limit as r → a in (2.7.14) gives us the formula for evaluating D n, namely,
115
2.7 External mixed problem of type II
∞
1 2 d ⌠ Dn = G C − 2 a 2n+2 G1 2 n d a ⌡ π
χn+1(ρ) dρ
. ρ (ρ − a ) n
2
2 1/2
(2.7.15)
a
Inversion of (2.7.12) equations
and differentiation
of (2.7.14)
lead to
the system of
G 1 f -n+1( r ) − G 2 f n+1( r ) = ψ-n+1( r ), G 1 f n+1( r ) − G 2 f -n+1( r ) = ψn+1( r ).
(2.7.16)
Here χ (ρ) dρ d -n+1 , (ρ2 − r 2)1/2 dρ ρn-1
∞
2 d ⌠ ψ-n+1( r ) = − 2 r 2n+1 r ⌡ d π r
χn+1(ρ) dρ
∞
2 d 2n+2 d ⌠ ψn+1( r ) = − 2 r dr ⌡ π dr
. ρ (ρ − r ) n
2
2 1/2
(2.7.17)
r
The following rule of differentiation was used during the above transformations: ∞
d ⌠ dρ ⌡ ρ
∞
f( x) dx = ρ⌠ 2 ( x − ρ2)1/2 ⌡ ρ
d[ f ( x )/ x ] . ( x 2 − ρ2)1/2
(2.7.18)
Solution of the system (2.7.16) is
f n+1( r ) =
f -n+1( r ) =
G 1ψn+1( r ) + G 2ψ-n+1( r ) G 21 − G 22
,
G 1ψ-n+1( r ) + G 2ψn+1( r ) G 21 − G 22
.
(2.7.19)
The set of expressions (2.7.10), (2.7.13), (2.7.15), (2.7.17), and (2.7.19) gives a complete solution to the problem. Since the shear traction is now known throughout the plane z =0, the tangential displacements inside the circle can be
116
CHAPTER 2.
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
defined by (2.7.2), and the normal displacement is defined by (2.7.4). We consider further the case of a penny-shaped crack in more detail, including the derivation of closed form solutions. Example: Penny-shaped crack. Let the crack of radius a be located in the plane z =0. Both crack faces are loaded by arbitrary shear tractions acting in opposite directions. The boundary conditions are antisymmetric, so the problem can be reduced to a mixed one for a half-space, with the boundary conditions: σ = 0,
for 0≤ρ<∞,
0≤φ<2π;
τ = τ(ρ,φ),
for 0≤ρ< a ,
0≤φ<2π;
u = 0,
for a ≤ρ<∞,
0≤φ<2π.
(2.7.20)
The shear tractions outside the crack, the tangential displacement inside, and the normal displacement in the plane z =0 are to be determined. The crack problem can be considered as a particular case of a more general one solved above. However, we can show that the crack problem has a simpler solution. Indeed, the integral equations (2.7.6) and (2.7.7) remain unchanged, but instead of (2.7.8) and (2.7.9) we have a
χ
n+1
(ρ) = −
aG
2n
x dx ⌠ 2 ⌠ 2 ρ ⌡ (ρ − x 2)1/2⌡
1x
τn+1(ρ0)+ G 2[2 n ρ2−(2 n +1) x 2]τ-n+1(ρ0) ρn0(ρ20 − x 2)1/2
n+1
x
0
2
dρ0, (2.7.21)
and a
χ
-n+1
(ρ) = −
aG
2n-2
x dx ⌠ 2 ⌠ n-1 2 ρ ⌡ (ρ − x 2)1/2⌡
ρn0(ρ20 − x 2)1/2
x
0
2 2 2 1ρ0τ-n+1(ρ0)+ G 2[(2 n −1)ρ0−2 n x ]τn+1(ρ0)
dρ0.
(2.7.22) The structure of integral equations indicates the possibility of expression the yet unknown τ outside the crack through the prescribed τ inside. Such a representation takes the form
τn+1(ρ) = −
2 πρn+1(ρ2 − a 2)1/2
a
⌠ ⌡
( a 2 − t 2)1/2τ ( t ) t n+2d t n+1 ρ2 − t 2
+
An ρn+1(ρ2 − a 2)1/2
,
0
τ
-n+1
(ρ) = −
2
a
⌠
πρn-1(ρ2 − a 2)1/2 ⌡
for n =0,1,2, ... ( a 2 − t 2)1/2τ ( t ) t nd t -n+1 ρ2 − t 2
, for n =1,2,3, ...
0
(2.7.23)
117
2.7 External mixed problem of type II
Here A n is the as yet unknown complex constant. (2.7.6) yields, after intermediate integration, ∞
⌡ x
2n+2
ρ
∞
τn+1( t ) t n+2d t
a
π −⌠ 2 2 1/2 + 2 A n (x − ρ ) ⌡ (x − t ) dx
2 G 1ρn+1 ⌠
Substitution of (2.7.23) in
2
2 1/2
0
a
2 2 −⌠ 2 n x − (2 n + 1) t τ ( t ) t n d t ⌠ 2n+2 2 -n+1 ( x 2 − t 2)1/2 ⌡ x ( x − ρ2)1/2 ⌡
+ 2 G 2ρ
n+1
dx
ρ
0
a
+ (2 n + 1) ⌠ ( a 2 − t 2)1/2 τ ( t ) t n d t = χ (ρ). -n+1 n+1
(2.7.24)
⌡
0
The following identity can be verified by a formal substitution of x by ρ t / x : ∞
t
1 x 2n+2d x dx ⌠ ⌠ . = (ρ t )2n+2 ⌡ (ρ2 − x 2)1/2( t 2 − x 2)1/2 ⌡ x 2n+2( x 2 − ρ2)1/2( x 2 − t 2)1/2 ρ
(2.7.25)
0
Interchanging the order of integration in (2.7.24), substituting (2.7.25), and yet another interchange lead to the expression
−
2G1 ρn+1
a
−
τn+1( t ) d t
⌠ ⌠ n 2 2 2 1/2 2 1/2 ( x ) ρ − ⌡ ⌡ t (t − x ) x
0
2G2
a
x 2n+2d x
a
⌠
ρn+1 ⌡
2 n ρ2 − (2 n + 1) x 2 (ρ2 − x 2)1/2
a
x dx ⌠
0
∞
+ ρ
dx πG A ⌠ 2n+2 2 ⌡ x ( x − ρ2)1/2 1 n
n+1
ρ
2n
τ
-n+1
(t) dt
n 2 2 1/2 ⌡ t (t − x )
x
118
CHAPTER 2.
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
a
+ 2 G 2(2 n + 1) ⌠ ( a 2 − t 2)1/2 τ ( t ) t nd t = χ (ρ). -n+1 n+1
(2.7.26)
⌡
0
Comparison of (2.7.26) and (2.7.21) shows that the equation is satisfied if A n is defined by a
An
2 = − (2 n + 1)( G 2/ G 1) ⌠ τ (ρ)ρn( a 2 − ρ2)1/2dρ. π ⌡ -n+1
(2.7.27)
0
Since τ
-n+1
is defined for n ≥1 only, we may conclude that A 0=0.
A similar
procedure of substitution of (2.7.23) in (2.7.7) leads to ∞
a
dx
2 G 1ρn-1 ⌠
⌡ x
2n-2
ρ
∞
+ 2 G 2ρn-1 ⌠
(x − ρ ) 2
2 1/2
τ-n+1( t ) t n d t
−⌠ 2 2 1/2 ⌡ (x − t ) 0
(2 n − 1) x 2 − 2 n ρ2
⌡
ρ
x (x − ρ ) 2n
2
2 1/2
a
d x −⌠
τn+1( t ) t n
⌡ ( x 2 − t 2)1/2
dt
0
a
1 ⌠ π + ( a 2 − t 2)1/2 τ ( t ) t n+2d t + A n = χ (ρ). n+1 -n+1 ax ⌡ 2
(2.7.28)
0
It may be noted that the integral ∞
∞
2 2 1/2 ⌠ (2 n −1 ) x − 2 n ρ d x = ⌠ d− ( x − ρ ) = 0, for n ≥1. x 2n+1( x 2 − ρ2)1/2 x 2n ⌡ ⌡ 2
2
ρ
ρ
(2.7.29) By using (2.7.25) and (2.7.29), equation (2.7.28) can be transformed into
−
2G1 ρn-1
a
a
x 2n-2d x
τ-n+1( t ) d t
⌠ ⌠ n-2 2 2 2 1/2 2 1/2 ( x ) ρ − ⌡ ⌡ t (t − x )
0
x
119
2.7 External mixed problem of type II
−
2G2 ρn-1
a
a
x 2n-2d x
⌠ ⌠ 2 2 1/2 x ( ) ρ − ⌡ ⌡
(2 n − 1) t 2 − 2 n x 2 t n( t 2 − x 2)1/2
τn+1( t ) d t = χ-n+1(ρ).
x
0
Comparison of the last expression with (2.7.22) proves that equation (2.7.7) is satisfied, and that (2.7.23) is indeed the solution to the crack problem. A closed form solution can be obtained by summation of (2.7.23) and (2.7.27), with the result
τ(ρ,φ) = −
−
2π a
1
⌠⌠ ⌡⌡
π (ρ − a ) 2
2
2 1/2
G 2e2 i φ
0
2π
a
( a 2 − ρ20)1/2τ(ρ0,φ0) ρ0dρ0dφ0
0
ρ2 + ρ20 − 2ρρ0cos(φ−φ0)
3 − (ρ0/ρ)e
i(φ-φ0)
2 ⌠ ⌠ − ρ20)1/2τ(ρ0,φ0)ρ0dρ0dφ0. i(φ-φ0) 2( a π G 1ρ (ρ − a ) ⌡ ⌡ [1 − (ρ /ρ)e ] 0 0 2
2
2
2 1/2
0
(2.7.30) One can notice that the first integral in (2.7.30) corresponds to the solution for the case of normal loading of a penny-shaped crack. Define the complex stress intensity factor as k (φ) = lim[(ρ − a )1/2τ(ρ,φ)]. ρ→ a
(2.7.31)
Substitution of (2.7.30) in (2.7.31) yields
k (φ) = −
2π a
⌠⌠
1
π2√2 a ⌡ ⌡ 0 0
( a 2 − ρ20)1/2τ(ρ0,φ0) ρ0dρ0dφ0 a 2 + ρ20 − 2 a ρ0cos(φ−φ0) (2.7.32)
2 iφ
−
G 2e
2π
a
3 − (ρ / a )e
i(φ-φ0)
0 2 ⌠ ⌠ − ρ20)1/2τ(ρ0,φ0) ρ0dρ0dφ0. i(φ-φ0) 2 ( a 2 2 ⌡ π G 1 a √2 a ⌡ [1 − (ρ0/ a )e ] 0 0
Since our definition of τ contains both x- and y- components, so will the expression for the stress intensity factor k = k x+i k y. If we require the expression for the radial and tangential components, we have to use the relationship
120
CHAPTER 2.
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
φ τzx + iτyz = (τzρ + iτφz)ei .
This allows us to rewrite (2.7.32) in terms of the second and third mode stress intensity factors as follows: e-i φ
2π a
⌠⌠ k2 + ik3 = − 2 π √2 a ⌡ ⌡ 0
( a 2 − ρ20)1/2τ(ρ0,φ0) ρ0dρ0dφ0 a 2 + ρ20 − 2 a ρ0cos(φ−φ0)
0
iφ
−
G 2e
2π
3 − (ρ0/ a )e
a
i(φ-φ0)
2 ⌠ ⌠ − ρ20)1/2 τ(ρ0,φ0) ρ0dρ0dφ0. i(φ-φ0) 2 ( a π G 1 a √2 a ⌡ ⌡ [1 − (ρ / a )e ] 0 0 2
2
0
(2.7.33) While using (2.7.33), one should remember that τ is still defined in terms of the cartesian coordinate components. In order to define the tangential displacements inside the crack directly in terms of the prescribed shear loading, equations (2.7.6) and (2.7.7) have to be rewritten for ρ≤ a . They will have a similar form, with a difference only in the limits of integration, namely, ∞
u n+1(ρ) = 2 G 1ρn+1 ⌠
⌡
a ∞
+ 2 G 2ρn+1⌠
⌡x
x 2 1/2
a
+
2G1 ρn+1
ρ
+
ρn+1
⌠ ⌡
2 n x 2 − (2 n + 1)ρ20 (x − 2
a
a
x 2n+2d x
ρ20)1/2
τ-n+1(ρ0)ρn0 dρ0
τn+1(ρ0) dρ0
⌠ ⌠ n 2 2 2 1/2 2 1/2 ⌡ (ρ − x ) ⌡ ρ0(ρ0 − x ) 0
2G2
( x 2 − ρ20)1/2
a
(x − ρ ) 2
⌠
τn+1(ρ0)ρn+2 0 dρ0
x 2n+2( x 2 − ρ2)1/2 ⌡
dx 2n+2
x
dx
x
ρ
a
x 2n d x
⌠ ⌠ 2 2 1/2 ( ) ρ − x ⌡ ⌡ 0
x
2 n ρ2 − (2 n + 1) x 2 ρn0(ρ20 − x 2)1/2
τ-n+1(ρ0) dρ0,
for n =0,1,2, ... ∞
u -n+1(ρ) = 2 G 1ρn-1 ⌠
dx
(2.7.34) x
⌠
2n-2 2 2 1/2 ⌡ x (x − ρ ) ⌡
a
a
τ-n+1(ρ0)ρn0dρ0 ( x 2 − ρ20)1/2
121
2.7 External mixed problem of type II
∞
x
dx
(2 n − 1) x 2 − 2 n ρ2
⌠ + 2 G 2ρ ⌠ 2n 2 2 1/2 x x − ρ ( ) ⌡ ⌡ n-1
a
+
a
ρ
2G1
a
x 2n-2d x
ρ20)1/2
τn+1(ρ0)ρn0 dρ0
τ-n+1(ρ0) dρ0
⌠ ⌠ n-2 2 2 2 1/2 2 1/2 ρ − x ( ) ⌡ ⌡ ρ0 (ρ0 − x )
ρn-1
0
+
(x − 2
x
ρ
2G2
⌠
ρn-1 ⌡
x 2n-2 d x
a
⌠
(2 n − 1)ρ20 − 2 n x 2
(ρ2 − x 2)1/2 ⌡
0
ρn0(ρ20 − x 2)1/2
x
τn+1(ρ0) dρ0,
for n =1,2,3, ... (2.7.35) The first two terms in (2.7.34) and (2.7.35) represent the displacement inside the crack due to the shear traction outside, while the remaining terms give the displacement caused by the shear tractions inside. Substitution of (2.7.23) in (2.7.34) yields, after integration with respect to ρ0, ∞
a
dx
u n+1(ρ) = −2 G 1ρn+1⌠
⌠
τn+1( t ) t n+2d t
2n+2 2 2 1/2 2 2 1/2 ⌡ x (x − ρ ) ⌡ (x − t ) 0
a ∞
− 2 G 2ρ
n+1
a
dx
⌠ 2n+2 2 ⌠ 2 1/2 ⌡ x (x − ρ ) ⌡
+
ρn+1
ρ
+
ρn+1
a
x 2n+2d x
2 1/2
τ-n+1( t ) t nd t
τn+1(ρ0) dρ0
⌠ ⌠ n 2 2 2 1/2 2 1/2 ⌡ (ρ − x ) ⌡ ρ0(ρ0 − x ) 0
2G2
(x − t ) 2
0
a
2G1
2 n x 2 − (2 n + 1) t 2
x
ρ
a
x 2nd x
⌠ ⌠ 2 2 1/2 ( ρ − x ) ⌡ ⌡ 0
x
2 n ρ2 − (2 n + 1) x 2 ρn0(ρ20 − x 2)1/2
τ-n+1(ρ0) dρ0.
Transform the third term in (2.7.36) by using (2.7.25). follows:
(2.7.36) The procedure is as
122
CHAPTER 2.
ρ
2G1
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
⌠ ⌠ n 2 2 2 1/2 2 1/2 ρ − x ( ) ⌡ ⌡ ρ0(ρ0 − x )
ρn+1
0
=
x
ρ
2G1
τn+1(ρ0) dρ0
⌠ ⌡
ρn+1
ρn0
0
a
+ ⌠ ⌡
ρn0
⌠
τn+1(ρ0) dρ0
⌡
ρn0
ρ
(ρ2 − x 2)1/2(ρ20 − x 2)1/2
x 2n+2d x
⌠ 2 2 1/2 2 2 1/2 ( x ) ( x ) ρ − ρ − ⌡ 0
ρn0
0
+ ⌠
⌠ ⌡
ρ
τn+1(ρ0) dρ0
ρn+1 ⌡ a
x 2n+2d x
0
ρ
2G1
ρ0
0
τn+1(ρ0) dρ0
ρ
=
τn+1(ρ0) dρ0
a
x 2n+2d x
∞
(ρρ0)2n+2d x
⌠ ⌡ x 2n+2( x 2 − ρ2)1/2( x 2 − ρ20)1/2
ρ
∞
(ρρ0)2n+2d x
⌠ ⌡ x 2n+2( x 2 − ρ2)1/2( x 2 − ρ20)1/2 ρ0
∞
= 2 G 1ρn+1 ⌠
a
dx
⌠ ⌡ x 2n+2( x 2 − ρ2)1/2 ⌡ a
a
+ ⌠
⌡ x ρ
2n+2
( x 2 − ρ20)1/2
0
x
dx (x − ρ ) 2
2 1/2
⌠ ⌡
τn+1(ρ0)ρn+2 0 dρ0
τn+1(ρ0)ρn+2 0 dρ0 (x − 2
0
ρ20)1/2
.
(2.7.37)
An analogous transformation of the last term in (2.7.36) leads to the identity 2G2
ρ
⌠
ρn+1 ⌡ 0
x 2nd x
a
⌠
(ρ2 − x 2)1/2 ⌡ x
2 n ρ2 − (2 n + 1) x 2 ρn0(ρ20 − x 2)1/2
τ-n+1(ρ0) dρ0
123
2.7 External mixed problem of type II
∞
a
dx ⌠ ⌠ 2n+2 2 ⌡ x ( x − ρ2)1/2 ⌡
= 2 G 2ρ
n+1
a
a
+ ⌠
⌡ x ρ
ρ20)1/2
(x − 2
0
2 n x 2 − (2 n + 1)ρ20
x
dx 2n+2
2 n x 2 − (2 n + 1)ρ20
(x − ρ ) 2
2 1/2
⌠ ⌡
(x − 2
0
τ-n+1(ρ0)ρn0 dρ0
τ-n+1(ρ0)ρn0 dρ0.
ρ20)1/2
(2.7.38)
Back substitution of (2.7.37) and (2.7.38) in (2.7.36) results in a
u n+1(ρ) = 2 G 1ρn+1 ⌠
τn+1(ρ0)ρn+2 0 dρ0
x
dx
⌠
⌡ x 2n+2( x 2 − ρ2)1/2 ⌡
ρ a
+ 2 G 2ρ
n+1
0
x
dx
⌠ ⌠ ⌡ x 2n+2( x 2 − ρ2)1/2 ⌡ ρ
( x 2 − ρ20)1/2
2 n x 2 − (2 n + 1)ρ20 (x − 2
0
ρ20)1/2
τ
-n+1
(ρ0)ρn0 dρ0,
for n =0,1,2, ..., and ρ≤ a .
(2.7.39)
A similar procedure can be applied to (2.7.35). (2.7.35) yields, after an integration with respect to ρ0,
Substitution of (2.7.23) in
∞
a
dx
u -n+1(ρ) = −2 G 1ρn-1 ⌠
τ-n+1( t ) t n d t
⌠ 2n-2 2 2 1/2 2 2 1/2 ⌡ x (x − ρ ) ⌡ (x − t )
a ∞
+ 2 G 2ρ ⌠ ⌡ n-1
0
(2 n − 1) x 2 − 2 n ρ2 x 2n( x 2 − ρ2)1/2
a
a
τ
0
a
π 1 ⌠ A n + ( a 2 − t 2)1/2τn+1( t ) t n+2 d t + ax ⌡ 2 0
+
2G1
ρ
⌠
ρn-1 ⌡ 0
x 2n-2d x
a
⌠
(ρ2 − x 2)1/2 ⌡ x
(t) dt
n+1 d x −⌠ ⌡ ( x 2 − t 2)1/2
τ-n+1(ρ0) dρ0 2 2 1/2 ρn-2 0 (ρ0 − x )
124
CHAPTER 2.
+
ρ
2G2
⌠
ρn-1 ⌡ 0
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
(2 n − 1)ρ20 − 2 n x 2
a
x 2n-2d x
⌠
(ρ2 − x 2)1/2 ⌡
ρn0(ρ20 − x 2)1/2
x
τn+1(ρ0) dρ0.
(2.7.40) The following identities may be established by using the procedures identical to those used for deriving (2.7.37) and (2.7.38) ρ
2G1
⌠
ρn-1 ⌡ 0
τ-n+1(ρ0) dρ0
a
x 2n-2d x
⌠
2 2 1/2 (ρ2 − x 2)1/2 ⌡ ρn-2 0 (ρ0 − x ) x
∞
= 2 G 1ρn-1 ⌠
τ
a
dx
(ρ0)ρn0
⌠ dρ ⌡ x 2n-2( x 2 − ρ2)1/2 ⌡ ( x 2 − ρ20)1/2 0 a
a
0
τ-n+1(ρ0)ρn0
x
dx
+ ⌠
-n+1
⌠
⌡ x 2n-2( x 2 − ρ2)1/2 ⌡ ( x 2 − ρ20)1/2 ρ
(2.7.41)
0
ρ
2G2
⌠
ρn-1 ⌡ 0
a
x 2n-2d x
⌠
(2 n − 1)ρ20 − 2 n x 2
(ρ2 − x 2)1/2 ⌡
ρn0(ρ20 − x 2)1/2
x
∞
= 2 G 2ρn-1 ⌠
a
(2 n − 1) x 2 − 2 n ρ2
⌡
dx ⌠
x 2n( x 2 − ρ2)1/2
⌡
a
a
+ ⌠
⌡ ρ
dρ0,
0
(2 n − 1) x 2 − 2 n ρ2 x (x − ρ ) 2n
2
2 1/2
x
dx ⌠
⌡
τ
n+1
τn+1(ρ0)ρn0 dρ0 ( x 2 − ρ20)1/2
τn+1(ρ0)ρn0 dρ0 (x − 2
0
(ρ0) dρ0
ρ20)1/2
.
The back substitution of (2.7.41) and (2.7.42) in (2.7.40) yields a
u
-n+1
(ρ) = 2 G 1ρn-1 ⌠
x
dx
τ-n+1(ρ0)ρn0
⌠ dρ ⌡ x 2n-2( x 2 − ρ2)1/2 ⌡ ( x 2 − ρ20)1/2 0 ρ
0
(2.7.42)
125
2.7 External mixed problem of type II
a
(2 n − 1) x 2 − 2 n ρ2
+ 2 G 2ρn-1 ⌠
⌡
x (x − ρ ) 2n
ρ
2
2 1/2
τn+1(ρ0)ρn0 dρ0
x
dx ⌠
(x −
⌡
2
0
ρ20)1/2
+ B n( a 2 − ρ2)1/2.
(2.7.43) Here a
Bn = a
-2n-1
⌠ t n( a 2 − t 2)1/2[τ ( t ) − (2 n + 1)( G 2/ G 1)τ ( t )] d t . n+1 -n+1 ⌡
(2.7.44)
0
Formulae (2.7.39), (2.7.43), and (2.7.44) give the tangential displacement of the crack faces in terms of the prescribed shear tractions. Note that the displacement vanishes outside the crack. The complex tangential displacement u can be represented in terms of its harmonics as ∞
u (ρ,φ) =
Σu
∞
n+1
(ρ)e
i ( n +1)φ
Σu
+
n=0
-n+1
(ρ)e
-i ( n -1)φ
,
(2.7.45)
n=1
where u n+1 and u -n+1 are defined by (2.7.39) and (2.7.44).
The summation will
be performed for the terms containing G 1 and G 2 separately. (2.7.39) in (2.7.45) and summation of the terms with G 1 gives 2π
G1
a
⌠ dφ ⌠ π ⌡ 0⌡
λ(ρρ0/ x 2, φ−φ0) d x ( x 2 − ρ2)1/2
ρ
0
τ(ρ0,φ0) ρ0dρ0
x
⌠ ⌡
( x 2 − ρ20)1/2
0
Substitution of
.
(2.7.46)
Here we should change the order of integration according to the scheme a
a
x
ρ
a
a
⌠ d x ⌠ dρ = ⌠ dρ ⌠ d x + ⌠ dρ ⌠ d x . 0 ⌡ ⌡ 0 ⌡ 0⌡ ⌡ ⌡ ρ
ρ
0
ρ0
0
(2.7.47)
ρ
Integration with respect to x in (2.7.46) can be performed as in (1.1.23), and yields G1
2π a
⌠ ⌠ 1 tan-1η τ(ρ ,φ ) ρ dρ dφ . 0 0 0 0 0 π ⌡⌡ R R 0 0
(2.7.48)
126
CHAPTER 2.
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
where R and η( x ) are defined by (2.2.14) and (1.1.6) respectively. We recall that, according to our convention, the abbreviation η stands for η( a ). Summation of the terms with G 2 in (2.7.39) and (2.7.43) is slightly more complicated, but generally it reduces to the series ∞
Σ
nz n =
n=0
z . (1 − z )2
The result of summation is 2π
G2
2 i φ0
⌠e
π ⌡
a
dφ0⌠
λ(ρρ0/ x 2, φ−φ0) d x
⌡
( x 2 − ρ2)1/2
ρ
0
-i φ
4 iq ρe 0 sin(φ − φ0) τ(ρ0,φ0) ρ0dρ0 . ×⌠ 1 + x 2(1 − ζ)(1 − ζ) ( x 2 − ρ20)1/2 ⌡ 0 x
(2.7.49) i(φ-φ0)
Here ζ=(ρρ0/ x )e , and ζ is a complex conjugate of ζ. The integral in (2.7.49), though looking formidable, is a perfect differential and can be computed as indefinite 2
-i φ
λ(ρρ0/ x 2, φ−φ0) d x
⌠ ⌡ ( x 2 − ρ2)1/2 ( x 2 − ρ20)1/2 -2 i φ0
= e
q η( x ) tan-1 + R qR
4 iq ρe 0 sin(φ − φ0) 1 + x 2(1 − ζ)(1 − ζ) -i φ0
2 i ρe
sin(φ − φ0) η( x )
x 2 q (1 − ζ)(1 − ζ)
.
(2.7.50) It is noteworthy, that (2.7.50) can be considered as a generalization of the integral representation for q 2/ R 3, given by (2.5.7). Indeed, such a representation can be obtained from (2.7.50) by taking a definite integral, namely, 2 i φ0
∞
2 π
⌠ ⌡
e
λ(ρρ0/ x 2, φ−φ0) d x
( x 2 − ρ2)1/2 ( x 2 − ρ20)1/2
max (ρ0,ρ)
-i φ
4 iq ρe 0 sin(φ − φ0) q . 1 + = 2 x (1 − ζ)(1 − ζ) qR
By changing the order of integration in (2.7.49) and integrating with respect to
127
2.7 External mixed problem of type II
x , according to (2.7.50), we obtain i φ0
2π a
G2
2 i ρe
sin(φ − φ )η
0 τ(ρ ,φ )ρ dρ dφ , ⌠ ⌠ q tan-1η + 2 0 0 0 0 0 π ⌡ ⌡ qR R a q (1 − t )(1 − t ) 0
(2.7.51)
0
where t is defined by (1.4.41). (2.7.44), with the result 2π a
G2
⌠⌠ 2 πa ⌡ ⌡ 0
2 i φ0
τ(ρ0,φ0) e
G 2 (3 − t ) τ(ρ0,φ0)
−
(1 − t )
The remaining step is the summation of B n
(1 − t )
G1
2
η ρ dρ dφ . 0 0 0
0
(2.7.52) Finally, the summation of (2.7.48), (2.7.51) and (2.7.52) leads to 2π a
2
G ⌠ ⌠ 1 tan-1η − 2 (3 − t ) η τ(ρ ,φ ) ρ dρ dφ u (ρ,φ) = 0 0 0 0 0 π ⌡⌡ R R G 21 a 2(1 − t )2 G1
0
0
2π a
G2
2 iφ
0 ⌠ ⌠ q tan-1η + η[( q / q ) − t e ] τ(ρ ,φ ) ρ dρ dφ . + 0 0 0 0 0 π ⌡ ⌡ qR R a 2(1 − t )(1 − t ) 0
0
(2.7.53) Since the normal tractions vanish in the plane z =0, the normal displacement w will be defined by (2.7.4) as 2π ∞
w (ρ,φ) = H α ℜ⌠ ⌠ ⌡⌡ 0
0
τ(ρ0,φ0) ρ0dρ0dφ0 ρei
φ
i φ0
,
− ρ0e
which is equivalent to the following series representation
e-i( n +1)φ ρ ⌠ τ (ρ0) ρn+1 w (ρ,φ) = 2π H α ℜ n+1 0 dρ0 -n ρ ⌡ n=0 0 ∞
Σ
128
CHAPTER 2.
a
− e ρ ⌠ ⌡ in φ n
ρ
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
τn+1(ρ0) ρn0
∞
dρ0 + ⌠ ⌡
τn+1(ρ0) ρn0
a
dρ0 ,
for ρ≤ a ; (2.7.54)
e-i( n +1)φ a ⌠ τ (ρ0) ρn+1 w (ρ,φ) = 2π H α ℜ n+1 0 dρ0 -n ρ ⌡ n=0 0 ∞
Σ
ρ
∞
in φ + ⌠ τ-n(ρ0) ρn+1 dρ0 − e ρn ⌠ 0 ⌡ ⌡
τn+1(ρ0) ρn0
ρ
a
dρ0 ,
for ρ≥ a .
(2.7.55) By substituting formulae (2.7.23) and (2.7.27) in (2.7.54) and (2.7.55), and performing the summation and integration, we obtain 2π a ( a 2 − ρ20)1/2 τ(ρ0,φ0) ρ0dρ0dφ0 w = H α ℜ ⌠ ⌠ q ⌡ a (1 − t )1/2 ⌡ 0 0
+
2π a
1 ( a 2 − ρ2)1/2 τ(ρ ,φ )eiφ0 dρ dφ , for ρ≤a ; ⌠⌠ − 1 0 0 0 0 0 G 1 a ⌡ ⌡ (1 − t )3/2 0 0 G2
and 2π a
2 w = H α ℜ ⌠ ⌠ π 0⌡ ⌡
( a 2 − ρ20)1/2 a (1 − t )1/2
a (1 − t ) tan 2 (ρ − a 2)1/2 1/2
-1
τ(ρ0,φ0) q
ρ0dρ0dφ0
0
2π a
a (1 − t )1/2 1 1 a -1 ⌠ ⌠ − tan sin-1 + ρ G 1 ⌡ ⌡ a (1 − t )3/2 a (ρ2 − a 2)1/2 G2
0
-i φ0
−
ρ0e
0
(ρ2 − a 2)1/2
q a 2(1 − t )
( a 2 − ρ2)1/2 τ(ρ ,φ ) eiφ0 dρ dφ , for ρ> a . 0 0 0 0 0
129
2.8 Inverse crack problem in elasticity
Exercise 2.7 1. Uniform shear tractions τ=τ0, where τ0 is a complex constant, are applied antisymmetrically inside a penny-shaped crack of radius a . Find the tangential displacements of the crack faces. Answer: u = 2τ0[( G 21 − G 22)/ G 1]( a 2 − ρ2)1/2. 2. In the example above find the shear tractions in the plane z =0 outside the crack. G 2 a 3e2 φ 2 -1 a a τ − τ Answer: τ(ρ,φ) = sin ( ) − 2 . π ρ (ρ − a 2)1/2 0 G 1ρ2(ρ2 − a 2)1/2 0 i
3. Subject to the conditions of the first example, find the normal displacement w in the plane z =0. -i i Answer: w (ρ,φ) = π H αρℜτ0e φ + ( G 2/ G 1)τ0e φ, for ρ≤ a ;
a a -i i w (ρ,φ) = 2 H αℜτ0e φ + ( G 2/ G 1)τ0e φρsin-1( ) − (ρ2 − a 2)1/2, for ρ≥ a . ρ ρ
2.8 Inverse crack problem in elasticity The usually considered elastic crack problems assume that the stress distribution on the crack faces is known, and the crack opening displacements are to be determined. Investigation of materials with rigid inclusions leads to another formulation of the crack problem, namely, the displacements are prescribed on the crack faces, and the stress distribution is to be determined. The problem so formulated is called the inverse crack problem. We shall consider two types of problem: the case of a smooth rigid inclusion (normal displacements are prescribed, tangential stresses vanish) is called the inverse crack problem of type I. The second type corresponds to the case when the normal stress is equal to zero over the plane z =0, and the antisymmetric tangential displacements are prescribed inside the crack. The stress distribution is to be determined in each case. Strictly speaking, the inverse crack problem does not belong to the class of the mixed problems, and its exact solution is known for a general crack. We show below how some specific results can be obtained by the new method. Smooth rigid inclusion problem. Consider a penny-shaped crack of radius a in a transversely isotropic space. Let the crack be opened by a rigid smooth inclusion. The crack opening displacements are prescribed as
130
CHAPTER 2.
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
w = w (ρ,φ),
for z = 0+,
0≤ρ≤ a ;
w = − w (ρ,φ),
for z = 0 ,
0≤ρ≤ a ;
Due to symmetry, the problem can be reduced to the one of a half-space, with the boundary conditions at the plane z =0 w = w (ρ,φ),
for ρ≤ a ,
0≤φ<2π;
w = 0,
for ρ≥ a ,
0≤φ<2π;
τ = 0,
for 0≤ρ<∞,
0≤φ<2π.
(2.8.1)
The general relationship between the crack opening displacement w and the applied pressure σ was established in (2.4.19), and in this case may be written as a
dx
w (ρ,φ) = 4 H ⌠
⌡ ( x 2 − ρ2)1/2
ρ
x
ρ dρ
ρρ
0 0 0σ(ρ ,φ). ⌠ 2 2 1/2 L x2 0 ⌡ ( x − ρ0) 0
(2.8.2) Since in our case the displacement w is known, and σ is unknown, we can interpret (2.8.2) as an integral equation. It can be solved exactly. The first operator to be applied is a
ρ dρ d ⌠ 1 L( t ) L , 2 2 1/2 d t ⌡ (ρ − t ) ρ t
with the result t
2π H ⌠
ρ0dρ0
ρ
2 2 1/2 ⌡ ( t − ρ0)
0
a
0 ρ dρ d 1 L σ(ρ0,φ) = −L( t ) ⌠ 2 2 1/2 L ρ w (ρ,φ). t d t ⌡ (ρ − t ) t
(2.8.3) The next operator to be applied is y
1 d ⌠ t dt L 2 2 1/2 L( t ). y d y ⌡ (y − t ) 0
The final result reads
131
2.8 Inverse crack problem in elasticity
y
a
ρ dρ t dt 1 1 d ⌠ 2 d ⌠ 1 σ( y ,φ) = − 2 L 2 2 1/2 L( t )d t 2 2 1/2 L ρ w (ρ,φ). y y d π Hy ⌡ (y − t ) ⌡ (ρ − t ) 0
t
(2.8.4) Formula (2.8.4) is valid inside the crack only. The normal traction outside the crack can be expressed directly in terms of the prescribed crack opening displacement due to the relationship between the normal stresses inside and outside the crack (2.4.17), which in this case takes the form a
2 ( a 2 − y 2)1/2 y ⌠ L σ( y ,φ) y d y , σ(ρ,φ) = − ρ π(ρ2 − a 2)1/2 ⌡ ρ2 − y 2
for ρ> a .
0
(2.8.5) Substitution of (2.8.4) in (2.8.5) yields, after integration with respect to y , a
L(1/ρ) a − t ρ2 − a 21/2d t ⌠ σ(ρ,φ) = 2 2 2 ρ2 − t 2 π H ρ (ρ − a 2)1/2 ⌡ 0
a ρ0dρ0 d 1 2 d ⌠ × t L( t ) L w (ρ0,φ). 2 2 1/2 dt d t ⌡ (ρ0 − t ) ρ0 t
Integration by parts in the last expression leads to a a ρ0dρ0 L(1/ρ) ⌠ d 1 xdx 2 ⌠ σ(ρ,φ) = L ( ) L w (ρ0,φ). x 2 2 2 3/2 2 2 1/2 d x ⌡ (ρ0 − x ) ρ0 π H ⌡ (ρ − x ) 0
x
(2.8.6) Expression (2.8.6) can also be represented in the form ρ0dρ0 L(1/ρ) d ⌠ xdx 2 d ⌠ 1 L ( x ) 2 2 2 1/2 2 2 1/2 L ρ w (ρ0,φ). ρ d d x 0 π Hρ ⌡ (ρ − x ) ⌡ (ρ0 − x ) a
σ(ρ,φ) = −
0
a
x
(2.8.7) Comparison of (2.8.7) and (2.8.4) indicates that the stress outside the crack can be represented in almost the same form as the stress inside, with the only difference in the upper limit of the first integral. By using the rule (1.3.9) and the condition w ( a ,φ)=0, expression (2.8.6) can be rewritten as
132
CHAPTER 2.
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
a a ρ0dρ0 1 ⌠ dx d x 2 w(ρ ,φ). ⌠ σ(ρ,φ) = 2 ρ L 0 π H ⌡ (ρ2 − x 2)3/2 ⌡ (ρ20 − x 2)1/2 dρ0 0 ρρ0 x
0
(2.8.8) Interchange of the order of integration and integration by parts yields 2π
a w (ρ0,φ0) ρ0dρ0dφ0 1 ⌠⌠ σ(ρ,φ) = − . 4π2 H ⌡ ⌡ [ρ2 + ρ20 − 2ρρ0cos(φ−φ0)]3/2 0
0
(2.8.9) The last expression can also be rewritten as 2π
a w (ρ0,φ0) ρ0dρ0dφ0 1 ⌠⌠ σ(ρ,φ) = − ∆ . 4π2 H ⌡ ⌡ [ρ2 + ρ20 − 2ρρ0cos(φ−φ0)]1/2 0 0
(2.8.10) Here ∆ is the two-dimensional Laplace operator. It will be shown later that formula (2.8.10) is of general nature: it is valid for an arbitrarily shaped flat crack all over the plane z =0. We can also express some integral characteristics in terms of the crack opening displacement. The resultant force P is defined by 2π a
P = ⌠ ⌠ σ(ρ,φ) ρdρdφ.
(2.8.11)
⌡⌡
0
0
Substitution of (2.8.4) in (2.8.11) leads to 2π
P = −
a
a
1 ⌠ xdx d ⌠ w (ρ,φ)ρ dρ dφ ⌠ 2 2 1/2 d x 2 2 1/2. π2 H ⌡ − ( a x ) ⌡ ⌡ (ρ − x ) 0
0
(2.8.12)
x
Integration by parts in (2.8.12) yields yet another representation 2π
a P = 2 ⌠ dφ ⌠ πH ⌡ ⌡ 2
0
0
a
a
(a
2
dx ⌠ − x 2)3/2 ⌡
w (ρ,φ)ρ dρ . (ρ2 − x 2)1/2
(2.8.13)
x
Several additional forms can be obtained by integration in (2.8.13) with respect
133
2.8 Inverse crack problem in elasticity
to x , and a consequent use of various integral representations for the complete elliptic integrals (see Exercise 2.8.2). One can also evaluate the resultant moment of the traction exerted by the rigid inclusion. Introducing the complex moment M = M +i M , we can deduce x
y
2π a
iφ M = −i⌠ ⌠ σ(ρ,φ)e ρ2 dρdφ.
(2.8.14)
⌡⌡ 0 0
Substitution of (2.8.4) in (2.8.14) yields 2π
a
i d ⌠ x 3d x iφ M = 2 ⌠ e dφ ⌠ 2 2 1/2 d x πH ⌡ ⌡ (a − x ) ⌡ 0
a
w (ρ,φ) dρ . (ρ2 − x 2)1/2
(2.8.15)
x
0
Yet another representation can be obtained from (2.8.15) (see Exercise 2.8.3). Example. It is of interest to consider displacements can be presented as an expansion
a
general
case
where
the
∞
w (ρ,φ) = ( a − ρ ) 2
2 1/2
Σw ρ
|n| in φ
n
e .
(2.8.16)
n=-∞
Substitution of (2.8.16) in (2.8.4) and (2.8.7) yields ∞
1 σ(ρ,φ) = 2√πH
Σ n=-∞
∞
1 σ(ρ,φ) = − 2π H
Γ(| n | + 3/2) φ w nρ|n|ein , Γ(| n | + 1)
Σ n=-∞
a 2|n|+3 w nein φ (2| n | + 3)ρ|n|+3
for ρ≤ a .
3 3 5 a2 F , +| n |; +| n |; 2, 2 2 2 ρ
for ρ> a .
(2.8.17) The Gauss hypergeometric function can be expressed in terms of elementary functions (Bateman and Erde´ lyi, 1955) 3 3 5 (−1)n(3 + 2 n ) dn (1−ζ)n 1−ζ1/2sin-1√ζ. 1 F , + n ; + n ; ζ = − 2 ζ 2 2 ζ n ! (1 − ζ)1/2 dζn
The traction in (2.8.17) becomes singular when ρ→ a . +
(2.8.18) Using integration by
134
CHAPTER 2.
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
parts in (2.8.6), the singular and nonsingular parts can be separated as follows: ∞
1 σ(ρ,φ) = − 2π H
Σ n=-∞
ρ a 2|n|+1 1, 1+| n |; 3+| n |; a 2w ein φ, − F 2 2 2 ρ|n|+1 (ρ2 − a 2)1/2 ρ2 n (2.8.19)
which yields the stress intensity factor as √a k1 = 2√2π H
∞
Σw a n
|n| in φ
e .
(2.8.20)
n=-∞
The hypergeometric function in (2.8.19) can be expressed in elementary functions: 1 1 3 2n + 1 dn (1 − ζ)n-1/2 sin-1√ζ. √1 − ζ F , +| n |; +| n |; ζ = (−1)n n 2 n! 2 2 dζ √ζ (2.8.21) The total force P and the tilting moment M can be determined from (2.8.12) and (2.8.15) respectively P =
πa2 w, 4H 0
M = −i
3π a 4 w 16 H -1
Inverse crack problem of type II. Let the tangential displacements be prescribed on the crack faces u = u (ρ,φ),
for ρ≤ a
z = 0+ ;
u = − u (ρ,φ),
for ρ≤ a
z = 0.
(2.8.22)
It is required to find the shear stress distribution in the plane z =0. Due to antisymmetry, the problem can be reduced to that of a half-space, subjected to the boundary conditions: u = u (ρ,φ),
for ρ≤ a ,
0≤φ<2π;
u = 0,
for ρ≥ a ,
0≤φ<2π;
σ = 0,
for 0≤ρ<∞,
0≤φ<2π.
(2.8.23)
The relationship between the tangential displacement u and the shear traction τ
135
2.8 Inverse crack problem in elasticity
was established in (2.7.39) and (2.7.43). Since in the inverse problems u is known, and τ is to be determined, we should treat these expressions as a set of integral equations to be solved for the two unknowns τn+1 and τ-n+1. Assume the solution in the form ρ
2n τn+1(ρ) = G 1 f n+1(ρ) + G 2 f -n+1(ρ) − n+1 ⌠ f -n+1( x ) x nd x , for n = 0, 1, 2 ... ρ ⌡ 0
ρ
f ( x) τ-n+1(ρ) = G 1 f -n+1(ρ) + G 2 f n+1(ρ) + 2 n ρn-1 ⌠ n+1n d x + C nρn-1, for n = 1, 2, 3 ... ⌡ x 0
(2.8.24) Here f is an as yet unknown complex function and C n is an as yet unknown complex constant. Substitution of (2.8.24) in equation (2.7.39) and subsequent integration results in a
u n+1(ρ) = 2( G 21 − G 22)ρn+1
x
n+2 ⌠ 2n+2 d2x 2 1/2 ⌠ f n+1(ρ2 0)ρ20 1/2dρ0 ⌡ x ( x − ρ ) ⌡ ( x − ρ0)
ρ
(2.8.25)
0
from which the value of f n+1 can be found as ρ
a
2 d ⌠ x 2n+3d x d ⌠ u n+1(ρ0) dρ0 f n+1(ρ) = − 2 2 π ( G 1 − G 22)ρn+2 dρ⌡ (ρ2 − x 2)1/2 d x⌡ ρ0n (ρ20 − x 2)1/2
(2.8.26)
x
0
Substitution of (2.8.24) in equation (2.7.43) gives, after simplification
u -n+1(ρ) = 2( G 21 − G 22)ρn-1
a
x
ρ
0
dx ⌠ ⌠ f -n+1(2ρ0)ρ20 1/2dρ0 + 2 D nρn-1( a 2 − ρ2)1/2 2n-2 2 2 1/2 ⌡ x ( x − ρ ) ⌡ ( x − ρ0) n
(2.8.27) Here D n is a complex constant defined by a
x
d x f (ρ) ρ dρ D n = G 2 2 nG 1 ⌠ 2n ⌠ n+12 2 1/2 ⌡ x ⌡ (x − ρ ) 0
0
n
136
CHAPTER 2.
a
+ 2 nG 2 ⌠ f -n+1(ρ)
⌡
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
ρn √π Γ( n ) -1 a 2n cosh (ρ) dρ + B n + G 1 C n 2Γ( n + 1/2) a
(2.8.28)
0
Now the purpose of introducing the constant C n in (2.8.24) becomes clear: we can choose C n so that D n vanish. In this case expression (2.8.27) can be inverted, with the result ρ
a
2 d ⌠ x 2n-1d x d ⌠ u -n+1(ρ0) dρ0 f -n+1(ρ) = − 2 2 2 2 1/2 π ( G 1 − G 22)ρn dρ⌡ (ρ2 − x 2)1/2 d x⌡ ρn-2 0 (ρ0 − x ) 0
(2.8.29)
x
We can substitute (2.8.26) and (2.8.29) in (2.8.28) in order to find the value of the as yet unknown constant C n from the condition D n = 0. The mathematical transformations involved are cumbersome though elementary. Some of the integrals are presented below. a
x
a
a
f (ρ) ρn dρ 1 2 n + 1⌠ u n+1(ρ) dρ − √π Γ( n ) ⌠ u (ρ) ρn-1dρ, ⌠ d2nx ⌠ n+12 2 1/2 = n+1 2 2 π( G 1 − G 2) 2 n ⌡ ρn+1 2 a 2n Γ( n + 1/2) ⌡ ⌡ x ⌡ (x − ρ ) 0
0
0
0
(2.8.30) a
a
√π Γ( n ) ⌠ f -n+1(ρ) ρncosh-1(a ) dρ = ⌠ u -n+1 (ρ) ρn-1dρ. 2 2 ρ π( G 1 − G 2) Γ( n − 1/2)⌡ ⌡ 0
(2.8.31)
0
Substitution of (2.8.24) in (2.7.44) gives the following expression for B n G 22 1 ⌠ B n = 2n+1 − 2 nG 2 f -n+1(ρ) + G 1 − (2 n + 1) f (ρ) G 1 n+1 a ⌡ a
0
ρ
ρ
0
0
G 22 n-1 fn+1( x ) 2 2 1/2 n n ⌠ − n+1 x f -n+1( x ) d x − 2 n (2 n + 1) ρ ⌠ d x ( a − ρ ) ρ dρ n G1 ρ ⌡ ⌡ x 2 nG 2
(2.8.32) The following integrals need to be computed
137
2.8 Inverse crack problem in elasticity
a
a
√π Γ( n + 1/2)⌠ 1 ⌠ f -n+1(ρ)ρ ( a − ρ ) dρ = u -n+1(ρ)ρndρ 2 2 Γ n ( ) π( G 1 − G 2) ⌡ ⌡ n
2
2 1/2
0
(2.8.33)
0
a
a
√π Γ( n + 1)⌠ a ⌠ f n+1(ρ)ρ ( a − ρ ) dρ = u n+1(ρ)ρn-1dρ 2 2 Γ( n + 1/2) π( G 1 − G 2) ⌡ ⌡ n
2
2 1/2
0
(2.8.34)
0
ρ
a
a
1 a √π Γ( n ) ⌠ u (ρ)ρn-1dρ ⌠ ( a 2 − ρ2)1/2dρ⌠ f -n+1( x ) x nd x = -n+1 2 2 ρ⌡ π( G 1 − G 2) Γ( n − 1/2) ⌡ ⌡ 0
0
0
a
√π Γ( n + 1/2) ⌠ − u -n+1(ρ)ρndρ Γ( n ) ⌡
(2.8.35)
0
ρ
a
⌠ρ ⌡
2n-1
0
(a − ρ ) 2
2 1/2
dρ⌠
f n+1( x )
⌡
xn
0
a
1 a 2n+1⌠ u n+1(ρ) dρ − dx = π( G 21 − G 22) 2 n ⌡ ρn+1 0
a
a √π Γ( n ) ⌠ − u n+1(ρ)ρn-1dρ 2Γ( n + 1/2)⌡
(2.8.36)
0
Substitution of (2.8.33-2.8.36) in (2.8.32) yields a
a
1 − 2 G √π Γ( n + 1)⌠ u (ρ)n-1dρ + G √π Γ( n + 1) ⌠ u (ρ)ρn-1dρ B n = 2n 2 -n+1 n+1 2 2 Γ( n − 1/2) 1 Γ( n + 1/2) π a ( G 1 − G 2) ⌡ ⌡ 0
− (2 n + 1) a
2a 2n G 2
G √π Γ( n ) (ρ) ⌠ u n+1n+1 dρ − C n 2 . G 1⌡ ρ G 1 2Γ( n + 1/2)
0
(2.8.37)
0
Substitution of (2.8.30), (2.8.31) and (2.8.37) in (2.8.28) gives a fairly simple formula for C n, namely,
138
Cn = −
CHAPTER 2.
a
4 G 2Γ( n + 3/2) π
3/2
⌠ ⌡
( G 21 − G 22) Γ( n )
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
u n+1(ρ) dρ
(2.8.38)
ρn+1
0
The problem, in principle, is now solved but a certain simplification is possible due to the following integral: ρ
f n+1( x )
⌠ n ⌡ x 0
ρ
1 2 ⌠ 2 d x 2 1/2 dx = − 2 2 2 2n π ( G 1 − G 2) ρ ⌡ (ρ − x )
0
d 2n+1⌠ u n+1(ρ0) dρ0 √π Γ( n + 3/2)⌠ u n+1(ρ) dρ × x n 2 2 1/2 − dx Γ( n + 1) ⌡ ρn+1 ⌡ ρ0 (ρ0 − x ) a
a
x
0
(2.8.39)
Note that the last term in (2.8.39) will cancel C n when substituted in (2.8.24). Finally, formulae (2.8.26), (2.8.29) and (2.8.38) lead to ρ
G 1 d x 2n+3 d x d u n+1(ρ0) dρ0 2 ⌠ ⌠ τn+1(ρ) = − 2 2 π ( G 1 − G 22) ρn+2 dρ⌡ (ρ2 − x 2)1/2 d x ⌡ ρ0n (ρ20 − x 2)1/2
a
x
0
ρ
d 1 ⌠ x 2n-1d x d ⌠ u -n+1(ρ0)dρ0 + G 2ρ 2 2 1/2 dρ ρ2n⌡ (ρ2 − x 2)1/2 d x ⌡ ρn-2 0 (ρ0 − x ) a
n
0
x
for ρ ≤ a , n = 0, 1, 2, ... ρ
(2.8.40)
a
2 G 1 d ⌠ x 2n-1 d x d ⌠ u -n+1(ρ0) dρ0 τ-n+1(ρ) = − 2 2 2 2 1/2 π ( G 1 − G 22) ρn dρ⌡ (ρ2 − x 2)1/2 d x ⌡ ρn-2 0 (ρ0 − x ) x
0
ρ
a
G d u n+1(ρ )dρ0 dx d + n2 ⌠ 2 2 1/2 x 2n+1⌠ n 2 0 2 1/2 ρ dρ ⌡ ( ρ − x ) d x ⌡ ρ0 (ρ0 − x ) 0
x
for ρ ≤ a , n = 1, 2, 3, ...
(2.8.41)
Expressions (2.8.40) and (2.8.41) are valid inside the crack only. In order to express the shear traction outside the crack, we recall formulae (2.7.23) and
139
2.8 Inverse crack problem in elasticity
(2.7.27) which relate one to the other. us to compute
An = −
4 G 2Γ( n + 3/2)
Substitution of (2.8.41) in (2.7.27) allows
a
⌠ u -n+1(ρ) ρndρ
(2.8.42)
π3/2( G 21 − G 22) Γ( n )⌡ 0
Finally, substitution of (2.8.40-2.8.42) in (2.7.23) yields a
a
2 G 1 d ⌠ x 2n+3 d x d ⌠ u n+1(ρ0) dρ0 τn+1(ρ) = − 2 2 π ( G 1 − G 22) ρn+2 dρ⌡ (ρ2 − x 2)1/2 d x ⌡ ρ0n (ρ20 − x 2)1/2 0
a
x
a
G2 d x 2n-1d x d ⌠ u -n+1(ρ0)dρ0 ⌠ + n 2 2 1/2 ρ dρ⌡ (ρ2 − x 2)1/2 d x⌡ ρn-2 0 (ρ0 − x ) 0
x
for ρ > a , n = 0, 1, 2, ...
a
(2.8.43)
a
2 G 1 d ⌠ x 2n-1 d x d ⌠ u -n+1(ρ0) dρ0 τ-n+1(ρ) = − 2 2 2 2 1/2 π ( G 1 − G 22) ρn dρ⌡ (ρ2 − x 2)1/2 d x⌡ ρn-2 0 (ρ0 − x ) 0
x
a
a
G d u n+1(ρ )dρ0 dx d + n2 ⌠ 2 2 1/2 x 2n+1⌠ n 2 0 2 1/2 ρ dρ⌡ (ρ − x ) d x ⌡ ρ0 (ρ0 − x ) 0
x
for ρ > a , n = 1, 2, 3, ...
(2.8.44)
Formulae (2.8.43) and (2.8.44) can be used to define the stress intensity factor directly in terms of the tangential displacements prescribed inside the crack. We recall that the stress intensity factor was given by (2.7.31). By using the properties ρ
d f( x) dx f(a) lim (ρ − a )1/2 ⌠ 2 2 1/2 = − , dρ⌡ (ρ − x ) √2 a ρ→ a 0
a
π d ⌠ f( x) dx f (ρ) 2 2 1/2 = − 2 lim 2 2 1/2 , d ρ ⌡ (x − ρ ) ρ→ a ρ→ a ( a − ρ ) lim
ρ
(2.8.45)
140
CHAPTER 2.
MIXED BOUNDARY VALUE PROBLEMS IN ELASTICITY
the appropriate harmonics of the stress intensity factor can be written as k n+1 = −
a G 1 u n+1(ρ) + G 2 u -n+1(ρ), lim ( a 2 − ρ2)1/2 π( G 21 − G 22)√2 a ρ→a
k -n+1 = −
a π( G 21 − G 22)√2 a
G u -n+1(ρ) + G 2 u n+1(ρ) lim 1 . ( a 2 − ρ2)1/2 ρ→ a
Their summation can be performed in an elementary manner, to give finally G 1 u (ρ, φ) + G 2e2 i φ u (ρ, φ) a lim . k (φ) = − ( a 2 − ρ2)1/2 π( G 21 − G 22)√2 a ρ→a
(2.8.46)
The simple expression (2.8.46) for the stress intensity factor will be very useful in the investigation of crack interaction since it is much easier to solve the integral equation involved in terms of the displacements than in terms of the stresses which are singular at the crack boundary. Expression (2.8.46) can be made more symmetric by introduction of the polar displacements
u (ρ) = u ρ + iu φ = e-i φ u ,
and
similar
shear
traction
τ(ρ) = τρz + i τφz = e-i φτ.
The
corresponding stress intensity factor k (ρ) will take the form
k
(ρ)
G 1 u (ρ)(ρ, φ) + G 2 u (ρ)(ρ, φ) a =− lim . ( a 2 − ρ2)1/2 π( G 21 − G 22)√2 a ρ→a
(2.8.47)
Note that k (ρ) is proportional to the combination k 2 + ik 3 of the second and the third mode stress intensity factors. Summation of (2.8.40-2.8.41) and (2.8.43-2.8.44) leads to another simple result τ=−
1 ⌠ ⌠ u d S + G Λ2⌠ ⌠ u d S . 2 2 G 1∆ 2 ⌡⌡R ⌡⌡R 2π ( G 1 − G 2) 2
S
(2.8.48)
S
It will be shown further, that the last expression is valid not just for a penny-shaped crack, but for a general flat crack all over the plane z = 0. Exercise 2.8 1.
The crack opening displacement w is prescribed by expression
141
2.8 Inverse crack problem in elasticity
w = w 0( a 2 − ρ2)1/2, with w 0 = const. plane z = 0. σ = w 0/(4 H ),
Answer: σ=− 2.
Find the normal stress distribution σ in the
for ρ ≤ a ;
w0 a a − sin-1( ) , for ρ > a 2π H ( a 2 − ρ2)1/2 ρ
Prove that the total force P exerted by an inclusion can be defined by 2π
a
a
1 w (ρ, φ)ρ dρ P = 2 ⌠ dφ⌠ ( a 2 − x 2)1/2d x ⌠ 2 2 2 2 1/2. π H⌡ ⌡ ⌡ ( a − ρ )(ρ − x ) 0
x
0
Hint : use (2.8.13) 3.
Prove that the tilting moment M can be defined by 2π
a
a
i x 2(3 a 2 − 2 x 2) ⌠ w (ρ, φ)dρ M = − 2 ⌠ ei φdφ ⌠ 2 2 3/2 d x 2 2 1/2. π H⌡ ⌡ (a − x ) ⌡ (ρ − x ) 0
0
Hint : integrate (2.8.15) by parts.
x
CHAPTER 3 MIXED-MIXED BOUNDARY VALUE PROBLEMS
The mixed-mixed problems in elasticity theory are among the most complicated due to the coupling between the normal and tangential parameters. We should mention the works of Mossakovskii (1954) and Ufliand (1956) among the first published exact solutions for the isotropic half-space, obtained by using various integral transforms. A more compact solution has been reported by Kapshivyi and Masliuk (1967), who used a special apparatus of p -analytical functions. The first elementary exact solution for a transversely isotropic elastic half-space was published in (Fabrikant, 1971a). We present here a general formulation of the internal and external mixed-mixed problems in terms of two-dimensional integral equations. Four kinds of exact solution to the internal axisymmetric problem are given, and yet another kind of solution is presented for the external axisymmetric problem. The action of a general loading on a flat circular bonded punch is considered in detail. A general solution to the non-axisymmetric internal and external problems is presented as a Fourier series expansion. The material in this chapter follows the results published in the papers (Fabrikant, 1971d, 1972, 1974b, 1975, 1976, 1986j).
3.1 General formulation of the problem Consider a transversely isotropic elastic half-space z ≥0. boundary conditions be prescribed on the plane z =0: u = u (ρ,φ),
for
0≤ρ≤ a ,
0≤φ<2π;
w = w (ρ,φ),
for
0≤ρ≤ a ,
0≤φ<2π;
σ = σ(ρ,φ),
for
a ≤ρ≤∞,
0≤φ<2π;
142
Let the following
143
3.1 General formulation of the problem
τ = τ(ρ,φ),
a ≤ρ≤∞,
for
0≤φ<2π.
(3.1.1)
The problem, so defined, will be called internal mixed-mixed . The system of governing integral equations is formulated due to (2.2.12) and (2.2.13), and is 2π a
H αℜ⌠ ⌠ ⌡ ⌡ 0
0
2π a
1 ⌠ ⌠ G 2 1⌡ ⌡ 0
2π a
τ(ρ ,φ ) ρ dρ dφ 0 0 0 0 0 iφ
iφ
ρe
− ρe 0
0
τ(ρ0,φ0) ρ0dρ0dφ0 R
0
2π a
− H α⌠ ⌠ ⌡ ⌡ 0 0
R
2π a
qR
0
σ(ρ0,φ0) ρ0dρ0dφ0 ρe
(3.1.2)
q τ(ρ0,φ0) ρ0dρ0dφ0
1 + G 2⌠ ⌠ 2 ⌡ ⌡
-i φ
= ω1(ρ,φ),
0
0
0
σ(ρ0,φ0) ρ0dρ0dφ0
+ H⌠ ⌠ ⌡ ⌡
-i φ
− ρ0e
= ω2(ρ,φ).
0
(3.1.3)
It is reminded that the notations q and R are defined by (2.2.5) and (2.2.14) respectively. Functions ω1 and ω2 are known from the boundary conditions (3.1.1): 2π ∞
ω1(ρ,φ) = w (ρ,φ) − H αℜ⌠ ⌠ ⌡ ⌡ 0 a
2π ∞
− H⌠ ⌠ ⌡ ⌡ 0
σ(ρ0,φ0) ρ0dρ0dφ0 R
τ(ρ0,φ0) ρ0dρ0dφ0 iφ
ρei φ − ρ0e
0
,
(3.1.4)
a
2π ∞
ω2(ρ,φ) = u (ρ,φ) + H α⌠ ⌠ ⌡ ⌡ 0
2 π ∞ τ( ρ , φ ) 0 0
1 − G1⌠ ⌠ 2 ⌡ ⌡ 0
a
a
ρ0dρ0dφ0 R
σ(ρ0,φ0) ρ0dρ0dφ0 ρe-i
φ
-i φ
− ρ0e
0
2 π ∞ q τ( ρ , φ ) 0 0
1 − G⌠ ⌠ 2 2⌡ ⌡ 0
ρ0dρ0dφ0
qR
.
a
(3.1.5) The external mixed-mixed boundary value problem for a transversely isotropic
144
CHAPTER 3
elastic half-space conditions are:
can
be
MIXED-MIXED BOUNDARY VALUE PROBLEMS
formulated
in
a
similar
manner.
u = u (ρ,φ),
for
a ≤ρ<∞,
0≤φ<2π;
w = w (ρ,φ),
for
a ≤ρ<∞,
0≤φ<2π;
σ = σ(ρ,φ),
for
0≤ρ≤ a ,
0≤φ<2π;
τ = τ(ρ,φ),
for
0≤ρ≤ a ,
0≤φ<2π.
The
boundary
(3.1.6)
The system of governing integral equations in this case takes the form 2π ∞
H αℜ⌠ ⌠ ⌡ ⌡ 0
τ(ρ0,φ0) ρ0dρ0dφ0 iφ
iφ
ρe
a
2π ∞
1 ⌠ ⌠ G 2 1⌡ ⌡ 0
− ρ0e
0
2π ∞
+ H⌠ ⌠ ⌡ ⌡ 0
τ(ρ0,φ0) ρ0dρ0dφ0 R
σ(ρ0,φ0) ρ0dρ0dφ0 R
2π ∞
− H α⌠ ⌠
⌡ ⌡ 0
2π ∞
a
ρe
q τ(ρ0,φ0) ρ0dρ0dφ0
1 + G 2⌠ ⌠ 2 ⌡ ⌡
σ(ρ0,φ0) ρ0dρ0dφ0 -i φ
(3.1.7)
a
0
a
= ω1(ρ,φ),
-i φ
− ρ0e
0
qR
a
= ω2(ρ,φ).
(3.1.8)
Functions ω1 and ω2 are known from the boundary conditions, and are 2π a
ω1(ρ,φ) = w (ρ,φ) − H αℜ⌠ ⌠ ⌡ ⌡ 0
2π a
− H⌠ ⌠
⌡ ⌡ 0
τ(ρ0,φ0) ρ0dρ0dφ0 iφ
ρe
0
σ(ρ0,φ0) ρ0dρ0dφ0 R
iφ
− ρ0e
0
,
(3.1.9)
0
2π a
ω2(ρ,φ) = u (ρ,φ) + H α⌠ ⌠ ⌡ ⌡ 0 0
σ(ρ0,φ0) ρ0dρ0dφ0 -i φ
ρe
-i φ
− ρ0e
0
145
3.2 Internal axisymmetric mixed-mixed problem
2 π a τ( ρ , φ ) 0 0
1 − G1⌠ ⌠ 2 ⌡ ⌡ 0
ρ0dρ0dφ0 R
2 π a q τ( ρ , φ ) 0 0
1 G⌠ ⌠ − 2 2⌡ ⌡ 0
0
ρ0dρ0dφ0
qR
.
0
(3.1.10) The exact closed form solution of these equations is not known at the moment, though we strongly believe that the new method is capable of furnishing such a solution. In the sections to follow we consider separately the axisymmetric case and the general one. The exact solution to both internal and external problems is obtained for the n -th harmonic, with the assumption that all the functions involved can be represented as Fourier expansions. Exercise 3.1 1.
Verify the derivation of (3.1.2−3.1.5)
2.
Verify the derivation of (3.1.7−3.1.10)
3. Derive the governing integral equations for the internal mixed-mixed problem in the case of an isotropic half-space.
3.2 Internal axisymmetric mixed-mixed problem In order to demonstrate versatility of our method, we present here four kinds of exact solution. The first kind is more convenient for the stress evaluation, while the second one has certain advantages for calculating the displacements outside the circle ρ= a . Since it is important to have one kind of solution easily transformed into another, the necessary relationships are established. Application of this technique to a bonded flat-ended circular punch under the action of a normal force and expanding in the radial direction is considered. The influence of an arbitrary axisymmetric normal and tangential tractions field, applied outside the punch, is investigated. The boundary conditions in the case of axial symmetry are u = u (ρ),
for
0≤ρ≤ a ,
0≤φ<2π;
w = w (ρ),
for
0≤ρ≤ a ,
0≤φ<2π;
σ = σ(ρ),
for
a ≤ρ≤∞,
0≤φ<2π;
τ = τ(ρ),
for
a ≤ρ≤∞,
0≤φ<2π.
(3.2.1)
146
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
The set of governing integral equations will take the form a σ(ρ )ρ dρ ρ 0 0 0 x d ⌠ 2 H −πα ⌠ τ(ρ0)dρ0 + 2⌠ = ω1(ρ), (ρ2 − x 2)1/2 ⌡ (ρ20 − x 2)1/2 ⌡ ⌡ x 0 ρ a
(3.2.2) a ρ ρ τ(ρ0) dρ0 x 2d x 2H ⌠ ⌠ ⌠ γ γ − π α σ ρ ρ ρ 2 ( ) d = ω2(ρ). 0 0 0 ρ 1 2⌡ (ρ2 − x 2)1/2 ⌡ (ρ20 − x 2)1/2 ⌡
x
0
0
(3.2.3) The functions ω1 and ω2 are known from the boundary conditions, and are defined by ∞ ∞ x σ(ρ )ρ dρ 0 0 0 x d ⌠ ω1(ρ) = w (ρ) + 2 H πα ⌠ τ(ρ0)dρ0 − 2⌠ , ( x 2 − ρ2)1/2 ⌡ ( x 2 − ρ20)1/2 ⌡ ⌡ a a a
(3.2.4) ∞
x
dx
τ(ρ0)
ρ20dρ0
⌠ ω2(ρ) = u (ρ) − 4 H γ1γ2ρ⌠ 2 2 2 1/2 2 2 1/2. x ( x ) − ρ ⌡ ⌡ ( x − ρ0) a
(3.2.5)
a
Multiply both sides of equation (3.2.2) by ρ( r 2−ρ2)-1/2dρ, integrate with respect to ρ from zero to r and differentiate with respect to r . The result is a r τ(ρ )dρ a σ(ρ )ρ dρ 0 0 0 0 0 2 ⌠ = χ ( r). ⌠ −α τ(ρ0)dρ0 + α r ⌠ 2 + 2 1/2 2 2 1/2 1 π ⌡ r − ρ ρ − r ( ) ( ) ⌡ ⌡ 0 0 0
(3.2.6)
r
0
Here r ω (ρ)ρ dρ 1 1 d ⌠ χ1( r ) = 2 . π H d r ⌡ ( r 2 − ρ2)1/2
(3.2.7)
0
A similar transformation can be applied to equation (3.2.3), with the result a r σ(ρ )ρ dρ τ(ρ0)dρ0 α 0 0 0 2 = χ ( r), ⌠ √γ1γ2r ⌠ − 2 2 1/2 2 2 1/2 2 π ( ρ − r ) √ γ γ ( r − ρ ) ⌡ ⌡ 0 0 2 2 r
0
(3.2.8)
147
3.2 Internal axisymmetric mixed-mixed problem
where r ω2(ρ)ρ2dρ 1 1 d ⌠ χ2( r ) = 2 . π H √γ1γ2 r d r ⌡ ( r 2 − ρ2)1/2
(3.2.9)
0
The set of modified integral equations (3.2.6) and (3.2.8) is solved below by four different methods. It is shown that all the solutions are consistent with each other, and give effectively the same solution. Solution of the first kind. Assume the solution to the set of equations (3.2.2) and (3.2.3) to be defined by ρ
1 d ⌠ σ(ρ) = ρ dρ ⌡
f 1( t ) t d t (ρ2 − t 2)1/2
ρ f 2( t ) d t d -1/2 ⌠ τ(ρ) = (γ1γ2) . dρ ⌡ (ρ2 − t 2)1/2
,
0
0
Here f 1 and f 2 are the as yet unknown stress functions.
(3.2.10) Substitution of (3.2.10)
in (3.2.6) yields, after integration, a f 2( t ) t d t α 2 2 2 1/2 ⌠ f ( r) + ( a − r ) 2 2 2 2 1/2 = χ1( r ) + b . π √γ1γ2 2 ⌡ ( t − r )( a − t ) 0
(3.2.11) Here the notation was introduced a f 2( t ) d t 2α ⌠ . b = π√γ1γ2 ⌡ ( a 2 − t 2)1/2
(3.2.12)
0
It will be shown later that we may assume b =0, without loss of generality. Hereafter the following identities are used r
a
a
⌠ 2 d x 2 1/2 d ⌠ 2f ( t ) d2t 1/2 = r ⌠ f 2( t ) d t 2 , ⌡ (r − x ) dx ⌡ (t − x ) ⌡ t(t − r ) 0
x
a
0
x
a
f(t) dt f(t) dt ⌠ 2 d x 2 1/2 d ⌠ = ( a 2 − r 2)1/2⌠ 2 2 1/2 2 2 2 2 1/2, ⌡ (x − r ) dx ⌡ (x − t ) ⌡ ( t − ρ )( a − t ) r
0
0
148
CHAPTER 3
a
a
dx d ⌠ ⌠ 2 2 1/2 d x ⌡ (x − r ) ⌡ r
MIXED-MIXED BOUNDARY VALUE PROBLEMS
π f ( r) f(t) dt , 2 1/2 = − 2 r (t − x ) 2
x
r
x
f(t) dt π f ( r ) − f (0) ⌠ 2 d x 2 1/2 d ⌠ = . 2 2 1/2 2 r ⌡ (r − x ) dx ⌡ (x − t ) 0
(3.2.13)
0
The identities (3.2.13) can be verified by using (1.3.2) and (1.3.9). substitution of (3.2.10) in (3.2.8) gives, after simplification, 2 α − f 1( r ) + r ( a 2 − r 2)1/2⌠ π √γ γ ⌡ 1 2
a
f 2( t ) d t ( t 2 − r 2)( a 2 − t 2)1/2
Now
= χ2( r ).
0
(3.2.14) Let
f 1( t ) = − f 1(− t ),
f 2( t ) = f 2(− t ).
(3.2.15)
These assumptions allow us to rewrite equations (3.2.11) and (3.2.14) in the form a f 1( t ) d t α 1 2 2 1/2 f ( r) + ( a − r ) ⌠ 2 2 1/2 = χ1( r ) + b , π √γ1γ2 2 ⌡ ( t − r)( a − t ) -a
a f 2( t ) d t α 1 2 2 1/2 ⌠ − f ( r) + ( a − r ) 2 2 1/2 = χ2( r ). π √γ1γ2 1 ⌡ ( t − r)( a − t ) -a
(3.2.16) By introducing the complex functions f = f 1+ if 2 and χ=χ1+ i χ2, the system depicted in (3.2.16) can be reduced to one singular integral equation, namely, a
−i
f(t) dt α 1 f ( r ) + ( a 2 − r 2)1/2 ⌠ 2 2 1/2 = χ( r ). π √γ1γ2 ⌡ ( t − r)( a − t ) -a
-i θ
iθ
(3.2.17)
Multiply both sides of equation (3.2.17) by ( a + r ) ( a − r ) ( r − y ) d r , where θ is an as yet unknown constant, and integrate with respect to r from − a to a . The -1
149
3.2 Internal axisymmetric mixed-mixed problem
result is α −i √γ1γ2
a
-i θ -i θ ⌠ a + r f ( r) d r − πf ( y )a + y a − y ⌡ a − r r − y
-a
a
a + y -i θ f(t) dt tanh(πθ) ⌠ − i(a − y ) 2 2 1/2 a − y ⌡ ( t − y )( a − t ) 2
2 1/2
-a
a
iθ
a + t + i tanh(πθ) ⌠ ⌡ a − t
a
-i θ ⌠ a + r χ( r) d r . r − y ⌡ a − r
f(t) dt = t − y
-a
-a
(3.2.18) Defining tanh(πθ) = α/√γ γ ,
(3.2.19)
1 2
equation (3.2.18) may be simplified as follows: -i θ
a + y −π a − y
a
f(t) dt f ( y ) + i tanh(πθ)( a 2 − y 2)1/2 ⌠ 2 2 1/2 π ⌡ ( t − y )( a − t ) -a
a
=
-i θ ⌠ a + r χ( r) d r . r − y ⌡ a − r
-a
(3.2.20) The singular integral may be eliminated from (3.2.20) by using (3.2.17), and the exact solution becomes available in the form a
1 a + y i θ ⌠ a + r -i θ χ( r ) d r f ( y ) = −cosh (πθ) i tanh(πθ)χ( y ) + . πa − y ⌡ a − r r − y 2
-a
(3.2.21) The following rule of interchanging the order of integration in singular integrals was employed (Muskhelishvili, 1946)
150
CHAPTER 3
a
a
dr ⌠ ⌡ r − y
-a
MIXED-MIXED BOUNDARY VALUE PROBLEMS
a
⌠ f ( r, t )d t = −π2 f ( y , y ) + ⌡ t − r
a
f ( r , t )d r ⌠ dt ⌠ . ⌡ ⌡ ( r − y )( t − r)
-a
-a
(3.2.22)
-a
The other integrals used here can be found in Appendix A3.1. Strictly speaking, the complete solution of (3.2.17) is given by (3.2.21) plus the term c ( a + y )i θ( a − y )-i θ, which represents the homogeneous solution of (3.2.17), with c being an arbitrary constant. The value of c is to be chosen to satisfy the condition b =0, where b is defined by (3.2.12). The appropriate integration of expression (3.2.21) shows that the condition b =0 is satisfied when c =0, and this is the reason why the final solution is written in the form (3.2.21). The general solution is now completed, and we can consider in more detail the case of a bonded axisymmetric punch, with no tractions applied outside the punch. The total force P may be obtained by integration a
a
f 1( t ) t d t
P = 2π⌠ σ(ρ)ρdρ = 2π⌠ 2 2 1/2 . ⌡ ⌡ (a − t ) 0
(3.2.23)
0
We recall that f 1 is an odd function, and that f 1=ℜ f .
Taking this into
consideration, we obtain, after substitution of (3.2.21) in (3.2.23) a
a + r-iθ χ( r)d r. ⌡ a − r
P = πcosh(πθ)ℜ ⌠
(3.2.24)
-a
The field of displacements outside the punch can be obtained by repeating the derivation of (3.2.2) and (3.2.3) for ρ> a , which results in a
a
dy ⌠ w (ρ) = 4 H ⌠ 2 2 1/2 ( ρ − y ) ⌡ ⌡ 0
for ρ> a ;
y
a
2H y 2d y ⌠ 2γ 1 γ 2 ⌠ u (ρ) = 2 2 1/2 ρ ρ − ( y ) ⌡ ⌡ 0
σ( x )d x , ( x − y 2)1/2 2
a
τ( x )d x α 2 1/2 − 2 P , (x − y ) 2
for
ρ> a .
y
(3.2.25) Substitution of (3.2.10) in (3.2.25) and integration with respect to x leads to
151
3.2 Internal axisymmetric mixed-mixed problem
a
a f (t)tdt 2 2 1/2 1 a y ( − ) ⌠ ⌠ w (ρ) = 4 H y d 2 2 1/2 2 2 2 2 1/2 , ⌡ (ρ − y ) ⌡ ( t − y )( a − t ) 0
2H u (ρ) = 2√γ γ ⌠ ρ 1 2⌡
a
0
0
a f ( t )d t 2 ( a 2 − y 2)1/2 y 2d y ⌠ α 2 2 1/2 2 2 2 2 1/2 − 2 P . (ρ − y ) ⌡ ( t − y )( a − t ) 0
(3.2.26) The singular integrals in (3.2.26) can be evaluated from (3.2.17) and (3.2.21), and the final result may be written as a
ℜ{ Z ( y )} 2 2 1/2 d y , ⌡ (ρ − y )
w (ρ) = 2 H ⌠ 0
a
H y ℑ{ Z ( y )} u (ρ) = √γ γ 2⌠ d y − P tanh(πθ), for ρ> a . 2 2 1/2 1 2 ρ ⌡ (ρ − y ) 0
(3.2.27) Here i a + y i θ Z ( y ) = πcosh (πθ)χ( y ) − tanh(πθ) π a − y 2
a
-i θ ⌠ a + r χ( r)d r . ⌡ a − r r − y
-a
(3.2.28) Formulae (3.2.10) and (3.2.21) are the main results of this section. Solution of the second kind. a F (t)tdt 1 1 d ⌠ , σ(ρ) = ρ dρ ⌡ ( t 2 − ρ2)1/2
ρ
Again, F
1
and F
2
Assume solution to the problem in the form
a F ( t )d t 2 1 d ⌠ . τ(ρ) = √γ γ dρ ⌡ ( t 2 − ρ2)1/2 1 2
(3.2.29)
ρ
are the as yet unknown stress functions.
Substitution of
(3.2.29) in (3.2.6) and (3.2.8) results in a
2 α ⌠ π√γ γ ⌡ 1 2 0
F ( t )d t 2
t
α r ⌠ + √γ γ ⌡ 2
1 2 0
a
F ( t )d t 2
t(t
2
− r) 2
−
π F ( r ) = χ ( r ), 1 2 1
152
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
a F ( t )d t 1 2 π α r ⌠ = χ ( r). − F 2( r ) − 2 2 2 π 2 √γ γ ⌡ t − r
(3.2.30)
1 2 0
Assuming that F ( t ) = F (− t ), 1
F ( t ) = − F (− t ),
1
2
(3.2.31)
2
equations (3.2.30) can be rewritten as a
α −F ( r) + 1 π√γ γ
2
⌠ ⌡
1 2
F ( t )d t t − r
= χ ( r ), 1
-a
a
α −F ( r) − 2 π√γ γ
F ( t )d t 1
⌠ = χ ( r ). 2 ⌡ t − r
1 2
(3.2.32)
-a
Introducing the complex functions F ( r ) = F ( r ) + iF ( r ), 1
2
χ( r ) = χ ( r ) + i χ ( r ), 1
2
(3.2.33)
the system (3.2.32) can be reduced to a single equation, namely,
F ( r) +
iα π√γ γ
1 2
a
F ( t )d t
⌠ = − χ( r ). ⌡ t − r
(3.2.34)
-a
Multiplication of (3.2.34) by ( a + r )-i θ( a − r )i θ( r − y )-1 and integration with respect to r from − a to a leads to a
-i θ -i θ -i θ ⌠ a + r F ( r)d r + i α −π2a + y F ( y ) + a + y πcoth(πθ) a − y i π√γ γ a − y ⌡ a − r r − y 1 2
-a
153
3.2 Internal axisymmetric mixed-mixed problem
a
a
a
F ( t )d t π a + t -i θ F ( t )d t a + r -i θ χ( r )d r = − ⌠ . × ⌠ − coth(πθ) ⌠ i ⌡ t − y ⌡ a − t t − y ⌡ a − r r − y -a
-a
-a
(3.2.35) Here formula (3.2.22) and the integrals from Appendix 3.1 were used. Taking, as before, tanh(πθ)=α/√γ γ , equation (3.2.35) can be simplified significantly, 1 2 namely, a
a
i F ( t )d t i a + y i θ ⌠ a + r -i θ χ( r )d r = − coth(πθ) . F ( y ) + coth(πθ) ⌠ π π a − y ⌡ a − r r − y ⌡t − y -a
-a
(3.2.36) The singular integral can be eliminated from (3.2.36) by using (3.2.34), and the final result is F ( y ) = cosh2(πθ)−χ( y ) +
iθ
i a + y tanh(πθ) π a − y
a
-i θ ⌠ a + r χ( r)d r . ⌡ a − r r − y
-a
(3.2.37) The general solution may be considered completed. Some additional results are presented for the case of a bonded punch. The total force, exerted by the punch is obtained by integration of σ: a
a
P = −2π⌠ F 1( t )d t = −π ⌠ F 1( t )d t .
⌡
⌡
0
(3.2.38)
-a
Substitution of (3.2.37) in (3.2.38) yields a
a + r-iθ χ( r)d r. ⌡ a − r
P = πcosh(πθ)ℜ ⌠ -a
As expected, this result is identical to (3.2.24). The displacements outside the punch can be found by substitution of (3.2.37) and (3.2.29) in (3.2.25), with the result
154
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
a
ℜ{ F ( y )} 2 2 1/2 d y , ⌡ (ρ − y )
w (ρ) = −2π H⌠ 0
a
H y ℑ{ F ( y )} u (ρ) = − √γ γ 2π⌠ 2 d y + P tanh(πθ), for ρ> a . 1 2 ρ ⌡ (ρ − y 2)1/2 0
(3.2.39) Comparison of (3.2.28) and (3.2.37) shows that Z ( y )=−π F ( y ), and this means that formulae (3.2.39) actually repeat (3.2.27). Solution of the third kind. Let the stresses in question be expressed through two new and as yet unknown functions q and q as follows 1
ρ
1 d ⌠ σ(ρ) = ρ dρ ⌡
a q ( t )d t 2 1 d ⌠ . τ(ρ) = √γ γ dρ ⌡ ( t 2 − ρ2)1/2
q (t)tdt 1
(ρ2 − t 2)1/2
2
,
1 2
0
ρ
(3.2.40) Substitution of (3.2.40) in (3.2.6) and (3.2.8) leads to the system a q (t)tdt a q (t)tdt 2 1 α ⌠ π 2 2 1/2⌠ − r) 2 2 + (a 2 2 2 2 1/2 = 2 χ1( r ), √γ γ ⌡ t − r ⌡ ( t − r )( a − t ) 1 2 0
0
−q ( r) − 2
α q ( r ) = χ ( r ). 1 2 √γ γ
(3.2.41)
1 2
Function q
2
can be expressed from the second equation (3.2.41) as
q ( r) = − 2
α q ( r ) − χ ( r ). 2 √γ1γ2 1
(3.2.42)
Substitution of (3.2.42) into the first equation (3.2.41) gives a
(a
2
− r) ⌠ 2 1/2
1
2 2 2 2 1/2 ⌡ ( t − r )( a − t ) 0
Here
q (t)tdt
a q (t)tdt 1 α2 ⌠ = ψ( r ). − 2 γ γ ⌡ t − r2 1 2 0
(3.2.43)
155
3.2 Internal axisymmetric mixed-mixed problem
π α ⌠ ψ( r ) = χ ( r) + 2 1 √γ γ ⌡
a
1 2 0
χ2( t ) t d t t 2 − r2
.
(3.2.44)
An exact solution of the integral equation (3.2.43) can be obtained in the following manner. Introduce the notation Y ( r ) = sinθln|
s
a + r | , a − r
Y ( r ) = cosθln|
c
a + r | . a − r
(3.2.45)
Multiply both sides of (3.2.43) by Y ( r )/( r 2− x 2) and integrate with respect to r c
from 0 to a .
The result is
a Y (t) Y ( x ) q ( t )d t s a 2 − x 21/2 s 1 π2 α2 π ⌠ 1− q ( x ) Y ( x ) + tanh(πθ) − − c 2 a 2 − t 2 x ( x 2 − t 2) 4 x 2 γ 1γ 2 1 ⌡ t 0
a q (t) Y (t) Y ( x) 1 s s πα2 d t = X ψ( x ). coth(πθ)⌠ 2 − − 2 c 2γ γ x t ⌡ x − t 1 2 0
(3.2.46) Hereafter the following integral operators are introduced a
X ψ( x ) = ⌠
⌡
c
Y ( r )ψ( r )d r c
r − x 2
2
a
X ψ( x ) = ⌠
,
s
0
⌡
Y ( r )ψ( r ) r d r s
r2 − x2
.
(3.2.47)
0
Again, assuming tanh(πθ)=α/√γ γ , equation (3.2.46) can be simplified as 1 2
π2 q ( x ) Y ( x )
a 2 2 1/2 q 1( t )d t a x π − + tanh(πθ) Y ( x )⌠ 1 − = x X ψ( x ). − 2 2 2 s c 2 a − t x 2 − t 2 4 x cosh (πθ) ⌡ 1
c
0
A similar procedure of multiplication of (3.2.43) by rY ( r )/( r s
simplification, to
2
(3.2.48) − x ) leads, after 2
156
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
π2 q 1( x ) Y s( x )
a q ( t )d t π a 2 − x 21/2 1 ⌠ − − tanh(πθ) Y c( x ) 1 − = Xsψ( x ). 2 a 2 − t 2 x 2 − t 2 4 x cosh2(πθ) ⌡ 0
(3.2.49) Equations (3.2.48) and (3.2.49) finally give the solution Y c( x ) 4 2 q 1( x ) = − 2 cosh (πθ) xY c( x )Xcψ( x ) + Y s( x )Xsψ( x ) + b 0 . x π (3.2.50) The last term in (3.2.50) represents the homogeneous solution, and b 0 is an arbitrary constant. If the stresses defined by (3.2.40) are to be nonsingular at ρ=0, then b 0 should be equal to zero. Substitution of (3.2.44) in (3.2.50) allows us to express the solution in terms of χ1 and χ2: q 1( x ) = −cosh2(πθ)
2
π
+
xY c( x )Xcχ1( x ) +
2 Y ( x )Xsχ1( x ) − tanh(πθ)χ2( x ) π s
2 2 xY c( x )Xs{χ2( x )/ x } − Y s( x )Xc{ x χ2( x )}. π π
(3.2.51)
Here a modified version of formula (3.2.22) was employed: a
a
a
a
2 ⌠ 2 d t 2 ⌠ f (2t , r)d r2 = − π f ( x2, x ) + ⌠ d r ⌠ 2 f ( t2, r)d2t 2 . 4x ⌡ t − x ⌡ r − t ⌡ ⌡ ( t − x )( r − t )
0
0
0
0
(3.2.52) The second stress function q 2 may be obtained from (3.2.42) and (3.2.51) in the form
2 tanh(πθ)[ xY c( x )Xcχ1( x ) + Y s( x )Xsχ1( x ) π
q 2( x ) = cosh2(πθ)
+ xY c( x )Xs{χ2( x )/ x } − Y s( x )Xc{ x χ2( x )}] − χ2( x ).
(3.2.53)
157
3.2 Internal axisymmetric mixed-mixed problem
The general solution is completed, and we can derive some additional results for the case of a bonded punch. The total force P may be obtained by integration of the first expression in (3.2.40): a
q 1( t ) t d t
P = 2π ⌠ 2 2 1/2 . ⌡ (a − t )
(3.2.54)
0
Substitution of (3.2.51) in (3.2.54) gives, after integration, a
P = 2πcosh(πθ)⌠ [χ ( r ) Y ( r ) + χ ( r ) Y ( r )]d r . 1 c 2 s
(3.2.55)
⌡
0
The result (3.2.55) coincides with (3.2.24) when χ
is an even function and χ
1
2
is odd. The displacements outside the punch may be found by substitution of (3.2.40) in (3.2.25), and are a
a
q (t)tdt
1 a 2 − y 21/2 dy ⌠ 2 w (ρ) = 4 H ⌠ 2 2 2 2 2 1/2 , ⌡ ρ − y ⌡ ( t − y )( a − t ) 0
0
a q ( y) ydy 2 1 ⌠ u (ρ) = − H √γ γ 2π 2 2 1/2 + P tanh(πθ) , for ρ> a . 1 2 ρ ⌡ (ρ − y ) 0
(3.2.56) One can prove that expressions (3.2.56) are in agreement with (3.2.28). Solution of the fourth kind. Let the solution be ρ
a
Q (t)tdt 1 1 d ⌠ σ(ρ) = , 2 ρ dρ ⌡ ( t − ρ2)1/2 ρ
1 d ⌠ τ(ρ) = d √γ1γ2 ρ ⌡
Q ( t )d t 2
(ρ − t 2)1/2 2
.
0
(3.2.57) By using the same methods as before, the following set of equations can be obtained α Q ( r) − Q ( r) = χ ( r) + b, 1 1 √γ1γ2 2
158
CHAPTER 3
a
(a − r ) ⌠ 2
2 1/2
Q 2( t )d t
2 2 2 2 1/2 ⌡ ( t − r )( a − t ) 0
MIXED-MIXED BOUNDARY VALUE PROBLEMS
a Q ( t )d t 1 α ⌠ π − 2 2 = 2 r χ2( r ), √γ γ ⌡ t − r 1 2 0
(3.2.58) where a Q 2( t )d t 2α ⌠ b = . π√γ γ ⌡ ( a 2 − t 2)1/2 1 2
(3.2.59)
0
It will be shown later, that Q 2 can always be chosen in such a way that b =0. Now Q 1 can be expressed from the first equation (3.2.58) and substituted in the second one, with the result a
(a
2
Q 2( t )d t
− r) ⌠ 2 1/2
a
Q 2( t )d t α ⌠ − = Ψ( r ). γ 1γ 2 ⌡ t 2 − r 2 2
2 2 2 2 1/2 ⌡ ( t − r )( a − t ) 0
0
(3.2.60) Here a
π α ⌠ Ψ( r ) = χ2( r ) − 2r √γ1γ2 ⌡
χ1( r ) + b t 2 − r2
dt.
(3.2.61)
0
Equation (3.2.60) is similar to (3.2.43), so its solution can be written down as Q 2( x ) = −
4 2 x2Y ( x )X Ψ( x ) + xY ( x )X Ψ( x ) + b Y ( x ). 2 cosh (πθ) c c s s 2 c π
(3.2.62) The last term in (3.2.62) represents the homogeneous solution, and b 2 is an arbitrary constant to be chosen from the condition b =0. Appropriate integration of (3.2.62), using the integrals from Appendix 3.1, allows us to define b 2 as follows a
4 b 2 = − 2cosh2(πθ) ⌠ Ψ( x ) Y c( x )d x . π ⌡ 0
Substitution of (3.2.63) in (3.2.62) yields
(3.2.63)
159
3.2 Internal axisymmetric mixed-mixed problem
4 2 2 2cosh (πθ) Y c( x )Xc{ x Ψ( x )} + xY s( x )XsΨ( x ) . π
Q 2( x ) = −
(3.2.64)
Function Q 2 can be expressed in terms of χ1 and χ2 by using (3.2.61) and (3.2.59) as follows: Q 2( x ) = − cosh2(πθ){tanh(πθ)χ1( x ) +
2 [ xY ( x )Xcχ1( x ) π s
− Y c( x )Xsχ1( x ) + Y c( x )Xc( x χ2( x )) + xY s( x )Xs(χ2( x )/ x )]}. (3.2.65) Consequently, from the first expression of (3.2.58), we have Q 1( x ) = −cosh2(πθ){χ1( x ) +
2 tanh(πθ)[ xY s( x )Xcχ1( x ) π
− Y c( x )Xsχ1( x ) + Y c( x )Xc( x χ2( x )) + xY s( x )Xs(χ2( x )/ x )]}. (3.2.66) The total force exerted by a bonded punch is defined by a
P = −2π⌠ Q 1( t )d t .
⌡
0
One can show that this result coincides with (3.2.55). the punch are a
The displacements outside
Q 1( y )d y
w (ρ) = −2π H ⌠ 2 2 1/2, ⌡ (ρ − y ) 0
a a Q 2( t )d t 2 2 1/2 a − y α 2H 2 u (ρ) = y dy ⌠ 2 2√γ γ ⌠ 2 2 2 1/2 − 2 P . ρ 1 2 ⌡ ρ2 − y 2 ⌡ ( t − y )( a − t ) 0
0
(3.2.67) Expressions (3.2.67) are in agreement with (3.2.27). Example 1. Consider a circular flat-ended punch of radius a , bonded to a transversely isotropic half-space, and acted upon by an axial force P . The boundary conditions (3.2.1) in this particular case will take the form
160
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
u (ρ) = 0,
w (ρ) = w 0 = const.,
σ(ρ) = 0,
τ(ρ) = 0,
for ρ< a ;
for ρ> a .
The stress functions are w0
a + y i θ coth(πθ)1 − cosh(πθ) , a − y πH
f( y) = i
2
w0
a + y i θ F ( y ) = − 2 cosh(πθ) , a − y πH w0
Q 2( y ) =
π2 H
cosh(πθ) Y c( y ),
π2 H
coth(πθ)[1− cosh(πθ) Y c( y )].
w 0cosh2(πθ)
q 1( y ) =
Q 1( y ) = −
w0
π2 H sinh(πθ)
q 2( y ) = −
Y s( y ),
w0 π2 H
cosh(πθ) Y s( y ),
One can notice that q 1=ℜ f , q 2=ℑ F , Q 1=ℜ F , and Q 2=ℑ f , and this is why we have only two different representations for the surface tractions, namely, ρ Y s( t ) t d t 1 d ⌠ σ(ρ) = 2 π H sinh(πθ) ρ dρ ⌡ (ρ2 − t 2)1/2
w 0cosh2(πθ)
0
a Y c( t ) t d t 1 d ⌠ , = − 2 cosh(πθ) ρ dρ ⌡ ( t 2 − ρ2)1/2 πH
w0
ρ
τ(ρ) = −
w 0cosh(πθ) π2 H α
ρ
d ⌠ dρ ⌡ 0
Y c( t )d t (ρ2 − t 2)1/2
= −
w 0cosh(πθ) π2 H √γ1γ2
a Y s( t )d t d ⌠ . dρ ⌡ ( t 2 − ρ2)1/2
ρ
The equivalence of these representations can be verified by using the identities
161
3.2 Internal axisymmetric mixed-mixed problem
ρ
a
Y c( t )d t
Y ( t )d t
s π ⌠ ⌠ = tanh( π θ ) 2 2 1/2 2 2 1/2 + 2cosh(πθ), ⌡ (t − ρ ) ⌡ (ρ − t )
(3.2.68)
ρ
0
ρ
Y s( t ) t d t
⌠ 2 2 1/2 ⌡ (ρ − t )
0
These and other relationship:
a Y c( t ) t d t πθ a ⌠ = − tanh(πθ) . cosh(πθ) ⌡ ( t 2 − ρ2)1/2
(3.2.69)
ρ
similar identities can be established by using the general
ρ
a
a
dx ⌠ 2f ( t ) t d t 2 + π lim[ tf ( t )]. 2 2 1/2 2ρ t→0 (ρ − x ) ⌡ t − x
2 ⌠ f ( x )d x ⌠ 2 2 1/2 = π ⌡ (x − ρ ) ⌡ ρ
0
0
(3.2.70) The relationship between the total force P and the punch settlement w 0 is
P =
2w0 aθ H tanh(πθ)
.
(3.2.71)
The surface displacements outside the punch are given by a Y c( x )d x 2 ⌠ w (ρ) = w cosh(πθ) 2 2 1/2, π 0 ⌡ (ρ − x ) 0
2 u (ρ) = w √γ γ cosh(πθ)⌠ π 0 1 2 ⌡
a
Y s( x ) x d x (ρ − x ) 2
2 1/2
−
πaθ 1 . cosh(πθ) ρ
0
(3.2.72) The integrals in (3.2.72) can be evaluated (Gradshtein and Ryzhik, 1963) a
I c(ρ) = ⌠
⌡ 0
Y c( x )d x (ρ2 − x 2)1/2
=
πθ sinh(πθ)
∞
Σ(−1)(2(kρ /+a 1)(− k1)!) k
2
2
-k-1/2
2
k =0
k
Π( θ
2
+ m 2),
m=1
(3.2.73)
162
CHAPTER 3
a
I s(ρ) = ⌠
Y s( x ) x d x
⌡ (ρ − x ) 2
2 1/2
π a θ2 = sinh(πθ)
MIXED-MIXED BOUNDARY VALUE PROBLEMS
∞
Σ
(−1)k(ρ2/ a 2 − 1)-k-1/2 (2 k + 1) k !( k + 1)!
k =0
0
Π( θ
2
+ m 2).
m=1
For real materials, the physical constant θ<1. material θ =
k
(3.2.74) For example, for an isotropic
1 ln(3 − 4ν). 2π
Since the Poisson coefficient ν≤0.5, materials. By using the approximation
(3.2.75) this
means
that
0≤θ<0.2
for
isotropic
k
( k !)
-2
Π( θ
2
+ m 2) = 1 + O(θ2),
m=1
the summation in (3.2.73) and (3.2.74) can be performed, and the results are I (ρ) ≈ c
a πθ sin-1( ), sinh(πθ) ρ
I (ρ) ≈ s
a2 a πθ2 2 (ρ − a 2)1/2ln1 − 2 + 2 a sin-1( ). sinh(πθ) ρ ρ
(3.2.76)
(3.2.77) Direct numerical computations show that the relative error of (3.2.76) is less than 6% when (ρ/ a )>1.01, and θ≤0.2; the relative error of (3.2.77) is less than 4%. The same accuracy can be obtained for (ρ/ a )>1.1, and θ≤0.3. The relative error of both (3.2.76) and (3.2.77) rapidly decreases when ρ increases, for example, the relative error is less than 4% and 2% respectively, for θ<0.9 and (ρ/ a )>3. Example 2. Consider the case where no forces act upon the punch bonded to a half-space, but the punch itself expands, so that the radial displacement is proportional to the radius, with k as the coefficient, namely, inside the circle ρ= a the boundary conditions are w (ρ) = w , 0
u (ρ) = k ρ.
Here w is the as yet unknown punch settlement. 0
The stress functions are
163
3.2 Internal axisymmetric mixed-mixed problem
f( y) =
coth(πθ) 2 ky a + y iθ 2 k (2 a θ + iy ) + w , iw i cosh( ) − − π θ 0 0 a − y √γ1γ2 π2 H √γ1γ2
cosh(πθ) 2k a + y i θ (2 a θ + iy ) , w + 2 a − y πH 0 √γ1γ2
F ( y) = −
q ( y) = 1
coth(πθ) 2k cosh( ) ( ) + [cosh( ) 2 ( )+ ( ) ] w π θ Y y π θ ( a θ Y y yY y ) − y , 0 s s c π2 H √γ1γ2
q ( y) = − 2
Q ( y) = − 1
Q ( y) = 2
cosh(πθ) 2k w Y ( y ) + [2 a Y ( y ) + yY ( y )] θ , 0 s s c π2 H √γ γ
1 2
cosh(πθ) 2k w Y ( y ) + [2 a θ Y ( y ) − yY ( y )] 0 c , c s π2 H √γ γ
1 2
coth(πθ) 2k w [1 − cosh( π θ ) Y ( y )] − cosh( π θ ) [2 a θ Y ( y ) − yY ( y )] . 0 c c s π2 H √γ γ
1 2
The stresses can be defined two ways ρ
q ( x) xdx
1 d ⌠ σ(ρ) = ρ dρ ⌡
1
(ρ2 − x 2)1/2
0
ρ
1 d ⌠ τ(ρ) = √γ γ dρ ⌡ 1 2
Q ( x) xdx 1 1 d ⌠ = , 2 ρ dρ ⌡ ( x − ρ2)1/2
Q ( x )d x 2
(ρ2 − x 2)1/2
0
a
ρ
a q ( x )d x 2 1 d ⌠ = . 2 √γ γ dρ ⌡ ( x − ρ2)1/2 1 2
ρ
The total force P is defined by P =
2aθ 2 ka θ coth(πθ) w + . H 0 √γ γ 1 2
When no force acts upon the punch, its normal displacement is equal to
(3.2.78)
164
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
w = −2 ka θ/√γ γ . 0
1 2
The force, needed to provide a zero normal displacement is P =
4 ka 2θ2 . Hα
The displacements outside the punch are a
w (ρ) = −2π H ⌠
Q 1( x )d x
2 2 1/2 ⌡ (ρ − x )
,
0
a
q ( x) xdx
2 1. u (ρ) = − H √γ1γ22π ⌠ 2 2 1/2 + P tanh(πθ) ⌡ (ρ − x ) ρ 0
One can calculate the stress at the point ρ=0 in elementary functions σ(0) =
coth(πθ)θ k w cosh(πθ) + [(1+4θ2)cosh(πθ) − 1], a 0 πH √γ1γ2
τ(0) = 0.
Discussion. One could notice that the four kinds of solution considered represent all combinations of the Abel type integrals with limits from 0 to ρ and from ρ to a . Solutions of the first and the second kind are more compact than the others, but they are only convenient to use when w (ρ) is an even function and u (ρ) is an odd one. The solution of the second kind is preferable when one is interested mainly in the displacements outside the punch, while the solution of the first kind has definite advantages when one is interested in the stress distributions. Formulae (3.2.10) are more convenient for numerical integration than (3.2.29), especially in the domain close to either ρ=0 or ρ= a : (i) the differentiation of the integrand can be performed in (3.2.10), thus avoiding numerical differentiation, which is less accurate; the differentiation in (3.2.29) is rather difficult because F ( a ) usually is not defined; (ii) formulae (3.2.10) allow us to determine easily the stress at ρ=0, directly through the stress function, namely, σ(0) =
π f ´(0), 2 1
τ(0) = f 2´(0)/√γ1γ2,
while the result of (3.2.29) is rather difficult to use, for example,
165
3.2 Internal axisymmetric mixed-mixed problem
a a F ( x ) − F (0) F 1( x ) x d x F 1(0) 1 d 1 1 ⌠ ⌠ σ(0) = lim dx − . = a x2 ( x 2 − ρ2)1/2 ρ→0 ρ dρ ⌡ ⌡ 0 ρ
(3.2.79) The solutions of the third and the fourth kind are more general because they do not require either w (ρ) to be an even function or u (ρ) to be an odd one. The same logic of preference is applicable here: the integral representations with limits from 0 to ρ are more convenient for evaluation of the stresses while the representations with limits from ρ to a are preferable for the displacement evaluation. It is of interest to establish relationships between the various kinds of solution. Some are obvious, due to the uniqueness of the solution, namely, f 1 = q 1,
f 2 = Q 2,
F 1 = Q 1,
F 2 = q 2.
(3.2.80) The other relationships may be found from (3.2.70) and the following identity ρ
a
a
f ( x )d x xdx 2 ⌠ ( a 2 − y 2)1/2 f ( y )d y ⌠ ⌠ − . = 2 2 1/2 π ⌡ ( x 2 − ρ2)1/2 ⌡ ( a 2 − x 2)1/2( y 2 − x 2) ⌡ (ρ − x ) ρ
0
0
(3.2.81) Comparison of (3.2.57) and (3.2.40) with (3.2.70) and (3.2.81) yields a
Q 1( t ) = −
( a 2 − y 2)1/2 yq 1( y )
2 ⌠ dy, π ⌡ ( a 2 − t 2)1/2( y 2 − t 2) 0
a
q 2( t ) = −
2 ⌠ t π ⌡
( a 2 − y 2)1/2 Q 2( y ) ( a 2 − t 2)1/2( y 2 − t 2)
dy,
0
(3.2.82) a
2
y Q ( y) 1 2 ⌠ q (t) = dy, 2 1 π t ⌡ ( y − t 2) 0
a
y q ( y) 2 2 ⌠ dy. Q (t) = 2 2 π ⌡ ( y − t 2) 0
(3.2.83) Comparison of (3.2.56) and (3.2.67) leads to slightly different expressions:
166
CHAPTER 3
a
MIXED-MIXED BOUNDARY VALUE PROBLEMS
( a 2 − t 2)1/2 yq 1( y )d y
2 ⌠ , π ⌡ ( a 2 − y 2)1/2( y 2 − t 2)
Q 1( t ) = −
0
a
2 ⌠ t π ⌡
q 2( t ) = −
( a 2 − t 2)1/2 Q 2( y )d y ( a 2 − y 2)1/2( y 2 − t 2)
.
0
(3.2.84) Expressions (3.2.84) differ from (3.2.82) by the term const/( a − t ) and 2 2 1/2 t ⋅const/( a − t ) respectively. One may notice from (3.2.57) and (3.2.40) that addition of these terms to Q 1 and q 2 respectively does not affect the stresses and 2
2 1/2
therefore, expressions (3.2.82) and (3.2.84) are equivalent. The same argument is applicable to q 1. Since the addition to q 1 of a term const/ t does not affect the solution, an alternative to the first expression of (3.2.83) may be suggested a
Q 1( y )d y
2 ⌠ q 1( t ) = t π ⌡
y2 − t2
.
(3.2.85)
0
Comparison of (3.2.39) and (3.2.26) between the complex stress functions
allows us to construct the relationship
a
2 2 1/2 ⌠ ( a2 − t2 )1/2 f ( y )d y . ⌡ (a − y ) (y − t)
1 F(t) = − π
(3.2.86)
-a
The inverse relationship takes the form a
1 f(t) = π
⌠ F ( y )d y . ⌡ y − t
(3.2.87)
-a
One can also deduce from (3.2.82) the following expression which is equivalent to (3.2.86): a
1 F(t) = − π
2 2 1/2 ) f ( y )d y ⌠ ( a2 − y2 1/2 . ⌡ (a − t ) (y − t)
-a
(3.2.88)
167
3.2 Internal axisymmetric mixed-mixed problem
Considerable simplifications occur when θ=0. In the case of an isotropic body this condition corresponds to the Poisson coefficient ν=1/2. The stress functions will be defined by a
1 f( y) = − π
⌠ χ( r)d r , ⌡ r − y
F ( y ) = −χ( y ),
-a
a χ ( r )d r 1 2 ⌠ q ( y) = − y 2 1 π ⌡ r − y2
q ( y ) = −χ ( y ), 2 2
0
a
2 ⌠ Q 2( y ) = − π ⌡
χ2( r ) r d r r2 − y2
Q 1( y ) = −χ1( y ).
0
The stress distributions are a χ1( x ) x d x 1 d ⌠ σ(ρ) = − , ρ dρ ⌡ ( x 2 − ρ2)1/2
a χ2( x )d x 1 d ⌠ τ(ρ) = − . √γ1γ2 dρ ⌡ ( x 2 − ρ2)1/2
ρ
ρ
The displacements outside the punch simplify as follows: a χ2( x ) x d x 1 ⌠ u (ρ) = 2π H √γ1γ2 , ρ ⌡ (ρ2 − x 2)1/2 0
a
χ1( x )d x
w (ρ) = 2π H ⌠ 2 2 1/2. ⌡ (ρ − x ) 0
(3.2.89) Notice that here the normal parameters are decoupled from the tangential ones, namely, the normal displacements affect the normal pressure, and the tangential displacements produce the shear tractions only. Substitution of (3.2.7) and (3.2.9) in (3.2.89) furnishes a direct relationship between the displacements inside and outside the circle ρ= a a
2 (ρ2 − a 2)1/2 ⌠ u ( x ) x 2d x u (ρ) = 2 2 1/2 2 2 , π ρ ⌡ ( a − x ) (ρ − x ) 0
for ρ> a ;
168
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
a
w( x) xdx 2 w (ρ) = (ρ2 − a 2)1/2 ⌠ 2 2 1/2 2 2 , π ⌡ ( a − x ) (ρ − x )
for ρ> a .
0
The last expressions demonstrate a certain mathematical similarity between the normal and tangential displacements.
Exercise 3.2 1. Prove that the settlement of a smooth flat circular punch is greater then or equal to that of a bonded punch. Hint : prove that (πθ)≥tanh(πθ). Note : a more general statement can be proven from the energy consideration. 2. An axisymmetric pressure σ=σ(ρ) is applied to the annulus b ≤ρ≤ c outside a flat bonded punch of radius a . Investigate its influence on the punch settlement w 0 and the traction distributions under the punch. Solution: The right hand side in the integral equations (3.2.2) and (3.2.3) will take the form ρ
ω1(ρ) = w 0 − 4 H ⌠
⌡ 0
c σ(ρ )ρ dρ 0 0 0 dx ⌠ 2 2 1/2 2 2 1/2, (ρ − x ) ⌡ (ρ0 − x )
ω2 = 0.
b
Now, from (3.2.7) and (3.2.9) c σ(ρ )ρ dρ 0 0 0 2 ⌠ χ1( r ) = 2 − , 2 2 1/2 π ⌡ (ρ0 − r ) πH
w0
χ2 = 0.
b
Substitution of the last expressions in the chosen kind of solution gives the stresses and the displacements. The total force can be defined as w0 aθ
c
P = 2cosh(πθ) − 2⌠ I c(ρ) σ(ρ)ρdρ, H sinh(πθ) ⌡ b
where I c is defined by (3.2.73).
When the punch is not subjected to a direct
169
3.2 Internal axisymmetric mixed-mixed problem
loading, its settlement will be c
2H sinh(πθ)⌠ I (ρ)σ(ρ)ρdρ. w = 0 aθ ⌡ c b
If no displacements are allowed for the punch, then the total force should be c
P = −4cosh(πθ)⌠ I (ρ) σ(ρ)ρdρ.
⌡
c
b
By using the approximation (3.2.76), the results can in many cases be expressed in elementary functions. For example, in the case when σ=σ =const, 0
πθσ 0 a a ⌠ I (ρ)σ(ρ)ρdρ ≈ c 2sin-1( ) − b 2sin-1( ) c 2sinh(πθ) c b ⌡ c
b
+ a [( c 2 − a 2)1/2 − ( b 2 − a 2)1/2].
3. Investigate the influence of radial tangential tractions τ=τ(ρ), applied at the surface of an annulus b ≤ρ≤ c , outside a bonded punch of radius a . Solution : We have from (3.2.2) and (3.2.3) c
ω (ρ) = w + 2π H α⌠ τ(ρ)dρ, 1
⌡
0
b
ρ c τ(ρ )dρ 2 0 0 1⌠ d x x ⌠ ω (ρ) = −4 H γ γ 2 2 1/2 2 2 1/2. 2 1 2 ρ ⌡ (ρ − x ) ⌡ (ρ0 − x ) 0
b
Substitution in (3.2.7) and (3.2.9) yields w
c
2 ⌠ χ ( r) = 2 + α τ(ρ)dρ, 1 π πH ⌡ 0
b
c
2 τ( x )d x . χ ( r ) = − √γ γ r⌠ 2 π 1 2 ⌡ ( x 2 − r 2)1/2 b
170
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
Since χ is an even function, and χ is an odd one, we can use any of the 1 2 four kinds of solution considered above.
The total force is given by (3.2.55):
c c aθ w + 2π H α⌠ τ(ρ)dρ − 2√γ γ ⌠ I (ρ)τ(ρ)dρ. P = 2cosh(πθ) 1 2 H sinh(πθ) 0 ⌡ ⌡ s b b
Here I
s
is defined by (3.2.74).
When the punch is not loaded, its settlement
will be c
c
sinh(πθ) w = 2 H −πα ⌠ τ(ρ)dρ + √γ γ ⌠ I (ρ)τ(ρ)dρ. 0 1 2 a θ ⌡ ⌡ s b
b
The value of the axial force to provide a zero punch displacement is c c πa θα ⌠ τ(ρ)dρ − √γ γ ⌠ I (ρ)τ(ρ)dρ. P = 4cosh(πθ) 1 2 sinh(πθ) ⌡ ⌡ s b b
4. A normal concentrated load P is applied at the point (0,0, z ) underneath a circular punch of radius a , bonded to a transversely isotropic elastic half-space. Find the traction distribution at the punch base and its settlement w . Answer: as an illustration, the solution is given according to (Fabrikant, 1971a). The stresses are defined by a ρ f (t)tdt f (t)tdt 1 2 1 d ⌠ ⌠ + σ(ρ) = , 2 2 1/2 2 ρ dρ ⌡ (ρ − t ) ( t − ρ2)1/2 ⌡
ρ
0
a ρ f ( t )d t f ( t )d t 1 2 1⌠ d α ⌠ + . τ(ρ) = − dρ γ γ ⌡ ( t 2 − ρ2)1/2 α⌡ (ρ2 − t 2)1/2 1 2
ρ
0
Here 2
Σ
cosh2πθ f (t) = 1 π2
k=1
m
tY ( t )sinhξ + z Y ( t )coshξ
( m − 1)sinhπθ k k
c
k
k s
z 2k
2
+ t
k
,
171
3.3 External axisymmetric mixed-mixed problem
coshπθ f 2( t ) = 2 π √γ1γ2
2
Σ k =1
γk
zkY c( t )sinhξk + tY s( t )coshξk
m k − 1
z 2k
+ t
2
, ξ = 2θtan-1( a ). zk k
The resultant of the stresses is 2
Σ
m sinhξ
γ (coshξ − 1)
k k k k . N = − P cothπθ + − m 1 k √γ1γ2( m k − 1) k =1
When no force is applied to the punch directly, its settlement due to the load P is 2
HP w = 2aθ
Σ
m sinhξ
γ (coshξ − 1)
k k k k . + − m 1 k √γ1γ2( m k − 1) k =1
Note : verify that in the case z →0 N =− P .
3.3 External axisymmetric mixed-mixed problem We choose this problem to demonstrate yet another kind of solution, which features two stress functions, introduced in such a way that they decouple the integral equations, so that each equation can be solved independently. The boundary conditions in the case of axial symmetry are u = u (ρ),
for
a ≤ρ≤∞,
0≤φ<2π;
w = w (ρ),
for
a ≤ρ≤∞,
0≤φ<2π;
σ = σ(ρ),
for
0≤ρ≤ a ,
0≤φ<2π;
τ = τ(ρ),
for
0≤ρ≤ a ,
0≤φ<2π.
The set of governing integral equations will take the form ∞
∞
x σ(ρ )ρ dρ 0 0 0 d x ⌠ ⌠ ⌠ τ(ρ )dρ + 2 2 H −πα = ω1(ρ), ⌡ 0 0 ⌡ ( x 2 − ρ2)1/2 ⌡ ( x 2 − ρ20)1/2 a ρ ρ
(3.3.1)
172
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
(3.3.2) ∞
2H 2γ γ ρ2⌠ ρ 1 2 ⌡
⌠ ⌠ ( ) d − π α σ ρ ρ ρ = ω2(ρ). 2 2 1/2 0 0 0 ⌡ ( x − ρ0) ⌡ x
dx x 2( x 2 − ρ2)1/2
ρ
ρ
τ(ρ )ρ20dρ 0 0
a
The functions ω
1
a
(3.3.3) are known from the boundary conditions, and are
and ω
2
defined by a
ω (ρ) = w (ρ) − 4 H⌠
⌡
1
0
a σ(ρ )ρ dρ 0 0 0 dx ⌠ , (ρ2 − x 2)1/2 ⌡ (ρ20 − x 2)1/2 x
(3.3.4) τ(ρ )dρ x 2d x 0 0 H ⌠ Hα ⌠ ⌠ ω (ρ) = u (ρ) − 4γ γ σ(ρ )ρ dρ . 2 2 1/2 2 2 1/2 + 2π ρ 2 1 2ρ 0 0 0 ⌡ (ρ − x ) ⌡ (ρ0 − x ) ⌡ a
a
0
a
0
x
(3.3.5) We shall seek the solution of the system (3.3.2) and (3.3.3) in the form ∞
f (t)tdt
1 d ⌠ σ(ρ) = ρ dρ ⌡
1
(t
ρ
ρ
2
− ρ)
2 1/2
ρ
f (t)tdt
, ⌡ (ρ − t )
+ ⌠
2
2
2 1/2
a ∞
f ( t )d t
f ( t )d t
C Da
d ⌠ τ(ρ) = C dρ 1⌡
(ρ2 − t 2)1/2
Here f
(3.3.6) are the as yet unknown functions, and C , C , and D are the
1
a
1
and f
2
2 1 + + C⌠ . 2 2 1/2 2 2⌡ (t − ρ ) ρ(ρ − a 2)1/2
ρ
1
2
constants to be determined. Substitution of (3.3.6) in (3.3.2) and (3.3.3) leads to two uncoupled equations which can be solved independently, provided that the constants are defined by C = α /γ γ , 1
1 2
C = −1/α. 2
The equations in question are ρ f ( t )d t πα2 1 -1 a 2 H − D sin ( ) − ⌠ 2 2 1/2 γγ ρ ( t ) ρ − ⌡ 12 a
(3.3.7)
173
3.3 External axisymmetric mixed-mixed problem
∞
∞ f (t)tdt 2 2 1/2 1 ( ) x − a ⌠ ⌠ + 2 d x 2 2 1/2 2 2 = ω1(ρ). ⌡ ( x 2 − ρ2)1/2 ⌡ (t − a ) (t − x )
ρ
a
∞ 2γ γ ∞ 2π H α 1 2⌠ dx ρ2 a 2( t 2 − x 2) − x 4( t 2 − a 2) ⌠ f ( t )d t 2 ρ πα2 ⌡ ( x 2 − ρ2)1/2( x 2 − a 2)1/2 ⌡ x 2( t 2 − a 2)1/2( t 2 − x 2)
ρ
a
ρ f (t)tdt 2 t ⌠ + ⌠ 1 f ( t )d t + aD − − 2 2 1/2 2 2 1/2 = ω2(ρ). 1 ⌡ ( t − a ) ⌡ (ρ − t ) ∞
a
a
(3.3.8) Let us solve the first equation (3.3.8). Divide both sides by ρ(ρ − r ) , integrate with respect to ρ from r to ∞, multiply the result by r , and differentiate with respect to r . The result is 2
∞
α2 ⌠ γ 1γ 2 ⌡
∞
f ( t )d t 1
t
2
− r
2
− ⌠
( r 2 − a 2)1/2 f ( t ) t d t 1
2 2 ⌡ r( t − a ) ( t − r )
a
2 1/2
2
2 1/2
= ψ ( r ). 1
a
(3.3.9) Here ∞
1 d ⌠ r ψ1( r ) = 2π H d r ⌡ r
ω (ρ)dρ
− α2 D lnr + a . γ γ 2 r r − a ρ(ρ2 − r 2)1/2 1 2 1
(3.3.10)
Equation (3.3.9) can be solved in a manner similar to that of (3.2.43). Multiply both sides of (3.3.9) by Y ( r )/( r 2− x 2) and integrate with respect to r from a to ∞.
c
We use in this section the notation Y
(3.2.45).
c
and Y , as it was defined in s
The result of integration is ∞ f 1( t )d t πa π2 α2 ⌠ 1 ( ) ( ) Y x f x − − 1 γγ c 4 x 2 2 x 2cosh(πθ)⌡ t ( t 2 − a 2)1/2 1 2 a
174
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
t ( x 2 − a 2)1/2 Y s( x )
∞
− tanh(πθ) ⌠ 2 2 2 1/2 ⌡ x (t − a )
Y s( t )
f ( t )d t
1 t t2 − x2
−
a
∞
π α2 + coth(πθ)⌠ 2 γ 1γ 2 ⌡
Y s( x ) x
Y s( t )
−
f 1( t )d t
t t2 − x2
∞ψ ( r ) Y ( r )d r 1 c
= ⌠
⌡
a
Again, expression tanh(πθ)=α/√γ1γ2,
r2 − x2
.
a
(3.3.11)
can
be
simplified
significantly,
(3.3.11) assuming
π2 Y c( x )
f ( t )d t Y s( x ) ∞ 2 2 1/2 1 t x a ( ) π − ⌠ 1 − f ( x) + tanh(πθ) 2 2 2 2 1/2 2 2 x ⌡ x( t − a ) t − x2 4 x cosh (πθ) 1 a
−
πa
∞
∞ψ ( r ) Y ( r )d r 1 c
f ( t )d t
⌠
= ⌠
1
2 x 2cosh(πθ) ⌡ t ( t 2 − a 2)1/2
⌡
a
r2 − x2
.
(3.3.12)
a
Multiplication of (3.3.9) by rY ( r )/( r 2− x 2) and transformations similar to those s
above lead to π2 Y s( x )
∞
f 1( t )d t 2 2 1/2 t ( x a ) π − f ( x) − tanh(πθ) Y ( x ) ⌠ 1 − c 2 4 x cosh2(πθ) 1 x ( t 2 − a 2)1/2 t 2 − x 2 ⌡ a
∞ψ ( r ) Y ( r ) r d r 1 s
= ⌠
⌡
r2 − x2
.
(3.3.13)
a
Equations (3.3.12) and (3.3.13) give the final solution ∞ψ ( r ) Y ( r )d r 1 c
4 f 1( x ) = 2cosh2(πθ) x xY c( x )⌠ π ⌡ a
r − x 2
2
∞ψ ( r ) Y ( r ) r d r 1 s
+ Y ( x)⌠ s
⌡
r − x 2
2
.
a
(3.3.14)
175
3.3 External axisymmetric mixed-mixed problem
Strictly
speaking,
we
should
have
added
a
term
BY c( x ),
representing
the
homogeneous solution, where B is an arbitrary constant. It will be shown further, that we may assume B =0, since the constant D , introduced earlier, actually plays the same role. The second equation in (3.3.8) can be solved in a similar manner. first stage, it is reduced to γ 1γ 2
∞
r ( r 2 − a 2)1/2 f 2( t )d t
∞
∞f
f 2( t )d t
At the
(t)tdt
⌠ − ⌠ 2 ⌠ 2 2 2 2 1/2 2 2 + 2 1/2 2 2 = ψ2( r ), α ⌡ (t − a ) (t − r ) ⌡ (t − a ) ⌡t − r a
a
a
(3.3.15) with ∞ ω (ρ)dρ 2 1 d ⌠ . r ψ2( r ) = 2 2 1/2 2π H α d r ⌡ ( ρ − r )
(3.3.16)
r
Its solution is ∞
4 f ( x ) = − 2sinh2(πθ) xY ( x⌠ 2 c ⌡ π
ψ2( r ) Y c( r )d r r2 − x2
a
∞
+ Y ( x)⌠ s ⌡
ψ2( r ) Y s( r ) r d r r2 − x2
.
a
(3.3.17) One could notice that terms of the form const./ρ were lost during transformation of the second equation (3.3.8), due to differentiation. This means that the solution (3.3.17) satisfies the second equation (3.3.8), except for the abovementioned terms. And here the role of the constant D becomes clear: it has to be chosen so that the equation be satisfied. We show below how this is done. Example. Let the exterior of a circle ρ= a be clamped, so that w = u =0, for ρ> a . A uniform pressure σ is applied inside the circle. The stress distribution 0
outside the circle and the displacements inside are to be determined. general solution described above yields for this particular case a
( a 2 − x 2)1/2d x ω (ρ) = −4 H σ ⌠ 2 2 1/2 , 1 0 ⌡ (ρ − x )
a2 ; 0ρ
ω (ρ) = π H ασ 2
0
ψ1( r ) = 2π H σ01 −
( r 2 − a 2)1/2 α2 D r + a ln , − 2 r γ γ r r − a 1 2
ψ2( r ) = 0;
The
176
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
2 coth(πθ)σ0[ tY s( t ) − 2 a θ Y c( t )] − D [1 − Y c( t )], π
f 1( t ) =
f 2( t ) = 0.
(3.3.18) Substitution of (3.3.18) in the second equation (3.3.8) defines the constant D : D =
2 a θσ0coth(πθ). π
(3.3.19)
Formulae (3.3.6), (3.3.18) and (3.3.19) give the complete solution for the traction distribution. The displacements inside the circle are given by
2 coth(πθ) ∞ tY s( t ) − a θ[1 + Y c( t )] ⌠ u (ρ) = 2π H ασ0 tdt π ρ ⌡ ( t 2 − ρ2)1/2 a − a θ( a 2 − ρ2)1/2 −
ρ , 2
2 2 1/2 ∞ ( t − a ) − tY c( t ) − a θ Y s( t ) w (ρ) = 4 H σ0 ⌠ dt ( t 2 − ρ2)1/2 ⌡ a
ρ
+ ⌠
⌡
π ( a 2 − t 2)1/2 d t − a θ tanh( π θ ) . 2 (ρ2 − t 2)1/2
0
The integrals above may be computed in elementary functions at the centre ρ=0, namely, u (0) = 0,
w (0) = 4π Ha θσ0
1 + cosh(πθ) . sinh(2πθ)
The reader who is interested in numerical results is referred to (Fabrikant, 1972), where the field of tractions and displacements was computed for steel, concrete and sandstone. The method of solution presented in this section is neither the only method available, nor the simplest one. The reader is encouraged to try various modifications of the approach presented in section 3.2.
177
3.4 Generalization for a non-homogeneous half-space
Exercise 3.3 1. Consider a transversely isotropic elastic half-space z ≥0. Let the exterior of a circle ρ= a be clamped, so that u = w =0, for ρ> a . An axisymmetric pressure σ(ρ) is applied in the annulus b ≤ρ≤ c , with c ≤ a . Find the traction distribution outside the circle and the displacements inside. Hint : use solutions presented in Exercise 3.2 as an example. 2. Subject to the conditions of the exercise above, prove that the stresses in the plane z =0 are in equilibrium. Note : this is generally not true in internal problems. 3. Consider a transversely isotropic elastic half-space z ≥0. Let the exterior of a circle ρ= a be clamped, so that u = w =0, for ρ> a . An axisymmetric tangential traction τ(ρ) is applied in the annulus b ≤ρ≤ c , with c ≤ a . Find the traction distribution outside the circle and the displacements inside. 4. Consider a transversely circle ρ= a be clamped, so outside the circle and the applied in the positive Oz y =0, z = b .
isotropic elastic half-space z ≥0. Let the exterior of a that u = w =0, for ρ> a . Find the traction distribution displacements inside due to a concentrated force P direction at the point with cartesian coordinates x =0,
3.4 Generalization for a non-homogeneous half-space Popov (1973) has considered an internal mixed-mixed problem for the case of
a
non-homogeneous
E 0=const, and 0≤κ<1. equation, interested only the given in
isotropic
half-space,
with
elastic
modulus
E = E z κ, κ
0
He reduced the problem to a generalized Abel integral
which he solved in terms of Jacobi polynomial expansion. The reader is referred to the original paper for details. We present here closed form solution of the generalized Abel integral equation, as it was (Fabrikant, 1976).
Consider the integral equation x
a
⌠ F ( t )d t + A ⌠ F ( t )d t = f ( x ), ⌡ ( x − t )κ ⌡ ( t − x )κ
b
for 0<κ<1, b ≤ x ≤ a .
x
(3.4.1) Here A is a known constant, f is a known function, and function F is to be determined. Make use of the following integral representations (Gradshtein and Ryzhik, 1963)
178
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
t
δ
( x − b ) ( t − b )1-δ-κ ⌠ ( y − b )κ-1( t − y )δ-1 = dy, κ B(κ,δ) ( x − y )δ+κ ⌡ (x − t) 1
b
(3.4.2) 1 ( t − x)
x
δ
( x − b ) ( t − b )1-δ-κ ⌠ ( y − b )κ-1( t − y )δ-1 dy. B(κ,1−δ−κ) ( x − y )δ+κ ⌡
= κ
b
(3.4.3)
Let us define the value of δ from the condition B(κ,1−δ−κ) = A. B(κ,δ)
(3.4.4)
By using the properties of Beta-functions, expression (3.4.4) can be simplified sin(πδ) = A. sin[π(δ+κ)] The value of δ can be found from the last expression as δ =
1 A sin(πκ) tan-1 . π 1 − A cos(πκ)
(3.4.5)
Substitute (3.4.2) and (3.4.3) in (3.4.1) and interchange the order of integration. The result is (x − b) B(κ,δ)
x
δ
a
κ-1 F ( t )d t ⌠ ( y − b ) κ+dδy ⌠ δ+κ-1 1-δ = f ( x ). ⌡ ( x − y) ⌡ ( t − b) ( t − y)
b
y
(3.4.6) The generalized Abel integral equation is now represented as a sequence of two Abel type operators, and each can be inverted. The solution is a
sin(πδ) sin[π(κ + δ)] d ⌠ ( r − b )1-κd r F ( t ) = −B(κ,δ) δ dt ⌡ π2( t − b )1-δ-κ (r − t) t
r
d ⌠ f ( x )d x × . d r ⌡ ( x − b )δ( r − x )1-δ-κ b
179
3.4 Generalization for a non-homogeneous half-space
(3.4.7) The form of solution given by (3.4.7) is not the only one which can be derived. Indeed, one can use the integral representations: a
( a − t )δ( a − x )1-δ-κ ⌠ ( a − y )κ-1( y − x )δ-1 = dy, B(δ,κ) ( y − t )δ+κ ⌡ ( x − t )κ 1
x
(3.4.8) a
1
= ( t − x )κ
( a − t )δ( a − x )1-δ-κ ⌠ ( a − y )κ-1( y − x )δ-1 dy. B(1−δ−κ,κ) ( y − t )δ+κ ⌡ t
(3.4.9) Substitution of (3.4.8) and (3.4.9) in (3.4.1) leads to a
y
( a − x )1-δ-κ ⌠ ( a − y )κ-1d y ⌠ ( a − t )δ F ( t )d t = f ( x ). 1-δ B(δ,κ) ( y − t )δ+κ ⌡ ( y − x) ⌡ x
b
(3.4.10) The solution will now take the form t
sin(πδ) sin[π(δ+κ)] d ⌠ ( a − r )1-κd r F ( t ) = −B(κ,δ) d t ⌡ ( t − r )1-κ-δ π2( a − t )δ b
a
d ⌠ f ( x )d x . × d r ⌡ ( a − x )1-δ-κ( x − r )δ r
(3.4.11) We leave it to the reader to establish the equivalence of the solutions (3.4.7) and (3.4.11). If one compares integral equations (3.2.30) and (3.4.1), the first impression is that they are so different that there is no way to relate them. This is not so. Consider a set of equations a
a
1 sign( r − t ) πκ α ⌠ 1 + −⌠ F ( t )d t + cot( ) κ κ 1 2 √γ γ ⌡ | r − t |κ ⌡ ( r + t ) |r − t| 1 2 0
0
180
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
ω (ρ)ρdρ
r
F ( t )d t = 2cos(πκ/2)Γ(1 − κ) ⌠ 1 − 2 2 1/2, κ π2 H ⌡ (r − ρ ) (r + t) 2 1
0
(3.4.12) a
− cot(
πκ α ⌠ 1 1 2 ) − κ F 1( t )d t + κ κ 2 √γ γ ⌡ | r − t | t (r + t) 1 2 0
a
1 sign( r − t ) 1 sign( t ) − − κ − F ( t )d t + ⌠ κ κ κ ⌡ (r + t) t t 2 |r − t| 0
r ω (ρ)dρ 2 2cos(πκ/2)Γ(1 − κ) r ⌠ 2 = 2 2 1/2 . π H √γ1γ2 ⌡ (r − ρ ) 0
(3.4.13) In the limiting case of κ→0, equations (3.4.12) and (3.4.13) transform into a r ω (ρ)ρdρ 1 α r t + 1 ⌠ F ( t ) ln| ⌠ 2 −⌠ F 1( t )d t + t , | d = 2 r − t π√γ1γ2 ⌡ 2 π H ⌡ ( r − ρ2)1/2 ⌡ r
0
0
0
(3.4.14) ω2(ρ)dρ α t2 1 ⌠ ⌠ ⌠ − F ( t ) ln 2 F ( t )d t = 2 r dt − 2 2 1/2. π√γ1γ2 ⌡ 1 | r − t 2| π H √γ1γ2 ⌡ 2 ⌡ (r − ρ ) a
r
0
r
0
0
(3.4.15) One can easily verify that differentiation of (3.4.14) and (3.4.15) with respect to r leads to (3.2.30). Thus, the connection is established. Introducing the complex function F = F 1+ iF 2, equations (3.4.12) and (3.4.13) may be unified as follows: r
πκ πκ iα F ( t )d t iα −1 + cot( )⌠ + 1 − cot( )⌠ κ 2 ⌡ ( r − t ) 2 ⌡ √γ1γ2 √γ1γ2 -a
r
a
F ( t )d t
κ ( t − r)
181
3.4 Generalization for a non-homogeneous half-space
2cos(πκ/2) Γ(1 − κ) ⌠ = π2 H √γ1γ2 ⌡
r
√γ1γ2ρω1(ρ) + ir ω2(ρ) ( r 2 − ρ2)1/2
dρ
0
a
+ 2i ⌠ ⌡
F 2( t )d t t
0
κ
πκ α cot( ) ⌠ − 2 ⌡ √γ1γ2
a
0
F 1( t )d t t
κ
.
(3.4.16)
We obtained the generalized Abel integral equation of the type (3.4.1), with b =− a , the solution of which is available in the forms (3.4.7) and (3.4.11), with the parameter δ, defined as κ + i θ, 2
(3.4.17)
and θ given by (3.2.19).
The stress function for a flat circular bonded punch is
δ = −
F(t) = −
sin(πκ) cosh(πθ) Γ(1 − κ) a + t i θ w 0( a 2 − t 2)κ/2 . 3 a − t π κH
The last result is in agreement with those of paragraph 3.2. The reader can derive several new modifications of the governing integral equations by using the integral representations min (ρ
⌠ ⌡
0
, ρ) x κ+1d x [(ρ − x )(ρ20 − x 2)](κ+1)/2 2
2
0
=
Γ(κ/2) Γ[(1 − κ)/2] 1 1 , − κ |ρ − ρ |κ 4√π (ρ + ρ0) 0
min (ρ
⌠ ⌡
0
, ρ) x κ-1d x [(ρ2 − x 2)(ρ20 − x 2)](κ+1)/2
0
=
Γ(κ/2) Γ[(1 − κ)/2] 1 1 . + κ κ 4√πρρ0 |ρ − ρ0| (ρ + ρ0)
182
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
Some additional representations can be obtained by simple addition, subtraction, integration or differentiation of those above. Exercise 3.4 1. Find a solution of the generalized Abel equation (3.4.1), with f ( x )= C =const. C sin[π(δ+κ)] Answer: F ( t ) = π( t − b )1-δ-κ( a − t )δ 2. Find a solution of the generalized Abel equation (3.4.1), with f ( x )= Cx, with C =const. C sin[π(δ+κ)][ t − b − δ( a − b ) + κ b ] Answer: F ( t ) = πκ( t − b )1-δ-κ( a − t )δ 3. Find a solution of the generalized Abel equation (3.4.1), with f ( x )= C x 2, with C =const. C sin[π(δ+κ)][2( t − b )( t + D ) + D 2 − δ( a 2 − b 2) + κ b 2] Answer: F ( t ) = . πκ(1 + κ)( t − b )1-δ-κ( a − t )δ Here D =κ b −δ( a − b ).
3.5 Effect of a shearing force and a tilting moment on a bonded circular punch. Consider a circular flat punch of radius a , bonded to a transversely isotropic elastic half-space z ≥0. The punch is subjected to a shearing force T , acting in the Ox direction, and a tilting moment M . We may assume, without loss of generality, that the moment is oriented along the Oy axis. We need to find the traction distribution under the punch, and to relate the translational ( u 0) and angular (δ) displacements of the punch to the applied loading parameters. The problem is considered in a separate section due to its practical importance.
The problem is characterized by the following boundary conditions on the plane z =0: u = u 0,
for 0≤ρ≤ a ,
0≤φ<2π;
w = −δρcosφ,
for 0≤ρ≤ a ,
0≤φ<2π;
σ = τ = 0,
for a ≤ρ<∞
0≤φ<2π.
(3.5.1)
The governing integral equations, due to (2.5.6), (3.1.2) and (3.1.3), take the
183
3.5 Effect of a shearing force and a tilting moment
form 2G1 ρ
⌠
⌠
ρ2 ⌡ (ρ2 − x 2)1/2 ⌡ 0
2π H α ⌠ − σ1(ρ0)ρ20dρ0 2 ρ ⌡
ρ0(ρ20 − x 2)1/2
x
2G ρ
ρ
τ2(ρ0)dρ0
a
x 4d x
τ0(ρ0)ρ0dρ0
a
ρ2 − 2 x 2
0
⌠ dx ⌠ + 2 2 1/2 = 0, ρ2 ⌡ (ρ2 − x 2)1/2 ⌡ (ρ0 − x ) 2
0
x
ρ
2G ⌠ 2
(ρ20 − 2 x 2) τ (ρ )dρ 2 0 0
a
dx ⌠ 2 1/2 ⌡ (ρ − x ) ⌡
ρ0(ρ20 − x 2)1/2
2
0
x
ρ
a
dx
a
+ 2π H α⌠ σ (ρ )dρ -1 0 0
⌡ ρ
τ0(ρ0)ρ0dρ0
⌠ + 2G ⌠ 2 2 1/2 2 2 1/2 = u 0, 1 ( x ) ρ − ⌡ ⌡ (ρ0 − x ) 0
x
dρ e-iφ ρ 0 i φ⌠ ⌠ 2π H α ℜ τ0(ρ0)ρ0dρ0 − ρe τ2(ρ0) ρ ⌡ ρ0 ⌡ 0 ρ a
ρ
+
a
x 2d x
4H⌠ ⌠ ρ ⌡ (ρ2 − x 2)1/2 ⌡ 0
x
σ1(ρ0)ei φ + σ-1(ρ0)e-i φ (ρ20 − x 2)1/2
dρ0 = −δρcosφ.
(3.5.2) The structure of equations (3.5.2) is such that we may assume that σ = σ . 1 -1 The solution may be represented in the form ρ
d⌠ f ( t )d t , σ1(ρ) = σ-1(ρ) = 2 dρ⌡ (ρ − t 2)1/2 0
a
C d ⌠ f(t)tdt D + , τ0(ρ) = − 2 ρ dρ ⌡ ( t 2 − ρ2)1/2 ( a − ρ2)1/2 ρ
184
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
f(t)tdt d 1⌠ 2 a 2 − ρ2 τ (ρ) = − C ρ − D . 2 dρ ρ2 ⌡ ( t 2 − ρ2)1/2 ρ2( a 2 − ρ2)1/2 a
ρ
(3.5.3) Here f is the as yet unknown stress function, and C and D are the constants to be determined. Substitution of (3.5.3) in the first two equations of (3.5.2) make them identities, provided that the following conditions are satisfied a
α C = , γγ
C D = ⌠ f ( t )d t , a⌡
1 2
(3.5.4)
0
a
π2 D f ( t )d t ( G 1 + G 2) + 2π H α⌠ 2 2 1/2 = u 0. 2 ⌡ (a − t )
(3.5.5)
0
Expressions (3.5.4) and (3.5.5) look contradictory: we have only two constants to satisfy three equations. This will be clarified further. An additional constant, representing the homogeneous solution, will appear in the expression for f . Substitution of (3.5.3) in the third equation (3.5.2) leads to a f(t)tdt 2π H α ⌠ ⌠ − C C f t t aD + ( )d + 2 ρ ⌡ ( t 2 − ρ2)1/2 ⌡ 0 ρ a
ρ
x 2( a 2 − x 2)1/2d x ⌠ (ρ2 − x 2)1/2 ⌡
8H⌠ + ρ⌡ 0
a
(t
2
f ( t )d t = −δρ. − x )( a 2 − t 2)1/2 2
0
Multiply the last expression by ρ2( r 2−ρ2)-1/2, and integrate with respect to ρ from 0 to r . The result is a
a
f ( t )d t α2 ⌠ f ( t )d t δ ( a 2 − r 2)1/2⌠ 2 = − . − 2 1/2 2 2 2 2 γ 1γ 2 ⌡ t − r πH ⌡ (a − t ) (t − r ) 0
Equation (3.5.6)
0
is similar
to (3.2.60),
(3.5.6) with the general solution given by
185
3.5 Effect of a shearing force and a tilting moment
(3.2.62).
The solution in this particular case is f(t) = −
δcosh2(πθ) tY s( t ) − θ aY c( t ) + AY c( t ). 2 sinh( ) πH πθ
(3.5.7)
The last term in (3.5.7) represents the homogeneous solution, with A being an arbitrary constant. Substitution of (3.5.7) in (3.5.4) and (3.5.5) allows us to define all the constants, namely, D =
πθα A, γ1γ2sinh(πθ) πθ( G + G )
1 2 δ a θα π2 H α -1. 1 + A = u0 + tanh(πθ)cosh(πθ) tanh(πθ)( G − G ) 1 2
(3.5.8) Formulae (3.5.7), (3.5.8) and (3.5.3) determine completely the traction distribution under the punch. Now we need to relate the applied loading to the punch displacements. Make use of the equilibrium conditions: 2π a
a
T = 2π⌠ τ0(ρ)ρdρ,
⌡
M = −⌠ ⌠ [σ1(ρ)ei φ + σ-1(ρ)e-i φ]ρ2cosφ dρdφ.
⌡ ⌡ 0
0
0
After carrying out the calculations, we obtain T = 4π2 A
aθ α , sinh(πθ) γ γ 1 2
M =
4δ a 3θ(1 + θ2) 4π2 a 2θ2 + A. 3 H tanh(πθ) cosh(πθ)
(3.5.9) Expressions (3.5.8) and (3.5.9) enable us to determine the displacements of the punch u0 =
1 (1 + 4θ2)tanh(πθ) 3Hα ( G − G ) T − M, π( G 1 + G 2) + 2 2 1 2 8a θ(1 + θ ) 4 a (1 + θ2) δ =
3Hα −T + M . 2 4 a (1 + θ ) a θ√γ γ 1 2
In the case of signs) with the sign convention. displacement of
2
(3.5.10) isotropy, formulae (3.5.10) are in agreement (except for some results of Ufliand (1967), who seems to have used a different It is noteworthy that the tilting moment produces translational the punch, even in the absence of the shearing force. The
186
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
shearing force, in turn, tilts the punch, even when no tilting moment is applied. We shall see further (section 5.11) that a similar situation takes place in the case of a finite friction between the punch and the elastic half-space. Exercise 3.5 1. A flat circular punch of radius a is bonded to a transversely isotropic elastic half-space z ≥0. A shearing force T is applied in the Oy direction. Find the tilting angle δ. 3Hα T , with tilting about Ox axis. Note that the angle is Answer: δ = 4 a 2(1 + θ2) positive. 2. Subject to the conditions above, find the tilting moment needed to prevent tilting. Answer: M = − a θ√γ γ T . 1 2
3. A tilting moment M about the axis Ox is applied to a flat circular punch, bonded to a transversely isotropic elastic half-space. Find the translational displacement u and its direction. 3Hα Answer: u = M , in the Oy direction. 2 4 a (1 + θ2) 4. A flat circular bonded punch is subjected to a shearing force T , acting in the Ox direction, and a tilting moment M about the Oy axis. Find the normal displacements outside the punch. 2 4δcosh(πθ) a x Y c( x ) + θ axY s( x ) ⌠ Answer: w (ρ,φ) = − dx πρ (ρ2 − x 2)1/2 ⌡
0
a xY ( x )d x s 1 πθ a + 4π HA tanh(πθ) − ⌠ 2 2 1/2 cosφ. ρcosh(πθ) ⌡ (ρ − x ) 0
5. Express the answer above in terms of the shearing force T and tilting moment M . a x 2 Y ( x ) + θ axY ( x ) 3 H αcosh(πθ) M c s Answer: w (ρ,φ) = − dx −T ⌠ 2 2 2 2 1/2 ⌡ πρ a (1 + θ )θ a √γ γ (ρ − x )
1 2
0
187
3.6 Non-axisymmetric internal mixed-mixed problem
a xY s( x )d x Hα cosh(πθ) ⌠ + T 1 − 2 2 1/2 cosφ. ρ πθ a ⌡ (ρ − x )
0
6. Express the normal displacement w outside the punch in terms of the stress function f . a 8 H a 2 x d x ( a 2 − x 2)1/2 f ( t )d t ⌠ ⌠ Answer: w (ρ,φ) = ρ ⌡ (ρ2 − x 2)1/2 ⌡ ( a 2 − t 2)1/2( t 2 − x 2) 0 0
a + 4π H α D cosφ. ρ
7. Express the tangential displacement u outside the punch in terms of the stress function f . Answer: u (ρ,φ) = u (ρ) + u (ρ)e2 i φ, 0
2
where α ⌠ u (ρ) = π( G − G ) 0 1 2 γ γ 1 2 ⌡
a
f ( x )d x -1 a 2 1/2 + π( G 1 + G 2) D sin (ρ), (ρ − x ) 2
0
1 u 2(ρ) = 22π H α⌠ f ( x )d{ x [( a 2 − x 2)1/2 − (ρ2 − x 2)1/2]} ρ ⌡ a
0
+ π( G + G ) Da (ρ2 − a 2)1/2. 1 2
8. Investigate the interaction of an arbitrary concentrated force, applied at some point inside the transversely isotropic half-space, and a flat bonded circular punch.
188
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
3.6 Non-axisymmetric internal mixed-mixed problem The general formulation of this problem is given in section 3.1, with the boundary conditions (3.1.1), and the governing integral equations (3.1.2−3.1.5). The closed form exact solution is not known at the moment. We assume that all the parameters involved can be expanded in a Fourier series. An exact solution for the n -th harmonic is presented below. The governing integral equations for the n -th harmonic are ρ
G 1 x 2τn+1(ρ0) + G 2[2 n ρ2 − (2 n +1) x 2]τ-n+1(ρ0)
a
2n
x dx 2 ⌠ ⌠ ρn+1 ⌡ (ρ2 − x 2)1/2 ⌡ 0
ρn0(ρ20 − x 2)1/2
x
dρ 0
ρ
2π H α ⌠ − ρn+1 ⌡
σn(ρ0)ρn+1 0 dρ0 = F n+1(ρ), for n ≥0.
0
ρ
G ρ20τ (ρ ) + G [(2 n −1)ρ20 − 2 n x 2]τ (ρ ) 1 -n+1 0 2 n+1 0
a
x 2n-2d x
2 ⌠ ⌠ ρn-1 ⌡ (ρ2 − x 2)1/2 ⌡
ρn0(ρ20 − x 2)1/2
x
0
σ (ρ )dρ -n 0 0
a
+ 2π H αρ ⌠ ⌡ n-1
= F -n+1(ρ),
ρn-1 0
ρ
dρ0
for n ≥1;
a τ (ρ )dρ e-in φ ρ n+1 0 0 n n in φ⌠ ⌠ 2π H α ℜ n τ (ρ )ρ0dρ − ρ e n 0 ρ ⌡ -n+1 0 ρ0 ⌡ 0 ρ
ρ
+
4H ⌠ ρn ⌡
a
x 2nd x (ρ − x ) 2
2 1/2
0
⌠ ⌡ x
-in in σ-n(ρ0)e φ + σn(ρ0)e φ 2 ρn-1 0 (ρ0
− x)
2 1/2
dρ = ℜ{ein φΦ (ρ)}, 0
n
for n ≥0. (3.6.1) Here the right hand sides are known from the boundary conditions, and are ∞
F
n+1
(ρ) = u
(ρ) − 2ρ
⌠ 2n+2 2d x 2 1/2 ⌡ x (x − ρ )
n+1
n+1
a
189
3.6 Non-axisymmetric internal mixed-mixed problem
x
⌠ ⌡
G 1ρ20τn+1(ρ0) + G 2[2 n x 2 − (2 n +1)ρ20]τ-n+1(ρ0) ρ20)1/2
(x − 2
a
∞
F -n+1(ρ) = u -n+1(ρ) − 2π H αρ
n-1
⌠ ⌡
σ-n(ρ0)dρ0
a ∞
x
dx ⌠ − 2ρn-1⌠ 2n 2 2 1/2 ⌡ x (x − ρ ) ⌡ a
ℜ{Φn(ρ)e
} = w n(ρ)e
in φ
x
dx ⌠ − 4 H ρn⌠ 2n 2 2 1/2 ( ) x x − ρ ⌡ ⌡ a
( x 2 − ρ20)1/2
+ w -n(ρ)e
∞
ρn-1 0
G 1 x 2τ-n+1(ρ0) + G 2[(2 n −1) x 2−2 n ρ2]τn+1(ρ0)
a
in φ
ρn0dρ0,
-in φ
σn(ρ0)e
in φ
in φ n ∞ τn+1(ρ0)dρ0 + 2π H α ℜ e ρ ⌠ ρn0 ⌡ a + σ-n(ρ0)e
-in φ
( x 2 − ρ20)1/2
a
ρn0dρ0,
ρn+1 0 dρ0.
(3.6.2) The case of axial symmetry ( n =0) was considered in detail in paragraph 3.2, and is not discussed here. The solution is sought for n ≥1. We may assume, without loss of generality, that the first two equations (3.6.1) are homogeneous. This can be achieved by addition of some special solutions to the parameters τ-n+1 and τn+1. These special solutions, satisfying the right hand sides of the first two equations (3.6.1), can be obtained from the results of section 2.6. Of course, this procedure will make the right hand side of the third equation (3.6.1) more complicated. Assume the solution of the set (3.6.1), transformed into homogeneous, in the form ρ
σn(ρ) = σ-n(ρ) =
t 2n-1d f n( t )
1 ⌠ ρn ⌡ (ρ2 − t 2)1/2 0
with
the first
two
equations
190
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
ρ
1 d ⌠ t 2n-2[(2 n − 1)ρ2 − 2 n t 2] = − n+1 f n( t )d t ; dρ ⌡ ρ (ρ2 − t 2)1/2 0
a
D nρn-1
f n( t ) t d t
d ⌠ + ; dρ ⌡ ( t 2 − ρ2)1/2 ( a 2 − ρ2)1/2
τ-n+1(ρ) = − C ρn-2
ρ
a a f n( t ) t d t d 1 d n 2n-2 ⌠ ⌠ τn+1(ρ) = C ρ y dy dρ ρ2n⌡ d y ⌡ ( t 2 − y 2)1/2
d 1 ⌠ + D nρ dρ ρ2n ⌡
ρ
a
n
d ⌠ = −Cρ dρ ⌡
a
n
ρ
ρ
y
t 2n-1d t ( a 2 − t 2)1/2
dy 2n-2 ⌠ f t t t ( ) d + (2 n − 1)⌠ 2n 2 ⌡ y ( y − ρ2)1/2 ⌡ n a
f n( t )d t t ( t 2 − ρ2)1/2
ρ
a
y
a
t 2n-1d t ρn-1 2n ⌠ − D n 2 . + ( a − ρ2)1/2 ρn+1 ⌡ ( a 2 − t 2)1/2 ρ
(3.6.3) Here f n is the as yet unknown complex stress function, and C , D n are the constants to be determined. In the following derivations we give in some cases two equivalent expressions of the same parameter, for the sake of convenience in the procedure of substitution into the governing equations (3.6.1). We present first some intermediate results related to the substitution of (3.6.3) in the first equation (3.6.1): a
τ
(ρ )dρ
⌠ n 2 2 1/2 ⌡ ρ0(ρ0 − x ) x
-n+1
0
0
πC = 2x
f ( x) n
x
a
− ⌠ ⌡
f ( t )d t n
t
2
+ π D, 2 ax n
x
(3.6.4)
191
3.6 Non-axisymmetric internal mixed-mixed problem
a
τn+1(ρ0)dρ0
πC = 2x
⌠ n 2 2 1/2 ⌡ ρ0(ρ0 − x )
f n( x ) x
x
a
2n − 1 ⌠ π a 2n-1 2n-2 + f ( t ) t d t D. − x 2n 2 x 2n+1 n ⌡ n x
(3.6.5) Substitution of (3.6.4) and (3.6.5) in the first equation (3.6.1) yields ρ
ρ
x 2n-2[2 n x 2 − (2 n −1)ρ2] 2π H α ⌠ πC ( G 1− G 2) n+1 ⌠ f n( x )d x − σ (ρ )ρn+1 2 2 1/2 n+1 0 dρ0 = 0, (ρ − x ) ρ ρ ⌡ ⌡ n 0 0
0
(3.6.6) provided that the following condition holds a
(2 n − 1) C ⌠ f ( t ) t 2n-2d t . D = n a 2n-1 ⌡ n
(3.6.7)
0
It is now easy to verify that substitution of the first expression (3.6.3) in (3.6.6) makes it an identity if C =
α . γγ
(3.6.8)
1 2
Here are some intermediary results related to the procedure of substitution of (3.6.3) in the second equation (3.6.1): a
τ
(ρ )dρ
-n+1 0 0 π ⌠ n-2 2 2 1/2 = 2 C f n( x ) + D n , ( x ) ρ ρ − ⌡ 0 0
(3.6.9)
x
a
⌠ ⌡
x
(2 n − 1)ρ20 − 2 n x 2 ρn0(ρ20
− x)
2 1/2
τ
n+1
(ρ )dρ = 0
0
π − C f ( x ) + D . n n 2
(3.6.10)
We used in transformations some general formulae from Appendix A3.3. Substitution of (3.6.9) and (3.6.10) in the second equation (3.6.1) reduces it to ρ
π ⌠ ρn-1 ⌡ 0
[ C ( G − G ) f ( x ) + ( G + G ) D ] x 2n-2d x 1
2
n
1
(ρ − x ) 2
2 1/2
2
n
a
+ 2π H αρ ⌠ ⌡ n-1
ρ
σ (ρ )dρ -n
0
ρn-1 0
0
= 0.
192
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
(3.6.11) Substitution of the first expression (3.6.3) in (3.6.11) makes it an identity, subject to (3.6.8), and an additional condition a
2Hα ⌠ x 2n-2d x √πΓ( n − 1/2) f ( x) = 0. ( G 1 + G 2) D n + 2n-2 2Γ( n ) a ( a 2 − x 2)1/2 ⌡ n
(3.6.12)
0
The conditions (3.6.7) and (3.6.12) might look contradictory. It will be shown further (see 3.6.14) that this is not so, because an additional constant will appear in the expression for f , corresponding to a homogeneous solution of the singular n
integral equation (3.6.14). By now we have satisfied the first two equations (3.6.1). Substitution of (3.6.3) in the third equation (3.6.1) requires the following transformation: ρ
dρ
a
⌠ 2n-1 2 2 1/2 ⌠ ⌡ ρ0 (ρ0 − x ) ⌡ x 0
0
t 2n-1d f ( t ) n
(ρ20
− t)
2 1/2
n
|t(a2
= ⌠ ( a 2 − x 2)1/2( a 2 − t 2)1/2 Q ( x , t )
⌡
n
0
0
+ ψ ( x , t ) ln
a
a | t 2 − x 2|1/2 t 2n-1d f ( t ). 2 1/2 2 2 1/2 n − x ) − x( a − t ) |
Here Q ( x , t ) is a polynomial in even negative powers of x and t . n
(3.6.13) Although we
cannot write the explicit expression for Q ( x , t ), it can be computed in elementary n
The explicit expression for ψ is n
manner for any particular n . n-1
1 ψn( x , t ) = π tx 2n-1
Σ k=0
Γ( n − k −1/2) Γ( k +1/2) x 2k . Γ( n − k ) Γ( k +1) t
Multiply both sides of the third equation (3.6.1) by ρn, differentiate with respect to ρ, divide the result by ρ2n-2( r 2−ρ2)1/2 and integrate with respect to ρ from 0 to r . This procedure allows us to split the kernel of the integral equation into two parts: a singular one and a degenerate one. The result takes the form a
f ( t )d t
(a − r ) ⌠ 2 2 2 2 1/2 ⌡ ( t − r )( a − t ) 2
n
2 1/2
0
α2 ⌠ − γ 1γ 2 ⌡
a
f ( t )d t n
t
2
− ρ2
= χ ( r ). n
0
(3.6.14)
193
3.6 Non-axisymmetric internal mixed-mixed problem
Here r
1 ⌠ dρ d n [ρ Φ (ρ)] χn( r ) = 2n-2 2 2 1/2 n 4π Hr ⌡ ρ ( r − ρ ) dρ 0
ρ
r
a
2⌠ dρ d ⌠ x 2n( a 2 − x 2)1/2d x ⌠ 2 2 1/2 ( a − t ) Q ( x , t ) t 2n-1d f ( t ) − 2 2 1/2 n n π r ⌡ ρ2n-2( r 2 − ρ2)1/2 dρ⌡ (ρ − x ) ⌡ 0
0
0
ρ
r
−
⌠ 2n-2 2dρ 2 1/2 ⌡ ρ (r − ρ )
2 π r 3/2
0
2 1/2 2n-2
t) t
a
dx ⌠ 2 2 1/2 2 2 1/2 ⌡ (ρ − x ) ( a − x ) 0
⌠ q (ρ, x , t )( a 2 ⌡ n
−
0
d f ( t ), n
(3.6.15) with n-2
q n(ρ, x , t ) =
Σ k=0
1 3 Γ( n − k −1/2)ρ2k x2 F (2− n + k , ; − n + k ; 2 ). 2 2 Γ( n − k ) t t
(3.6.16)
Note that the hypergeometric function in (3.6.16) is, in fact, a polynomial, and that all the integrals with respect to x and ρ in the degenerate part of the kernel (3.6.15) are computable in elementary function for any n . The integral equation (3.6.14) was solved in paragraph 3.2, and the solution is a
4 f n( t ) = − 2cosh2(πθ) t tY c( t )⌠ π ⌡
χn( r ) Y c( r )d r r2 − t 2
0
a
+ Y ( t )⌠ s
⌡
χn( r ) Y s( r ) r d r r − t 2
2
+ A Y ( t ). n c
(3.6.17)
0
The last term in (3.6.17) represents the homogeneous solution, with A being an n arbitrary constant.
Its value, along with the constant D and others which appear n
due to the degenerate part of the kernel, can be found from the appropriate system of linear algebraic equations and the conditions (3.6.7) and (3.6.12). The general solution may be considered completed. The main handicap of the solution is the necessity for solving a set of linear algebraic equations whose order increases with n , thus making the exact solution for higher harmonics very cumbersome. We are not aware of any other solution for a transversely
194
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
isotropic solid. The corresponding problem for an isotropic body was solved by Ufliand (1967) who used the method of Mehler-Fok integral transform. It has the same hindrance: one needs to solve a set of linear algebraic equations, whose order increases with n . Example. Consider the action of outside a flat circular punch of radius elastic half-space z ≥0. We may assume, is applied at the point ρ= b , φ=0 ( b > a ).
a normal concentrated load P applied a , bonded to a transversely isotropic without loss of generality, that the force Thus, the boundary conditions are
for ρ≤ a ;
u = w =0, σ = P δ(ρ− b )δ(φ−0)/ρ,
τ = 0,
for ρ> a .
The boundary conditions yield F
n+1
(ρ) = 0,
for n ≥0;
F
-n+1
= − PH α
ρn-1 , bn
for n ≥1;
ρ
4 PH ⌠ x 2nd x . Φn(ρ) = − π(ρ b )n ⌡ (ρ2 − x 2)1/2( b 2 − x 2)1/2 0
(3.6.18) We present now the explicit solutions for some specific values of n . In the axisymmetric case n =0, the following results may be obtained: ρ
f 0( t ) t d t
1 d ⌠ σ (ρ) = 0 ρ dρ ⌡
(ρ2 − t 2)1/2
,
0
a f 0( t )d t α d ⌠ τ (ρ) = − , 1 γ γ dρ ⌡ ( t 2 − ρ2)1/2 1 2
ρ
a Y ( r )d r c 2 2 ⌠ f ( t ) = 3 P cosh (πθ) tY ( t ) 2 2 0 c ⌡ ( b − r )1/2( r2 − t 2) π 0
a
Y ( r) rdr
s + Y ( t )⌠ 2 2 1/2 2 2 . s ⌡ ( b − r ) ( r − t ) 0
The results for n =1 are:
(3.6.19)
195
3.6 Non-axisymmetric internal mixed-mixed problem
ρ
d ⌠ σ1(ρ) = dρ ⌡
f 1( t )d t (ρ2 − t 2)1/2
,
0
a
f (t)tdt D 1 1 α 1 d ⌠ + , τ0(ρ) = − 2 2 1/2 2 γ1γ2 ρ dρ ⌡ ( t − ρ ) ( a − ρ2)1/2 ρ
a f (t)tdt 1 α d 1 ⌠ − D 2 a 2 − ρ2 , τ2(ρ) = − ρ 1ρ2( a 2 − ρ2)1/2 dρ ρ2 ⌡ ( t 2 − ρ2)1/2 γ 1γ 2
ρ
t f ( t ) + A Y ( t ), 1 c b 0
f (t) = 1
πθ( G 1 + G 2) Pθ cosh(πθ) -1, I (b) 1 + D = − − 1 1 s πaθ tanh(πθ)( G 1 − G 2) π b √γ γ 1 2
A
1
= −
P cosh2(πθ) cosh(πθ) D √γ γ . [ I s( b ) − 2 a θ I c( b )] + 3 1 1 2 πθ π θ ab
(3.6.20) We recall that the notations I c,s are defined by (3.2.73) and (3.2.74) respectively. The case of n =2 is more cumbersome: ρ
1 d ⌠ σ2(ρ) = 3 ρ dρ ⌡
4 t 2 − 3ρ2 2 t f ( t )d t ), 2 (ρ2 − t 2)1/2
0
a f (t)tdt 2 d α ⌠ τ (ρ) = − − D ( a 2 − ρ2)1/2, 2 -1 2 dρ γ1γ2⌡ ( t − ρ2)1/2
ρ
d 1 α ⌠ ρ2 − 2 t 2 2 a 2 + ρ2 2 2 1/2 f t t t D a τ3(ρ) = ρ 4 ( ) d + ( − ρ ) , 2 dρ ρ γ1γ2 ⌡ ( t 2 − ρ2)1/2 2 3 a
2
ρ
196
CHAPTER 3
4 f ( t ) = − 2 cosh2(πθ) t tY ( t )⌠ 2 c ⌡ π
a
MIXED-MIXED BOUNDARY VALUE PROBLEMS
χ ( r ) Y ( r )d r 2
c
r − t 2
2
0
+ A Y (t) + 2 c
a
+ Y ( t )⌠ s ⌡
χ ( r) Y ( r) rdr 2
s
r − t 2
2
0
2coth(πθ)2θ 1 Y c( t ) − Y ( t ) B . πa a t s 2
The constant B corresponds to the degenerate part of the kernel: 2 a
2t2 − a2 ⌠ B2 = 2 2 1/2 f 2( t )d t . ⌡ (a − t ) 0
All the constants are determined as follows: D2 =
π a 3θ 1 − 2θ2 3α 2 a θ2(1 + 4θ2)cosh(πθ) L A − B , + 2 sinh(πθ) 2 3 γ 1γ 2 a 3 1 3sinh2(πθ)
A2 =
cosh(πθ) 4θ(1 + 2θ2) L − B , 2 sinh(πθ) 2π a 2θ2 3
3π( G 1 + G 2) 2 Pa 2 + L 1 − θ2 B 2 = 2θ 2 + L2 + L1 4a(G − G ) 34 1 2 2πb πθ(1 + θ2)( G 1 + G 2) -1 θ , + (1 − 2θ2) 1 + 4( G 1 − G 2) tanh(πθ)( G 1 − G 2) sinh(πθ) ( G 1 + G 2)
a
L1
2cosh2(πθ) ⌠ 1 − 2θ2 = 2 a θ r 2 + a 2 Y ( r) πsinh(πθ) ⌡ 3 c 0
+ r (2 a 2θ2 − r 2) Y s( r ) χ2( r )d r ,
197
3.6 Non-axisymmetric internal mixed-mixed problem
a
4 = cosh(πθ)⌠ π ⌡
L2
0
21 r2 2 a 4 − θ + 2 Y c( r) + θ arYs( r)χ2( r)d r, a
4 = cosh(πθ) ⌠ [( r 2 − 2 a 2θ2) Y c( r ) + 2θ arY s( r )]χ2( r )d r , π ⌡
L3
0
χ2( r ) = −
1 1 P 2 2 2 1/2 + 2 2 1/2 . 2π b ( b − r ) b + (b − r )
In order to provide zero displacements, the punch should be loaded by a normal force N , a shearing force T , directed in the negative Ox direction, and a tilting moment M about the Oy axis. Their relationship with the force P can be established by the statics equations, namely, a
a
α ⌠ T = 2π⌠ τ0(ρ)ρdρ = 2π f 1( t )d t + D 1 a , γ γ ⌡ 1 2 ⌡ 0
0
a
a
f 0( t ) t d t
N = 2π⌠ σ0(ρ)ρdρ = 2π⌠
2 2 1/2 ⌡ (a − t )
⌡ 0
,
0
2π a
M = −⌠ ⌠ [σ1(ρ)e
⌡ ⌡
0
a iφ
-i φ
+ σ-1(ρ)e ]ρ cosφdρdφ = 2π⌠ 2
⌡
0
a2 − 2t2 f ( t )d t . ( a 2 − t 2)1/2 1
0
Performing the integrations, we get a
T = 4π aD 1,
2P N = − cosh(πθ)⌠ π ⌡
Y c( r )d r ( b 2 − r 2)1/2
,
0
a
4P M = cosh(πθ)⌠ πb ⌡
r 2 Y c( r ) + θ arY s( r ) ( b 2 − r 2)1/2
d r + 4π a 2θ√γ1γ2D 1.
0
(3.6.21)
198
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
If the punch is not loaded then it will undergo the translational, normal, and angular displacements. Their values can be determined from (3.2.71), (3.5.10) and (3.6.21). Exercise 3.6 1. Subject to the conditions in the example above (page 200), find the normal traction at the punch centre. a [ aY ( x ) + 2θ xY ( x )]d x c s P cosh2(πθ)⌠ πθ Answer: σ(0) = 2 2 2 1/2 2 2 1/2 − sinh(πθ) . π ab ⌡ [ b + ( b − x ) ]( b − x ) 0
2. Subject to the conditions in the example above, find the shear traction at the punch centre. a Y ( x )d x D πθ A s 1 1 Pα 2 ⌠ b Answer: τ(0) = 1 − coth( π θ ) + + . 2 2 1/2 π a 2π b 2γ γ √γ γ ⌡ x( b − x ) 1 2
1 2
0
3. Subject to the conditions in the example above, find the traction distribution under the punch in the limiting case α=0. Answer: σ (ρ) = − n
P ( b 2 − a 2)1/2 ρn, 2 2 2 1/2 2 2 π ( a − ρ ) ( b − ρ ) b
τ (ρ) = 0. n
4. No loading is applied to a flat circular punch of radius a bonded to a transversely isotropic half-space. A normal concentrated force P is applied at the point ρ= b , φ=π/2. Find the normal component w of the punch displacement. a Y ( r )d r c PH sinh(πθ) ⌠ Answer: w = 2 2 1/2. πθ a ⌡ (b − r ) 0
Hint : use (3.2.71) and (3.6.21) 5. Subject to the conditions of Exercise 4, find the tangential component u of the punch displacement and its direction. a x 2 Y ( x ) + θ axY ( x ) c s 3 PH αcosh(πθ) ⌠ 1 Answer: u = d x + PH α 1 2 2 2 2 1/2 b (b − x ) π b a (1 + θ ) ⌡ 0
a xY ( x )d x s cosh(πθ) ⌠ − 2 2 1/2 , πθ a ⌡ (b − x ) 0
in the Oy direction.
199
3.6 Non-axisymmetric internal mixed-mixed problem
6. Subject to the conditions of Exercise 4, find the tilting angle δ of the punch and its direction. a Y c( x ) x 2 + θ axY s( x ) 3 PH sinh(πθ) ⌠ Answer: δ = dx, in the positive π a 3 b θ(1 + θ2) ⌡ ( b 2 − x 2)1/2 0
direction about the Ox axis. 7. No loading is applied to a flat circular punch of radius a bonded to a transversely isotropic half-space. A tangential concentrated force T is applied at the point ρ= b , φ=0 in the positive Ox direction. Find the normal component w of the punch displacement. TH sinh(πθ) πθ a , with I s defined by (3.2.74). √γ1γ2 I s( b ) − cosh(πθ) πθ ab Hint : use (3.2.72) and the reciprocal theorem. Answer: w =
8. Subject to the general boundary conditions (3.1.1), find the displacements for ρ> a , expressed in terms of the stress function f n. (α/γ1γ2)( G 1 − G 2) f n( x ) + ( G 1 + G 2) D n
a
π Answer: u -n+1(ρ) = n-1 ⌠ ρ ⌡
(ρ − x ) 2
2 1/2
tangential
x 2n-2d x ,
0
a
2π H α ⌠ u n+1(ρ) = f ( x )d{ x 2n-1[( a 2 − x 2)1/2 − (ρ2 − x 2)1/2]} ρn+1 ⌡ n 0
+ π( G 1 + G 2) D n a 2n-1
(ρ2 − a 2)1/2 . ρn+1
9. Subject to the general boundary conditions (3.1.1), find displacements for ρ> a , expressed in terms of the stress function f n. Answer: ℜ{ w n(ρ)e
in φ
+ w -n(ρ)e
-in φ
} = 2π3/2 H α
normal
Γ( n ) a 2n-1 in φ n ℜ{ D n e } Γ( n + 1/2)ρ ρ
a a 2n dρ0 d x x 8H in φ ⌠ + n ℜ e ⌠ ⌠ (ρ2 − x 2)1/2 ⌡ ρ2n-1 (ρ20 − x 2)1/2 ⌡ ρ ⌡ 0 0 x 0
10.
the
0
t 2n-1d f n( t ) (ρ20 − t 2)1/2
.
Investigate the interaction of an arbitrary tangential force with a bonded
200
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
axisymmetric punch.
3.7 Non-axisymmetric external mixed-mixed problem The general formulation of the problem is given in section 3.1, with the boundary conditions (3.1.6), and the governing integral equations (3.1.7−3.1.10). We assume that all the parameters involved can be expanded in a Fourier series. An exact solution for the n -th harmonic is presented below. The governing integral equations for the n -th harmonic are ∞
2ρn+1⌠
xG
dx
⌡x
2n+2
ρ
⌠ ⌡
(x − ρ ) 2
2 1/2
ρ2τ (ρ ) 1 0 n+1 0
+ G [2 n x 2 − (2 n +1)ρ20]τ (ρ ) 2 -n+1 0 (x − 2
a
ρn0dρ0
ρ20)1/2
ρ
2π H α ⌠ − ρn+1 ⌡
σn(ρ0)ρn+1 0 dρ0 = F n+1(ρ), for n ≥0.
a
∞
2ρn-1⌠
xG
dx
⌡ x (x − ρ ) 2n
2
⌠ ⌡
2 1/2
ρ
1
x 2τ-n+1(ρ0) + G 2[(2 n −1) x 2 − 2 n ρ2]τn+1(ρ0) (x − 2
a
∞
+ 2π H αρ ⌠ ⌡ n-1
σ-n(ρ0)dρ0 ρn-1 0
ρ
= F (ρ), -n+1
ρ20)1/2
ρn0dρ0
for n ≥1;
∞ τ (ρ )dρ e-in φ ρ n+1 0 0 in φ n n ⌠ 2π H α ℜ n ⌠ τ (ρ )ρ0dρ − ρ e n -n+1 0 0 ρ0 ⌡ ρ ⌡ a ρ ∞
+ 4 H ρn⌠
dx
x σ (ρ )e-in φ -n 0
⌠
⌡ x 2n( x 2 − ρ2)1/2 ⌡ ρ
a
+ σ (ρ )ein φ n 0
( x 2 − ρ20)1/2
in φ ρn+1 0 dρ = ℜ{e Φ (ρ)}, 0
n
for n ≥0. (3.7.1) The right hand sides in (3.7.1) are known from the boundary conditions, and are a
F n+1(ρ) = u n+1(ρ) +
2π H α⌠ σ (ρ )ρn+1dρ0 ρn+1 ⌡ n 0 0 0
201
3.7 Non-axisymmetric external mixed-mixed problem
a
−
a
2n
2 ⌠ x dx ⌠ 2 ρ ⌡ (ρ − x 2)1/2 ⌡
G 1 x 2τn+1(ρ0) + G 2[2 n ρ2 − (2 n +1) x 2]τ-n+1(ρ0) ρn0(ρ20 − x 2)1/2
n+1
x
0
dρ0,
a
2 x 2n-2d x F -n+1(ρ) = u -n+1(ρ) − n-1 ⌠ 2 ρ ⌡ (ρ − x 2)1/2 0
a
×⌠
G 1ρ20τ-n+1(ρ0) + G 2[(2 n −1)ρ20−2 n x 2]τn+1(ρ0) ρn0(ρ20 − x 2)1/2
⌡
x
ℜ{Φn(ρ)e
in φ
} = w n(ρ)e
in φ
+ w -n(ρ)e
a
a
x 2nd x 4H ⌠ − n⌠ 2 ρ ⌡ (ρ − x 2)1/2 ⌡ x
0
The not loss can
-in φ
dρ0,
e-in φ a + 2π H α ℜ n ⌠ τ-n+1(ρ0)ρn0dρ0 ρ 0⌡
σn(ρ0)e
in φ
+ σ-n(ρ0)e
2 2 1/2 ρn-1 0 (ρ0 − x )
-in φ
dρ0.
(3.7.2) case of axial symmetry ( n =0) was considered in detail in section 3.3, and is discussed here. The solution is sought for n ≥1. We may assume, without of generality, that the first two equations (3.7.1) are homogeneous. This be achieved by the addition of some special solutions to the parameters τ-n+1
and τn+1.
These special solutions, satisfying the right hand sides of the first two
equations (3.7.1), can be obtained from the results of section 2.7. Of course, this procedure will make the right hand side of the third equation (3.7.1) more complicated. Assume the solution of (3.7.1), with the first two equations transformed into homogeneous ones, in the form ∞
df (t)
n σn(ρ) = σ-n(ρ) = ρ ⌠ 2n 2 , ⌡ t ( t − ρ2)1/2 n
ρ
202
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
ρ f ( t )d t Dn n C d ⌠ τn+1(ρ) = n − n+1 2 , ρ dρ ⌡ (ρ2 − t 2)1/2 ρ (ρ − a 2)1/2 a
y ρ f n( t )d t C d 2n⌠ d y d ⌠ τ-n+1(ρ) = n ρ 2n 2 2 1/2 ρ dρ ⌡ y dy ⌡ (y − t )
a
a
ρ d 2n⌠ dt + n ρ . t 2n+1( t 2 − a 2)1/2 ρ dρ ⌡ a
Dn
Here f
n
is the as yet unknown complex stress function, and C , D
n
(3.7.3) are the
constants to be determined. Substitution of (3.7.3) in the first two equations (3.7.1) satisfies them identically, provided that the following conditions hold α C = , γ 1γ 2
∞
D = −2 n C a n
2n+1
f (t) n
⌠ 2n+1 d t . ⌡ t
(3.7.4)
a
∞
π3/2Γ( n + 1) D + 2π H α a 2n+2 ⌠ x f ( x ) ( G 1 + G 2) 2Γ( n + 3/2) n ⌡ n a
x
dx = 0. − ⌠ f n( t )d t 2n+2 2 x x ( − a 2)1/2 ⌡
(3.7.5)
a
Some general formulae from Appendix A3.3 were used in transformations. The conditions (3.7.4) and (3.7.5) might appear contradictory. It will be shown below (see 3.7.11) that this is not so, because an additional constant will appear in the expression for f , due to a homogeneous solution of the integral equation (3.7.8). n
So far, we have satisfied the first two equations (3.7.1). Substitution of (3.7.3) in the third of equations (3.7.1) requires the following transformation:
203
3.7 Non-axisymmetric external mixed-mixed problem
x
dρ ρ2n+1 0 0
⌠ 2 2 1/2 ⌡ ( x − ρ0) a
∞
∞
df (t)
n ⌠ = ⌠ ( x 2 − a 2)1/2( t 2 − a 2)1/2 Q ( x , t ) 2n 2 2 1/2 n ⌡ t ( t − ρ0) ⌡ ρ a
0
|( x 2 − a 2)1/2 + ( t 2 − a 2)1/2| + ψ ( x , t ) ln n | t 2 − x 2|1/2
df (t) n
t 2n
.
Here Q ( x , t ) is a polynomial in even powers of x and t . n
(3.7.6) Although we cannot
write the explicit expression for Q ( x , t ), it can be computed in elementary n
manner for any particular n . x 2n ψn( x , t ) = π
n
The explicit expression for ψ is n
Σ k=0
Γ( n − k +1/2) Γ( k +1/2) t 2k . Γ( n − k +1) Γ( k +1) x
(3.7.7)
Divide both sides of the third of equations (3.7.1) by ρn, differentiate with respect to ρ, multiply the result by ρ2n/(ρ2− r 2)1/2 and integrate with respect to ρ from r to ∞. This procedure allows us to split the kernel of the integral equation into two parts: a singular one and a degenerate one. The result takes the form ∞ ∞ f (t)tdt f ( t )d t n n ( r 2 − a 2)1/2 ⌠ α2 ⌠ + = χ ( r ). − 2 2 2 2 1/2 n r γ1γ2 ⌡ t 2 − ρ2 ⌡ ( t − r )( t − a ) a
a
(3.7.8) Here ∞ Φ (ρ) 1 ⌠ d n ρ2ndρ χn( r ) = 4π H ⌡ (ρ2 − r 2)1/2 dρ ρn r
∞
∞
2 ⌠ d ⌠ ρ2ndρ − 2 2 1/2 π ⌡ (ρ − r ) dρ ⌡ ρ
r
∞
( x 2 − a 2)1/2d x ⌠ 2 ( t − a 2)1/2 Q ( x , t ) n x 2n( x 2 − ρ2)1/2 ⌡
df (t) n
t 2n
a
∞ ∞ ∞ df (t) n 2 ⌠ d x ρ2n-1dρ 2 2 1/2 ⌠ ⌠ + 3/2 q (ρ, x , t )( t − a ) 2 2 1/2 2 2 1/2 2 2 1/2 2n-2 , n π ⌡ (ρ − r ) t ⌡ (x − ρ ) (x − a ) ⌡ r
ρ
a
204
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
(3.7.9) with n-1
q (ρ, x , t ) = n
Σ k=0
t2 Γ( n − k +1/2) t 2k 1 1 F ( 1− n + k , ; − n + k ; 2 ) . Γ( n − k +1) ρ 2 2 x
(3.7.10) Note that the hypergeometric function in (3.7.10) is, in fact, a polynomial, and that all the integrals with respect to x and ρ in the degenerate part of the kernel (3.7.9) are computable in elementary function for any n . The integral equation (3.7.8) was solved in paragraph 3.3, and the solution is ∞
4 f ( t ) = 2cosh2(πθ) t tY ( t )⌠ n c ⌡ π
χ ( r ) Y ( r )d r n
c
r2 − t 2
a
∞
+ Y ( t )⌠ s ⌡
χ ( r) Y ( r) rdr n
s
r2 − t 2
+ A Y ( t ). n c
a
(3.7.11) The last term in (3.7.11) represents the homogeneous solution, with A n being an arbitrary constant.
Its value, along with the constant D and others which appear n
due to the degenerate part of the kernel, can be found from the appropriate system of linear algebraic equations and the conditions (3.7.4) and (3.7.5). The general solution may be considered completed. Example. Consider the action of a normal concentrated load P applied at some point inside the circle ρ= a , with the rest of the plane z =0 being clamped. We may assume, without loss of generality, that the force is applied at the point ρ= b , φ=0 ( b < a ). Thus, the boundary conditions are for ρ> a ;
u = w =0, σ = P δ(ρ− b )δ(φ−0)/ρ,
τ = 0,
for ρ< a .
We determine in this case F -n+1(ρ) = 0,
for n ≥1;
F n+1 = PH α
bn , ρn+1
for n ≥0;
∞
4 dx Φ (ρ) = − PH (ρ b )n ⌠ 2n 2 . 2 n π ⌡ x ( x − ρ )1/2( x 2 − b 2)1/2 ρ
(3.7.12) We present now the explicit solutions for some specific values of n . In the axisymmetric case n =0, the following results may be obtained:
205
3.7 Non-axisymmetric external mixed-mixed problem
∞
∞
d f 0( t )
σ0(ρ) = ⌠ ⌡ ( t 2 − ρ2)1/2 ρ
ρ
α d ⌠ τ1(ρ) = γ1γ2 dρ ⌡
f 0( t ) t d t
1 d ⌠ = ρ dρ ⌡
( t 2 − ρ2)1/2
ρ
f 0( t )d t (ρ2 − t 2)1/2
−
D0 ρ(ρ2 − a 2)1/2
,
,
a
∞
2P f 0( t ) = 3 cosh2(πθ)⌠ π ⌡
Y c( t ) rY c( r ) + tY s( t ) Y s( r )
dr
( r 2 − t 2)( r 2 − b 2)1/2
a
− D0
γ 1γ 2 aα
[1 − Y c( t )], ∞
Y s( r )d r
2 P αsinh(πθ) 1 − coth(πθ)⌠ D0 = − 2 π 4π θγ1γ2 ⌡
. (r − b ) 2
2 1/2
a
The results for the first harmonic ( n =1) are ∞
df (t)
1 σ1(ρ) = σ-1(ρ) = ρ⌠ 2 2 , ⌡ t ( t − ρ2)1/2
ρ
ρ
f 1( t )d t
α 1 d ⌠ τ2(ρ) = γ1γ2 ρ dρ ⌡
(ρ2 − t 2)1/2
−
D1 ρ2(ρ2 − a 2)1/2
,
a
ρ
y ρ f 1( t )d t dy d ⌠ dy , ⌠ + D 3 2 2 1/2 1 y 2 d y ⌡ ( y 2 − t 2)1/2 ⌡ y ( y − a )
1 d 2 α ⌠ τ0(ρ) = ρ ρ dρ γ1γ2 ⌡
a
a
a
∞χ ( r ) Y ( r )d r 1 c
4 f 1( t ) = 2cosh2(πθ) t tY c( t )⌠ π ⌡ a
r − t 2
2
∞χ ( r ) Y ( r ) r d r 1 s
+ Y s( t )⌠
⌡ a
r − t 2
2
+ A Y ( t ), 1 c
206
χ ( r) = 1
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
1 1 Pb + 21 − ( r2 − a 2)1/2B . + 2π r ( r 2 − b 2)1/2 π r 1 r + ( r 2 − b 2)1/2
The constants D , A , and B are to be determined from the set of linear 1 1 1 algebraic equations ∞
2α a 3 ⌠ D1 = − γ 1γ 2 ⌡
∞
f ( t )d t 1
t3
,
B
a
1
= ⌠
⌡
( t 2 − a 2)1/2 d f ( t ), 1 t2
a
∞
x
2π dx ( G + G ) D + 2π⌠ xf ( x ) − ⌠ f ( t )d t 4 2 − − Pb = 0. 2 1 1 1 3Hα 1 x ( x − a 2)1/2 ⌡ ⌡ a
a
It should be noted that the system of tractions in the clamped part is such that its resultant vector is exactly equal to P . This can be shown by a direct integration. Exercise 3.7 1. Find the solution of the example above (page 211) in the limiting case b =0. Answer: the only non-zero stress function is 2cosh (πθ) Y ( t ) sinh(2πθ)[1 − Y ( t )] s c P f (t) = . − 0 2π 2πθ[1 + cosh(πθ)] a πsinh(πθ) t 2
Find the solution of the example above in the limiting case α=0. P ( a 2 − b 2)1/2 b n, Answer: σ (ρ) = − 2 2 τn = 0. 2 1/2 2 2 n π (ρ − a ) (ρ − b ) ρ
2.
3. Prove that the tractions in the plane z =0 are in equilibrium when the exterior is clamped. Note : this is not the case in internal problems. 4. Try to find an exact closed form solution to the mixed-mixed boundary value problem.
207
Appendix A3.1
Appendix A3.1 Some integrals, related to solving internal mixed-mixed problem, are presented here. The notations Y c,s are defined by (3.2.45). It is assumed that 0< r < a . Integrals, containing Y c: a
Y c( x )d x
π ⌠ 2 2 = − 2 r coth(πθ) Y s( r ), ⌡ x − r 0
a
Y ( x )d x
2 2 1/2 c π ⌠ a 2 − r 2 2 2 = − 2 r tanh(πθ) Y s( r ), ⌡ a − x x − r 0
a
⌠ ⌡
Y c( x ) x 2 d x x − r 2
2
πθ a π − r coth(πθ) Y s( r ), sinh(πθ) 2
=
0
a
⌠ ⌡
Y c( x ) x 4 d x x2 − r2
2 π 3 πθ a 2 2 1 − 2θ r + a = − r coth(πθ) Y s( r ) + , 2 sinh(πθ) 3
0
a
⌠ ⌡
( a 2 − x 2)1/2 Y c( x )d x x − r 2
2
= −
π 2 π ( a − r 2)1/2tanh(πθ) Y s( r ) − , 2r 2cosh(πθ)
0
a
⌠ ⌡
x 2( a 2 − x 2)1/2 Y c( x )d x x2 − r2
=
πa2 1 ( + θ2) cosh(πθ) 4
0
− a
π 2 r2 r ( a − r 2)1/2tanh(πθ) Y s( r ) + , 2 cosh(πθ)
x 2 Y c( x )d x
π π r ⌠ Y ( r) , 2 2 1/2 2 2 = 2cosh(πθ) − 2 tanh(πθ) 2 ( a − r 2)1/2 s ⌡ (a − x ) (x − r ) 0
208
CHAPTER 3
a
MIXED-MIXED BOUNDARY VALUE PROBLEMS
a
⌠ Y ( x )d x = πθ a , sinh(πθ) ⌡ c
2 3 ⌠ x 2 Y ( x )d x = πθ(1 − 2θ ) a , c 3sinh(πθ) ⌡
0
0
a
2 ⌠ ( a − x ) Y ( x )d x = πa (1 + θ2), c cosh(πθ) 4 ⌡ 2
2 1/2
0
a
x 2 Y c( x )d x
a
1 − Y c( x )
a
Y c( x )d x
π ⌠ 2 2 1/2 = 2cosh(πθ), ⌡ (a − x )
0
πa 1 2 ⌠ 2 2 1/2 = cosh(πθ)(4 − θ ), a x ( − ) ⌡ 2
0
⌠ ⌡
x
2
dx =
πθcoth(πθ) − 1 . a
0
Integrals, containing Y s: a
xY s( x )d x
π π ⌠ 2 2 = 2 coth(πθ) Y c( r ) − 2sinh(πθ), ⌡ x − r 0
a
Y s( x )d x
π ⌠ 2 2 = 2 coth(πθ) ⌡ x( x − r )
Y c( r ) − 1 r2
,
0
a
⌠ ⌡
x 3 Y s( x )d x x2 − r2
=
πθ2 a 2 πr2 + [cosh(πθ) Y c( r ) − 1], sinh(πθ) 2sinh(πθ)
0
a
Y ( x )d x
Y ( r)
s c 1 π ⌠ = tanh( ) π θ 2 2 1/2 2 2 2 2 2 1/2 − a , ( a − r ) 2r ⌡ x( a − x ) ( x − r ) 0
209
Appendix A3.1
a
x 3 Y s( x )d x
πθ a πr ⌠ tanh(πθ) Y c( r ), 2 2 1/2 2 2 = cosh(πθ) + 2 2( a − r 2)1/2 ⌡ (a − x ) (x − r ) 2
0
a
x Y ( x )d x
2 2 1/2 s π ⌠ a 2 − r 2 2 2 = 2 tanh(πθ) Y c( r ), ⌡ a − x x − r 0
a
⌠ ⌡
x ( a 2 − x 2)1/2 Y s( x )d x x2 − r2
=
π πθ a tanh(πθ) ( a 2 − r 2)1/2 Y c( r ) − , 2 cosh(πθ)
0
a
a
2 2 ⌠ x Y ( x )d x = πθ a , sinh(πθ) ⌡ s
2 2 4 ⌠ x 3 Y ( x )d x = πθ (2 − θ ) a , s 3sinh(πθ) ⌡
0
a
0
x Y s( x )d x
πθ a ⌠ 2 2 1/2 = cosh(πθ), ⌡ (a − x ) 0
a
⌠ ⌡
( a 2 − x 2)1/2 Y s( x )d x x
=
x 3 Y s( x )d x
a
πθ(5 − 4θ2) a 3 ⌠ = , 2 2 1/2 6cosh(πθ) ⌡ (a − x )
0
π πθ a a tanh(πθ) − , 2 cosh(πθ)
0
a
Y s( x )d x
π ⌠ 2 2 1/2 = 2 a tanh(πθ), ⌡ x( a − x ) 0
a
Y c( x )d x
a
Y ( x) xdx
a
2 3 ⌠ x ( a 2 − x 2)1/2 Y ( x )d x = πθ(1 + 4θ ) a , s 6cosh(πθ) ⌡
0
πsinh[2θtan-1( a / r )] ⌠ 2 = , 2 2 r sinh(πθ) ⌡ x + r 0
s π{cosh[2θtan-1( a / r )] − 1} ⌠ 2 = , 2 2sinh(πθ) ⌡ x + r 0
210
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
Integrals, containing combination Y c+ iY s: a
iθ π dx 1 − cosh(πθ) a + riθ, ⌠ a + x = i sinh(πθ) a − r ⌡ a − x x − r
-a
a
iθ π i coth(πθ) a + r i θ dx ⌠ a + x = , a + r a − r ⌡ a − x ( a + x )( x − r)
-a
a
iθ π i tanh(πθ) a + r i θ dx ⌠ a + x = , 2 2 1/2 ( a 2 − r 2)1/2 a − r ⌡ a − x ( x − r)( a − x )
-a
a
2 2 1/2 iθ iθ ⌠ a + x ( a − x ) d x = πi tanh(πθ)( a 2 − r2)1/2a + r − π(2 ia θ + r), cosh(πθ) x − r a − r ⌡ a − x
-a
a
iθ
⌠ a + x ⌡ a − x
dx =
2πθ a , sinh(πθ)
-a
a
2 3 iθ ⌠ a + x x 2d x = 2πθ(1 − 2θ ) a , 3sinh(πθ) ⌡ a − x
-a
a
iθ dx iπ ⌠ a + x = − , sinh(πθ) ⌡ a − x x + a
-a
a
iθ π a ( i + 2θ) xdx ⌠ a + x = , sinh(πθ) ⌡ a − x x + a
-a
a
2 2 iθ ⌠ a + x x d x = 2 i πθ a , sinh(πθ) ⌡ a − x
-a
211
Appendix A3.2
a
iθ π a 2[2θ + i (1 − 2θ2)] x 2d x ⌠ a + x . = − sinh(πθ) ⌡ a − x x + a
-a
Appendix A3.2 Some integrals, used in solving the external mixed-mixed problem, are presented here. It is assumed that a < r <∞. Integrals containing Y c: ∞
Y c( x )d x
π ⌠ 2 2 = 2 r coth(πθ) Y s( r ), ⌡ x − r a
∞
Y c( x )d x
π πθ ⌠ 2 2 , 2 = 3 coth(πθ) Y s( r ) − 2 ar sinh(πθ) 2r ⌡ x (x − r ) a
∞
xY c( x )d x
∞
Y ( x )d x
∞
Y c( x )d x
Y s( r )
π ⌠ 2 2 1/2 2 2 = 2 tanh(πθ) 2 ( r − a 2)1/2 ⌡ (x − a ) (x − r ) a
Y ( r)
c s π 1 ⌠ 2 2 1/2 2 2 = 2 − a cosh(πθ) + tanh(πθ) 2 2 1/2 , − − − x ( x a ) ( x r ) 2 r ( r a ) ⌡ a
π 1 ⌠ 3 2 2 1/2 2 2 = 4 − a cosh(πθ) 2r ⌡ x (x − a ) (x − r ) a
+ tanh(πθ)
Y s( r )
− π[(1/4) − θ2], ( r 2 − a 2)1/2 a 3 r 2cosh(πθ)
212
∞
⌠ ⌡
CHAPTER 3
( x 2 − a 2)1/2 Y ( x )d x c x( x − r ) 2
2
=
MIXED-MIXED BOUNDARY VALUE PROBLEMS
π a 2 2 1/2 2 cosh(πθ) + tanh(πθ)( r − a ) Y s( r ) , 2r
a
∞Y
⌠ ⌡
c
( x )d x x2
πθ = , a sinh(πθ)
a
∞
∞
Y c( x )d x
π ⌠ 2 2 1/2 = 2 a cosh(πθ), ⌡ x( x − a )
a
Y c( x )d x
π[(1/4) − θ2] ⌠ 3 2 = , 2 1/2 a 3cosh(πθ) ⌡ x (x − a ) a
∞
⌠ [1 − Y ( x )]d x = a [πθcoth(πθ) − 1], c ⌡ a
∞
⌠ [1 − Y ( x )] 2 x d x 2 1/2 = πθ a tanh(πθ), c (x − a ) ⌡ a
∞
xY ( x )
c d x = a [πθtanh(πθ) − 1], ⌠ 1 − 2 2 1/2 ( ) x − a ⌡ a
∞
⌠ ( x 2 − a 2)1/2d[ Y ( x )] = πθ a tanh(πθ), c ⌡ a
∞
⌠ 2 x 2 1/2 − 1 Y ( x )d x = 2πθ a , sinh(2πθ) c ⌡ ( x − a ) a
∞
2 2 1/2 2πθ a πa ⌠ 1 − ( x − a ) Y ( x )d x = , − c sinh(2 x 2cosh( π θ ) πθ) ⌡ a
213
Appendix A3.2
Integrals containing Y : s ∞
⌠ ⌡
x Y s( x )d x x − r 2
2
=
π coth(πθ)[1 − Y ( r )], c 2
a
∞
Y s( x )d x
π 1 ⌠ 2 2 = 2 sinh(πθ) − coth(πθ) Y c( r ) , − x ( x r ) 2 r ⌡ a
∞
Y s( x )d x
∞
x 2 Y s( x )d x
∞
Y ( x )d x
πtanh(πθ) ⌠ Y ( r ), 2 2 1/2 2 2 = − 2 r ( r 2 − a 2)1/2 c ⌡ (x − a ) (x − r ) a
π r ⌠ 2 2 1/2 2 2 = 2 tanh(πθ) 1 − 2 2 1/2 Y c( r ) , (r − a ) ⌡ (x − a ) (x − r ) a
s π tanh(πθ) θ , ⌠ 2 2 2 1/2 2 2 = − 2 2 2 1/2 Y c( r ) + 2 2 ( ) cosh( ) x ( x a ) ( x r ) r r r a a − π θ − − ⌡ a
∞
⌠ ⌡
( x 2 − a 2)1/2 Y ( x )d x s x2 − r2
=
( r 2 − a 2)1/2 π tanh(πθ)1 − Y c( r ), 2 r
a
∞
⌠ Y ( x ) d x = π tanh(πθ), 2 2 x ⌡ s a
∞
2 2 ⌠ Y ( x ) d x5 = πθ 4(2 − θ ), 3 a sinh(πθ) x ⌡ s a
∞
Y s( x )d x
πθ ⌠ 2 2 , 2 1/2 = 2 a cosh(πθ) ⌡ x (x − a ) a
∞
2 ⌠ Y ( x ) d x3 = 2 πθ , x a sinh(πθ) ⌡ s
a
∞
Y s( x )d x
π ⌠ 2 2 1/2 = 2 tanh(πθ), ⌡ (x − a ) a
214
CHAPTER 3
MIXED-MIXED BOUNDARY VALUE PROBLEMS
∞
⌠ ( x 2 − a 2)1/2d[ Y ( x )] = πa 2tanh(πθ)(θ2 − 1/4), s ⌡ a
∞
2θ ⌠ ( x 2 − a 2)1/2 Y ( x ) d x2 = πtanh(πθ) − s cosh(πθ) 2 x ⌡ a
∞
2 ⌠ ( x 2 − a 2)1/2 Y ( x ) d x4 = 2πθ(2θ + 1/4) s x 3 a cosh(πθ) ⌡ a
∞
2 8θ2 ⌠ 2 x 2 1/2 − 1 Y ( x ) x d x = πa tanh(πθ) + , 4 sinh(2πθ) s ⌡ ( x − a ) a
Integrals containing both Y c and Y s: ∞
⌠ [ x Y ( x ) − 2 a θ Y ( x )] d x = πa 2θ2coth(πθ), s c ⌡ a
∞
2 ⌠ [ x Y ( x ) − 2 a θ Y ( x )] 2 x d x 2 1/2 = πa tanh(πθ)(1 + 4θ2), s c 4 (x − a ) ⌡ a
∞
2 2 1/2 ⌠ 1 − ( x − a ) [ x Y ( x ) − 2 a θ Y ( x )] d x s c x ⌡ a
=
2πθ2 a 2 πa2 πθ a 2 + tanh(πθ) − . sinh(2πθ) 4 cosh(πθ)
215
Appendix A3.3
Appendix A3.3 Some formulae, related to transformation and computation of integrals, are presented here. Transformations of the limits of integration: ρ
⌠ ⌡ 0
a
a
2 ⌠ dy f ( x )d x ⌠ 2 2 1/2 = π ρ 2 2 (ρ − x ) ρ − y ⌡ ⌡ 0
f ( x )d x ( x − y 2)1/2 2
y
a
a
y
2 2 f ( x )d x ydy f ( x )d x 2 1/2⌠ ⌠ ⌠ 2 2 1/2 = π( a − ρ ) 2 2 2 2 1/2 2 2 1/2 ⌡ (x − ρ ) ⌡ ( y − ρ )( a − y ) ⌡ ( y − x ) ρ
0
ρ
0
a
2 ⌠ dy ⌠ xf2 ( x )d x2 + π lim[ xf ( x )], = 2 2 1/2 2ρx→0 π ⌡ (ρ − y ) ⌡ x − y 0
0
a
a
y
f ( x )d x 2 2 ydy f ( x )d x 2 1/2⌠ ⌠ ⌠ 2 2 1/2 = π(ρ − a ) 2 2 2 2 1/2 2 2 1/2 ⌡ (ρ − x ) ⌡ (ρ − y )( a − y ) ⌡ ( y − x ) 0
0
0
Simplification of two consecutive integrals: a
a
a
x 2d x d ⌠ f ( y )d y π ⌠ ⌠ f ( y )d y 2 2 1/2 d x 2 2 1/2 = − 2 ρ f (ρ) + ⌡ (x − ρ ) ⌡ ⌡ (y − x ) ρ
ρ
x
−
a 2 2 1/2 2 1/2 lim[ f ( r )( a − r ) ] ; (a − ρ ) r→a 2
a
a
f ( y )d y π f (ρ) ⌠ 2 2 d x 2 1/2 d ⌠ = − 2 + ⌠ 2 2 1/2 d 2 x ρ ( ) ( ) x x y x − ρ − ρ ⌡ ⌡ ⌡ ρ
x
ρ
a
f ( y )d y . y3
216
CHAPTER 3
x
a
MIXED-MIXED BOUNDARY VALUE PROBLEMS
a
( x 2 − b 2)1/2 + ( y 2 − b 2)1/2 f ( y )d y ⌠ 2 t d t 2 1/2 ⌠ ⌠ = ( )ln dy, f y 2 2 1/2 | x 2 − y 2|1/2 ⌡ (x − t ) ⌡ (y − t ) ⌡ b
t
a
b
t
a
( a 2 − x 2)1/2 + ( a 2 − y 2)1/2 tdt f ( y )d y ⌠ ⌠ = ( )ln dy, f y − x 2)1/2 ⌡ ( t 2 − y 2)1/2 | x 2 − y 2|1/2 ⌡
⌠ 2 ⌡ (t x
b
b
a
a
f ( r) rdr π ⌠ 2 d x2 (1+κ)/2 d ⌠ 2 2 (1-κ)/2 = − 2cos(πκ/2) f (ρ), d x ⌡ (x − ρ ) ⌡ (r − x ) ρ
x
ρ
x
f ( r) rdr π f (ρ) − 1 lim[ rf( r)] ⌠ 2 d x2 (1+κ)/2 d ⌠ = 2 2 (1κ)/2 2cos(πκ/2) dx ⌡ (x − r ) ρ r→0 ⌡ (ρ − x ) 0
0
x
a
a
f ( y )d y f ( y )d y 2 2 1/2 ⌠ ⌠ 2 d t 2 1/2 d ⌠ 2 2 1/2 = ( x − b ) 2 2 2 2 1/2, t d ⌡ (x − t ) ⌡ (y − t ) ⌡ ( y − x )( y − b ) b
t
b
a
t
⌠ 2 ⌡ (t
a
f ( y )d y f ( y )d y dt d ⌠ 2 − x 2)1/2⌠ 2 1/2 d t 2 2 1/2 = ( a 2 2 2 2 1/2, − x) ⌡ (t − y ) ⌡ ( y − x )( a − y )
x
b
x
b
a
a
d ⌠ tdt f ( y )d y x ( y 2 − b 2)1/2 f ( y )d y ⌠ ⌠ = , d x ⌡ ( x 2 − t 2)1/2 ⌡ ( y 2 − t 2)1/2 ( x 2 − b 2)1/2 ⌡ y2 − x2 b
t
a
b
t
a
tdt f ( y )d y x d ⌠ ( a 2 − y 2)1/2 f ( y )d y ⌠ ⌠ = , d x ⌡ ( t 2 − x 2)1/2 ⌡ ( t 2 − y 2)1/2 ( a 2 − x 2)1/2 ⌡ y2 − x2 x
b
b
a
x
a
dx d ⌠ f ( y )d y f ( y )d y 2 2 1/2⌠ ⌠ 2 2 1/2 d x 2 2 1/2 = (ρ − a ) 2 2 2 2 1/2, ⌡ (ρ − x ) ⌡ (x − y ) ⌡ (ρ − y )( a − y ) 0
0
0
217
Appendix A3.3
∞
⌠ ⌡ a
∞
∞
xdx ⌠ 2 ( x − ρ2)1/2 ⌡
df(t) = ⌠ 2 ( t − x 2)1/2 ⌡
x
a
∞
dx d ⌠ 2 2 1/2 d x (x − ρ ) ⌡
f(t)tdt ( t − x 2)1/2 2
x
∞
= (a
2
− ρ) ⌠ 2 1/2
⌡ (t
2
f(t)tdt . − a 2)1/2( t 2 − ρ2)
a
Computation and/or transformation of some integrals: a
a
π( a 2n − x 2n) t 2n-1d t ⌠ 2n+1 2dρ 2 1/2 ⌠ = , 2 2 1/2 4 nax 2n+1 ⌡ ρ (ρ − x ) ⌡ ( a − t ) ρ
x
a
a
π( a 2n-1 − x 2n-1) t 2n-1d t ⌠ 2n-1 2dρ 2 1/2 ⌠ , = 2 2 1/2 2(2 n − 1) x 2n-1 ⌡ ρ (ρ − x ) ⌡ ( a − t ) ρ
x
a
a
⌠ 2n-1 2 ⌡ ρ (a x
=
ρ2n-1dρ 1 dρ ⌠ = − ρ2)1/2(ρ2 − x 2)1/2 ( ax )2n-1 ⌡ ( a 2 − ρ2)1/2(ρ2 − x 2)1/2 x
π √πΓ( n − 1/2) 1 x2 1 3 x2 ( − − ) ( − − ), F n F n n 1 , ; 1; 1 = 1 , ; ; 2 2 2 a2 a2 2 ax 2n-1Γ( n ) 2 ax 2n-1
a
a
π( a 2 − x 2) a 2 − ρ2)1/2dρ 1 (ρ2 − x 2)1/2ρ2n-1dρ ⌠ (2n+1 ⌠ = = 2 2 1/2 4 ax 2n+1 a 2n-1 x 2n+1 ⌡ ( a 2 − ρ2)1/2 ⌡ ρ (ρ − x ) x
x
× F (1− n ,
1 x2 √πΓ( n + 1/2)( a 2 − x 2) 1 1 x2 ; 2; 1− 2) = F ( 1 − n , ; − n ; ) 2 2 2 2 n ! ax 2n+1 a a2
a
a
ρ2 − x 2)1/2dρ 1 ( a 2 − ρ2)1/2ρ2n-1dρ ⌠ (2n+1 ⌠ = 2 2 1/2 (ρ2 − x 2)1/2 a 2n+1 x 2n-1 ⌡ ⌡ ρ (a − ρ ) x
x
218
CHAPTER 3
= ρ
⌠ ⌡ a
MIXED-MIXED BOUNDARY VALUE PROBLEMS
x2 π( a 2 − x 2) 3 F n ; 2; 1 ( 1 − , − ). 2 a2 4 a 3 x 2n-1 ρ
x 2nd x = a 2n+1ρ2n⌠ (ρ2 − x 2)1/2 ⌡
x
2n+1
dx , ( x − a 2)1/2 2
a
a
a
x 2nd x dx 2n+1 2n⌠ ⌠ ρ . 2 2 1/2 = a 2n+1 2 ⌡ (x − ρ ) ⌡ x ( a − x 2)1/2 ρ
ρ
a
(a − x ) (a − t ) dρ ⌠ 2n-1 2 2 1/2 2 2 1/2 = x t a 2 x 2n-2( x 2 − t 2) ( ) ( ) ρ ρ − ρ − ⌡ 2
2 1/2
2
2 1/2
x
n-1
Σ m=0
a 2 − x 2m a2
-1 t 2( a 2 − x 2) dm Γ( n ) m-1/2 sin √ζ . (1 ) , with = ζ − − ζ × a 2( x 2 − t 2) √ζ Γ( n − m )( m !)2 dζm
ρ
∞
1 dx ⌠ 2n+1 2 ⌠ 2 1/2 2 2 = 2n+1 2n+2 a ρ ⌡ ⌡ x (x − a ) (x − ρ )
x 2n+2d x (ρ2 − x 2)1/2( a 2 − x 2)
0
a
n
Σ
1 π a 2n+1 √π = 2n+1 2n+2 2 2 1/2 − 2 2( a − ρ ) a ρ
k=1
Γ( k +1/2) 2(n-k) 2k a ρ , Γ( k +1)
Rules of interchanging the order of integration: a
r
a
a
f (ρ)ρdρ F ( r) rdr ⌠ F ( r)d r d ⌠ ⌠ f (ρ)dρ d ⌠ + 2 2 1/2 = − 2 d r d ρ ⌡ ( r − ρ2)1/2 ⌡ ⌡ (r − ρ ) ⌡ 0
0
0
ρ
d ⌠ F ( r) rdr F ( r) rdr ⌠ limρ f (ρ) + lim f ( ) ρ , dρ ⌡ ( r 2− ρ2)1/2 ( r 2− ρ2)1/2 ρ→0 ρ→a ⌡ ρ ρ a
a
219
Appendix A3.3
∞
∞
⌠ F (ρ)dρ d ⌠ dρ ⌡ ⌡ ρ
a
a
∞
x
f( x) xdx ⌠ f ( x )d x d ⌠ F2 (ρ)ρd2ρ1/2, 2 2 1/2 = − dx ⌡ (x − ρ ) (x − ρ ) ⌡ a
a
a
a
a
2 ⌠ 2 d t 2 ⌠ f (2t , r)d r2 = − π f ( x2, x ) + ⌠ d r ⌠ 2 f ( t2, r)d2t 2 , 4x ⌡t − x ⌡ r − t ⌡ ⌡ ( t − x )( r − t ) 0
0
a
0
a
0
a
a
2 ⌠ d t ⌠ f (2t , r)d r2 = − π f (0,0) + ⌠ d r ⌠ f (2 t , r)d t2 . 4 ⌡ ⌡ r − t ⌡ ⌡ (r − t ) 0
0
0
0
CHAPTER 4 APPLICATIONS IN FRACTURE MECHANICS
The great majority of the punch and crack problems solved deals with the stresses and displacements in the plane z =0 only. Some solutions of this kind have been presented in previous chapters. There are just a few complete solutions published (Sneddon, 1951; Elliott, 1949; Westmann, 1965), where explicit expressions are given for the field of displacements and stresses for the simplest axisymmetric problems (a circular punch and a penny-shaped crack). The explicit expressions for the field of displacements due to an elliptic crack can be found in (Kassir and Sih, 1975). Knowledge of complete solutions is indispensable for consideration of more complicated problems of crack interactions, influence of external loads on punches and cracks, etc. We present in this chapter a complete solution to the problem of a penny-shaped crack in a transversely isotropic elastic space, subjected to an arbitrary normal and tangential loading. All the relevant Green’s functions are given explicitly in terms of elementary functions. An approximate analytical solution is given for a flat crack of arbitrary shape. The solution’s accuracy is high, which is mainly due to the fact that it becomes exact in the case of an elliptical crack. The derivation of non-singular governing integral equations enables us to consider a very close interaction of coplanar cracks. Some of the material presented in this Chapter is still unpublished. The rest follows the papers (Fabrikant, 1987a, 1987b, 1987f, 1987g, 1988b, 1989).
4.1 Flat crack under arbitrary normal loading A general solution to some mixed problems in terms of three harmonic functions was given in Chapter 2. We show here that in the case of a flat crack under normal loading, the general solution can be expressed through just
220
221
4.1 Flat crack under arbitrary normal loading
one such function. Consider a transversely isotropic elastic space weakened by a flat crack S in the plane z =0, with arbitrary pressure p applied to the crack faces. Due to symmetry, the problem can be formulated as follows: find the solution to the set of differential equations (2.1.3) for a half-space z ≥0, subject to the mixed boundary conditions on the plane z =0: σz = − p ( x,y), for ( x,y)∈ S ;
w = 0, for ( x,y)∉ S ;
τz = 0 , for −∞< ( x,y) <∞.
(4.1.1)
These conditions can be satisfied by a representation in terms of one harmonic function. Let us put, according to (2.1.13), F ( z ) = c F ( z ), 1
1
F ( z ) = c F ( z ),
1
2
2
2
F ( z) = 0 . 3
(4.1.2)
Expressions of the type F ( z ) and F ( z ), etc., everywhere in the book should be 1
1
understood as F ( x,y,z) and F ( x,y,z ) respectively. 1
1
The substitution of (4.1.2) and
the last of expressions (2.1.12) in the third condition (4.1.1) yields: c = −c γ / m γ 1
2 1
(4.1.3)
1 2
We can represent the function F as the potential of a simple layer, i.e. ω( N ) d S F (ρ,φ, z ) ≡ F ( z ) = ⌠ ⌠ , ⌡ ⌡ R(M,N)
(4.1.4)
S
where ω stands for the crack face displacement w ( x,y,0), R(M,N) is the distance between the points M (ρ,φ, z ) and N ( r ,ψ,0), and the integration is taken over the crack domain S . Expression (4.1.4) satisfies the second condition (4.1.1) identically, due to the well known property of the potential of a simple layer. Inside the crack the same property gives: ∂F = −2πω = −2π w ( x,y,0) ∂ z z=0
(4.1.5)
Now expressions (4.1.2), (4.1.4), (4.1.5), and (2.1.6) give the second equation for c and c : 1
2
− m c /γ − m c /γ = 1/2π 1 1
1
2 2
2
(4.1.6)
222
CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
The constants c 1 and c 2 are determined from (4.1.3) and (4.1.6) as c1 = −
γ1
c2 = −
,
2π( m 1 − 1)
γ2 2π( m 2 − 1)
.
(4.1.7)
The potential functions will be given by F 1(z) = −
γ1 2π( m 1 − 1)
F 2(z) = −
F ( z 1),
γ2 2π( m 2 − 1)
F ( z 2).
(4.1.8) The substitution of (4.1.8) and (2.1.12) in the first condition (4.1.1) leads to the governing integral equation: p ( N 0) = −
1 ω( N ) d S ∆⌠ ⌠ , 2 4π H ⌡ ⌡ R ( N0,N)
(4.1.9)
S
where, as before, R ( N0,N) stands for the distance between two points N 0 and N , and both N 0, N ∈ S .
The following identities were used:
( m 1 − 1)/( m 1 + 1) = 2π A 44 H (γ1 − γ2).
m 1 m 2 = 1,
We next consider the penny-shaped crack in more detail. case of a general crack in section 4.8 below.
(4.1.10) We shall return to the
Green’s functions for a penny-shaped crack. An exact solution in elementary functions is possible when the crack is circular. Let a be the radius of the crack. The governing integral equation (4.1.9) can be rewritten in polar coordinates as follows (see section 2.8) ρ
1 1 d ⌠ L( ) p (ρ,φ) = − 2 ρ dρ ⌡ π Hρ
x dx (ρ − x ) 2
2 1/2
L( x 2)
0
ρ0dρ0 d ⌠ 1 × L( ) ω(ρ0,φ). ρ0 d x ⌡ (ρ 2 − x 2)1/2 0 a
x
(4.1.11) The integral operator inverse to (4.1.11) is defined by (2.8.2)
223
4.1 Flat crack under arbitrary normal loading
x ρ dρ ρρ 0 0 0 d x p (ρ , φ). ⌠ ⌠ ω(ρ, φ) = 4 H L 2 2 1/2 2 2 1/2 2 0 ( x − ρ ) ( x − ρ ) x ⌡ ⌡ 0 a
ρ
(4.1.12)
0
Another form of solution can be obtained from (1.4.33), namely, 2π a p (ρ , φ ) 0 0
2 ⌠ ⌠ ω = H π ⌡ ⌡ 0
R
η tan-1 ρ0dρ0dφ0, R
(4.1.13)
0
where R =[ρ2 + ρ20 − 2ρρ cos(φ−φ )]1/2, 0 0
η = ( a 2 − ρ2)1/2 ( a 2 − ρ20)1/2/ a .
(4.1.14) In this chapter, we do not restrict attention to the plane z =0: our purpose is obtaining a complete solution. We shall call F (ρ,φ, z ), as defined by (4.1.4), the main potential function since both functions F 1 and F 2 become easily available, when F is found. The substitution of (4.1.13) in (4.1.4) allows us to express the main potential function as follows: 2π a
2 F (ρ,φ, z ) = H⌠ ⌠ K (ρ,φ, z ; ρ ,φ ) p (ρ ,φ )ρ dρ dφ , 0 0 0 0 0 0 0 π ⌡ ⌡ 0
(4.1.15)
0
where the Green’s function K reads: K ( M;N ) = K (ρ,φ, z ; ρ ,φ ) 0 0 0 2π a
( a 2 − r 2)1/2( a 2 − ρ20)1/2 1 rd rdψ . = ⌠ ⌠ tan-1 a R ( N,N ) R ( M,N) ⌡ ⌡ R ( N,N0) 0 0
0
(4.1.16) Here R (⋅,⋅) denotes the distance between respective points: M (ρ,φ, z ), N ( r ,ψ,0), and N (ρ ,φ ,0). Although we can not compute the integral in (4.1.16) in elementary 0 0 0 functions, all its derivatives can be expressed in elementary functions, due to the fundamental integral established in section 1.6. Making use of (1.6.19), one can write:
224
CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
∂K h 2π , tan-1 = − R ( M,N ) R ( M,N ) ∂z 0 0
(4.1.17)
where h = ( a 2 − l 21)1/2( a 2 − ρ20)1/2/ a , (4.1.18) and the contraction l everywhere in the book stands for l ( a ), as defined by 1
1
(0.18). Note that h tends to η, as defined by (4.1.14), for z →0 and ρ< a . Expressions (4.1.15) and (4.1.17) allow us to write: 2π a
∂F h 1 p (ρ ,φ )ρ dρ dφ . tan-1 = −4 H⌠ ⌠ 0 0 0 0 0 ∂z R ( M,N ) R ( M,N ) ⌡ ⌡ 0 0 0
(4.1.19)
0
The integral in (4.1.19), although looking difficult to compute even for p =const, can be expressed in elementary functions for any polynomial loading. This becomes evident, if we use the equivalent representations through the L-operator (see 1.4.31): l2
∂F = −8π H ⌠ ∂z ⌡
dx
l (0) 2
( x 2 − ρ2)1/2
g(x)
ρ dρ
ρρ
0 0 0p (ρ ,φ). ⌠ 2 2 1/2 L x2 0 ⌡ [ g ( x ) − ρ0] 0
(4.1.20) Here z2 1/2, g( x) = x 1 + 2 2 ρ − x and the contraction l (0.14).
2
(4.1.21)
everywhere in the paper stands for l ( a ), as defined by 2
Using the change of variables x = l 2( t ), t = g ( x ), expression (4.1.20) can be
rewritten as follows: a t dl (t) ρ dρ ρρ 2 0 0 0 ∂F p (ρ ,φ). ⌠ ⌠ = −8π H L 2 2 1/2 2 2 1/2 2 ∂z l 2( t ) 0 ⌡ [ l 2( t ) − ρ ] ⌡ ( t − ρ0) 0
0
(4.1.22) Since the function F vanishes at infinity, it can be determined from (4.1.22) in the form
225
4.1 Flat crack under arbitrary normal loading
z
a
ρ dρ
t
d l 2( t )
ρρ
0 0 0 p (ρ ,φ). ⌠ 2 F (ρ,φ, z ) = −8π H ⌠ d z ⌠ 2 L 2 1/2 2 1/2 l 22( t ) 0 ⌡ ⌡ [ l 2( t ) − ρ ] ⌡ ( t − ρ0) ∞
0
0
(4.1.23) By using the property ∂ l 2( t ) ∂t
= −
[ l 22( t ) − ρ2]1/2 ∂ l 1( t ) ∂z
[ρ2 − l 21( t )]1/2
,
which is a consequence of formulae (A4.1.28) and (A4.1.29) from Appendix A4.1, expression (4.1.23) can be modified as follows: a
ρ0dρ0
t
F (ρ,φ, z ) = 8π H⌠ d t ⌠
⌡
⌡ (t −
0
ρ20)1/2
2
0
l1(t)
y 2ρ0
p (ρ ,φ). ⌠ 2 2 1/2 L 0 t 2ρ ⌡ (ρ − y ) dy
0
(4.1.24) Expression (4.1.24) is convenient for exact evaluation of the potential function F and proves that it can be expressed in elementary functions for arbitrary polynomial loading. A simple change of variables gives another formula, equivalent to (4.1.24): ∞
ρρ0 dx ⌠ ⌠ ⌠ F (ρ,φ, z ) = 8π H t d t 2 2 1/2 ⌡ 2 2 1/2 L 2 p (ρ0,φ). x ⌡ ⌡ ( t − ρ0) l ( t ) x ( x − t ) 0 0 2 a
t
ρ0dρ0
(4.1.25) We can proceed now with the remaining derivatives of the Green’s function K , defined by (4.1.16). Differentiation of (4.1.16) yields: 2π a
Λ K (ρ,φ, z ;ρ0,φ0) = −⌠ ⌠
⌡ ⌡ 0
ρei
φ 3
− r ei
R ( M,N )
ψ
( a 2 − r 2)1/2( a 2 − ρ20)1/2
tan-1
0
This integral is computed in Appendix A4.3.
a R ( N,N0)
rd rdψ . R ( N,N0)
(4.1.26) By using (A4.3.11), one can write:
( a 2 − ρ20)1/2 2π z h s -1 , Λ K (ρ,φ, z ; ρ0,φ0) = tan − tan-1 2 2 1/2 R0 q R 0 s (l2 − a ) (4.1.27)
226
CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
where Λ is given by (2.1.5), h is defined by (4.1.18), and -i φ
q = ρe
-i φ
− ρ0e
0
-i(φ-φ0) 1/2
s = ( a 2 − ρρ0e
,
)
,
R 0 = R ( M,N0) = [ρ2 + ρ20 − 2ρρ0cos(φ−φ0) + z 2]1/2 .
(4.1.28)
The other derivatives, which will be needed for the complete solution, are: ρ2 − l 21 z h h z 2 ∂2 -1 K (ρ,φ, z ; ρ0,φ0) = 2π 3 tan − 2 , − R 0 R0 z [ R 20 + h 2] l 22 − l 21 R0 ∂ z2
iφ
ρe ∂ Λ K (ρ,φ, z ; ρ ,φ ) = 2π 0 0 ∂z
+
iφ
− ρe 0
iφ
ρe
+ 2 2 R 0 + h l 22 − l 21
−
− z)
q 2 R 30
2
iφ
− ρe 0
0
R 20
( a − Λ2 K (ρ,φ, z ; ρ ,φ ) = 2π 0 0 q s z (3 R 20
h tan-1 R 0
ρe
2
0
R 30
iφ
h
(4.1.29)
ρ20)1/2
iφ
2 q
−
ρe 0
,
(4.1.30)
0
s tan-1 ( l 22 − a 2)1/2 s2 iφ
h tan-1 + R 0
( a 2 − ρ20)1/2( l 22 − a 2)1/2ρ0e
iφ q ρ2e2 − 2 − 2 R 0 + h 2 q R 20 ( l 2 − l 21)( l 22 − ρ2)
zh
-i(φ-φ0)
q s 2[ l 22 − ρρ0e
.
0
]
(4.1.31)
This concludes the general solution to the problem of a penny-shaped crack subjected to an arbitrary pressure. Formulae (4.1.17) and (4.1.24−4.1.31) are the main results of this section.
1.
Exercise 4.1 Prove the identity (4.1.10).
227
4.2 Point force loading of a penny-shaped crack
2.
Establish (4.1.19).
3.
Verify the derivation of (4.1.27) −(4.1.31).
4. Find the Green’s functions for a semi-infinite plane crack in a transversely isotropic space, subjected to a normal loading. Hint : consider the limiting case of (4.1.24−4.1.31), when the radius a →∞, and the coordinate origin moves from the circle centre to its boundary.
4.2 Point force loading of a penny-shaped crack Consider a penny-shaped crack opened by two equal concentrated forces P applied in opposite directions at the point (ρ ,φ ,0±), ρ < a . Formulae (2.1.6), 0
0
0
(2.1.12), (4.1.8), (4.1.17), and (4.1.24−4.1.31) give a complete solution for the field of displacements and stresses in elementary functions, namely, γ γ 1 2 2 u = HP f (z ) + f (z ) , π m − 1 1 2 m1 − 1 1 1 2
(4.2.1)
m m 1 2 2 w = HP f (z ) + f (z ) , π m − 1 2 2 m1 − 1 2 1 2
(4.2.2)
γ1 1 σ1 = 2 f (z ) 2 − γ 3 1 π (γ − γ ) ( m + 1)γ3 1 1 2 1 2P
γ2 1 − − f ( z ) , γ 3 2 ( m 2 + 1)γ23 2
(4.2.3)
γ γ 1 2 4 σ = HA P f (z ) + f (z ) , 2 66 4 1 π m − 1 4 2 m − 1 1
σ = z
P γ f (z ) − γ f (z ) , 2 3 2 π (γ − γ ) 1 3 1 2
1
2
(4.2.4)
2
(4.2.5)
228
CHAPTER 4
τz =
APPLICATIONS IN FRACTURE MECHANICS
P f (z ) − f (z ) , 5 2 π (γ 1 − γ 2) 5 1
(4.2.6)
2
where 2 2 1/2 s z h 1 ( a − ρ0) -1 f 1( z ) = − tan-1 2 tan , R0 R0 q s ( l 2 − a 2)1/2
f 2( z ) =
(4.2.7)
h 1 tan-1 , R0 R 0
(4.2.8)
ρ2 − l 21 z z2 h -1 h − 2 f 3( z ) = − 3 tan + R0 R0 R0 z ( R 20 + h 2) l 22 − l 21
f 4( z ) =
+
( a 2 − ρ20)1/2 q s
z (3 R 20 − z 2) q 2 R 30
iφ
ρ0e
s2
(4.2.9) 0
−
s 2 tan-1 2 ( l 2 − a 2)1/2 q iφ
h tan-1 − R 0
,
( a 2 − ρ20)1/2( l 22 − a 2)1/2ρ0e -i(φ-φ0)
q s 2[ l 22 − ρρ0e
0
]
iφ ρ2e2 q , + 2 2 2 − 2 2 2 2 R0 + h qR0 ( l 2 − l 1)( l 2 − ρ )
zh
iφ
ρe f 5( z ) = −
iφ
− ρ0e R 30
0
(4.2.10)
iφ
ρe
iφ
h h ρe tan-1 + 2 + R 0 R 0 + h 2 l 22 − l 21
It is reminded that R 0=[ρ + 2
ρ20
R 20
− 2ρρ0cos(φ−φ0) + z ] . 2 1/2
(4.2.5) simplifies when z =0 and ρ> a , namely, σz =
P
( a 2 − ρ20)1/2
π2 (ρ2 − a 2)1/2[ρ2 + ρ20 − 2ρρ0cos(φ−φ0)]
iφ
− ρ0e
.
0
.
(4.2.11) The expression for σz
(4.2.12)
229
4.3 Concentrated load outside a circular crack
Defining the stress intensity factor k 1 = lim{(ρ − a )1/2σz} , ρ→ a
the following result may be obtained from (4.2.12):
k = 1
( a 2 − ρ20)1/2
P
π2(2 a )1/2 a 2 + ρ20 − 2 a ρ0cos(φ−φ0)
.
(4.2.13)
One can write for an arbitrarily distributed pressure:
k = 1
2π a
1 π (2 a ) 2
1/2
⌠ ⌠ ⌡ ⌡ 0 0
( a 2 − ρ20)1/2 p (ρ ,φ ) ρ dρ dφ 0 0 0 0 0 a 2 + ρ20 − 2 a ρ0cos(φ−φ0)
,
which corresponds to the well known result (Cherepanov, 1974).
1.
Exercise 4.2 Derive the solution (4.2.1) −(4.2.6) for the case of an isotropic body.
2.
Verify the derivation of (4.2.7) −(4.2.11).
4.3 Concentrated load outside a circular crack Consider a transversely isotropic space weakened by a penny-shaped crack of radius a in the plane z =0. Let a concentrated force P be applied at an arbitrary point (ρ,φ, z ) in the Oz direction. The crack faces are stress-free. Let us find the crack opening displacement and the opening mode stress intensity factor k 1. Consider the second system in equilibrium: two unit concentrated forces Q applied normally to the crack faces in opposite directions at the point (ρ0,φ0,0±). Denote the normal displacement in the space due to the forces Q by w ; while Q w P is the crack opening displacement due to force P . reciprocal theorem to the two systems yields Qw P = Pw Q ,
Application of the
230
CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
which gives the crack opening displacement m2 m1 2 f (z ) , f (z ) + w P(ρ0,φ0) = HP π m2 − 1 2 2 m1 − 1 2 1 with f 2 defined by (4.2.8).
(4.3.1)
The stress intensity factor can be determined by
w P(ρ0,φ0) 1 lim k 1(φ0) = 8π H ρ0→a ( a − ρ )1/2 0 m
m
1 2 = ( ) + f z f 6( z 2) , 1/2 2 6 1 m2 − 1 2(2 a ) π m 1 − 1
P
where f 6( z ) = ( a 2 − l 21)1/2/ r 2a ,
r 2a = ρ2 + a 2 − 2ρ a cos(φ−φ0) + z 2. (4.3.2)
The stress intensity factor vanishes as z tends to zero for ρ≥ a . In the case of an isotropic body, expression (4.3.1) transforms into w P(ρ0,φ0) =
P 1 − ν h tan-1 R 2 R 0 πµ 0
z2 1 z2 -1 h tan . − − − 2 R 20 + h 2 l 22 − l 21 R 0 R 20 R 30 h
ρ2 − l 21
Here µ is the shear modulus, and ν is Poisson’s ratio. expression for the stress intensity factor will take the form: k 1(φ0) =
P ( a 2 − l 21)1/2 (8 a )1/2π2 r a2
(4.3.3)
The corresponding
2 ρ2 − l 21 1 1 + z − . 2 2 2 1 − ν ra 2( l 2 − l 1)
(4.3.4) In the case of axial symmetry ρ=0, and formulae (4.3.3−4.3.4) simplify as follows:
231
4.4 Plane crack under arbitrary shear loading
a 2 − ρ20 1/2 P z2 1 − ν -1 tan w P(ρ0,φ0) = 2 + 2 ρ20 + z2 π µ (ρ20 + z 2)1/2 (ρ0 + z 2)3/2
+
k1 =
, 2(ρ20 + z 2)( a 2 + z 2) z 2( a 2 − ρ20)1/2
1 Pa 1/2 z2 1 + , 1 − ν a 2 + z 2 2π2√2( a 2 + z 2)
which is in agreement with the results considered the axisymmetric case only.
1.
reported
by
Collins
(1962),
who
Exercise 4.3 Verify (4.3.2)
2. Derive a complete solution for the case of an arbitrary point force applied outside a penny shaped crack in a transversely isotropic space.
4.4 Plane crack under arbitrary shear loading Consider a transversely isotropic elastic space weakened by a flat crack S in the plane z =0, with arbitrary shear loading applied to the crack faces antisymmetrically. The problem can be formulated as follows: find the solution to the set of differential equations (2.1.3) for a half-space z ≥0, subject to the mixed boundary conditions on the plane z =0: τz = −τ( x,y), for ( x,y)∈ S ;
u = 0, for ( x,y)∉ S ;
σz =0, for −∞<( x,y)<∞.
(4.4.1)
It is no longer possible to present the solution in the form (4.1.2). complicated representation is necessary, namely, F 1 = c 1(Λχ1 + Λχ1) ,
F 2 = c 2(Λχ2 + Λχ2) ,
A more
F 3 = c 3(Λχ3 − Λχ3).(4.4.2)
Here c 1, c 2 and c 3 are the as yet unknown constants; χ1, χ2, and χ3 are the as yet unknown complex harmonic functions.
A bar indicates the complex conjugate
232
CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
Introducing the notation z = z /γ , for k =1,2,3, we
value throughout this book.
k
k
assume also that χ ( z ) = χ( z ), 1
χ ( z ) = χ( z ),
1
2
χ ( z ) = χ( z ).
2
3
(4.4.3)
3
This assumption will allow us to reduce the problem to finding just one harmonic function which is much easier than searching for three. By substituting (4.4.3) into the third equation (2.1.12), we obtain the first equation for the constants, namely, c + m c = 0. 1
(4.4.4)
2 2
The third condition in (4.4.1) is thus satisfied. yields
Substitution of (4.4.2) in (2.1.6)
u = c (Λ2χ + ∆χ ) + c (Λ2χ + ∆χ ) + i c (Λ2χ − ∆χ ), 1
1
1
2
2
2
3
3
3
where the differential operators Λ and ∆ are defined by (2.1.5). equation (4.4.5) transforms into u = ( c + c + i c )Λ2χ + ( c + c − i c )∆χ. 1
2
3
1
2
3
(4.4.5) When z =0,
(4.4.6)
It is convenient to assume c + c + i c = 0. 1
2
(4.4.7)
3
This assumption simplifies (4.4.6) as follows u = ( c + c − i c )∆χ, 1
2
(4.4.8)
3
and makes it possible to represent χ( M ) = ⌠ ⌠ ln[ R ( M,N ) + z ] u ( N ) d S
⌡⌡
N
.
(4.4.9)
S
The representation (4.4.9) satisfies the second condition (4.4.1) identically, and inside the crack the following equation becomes valid c + c − i c = 1/2π. 1
2
3
(4.4.10)
233
4.4 Plane crack under arbitrary shear loading
The solution of the set of equations (4.4.4), (4.4.7), and (4.4.10) gives 1 , 4π( m 1 − 1)
c1 = −
c2 = −
1 , 4π( m 2 − 1)
c3 =
i . (4.4.11) 4π
Substitution of (4.4.2) and (4.4.11) in the last of expressions (2.1.12) gives the following expression for the tangential stress: A 44 ∂ m 1 + 1 (Λ2χ + ∆χ ) τz = − 1 1 4π ∂ z m − 1 1 m2 + 1
+
m2 − 1
(Λ2χ + ∆χ ) + (Λ2χ − ∆χ ) . 2 2 3 3
(4.4.12) Expression (4.4.12) simplifies for z =0 m + 1
A 44
m + 1
1 1 ∂χ ∂χ 1 2 + Λ2 + τz = − − ∆ . 4π ( m − 1)γ γ 3 ∂z γ 3 ∂ z ( m 2 − 1)γ2 1 1 Finally, satisfaction of the integro-differential equation: τ( N 0 ) = −
1 2π
2
( G 21
−
G 22)
first
condition
(4.4.1)
yields
the
(4.4.13) governing
G ∆⌠ ⌠ u ( N) d S + G Λ2⌠ ⌠ u ( N) d S , 2 1 ⌡ ⌡ R ( N,N0) N ⌡ ⌡ R ( N,N0) N S
S
(4.4.14) where the elastic constants G 1 and G 2 are defined by (2.1.9). Green’s functions in the case of shear loading. The integro-differential equation (4.4.14) was solved exactly for a penny-shaped crack in section 2.7. The closed form solution is (see 2.7.53) 2π a
G1
2
G ⌠ ⌠ 1 tan-1 η − 2 (3 − t ) η τ(ρ ,φ ) ρ dρ dφ u (ρ,φ) = 0 0 0 0 0 π ⌡ ⌡ R R G 21 a 2(1 − t )2 0
2π a
G2
0
2 iφ
0 ⌠ ⌠ q tan-1 η + η [( q/q) − t e ] τ(ρ ,φ ) ρ dρ dφ , + 2 0 0 0 0 0 R π ⌡ ⌡ Rq a (1 − t )(1 − t ) 0
0
234
CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
(4.4.15) where R and η are defined by (4.1.14), q is defined by (4.1.28), a bar indicates the complex conjugate value, and ρρ t =
0
a
2
i(φ-φ0)
e
.
(4.4.16)
The potential functions can be found by substitution of (4.4.15) and (4.4.9) in (4.4.2) and evaluation of the resulting integrals. This looks at first somewhat difficult, nevertheless, it will be shown here that all the Green’s functions can be expressed in elementary functions. Note the following property: (3 − t ) η 1 η tan-1( ) − 2 R R a (1 − t )2
Λ
2 iφ
q η η [( q/q) − t e 0] . = −Λ tan-1 + 2 R Rq a (1 − t )(1 − t )
(4.4.17)
Introduce the following notation: ( a 2 − r 2)1/2( a 2 − ρ20)1/2 1 -1 , tan E ( N, N ) = 1 0 ( ) R ( N,N ) a R N,N 0 0 i (ψ-φ0)
(3 a 2 − r ρ e E ( N, N ) = 2
) ( a 2 − r 2)1/2( a 2 − ρ20)1/2
0
0
iφ
r ei ψ − ρ e E ( N, N ) = 3
0
,
i (ψ-φ0) 2
a( a2 − rρ e
0
0
0
-i φ
R ( N,N )( r e-i ψ − ρ e 0
)
0
a ( a 2 − r 2)1/2( a 2 − ρ20)1/2
0
)
( a 2 − r 2)1/2( a 2 − ρ20)1/2
tan -1
a R ( N,N ) 0
iφ
r ei ψ − ρ e
0
rρ
0 i (ψ+φ0) + e . − ψ φ ψ φ i i ( ) ( ) 2 -i φ 0 a ( a 2 − r ρ e 0 )( a 2 − r ρ e ) r e-i ψ − ρ e 0 0 0 0
Here the points N and N
0
0
(4.4.18) are characterized by the cylindrical coordinates ( r ,ψ,0)
235
4.4 Plane crack under arbitrary shear loading
and (ρ0,φ0,0) respectively.
Note the following relationships of symmetry:
E 1( N , N 0) = E 1( N 0, N ),
E 2( N , N 0) = E 2( N 0, N ),
E 3( N , N 0) = E 3( N 0, N ).
(4.4.19)
Let R ( M,N) denote the distance between the points M (ρ,φ, z ) and N ( r ,ψ,0). using (4.4.17) one may write dS
By
dS
N N ⌠ ⌠ Λ[ E ( N, N ) − E ( N, N )] ⌠ ⌠ ΛE ( N, N ) − . = 1 0 2 0 0 R ( M,N ) R ( M,N) ⌡⌡ ⌡⌡ 3 S
S
(4.4.20) Integration by parts in (4.4.20) leads to an important property:
⌠ ⌠ [ E ( N, N ) − E ( N, N )]Λ 1 d S = −⌠ ⌠ E ( N, N )Λ 1 d S . 0 2 0 0 R ( M,N) N R ( M,N) N ⌡⌡ 1 ⌡⌡ 3 S
S
(4.4.21) Two more properties can be obtained by applying Λ and Λ to both sides of (4.4.21), namely,
⌠ ⌠ [ E ( N, N ) − E ( N, N )]Λ2 1 d S = −⌠ ⌠ E ( N, N )∆ 1 d S , 0 2 0 0 R ( M,N ) N R ( M,N) N ⌡⌡ 1 ⌡⌡ 3 S
S
(4.4.22)
⌠ ⌠ [ E ( N, N ) − E ( N, N )]∆ 1 d S = −⌠ ⌠ E ( N, N )Λ2 1 d S . 0 2 0 0 R ( M,N) N R ( M,N) N ⌡⌡ 1 ⌡⌡ 3 S
S
(4.4.23) The properties (4.4.21−4.4.23) will allow us to substitute the evaluation of various integrals involving E 3, which look very formidable, by evaluation of integrals involving expressions E 1 and E 2, some of which have already been computed
(4.1.27−4.1.31), and the remaining can be evaluated relatively easy (see Appendix A4.4). Introduce the notation X = Λχ + Λχ ,
Y =Λχ − Λχ .
(4.4.24)
In order to obtain the complete solution, we shall need the following expressions
236
CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
for various derivatives of X and Y : the tangential displacements are defined by Λ X and Λ Y ; the normal displacements by ∂ X /∂ z ; the field of stresses may be computed through ∂2 X /∂ z 2, Λ2 X , Λ2 Y , Λ(∂ X /∂ z ), Λ(∂ Y /∂ z ). All the Green’s functions involved can be expressed as various derivatives of two fundamental functions, namely, K 1( M , N 0) = ⌠ ⌠ E 1( N , N 0) ln[ R ( M,N ) + z ] d S N ,
⌡⌡ S
K 2( M , N 0) = ⌠ ⌠ E 2( N , N 0) ln[ R ( M,N ) + z ] d S N .
(4.4.25)
⌡⌡ S
Rewrite formula (4.4.15) as
u( N) =
G1
G 22
π⌡ ⌡
G 21
⌠ ⌠ [ E ( N, N ) − 1 0
E 2( N , N 0)] τ( N 0) d S N
0
S
+
G2
⌠ ⌠ E ( N , N ) τ( N ) d S . 3 0 0 N
π⌡ ⌡
0
S
By substituting (4.4.9) and (4.4.26) in (4.4.24) (4.4.21−4.4.23), we obtain the following results:
X =
G1 − G2 π
G
Y =
π
using
the
(4.4.26) properties
G
Λ⌠ ⌠ K + 2 K τ d S + Λ⌠ ⌠ K + 2 K τ d S , G 1 2 G 1 2 ⌡⌡ 1 ⌡⌡ 1 S
G1 + G2
and
S
G
G
−Λ⌠ ⌠ K − 2 K τ d S + Λ⌠ ⌠ K − 2 K τ d S . G 1 2 G 1 2 ⌡⌡ 1 ⌡⌡ 1 S
S
We shall only need the following derivatives of X and Y for the complete solution: ΛX =
G1 − G2 π
G
G
− ∂2 ⌠ ⌠ K + 2 K τ d S + Λ2⌠ ⌠ K + 2 K τ d S , G 1 2 G 1 2 ∂ z2 ⌡ ⌡ 1 ⌡⌡ 1 S
S
237
4.4 Plane crack under arbitrary shear loading
(4.4.27) ΛY =
G1 + G2
G2
∂2 ⌠ ⌠ K − K 2 τ d S + Λ2⌠ ⌠ K 1 − K 2 τ d S , 2 1 G G ⌡⌡ ∂ z ⌡ ⌡ 1 1
π
S
∂X = ∂z
G2
S
(4.4.28)
G1 − G2
G2
G2
∂ ⌠⌠ K + K τ d S + Λ⌠ ⌠ K + K τ d S , Λ 1 2 ∂z ⌡ ⌡ 1 G 1 2 G ⌡ ⌡ 1
π
S
S
(4.4.29) G + G G G ∂Y 1 2 ∂ 2 2 ⌠ ⌠ ⌠ ⌠ = −Λ K − K τ dS + Λ K − K τ d S , 1 2 G 2 G ∂z π ∂z ⌡ ⌡ 1 ⌡⌡ 1 1 S
∂ ΛX = ∂z
G − G 1
S
(4.4.30) 2
π
G G 2 2 ∂ ∂2 ⌠ ⌠ 2⌠ ⌠ K + K τd S + Λ K + K τd S , − 2 1 2 ∂ z ∂ z ⌡ ⌡ 1 G 2 G ⌡⌡ 1 1 S
S
(4.4.31) ∂ ΛY = ∂z
G + G 1
π
2
G G 2 2 ∂ ∂2 ⌠ ⌠ 2⌠ ⌠ d + K − K τ S Λ K − K τ d S , 2 1 2 1 ∂ z ∂ z ⌡ ⌡ G G 2 ⌡ ⌡ 1 1 S
∂2 X ∂ z2
G − G =
1
S
(4.4.32) 2
π
G G 2 2 ∂2 ⌠ ⌠ ⌠ ⌠ Λ K + K τ d S + Λ K + K τ d S , 2 1 2 1 G G 2 ⌡⌡ ∂z ⌡ ⌡ 1 1 S
S
(4.4.33) G − G ΛX = 2
1
2
π
G
G
2 2 ∂2 Λ− 2⌠ ⌠ K + K τ d S + Λ2⌠ ⌠ K + K τ d S , 1 2 G 2 G ∂z ⌡ ⌡ 1 ⌡⌡ 1 1 S
S
(4.4.34) G + G Λ2 Y =
1
π
2
G
G
2 2 ∂2 Λ 2⌠ ⌠ K − K τ d S + Λ2 ⌠ ⌠ K − K τ d S , 1 2 1 2 G G ⌡⌡ ∂ z ⌡ ⌡ 1 1 S
S
(4.4.35) All the integrals in (4.4.27−4.4.35) are computed in elementary functions in Appendix A4.4 and in (4.1.27−4.1.31). The results above may be applied to solving the problem of a tangential point force loading of a penny-shaped crack. The solution will give us all the
238
CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
Green’s functions, related to the case. Consider an infinite transversely isotropic solid weakened in the plane z =0 by a penny-shaped crack of radius a . Let two equal concentrated forces T = T + iT be applied to the crack faces antisymmetrically x
±
at the point N (ρ ,φ ,0 ). 0
0
0
y
The previously obtained results give the complete
solution in elementary functions: γγH u =
2
G G 2 2 1 f ( z ) T f (z ) T + f (z ) + − f (z ) + G 8 k G 7 k m − 1 2 k 16 k 1 1 k
Σ
1 2
π
k=1
G G 2 2 β + f (z ) − f (z ) T + f (z ) − f ( z ) T , G 7 3 G 8 3 π 2 3 16 3 1 1
m
2
2 w = Hγ γ ℜ 1 2 π
Σ
G
f ( z ) + 2 f ( z ) T , ( m − 1)γ 1 k G 9 k k k 1
k=1
2γ1γ2 σ = ℜ 2 1 π ( γ 1 − γ 2 )
(4.4.36)
k
(4.4.37)
G 1 1 2 T , (−1) ( ) + ( ) f z f z − G 10 k γ23( m k + 1) γ2k 5 k 1 k=1 2
Σ
k+1
(4.4.38) 2
σ = − 2
2 1 f (z ) + f ( z ) T m − 1 5 k G 13 k
G
Σ
2 A Hγ γ π 66 1 2
k=1
k
1
G G 2 2 1 + f (z ) + f ( z ) T − 2 −f ( z ) + f ( z ) T 5 3 G 12 k G 13 3 11 k π γ 1 1 3 G 2 + f (z ) − f ( z ) T , G 12 3 11 3 1
γ 1γ 2 σ = ℜ 2 z π ( γ 1 − γ 2 ) γγ τ = z
2
1 2
2π2(γ − γ ) 1
2
2
Σ k=1
(4.4.40)
G G 2 2 (−1)k f (z ) + f ( z ) T + −f ( z ) + f ( z ) T G 14 k G 15 k γ 3 k 4 k k 1 1
Σ k=1
G 2 (−1) f (z ) + f ( z ) T , G 10 k 5 k 1 k+1
(4.4.39)
239
4.4 Plane crack under arbitrary shear loading
G G 2 2 1 T . + ( ) ( ) + ( ) + f ( z ) f z T f z f z − G 15 3 G 14 3 4 3 2π2 3 3 1 1
(4.4.41)
Here ℜ indicates the real part, the elastic coefficients are defined by (2.1.9), and the functions f with subindex less than 6 are given by formulae (4.2.7−4.2.11), and the remaining functions are computed in Appendix A4.4, namely,
f ( z) = 7
ha 2 3 s s 2
t
−
2
− a t
l 22
2
3( l 22 − a 2)1/2
−
s
s tan-1 2 2 1/2 , ( l 2 − a )
3
(4.4.42) (a − (ζ − 1)1/2 -1 1 1 2 2 1/2 -1 f ( z ) = ( a − ρ0) tan − tan 1/2 1/2 8 (ζ − 1) a (ζ − 1) q q 2
l 21)1/2
2 2 1/2 ei φ ( a − l 1) ρ2 − 1 , 1 + − i(φ-φ ) a ρ l 22 − ρρ e 0 0
iφ
f ( z ) = −ρe 9
−
f ( z) = − 10
1 f ( z) = 11 q
( a 2 − ρ20)1/2 a ( l 22 − a 2)1/2 1 -1 a t sin (l ) + i(φ-φ ) a3 (1 − t )( l 22 − ρρ e 0 ) 2 0
1/2 1 -1 a (1 − t ) , tan ( l 22 − a 2)1/2 t (1 − t )3/2
(4.4.44)
h ρei φ(3 l 22 − a 2 t ) ( l 22 − l 21)( l 22 − a 2 t )2 3 R 40
+
6 R 20 z 2 R 30 q 2
2 iφ
+
(4.4.43)
3ρ20e
s4
0
− z
,
(4.4.45) iφ
4
z 8 h tan-1( ) − ( a 2 − ρ20)1/2 2 − R sq 0 iφ
iφ
s tan-1 − e 2e + 3 ρ ρ ( l 22 − a 2)1/2 q
4ρ e 0
s2 q
0
240
CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
( a 2 − l 21)1/2 3(ζ − 1)1/2 -1 1 -1 tan − 1/2 2 1/2 − tan q (ζ − 1) a (ζ − 1) iφ
2 iφ
+
−
2ρ0e
ha e
ρs2
0
iφ
iφ
ρ0e
0
+ 2 − − ρ s q 2e
s2
2
2 2 2 ( l 2 − a ) t − q l 22 − a 2 t
3 iφ ei φ( l 22 − ρ2) e q z2 q ρ 2 i φ + + 2e , − + h 2 l 22 − l 21 R 20 q ρq
h R 20
(4.4.46)
( a 2 − l 21)1/2 3(ζ − 1)1/2 -1 1 1 2 2 1/2 -1 − tan ( a − ρ0) tan f ( z) = 12 (ζ − 1)1/2 a (ζ − 1)1/2 q q2
−
+
e2 i φ( a 2 − l 21)1/2 −
a ( l 22 ei φ3 ρ q
+
l 21) 2ei φ ρ
l 22 + ρ2
2ρ2( l 22 − a 2)
+ i(φ-φ0) 2 + 1 2 l 22 − ρρ ei(φ-φ0) ( l 2 − ρρ e ) 0 0 ( a 2 − l 21)1/2
−
a
l 22 + 2ρ2 1 ei φ + 2 + ( ) , ρ q ( l 22 − ρρ ei(φ-φ0) ) q 0 (4.4.47)
a 2 iφ f ( z ) = − h 2 ρ e 0 13 0 s
15( l 22
+
f ( z) = 14
5 s 2( l 22 − a 2 t )
s5
+
i(φ-φ0)
ρρ e +
0
s − 15 tan-1 2 ( l 2 − a 2)1/2 s4 φ ρei (3 l 22 − a 2 t )
+ , ( l 22 − l 21)( l 22 − a 2 t )2 ( l 22 − a 2 t )2 2t
( a 2 − ρ20)1/2 a 3(1 − t )
− a)
2 1/2
a ( l 22 − a 2)1/2
2 i(φ-φ0) 2 2 ( l 2 − l 1)( l 2 − ρρ0e )
(2 l 22 + l 21 t − 3ρ2) i(φ-φ0)
l 22 − ρρ e 0
3( l 22 − l 21 t ) 1 − t
1/2 3 -1 a (1 − t ) − tan , ( l 22 − a 2)1/2 (1 − t )3/2
(4.4.48)
241
4.4 Plane crack under arbitrary shear loading
2 2 iφ
ρe
(4.4.49) (a
f ( z) =
2
−
ρ20)1/2( l 22
− a)
2 1/2
(3 l 22
i(φ-φ0)
− ρρ e 0
) ,
i(φ-φ0) 2
l 22( l 22 − l 21)( l 22 − ρρ e
15
0
)
iφ
ρ0e
+ z z 1 h tan-1( ) + ( a 2 − ρ20)1/2 2 f ( z) = 16 R s s q 0 R0q R 20
2
(4.4.50)
0
−
2 -1 s tan 2 2 1/2 (l2 − a ) q
( a 2 − l 21)1/2 iφ iφ (ζ − 1)1/2 -1 1 -1 + e − e ha 2. + tan tan − ρ (ζ − 1)1/2 q a (ζ − 1)1/2 ρs2
(4.4.51) The solution (4.4.36−4.4.51) presents, in fact, the explicit expressions for the Green’s functions, and allows us to write a complete solution for the case of arbitrary tangential loading in quadratures. The general results simplify significantly for z =0, namely, G
2
G 1 tan-1 η − 2 (3 − t ) η T u = R π R G 21 a 2(1 − t )2 1
G
2 iφ
q η [( q/q) − t e 0] -1 η + tan + 2 T, R π Rq a (1 − t )(1 − t ) 2
for ρ< a ,
(4.4.52)
-i φ
G e 0 a2 1 1 2 2 2 1/2 w = H α( a − ρ0) ℜ + T − qs G ρ s3 a 1 0
for ρ≤ a , -i φ
G a 2e 0 1 -1 2 2 s s 2 2 1/2 + tan-1 2 w = H α( a − ρ0) ℜ tan 2 2 1/2 3 qs G ρs π (ρ − a ) (ρ − a 2)1/2 1 0 -i φ
(ρ2 − a 2)1/2 e 0 -1 a sin , T − − ( ) aρ ρ s2 q 0
for
ρ> a
(4.4.53)
242
CHAPTER 4
2 σ = 2 ℜ2π H A γ γ − 1 66 1 2 π
γ + γ 1
γγ
1 2
2
APPLICATIONS IN FRACTURE MECHANICS
η 1 tan-1(η ) + 1 2 R qR a (1 − t )(1 − t )q
G ηρe-i φ(3 − t ) 2 ρe-i φ T , + 2 + G ( a 2 − ρ2) a 2(1 − t )2 a − ρ2 1
σ = 0,
for ρ< a ,
for ρ> a ,
1
(4.4.54)
G 2 1 1 σ = 2 2π A H γ γ + f (0) T f (0) T + 2 66 1 2 5 γ G 12 π 3 1
G 2 1 f (0) T + f (0) T , + 2π A H γ γ − G 13 γ 11 66 1 2 3 1 σ = 0,
(4.4.55)
z
τ = z
( a 2 − ρ20)1/2
G e2 i φ(3ρ − ρ e
i(φ-φ0)
T
+
π2(ρ2 − a 2)1/2 R 2
2
G
1
0
)T
i(φ-φ0) 2
ρ(ρ − ρ e 0
)
,
for ρ> a .
(4.4.56) Here R and η are defined by (4.1.14). The second and the third mode stress intensity factors can be expressed through the decomposition τ(n)=τ + i τ , which is (n) i φ
related to τ by a relationship τ =τ e . z
z
zn
tz
Introducing the complex stress intensity
factor k + ik = lim{(ρ − a )1/2τ e-i φ}, 2
3
ρ→ a
z
one gets from (4.4.56)
k + ik = 2
3
( a 2 − ρ20)1/2 π2(2 a )1/2
T e-i φ ρ20 + a 2 − 2 a ρ0cos(φ−φ0)
(4.4.57)
243
4.5 Penny-shaped crack under uniform pressure
G 2 ei φ(3 a − ρ0e
i(φ-φ0)
+
G1
)T
. )
i(φ-φ0) 2
a ( a − ρ0e
(4.4.58) In the general case of arbitrarily distributed loading, the stress intensity factor takes the form 2π a
k 2 + ik 3 = ⌠ ⌠
⌡ ⌡ 0
( a 2 − ρ20)1/2 π2(2 a )1/2
0
i(φ-φ0)
+
G 2 (3 a − ρ0e G1
) ei
φ
0 0 ρ20 + a 2 − 2 a ρ0cos(φ−φ0)
τ(ρ0,φ0)
i(φ-φ0) 2
a ( a − ρ0e
e-i φτ(ρ ,φ )
)
ρ dρ dφ , 0 0 0
(4.4.59)
which is in agreement with (2.7.30).
1.
Exercise 4.4 Derive the general solution in the case of isotropy.
2.
Find the isotropic equivalent of (4.4.14).
3.
Verify the property (4.4.17).
4.
Derive the solution (4.4.36) −(4.4.51).
5.
Rederive the solution (4.4.36) −(4.4.51) for the case of isotropy.
4.5 Penny-shaped crack under uniform pressure Let a penny-shaped crack of radius a be opened by the pressure p =const. In this case one gets from (4.1.12) ω(ρ,φ) = 4 Hp ( a 2 − ρ2)1/2.
(4.5.1)
The potential function F can be obtained by substitution of (4.5.1) in (4.1.4). The integral can be computed in elementary functions (A4.1.4), giving
244
CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
a F = 2π Hp (2 a 2 + 2 z 2 − ρ2)sin-1( ) − l2
2 a 2 − 3 l 21
( l 22 − a 2)1/2.
a
(4.5.2) The complete solution can be expressed through various derivatives of the potential function, as prescribed by formulae (2.1.6) and (2.1.12). All the derivatives are given in Appendix A4.1. The solution is: a ( l 22k − a 2)1/2 γk -1 a u = 2 Hp ρe m − 1sin (l ) − l 22k k 2k k=1 2
iφ
Σ
,
(4.5.3)
mk ( a 2 − l 2 )1/2 − zsin-1( a ), w = 4 Hp 1k mk − 1 l 2k k=1 2
Σ
γ2k − ( m k + 1)γ23
2
σ1 = 8 Hp A 66
Σ
γk( m k − 1)
k=1
σ = −8 HpA66a e 2
( l 22k − l 21k)
− a) γk m − 1 2 2 l 2k( l 2k − l 21k) k k=1 2
2 iφ
a ( l 22k − a 2)1/2
Σ
l 21k( l 22k
2 1/2
− sin-1(
(4.5.4)
a ) , l 2k
(4.5.5)
,
(4.5.6)
k+1 a ( l 22k − a 2)1/2 -1 a (−1) γ sin , σz = − ( ) k π(γ − γ ) l 2k l 22k − l 21k 1 2 k=1
(4.5.7)
2 pa 2ρei φ τz = π(γ − γ )
(4.5.8)
2
Σ
2p
1
2
k ( a 2 − l 21k)1/2 (−1) 2 2 , l 2k( l 2k − l 21k) k=1 2
Σ
Here the notation was introduced l 1k =
1 {[( a + ρ)2 + z 2k]1/2 − [( a − ρ)2 + z 2k]1/2}, 2
l 2k =
1 {[( a + ρ)2 + z 2k]1/2 + [( a − ρ)2 + z 2k]1/2}, 2
z k = z /γ k ,
for k =1,2.
(4.5.9)
245
4.5 Penny-shaped crack under uniform pressure
The problem was first solved by Elliott (1949) by the integral transform method. Our results are essentially in agreement with those of Elliott, who expressed them in the form of integrals involving Bessel functions, namely, ∞
Cm n
ρ z x n-2cos( x ) J m( x ) exp(− x ) d x , a a
= ⌠
⌡ 0
∞
Sm n
z ρ x n-2sin( x ) J m( x ) exp(− x ) d x . a a
= ⌠
⌡ 0
These integrals can be computed in elementary functions, and the results in our notation are S 01
C 03
S 03
C 11
a = sin ( ), l2 -1
=
=
=
S 21 =
C 12 =
C 13
=
S 02
=
a ( a 2 − l 21)1/2 l 22 − l 21
C 02
,
a ( a 2 − l 21)1/2[ l 42 + a 2(ρ2 − 2 a 2 − 2 z 2)] ( l 22 − l 21)3 a ( l 22 − a 2)1/2[ a 2(2 a 2 + 2 z 2 − ρ2) − l 41] ( l 22 − l 21)3 z [ a − ( a 2 − l 21)1/2] ρ( a 2 − l 21)1/2
,
z [ a − ( a 2 − l 21)1/2]2 ρ (a 2
2
−
l 21)1/2
a − ( a 2 − l 21)1/2 ρ
S 11
−
=
,
,
ρ a l 1( l 22 − a 2)1/2
a l 1( a 2 − l 21)1/2 l 2( l 22 − l 21)
a 2ρ( l 22 − a 2)1/2( l 22 + 3 l 21 − 4 a 2) ( l 22 − l 21)3
l 22 − l 21
a − ( a 2 − l 21)1/2
S 12 =
,
=
a ( l 22 − a 2)1/2
,
l 2( l 22 − l 21)
,
,
,
,
246
CHAPTER 4
S 13
C 22
S 22
=
=
=
APPLICATIONS IN FRACTURE MECHANICS
a 2ρ( a 2 − l 21)1/2(3 l 22 + l 21 − 4 a 2) ( l 22 − l 21)3 2 a [( l 22 − a 2)1/2 − z ] ρ2 2 a [ a − ( a 2 − l 21)1/2)] ρ2
,
a ( l 22 − a 2)1/2
−
l 22 − l 21
,
a ( a 2 − l 21)1/2
−
l 22 − l 21
,
a l ( a 2 − l 21)1/2 a − ( a 2 − l 21)1/2 a 2 1 C 23 = − ρ ρ l ( l 22 − l 21) 2
−
a ( a 2 − l 21)1/2[ l 42 + a 2(ρ2 − 2 a 2 − 2 z 2)] ( l 22 − l 21)3
.
There are some misprints (or errors) in Elliott’s paper. For example, according to his formula (4⋅2⋅5), the tangential displacement u vanishes on the plane z =0 which cannot be correct; there are some missing terms and obvious misprints in formula (4⋅2⋅6). In the limiting case of γ →γ →γ →1, m → m →1, H =(1−ν)/2πµ, A = A =µ, 1
2
3
1
2
44
formulae (4.5.3−4.5.8) give the complete solution for an isotropic body. using the L’Hoˆ pital rule, one obtains
66
By
a ( l 22 − a 2)1/2 2 a 2| z |( a 2 − l 21)1/2 p ρei φ -1 a u = (1−2ν) − sin ( ) + , 2 2 2 2πµ l l 22 l ( l − l ) 2 2 1 2
(4.5.10) w =
2p z a 2(1 − ν) ( a 2 − l 21)1/2 − z sin-1( ) l πµ | z| 2
a + z sin-1( ) − l 2
a ( l 22 − a 2)1/2 l 22 − l 21
, (4.5.11)
247
4.5 Penny-shaped crack under uniform pressure
a ( l 22 − a 2)1/2 2p a (1 + 2ν) 2 − sin-1( ) σ1 = 2 l π l2 − l1 2
+
az 2[ l 41 + a 2(2 a 2 + 2 z 2 − 3ρ2)]
a l 21e2 i φ( l 22 − a 2)1/2
σ2 =
2p π
σz =
2p π
,
( l 22 − l 21)3( l 22 − a 2)1/2
1−2ν +
l 22( l 22 − l 21)
(4.5.12)
z 2[ a 2(6 l 22 − 2 l 21 + ρ2) − 5 l 42] ( l 22 − l 21)2( l 22 − a 2)
, (4.5.13)
a ( l 22
− a)
2 1/2
l 22 − l 21
a − sin-1( ) − l
az
2
[ l 41
2
+ a (2 a + 2 z − 3ρ )] 2
2
2
( l 22 − l 21)3( l 22 − a 2)1/2
, (4.5.14)
iφ
2p τz = − π
2
z l e ( l 22 − a 2)1/2[ a 2(4 l 22 − 5ρ2) + l 41] 1 l ( l 22 − l 21)3 2
.
(4.5.15)
This problem was first solved by Sneddon (1951), using the integral transform method. He was seemingly unable to compute the potential function (4.5.2), so he resorted to differentiation under the integral sign, with a subsequent computation of various integrals involving Bessel functions. His final results are given as elementary functions of the four parameters r = (1 + ( z/a )2)1/2, θ = tan-1( a/z ),
R 2 = [(ρ/ a )2 + ( z/a )2 − 1]2 + 4( z/a )2, 2 az φ = tan-1 2 2 2 . ρ + z − a
This choice of parameters is not the best possible. Here is one illustration. The expression for S 01 in Sneddon’s notation takes the form (Sneddon, 1951, p.497) r sinθ + √ R sin(φ/2) , S 01 = tan-1 rcosθ + √R cos(φ/2) with the limitation ρ≠0, and no indication of what the result would be if ρ=0. In our notation the corresponding result is sin-1( a / l ), with no limitations attached. 2
248
CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
Introduction of Sneddon’s parameters r and θ seems to be unnecessary. There exist relationships between his parameters R and φ, and our l and l , namely, 1 2 R 2 = ( l 22 − l 21)/ a 2,
sin(φ/2) = ( a 2 − l 21)1/2/( l 22 − l 21)1/2.
These relationships may be used to compare the solutions, which are in good agreement, except for some misprints: factor ζ is missing in Sneddon’s formula (139, p. 496), and the last term in his formula (145, p. 499) should read − S 01, rather than + S 00.
1.
Exercise 4.5 Establish the result (4.5.3) −(4.5.8).
2.
Derive (4.5.10) −(4.5.15).
3. Derive the expressions for the polar components of stresses and displacements equivalent to (4.5.10) −(4.5.15). 4. Find the complete solution in the case where the penny-shaped crack is loaded by a normal stress, whose magnitude is proportional to the x -coordinate.
4.6 Penny-shaped crack under uniform shear loading Consider a circular crack of radius a in a transversely isotropic elastic space, subjected to a uniform shear loading τ, where τ is a complex constant. The solution of the integro-differential equation (4.4.14) in this case is
u =
2( G 21 − G 22) G
τ( a 2 − ρ2)1/2,
for z =0 and ρ≤ a .
(4.6.1)
1
Substitution of (4.6.1) in (4.4.9) leads to the integral 2( G 21 − G 22) G
1
2π a
τ ⌠ ⌠ ( a 2 − ρ20)1/2 ln( R
⌡ ⌡
0
0
+ z ) ρ dρ dφ , 0
0
0
0
which has been computed in Appendix A4.1, with all the necessary derivatives. The complete solution will take the form:
249
4.6 Penny-shaped crack under uniform shear loading
u =
2( G 21 − G 22) G
1 m − 1f 17( z1)τ + f 18( z1)τ 1
1
+
1 f ( z ) τ + f ( z ) τ + f ( z ) τ − f ( z ) τ , 18 2 17 3 18 3 m − 1 17 2 2
−
G 22
(4.6.2) w =
G 21
2G
2
(τρei φ + τρe-i φ)
1
m
Σ k=1
a ( l 22k
sin-1( a ) − m − 1 l k 2k k
− a)
2 1/2
l 22k
1, γk (4.6.3)
σ1 =
−
2( G 21
G
G 22) A 66
2
iφ
a ( τe
-i φ
+ τe )
1
γ23( m k
Σ k=1
γ2k
+ 1) −
γ2k( m k − 1)
l (a
2
1k
−
l 21k)1/2
l ( l 22k − l 21k) 2k
, (4.6.4)
σ2 =
−
2( G 21
G
τz =
1 m − 1f 19( z1)τ + f 20( z1)τ 1
1
1 f ( z ) τ + f ( z ) τ + f ( z ) τ − f ( z ) τ , 20 2 19 3 20 3 m − 1 19 2 2
+
σz =
G 22) A 66
2γ γ β(τe-i φ + τei φ) 1 2 π G 1(γ 1 − γ 2) 2γ γ 1 2 πG1
(4.6.5)
β γ − γ 1 2
2
Σ(−1) k=1
2
k
a l ( a 2 − l 21k)1/2 1k l ( l 22k − l 21k) 2k
,
(4.6.6)
(−1)k [ f ( z ) τ + f ( z ) τ] + H [ f ( z ) τ − f ( z ) τ] . 22 k 21 3 22 3 γk 21 k
Σ k=1
(4.6.7) Here 2 iφ
f ( z) = e
2 a 3 − ( l 21 + 2 a 2)( a 2 − l 21)1/2 3ρ2
17
a f ( z ) = z sin-1( ) − ( a 2 − l 21)1/2, 18 l 2
,
(4.6.8)
(4.6.9)
250
CHAPTER 4
a l 1( a 2 − l 21)1/2
3 iφ
f 19( z ) = e
l 2( l 22
−
l 21)
a l 1( a 2 − l 21)1/2
iφ
f 20( z ) = e
l 2( l 22 − l 21)
a f 21( z ) = −sin ( ) + l2 -1
2 iφ
e f 22( z ) =
APPLICATIONS IN FRACTURE MECHANICS
iφ
− 4e f 17( z ),
,
(4.6.11)
a ( l 22 − a 2)1/2 l 22 − l 21
a l 21( l 22 − a 2)1/2 l 22( l 22 − l 21)
(4.6.10)
,
(4.6.12)
,
(4.6.13)
A complete solution to this problem for the case of isotropy can be found in (Westmann, 1965). In the case of isotropy, formulae (4.6.2−4.6.7) transform into
u =
a (−5 + 4ν) z sin-1( ) + 4(1 − ν)( a 2 − l 21)1/2τ πµ(2 − ν) l2 1
+
za ( l 22 − a 2)1/2 l 22 − l 21
l 21
τ + τe2 iφ , l 22
(τe + τe )ρ1−2ν -1 a w = sin ( ) − πµ(2 − ν) 2 l2 iφ
(4.6.14)
a ( l 22 − a 2)1/2
-i φ
l 22
+
za 2( a 2 − l 21)1/2 l 22( l 22 − l 21)
, (4.6.15)
a l 1( a − iφ -i φ 2(τe + τe ) σ1 = −2(1 + ν) π(2 − ν) l 2( l 22 − l 21) 2
l 21)1/2
+
z l 1( l 22 − a 2)1/2[ a 2(4 l 22 − 5ρ2) + l 41] l 2( l 22 − l 21)3
,
(4.6.16)
251
4.6 Penny-shaped crack under uniform shear loading
a l 1( a 2 − l 21)1/2 2e 4(1 − ν) σ2 = − τ π(2 − ν) l 2( l 22 − l 21) iφ
+
z l 1( l 22 − a 2)1/2 4 a 2 l 2( l 22 − l 21) l 22
2 iφ
τe
−
a 2(4 l 22 − 5ρ2) + l 41 ( l 22 − l 21)2
(τ + τe2 ), iφ
(4.6.17) iφ
-i φ
2(τe + τe ) σz = − π(2 − ν)
z l 1( l 22
− a ) [a 2 1/2
2
− 5ρ ) +
(4 l 22
2
l 41] ,
l 2( l 22 − l 21)3
(4.6.18)
a ( l 22 − a 2)1/2 2 a τz = (2 − ν) − sin-1( ) 2 2 π(2 − ν) l l2 − l1 2
+
+
z ( a 2 − l 21)1/2[ l 41 + a 2(2 a 2 + 2 z 2 − 3ρ2)] ( l 22
−
l 21)3
τ + ν a ( l 2 − a 2)1/2 2
z ( a 2 − l 21)1/2[ a 2(6 l 22 − 2 l 21 + ρ2) − 5 l 42] ( l 22 − l 21)2
2 iφ
l 2e 1 τ . l 22( l 22 − l 21)
(4.6.19)
Exercise 4.6 1.
Investigate a penny-shaped crack of radius a in a transversely isotropic iφ
elastic body, subjected to the shear loading τ= c ρe , where c =const. Answer: the main potential function will be proportional to the integral 2π a
iφ
I = ⌠ ⌠ ln( R 0 + z )( a 2 − ρ20)1/2e
⌡ ⌡ 0
0
ρ20 dρ0dφ0
0
=
4 19 2 π iφ ρe ( a 2 − l 21)1/2 a 2 + 7ρ2 − l − 4 l 22 8 3 1 3
252
CHAPTER 4
+
2(8 a 4 + 4 a 2 l 21 + 3 l 41) 15ρ2
APPLICATIONS IN FRACTURE MECHANICS
5 + z(4 a 2 − 3ρ2 + 4 z2)sin-1( a ) − 16 a . l2 15ρ2
It might be a good exercise for the reader to perform the differentiation and write a complete solution. As an example, here are the first two derivatives l 21 l 21 a ∂I π = ρei φ a ( l 22 − a 2)1/215 2 − 12 − 2 2 + sin-1( )4 a 2 − 3ρ2 + 12 z 2, 8 l ∂z a l 2 2 2 ∂2 I i φ 2 2 1/2a + 3 zsin-1( a ). = e ( ) a l π ρ − 2 1 2 − 3 l l2 ∂z 2
Note : ∂ I /∂ z is proportional to the main potential function for the case of linear normal loading of a penny-shaped crack. 2. Find the expressions for the energy release rate by using a procedure similar to the one employed by Kassir and Sih (1968). Answer: G = 2π2 H k 21, G2 = π2( G 1 − G 2) k 22, G3 = π2( G 1 + G 2) k 23. 1 3. Find the Green’s functions for a semi-infinite plane crack in a transversely isotropic space, subjected to a shear loading. Hint : consider the limiting case of (4.4.36−4.4.41), when the radius a →∞, and the coordinate origin moves from the circle centre to its boundary.
4.7 Asymptotic behavior of stresses and displacements near the crack rim Kassir and Sih (1975) have derived expressions relating the stress intensity factors to the field of stresses and displacements in the immediate vicinity of the crack edge, by using their very complicated solution for an elliptical crack. Their derivation looks so complicated that nobody so far was capable to repeat it and verify its accuracy. There is a notion in fracture mechanics that the asymptotic behavior is defined completely by three stress intensity factors, and is invariant for any crack, with a smooth boundary. If this is so, then we can obtain the same results from the much simpler solution for a penny-shaped crack. The results, presented here, are simpler than those of Kassir and Sih, and obtained in a simple manner.
253
4.7 Asymptotic behavior of stresses and displacements
Asymptotic behavior for mode I loading. Though formulae (4.5.3−4.5.8) are valid for a penny-shaped crack subjected to a uniform pressure only, they can be used to obtain some general results which are valid for an arbitrary crack with a smooth boundary subjected to arbitrary loading, in other words, we can explore the asymptotic behavior of displacements and stresses near the rim of a general crack. Introduce the local system of spherical coordinates ( r ,θ,φ), with the coordinate origin at the crack rim. The following asymptotics are valid for the main parameters used in (4.5.3−4.5.8): ρ = a + r cosθ, z = r sinθ,
l 1k ≈ a −
1 rS 2k, 2
l 22k − l 21k ≈ 2 arQk,
l 2k ≈ a +
1 rT 2k, 2
( a 2 − l 21k)1/2 ≈ ( ar )1/2 S , k
a r sin-1( ) ≈ −( )1/2 T . k l2 a
( l 22k − a 2)1/2 ≈ ( ar )1/2 T , k
(4.7.1)
Here the notation was introduced: Q k = [cos2θ + (1/γ2k)sin2θ]1/2 T k = [ Q k + cosθ]1/2,
S k = [ Q k − cosθ]1/2,
for k = 1,2,3.
(4.7.2)
Introducing the opening mode stress intensity factor k1 =
p √2 a , π
(4.7.3)
the following asymptotic expressions can be derived by substitution of (4.7.1) and (4.7.3) in (4.5.3−4.5.8): γ1 T 1 γ2 T 2 + 0(1), + u = u n = −2π Hk 1√2 r 1 1 m − m − 1 2 m S
m S
1 1 2 2 + 0( r), + w = 2π Hk 1√2 r 1 1 m − m − 1 2
2
2 σ1 = 2π A 66 Hk 1( )1/2 r
γ2k − ( m k + 1)γ23 T k
Σ k=1
(4.7.4)
γk( m k − 1)
Qk
+ 0(1),
(4.7.5)
(4.7.6)
254
CHAPTER 4
γT
2
Σ(m
2 σ = −2π A Hk ( )1/2 2 66 1 r
k=1
k
σ = z
k k
k
γT
− 1) Q
+ 0(√ r ),
(4.7.7)
k
γT
1 1 − 2 2 + 0(1), Q √2 r (γ − γ ) Q 1 2 1 2 1
k
τ = τ z
APPLICATIONS IN FRACTURE MECHANICS
S
(4.7.8)
S
1 − 2 + 0(√r), = − Q √2 r (γ − γ ) Q 1 2 1 2
zn
1
(4.7.9)
These results were computed for φ=0. This assumption allows us to avoid a cumbersome axis transformation, without loss of generality. The parameter σ in this case is interpreted as the sum σ
+ σ , and σ
n
t
2
= σ
n
1
− σ + 2iτ . t
By
nt
taking the sum and the difference of (4.7.6) and (4.7.7), one can get k
T
T
− 1 + 2 , σ = n γQ √2 r (γ − γ ) γ1 Q 1 2 2 1 2 1
[2γ21 − ( m + 1)γ23] T
2 σ = π A Hk ( )1/2 t 66 1 r
1
γ ( m − 1) Q 1
1
1
[2γ22 − ( m + 1)γ23] T +
2
γ ( m − 1) Q 2
1
2
2
2
.
Formulae (4.7.4−4.7.9) are essentially in agreement with the results of Kassir and Sih (1975), except for some misprints, for example, one should read √ n and 1
√ n 2 instead of n 1 and n 2 in the denominator of the terms in curly brackets of their formula (8.94c). In order to compare our results with those of Kassir and Sih, one should keep in mind that their definition of the stress intensity factor is √2times greater than ours, their notation n corresponds to our γ2k; Kassir and Sih k
seem to have not noticed the properties (4.1.10) and the relationship S =(sinθ)/(γ T ), which in some cases can be used to simplify their results k
k k
significantly.
For example, they have an expression [ A m 13
k
− A γ2k]/[ A ( m 11
1)] in formula (8.95a), without realizing that it is equal to −1 for k =1,2.
44
k
+
The asymptotic behavior of the displacements and stresses near the crack edge in an isotropic body can be found from either (4.7.4−4.7.9) or (4.5.10−4.5.15). The result is
255
4.7 Asymptotic behavior of stresses and displacements
u = un =
w =
σ = 1
σ2 =
σz =
k 1√ r µ
k 1√ r 2µ
θ θ cos( ) 2(1 − ν) − cos2( ) + 0(1), 2 2
θ θ sin 2(1 − ν) − cos2( ) + 0( r ), 2 2
k1
θ θ 3θ cos 1 + 2ν − sin sin + 0(1), 2 2 2 √r k1
3θ θ θ cos 1 − 2ν − sin sin + 0(√ r ), 2 2 2 √r k1
3θ θ θ cos 1 + sin sin + 0(1), 2 2 2 √r
τz = τzn =
k1
3θ sinθ cos + 0(√ r ), 2 2√ r
(4.7.10)
(4.7.11)
(4.7.12)
(4.7.13)
(4.7.14)
(4.7.15)
which is in agreement with the results given in (Sih and Liebowitz, 1968). Asymptotic behavior for mode II and III loading. We can derive again some results of general nature, namely, the asymptotic behavior of the field of stresses and displacements in the neighbourhood of the edge of a flat crack with a smooth boundary. We recall that at φ=0 the decompositions u = u x+ iu y and τz=τzx+ i τyz are equal to u (n)= u n+ iu t and τ(n)=τzn+ i τtz respectively; σ1 is understood as σn+σt, and σ2=σn−σt+2 i τnt.
This will allow us to avoid a cumbersome axis
transformation. The complex stress intensity factor, introduced in (4.4.57), can be expressed through the prescribed shear loading τ as G2 √2 a k = k 2 + ik 3 = τ + τ, π G1 and its inversion gives
(4.7.16)
256
CHAPTER 4
τ =
π G 1( k G 1 − k G 2) √2 a ( G 21 − G 22)
APPLICATIONS IN FRACTURE MECHANICS
.
(4.7.17)
Substitution of (4.7.1) and (4.7.17) in (4.6.2−4.6.13) yields k 2γ1γ2√2 r
S
S
1 2 − + + u n + iu t = A 44(γ1 − γ2) m 1 + 1 m 2 +1
i k 3γ3√2 r A 44
S 3, (4.7.18)
k 2γ1γ2√2 r
m T
m T
1 1 2 2 − , + w = A 44(γ1 − γ2) γ1( m 1 + 1) γ2( m 2 + 1)
2
σ1 = 2π k 2γ1γ2 H A 66
2 ( )1/2 r
(4.7.19)
γ23( m k + 1) − γ2k S k
Σ
γ2k( m k − 1)
k=1
Qk
,
(4.7.20)
S1 S2 ik 3√2S 3 2 1/2 + , σ2 = 2π k 2γ1γ2 H A 66( ) − ( m − 1) Q r ( m 1 − 1) Q 1 γ 3√ r Q 3 2 2 (4.7.21)
k 2γ 1γ 2
S
S
1 − 2 , σz = − Q 2 √2 r (γ1 − γ2) Q 1 τzn
k 2γ 1γ 2
T
(4.7.22)
T
ik T
1 − 2 + 3 3. + iτ = − tz γ 2 Q 2 √2 r (γ1 − γ2) γ1 Q 1 √2 r Q 3
(4.7.23)
By taking the sum and the difference of (4.7.20) and (4.7.21), one gets k 2γ 1γ 2
S
S
1 − 2 , σn = √2 r (γ1 − γ2) γ21 Q 1 γ22 Q 2 k 2γ 1γ 2
S
(4.7.24)
S
1 1 1 1 1 2 σt = (γ − γ ) 2 − 2Q − 2 − 2Q √2 r γ1 γ3 1 γ2 γ3 2 2 1 1
257
4.7 Asymptotic behavior of stresses and displacements
S1 S2 , + 2π H A + 66Q Q 1 2 k3S3
τnt = −
√2 r γ3 Q 3
(4.7.25)
.
(4.7.26)
Our formulae (4.7.18−4.7.26) are in relatively good agreement with similar results of Kassir and Sih (1975), except for formula (8.96b, p.371) for u which should t
correspond to the imaginary part of our (4.7.18). Formula by Kassir and Sih (8.96b) seems to be in error because it implies that u depends on k , γ , and t 2 1
γ2, which is wrong: our result relates u t to k 3 and γ3 only, as it should be.
There are several misprints in their formulae (8.96a) and (8.96c). The remaining formulae are in agreement, though the formulae by Kassir and Sih (1975) look more complicated than ours, mainly because they did not notice the properties (4.1.10) which could make some expressions much simpler.
1.
Exercise 4.7 Establish (4.7.1).
2.
Derive (4.7.4) −(4.7.9).
3.
Derive (4.7.18) −(4.7.23).
4. Find the asymptotic behavior of the stresses and displacements near the crack rim for the case of mode II and III loading in isotropic bodies.
Answer: u n + iu t =
w = k
2
k
σ = 1
√r θ θ sin 2(1 − ν) + cos2( ) k + 2 i k , 3 2 2 2 µ
√r θ θ cos( ) −(1 − 2ν) + sin2( ), 2 2 µ
3θ θ θ sin( )−2(1 + ν) − cos( )cos( ), 2 2 2 √r 2
258
CHAPTER 4
σ2 =
σz =
APPLICATIONS IN FRACTURE MECHANICS
sin(θ/2) 3θ θ −2(1 − ν) − cos(2)cos( 2 )k 2 − 2 ik 3, √r
k2 2√ r
sinθ cos(
τzn + i τtz =
3θ ), 2
cos(θ/2) 3θ θ + 1 sin sin − ( ) ( ) k ik . 3 2 2 2 √r
Note : compare with the results presented in Sih and Liebowitz (1968). k S 3 3 . τ = − nt √2 r γ Q 3
(4.7.26)
3
Our formulae (4.7.18−4.7.26) are in relatively good agreement with similar results of Kassir and Sih (1975), except for formula (8.96b, p.371) for u which should t
correspond to the imaginary part of our (4.7.18). Formula by Kassir and Sih (8.96b) seems to be in error because it implies that u depends on k , γ , and γ , which is wrong: our result relates u to k 2
t
t
3
and γ
3
2
1
only, as it should be.
There are several misprints in their formulae (8.96a) and (8.96c). The remaining formulae are in agreement, though the formulae by Kassir and Sih (1975) look more complicated than ours, mainly because they did not notice the properties (4.1.10) which could make some expressions much simpler.
1.
Exercise 4.7 Establish (4.7.1).
2.
Derive (4.7.4) −(4.7.9).
3.
Derive (4.7.18) −(4.7.23).
4. Find the asymptotic behavior of the stresses and displacements near the crack rim for the case of mode II and III loading in isotropic bodies.
Answer: u + iu = n
t
√r θ θ sin 2(1 − ν) + cos2( ) k + 2 i k , 3 2 2 2 µ
259
4.8 Flat crack of general shape
w = k2
θ θ √r cos( ) −(1 − 2ν) + sin2( ), µ 2 2
k2
σ1 =
θ θ 3θ sin( )−2(1 + ν) − cos( )cos( ), 2 2 2 √r
σ2 =
sin(θ/2) 3θ θ −2(1 − ν) − cos(2)cos( 2 )k 2 − 2 ik 3, √r k
σ = z
τ
zn
2
2√ r
sinθ cos(
+ iτ = tz
3θ ), 2
cos(θ/2) 3θ θ 1 − sin(2) sin( 2 )k 2 + ik 3. √r
Note : compare with the results presented in Sih and Liebowitz (1968).
4.8 Flat crack of general shape The general method is applied here to the analysis of an elastic space weakened by a flat crack of arbitrary shape under the action of a uniform normal pressure. A simple yet accurate relationship is established between the crack face displacements and the applied pressure for an arbitrary flat crack. Specific formulae are derived for a crack in the shape of a polygon, a rectangle, a rhombus, a cross, a circular sector and a circular segment. All the formulae are checked against the solutions known in the literature, and their accuracy is confirmed. A similar approach can be used for the analysis of a crack under a general polynomial loading. The material in this section follows the paper (Fabrikant, 1987b). Theory. Consider an elastic space weakened in the plane z =0 by a flat crack occupying the domain S whose boundary is given in polar coordinates as ρ = a (φ).
(4.8.1)
Let a uniform pressure p be applied normally to the crack faces in opposite directions. The governing integral equation in this case is given by (4.1.9). The approach is based on the integral representation of the reciprocal of the
260
CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
distance between two points established in (1.1.27). Substitution of (1.1.27) into (4.1.9) gives, after interchanging the order of integration
ρ
σ(ρ,φ) = −
2π
dx
a (φ0)
1 ⌠ ⌠ dφ ⌠ 3 ∆ 2 2 1/2 0 ⌡ 2π H ⌡ (ρ − x ) ⌡ 0
0
x
λ(
x2 , φ−φ ) 0 ρρ0
(ρ20 − x 2)1/2
w (ρ0,φ0)ρ0dρ0.
(4.8.2) It is noteworthy that the change of the order of integration which led to (4.8.2) is valid inside the circle ρ≤min{ a (φ)} only, and this is one of the reasons why the accuracy generally deteriorates for domains with the aspect ratio very far away from unity. Nevertheless, one can obtain from (4.8.2) the exact solution for an ellipse and sufficiently accurate formulae for various crack shapes as will be demonstrated further on. Let the normal displacements of the crack face be w =
1/2 δ 2 a (φ) − ρ2 , a (φ)
(4.8.3)
where δ is a constant to be defined. Now substituting (4.8.3) in (4.8.2), we can verify how close to a constant will be the traction σ producing the displacements (4.8.3). Integration with respect to ρ gives 0
σ(ρ,φ) = −
δ ∆ 8π2 H
∞
Σ n=-∞
ρ
2 πa 2 ( φ ) 0
xdx
⌠ ⌠ (x )|n| 2 2 1/2 ⌡ ρ (ρ − x ) ⌡
0
|n| 1 x 2 i(φ-φ0) e dφ . × F 2− , ; 2; 1− 2 0 2 2 a (φ0)
0
− x2
a 2(φ0)
(4.8.4)
Here F stands for the Gauss hypergeometric function. Further evaluation of the normal traction can be done separately for each value of n . The zero th term has the form σ0 =
δB , 8π H
(4.8.5)
where the notation 2π
dφ B = ⌠ ⌡ a (φ) 0
(4.8.6)
261
4.8 Flat crack of general shape
was introduced. It is clear that the value of the integral in (4.8.6) will depend not only on the domain contour but also on the location of the coordinate origin. The following argument might be useful for establishing certain rules in this regard. According to (4.8.3), the coordinate origin location corresponds to the point where the crack face displacement attains its maximum. We shall call this point the crack centre. In the case of a crack domain with one axis of symmetry, we may conclude from physical considerations that this point should be located at the axis. When this domain possesses two axes of symmetry the location of the crack center is at their intersection, i.e. at the center of gravity of the domain. It is noteworthy that the integral (4.8.6) attains its minimum in this case. One can extend this rule to a general crack, namely, the crack center should be identified with the point inside S where the integral (4.8.6) reaches its minimum. Direct computations for various domains indicate that this minimum is, in general, sufficiently flat, so that in many cases one may locate the crack center at the center of gravity, without any significant loss in accuracy. We shall discuss this in more detail further on when considering the domain S in the shape of a circular segment and sector. It is important to note that the second harmonic is equal to zero for an arbitrary contour, and that all the odd harmonics will be zero if the expression for a (φ) does not contain odd harmonics. Here is the expression for the fourth harmonic 2πcos4(φ
4δ
ρ⌠ σ4 = − 5π2 H ⌡ 0
− φ0)dφ0
a 2(φ )
(4.8.7)
0
The investigation of the fourth and further harmonics shows that their amplitude decreases for general domains, and they vanish in the case of an ellipse. If we assume that p ≈σ then the remaining harmonics may be called the solution error. 0
This implies the establishment of the following relationship between the applied traction p and the maximum displacement of the crack face p =
δB 8π H
(4.8.8)
One can verify that in the case of an ellipse, the solution given by (4.8.3) and (4.8.8) is exact . We expect it to be reasonably accurate for a crack of general shape. This assumption will be justified in the next Section where several particular crack configurations are considered. We also expect (4.8.3) to be sufficiently accurate in the neighbourhood of the crack center, though the relative error might be quite significant close to the boundary.
262
CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
The crack energy can be defined as W = ⌠ ⌠ σ w dS.
(4.8.9)
⌡⌡ S
Feeding (4.8.3) and (4.8.8) in (4.8.9) yields 16π Hp 2 A , 3B
W =
(4.8.10)
where A is the crack area. δav =
Introduce the average displacement δ
av
as
1⌠ ⌠ w dS. A⌡ ⌡ S
Substitution of (4.8.3) in the last expression gives δ
av
=
2 δ. 3
Define the dimensionless parameter τ in the form δ τ =
av
2π Hp √ A
=
W . 2π Hp 2 A 3/2
(4.8.11)
The physical meaning of τ can be defined either as a ratio of the average displacement to a certain fixed displacement, or as a ratio of the crack energy to a certain fixed one. Both forms (4.8.11) lead to the following expression for τ: τ =
8 . 3√ A B
(4.8.12)
One can deduce that the value of τ does not depend on the size of the domain S , and is determined by its shape only. It attains its maximum in the case of a circle, so that 0≤τ≤4/(3π3/2)=0.2394. Tabulation of the coefficient τ for various crack shapes might prove very useful since its knowledge allows us to find the maximum (or average) crack face displacement and the crack energy by using (4.8.11). It might seem more logical to define τ as the ratio of the given crack energy to the energy of a circular crack having the same area. In this case the value of τ would vary between zero and unity. The main reason for the definition (4.8.12) was the desire to preserve the bridge between the crack
263
4.8 Flat crack of general shape
problems in mechanics and the mathematically equivalent problems in electrostatics. The value of τ defined by (4.8.11) corresponds exactly to the coefficient of electrical polarizability in the theory of wave propagation through small apertures (Bethe 1944). We did not find in the mechanics literature any report containing numerical data for nonelliptic cracks which could be compared with the theory of this paper. The situation is slightly better in electrical sciences. Cohn (1952) has measured the coefficient of electrical polarizability of several aperture configurations experimentally. Numerical solution to the same problem was given by De Meulenaere and Van Bladel (1977), and by Okon and Harrington (1981). These numerical and experimental data will be used to estimate the accuracy of the proposed theory. An empirical formula for the coefficient of electrical polarizability was proposed by Fikhmanas and Fridberg (1973). This formula in our notation reads τ =
8√ A , 3π L
(4.8.13)
where L stands for the perimeter of the domain S . Formula (4.8.13) is also exact for an ellipse. It is of interest to compare its performance with our (4.8.12). Several crack shapes are considered for this purpose. A high degree of accuracy of formula (4.8.12) is confirmed by comparison with available numerical solutions. Example 1: Polygon. Consider a flat crack in the shape of a polygon with n sides, with the only limitation that the function a (φ) describing its boundary be continuous and single-valued. The origin of the coordinate system is located at the crack center as it was defined earlier. Let us number the polygon sides in a counter-clockwise direction from 1 to n , a being the length k
of the k th side.
The apex, at which the sides a and a intersect, is k k+1
numbered k +1. It is clear that the value of the index n +1 is to be understood as 1. Denote the distance from the crack center to the k th apex as b . Let k
Ak
be the area of the triangle formed by a , b and b , the total area A of k k k+1
the polygon being equal to the sum of A . k
Then formulae (4.8.6) and (4.8.12)
yield the following expression for the coefficient τ: 8 τ = 3√ A
n
Σ k=1
a 2k 4 A 2k
a 2k 1 1/2 1 1/2-1 + − 2 − . 4 A 2k bk b 2k+1
In the case of a regular polygon formula (4.8.14) simplifies to
(4.8.14)
264
CHAPTER 4
τ =
4√cot(π/ n ) 3 n 3/2sin(π/ n )
APPLICATIONS IN FRACTURE MECHANICS
.
(4.8.15)
Formula (4.8.13) gives for a regular polygon τ =
4 cot(π/ n ) 1/2 . n 3π
(4.8.16)
It is of interest to compare the numerical results due to (4.8.15) and (4.8.16). Here the relevant computations are presented
n= formula (4.8.15) τ= formula (4.8.16) τ= discrepancy (%)
3 0.2251 0.1862 17.3
4 0.2357 0.2122 10.0
5 0.2380 0.2227 6.5
6 0.2388 0.2280 4.5
7 0.2391 0.2312 3.3
8 0.2392 0.2331 2.6
∞ 0.2394 0.2394 0.0
While both formulae in the limiting case n →∞ give the same exact result for a circle, their discrepancy for small n is quite significant, so it is important to establish which one is more accurate. We have not found any data for an equilateral triangle. If one takes the experimental result by Cohn for a square τ=0.2274 as exact, then our formula (4.8.15) is in error by 3.6% while formula (4.8.16) due to Fikhmanas and Fridberg is in error by 6.7%. The numerical result due to Okon and Harrington for a square is 0.2258 which also favours our formula. In the case of a regular hexagon, the result by Okon and Harrington is 0.2375, so that our result differs by 0.5% only, while the error of (4.8.16) is 4%. It is noteworthy that the value of τ does not change significantly in the whole range 3≤ n <∞. We can also compare the normal displacements along a central line of a hexagon perpendicular to its side, given by (4.8.3) with numerical data due to Okon and Harrington (1981). Here are the results ( w ∗ stands for w /2π Hp √ A )
ρ/ a = Okon et al w ∗= formula (4.8.3) w ∗= discrepancy (%)
0. 0.351 0.357 -1.7
0.1667 0.346 0.352 -1.7
0.3333 0.331 0.3366 -1.4
0.5000 0.305 0.3092 -1.4
0.6667 0.263 0.266 -1.2
0.8333 0.210 0.1973 6.0
As we expected, the agreement is good, except for the points very close to the boundary. Example 2: Rectangle. Consider a rectangular crack, a and b being its semiaxes along the axes Ox and Oy respectively. Introduce the aspect ratio ε= b / a ≤1. Formula (4.8.15) in this case reduces to
265
4.8 Flat crack of general shape
τ =
√ε . 3(1 + ε2)1/2
(4.8.17)
Formula (4.8.13) in this case gives τ =
4√ε . 3π(1 + ε)
(4.8.18)
We present below the results of computations due to (4.8.17) and (4.8.18) compared with the experimental results of Cohn If one assumes the results of
ε= experiment τ= formula (4.8.17) τ= discrepancy (%) formula (4.8.18) τ= discrepancy (%)
0.1000 0.1202 0.1049 12.7 0.1220 -1.5
0.1500 0.1411 0.1277 9.5 0.1429 -1.3
0.2000 0.1565 0.1462 6.6 0.1582 -1.1
0.3000 0.1789 0.1749 2.3 0.1788 0.1
0.5000 0.2093 0.2108 -0.7 0.2001 4.4
0.7500 0.2251 0.2309 -2.6 0.2100 6.7
1.0000 0.2274 0.2357 -3.7 0.2122 6.7
Cohn to be exact then our formula performs better for ε≥0.5 while the formula by Fikhmanas and Fridberg is more accurate for ε<0.5. If instead we take the numerical results received in a personal communication from De Smedt as correct then the conclusion might be different. For example, his value of τ for ε=0.1 is 0.1142; now our result is in error by 8% while the result by Fikhmanas and Fridberg is in error by -7%. At this moment nobody seems to know which estimate is correct. We can also compare the dimensionless displacements w ∗ due to (4.8.3) with the numerical results received in a personal communication from De Smedt for a rectangle with aspect ratio ε=0.5 (as before, w ∗ stands for w /2π Hp √ A ). Here are the data computed along the axis Ox for y/b =0.025. The
x/a = De Smedt w *= formula (4.8..3) w *= Discrepancy (%)
0.0250 0.3161 0.3158 0.1
0.2250 0.3118 0.3081 1.2
0.4250 0.2989 0.2862 4.2
0.6250 0.2713 0.2469 9.0
0.8250 0.2107 0.1787 15.2
0.9750 0.0852 0.0703 17.5
agreement is not bad except for the zone x/a >0.625. Here are the data computed along the axis Oy for x/a =0.025. We observe here a good agreement
y/b = De Smedt w *= formula (4.8.3) w *= Discrepancy (%)
0.0250 0.3161 0.3158 0.1
0.1250 0.3067 0.3062 0.2
0.2250 0.2836 0.2824 0.4
0.3250 0.2424 0.2403 0.8
0.4250 0.1690 0.1666 1.4
0.4750 0.0976 0.0987 -1.2
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APPLICATIONS IN FRACTURE MECHANICS
Consider the case where the domain S is a Example 3: Rhombus. rhombus, a and b being its semiaxes along the axes Ox and Oy respectively. Introduce the aspect ratio ε= b / a ≤1. Formula (4.8.14) in this case reduces to τ =
√2ε . 3(1 + ε)
(4.8.19)
The result due to Fikhmanas and Fridberg is τ =
2√2ε . 3π(1 + ε2)1/2
(4.8.20)
We did not find in the mechanics literature any result related to a crack with a rhombus planform. The coefficient of electrical polarizability for a diamond with the aspect ratio ε=0.5 was found numerically by Okon and Harrington as τ=0.2082. Our result is 0.2222 (discrepancy 6.7%) while formula (4.8.20) gives 0.1898 (discrepancy 8.9%). We have received two sets of data in personal communications from De Smedt and Lee. Here are the data received as compared to formulae (4.8.19) and (4.8.20) The data received from Lee is given
ε= De Smedt τ= formula (4.8.19) τ= Discrepancy (%) formula (4.8.20) τ= Discrepancy (%)
0.100 0.111 0.136 -21.9 0.094 15.1
0.200 0.151 0.176 -16.4 0.132 12.8
0.333 0.182 0.204 -12.0 0.164 9.8
0.500 0.204 0.222 -9.0 0.190 6.9
0.800 0.219 0.234 -6.8 0.210 4.4
1.000 0.221 0.236 -6.6 0.212 4.1
as a function of the angle α=tan-1ε We have presented both sets of data in
α(deg.)= Lee τ= formula (4.8.19) τ= Discrepancy (%) formula (4.8.20) τ= Discrepancy (%)
10. 0.147 0.168 -14.2 0.124 15.8
15. 0.174 0.192 -10.6 0.150 13.7
20. 0.193 0.209 -8.1 0.170 11.8
25. 0.207 0.220 -6.3 0.186 10.1
30. 0.216 0.227 -5.2 0.197 8.5
40. 0.226 0.235 -3.8 0.211 6.9
45. 0.228 0.236 -3.6 0.212 6.7
order to underline the fact that there is no really reliable data as yet. The first set of data suggests that the formula by Fikhmanas and Fridberg is the more accurate, while the second set favours ours. It is noteworthy that formula (4.8.19) seems to give the upper bound, and formula (4.8.20) provides the lower bound, their average being very close to the numerical data. We can also compare the normal displacements due to our (4.8.3) with a
267
4.8 Flat crack of general shape
ρ/ a = Okon et al. w ∗= formula (4.8.3) w ∗= discrepancy (%)
0. 0.335 0.3333 0.5
0.3333 0.304 0.3142 -3.4
0.6667 0.257 0.2484 3.3
mainly due to the assumption of a square root singularity in (4.8.3) which does not hold for domains with sharp angles. Example 4: Circular segment. Let the radius r and the angle 2α be the segment parameters. Direct numerical computations show that the crack center can be identified with the center of gravity, with an error comparable with the accuracy of the theory presented. The location of the center of gravity is defined by x = kr, where c
2 sin3α . k = 1 3(α − sin 2α) 2 The equation of the segment boundary with respect to its center of gravity takes the form a (φ) = r [− k cosφ + (1 − k 2sin2φ)1/2], and a (φ) = r
k − cosα , cos(π − φ)
where γ = tan-1(sinα/( k − cosα)). gives τ =
for 0≤φ≤π−γ or π+γ≤φ<2π,
for π−γ≤φ≤π+γ,
(4.8.21)
Feeding of (4.8.21) in (4.8.6) and (4.8.12)
4 sinγ k sinγ + E(π−γ, k ) + −1. 2 1 k − cosα 1 − k 3(α − sin2α)1/2 2
(4.8.22)
where E(⋅,⋅) stands for the incomplete elliptic integral of the second kind. formula due to Fikhmanas and Fridberg gives 1 sin2α)1/2 2 . 3π(α + sinα)
The
4(α − τ =
(4.8.23)
The coefficient of electrical polarizability for a semi-circle was computed by Okon and Harrington as τ=0.2161. Our result due to (4.8.22) is τ=0.2163 which is practically identical to the previously mentioned one. The result due to
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APPLICATIONS IN FRACTURE MECHANICS
(4.8.23) is τ=0.2069 (discrepancy 4.3%). An additional confirmation of correctness of the new method can be obtained by observing the plot of the electrical polarizability density distribution for a semi-circle presented by Okon and Harrington (1981). Its maximum is located at a distance ≈0.47 r from the circle’s center. Our definition of the crack center requiring the minimization of the integral (4.8.6) gives its coordinate at 0.48 r which is very close. The center of gravity of the semi-circle is located at 0.42 r . Example 5: Circular sector. Let r and 2α be its radius and the polar angle. The crack center is assumed to be located on the axis of symmetry at a distance kr from the circle’s center. Numerical computations show that the crack center may be located at the center of gravity for 0.1π<α<0.6π. In this case the value of k is defined by k = 2sinα/(3α). In the range α<0.1π or α>0.6π, the value of k should be found from the minimum condition for the integral (4.8.6). Repetition of the procedure described in the previous paragraph leads to the following result τ =
cosα + cos(α − γ) −1 4 k sinγ + E(γ, k ) + . k sinγ 3√α 1 − k2
Here, γ = tan-1(sinα/(cosα − k )). reads τ =
(4.8.24)
The formula due to Fikhmanas and Fridberg
4√α . 3π(1 + α)
(4.8.25)
Note that neither (4.8.24) nor (4.8.25) reduce to the exact value for a circle when α=π. This is due to the fact that we do not really have the case of a penny-shaped crack when α approaches π: we have a circular crack which has its faces bonded along the radius φ=π. This case has not been considered by other authors so we cannot say which formula is more accurate. Okon and Harrington obtained in the case of a quadrant τ=0.2269, formula (4.8.24) gives τ=0.2308 (discrepancy 1.7%), and formula (4.8.25) gives τ=0.2107 (discrepancy 7%). It is noteworthy that the value of τ for a quadrant is greater than that for a semi-circle. The general impression is that our theory in the particular cases of a circular sector and segment provides the upper bound for τ while the formula due Fikhmanas and Fridberg gives the lower bound. Consider a crack configuration obtained by an Example 6: Cross. orthogonal intersection of two equal rectangles with sides 2 a and 2 b . Introduce the aspect ratio as ε= b / a ≤1. The area can be expressed as A = 4 a 2ε(2 − ε) , The following expression can be obtained for τ:
269
4.8 Flat crack of general shape
τ =
√2ε 6√2 − ε{[2(1 + ε2)]1/2 − 1}
(4.8.26)
The formula due to Fikhmanas and Fridberg is τ =
2√ε(2 − ε) . 3π
(4.8.27)
Here, we present the results given by formulae (4.8.26) and (4.8.27) compared to the experimental results of Cohn and the numerical results by De Meulenaere and Van Bladel (1977), and those received in personal communication from De Smedt
ε= experimental τ= De Meulenaere τ= De Smedt τ= formula (4.8.26) τ= formula (4.8.27) τ=
0.1000 0.0942 – 0.0835 0.1284 0.0925
0.2000 0.1333 – 0.1183 0.1777 0.1273
0.3000 0.1609 – – 0.2078 0.1515
0.4000 – 0.19 0.1767 0.2252 0.1698
0.6000 – 0.22 0.2084 0.2376 0.1944
0.8000 – 0.23 0.2193 0.2372 0.2079
1.0000 0.2274 0.238 0.2212 0.2357 0.2122
We did not compute the discrepancy since the data disagreement is too large thus making all the data not very reliable. The general impression is that our (4.8.26) gives the upper bound for τ while the formula due to Fikhmanas and Fridberg provides the lower bound. This conclusion might be wrong if the numerical results received in the personal communication from De Smedt are correct. For example, his result for ε=0.1 is τ=0.08347 which differs from the experimental result by 11%. All this proves one point: the existing numerical methods are too crude and there is a need to develop some new and more reliable numerical methods. It should be noted that the function defined by (4.8.26) is not monotonic: a relatively flat maximum is observed for ε≈0.7. The remaining data are monotonic. We have no rigorous proof to claim that the quantitative behavior of (4.8.26) is correct while the other data behavior is not, but we can indicate that the value of τ for a quadrant is also greater than that for a semi-circle, and this is mainly due to the fact that the shape of a quadrant is more close to the shape of a circle than that of a semi-circle. A similar statement can be made about a cross with the aspect ratio ε≈0.7 as compared to a square. Discussion. The majority of the examples considered indicate that the exact result is sandwiched between the results given by our (4.8.12) and by the formula due to Fikhmanas and Fridberg (4.8.13). In this sense the formulae act as upper and lower bounds respectively, which leads to a conjecture: for an arbitrary contour one of the inequalities holds, namely, either τ12≤τexact≤τ13, or τ12≥τexact≥τ13.
We can indicate one way to disprove the conjecture.
A look at
the table related to a rectangle in the previous Section indicates that our formula
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CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
(4.8.17) should give the lower bound for small aspect ratio ε, and it should give the upper bound for ε close to unity, and vice-versa for the formula (4.8.18). This means that there should be a value of ε for which both formulae (4.8.17) and (4.8.18) are exact , and give the same result. By equating (4.8.17) and (4.8.18), one gets the value of the aspect ratio ε1,2 =
π2 ± 4( 2π2 − 16 )1/2 16 − π2
,
which yields ε =0.3482 and ε =2.8716, with an obvious property ε ε =1. 1 2 1 2
The
corresponding value of τ is 0.18576. From the table of the previous Section, one can see that for ε=0.3 the exact value of τ should be between 0.1749 and 0.1788. If we could be sure that the experimental value 0.1789 is exact then this would disprove the conjecture, but at the moment none of the existing experimental or numerical techniques can offer such an accuracy. This can be achieved on the basis of the method for accurate evaluation of singular two-dimensional integrals presented in (Fabrikant, 1986e). A significant effort is required to prove or to disprove the conjecture, and is left to the interested reader. The accuracy of formula (4.8.12) can be improved by taking into consideration the fourth harmonic (4.8.7) in combination with the variational approach (Noble, 1960). The following functional is stationary at the exact solution of (4.1.9) I ( w ) = 2⌠ ⌠ σ( M ) w ( M )d S
⌡⌡ S
M
+
1 ⌠⌠ ∆ ⌠ ⌠ w( N) d S d S . w ( M ) ⌡ ⌡ R ( M,N) N M 4π2 H⌡ ⌡ S
(4.8.28)
S
Taking −
1 ⌠ ⌠ w( N) dS ≈ σ + σ , ∆ N 0 4 4π2 H ⌡ ⌡ R ( M,N) S
where σ and σ are defined by (4.8.5) and (4.8.7) respectively, and substituting 0 4 them in (4.8.28), we obtain a functional which can be considered as a function of δ. From the extremum condition ∂I = 0 ∂δ one finally gets
271
4.8 Flat crack of general shape
τ =
8 , 3 B √ A (1 − c )
(4.8.29)
where c =
3( F E + F E ) c c s s 5 AB
,
and the following geometrical characteristics were introduced 2π
2π
cos4φ dφ , F = ⌠ c ⌡ a 2(φ)
sin4φ dφ , F = ⌠ s ⌡ a 2(φ)
0
0
2π
2π
E = ⌠ a 3(φ)cos4φ dφ,
E = ⌠ a 3(φ)sin4φ dφ.
⌡
c
s
0
⌡ 0
The results of computations due to (4.8.29) for a rectangle are presented below against the experimental results of Cohn Comparison of this table with a
ε= Cohn τ= formula (4.8.29) τ= discrepancy (%)
0.1000 0.1202 0.1054 12.3
0.1500 0.1411 0.1290 8.6
0.2000 0.1565 0.1484 5.2
0.3000 0.1789 0.1785 0.2
0.5000 0.2093 0.2125 -1.5
0.7500 0.2251 0.2257 -0.3
1.0000 0.2274 0.2278 -0.2
corresponding one presented earlier indicates that the variational approach does improve the accuracy, though the improvement is still not sufficient for small ε. There is no proof that the variational approach will always improve the accuracy. On the contrary, one can find quite a few examples when the accuracy deteriorates. This can usually be observed for domains with a very small aspect ratio ε. It is up to the user to decide whether the more cumbersome computations are worth somewhat better accuracy. In this section we have considered in detail only the case of a uniform crack pressure. Some considerations can be presented for a general case. It is known (see, for example, 4.1.13) that in the case of a penny-shaped crack the following relationship can be established between the displacements w and the internal pressure σ 2π a [( a 2 − ρ2)( a 2 − ρ20)]1/2 σ(ρ0,φ0) 2 ⌠ ⌠ -1 tan w (ρ,φ) = H ρ dρ dφ , 0 0 0 π ⌡ ⌡ aR R 0
0
272
CHAPTER 4
where R =ρ2 + ρ20 − 2ρρ cos(φ−φ ). 0 0
APPLICATIONS IN FRACTURE MECHANICS
The following generalization of the last
formula seems to be natural 2π a (φ0)
2 w (ρ,φ) = H ⌠ π ⌡ 0
[( a 2(φ) − ρ2)( a 2(φ ) − ρ20)]1/2 0
⌠ tan-1 ⌡
a (φ) R
σ(ρ0,φ0) R
ρ0dρ0dφ0.
0
Though a complete investigation there is reason to believe that loading of cracks whose aspect example. Let us compute the
of (4.8.30) it will be ratio is not displacement
(4.8.30) is beyond the scope of this book, sufficiently accurate for a general far away from unity. Here is an w at the centre of an elliptical 0
crack ( a and b are the semiaxes of the ellipse, a > b ) under a uniform pressure p . The result due to (4.8.30) is w 0=(8/π) HpbK(k) , where K(k) is the complete elliptic integral w 0=2π Hpb/E(k) .
of the first kind and k =(1− b 2/ a 2)1/2. The exact result is Both results are close to each other for small k , coinciding in
the case of a circle ( k =0). Direct computation shows that the error of the approximate expression does not exceed 7% for an ellipse with aspect ratio b/a ≥0.5. In the case of a square the dimensionless displacement w * at its centre can be computed from (4.8.30) as (4/π2)ln(1+√2 )=0.3572, while a similar result due to the experimental value of Cohn is (3/2)τ=0.3411, and the discrepancy does not exceed 5%. Of course, these examples do not prove anything conclusively, but they make it quite clear that expression (4.8.30) is worth to investigate further. A similar statement can be made about the following expression 2πa (φ0)
σ(ρ,φ) = −
2 ⌠ ⌠ π[ρ − a 2(φ)]1/2⌡ ⌡ 2
0
0
[ a 2(φ ) − ρ20]1/2σ(ρ ,φ )ρ dρ dφ 0 0 0 0 0 0 ρ2 + ρ20 − 2ρρ0cos(φ−φ0)
,
which gives the normal stress distribution outside a circular crack directly through its values inside the crack. The future research will show whether the last expression is useful for a general crack. Investigation of the stress intensity this section, but we can show that some results which are close to (Weaver, relationship between the limiting values the stress intensity factor, namely,
factors was beyond the main scope of simple formulae may be derived to give 1977). Make use of the asymptotic of the crack opening displacements and
w = 2 H (2π r )1/2 k 1. Consider a rectangular crack of dimensions 2 a and 2 b ( a>b ), subjected to a
4.9 General crack under uniform shear
uniform pressure p . By substituting (4.8.3) in the last expression, the following formulae are obtained for the dimensionless stress intensity factor κ= k 1/[ p (π b )1/2]: along the shorter side κ=3τ; along the longer side κ=3τ( a/b )1/2. In the case of a square our formulae give κ=3×0.2357=0.7071. The result due to Weaver is about 0.73, with the discrepancy of about 3%. In the case of b/a =0.3, our result at the middle of the longer side is κ=0.98 which is very close to the result of Weaver. Expression (4.8.17) indicates that the limiting value of the stress intensity factor as ( b/a )→0 is κ→1, as it should be in the case of an infinite strip slit. The only discrepancy with the results of Weaver (1977) is the value of the dimensionless stress intensity factor along the shorter side: according to our formula it should decrease with b/a , tending to zero as ( b/a )1/2; in the paper by Weaver (1977) its value increases. The reader is referred also to the paper (Fabrikant, 1987f) where some inconsistencies of (Kassir and Sih, 1975) in defining the stress intensity factor for elliptical cracks are pointed out. The mathematically equivalent problem of sound penetration through an aperture of general shape in a soft screen is solved in (Fabrikant, 1988d). The same apparatus is used in the investigation of electrical polarizability of small apertures of general shape (Fabrikant, 1987c).
1.
Exercise 4.8 Derive (4.8.5).
2.
Establish (4.8.14).
3. Try to prove or disprove the conjecture, expressed in the section ’Discussion’ above. 4. Find the domain of validity of formula (4.8.30) for an elliptical crack. (Find the ratio of ellipse semiaxes, for which the error does not exceed 5%.) 5. Solve the problem of a general flat crack subjected to normal loading, with its magnitude proportional to the x -coordinate. (Bending of an elastic space, with a flat crack of general shape).
4.9 General crack under uniform shear Let the crack boundary be described in polar coordinates by the equation We can always ρ= a (φ), where a (φ) is a single-valued continuous function. choose the coordinate axis orientation so that the first harmonic will vanish from the Fourier expansion of a (φ). An approximate analytical solution of (4.4.14) for a general crack can be obtained by the method used in previous section. The
273
274
CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
method uses the following representation
ρ
2π
dx
λ(
a (φ0)
⌠ ⌠ u( N) dS = ⌠ ⌠ dφ ⌠ 2 2 1/2 ⌡ ⌡ R ( N,N0) N ( ) ρ − x ⌡ 0 ⌡ ⌡ 0
0
S
x2 , φ−φ ) 0 ρρ0
(ρ20 − x 2)1/2
x
Consider the case of a uniform shear loading. have the form
u (ρ0,φ0)ρ0dρ0.
(4.9.1) Let the tangential displacements
u (ρ,φ) = u 0[ a 2(φ) − ρ2]1/2/ a (φ) ,
(4.9.2)
where u is an as yet unknown complex constant. 0
Substitution of (4.9.2) in
(4.9.1) yields, after integration and retaining the first two harmonics only, 2π
⌠ ⌠ u( N) dS ≈ ⌠ 2 a ( φ ) 0 ⌡ ⌡ R ( N,N0) N 16π( G 21 − G 22) ⌡ 0 S u0
−
ρ2 1 + 3cos2(φ−φ )dφ . 0 0 a (φ0)
(4.9.3) By substituting (4.9.3) in (4.4.14) and performing necessary differentiation, we obtain the relationship between the shear loading and the amplitude of the tangential displacements, namely, τ =
1 4π( G 21
−
G 22)
u G B + 3u G B , 0 2 2 0 1 1
(4.9.4)
where 2π
B1
dφ = ⌠ , ⌡ a (φ) 0
2π
e2 i φdφ . B2 = ⌠ ⌡ a (φ)
(4.9.5)
0
Equation (4.9.4) can be solved, to give
u0 =
4π( G 21 − G 22) G 21 B 21 − 9 G 22 B 22
[ G B τ − 3 G B τ]. 1 1 2 2
(4.9.6)
275
4.10 Close interaction of pressurized coplanar circular cracks
The integrals in (4.9.5) can be computed easily for various crack shapes. For example, a rectangular crack with sides 2 a and 2 a is characterized by the 1
2
values
B
1
=
4( a 21 + a 22)1/2 aa
1 2
,
B
2
=
4( a 22 − a 21) 3 a a ( a 21 + a 22)1/2
.
(4.9.7)
1 2
It is noteworthy that despite the fact that the integral representation (4.9.1) is valid inside a circle ρ≤min{ a (φ)} only, and despite the approximate nature of (4.9.3), the solution given by (4.9.4−4.9.5) is exact for an ellipse. We did not find in the literature any reliable data related to a non-elliptical crack under shear loading, therefore it is difficult to say how accurate the solution is for various crack shapes.
1.
Exercise 4.9 Derive (4.9.3).
2.
Establish (4.9.6).
3.
Consider the case of semicircular crack.
4.
Solve the problem for a cross-shaped crack
5.
Solve the problem of a general flat crack subjected to torsion.
4.10 Close interaction of pressurized coplanar circular cracks The general method is applied here for the stress analysis of an elastic space weakened by several arbitrarily located coplanar circular cracks under the action of an arbitrary normal pressure. The governing integral equations are derived, which have definite advantages over other methods: the equations are non-singular, the iteration procedure is rapidly convergent even for very close interactions; there is no need to solve the integral equations if one is interested only in obtaining the upper and the lower bounds for the quantities of interest. In the case of the cracks which are far apart, these bounds are so close that they provide, in fact, a sufficiently accurate solution to the problem. The method allows us also to obtain a practically exact numerical solution to the problem of very close interactions. Several illustrative examples are considered.
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CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
Theory. Consider an elastic space weakened in the plane z =0 by n arbitrarily located circular cracks. The cracks do not intersect. Let the centre of the k th crack be located at the point with Cartesian coordinates x and y , its radius be denoted by a . k
k
Let an arbitrary pressure σ
k
be applied normally to
k
the crack faces in opposite directions. The set of governing integral equations can be written, due to (4.1.9), in the form n
1 σ (M ) = − ∆ i i 4π2 H
w (M )
Σ k=1
⌠ ⌠ k k d S , for i = 1,2, ... ⌡ ⌡ R ( M i , M k) k
,n,
Sk
(4.10.1) where ∆ is the two-dimensional Laplace operator, S
k
is the k th crack domain,
R(M ,M ) stands for the distance between the points M i
k
M ∈ S ); w k
k
k
i
and M , ( M ∈ S k
i
i
and
denotes the normal displacements of the crack face (an unknown
function), σ stands for the normal traction acting inside the i th crack (a known i
function). We can single out, without loss of generality, the crack number 1, and consider the set of cracks in the local polar system of coordinates with the origin coinciding with the centre of the first crack. By using the integral representation (4.1.11), the first equation from the set (4.10.1) can be rewritten as ρ
xdx 1 1 d ⌠ 2 σ (ρ,φ) = − 2 ( ) L 2 2 1/2 L( x ) 1 d ρ ρ π Hρ ⌡ (ρ − x ) 0
ρ dρ 1 0 0 d 1 1 × ⌠ L( ) w 1(ρ0,φ) − 2 d x (ρ0 − x 2)1/2 ρ 4π2 H 0 ⌡ a
x
w (M )
n
Σ
k k dS . ∆⌠ ⌠ k R(M ,M ) ⌡ ⌡ 1 k k=2 Sk
(4.10.2) Since the integrals under the summation sign in (4.10.2) are non-singular, the differentiation can be performed under the integral sign, with the result ρ
1 1 d ⌠ σ (ρ,φ) = − 2 L( ) 1 dρ ⌡ ρ π Hρ
xdx (ρ − x ) 2
2 1/2
L( x 2 )
0
ρ dρ 1 0 0 d 1 1 × ⌠ L( ) w 1(ρ0,φ) − 2 2 d x (ρ0 − x )1/2 ρ0 4π2 H ⌡ a
x
w (M )
n
Σ
⌠⌠ k k dS . k ⌡ ⌡ R 3( M , M ) 1 k k=2 Sk
277
4.10 Close interaction of pressurized coplanar circular cracks
(4.10.3) Formula (4.1.12) allows us to express w 1 from (4.10.3) by constructing an inverse operator, namely, a
x ρ0dρ0 ρρ0 d x ⌠ ⌠ w 1(ρ,φ) = 4 H 2 2 1/2 2 2 1/2 L( 2 ) σ1(ρ0,φ) x ⌡ ( x − ρ ) ⌡ ( x − ρ0) 1
ρ
+
0
( a 21 − ρ2)1/2 π2
n
Σ k=2
w ( ρ , φ ) ρ dρ dφ
k 0 0 0 0 0 ⌠⌠ . 2 2 1/2 2 2 ⌡ ⌡ (ρ0 − a 1) [ρ + ρ0 − 2ρρ0cos(φ−φ0)] Sk
(4.10.4) Here we used the following integral representation for 1/ R
ρ
3
x2 λ( , φ−φ0) x 2d x ρ0
2 1 d⌠ 1 2 2 3/2 = πρL(ρ) dρ 2 2 1/2 2 2 3/2, (ρ + ρ0 − 2ρρ0cos(φ−φ0)) ⌡ (ρ − x ) (ρ0 − x ) 0
for ρ >ρ. 0
(4.10.5)
The representation (4.10.5) allows us to compute various integrals involving the Abel and the L-operators. For example, the following integral is an immediate consequence of (4.10.5) x
⌠ 2 ρ dρ2 1/2 L(ρ)[ρ2 + ρ20 − 2ρρ cos(φ−φ )]-3/2 0 0 ⌡ (x − ρ ) 0
=
x x2 λ ( , φ−φ0). ρ0 (ρ20 − x 2)3/2
A similar procedure can be applied to the remaining n −1 cracks, and the additional n −1 equations of the type (4.10.4) can be obtained. Note that each such equation is valid in a local system of polar coordinates related to a certain crack. The set of equations (4.10.4) can be solved numerically by iteration. Here we show that one can obtain the upper bound, the lower bound and a reasonably accurate central estimation for all the quantities of interest without solving the integral equations (4.10.4). Since w does not change sign in S , we k
k
can apply the mean value theorem to the second integral in (4.10.4), with the
278
CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
result a
x ρ0dρ0 ρρ0 d x σ (ρ ,φ) ⌠ ⌠ w 1(ρ,φ) = 4 H 2 2 1/2 2 2 1/2 L 2 x 1 0 ⌡ ( x − ρ ) ⌡ ( x − ρ0) 1
ρ
+
0
( a 21 − ρ2)1/2 π2
Vk
n
Σ(r k=2
2 1k
− a 21)1/2 [ρ2 + r 21k − 2ρ r 1kcos(φ−φ1k)]
. (4.10.6)
Here V denotes half of the volume of the opened k th crack k V k = ⌠ ⌠ w kd S k,
⌡⌡ Sk
and though the exact location of the point R with the polar coordinates 1k
( r ,φ ) is unknown, the fact of belonging to the domain S limits the possible 1k 1k k
variation of the quantities of interest and allows us to obtain the upper and the lower bounds as well as a sufficiently accurate central estimation which will be discussed further. The symmetry considerations can also be used to sharpen the estimations. It will be shown further that these bounds can be so close in the case of cracks remote from each other, that they provide, in fact, a sufficiently accurate solution to the problem. The crack volume 2 V domain S 1.
1
can be estimated by integration of (4.10.4) over the
The result is
2π a
1
V 1 = 4 H⌠ ⌠ σ(ρ,φ)( a 21 − ρ2)1/2ρdρdφ ⌡ ⌡ 0 0
a
n
Σ
2 + π
a
1 1 -1 w (ρ,φ)ρdρdφ. ⌠⌠ 2 2 1/2 − sin ρ k ⌡ ⌡ (ρ − a 1) k=2 Sk
(4.10.7) We can use again the mean value theorem, with the result
279
4.10 Close interaction of pressurized coplanar circular cracks
2π a
n
1
Σ
a
a
1 1 2 V . V 1 = 4 H⌠ ⌠ σ1(ρ,φ) ( a 21 − ρ2)1/2ρdρdφ + − sin-1 2 2 1/2 π ρ k (ρ1k − a 1) ⌡ ⌡ 1k k=2 0 0
Again, the point with the polar coordinate ρ1k belongs to S k
(4.10.8) thus limiting the
possible variation. Integration of the remaining n −1 equations over the area of each crack provides finally a system of n linear algebraic equations which can be solved with respect to the unknowns V k. Their feeding back into (4.10.6) gives the complete solution to the problem. Although the exact values of the coordinates r ik , φik and ρik are not known, we can always use their maximum and minimum values in order to obtain the upper and the lower bounds for all the quantities of interest. Defining the stress intensity factor as k k(φ) =
lim {(ρ − a k)1/2σk(ρ,φ)} , ρ→ a k
one can get an equivalent expression through the crack face displacements w k(ρ,φ)
1 k k(φ) = 4π H
lim 1/2 . a − ρ ( ) ρ→ a k k
(4.10.9)
Substitution of (4.10.4) in (4.10.9) gives 2π a
k (φ) = 1
1 ⌠ ⌠ π √2 a 1 ⌡ ⌡ 2
0
+
√a1 2√2π3 H
0
n
Σ k=2
( a 21 − ρ2)1/2σ (ρ,φ ) ρdρdφ 1 0 0
1
ρ2 + a 21 − 2ρ a 1cos(φ − φ0) w (ρ, φ ) ρdρdφ
k 0 0 ⌠⌠ . ⌡ ⌡ (ρ2 − a 21)1/2[ρ2 + a 21 − 2ρa 1cos(φ−φ0)] Sk
By using again the mean value theorem, the following expression for the stress intensity factor can be obtained
280
CHAPTER 4
2π a
k (φ) = 1
1 ⌠ ⌠ π √2 a ⌡ ⌡ 1 √a
( a 21 − ρ2)1/2σ (ρ,φ ) ρdρdφ 1 0 0 ρ2 + a 21 − 2ρ a cos(φ−φ ) 1 0
2
0
+
1
0
Vk
n
1
2√2π3H
Σ k=2
APPLICATIONS IN FRACTURE MECHANICS
( r 21k − a 21)1/2 [ r 21k + a 21 − 2 r a cos(φ − φ )] 1k 1 1k
.
(4.10.10) Similar expressions can be derived for the remaining n −1 cracks. The first term in (4.10.10) presents the stress intensity factor of a solitary crack opened up by an arbitrary normal pressure σ , the remaining terms display the influence of 1
interacting cracks. interacting coplanar solitary crack under Mura (1983) to be factor along part of
It is clear from (4.10.10) that the stress intensity factor of cracks is always greater than the stress intensity factor of a similar pressure. This proves the results of Mastrojannis and incorrect, since they report a decrease in the stress intensity the boundary.
Example 1: Two cracks. Consider the case of two coplanar circular cracks with the radii a and a under the action of a uniform normal pressure σ =p 1
1
and σ = p 2
1
2
2
respectively.
Let l be the distance between their centres.
Equations (4.10.8) in this case will take the form a a 1 1 8 3 2 -1 , − π a 1 Hp + V 2 sin 2 1/2 1 ρ12 3 π 2(ρ12 − a 1)
V1 =
V2
a2 a2 8 3 2 -1 . = π a 2 Hp 2 + V 1 2 − sin ρ 3 π (ρ21 − a 22)1/2 21
The solution is
V
1
8 = πH 3
a 31 p + c a 32 p 1
12
1 − c 12 c 21
2
,
V
2
8 = πH 3
c a 31 p 21
1
+ a 32 p
1 − c 12 c 21
2
,
where
c
12
a a 1 1 2 -1 , = − sin 2 2 1/2 π(ρ12 − a 1) ρ 12
c 21
a a 2 2 2 -1 . = − sin 2 2 1/2 π(ρ21 − a 2) ρ 21
(4.10.11)
281
4.10 Close interaction of pressurized coplanar circular cracks
Here ρ12 varies between l − a 2 and l + a 2, and l − a 1≤ρ21≤ l + a 1.
(4.10.12) The upper bound
for V k corresponds to ρ12= l − a 2 and ρ21= l − a 1; the lower bound is achieved at
ρ12= l + a 2
and
ρ21= l + a 1.
The
estimations
can
be
sharpened
reciprocal theorem which implies the identity c 12 a 32= c 21 a 31.
by
using
the
This identity narrows
the range of admissible variation for ρ (in the case of unequal cracks only) thus making the estimations sharper. In this vein the case of two equal cracks should be considered as the least accurate. We shall also consider the central estimation which corresponds to ρ12=ρ21= l . It will be shown that the central estimation gives a sufficiently accurate solution even for relatively close crack interactions. Formulae (4.10.11−4.9.12) simplify in the case of equal cracks as a 1= a 2= a , and if p 1= p 2= p , then
V 1 = V 2 =V =
V0 1 − c
,
c =
a a 2 − sin-1 . 2 2 1/2 π (ρ − a ) ρ
(4.10.13)
Here V 0=(8/3)π Ha 3 p stands for a half of the volume of a solitary crack, and ρ
varies between l − a and l + a . Note that in the case of a uniform pressure, the crack energy W is proportional to its volume, namely, W = pV . It is clear from (4.10.13) that the crack interaction increases their energy. Substitution of (4.10.13) in (4.10.10) and use of the mean value theorem yield the following expression for the stress intensity factor
2ε3 K (φ) = K 01 + , {3π(1 − ε2)1/2 − 6[ε − (1 − ε2)1/2sin-1ε]}[1+ε2−2εcosφ] (4.10.14) where K 0= p √2 a /π corresponds to the stress intensity factor for an isolated crack under the action of a uniform normal pressure p . Here the upper bound for the stress intensity factor is given by ε= a /( l − a ), the lower bound corresponds to ε= a /( l + a ), and the central estimation is defined by ε= a / l . Now we need an accurate numerical solution in order to estimate the accuracy of the approximate formulae derived. Assume the crack face displacements in the form w (ρ,φ) = 4 Hp ( a 2 − ρ2)1/2 f (ρ,φ), where f is as yet unknown function.
(4.10.15)
It may be called the interaction function
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CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
since it is equal to the ratio of the crack opening displacements to those of a solitary crack. The values of f ( a ,φ) are equal to the ratio of the stress intensity factor of interacting cracks to the stress intensity factor of a solitary crack. We shall call f ( a , φ) the interaction factor . Substitution of (4.10.15) in (4.10.4) gives a convenient expression for the procedure of iteration
w (ρ, φ) = 4 Hp ( a 2 − ρ2)1/2 1
2π a
+
( a 2 − r 20)1/2 f(r0, ψ 0) r0 d r0 dψ0
1⌠ ⌠ . π2⌡ ⌡ ( l 2 + r 20 + 2 lr 0cosψ0 − a 2)1/2[ r 2 + r 20 + 2 rr0cos(ψ+ψ0)] 0
0
(4.10.16) Here we have introduced the new variables r =(ρ + l −2 l ρcosφ) , ψ=sin [(ρ/ r )sinφ]. The integral in (4.10.16) has a logarithmic singularity for r = l − a , ψ=0, as l →2 a , therefore the procedure of iteration might not be convergent for l very close to 2 a . The limiting value of l can be found by analyzing the integral operator 2
2π a
Z(f) =
2
1/2
-1
( a 2 − r 20)1/2 f(r0, ψ 0) r0 d r0 dψ0
1⌠ ⌠ . π2⌡ ⌡ ( l 2 + r 20 + 2 lr 0cosψ0 − a 2)1/2[( l − a )2 + r 20 +2( l − a ) r 0cosψ0] 0
0
(4.10.17) According to the Banach’s theorem, it is sufficient to prove that the integral operator (4.10.17) is a contraction operator. We define the distance in the class of continuous functions by δ( f,f 1) = max| f (ρ,φ) − f 1(ρ,φ)|. We assess the value of | Z ( f ) − Z ( f 1)| 2π a
=
1⌠ ⌠ π2⌡ ⌡ ( l 2 + r 2 + 2 lr cosψ − a 2)1/2[( l − a )2 + r 2 + 2( l − a ) r cosψ] 0
≤
( a 2 − r 2)1/2 | f(r, ψ ) − f1(r, ψ ) | r d r dψ
0
δ( f,f 1) 2π
⌠ ⌠
a
( a 2 − r 2)1/2 r d r dψ
π2 ⌡ ⌡ ( l 2 + r 2 + 2 lr cosψ − a 2)1/2[( l − a )2 + r 2 + 2( l − a ) r cosψ] 0
0
283
4.10 Close interaction of pressurized coplanar circular cracks
2δ( f,f ) <
1
π
a
( a 2 − r 2)1/2 r d r
2 a δ( f,f )
⌠ < . 2 2 1/2 2 2 π[( l − a )2 − a 2]1/2 ⌡ [( l − r) − a ] [( l − a ) − r ] 1
0
The integral operator (4.10.17) will be a contraction operator if 2a π[( l − a )2 − a 2]1/2
< 1 ,
indicating that the iteration procedure will be convergent for l >2.18 a which corresponds to a fairly close interaction. The estimation given above is crude. Direct computations show that the iteration procedure converges even for l =2.0005 a (which corresponds to the case when the shortest distance between the cracks is equal to 0.0005 of its radius), and converges rapidly: the first iteration with f ≡1 has the maximum relative error less than 2%, and the sixth iteration may be considered practically as an exact solution since the maximum relative error becomes less than 10-7. The accuracy of the first iteration improves as the distance between cracks increases. For example, the first iteration for l =10 is practically exact with maximum relative error less than 10-7. We could not go closer than l =2.0005 a , not because of non-convergence, but because the standard subroutine DBLIN from IMSL library, which was used to compute the integrals, failed giving terminal errors. Though we do not have a rigorous proof, it seems probable that the iteration procedure is theoretically convergent for an arbitrarily small distance between cracks. The values of the interaction function f (ρ,φ) are presented in (Fabrikant, 1987g) for various ratio of l/a . We limit ourselves here to just one abbreviated table, related to the closest interaction considered, with l =2.0005 a . The reader is referred to the original paper for additional data. The first line in Table 4.10.1 gives the ratio of the stress intensity factor of the interacting cracks to the stress intensity factor of a solitary crack under the same uniform load. All the -6 computations were made with the relative error not exceeding 10 . It was established that Collins’ (1963) formulae are accurate within 1% for l >2.5 a . The relative error of the central estimation corresponding to (4.10.14) does not exceed 2% for l >2.5 a . One can also notice that the central estimation is always slightly below the exact value, thus giving in the case of two cracks a very close lower bound for the quantities of interest. The same can be said about the formulae by Collins (1963). The accuracy of the central estimation deteriorates rapidly as l decreases, for example, the maximum error in the stress intensity factor for l =2.2 (the distance between cracks is 0.2 of its radius) is about 10%. The accuracy of the central estimation of the crack energy, corresponding to (4.10.13) is much better, and is discussed in more detail in the
284
CHAPTER 4
Table 4.10.1. ρ
φ
1.00000 0.91667 0.83333 0.75000 0.66667 0.58333 0.50000 0.41667 0.33333 0.25000 0.16667 0.08333 0.00000
APPLICATIONS IN FRACTURE MECHANICS
The interaction function for l =2.0005 a 0.
30.
60.
90.
120.
150.
180.
2.77577 1.42135 1.27366 1.20176 1.15812 1.12865 1.10743 1.09146 1.07903 1.06913 1.06108 1.05442 1.04884
1.18634 1.16479 1.14560 1.12870 1.11396 1.10117 1.09012 1.08058 1.07234 1.06519 1.05899 1.05358 1.04884
1.06686 1.06729 1.06728 1.06682 1.06593 1.06464 1.06299 1.06103 1.05883 1.05646 1.05396 1.05141 1.04884
1.03605 1.03761 1.03916 1.04067 1.04214 1.04353 1.04482 1.04597 1.04696 1.04777 1.04836 1.04872 1.04884
1.02471 1.02614 1.02766 1.02928 1.03102 1.03287 1.03483 1.03692 1.03911 1.04142 1.04382 1.04630 1.04884
1.02009 1.02137 1.02277 1.02431 1.02601 1.02788 1.02996 1.03228 1.03486 1.03775 1.04101 1.04468 1.04884
1.01881 1.02003 1.02139 1.02289 1.02455 1.02642 1.02850 1.03086 1.03353 1.03657 1.04007 1.04411 1.04884
next section. In the examples to follow we shall consider only the central estimation for all the quantities. Example 2: Infinite row of equal cracks. Let the crack radius be a , and the distance between the adjacent crack centres be l . The cracks are opened by a uniform pressure p . The central estimation for the crack opening volume 2 V can be defined, according to (4.10.8), by a single equation ∞
8 4 V = V π Ha3p + π 3
Σ k=1
a -1 a , 2 1/2 − sin kl (k l − a ) 2 2
with the result V0
V = ∞
4 1 − π
Σ k=1
,
(4.10.18)
a -1 a 2 1/2 − sin kl (k l − a ) 2 2
where V 0=(8/3)π Hp is the corresponding result for the case of a solitary crack. The crack opening displacement will take the form, according to (4.10.6) w (ρ,φ) = ( a 2 − ρ2)1/24 pH
285
4.10 Close interaction of pressurized coplanar circular cracks
∞
2V + 2 π
Σ(k l
2 2
k=1
ρ2 + ( kl )2
, − a ) [ρ + ( kl ) − 2(ρ kl ) cos2φ] 2 1/2
4
4
2
and, as an immediate consequence of the previous expression, the stress intensity factor will be defined by ∞
Σ
V a 2 + ( kl )2 , K (φ) = K 01 + 2 2 2 2 1/2 4 4 2 2π pH ( k l − a ) [ a + ( kl ) − 2( akl ) cos2φ] k=1 where K stands, as before, for the stress intensity factor of a solitary crack. 0 Example 3: Polygonal configuration. Consider a circular crack of radius b , its centre coinciding with the centre of a regular polygon, surrounded by n cracks of radius a , with their centres located at the apices of the polygon. Let l be the distance from the polygon centre to its apex. Let a uniform pressure p c open up the central crack, and a uniform pressure p act inside the cracks located at the polygon apices. Due to the system symmetry, the crack opening volume can be defined by just two equations Vc =
V =
8 2 b b , π Hb3p c + − sin-1 nV 2 3 π l ( l − b 2)1/2 8 2 a a π Ha3p + − sin-1 Vc 2 2 1/2 3 π l (l − a ) n
2 + V π
Σ k=2
a -1 a , 2 2 1/2 − sin lk (lk − a )
where V and V denote half of the volume of the central crack and the apex c
crack respectively, and l =2 l sin(π k/n ). k
Vc =
8 πH 3
b 3 p c c 22 + a3p c 12 c 11 c 22 − c 12 c 21
The solution is
,
V =
8 πH 3
b 3 p c c 21 + a3p c 11 c 11 c 22 − c 12 c 21
,
(4.10.19) where
286
CHAPTER 4
n
c
11
= 1,
c
22
Σ(l
2 = 1 − π
k=2
c
21
=
2 k
APPLICATIONS IN FRACTURE MECHANICS
a -1 a , 2 1/2 − sin l − a) k
2 a a , − sin-1 l π ( l 2 − a 2)1/2
c
12
=
2 b b n 2 − sin-1 . 2 1/2 l π ( l − b )
The central crack opening displacements are, according to (4.10.6), w (ρ, φ) = 4 Hp ( b 2 − ρ2)1/21 c
c
n
V + 2 4π Hp ( l 2 − b 2)1/2 c
Σ k=1
ρ + l 2
2
1 − 2ρ l cos(φ − φ )
(4.10.20)
k
Assuming φ =2π k/n , the summation in (4.10.20) can be performed, with the result k
w (ρ,φ) = 4 Hp ( b 2 − ρ2)1/21 c
c
+
nV l 2n − ρ2n . 2 2 2 1/2 2 2 2n 2n n n 4π Hp ( l − b ) ( l − ρ ) (ρ + l − 2ρ l cosn φ) c
The stress intensity factor for the central crack takes the form nV l 2n − b 2n . K = K 1 + 2 2 3/2 2n 2n n n 2 c 0 4π Hp ( l − b ) ( b + l − 2 b l cosn φ) c Due to symmetry of the system, all the apex located cracks will have the same characteristics. The following expressions are valid in a local system of polar coordinates, with its origin at the crack centre and the polar axis coinciding with the line connecting the polygon centre with its apex. The crack opening displacements are V c 1 w (ρ,φ) = 4 Hp ( a − ρ ) 1 + 2 2 2 1/2 2 4π Hp ( l − a ) (ρ + l 2 + 2 l ρcosφ) 2
2 1/2
287
4.10 Close interaction of pressurized coplanar circular cracks
n
+
Σ k=2
, [ρ2 + l 2k + 2 l kρcos(φ + ψk)] V
( l 2k − a 2)1/2
where V and V c are defined by (4.10.19); l k=2 l sin[π( k −1)/ n ], and ψk=2π( k −1)/ n . The stress intensity factor can be written as Vc 1 K = K0 1 + 4π2 Hp ( l 2 − a 2)1/2 ( a 2 + l 2 + 2 la cosφ) n
+
Σ
. [ a 2 + l 2k + 2 al kcos(φ + ψk)] V
( l 2k − a 2)1/2
k=2
Discussion. It is of interest to compare our results to those available in the literature. The paper by Collins (1963), though published 23 years ago, seems to be still the most reliable source. He did not give the stress intensity factor explicitly but it could be derived easily for the case of two equal cracks, and it reads in our notation
30ε7 12ε8 2ε 3 8ε 5 4ε 6 K = K 0 1 + + + + + 7π 3π 5π 5π2 9π2
+
4ε 4 2ε 3 4ε 5 18ε2 2ε 3 1 + 3ε 2 + + 9ε4cosφ + 1 + + cos2φ 3π 3π 3π 5 3π
4ε 6 21ε2 4ε 7 4ε 8 + 1 + cos3φ + cos4φ + cos5φ. 3π 5 3π 3π
(4.10.21)
Our result (4.10.14), if expanded in series, reads 2ε 3 K = K 01 + (1 + 2 3π
∞
Σ
εkcosk φ)(1 +
k=1
3 2 2ε 3 7 4 ε + + ε + ... ). 2 3π 4
(4.10.22)
Comparison of (4.10.21) with (4.10.22) reveals quite a few common terms. Though Collins himself assumed that his results are valid only for the cracks whose radius is much smaller than the distance between their centres, the
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CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
numerical results indicate that (4.10.21) is accurate within 1% for l ≥2.5 a which corresponds to the shortest distance reported in the literature. Direct computations show that the central estimation corresponding to (4.10.14) does not differ from Collins’ (4.10.21) by more than 3% in the whole range 2 a < l <∞, and differs by less than 0.9% for l >2.5 a . The stress intensity factor, due to Andreikiv and Panasiuk (1971), is K = K (1 + 0
2ε4 2ε3 cosφ). + 3π 3π
One should notice that a factor 2 is missing in the third term of their result. Collins (1963) gave the following expression for the crack energy of two equal cracks 18ε7 32ε8 2ε 3 6ε 5 4ε 6 + ... + + + + W = W 1 + 0 7π 3π 5π 15π2 9π2 where W =(8/3)π Ha3p2 is the energy of a solitary crack. 0
,
Our expression for the
crack energy is W W =
0
1 − c
,
(4.10.23)
where c is defined according to (4.10.13) as c =
2 ε − sin-1ε . 2 1/2 π (1 − ε )
Series expansion of (4.10.23) results in 2ε 3 3ε 5 4ε 6 15ε7 + + + + ... W = W 1 + 0 3π 5π 28π 9π2
.
which is very close to Collins’ result, ours being slightly lower. Table 4.10.2 displays the values of W/W . It confirms that the error of the central estimation 0
(4.10.23) is under 3% even for a close interaction when the shortest distance between the cracks is equal to 0.001 of their radius. Collins’ results are presented in order to emphasize the accuracy of our numerical solution. The energy increase due to the crack interaction is relatively small even for very close interactions; this is mainly due to the sharp localization of interaction effects (see, for example, Table 4.10.1.)
289
4.10 Close interaction of pressurized coplanar circular cracks
Table 4.10.2.
Crack energy increase (two equal cracks).
l/a
upper bound
lower bound
central estimation
result of Collins
exact result
error (%)
2.001 2.005 2.010 2.020 2.050 2.100 2.200 2.500 3.000 5.000 10.000
– – – – – 2.965366 1.498350 1.117132 1.035432 1.003526 1.000294
1.008800 1.008762 1.008714 1.008621 1.008349 1.007921 1.007150 1.005370 1.003526 1.001009 1.000161
1.035366 1.035105 1.034783 1.034152 1.032352 1.029637 1.025091 1.016120 1.008809 1.001764 1.000214
1.046308 1.045921 1.045444 1.044510 1.041861 1.037908 1.031413 1.019160 1.009900 1.001834 1.000216
1.062976 1.061694 1.060378 1.058066 1.052411 1.045276 1.035399 1.020067 1.010035 1.001835 1.000216
2.6 2.5 2.4 2.3 1.9 1.5 1.0 0.4 0.1 0.0 0.0
It is also of interest to compare the upper bound for the stress intensity factor defined by (4.10.14) with the upper bound derived by Ioakimidis (1982). His result for two equal cracks reads in our notation K K = 1 −
0
(4.10.24) 3
2a 3π( l − 2 a )
If we expand (4.10.13) in power series, retaining the first term only, the result is V0
V = 1 −
(4.10.25) 3
2a 3π( l − a )
Comparison of (4.10.24) with (4.10.25) explains why our estimation is sharper: we have in the denominator l − a while Ioakimidis has l −2 a . Here is a numerical example. For l =3 a the exact result reads K =1.0234 K 0. Our upper bound gives K =1.127 K 0 with an error of 10%, while the result of (4.10.24) is K =1.269 K 0, with an error of about 25%. Collins (1963) gave the following expression for the crack energy in the case of an infinite row of equal cracks
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CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
W = W (1 + 0.5102ε3 + 0.7921ε5 + 0.2603ε6 + 1.6507ε7 + 0.8083ε8 + ...) 0
(4.10.26) Our expression due to (4.10.18) is W W = ∞
1 −
Σ(k
4 π
k=1
2
0
(4.10.27)
ε -1 ε 2 1/2 − sin k − ε)
The value of W/W , computed on the basis of (4.10.26) and (4.10.27), can be 0 found in (Fabrikant, 1987g). Our (4.10.27) does not differ from (4.10.26) by more than 2% in the whole range of 2.001 a ≤ l <∞. Of course, this does not mean that the central estimation is so accurate; it means only that our simple approximate solution is almost as accurate as a very complicated one by Collins. We are not aware of any result in the mechanics literature to compare with our results for the polygonal configuration. Some well known results can be simplified significantly by using the method of computation of various integrals involving distances between several points. For example, here is the set of governing integral equations derived by Panasiuk et al (1986, p. 83) for the problem of interaction of N coplanar thin spheroidal inclusions: N
1 w (x , y ) + HΓw (x , y ) − n n n n n n 4π2
Σ k=1
⌠ ⌠ w ( x , y )Γ[( x − x )2 n k ⌡⌡ k k k Sk
k≠n
+ ( y n − y k)2]-3/2 d x kd y k = f n( x n, y n),
for n = 1,2, ...
, N.
(4.10.28) Here f n is a known function, w n is the normal displacement at the boundary of
the n th inclusion, S n is its median crossection, ( x n, y n)∈ S n and the integral operator Γ is defined by
ΓΦ( x k, y k) = ⌠ ⌠
Φ(ξ,η) dξdη
⌡ ⌡ [( x k − ξ)2 + ( y k − η)2]1/2 Sk
291
4.10 Close interaction of pressurized coplanar circular cracks
−
1 dξ dη 2 2 2 2 π ⌠ ⌠ {[ξ + η − a k][( x k − ξ)2 + ( y k − η)2]}1/2
⌡⌡ Sk
×⌠ ⌠
Φ(ξ1,η1)( a 2k − ξ21 − η21)1/2
⌡ ⌡ (ξ − ξ )2 + (η − η )2 1 1
dξ1dη1.
(4.10.29)
Sk
Here a k is the radius of median crossection of the k th inclusion, and S k denotes the area outside S k.
The double and quadruple integrals in (4.10.29), which is a
kernel of the integral equation (4.10.28), make any numerical solution next to impossible. Panasiuk et al. (1986) have managed to give an approximate solution for the case when the inclusions are far apart, which is of little practical value, since there is almost no interaction at such distances between the inclusions. Let us show that (4.10.28) can be simplified so that its kernel be presented in elementary functions. Making use of the integral 2π ∞
( a 2 − ρ20)1/2
1
r d r dψ
⌠ ⌠ 2 ⌡ ⌡ ( r − a 2)1/2 ρ20 + r2 − 2ρ0 rcos(φ0−ψ) [ρ2 + r2 − 2ρrcos(φ−ψ)]1/2 0 a
( a 2 − ρ2)1/2( a 2 − ρ20)1/2 2π π -1 , − tan = R 2 aR one can change the order of integration in the second integral of (4.10.29), perform the integration in S k, and the integral operator Γ simplifies in polar coordinates significantly, namely, ΓΦ(ρ,φ) =
η 2 ⌠⌠ 1 tan-1( ) Φ(ρ0,φ0) ρ0dρ0dφ0, π ⌡⌡ R R Sk
which is much simpler than (4.10.29). It is reminded that R and η are defined by (4.1.14). We can also compute Γ[( x n − x k)2 + ( y n − y k)2]-3/2 in elementary functions.
Indeed, one may obtain from (1.6.19)
292
CHAPTER 4
2π a
APPLICATIONS IN FRACTURE MECHANICS
ρ dρ dφ
0 0 0 ⌠ ⌠ 1 tan-1(η ) R [ρ20 + r 2 − 2 r ρ cos(φ −ψ)]3/2 ⌡ ⌡ R 0 0 0 0
=
2π( a 2 − ρ2)1/2 , ( r 2 − a 2)1/2[ r 2 + ρ2 − 2 r ρcos(φ−ψ)]
for r > a .
The above results simplify (4.10.28) so significantly that now it can be easily solved by iteration numerically or analytically.
1.
Exercise 4.10 Derive (4.10.4).
2. Rederive (4.10.4) by using an alternative approach: the results of section 4.2 combined with the reciprocity theorem. 3. Investigate the convergence of an iterative process, applied to the integral equation (4.10.16). 4.
Consider the interaction of a crack with a microcrack.
5. Consider the interaction of two cracks due to the bending of an elastic space. Assume the space stretched so that cracks do not close due to the bending.
4.11 Close interaction of coplanar circular cracks under shear loading In this section, the general method is applied to the stress analysis of an elastic space weakened by several arbitrarily located coplanar circular cracks subjected to an arbitrary shear loading. The problem is reduced to a set of Fredholm integral equations. The number of equations is equal to the number of cracks, and can be reduced in the case of a symmetrical configuration. The equations are non-singular. It is shown that the iteration procedure is convergent, and the convergence is so rapid that a practically exact numerical solution can be obtained even for very closely located cracks. One can get an approximate analytical solution without having solved the integral equations. It provides sufficiently accurate estimations for the quantities of interest, like the stress intensity factor, the crack energy, the crack face displacement, etc. The cases of two cracks and an infinite row of equal cracks are considered as illustrative
293
4.11 Close interaction of coplanar circular cracks
examples. Theory. Consider an elastic space weakened in the plane z =0 by n arbitrarily located circular cracks. The cracks do not intersect. Let the centre of the k th crack be located at the point with Cartesian coordinates x and y , k
and
its
radius
be
denoted
by
a k.
We
introduce
the
complex
displacement u = u x+ iu y, and the complex shear stress τ=τzx+ i τyz.
k
tangential
Let an arbitrary
skew-symmetric shear traction τk be applied to the k -th crack faces.
We can
single out, without loss of generality, the crack number 1, and consider the set of cracks in the local polar system of coordinates with the origin coinciding with the centre of the first crack. In order to be able to use the reciprocal theorem, we need to consider the second set of tractions applied to the same crack configuration. We apply two unit concentrated forces T to both faces of the x
first crack in opposite directions at the point with polar coordinates (ρ,φ), and parallel to the axis Ox . We also apply shear tractions q and q to the kx ky remaining cracks. These tractions are chosen so as to provide zero displacement discontinuity at the crack faces, so that the whole system would behave as a single crack (number one) in an infinite body. This choice will allow us to use the Green’s functions for an isolated circular crack derived in section 4.4. The following integral equation can be obtained by using the reciprocal theorem: n
u 1x +
Σ k=2
⌠ ⌠ q u dS + ⌡ ⌡ kx kx k Sk
n
Σ k=2
⌠ ⌠ q u d S = ⌠ ⌠ (τ u + τ u )d S . 1y yT 1 ⌡ ⌡ ky ky k ⌡ ⌡ 1x xT x x Sk
S1
(4.11.1) Here q kx and q ky stand for the shear tractions in the k th crack domain due to a pair of unit concentrated forces applied at an arbitrary point of the first crack in the direction parallel to the Ox axis; u and u are the tangential xT yT x
x
displacements of the first crack face due to the same unit forces; and u 1x, u kx, and u ky are the as yet unknown tangential displacements of the first and the k th crack faces respectively. Similar considerations, with the unit concentrated forces T y applied parallel to the Oy direction, yield the second integral equation n
u 1y +
Σ k=2
⌠ ⌠ s u dS + ⌡ ⌡ kx kx k Sk
n
Σ ⌠⌡ ⌠⌡ s k=2
Sk
u d S k = ⌠ ⌠ (τ1x u xT
ky ky
⌡⌡ S1
y
+ τ u )d S . 1y yT 1 y
(4.11.2) The meaning of the notation in (4.11.2) is similar to that in (4.11.1). All the integrals in (4.11.1) and (4.11.2) are evaluated on one side of the relevant crack.
294
CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
Now we need the explicit expressions for the quantities q kx, q ky, s kx, s ky, u xT , x
u yT , u xT , u yT . x
y
These expressions were derived in section 4.4, and are
y
q kx= −ζ − ℜ Z , u xT u xT
x
y
q ky = −ℑ Z = s kx ,
= ζ1 − ℜ Z 1 + ζ2 + ℜ Z 2, = ℑZ1 + ℑZ2 ,
u yT
u yT
s ky = −ζ + ℜ Z ,
x
(4.11.3)
= −ℑ Z 1 + ℑ Z 2 ,
= ζ1 −ℜ Z 1 − ζ2 − ℜ Z 2 ,
y
(4.11.4)
where ℜ and ℑ stand for the real and the imaginary part, and ζ =
ζ1 =
ζ2 =
( a 21 − ρ2)1/2 π2(ρ20 − a 21)1/2 R 2 G1 πR G 2ξ πR
-1
tan
,
η( a 1) R
-1
tan
η( a 1) R
Z =
,
iφ
G 1 π2(ρ20 − a 21)1/2
Z1 =
-i φ
e 0(3ρ0e
( a 21 − ρ2)1/2
G2
πG1
ρ0(ρe
Z2 =
− ρ0e
, 0 2
)
,
a 21(1 − t 1)2
2 iφ
,
-i φ
− ρe ) -i φ
-i φ
G 22 (3 − t 1) η( a 1)
0
G 2 η( a 1)(ξ − t 1e
0
)
π a 21 (1 − t 1)(1 − t 1)
.
(4.11.5)
Here R =[ρ2 + ρ20 − 2ρρ0cos(φ−φ0)]1/2, t 1=(ρρ0/ a 21)e
i(φ-φ0)
η( x ) = ( x 2 − ρ2)1/2( x 2 − ρ20)1/2/ x , iφ
ξ = (ρe
,
iφ
-i φ
− ρ0e 0)/(ρe
-i φ
− ρ0e
(4.11.6) 0
).
(4.11.7)
We multiply equation (4.11.2) by the imaginary unit i , add the result to (4.11.1) and, after substitution of (4.11.3), (4.11.4) and (4.11.5), obtain
u 1(ρ,φ) =
G 1 2π
a
⌠ ⌠ 0 ⌡
π⌡
0
1
1 tan-1 R
η( a 1) R
−
G 22 (3 − t 1) η( a 1) G 21
a 21(1
τ (ρ ,φ )ρ dρ dφ − t 1) 1 0 0 0 0 0 2
295
4.11 Close interaction of coplanar circular cracks
+
G 2 2π
a
⌠ ⌠ ⌡ 0
1
π⌡
0
+
( a 21−ρ2)1/2 π2
2 iφ
η( a 1)
η( a 1) (ξ − t 1e
0
)
ξ tan-1 + τ ( ρ , φ ) ρ dρ dφ 2 R R a 1(1 − t 1)(1 − t 1) 1 0 0 0 0 0 n
Σ k=2
-i φ
u k(ρ0,φ0)
⌠⌠ ⌡⌡ Sk
G 2 u k(ρ0,φ0) (3ρ0e
+
R2
0
-i φ − ρe )e -i φ
G1
-i φ ρ0(ρe − ρ0e
iφ
0 2
)
0
ρ dρ dφ
0 0 0. (ρ20−a 21)1/2
(4.11.8) The first two integrals in (4.11.8), though looking formidable, can be computed exactly and expressed in elementary functions for any polynomial loading. They give the tangential displacements of the first crack, as if it were an isolated crack, under the prescribed loading τ . The remaining integrals represent the 1
influence of the other cracks. A similar procedure can be applied to the remaining n −1 cracks, and the additional n −1 equations of the type (4.11.8) can be obtained. Note that each such equation is valid in a local system of polar coordinates related to a certain crack. The equations are non-singular. They can be solved numerically by iteration. As will be shown further, the convergence is so rapid, that the first iteration has an error less than 2.5% even for a very close interaction when two cracks are separated only by 0.01 of their radius. In the case of a uniform shear loading τ=τ = const, the set of equations 0 (4.11.8) simplifies as follows u 1 = 2τ0
G 21 − G 22 G1
(a
2
− ρ)
2 1/2
-i φ
+
G 2 u k(ρ0,φ0) (3ρ0e G1
-i φ
ρ0(ρe
+
0
( a 21 − ρ2)1/2 π2 iφ
− ρ0e
0 2
)
Σ k=2
− ρe-i φ) e -i φ
n
0
u k(ρ0,φ0)
⌠⌠ ⌡⌡ Sk
R2
ρ dρ dφ
0 0 0 2 2 1/2 . (ρ0 − a 1)
(4.11.9)
In some cases we can obtain a sufficiently accurate analytical solution of the set (4.11.8) by applying the mean value theorem. The result is:
296
CHAPTER 4
G u (ρ,φ) =
2π a
1
⌠ ⌠
η( a 1)
1
G 22 (3 − t 1) η( a 1)
1 tan-1 − R R G 21
π ⌡ 0 ⌡
1
APPLICATIONS IN FRACTURE MECHANICS
τ (ρ ,φ )ρ dρ dφ − t 1) 1 0 0 0 0 0
a 21(1
2
0
2π a
G2
⌠ ⌠ + π ⌡ 0 ⌡
1
ξ tan-1 R
η( a 1) R
0
n
+
( a 21 − ρ2)1/2
Σπ (ρ 2
k=2
2 k
−
a 21)1/2
2 iφ
+
η( a 1) (ξ − t 1e
0
)
τ (ρ ,φ ) ρ dρ dφ 1 0 0 0 0 0
a 21(1 − t 1)(1 − t 1)
-i φ
Uk
G 2 U k(3ρke
+ G1 ρ2 + ρ2k − 2 ρ ρkcos(φ − φ k)
k
− ρe-i φ)e -i φ
ρk(ρe-i φ − ρke
iφ
k 2
)
k
. (4.11.10)
Here U k = ⌠ ⌠ u kdSk ,
⌡⌡ Sk
and ρ and φ are the polar coordinates of a point inside S . k k k
Though the exact
location of the point is unknown, the fact of belonging to the domain S limits k the possible variation of the quantities of interest and allows the construction of upper and lower bounds as well as a sufficiently accurate central estimation which corresponds to the assumption of ρk and φk being located at the centre of the k th crack. The symmetry considerations can also be used to sharpen the estimates. It will be shown further that the central estimation provides in some cases a sufficiently accurate solution to the problem. The value of U 1 can be estimated by integration of (4.11.8) over the domain S . 1
The result is
G 21 − G 22 2π 1 ⌠ ⌠ τ(ρ,φ)( a 2 − ρ2)1/2ρdρdφ + 2 U1 = 2 1 G1 ⌡ π a
0
⌡ 0
a1 ⌠ ⌠ ⌡ ⌡ (ρ2 − a 21)1/2 k=2 n
Σ
Sk
297
4.11 Close interaction of coplanar circular cracks
a − sin-1
1
u (ρ,φ) + k
ρ
G G
2
a 31
1
u (ρ,φ)e2 i φ k ρ2(ρ2 − a 21)1/2
ρdρdφ.
(4.11.11)
We can use again the mean value theorem, with the result for the central estimation
U1 = 2
2π a
G 21 − G 22 G
⌠ ⌠ ⌡ ⌡ 0
1
1
τ(ρ,φ)( a 21 − ρ2)1/2ρdρdφ
0
2 iφ n
+
2 π
Σ k=2
a1
2G
a1
n
Σ
U ke
1k
U + . − sin-1 a3 l 1k k πG1 1 ( l 21k − a 21)1/2 l 21k( l 21k − a 21)1/2 k=2 2
(4.11.12) Here ( l ,φ ) are the polar coordinates of the k th crack centre, with respect to 1k 1k the system of coordinates having its origin at the centre of the first crack. Integration of the remaining n -1 equations over the area of each crack provides finally a system of n linear algebraic equations which can be solved with respect to the unknowns U k. Their feeding back into (4.11.10) gives the complete solution to the problem. Define the stress intensity factor at the edge of the first crack as K (φ) = 1
lim {(ρ − a )1/2τ (ρ,φ)}. 1 1 ρ→ a 1
An important feature of the present method is the possibility to compute the stress intensity factor directly through the displacements (see the derivation of (2.8.46) for details):
K 1(φ) = −
a1 π( G 21
−
G 22)√2 a 1
G 1 u 1(ρ,φ) + G 2e2 i φ u 1(ρ,φ)
lim ρ→ a 1
( a 21
− ρ)
2 1/2
.
(4.11.13)
The stress intensity factor for the remaining cracks can be defined in a similar manner. Substitution of (4.11.8) in (4.11.13) yields, after simplification,
298
CHAPTER 4
2π a
1 K 1(φ) = − 2 ⌠ ⌠ π √2 a 1 ⌡ 0 0⌡ G 2 ei φ
2π a
⌠ ⌠ G1 a1 ⌡ 0 ⌡
+
1
1
( a 21 − ρ20)1/2 τ(ρ0,φ0) ρ0dρ0dφ0 a 21 + ρ20 − 2 a 1ρ0cos(φ − φ0)
-i φ
a1
−
-i φ
-i φ
− ρ0e
3 a 1e [ a 1e
0
APPLICATIONS IN FRACTURE MECHANICS
n
π3( G 21 − G 22) √2 a 1
-i φ
− ρ0e
Σ k=2
0
0 2
( a 21 − ρ20)1/2 τ(ρ0,φ0) ρ0dρ0dφ0
]
G G ⌠ ⌠ 1 + 2 Θ e2 i φ u (ρ ,φ ) G1 1 k 0 0 ⌡ ⌡ R 21 Sk 2
2 iφ ρ0dρ0dφ0 e + G2 + Θ1 u k(ρ0,φ0) . R 21 (ρ20 − a 21)1/2
(4.11.14)
Here -i φ
Θ1 =
3ρ0e -i φ
ρ0e
0
0
-i φ
− a 1e
-i φ
( a 1e
-i φ
− ρ0e
R 1 = [ a 21 + ρ20 − 2ρ0 a 1cos(φ − φ0)]1/2.
, 0 2
)
The first two integrals in (4.11.14) give the stress intensity factor for an isolated crack, while the remaining integrals represent the influence of the other cracks. We shall see next how these general expressions can be applied to some specific problems. Example: Two cracks. Consider the case of two coplanar circular cracks with radii a 1 and a 2 under the action of a uniform shear loading τ1 and τ2 respectively. Let l be the distance between their centres. As was established in the previous Section, the problem is reduced to evaluating the integral characteristics U 1 and U 2. We consider the central estimation only. Equations (4.11.12) in this case will take the form
U1 =
4 3 πa τ 3 1 1
G 21 − G 22 G1
+
a1
a1
2G2
a 31 U 2
2 + − sin-1 U , π 2( l − a 21)1/2 π G 1 l 2( l 2 − a 21)1/2 l
299
4.11 Close interaction of coplanar circular cracks
G 21 − G 22
a2
a2
2G2
a 32 U 1
2 + U − sin-1 . l π 1( l − a 22)1/2 π G 1 l 2( l 2 − a 22)1/2 (4.11.15) Strictly speaking, the mean value theorem is not applicable in this case, since the imaginary part of u does change sign inside the crack. A numerical evidence will be presented later which justifies neglect of the imaginary part. Under this assumption, the solution is
U2 =
4 3 πa τ 3 2 2
G1
U1 =
4 π 3
4 U2 = π 3
+
3 3 G 21 − G 22 a 1 τ1 + c 12 a 2 τ2
1 − c c 12 21
G1
3 3 G 21 − G 22 c 21 a 1 τ1 + a 2 τ2
1 − c c 12 21
G1
,
,
(4.11.16)
where
c 12
a1 a1 G2 a 31 2 , = + − sin-1 π ( l 2 − a 21)1/2 l G 1 l 2( l 2 − a 21)1/2
c 21
a2 a2 G2 a 32 2 . = + − sin-1 π ( l 2 − a 22)1/2 l G 1 l 2( l 2 − a 22)1/2
(4.11.17)
It will be shown that the central estimation gives a sufficiently accurate solution even for relatively close crack interactions. Formulae (4.11.16−4.11.17) simplify in the case of equal cracks as a = a = a , 1 2
and if τ =τ =τ then 1 2 0
U U = U =U = 1
2
0
1 − c
,
G2 a3 2 a -1 a . + c = − sin 2 2 1/2 2 2 2 1/2 π ( l − a ) l G1 l (l − a )
(4.11.18)
300
CHAPTER 4
Here U 0=(4/3)π a 3τ0( G 21− G 22)/ G 1. crack energy W
APPLICATIONS IN FRACTURE MECHANICS
Note that in the case of a uniform loading, the
is proportional to U , namely, W =τ0 U .
It is clear from
(4.11.18) that the crack interaction increases their energy when the applied loadings act in the same direction, otherwise, their energy decreases. The crack face displacements will take the form, according to (4.11.9) and (4.11.10), u 1(ρ,φ) = ( a 21 − ρ2)1/2
+
u 2(ρ,φ) =
U2 π2( l 2 − a 21)1/2
( a 22
+
G 21 − G 22 2τ1 G1
− ρ)
2 1/2
G 2 3 l − ρe-i φ 1 , + 2 2 -i φ 2 G 1 l ( l − ρe ) ρ + l − 2ρl cosφ
G 21 − G 22 2τ2 G1
U1 π2( l 2 − a 22)1/2
G 2 3 l + ρe-i φ 1 . + 2 2 φ -i 2 G ρ + l + 2ρl cosφ 1 l ( l + ρe )
(4.11.19)
We recall that each expression in (4.11.19) is valid in a local system of polar coordinates, with the coordinate origin located at the centre of the respective crack. Substitution of (4.11.19) into (4.11.13) yields the expression for the stress intensity factor. Now we need an accurate numerical solution in order to estimate the accuracy of the approximate formulae derived. For the sake of simplicity, consider the case of two equal cracks. Assume the crack face displacements in the form u (ρ,φ) = 2τ0
G 21 − G 22
where f is an as yet interaction function since displacements to those of the stress intensity factor
G1
( a 2 − ρ2)1/2 f (ρ,φ),
(4.11.20)
unknown complex function. It may be called the it is equal to the ratio of the interacting crack face an isolated crack. The values of f(a , φ) are related to of interacting cracks through
301
4.11 Close interaction of coplanar circular cracks
√2 a [ G τ f ( a ,φ) + G τ f ( a ,φ)e2 i φ]. 1 0 2 0 πG
K (φ) = −
(4.11.21)
1
In the case when τ
is a real constant, we can neglect the imaginary part of f ,
0
and the stress intensity factor can be expressed as follows: K (φ) = K (φ) f ( a ,φ),
(4.11.22)
0
where K (φ) = − 0
√2 a τ [ G + G e2 i φ] 2 πG1 0 1
is the stress intensity factor for an isolated crack. We shall call f(a ,φ) the interaction factor , since its value is approximately equal to the ratio of the stress intensity factor of an interacting crack to that of an isolated crack. Substitution of (4.11.20) into (4.11.9) gives a convenient expression for the procedure of iteration u (ρ,φ) = 2τ
G 21 − G 22 G
0
2π a
( a 2 − ρ2)1/2 1
1
f(r , ψ ) 0 0 1⌠ ⌠ + 2 π ⌡ ⌡ ( l2 + r20 + 2lr cosψ0 − a 2)1/2 r2 + r20 +2rr cos(ψ − ψ0) 0
( a 2 − r 20)1/2 r 0d r 0dψ0
0
0
-i ψ
G f ( r ,ψ ) (2 l + r e +
G
2
0
1
( r 20
0
+ l
0
2
0
0
iψ
− r e-i ψ)( l + r 0e -i ψ
+ 2 lr cosψ ) ( r e 0 0 0
0
0
-i ψ 2
− re
)
)
.
(4.11.23)
r =(ρ2+ l 2−2 l ρcosφ)1/2, Here we have introduced the new variables ψ=π−sin-1[(ρ/ r )sinφ]. The integral in (4.11.23) has a logarithmic singularity for r = l − a , ψ=0, as l →2 a , therefore the procedure of iteration might not be convergent for l very close to 2 a . Direct computations show that the iteration procedure converges for l =2.01 a ( which corresponds to the case when the shortest distance between the cracks is equal to 0.01 of its radius), and converges rapidly: the first iteration with f ≡1 has the maximum relative error less than 2.5%, and the sixth iteration may be considered practically as an exact solution,
302
CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
since the error becomes less than 10-7. The accuracy of the first iteration improves as the distance between cracks increases. For example, the first iteration for l =10 is practically exact with maximum relative error less than 10-7. We could not go closer than l =2.01 a , not because of non-convergence, but because the standard subroutine DBLIN from the IMSL library, which was used to compute the integrals, failed, giving terminal errors. Though we do not have a rigorous proof, it seems probable that the iteration procedure is theoretically convergent for arbitrarily small distance between cracks. In the case when G =0 (for an isotropic body this condition is equivalent 2
to the Poisson ratio ν=0), the crack interaction due to a shear loading is the same as the crack interaction due to a normal loading (compare (4.1.9) with (4.4.14)). The maximum value of the ratio G / G for an isotropic body is 1/3, 2
1
and this value was taken in all the numerical computations, in order to expose the maximum possible difference between the crack interaction due to a normal loading, and the crack interaction due to a shear loading. Some values of the interaction function f (ρ,φ) are presented in Tables 4.11.1 and 4.11.2, for the closest interaction considered, corresponding to l/a =2.01. The reader is referred to the original paper (Fabrikant, 1989) for the complete data.
Table 4.11.1. ρ
φ 1.00 0.75 0.50 0.25 0.00
0
15
30
45
90
135
180
2.14189 1.32715 1.19220 1.13108 1.09668
1.52935 1.28240 1.18283 1.12916 1.09668
1.24944 1.20537 1.16050 1.12389 1.09668
1.14303 1.14660 1.13539 1.11649 1.09668
1.06042 1.07271 1.08451 1.09335 1.09668
1.04512 1.05393 1.06514 1.07930 1.09668
1.04195 1.04976 1.06027 1.07499 1.09668
Table 4.11.2.
ρ
φ 1.00 0.75 0.50 0.25 0.00
The interaction function (real part) for l =2.01 a
The interaction function (imaginary part) for l =2.01 a ]
0
15
30
45
90
135
180
0.00000 0.00000 0.00000 0.00000 0.00000
-0.11448 -0.04250 -0.01526 -0.00458 0.00000
-0.09508 -0.05420 -0.02474 -0.00834 0.00000
-0.06835 -0.04922 -0.02774 -0.01082 0.00000
-0.02738 -0.02518 -0.01983 -0.01107 0.00000
-0.01092 -0.01053 -0.00918 -0.00609 0.00000
0.00000 0.00000 0.00000 0.00000 0.00000
303
4.11 Close interaction of coplanar circular cracks
All the computations were made with a relative error not exceeding 10-6. The first line in Table 4.11.1 is approximately equal to the ratio of the stress intensity factor of interacting cracks to the stress intensity factor of an isolated crack under the same uniform load. The data in Table 4.11.2 justify our neglect of the imaginary part in the interaction function when deriving the approximate analytical solutions: the maximum value of the imaginary part is less than 8% of the corresponding real part, and it reduces significantly, when the distance between the cracks increases. The data in the tables are presented for 0≤φ≤π. The following rules should be applied if one is interested in the value of the interaction function for φ>π: ℜ f (ρ,φ)=ℜ f (ρ,2π−φ) and ℑ f (ρ,φ)=−ℑ f (ρ,2π−φ). Comparison with the results given in section 4.10 shows that the crack interaction due to a shear loading (when acting on both cracks in the same direction) is stronger than the interaction due to a normal loading. For example, the maximum value of the interaction factor for the case l =2.01 a and G /G =1/3 is 2
1
2.1419 (Table 4.11.1) while the corresponding value in the case of a normal loading is 1.8613. We can now assess the accuracy of the analytical solution given (4.11.18−4.11.19). We can compute the exact value of the crack energy W using the computed data and compare it with the approximate values due (4.11.18). The ratio W/W ( W is the energy of an isolated crack) is given 0
The approximate value was computed as W/W =1/(1− c ), where c 0
Table 4.11.3.
Table 4.11.3.
0
by by to in
The ratio W / W 0 due to exact and approximate solutions.
l/a
10.0
3.0
2.5
2.1
2.05
2.02
2.01
exact approx. error (%)
1.00043 1.00043 0.0
1.01940 1.01736 0.2
1.03817 1.03165 0.6
1.08367 1.05802 2.4
1.09621 1.06333 3.0
1.10603 1.06684 3.5
1.11000 1.06808 3.8
is defined by (4.11.18). The agreement is very good for l/a =≥2.5. Even for a very close interaction ( l/a =2.01) the relative error does not exceed 4%; of course, this is mainly due to the fact that the increase in the crack interaction energy is rather small. This should be attributed to a sharp localization of the interaction effects (see Table 4.11.1). The analytical expression for according to (4.11.19), in the form f (ρ,φ) = 1 +
G1
the
U
2τ ( G 21 − G 22) π2( l 2 − a 2)1/2 0
interaction
function
1 2 2 ρ + l − 2ρl cosφ
can
be
written,
304
CHAPTER 4
+
APPLICATIONS IN FRACTURE MECHANICS
3 l − ρe-i φ
. G 1 l ( l − ρe-i φ)2 G2
We have computed only the interaction factor f ( a ,φ) due to the last formula and compared it with the exact values in Table 4.11.4. The relative error of the
Table 4.11.4.
l/a
φ(deg.)=
Comparison of exact and approximate solutions for the interaction factor.
0
30
60
90
120
150
180
Real exact approximate 2.50 error (%) Imag. exact approximate error (%)
1.11020 1.07927 2.8 0.00000 0.00000 0.0
1.07479 1.05793 1.6 -0.01875 -0.01439 23.3
1.04136 1.03465 0.6 -0.01485 -0.01210 18.5
1.02764 1.02393 0.4 -0.00936 -0.00782 16.5
1.02198 1.01925 0.3 -0.00547 -0.00461 15.7
1.01959 1.01722 0.2 -0.00254 -0.00215 15.4
1.01892 1.01664 0.2 0.00000 0.00000 0.0
Real exact approximate 2.05 error (%) Imag. exact approximate error (%)
1.63484 1.21014 26.0 0.00000 0.00000 0.0
1.21826 1.12652 7.5 -0.07834 -0.04268 45.5
1.08826 1.06405 2.2 -0.04299 -0.02834 34.1
1.05542 1.04226 1.2 -0.02416 -0.01667 31.0
1.04397 1.03379 1.0 -0.01359 -0.00951 30.0
1.03945 1.03029 0.9 -0.00622 -0.00437 29.7
1.03821 1.02931 0.9 0.00000 0.00000 0.0
central estimation of the real part of the interaction factor does not exceed 3% for l >2.5 a . Though the relative error of the imaginary part is large, this is due to the fact that the imaginary part constitutes a small percentage of the real part; the absolute error is very small, and we can consider the analytical solution (4.11.18−4.11.19) sufficiently accurate when the distance between interacting cracks is not less than half of their radius. The accuracy of the central estimation deteriorates rapidly as l decreases. One can also notice that the central estimation is always slightly below the exact value thus giving a very close lower bound for the quantities of interest in the case of two interacting cracks. Infinite row of equal cracks. Let the crack radius be a , and the distance between the adjacent crack centres be l . The cracks are subjected to a uniform shear loading τ. The central estimation for the integral characteristic U can be defined, from (4.11.12), by a single equation
305
4.11 Close interaction of coplanar circular cracks
G 21 − G 22
4 U = π a 3τ 3
G
1
Σ
. G k2l2 ( k2l2 − a 2)1/2 1 G
+
a -1 a U 2 2 2 1/2 − sin kl (k l − a ) k=1 ∞
4 + π a3U
2
If we neglect the imaginary part of U then the solution is U 0 U =
,
G a3 2 a -1 a − sin + 22 kl G k2l2 ( k2l2 − a 2)1/2 ( k l − a 2)1/2 1 k=1 ∞
Σ
4 1 − π
where U =(4/3)π a 0
3
τ( G 21− G 22)/ G 1
(4.11.24) corresponds to the case of an isolated crack. The
crack face displacement will take the form, according to (4.11.9) and (4.11.10), u (ρ,φ) = ( a 2 − ρ2)1/2
G 21 − G 22 2τ G 1
G 3 k2l2 − ρ2e-2 i φ 2 1 2(ρ2 + k2l2) , + G ( k2l2 − ρ2e-2 i φ)2 ( k2l2 − a 2)1/2 ρ4 + k4l4 − 2ρ2 k2l2 cos2φ 1 k=1 ∞
U + 2 π
Σ
(4.11.25) and substitution of (4.11.25) in (4.11.13) will give the expression for the stress intensity factor. Discussion. It is of interest to compare our results with those available in the literature. We have found only one paper (Fu and Keer, 1969) where the problem of two interacting coplanar circular cracks was considered by a method similar to that of Collins (1963). Only the case when the distance between the crack centres l is much greater than the crack radius a (ε= a/l << 1) was considered. Fu and Keer considered in detail two equal cracks subjected to a uniform shear loading, acting on both cracks in the same direction horizontally (Case a ), and acting in opposite directions (Case b ). There are several points in their solution which seem to be incorrect. One of the results (Fu and Keer, 1969, p. 371) states that the absolute value of the m th harmonic ( m =1,2,3, ...) of the vertical displacement u is equal to the corresponding harmonic of the y
306
CHAPTER 4
horizontal displacements u x.
APPLICATIONS IN FRACTURE MECHANICS
This cannot be true, since the vertical displacements
depend on the ratio G2/G1 (in the isotropic case this ratio is equal to ν/(2−ν),
where ν is the Poisson coefficient), and the vertical displacements vanish when G 2=0 (ν=0)). It is also possible to compare the expressions for the increase in the strain energy of deformation W per crack. The expression, given by Fu and Keer (1969), reads in our notation: W = W 0 1
4 3 ε 3π
(
16 6 2 3 + ) ε5 + ε , 5π 5 9π2
(4.11.26)
where ε= a/l ; the sign corresponds to the cases (a) and (b) respectively, and 3 2 2 2 W =(4/3)π a τ ( G 1− G 2)/ G stands for the energy of an isolated crack. Note an 0
1
obvious misprint in (4.11.26): the plus sign should correspond to the case (a) and minus to the case (b). Our expression for the crack energy is
W =
W0 1 − c
,
(4.11.27)
where c is defined according to (4.11.18) as G2 a3 2 a -1 a . + − sin c = π ( l 2 − a 2)1/2 l G 1 l 2( l 2 − a 2)1/2
(4.11.28)
Series expansion of (4.11.27) results in G
G
2 2 2ε3 1 3 ε5 + ( + ) + W = W 0 1 + ( ) 5 π 3 π G1 G1
G2 G2 4ε6 1 3ε7 5 2 + 2 ( + + ) + ( ) + ... 3 4π 7 G1 G1 π
.
(4.11.29)
There is a definite disagreement between (4.11.26) and (4.11.29): each term in (4.11.29) depends on the ratio G2/G1 while each term in (4.11.26) is not dependent of the elastic constants. In the case of an isotropic body G2/G1 =ν/(2−ν) (as it should), and we can observe an agreement between (4.11.26)
and (4.11.29) only for ν=1/2 which is just a coincidence.
Collins (1963) gave
307
Appendix A4.1
the following expression for the case of two cracks subjected to a normal loading: W = W 0 1 +
32ε8 4ε6 18ε7 6ε5 2ε3 + + ... + + + 7π 5π 3π 15π2 9π2
.
(4.11.30) As was noticed before, in the case when G 2=0 (ν=0) the interaction of cracks subjected to a shear loading is mathematically equivalent to the interaction under a normal loading, which means that both (4.11.26) and (4.11.29) should be in agreement with (4.11.30) for ν=0. One can see that this is not the case for expression (4.11.26).
1.
Exercise 4.11 Derive (4.11.8).
2.
Establish (4.11.11).
3.
Investigate convergence of the procedure of iteration, applied to (4.11.8).
4.
Consider the case of a polygonal configuration of identical cracks.
Appendix A4.1 Here the main potential function is given, together with selected partial derivatives. We define the potential function by 2π a
Ψ = ⌠ ⌠ ( a 2 − ρ20)1/2 ln( R 0 + z ) ρ0dρ0dφ0 ,
⌡ ⌡ 0 0
where R 0=[ρ2 + ρ20 − 2ρρ0cos(φ−φ0) + z 2]1/2.
The integral can be computed in
elementary functions: Ψ =
π 2 1 10 2 a z (2 a 2 − ρ2 + z 2)sin-1( ) + ( a 2 − l 21)1/2(5ρ2 − a 2 3 3 3 l2 − 2 l 22 −
11 2 4 3 l) + a ln[ l 2 + ( l 22 − ρ2)1/2] 3 1 3
The following derivatives may be computed:
(A4.1.1)
308
CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
a ∂Ψ = π x − z sin-1 + ( a 2 − l 21)1/2(1 − l ∂x 2
l 21 + 2 a 2
a ∂Ψ = π y − z sin-1 + ( a 2 − l 21)1/2(1 − l ∂y 2
l 21 + 2 a 2
a ∂Ψ π = (2 a 2 + 2 z 2 − ρ2) sin-1 − 2 l ∂z 2
3ρ2
3ρ2 2 a 2 − 3 l 21 a
) +
2 a 3 3ρ2
(A4.1.2)
) +
2 a 3 3ρ2
(A4.1.3)
( l 22 − a 2)1/2
l 21 + 2 a 2 2 a 3 -1 a 2 2 1/2 + ( a − l 1) (1 − + ΛΨ = πρe − z sin ) l 3ρ2 3ρ2 2 iφ
(A4.1.4)
(A4.1.5)
a ∂2Ψ = π− z sin-1 + ( a 2 − l 21)1/2 l ∂x2 2 2x2 + (1 − 2 ) ρ
2 a 3 − ( l 21 + 2 a 2) ( a 2 − l 21)1/2 3ρ2
(A4.1.6)
(A4.1.7)
∂2Ψ -1 a + ( a 2 − l 21)1/2 2 = π − z sin l ∂y 2 2y2 + (1 − 2 ) ρ ∂2Ψ = −2π xy ∂x∂y 2 iφ
Λ2Ψ = −2πe
2 a 3 − ( l 21 + 2 a 2) ( a 2 − l 21)1/2 3ρ
2
2 a 3 − ( l 21 + 2 a 2) ( a 2 − l 21)1/2 3ρ4 2 a 3 − ( l 21 + 2 a 2) ( a 2 − l 21)1/2 3ρ2
a + ( a 2 − l 21)1/2 ∆Ψ = 2π − z sin-1 l 2
(A4.1.8)
(A4.1.9)
(A4.1.10)
309
Appendix A4.1
∂2Ψ zsin-1 a − ( a 2 − l 2)1/2 2 = 2π 1 l ∂z 2 a ∂ + ΛΨ = πρei φ −sin-1 l ∂z 2
(A4.1.11)
a ( l 22 − a 2)1/2 l 22
(A4.1.12)
1 + 2 a x l 22( l 22 − l 21)
(A4.1.13)
a ( l 22 − a 2)1/2 2a2y2 ∂Ψ -1 a = π −sin + 1 + 2 2 2 l l 2( l 2 − l 21) l ∂ y 2∂ z 2 2
(A4.1.14)
∂3Ψ -1 a = π −sin l + 2 ∂x ∂z 2
a ( l 22 − a 2)1/2 l 22
2 2
3
∂3Ψ -1 a = 2 π sin l − 3 ∂z 2
a ( l 22 − a 2)1/2
∂3Ψ = 2π ∂x∂y∂z ∂3Ψ = 2π ∂ x∂ z2 ∂3Ψ = 2π ∂ y∂ z2
(A4.1.15)
a 3 xy ( l 22 − a 2)1/2
(A4.1.16)
l 42( l 22 − l 21)
a 2 x ( a 2 − l 21)1/2
(A4.1.17)
l 22( l 22 − l 21) a 2 y ( a 2 − l 21)1/2
(A4.1.18)
l 22( l 22 − l 21)
∂ Λ2Ψ = 2πρ2e2 i φ ∂z
a 3( l 22 − a 2)1/2
(A4.1.19)
l 42( l 22 − l 21)
a ∂ + ∆Ψ = 2π −sin-1 l ∂z 2
a ( l 22 − a 2)1/2 l 22
( a 2 − l 21)1/2
∂ 2 iφ 2 ΛΨ = 2π a ρe l 22( l 22 − l 21) ∂z 2
l 22 − l 21
−
l 21
(A4.1.20)
(A4.1.21)
310
CHAPTER 4
iφ
Λ∆Ψ = − 2π a ρe 2
APPLICATIONS IN FRACTURE MECHANICS
( a 2 − l 21)1/2
(A4.1.22)
l 22( l 22 − l 21)
a l ( a 2 − l 21)1/2 4[( l 21 + 2 a 2)( a 2 − l 21)1/2 − 2 a 3] 1 + Λ3Ψ = −2πe3 i φ 3 3ρ l ( l 22 − l 21) 2 (A4.1.23) ∂4Ψ = −2π ∂ z4
za [ l 41
+ a (2 a + 2 z − 3ρ )] 2
2
2
(A4.1.24)
( l 22 − l 21)3( l 22 − a 2)1/2 l ( l 22 − a 2)1/2
1 ∂Ψ 3 = −2π l ( l 22 − l 21)3 ∂ρ∂ z 4
2
[ a 2(4 l 42 − 5ρ2) + l 41]
(A4.1.25)
2
∂2 2 2 2 iφ 3 a z 2 Λ Ψ = 2πρ e ∂z
a 2(6 l 22 − 2 l 21 + ρ2) − 5 l 42
(A4.1.26)
l 42( l 22 − l 21)3( l 22 − a 2)1/2
The following identities were used in the derivation of (A4.1.1−A4.1.26): l l
1 2
= a ρ,
l 21 + l 22 = a 2 + ρ2 + z 2,
(A4.1.27)
( l 22 − ρ2)1/2( l 22 − a 2)1/2 = z l ,
( a 2 − l 21)1/2(ρ2 − l 21)1/2 = z l ,
( a 2 − l 21)1/2( l 22 − a 2)1/2 = za ,
( l 22 − ρ2)1/2(ρ2 − l 21)1/2 = z ρ.
2
∂l
∂l
zl 1 = −2 , ∂z l 2 − l 21
∂l
1
1
∂ρ
al =
2
− ρl
l 22 − l 21
(A4.1.28) 2
∂z 1
=
1
zl =
l 22
ρ( a 2 − l 21) l ( l 22 − l 21) 1
2
− l 21
, ∂l
,
2
∂ρ
ρl =
2
− al
l 22 − l 21
1
=
ρ( l 22 − a 2) l ( l 22 − l 21)
.
2
(A4.1.29)
311
Appendix A4.2
Appendix A4.2 Here we present some indefinite integrals of expressions containing l 1 and l 2.
⌠ ( l 2 − a 2)1/2d z = ( a 2 − l 2)1/2 1 ⌡ 2
l 22 − 2 a 2
+
2a
⌠ ( l 2 − a 2)1/2 l 2d z = −a ( a 2 − l 2)1/2 1 1 ⌡ 2
ρ2 ln[ l 2 + ( l 22 − ρ2)1/2], 2
l 21 + 2 a 2
(A4.2.1)
+ a 2ρ2ln[ l 2 + ( l 22 − ρ2)1/2],
3
(A4.2.2)
⌠ ( a 2 − l 2)1/2d z = 1 ⌡ ⌠ ( a 2 − l 2)1/2 1 ⌡
2a − 2
l 21
2a
( l 22 − a 2)1/2 +
ρ2 a sin-1( ), 2 l2
(A4.2.3)
l 21(2 l 21 + 3ρ2) 3 a l 21d z = − ( l 22 − a 2)1/2 + ρ2( ρ2 − a 2) sin-1( ), 8a 8 l2 (A4.2.4) l 21
2
8a 2 ⌠ ( l 2 − a 2)1/2 − l 21)1/2 1 − − 2 2 dz = a(a ⌡ 15ρ2 l2
⌠ ⌡
( a 2 − l 21)1/2 l 22
−
l 21
( a 2 − l 21)1/2
⌠ ⌡ l 22( l 22
2
4a +
3 l 21
15 l 22
,
(A4.2.5)
a d z = −sin-1( ), l2
1 dz = 2 − l 1) 2a2
(A4.2.6)
a ( l 22 − a 2)1/2 l 22
a − sin-1( ) , l2
(A4.2.7)
⌠ sin-1( a ) d z = zsin-1( a ) − ( a 2 − l 2)1/2 + a ln[ l + ( l 2 − ρ2)1/2], 1 2 2 l2 l2 ⌡ ⌠ zsin-1( a ) d z = 1 (2 a 2 + 2 z2 + ρ2) sin-1( a ) + ( l 2 − a 2)1/2 2 4 l2 l2 ⌡
(A4.2.8)
2 a 2 + l 21 4a
, (A4.2.9)
⌠ z sin ( a ) d z = 1 z3sin-1( a ) + 1 ( a 2 − l 2)1/2(3 l 2 + 6ρ2 + 8 a 2 − 2 l 2) 1 2 1 3 18 l2 l2 ⌡ 2
-1
312
CHAPTER 4
−
APPLICATIONS IN FRACTURE MECHANICS
1 a (3ρ2 + 2 a 2) ln[ l 2 + ( l 22 − ρ2)1/2]. 6
(A4.2.10)
The integration in (A4.2.1−A4.2.10) was performed by parts, with a consequent change of variables: z =( a 2 − l 21)1/2(ρ2 − l 21)1/2/ l 1 or z =( l 22 − a 2)1/2( l 22 − ρ2)1/2/ l 2. z (2 a 2 − l 2)
1 ⌠ ρ sin-1( a ) dρ = ρ sin-1( a ) + , 2 2 l2 l2 2 ⌡ 2( a − l 1)1/2 2
(A4.2.11)
z ρ(2 a 2 − l 2)
1 ⌠ ρ2sin-1( a )dρ = ρ sin-1( a ) + 2 2 1/2 l2 l2 3 ⌡ 6( a − l 1) 3
l2 l1 1 1 2 2 -1 2 2 -1 − + a ( a − 3 z ) cosh 2 z (3 a − z )sin 2 , 6 6 ( a + z 2)1/2 ( a + z 2)1/2 (A4.2.12) ⌠ a sin-1( a ) d a = 1(2 a 2 + 2 z2 − ρ2)sin-1( a ) l2 4 l2 ⌡ + l 1(ρ2 − l 21)1/2 − 2 z ( a 2 − l 21)1/2.
(A4.2.13)
The integration in (A4.2.11−A4.2.12) was performed by parts, with a consequent change of variables: ρ= y [1+ z 2/( a 2− y 2)]1/2, which corresponds to the substitution l 2= y . A similar remark is valid for formula (A4.2.13).
Appendix A4.3 Two important integrals are computed here. 2π a
iφ
iψ
1 ⌠ ⌠ ρe − re 2π ⌡ ⌡ R 3( M,N) 0
0
tan -1
The first integral is
( a 2 − r 2)1/2( a 2 − ρ20)1/2 a R ( N,N0)
r d r dψ . R ( N,N0)
(A4.3.1)
The integral (A4.3.1) can be computed indirectly by using (40), which leads to an equivalent expression:
313
Appendix A4.3
z
1 h d z ⌠ Λ tan-1 R ( M,N ) R ( M,N ) ⌡ 0 0
(A4.3.2)
∞
Let us make use of the following identities: φ
h ρei , l 22 − l 21
Λh = − Λ
tan-1
h R ( M,N0)
1
R ( M,N0) iφ
ρe = −
iφ
− ρ0e
0
iφ
h h ρe tan-1 − 2 + R 0 R 0 + h 2 l 22 − l 21
R 30
The notation R
0
(A4.3.3)
iφ
ρe
iφ
− ρ0e
0
.
R 20
(A4.3.4) in this Appendix is used as a contraction for R ( M,N ). The 0
substitution of (A4.3.4) in (A4.3.2) yields, after integration by parts: z
h 1 d z ⌠ Λ tan-1 R M,N R M,N ( ) ( ) ⌡ 0 0 ∞
z
1
= − ρe-i
z
− ⌠
φ
-i φ
− ρ0e
0
2 2 ⌡ l2 − l1
∞
ρei
iφ
ρe
z tan-1 h − ⌠ z ∂ tan-1 h d z R R R0 ⌡ R 0 ∂ z 0 0
+
∞
φ
iφ
− ρ0e R 20
0
hdz . R 20 + h 2
(A4.3.5)
The following identities are to be used now: h (ρ2 − l 21) ∂h = , ∂z z ( l 22 − l 21)
(A4.3.6)
314
CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
hR 0 ρ2 − l 21 h z2 ∂ -1 tan = . − 2 ∂z R 0 z ( R 20 + h 2) l 22 − l 21 R0
(A4.3.7)
The substitution of (A4.3.7) in (A4.3.5) allows us to proceed: 1
−
-i φ
-i φ
ρe
− ρ0e
z
iφ
0
iφ
ρe
ρe
− ⌠
z ρ2 − l 21 z h h z2 -1 ⌠ d − z R tan R − R 20 + h 2 l 22 − l 21 R 20 ⌡ 0 0 ∞
+
2 2 ⌡ l2 − l1
∞
z
h dz
+ ⌠
iφ
− ρ0e
z
hdz = − R 20 + h 2
R 20
ρ2 − l 21
2 2 2 2 ⌡ ( R 0 + h )( l 2 − l 1) ∞
0
-i φ
ρe
-i φ
− ρe
iφ
− ρe 0
h tan-1 R 0
z -i φ
(ρe
-i φ
− ρe 0
0
)R
0
0
iφ z2 iφ 0 − ⌠ ρ − ρ + e e 2 2 2 φ -i 0 ⌡ R 0( R 0 + h ) ρe-iφ − ρ e 0 ∞ 0
hdz
i(φ-φ )
=
1 -i φ
ρe
-i φ
− ρe 0
0
z ρρ e 0 − l 21 z h hdz 0 -1 ⌠ − − tan + 1 R R 0 R 20 + h 2 l 22 − l 21 ⌡ 0 ∞ i(φ-φ )
=
1 -i φ
ρe
-i φ
− ρe 0
0
z ρρ e 0 − l 22 z h 0 -1 h d z − R tan R + ⌠ R 20 + h 2 l 22 − l 21 ⌡ 0 0 ∞
(A4.3.8) Taking into consideration the identity i(φ-φ0)
R 20 + h 2 = ( l 22 − ρρ e 0
-i(φ-φ0)
) ( l 22 − ρρ e 0
)/ l 22 ,
the integral in (A4.3.8) can be transformed as follows:
(A4.3.9)
315
Appendix A4.3
( a 2 − ρ20)1/2
h l 22 d z
⌠ = ⌡ ( l 2 − l 2)( l 2 − ρρ e-i(φ-φ0) ) 2 1 2 0
a
( a 2 − l 21)1/2 l 22 d l 2
⌠ ⌡ z l ( l 2 − ρρ e-i(φ-φ0) ) 2 2 0
l dl
= (a
−
2
ρ20)1/2
2 2 ⌠ = ⌡ ( l 2 − a 2)1/2( l 2 − ρρ e-i(φ-φ0) ) 2 2 0
( a 2 − ρ20)1/2
-1
tan
s
( l 22 − a 2)1/2 s
,
(A4.3.10) -i(φ-φ0) 1/2
where s =( a 2 − ρρ e
) .
0
Finally, formulae (A4.3.8) and (A4.3.10) allow us
to compute the original integral (A4.3.1): 2π a
φ
i i 1 ρe − r e ⌠ ⌠ 2π ⌡ ⌡ R 3( M,N ) 0
-i ρe
It
is
φ
-i φ
− ρe
( a 2 − r 2)1/2( a 2 − ρ20)1/2 a R ( N,N ) 0
0
( a 2 − ρ20)1/2 s
tan-1
s ( l 22 − a 2)1/2
−
r d r dψ R ( N,N0)
z h . tan-1 R ( M,N ) R ( M,N ) 0
0
0
reminded
2ρρ cos(φ−φ ) 0
tan -1
0
1
=
ψ
0
that
h
+ z 2]1/2.
is
defined
The
R ( M,N )=[ρ
by
(4.1.18),
and
right-hand
side in
(A4.3.11)
2
0
(A4.3.11) + ρ20 −
simplifies in
the
limiting case of ρ →ρ and φ →φ, namely, 0
ρei 2( a
2
φ
0
( a 2 − ρ2)1/2 1
− ρ)
2 1/2
2 2 1/2 ( l 22 − a 2)1/2 ( a − ρ ) . tan-1 2 2 1/2 − 2 2 (l2 − a ) l2 − ρ
(A4.3.12) The second integral to be computed is: R
z 0 h I2 = ⌠ + tan-1 d z . 3 ⌡ R0 h R 0 We proceed with integration by parts.
The result is
zdz 1 h dz d h − tan-1 + ⌠ tan-1 . 2 R0 R 0 R 0 ⌡ R0h ⌡ R0 dz
I2 = ⌠
(A4.3.13)
(A4.3.14)
316
CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
We modify (A4.3.7) as follows: R0 h (ρ2 − l 21) z z d -1 h tan = 2 + − dz h hR R 0 R 0 + h 2 z ( l 22 − l 21) 0
=
R 0( l 22 − a 2)1/2( l 42 − ρ2ρ20) ρ20)1/2( l 22
(a − 2
−
l 21) l 22( R 20
2
+ h)
−
z . hR 0
(A4.3.15)
By substituting (A4.3.15) in (A4.3.14), and taking into consideration (A4.3.9) and (A4.1.29), we obtain I
2
= −
1 -1 h tan R R 0
0
+
( l 42 − ρ2ρ20) d l
2 ⌠ ( a 2 − ρ20)1/2 ⌡ ( l 2 − ρ2)1/2( l 2 − ρρ ei(φ-φ0) )( l 2 − ρρ e-i(φ-φ0) ) 2 2 2 0 0
1
(A4.3.16) The integral in (A4.3.16) is elementary, i.e.
I
2
= −
1 h 1 2 2 1/2 tan-1 + ln[ l 2 + ( l 2 − ρ ) ] R0 R 0 ( a 2 − ρ20)1/2
a (ζ − 1)1/2 1 a (ζ − 1)1/2 ρ i(φ-φ0) -1 − − 2 2 1/2 , ζ = ρ e 1/2tan 2 2 1/2 1/2tan ( a − l 1) ( a − l 1) (ζ − 1) (ζ − 1) 0 1
-1
(A4.3.17) Since the integration was indefinite, we might have lost a function of the variables, other than z . This function can be found from the condition that the result of integration should not have a logarithmic singularity at ρ=0 or at q =0. The functions eliminating such a singularity are tan-1[(ζ − 1)1/2] and tan-1[(ζ − 1)1/2]. The final result can now be represented in the form R
⌠ z 0 + tan-1 h d z ⌡ R 30 h R 0
317
Appendix A4.4
= −
h 1 1 2 2 1/2 tan-1 + ln[ l 2 + ( l 2 − ρ ) ] R0 R 0 ( a 2 − ρ20)1/2
a (ζ − 1)1/2 1 -1 -1 1/2 − 2ℜ − ζ − tan tan ( 1) , (ζ − 1)1/2 ( a 2 − l 21)1/2
ζ =
ρ i(φ-φ0) e ρ0
The last expression proves the correctness of formula (5.1.13).
Appendix A4.4 Some integrals related to the problem of a penny-shaped crack under shear loading are presented here, without derivation. The first integral, which can be computed directly, is: 2π a
i (ψ-φ0)
(3 a 2 − r ρ0e
) ( a 2 − r 2)1/2( a 2 − ρ20)1/2
⌠ ⌠ ⌡ ⌡ ( a 2 − rρ ei (ψ-φ0))2 0 0
a R ( M,N )
r d r dψ
0
( a 2 − ρ20)1/2 2 ( l 22 − a 2)1/2[ l 21(4 − t ) − 3 a 2] ρ -1 a = π t sin (l ) + a3 a (1 − t )2 2
1/2 ρ2 1 3z2 2 -1 a (1 − t ) . + + a (3 − 2 t ) − tan t ( l 22 − a 2)1/2 (1 − t )3/2 1 − t
Here t is defined by (4.4.16). (A4.4.1) yields 2π a
(A4.4.1) Application of the operator Λ to both sides of
i (ψ-φ0)
(3 a 2 − r ρ0e
) ( a 2 − r 2)1/2( a 2 − ρ20)1/2
⌠ ⌠ ⌡ ⌡ ( a 2 − rρ ei (ψ-φ0))2 0 0
3
a R ( M,N)
(ρei φ − r ei ψ) r d r dψ
0
iφ
= −2πρe
a ( l 22 − a 2)1/2 ( a 2 − ρ20)1/2 1 -1 a t sin (l ) + i(φ-φ ) a3 (1 − t )( l 22 − ρρ0e 0 ) 2
318
CHAPTER 4
−
APPLICATIONS IN FRACTURE MECHANICS
1/2 1 -1 a (1 − t ) tan . ( l 22 − a 2)1/2 t (1 − t )3/2
(A4.4.2)
Another application of Λ to both sides of (A4.4.2) gives 2π a
i (ψ-φ0)
(3 a 2 − r ρ e 0
) ( a 2 − r 2)1/2( a 2 − ρ2)1/2 0
⌠ ⌠ ⌡ ⌡ ( a 2 − rρ ei (ψ-φ0))2 0 0
3(ρei φ − r ei ψ)2 r d r dψ
5
a R ( M,N )
0
ρ2e2 i φ( a 2 − ρ20)1/2( l 22 − a 2)1/2(3 l 22 − ρρ e
i(φ-φ0)
0
= 2π
i(φ-φ0) 2
l 22( l 22 − l 21)( l 22 − ρρ e 0
)
) .
(A4.4.3)
Differentiation with respect to z of both sides of (A4.4.1) results in 2π a
i (ψ-φ0)
(3 a 2 − r ρ e 0
) z ( a 2 − r 2)1/2( a 2 − ρ2)1/2 0
⌠ ⌠ ⌡ ⌡ ( a 2 − rρ ei (ψ-φ0))2 0 0
a R 3( M,N )
r d r dψ
0
= 2π
ha 2 3 s s 2
2
t
−
l 22
−
− a t 2
3( l 22 − a 2)1/2 s
Here h is defined by (4.1.18). with respect to z yields 2π a
3
s . tan-1 2 ( l 2 − a 2)1/2
(A4.4.4)
Another differentiation of both sides of (A4.4.4)
i (ψ-φ0)
(3 a 2 − r ρ e
) ( a 2 − r 2)1/2( a 2 − ρ2)1/2 0
⌠ ⌠ ⌡ ⌡ ( a 2 − rρ ei (ψ-φ0))2 0 0 0
a R 3( M,N )
1 − 3 z2 rd rdψ R 2( M,N )
0
( a 2 − ρ20)1/2
a ( l 22 − a 2)1/2
= 2π 2 3 i(φ-φ0) 2 2 a (1 − t ) ( l 2 − l 1)( l 2 − ρρ0e ) i(φ-φ0)
ρρ e +
0
(2 l 22 + l 21 t − 3ρ2) i(φ-φ0)
l 22 − ρρ e 0
3( l 22 − l 21 t ) 1 − t
1/2 3 -1 a (1 − t ) − tan . ( l 22 − a 2)1/2 (1 − t )3/2
319
Appendix A4.4
Application of the operator Λ to both sides of (A4.4.4) yields 2π a
(A4.4.5)
i (ψ-φ0)
(3 a 2 − r ρ0e
) z ( a 2 − r 2)1/2( a 2 − ρ20)1/2
⌠ ⌠ ⌡ ⌡ ( a 2 − rρ ei (ψ-φ0))2 0 0
iφ
3(ρe
5
a R ( M,N )
− r ei ψ) r d r dψ
0
= 2π
h ρei φ(3 l 22 − a 2 t ) ( l 22 − l 21)( l 22 − a 2 t )2
.
(A4.4.6)
A different result is obtained if Λ is applied to a complex conjugate of expression (A4.4.4), namely, 2π a
- i (ψ-φ0)
(3 a 2 − r ρ0e
) z ( a 2 − r 2)1/2( a 2 − ρ20)1/2
⌠ ⌠ ⌡ ⌡ ( a 2 − rρ e- i (ψ-φ0))2 0 0
a R 5( M,N )
3(ρei φ − r ei ψ) r d r dψ
0
2 2 1/2 a 2 i φ 15( l 2 − a ) s 0 -1 − 15 = 2π h 2 ρ e tan 0 ( l 22 − a 2)1/2 s5 s4 s
+
5 s 2( l 22 − a 2 t )
+
ρei φ(3 l 22 − a 2 t )
+ . ( l 22 − a 2 t )2 ( l 22 − l 21)( l 22 − a 2 t )2 2t
Integration of both sides of (A4.4.1) with respect to z gives 2π a
i (ψ-φ0)
(3 a 2 − r ρ0e
) ( a 2 − r 2)1/2( a 2 − ρ20)1/2
⌠ ⌠ ⌡ ⌡ ( a 2 − rρ ei (ψ-φ0))2 0 0
a
ln[ R ( M,N ) + z ] r d r dψ
0
= π( a − 2
−
ρ20)1/2
( a 2 − l 21)1/2 z2 2 2 1/2 2ln[ + ( ) ] 2 + + ρ2 l l − ρ − − 2 3 2 1 − t a (1 − t )
1 2 zζ a z z2 + a 2(3 − 2 t − ζ) ( l 1 + 2 a 2) + sin-1( ) + 3 3 a l2 a (1 − t )3/2 1 − t
(A4.4.7)
320
CHAPTER 4
APPLICATIONS IN FRACTURE MECHANICS
( a 2 − l 21)1/2 1/2 1 a − t (1 ) + 2(ζ − 1)1/2tan-1 − tan-1 . ×tan-1 2 ( l 2 − a 2)1/2 (ζ − 1)1/2 a (ζ − 1)1/2
(A4.4.8) Here ζ is defined by (A4.3.17), and the bar, as usual, indicates the complex conjugate value. A similar integration with respect to z of (A4.4.3) yields 2π a (3 a 2
i (ψ-φ0)
− r ρ0e
) ( a 2 − r 2)1/2( a 2 − ρ2)1/2 0
⌠ ⌠ ⌡ ⌡ ( a 2 − rρ ei (ψ-φ0))2 0 0
a
Λ2{ln[ R ( M,N) + z ]} r d r dψ
0
( a 2 − l 21)1/2 (ζ − 1)1/2 -1 1 2π 2 2 1/2 -1 − tan = ( a − ρ0) − tan (ζ − 1)1/2 a (ζ − 1)1/2 q q
2 2 1/2 ρ2 ei φ ( a − l 1) − 1 . + 1 + i(φ-φ ) ρ a l 22 − ρρ0e 0
(A4.4.9)
It is reminded that q is defined by (4.1.28), and Λ2ln[ R ( M,N) + z ] = −(ρei φ − r ei ψ)2
2 R ( M,N) + z R 3( M,N) [ R ( M,N) + z ]2
Yet another application of the operator Λ to (A4.4.9) gives 2π a
i (ψ-φ0)
(3 a 2 − r ρ e 0
) ( a 2 − r 2)1/2( a 2 − ρ20)1/2
⌠ ⌠ ⌡ ⌡ ( a 2 − rρ ei (ψ-φ0))2 0 0
a
Λ3{ln[ R ( M,N ) + z ]} r d r dψ
0
( a 2 − l 21)1/2 3(ζ − 1)1/2 -1 1 2π 2 2 1/2 -1 − tan = ( a − ρ0) tan 2 1/2 1/2 (ζ − 1) a (ζ − 1) q q
−
2 e i φ( a 2 − l 21)1/2
a ( l 22 − l 21)
(A4.4.10)
l 22 + ρ2
2ρ2( l 22 − a 2)
+ i(φ-φ0) 2 + 1 2 l 22 − ρρ ei(φ-φ0) ( l 2 − ρρ0e ) 0
321
Appendix A4.4
ei φ 3
+
ρ q
+
2ei φ ρ
( a 2 − l 21)1/2
−
a
1 ei φ + 2 ( + ) . ρ q ( l 22 − ρρ ei(φ-φ0) ) q 0 l 22 + 2ρ2
(A4.4.11) Here iφ
Λ ln[ R ( M,N ) + z ] = (ρe 3
iψ 3
− re )
8 R 2( M,N ) + 9 R ( M,N ) z + 3 z 2 R 5( M,N ) [ R ( M,N ) + z ]3
.
Formula (4.1.27) can be used to obtain some additional results. (4.1.27) with respect to z gives
(A4.4.12) Integration of
2π a
( a 2 − r 2)1/2( a 2 − ρ20)1/2 r d r dψ ρei φ − r ei ψ ⌠ ⌠ tan-1 ( )[ ( ) + ] ( ) R M,N R M,N z a R N,N R ( N,N0) ⌡ ⌡ 0 0
0
=
2π z s -1 h 2 2 1/2 -1 R 0tan (R ) − ( a − ρ0) tan 2 2 1/2 ( ) q s l a − 2 0
− (ζ − 1)
1/2
( a 2 − l 21)1/2 1 -1 tan-1 1/2 − tan 1/2 . (ζ − 1) a (ζ − 1)
(A4.4.13)
The following indefinite integrals were used here s s d z = z tan-1 ⌠ tan-1 2 2 1/2 2 ( l 2 − a ) ( l 2 − a 2)1/2 ⌡ + s ln[ l
2
a (ζ − 1)1/2 + ( l 22 − ρ2)1/2] + (ζ − 1)1/2 tan-1 2 , ( a − l 21)1/2
⌠ z tan-1 h d z = R tan-1 h − ( a 2 − ρ2)1/2−ln[ l + ( l 2 − ρ2)1/2] 0 2 0 2 ⌡ R0 R 0 R 0 ( a 2 − l 2)1/2
( a 2 − l 2)1/2
1 1 + (ζ − 1)1/2tan-1 . + (ζ − 1) tan 1/2 1/2 a (ζ − 1) a (ζ − 1) 1/2
-1
By applying the operator Λ to both sides of (A4.4.13), one gets
(A4.4.14)
322
CHAPTER 4
2π a
iφ
(ρe
⌠ ⌠ ⌡ ⌡ 0
iψ
− r e )2[2 R ( M,N ) + z ]
3
R ( M,N ) [ R ( M,N ) + z ]
APPLICATIONS IN FRACTURE MECHANICS
( a 2 − r 2)1/2( a 2 − ρ20)1/2
tan -1
2
a R ( N,N0)
0
iφ
R 2 + z2 z ρ0e 2π 0 -1 h 2 2 1/2 = tan ( ) + ( a − ρ0) R0 s s2 q R0q
rd rdψ R ( N,N0)
0
−
2 -1 s tan 2 2 1/2 ( l 2 − a ) q
( a 2 − l 21)1/2 iφ iφ (ζ − 1)1/2 -1 1 e e ha 2 -1 − tan − + + tan , ρ ρs2 q a (ζ − 1)1/2 (ζ − 1)1/2
(A4.4.15)
and yet another application of Λ to (A4.4.15) yields 2π a
iφ iψ 3 2 2 ⌠ ⌠ (ρe − re ) [8 R ( M,N) + 9 R ( M,N) z + 3 z ] R 5( M,N ) [ R ( M,N ) + z ]3 ⌡ ⌡ 0
0
( a 2 − r 2)1/2( a 2 − ρ20)1/2
× tan -1
2π = q
a R ( N,N0) 3 R 40 + 6 R 20 z 2 − z 4 R 30 q 2
rd rdψ R ( N,N0) iφ
iφ
4ρ0e
z 8 − ( a 2 − ρ20)1/2 − s q 2
iφ
e 2e h tan ( ) + ( a 2 − ρ20)1/2 ρ ρ R0 -1
2
s q
2 iφ
0
+
3ρ20e s
4
0
s tan-1 2 ( l 2 − a 2)1/2
( a 2 − l 21)1/2 3(ζ − 1)1/2 -1 1 -1 − tan 2 1/2 − tan 1/2 q (ζ − 1) a (ζ − 1) 2 iφ
+
ha e
ρs2
iφ
2ρ0e s2
0
iφ
2e − ρ
iφ
ρ0e
− + 2 s q 2
0
+
2 2 2 ( l 2 − a ) t − q l 22 − a 2 t
3
q
323
Appendix A4.4
3 iφ
q ρe − 2 + 2 R 0 + h l 22 − l 21 h
ei φ( l 22 − ρ2) ρq
−
z2 q R 20 q
+ 2e2 i φ. (A4.4.16)
CHAPTER 5 APPLICATION TO CONTACT PROBLEMS
The various elastic contact problems, solved in Chapters 2 and 3, deal primarily with the stresses and displacements in the plane z =0. We concentrate here on the complete solutions. The solution is called complete when explicit expressions are given for the field of stresses and displacements in the whole half-space. The complete solution, combined with the reciprocal theorem, enables us to solve more complicated problems of punch interactions, influence of external loads on punches, etc. An approximate analytical solution is given to the non-classical punch problem. Significant part of the material presented here is as yet unpublished. The rest follows the papers (Fabrikant, 1974a, 1986a, 1986c, 1986i, 1987h).
5.1 Contact problem for a smooth punch. A smooth punch is pressed against a transversely isotropic elastic half-space z ≥0 by a normal force P . The term ‘smooth’ is used to identify a punch which does not exert any shear traction at its base. Let S denote the domain of contact. The mixed boundary conditions on the plane z =0 are: σz = 0,
for (ρ,φ)∉ S ;
w = ω(ρ,φ),
for (ρ,φ)∈ S ,
τz = 0, for −∞<( x,y)<∞. As in the previous chapter, we may assume again that F 1( z ) = c 1 F ( z 1),
F 2( z ) = c 2 F ( z 2),
(5.1.1)
F 3( z ) = 0.
Substitution of (5.1.2) and (2.1.12) in the third condition (5.1.1) yields:
324
(5.1.2)
325
5.1 Contact problem for a smooth punch.
c = − c γ 1/ m 1γ 2. 1 2
(5.1.3)
Now we need to define the main potential function F so that its second z -derivative vanishes at z =0 outside the domain of contact. Comparison with (4.1.4) gives F (ρ,φ, z ) = F ( M ) = ⌠ ⌠ ln[ R ( M,N ) + z ] σ( N ) d S . N
⌡⌡
(5.1.4)
S
The simple layer potential property yields inside the domain of contact: ∂2 F = −2πσ, ∂ z 2 z=0
(5.1.5)
which, combined with (2.1.12), gives the second equation for the constants c and 1
c
2
2π A 44[(1 + m 1) c + (1 + m 2) c ] = 1. 1 2
(5.1.6)
Equations (5.1.3) and (5.1.6) determine the constants c = H γ1/( m 1 − 1), 1
c = H γ2/( m 2 − 1). 2
(5.1.7)
The simplifications made in (5.1.7) are due to the properties (4.1.10). Finally, substitution of (5.1.2), (5.1.4) and (5.1.7) in the second of equations (2.1.6) yields for z =0 the well known governing integral equation of elastic contact problem for a smooth punch: ω( N 0) = H ⌠ ⌠
σ( N ) dS . ⌡ ⌡ R ( N,N0) N
(5.1.8)
S
An approximate analytical solution of (5.1.8) for a punch of arbitrary shape will be given later. We consider in more detail the case of a circular punch. The integral equation (5.1.8) can be rewritten for a circular domain of contact of radius a as follows (c.f. 1.4.5) ρ
a
ρ dρ
2
0 0 x σ(ρ ,φ) = ω(ρ,φ) ⌠ 4H ⌠ 2 L 2 1/2 2 2 1/2 0 ρρ0 ⌡ (ρ − x ) ⌡ (ρ0 − x ) 0
dx
x
326
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
Its exact closed form solution is (c.f. 1.4.10) a
xdx 1 d ⌠ σ(ρ,φ) = − 2 L (ρ) 2 dρ ⌡ ( x − ρ2)1/2 π Hρ ρ
ρ0dρ0 1 d⌠ L (ρ0) ω(ρ0,φ). × L 2 x d x⌡ ( x 2 − ρ20)1/2 x
0
The potential function F can be found in two stages. from (5.1.4):
(5.1.9) First of all, one has
2π a
σ( r ,ψ) r d r dψ ∂F = ⌠⌠ ∂z ⌡ ⌡ [ρ2 + r2 − 2 rρcos(φ−ψ) + z2]1/2 0
(5.1.10)
0
Substitution of (5.1.9) in (5.1.10) yields, after integration (c.f.
1.4.21)
2π a
R0 h z 1 ∂F = 2 ⌠⌠ + tan-1 3 ω(ρ0,φ0)ρ0dρ0dφ0. ∂z R 0 R 0 π H ⌡ ⌡ h 0
(5.1.11)
0
Here R 0=[ρ2 + ρ20 − 2ρρ0cos(φ−φ0) + z 2]1/2, and h is defined by (4.1.18). The next integration of (5.1.11) with respect to z gives the expression for the potential function directly in terms of the prescribed displacement under the punch, namely, 2π a
F (ρ,φ, z ) =
1 ⌠⌠ K(ρ,φ, z ; ρ0,φ0) ω(ρ0,φ0)ρ0dρ0dφ0, πH ⌡⌡ 2
0
0
where K(ρ,φ, z ; ρ0,φ0) = −
1 -1 h 1 tan + ln[ l 2 + ( l 22 − ρ2)1/2] 2 2 1/2 R0 R 0 ( a − ρ0)
(5.1.12)
327
5.1 Contact problem for a smooth punch.
1 a (ζ − 1)1/2 -1 -1 1/2 tan tan ( 1) − 2ℜ − ζ − , (ζ − 1)1/2 ( a 2 − l 21)1/2
ζ =
ρ i(φ-φ0) e ρ0
(5.1.13) The details of integration are presented in Appendix A4.3. Now all the Green’s functions related to the field of displacements and stresses can be obtained in elementary functions from (5.1.13) by differentiation R0 z h ∂K = 3 + tan-1 , ∂z R 0 R0 h
(5.1.14)
( a 2 − l 21)1/2 q R0 h 1 -1 + tan ΛK = 3 − 1 − a R 0 R0 h hq
a (ζ − 1)1/2 -1 -1 1/2 tan tan ( 1) , − − ζ − ( a 2 − l 21)1/2 a (ζ − 1)1/2 (a − 2
l 21)1/2
ρ2 − l 21 1 3 z 2R 0 1 h z2 ∂2K -1 , + tan + − 2 2 = 3 1 − 2 2 2 2 2 R 0 R 0 R 0 h h ( R 0 + h )R 0 l2 − l1 ∂z 3 zq R 0 h z q ρei φ ∂K -1 = − 5 + tan + + , Λ ∂z R 0 R0 h h ( R 20 + h 2) l 22 − l 21 R 20
(5.1.15)
(5.1.16)
(5.1.17)
R l 22 − ρ2 2 h 3q2 0 1 q -1 − e2 i φ + tan + ΛK = − 5 2 2 2 2 2 R 0 R0 h h ( R 0 + h )R 0 l2 − l1 2
a (ζ − 1)1/2 3 1 1 1 -1 -1 1/2 + 2 − 1 + tan tan ( 1) − ζ − ( a 2 − l 21)1/2 (ζ − 1)1/2 ( a 2 − ρ20)1/2 q h
+
ei φ a 3ζ 1 − . ( a 2 − l 21)1/2( a 2ζ − l 21) ρ q ( a 2 − ρ20)1/2
The following identities were used in the simplification of (5.1.14−5.1.18): ( l 22 − ρ20)( a 2 − l 21) + a 2 qq = a 2( R 20 + h 2), ( a 2ζ − l 21)( a 2ζ − l 21) = ( R 20 + h 2) l 21 a 2/ρ20,
(5.1.18)
328
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
( l 22 − ρ20)( qq + z 2) − qq( a 2 − ρ20) = ( l 22 − a 2)( R 20 + h 2). (5.1.19) Formulae (5.1.13−5.1.18) are the main new results of this section. They give a complete solution for the contact problem under consideration when combined with formulae (2.1.6), (2.1.12), (5.1.2) and (5.1.7). When the prescribed displacement can be expressed as a sum of powers in x and y , the complete solution is elementary. Some particular examples will be considered further.
1.
Exercise 5.1 Derive 5.1.13.
2.
Verify (5.1.14) −(5.1.18).
3.
Prove the identities (5.1.19).
4.
Derive the general solution for the case of isotropy.
5.2 Flat centrally loaded circular punch. We consider a transversely isotropic elastic half-space z ≥0, indented by a rigid circular punch of radius a . The punch loading is statically equivalent to a centrally applied normal force P . Denote the punch settlement by ω=const. Solution of the integral equation (5.1.8) is given by (5.1.9), and in this particular case takes the form σ(ρ,φ) =
ω π H(a 2
2
− ρ2)1/2
.
(5.2.1)
Integration of the last expression over the circle ρ≤ a relates the punch settlement ω to the total force P as P =
2ω a . πH
The substitution of (5.2.1) in (5.1.4) leads to an integral which can be evaluated by differentiation of the expression for the main potential function (A4.1.1) with respect to a . The result is F =
2ω a z sin-1( ) − ( a 2 − l 21)1/2 + a ln[ l 2 + ( l 22 − ρ2)1/2]. πH l2
(5.2.2)
329
5.2 Flat centrally loaded circular punch.
The appropriate differentiation of the potential function (5.2.2) gives the complete solution: 2ω a ei φ u = πρ
Σm k=1
2
Σm
2ω w = π
k=1
σ1 = −
k
mk k
4ω A 66 π
− 1
Σ
π
sin-1(
a
,
(5.2.3)
a ), l 2k
(5.2.4)
γk( m k − 1)
k=1
σ2 =
1 − − 1
γ2k − ( m k + 1)γ23 ( a 2 − l 21k)1/2
2
4ω A 66e2 i φ
( a 2 − l 21k)1/2
γk
2
l 22k − l 21k
,
(5.2.5)
( a 2 − l 21k)1/2 ( m k − 1) l 22k − l 21k k=1 γk
2
Σ
( a 2 − l 21k)1/2 2a , − 2 1 − a ρ
(5.2.6) σz =
τz =
ω π2 H (γ1 − γ2) ωei φ π2 H (γ1 − γ2)
(a − 2
2
Σ
(−1)kγk
k=1
2
Σ(−1)
k
l 21k)1/2
l 22k − l 21k
,
l 1k( l 22k − a 2)1/2
k=1
l 2k( l 22k − l 21k)
(5.2.7)
.
(5.2.8)
This problem was first solved by Elliott (1949). His solution is essentially in agreement with ours. The fact that he uses the notation ρ as an elastic parameter and as a polar radius simultaneously, somewhat complicates the comparison. Introduce the stress intensity factor as k = lim{( a − ρ)1/2σz}, for z =0. 1
ρ→ a
Substitution of (5.2.7) in (5.2.9) yields
(5.2.9)
330
CHAPTER 5
k = − 1
APPLICATION TO CONTACT PROBLEMS
ω P a 1/2 = − ( ) . 2 π2 π H √2 a
(5.2.10)
2
The asymptotic behavior of the field of stresses and displacements near the punch edge can be derived by substitution of (4.7.1) and (5.2.10) in (5.2.3−5.2.8). The result is: γ1 S 1 γ2 S 2 + 0(1), + u = u n = 2π Hk 1√2 r m 2 − 1 m 1 − 1
(5.2.11)
m1T1 m2T2 + 0(1), w = −2π Hk 1√2 r + m 2 − 1 m 1 − 1
(5.2.12)
2
2 σ1 = 2π A 66 Hk 1( )1/2 r
γ2k − ( m k + 1)γ23 S k
Σ
γk( m k − 1)
k=1
γk S k
2
2 σ2 = −2π A 66 Hk 1( )1/2 r
Σ(m k=1
σz =
k1
γS
k
− 1) Q k
Qk
+ 0(1),
+ 0(√ r ),
τz = τzn
(5.2.14)
γS
1 1 − 2 2 + 0(1), Q2 √2 r (γ1 − γ2) Q 1 k1
(5.2.13)
T
(5.2.15)
T
1 − 2 + 0(√r), = − Q 2 √2 r (γ1 − γ2) Q 1
(5.2.16)
Comparison of (5.2.11−5.2.16) with (4.7.4−4.7.9) indicates that they become identical if one substitutes formally θ by π−θ. This is easy to explain. The punch problem is mathematically equivalent to an external crack problem. There exist a notion that the asymptotic behavior of stresses and displacements near the edge of an arbitrary flat crack with a smooth boundary is completely defined by three stress intensity factors. This means that the asymptotics of an internal crack and an external one should be the same. The system of local axes was introduced in previous section so that the angle θ was measured from the direction outside the penny-shaped crack. In the case of the punch problem the axis On should be directed into the circle, this will make the expressions (5.2.11−5.2.16) identical to (4.7.4−4.7.9).
331
5.2 Flat centrally loaded circular punch.
Exercise 5.2 1. Find the complete solution to the problem of a smooth flat centrally loaded circular punch in the case of an isotropic half-space. Answer: a − ( a 2 − l 21)1/2 z l 1( l 22 − a 2)1/2 ωei φ + u = −(1 − 2ν) , ρ π(1 − ν) l 2( l 22 − l 21)
z ( a 2 − l 21)1/2 2ω -1 a , sin ( ) + w = l2 π 2(1 − ν)( l 22 − l 21) ( a 2 − l 21)1/2 z 2[ l 41 − a 2(2 a 2 − ρ2 + 2 z 2)] 2ωµ (1+2ν) , + σ1 = − π(1 − ν) ( a 2 − l 21)1/2( l 22 − l 21)3 l 22 − l 21 ( a 2 − l 21)1/2 2[ a − ( a 2 − l 21)1/2] 2ωµe2 i φ σ2 = −(1 − 2ν) 2 − π(1 − ν) l 2 − l 21 ρ2
−
az ( l 22 − a 2)1/2[2 l 41 + ρ2( l 21 + 3 l 22 − 6 a 2)] l 22( l 22 − l 21)3
,
z 2[ l 41 + a 2(ρ2 − 2 a 2 − 2 z 2)] ( a 2 − l 21)1/2 2ωµ + σz = − , π(1 − ν) ( a 2 − l 21)1/2( l 22 − l 21)3 l 22 − l 21
z ρei φ( a 2 − l 21)1/2(3 l 22 + l 21 − 4 a 2) 2ωµ . τz = − π(1 − ν) ( l 22 − l 21)3 Hint : use formulae (5.2.3−5.2.8). 2.
Rewrite the result of the Exercise 1 in polar coordinates.
Answer: a − ( a 2 − l 21)1/2 z l 1( l 22 − a 2)1/2 ω + u = −(1 − 2ν) , ρ π(1 − ν) ρ l 2( l 22 − l 21)
332
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
a − ( a 2 − l 21)1/2 ( a 2 − l 21)1/2 2ωµ (1 − 2ν) − σ = ρ π(1 − ν) l 22 − l 21 ρ2
−
az ( l 22 − a 2)1/2[ l 41 − l 42 + ρ2( l 21 + 3 l 22 − 4 a 2)]
,
l 22( l 22 − l 21)3
a − ( a 2 − l 21)1/2 2ωµ σ = − (1 − 2ν) φ π(1 − ν) ρ2
+
( a 2 − l 21)1/2[ a 2 − (1 − 2ν) l 22] l 22( l 22 − l 21)
,
z ρ( a 2 − l 21)1/2(3 l 22 + l 21 − 4 a 2) 2ωµ τ = − . ρz π(1 − ν) ( l 22 − l 21)3
5.3 Inclined circular punch on an elastic half-space The case of a flat circular punch pressed against a transversely isotropic elastic half-space by a non-centrally applied force P can be considered as a superposition of two problems: that of a centrally loaded punch, which was considered earlier, and a punch subjected to a tilting moment M which will be considered here. Let the displacements under the punch be ω = bxy − byx.
(5.3.1)
Here b x and b y are the tilting angles about the axes Ox and Oy respectively. Introduce the complex tilting angle b = b x + ib y.
(5.3.2)
Expression (5.3.1) can now be rewritten as ω(ρ,φ) = ℑ{ b ρei φ}.
(5.3.3)
333
5.3 Inclined circular punch on an elastic half-space
Here ℑ indicates the imaginary part. 2ℑ{ b ρei φ}
σ(ρ,φ) =
π2 H ( a 2 − ρ2)1/2
The substitution of (5.3.3) in (5.1.9) yields
.
(5.3.4)
Evaluation of the potential function (5.1.4) leads to the integral i φ0
2π a
ρ0e
⌠⌠ ln( R 0 + z ) ρ0dρ0dφ0. ⌡ ⌡ ( a 2 − ρ20)1/2 0
(5.3.5)
0
Taking into consideration that iφ ρe
(a
2
− ρ)
2 1/2
= −Λ( a 2 − ρ2)1/2,
the integral in (5.3.5) can be evaluated by parts, with the result given by (A4.1.5). The potential function takes the form: 2ℑ{ b ρei φ} a F = z sin-1 − ( a 2 − l 21)1/2(1 − l2 πH
l 21 + 2 a 2 3ρ2
) −
2 a 3 . 3ρ2
(5.3.6) All the necessary derivatives are readily available from Appendix A4.1, and we can write the complete solution:
a b z ksin-1( ) − ( a 2 − l 21k)1/2 mk − 1 l 2k k=1 2
2i u = π
Σ
γk
2 iφ
− be
2 a 3 − ( l 21k + 2 a 2)( a 2 − l 21k)1/2
2
2 w = ( bxy − byx) π
Σ k=1
σ1 = −
8 A 66 π
,
3ρ2 mk
sin-1( a ) − m k − 1 l 2k 2
( bxy − byx) a2
Σ k=1
(5.3.7)
a ( l 22k − a 2)1/2 l 22k
,
γ2k − ( m k + 1)γ23 ( a 2 − l 21k)1/2 γk( m k −1)
l 22k( l 22k − l 21k)
(5.3.8)
, (5.3.9)
334
CHAPTER 5
iφ
σ2 =
a l 1k( a 2 − l 21k)1/2 b m k − 1 l 2k( l 22k − l 21k) γk
2
2 A 66 i e
Σ
π
k=1
2 iφ
− be
σz =
APPLICATION TO CONTACT PROBLEMS
4[( l 21k + 2 a 2)( a 2 − l 21k)1/2 − 2 a 3]
3ρ
3
2( b x y − b y x ) π2 H (γ1 − γ2)
a 2( a 2 − l 21k)1/2
2
Σ(−1) γ k
k=1
k
l 22k( l 22k − l 21k)
+
a l 1k( a 2 − l 21k)1/2 l 2k( l 22k
−
l 21k)
,
,
(5.3.10)
(5.3.11)
a ( l 22k − a 2)1/2 -1 a τz = 2 b sin (l ) − 2 2 π H (γ 1 − γ 2) l 2k − l 1k 2k k=1 2
Σ
i
2 iφ
+ be
a l 21k( l 22k − a 2)1/2 l 22k( l 22k − l 21k)
.
(5.3.12)
Note. Some of the integrals involving special functions can now be computed simply by comparison of the solutions obtained by the integral transform method with the corresponding solution given by the present method. For example, comparison of (4.1.24) with formulae (1.42) and (1.43) of (Kassir and Sih, 1975) leads to ∞
l1
2n ⌠ J ( as) J (ρs) e-sz d s = 2 ( a ρ)-n⌠ 2 x d x2 1/2. n πa √s ⌡ n+1/2 ⌡ (ρ − x ) 1/2
0
0
We have verified numerically that the last formula is correct for non-integer n as well. An enormous amount of material on Bessel functions exists in the literature, so it is difficult to claim that the last result is new, but there is some chance that it is new, since our notation l 1 and l 2 does not seem to have been used before. Here is another example of just how useful this notation is. Suppose that we need to compute the integral
335
5.3 Inclined circular punch on an elastic half-space
∞
⌠ sinax J 1(ρx ) e-zx d x2 . x ⌡
(5.3.13)
0
We could not find this integral in the tables, but we have found another one (Gradshtein and Ryzhik, 1963, formula 6.752.2) ∞
⌠ sinax J 1(ρx ) e-zx d x = a (1 − r), x ρ ⌡
(5.3.14)
0
where the parameter r is a positive root of the equation a2 =
ρ2 z2 . − 1 − r2 r2
(5.3.15)
The original integral (5.3.13) can be computed by integration of both sides of (5.3.14) with respect to z . We need to get an explicit expression for r from the fourth order algebraic equation (5.3.15), substitute the result in (5.3.14) and integrate the result with respect to z which does not seem possible at first. The introduction of the parameters l 1 and l 2 allows us to find the positive root of (5.3.15) in a very simple form, namely, r =( a 2 − l 21)1/2/ a . be performed using (A4.2.3), and the final result is ∞
⌠ sinax J 1(ρx )e ⌡
-zx
dx = x2
(2 a 2 − l 21)( l 22 − a 2)1/2 − 2 a 2 z 2aρ
+
The z -integration can
ρ a sin-1( ). 2 l2
0
(5.3.16)
Exercise 5.3 1. Consider the interaction of a concentrated load P 0, applied at an arbitrary point (ρ,φ, z ) in the z -direction, with a flat circular punch of radius a . Solution : One can deduce from (5.2.4) that the normal displacement w at the point (ρ,φ, z ), due to a unit force applied to the punch, is
336
CHAPTER 5
2
H w = a
Σm k=1
mk − 1
k
sin-1(
APPLICATION TO CONTACT PROBLEMS
a ). l 2k
Application of the reciprocal theorem immediately gives the punch settlement ω due to the load P 0 2
H P ω = a 0
Σm k=1
mk k
− 1
sin-1(
a ). l 2k
The tilting angle δ of the punch can be obtained in a similar manner from (5.3.8). The result is δ =
3H 2a
3
2
P0
Σ k=1
mk
ρsin-1( a ) − m k − 1 l 2k
l 1k( l 22k − a 2)1/2 l 2k
.
All the other parameters of interest can be found in a similar manner. 2. Investigate the problem of interaction between a flat circular punch and an arbitrarily located tangential force T = T x+i T y.
5.4 Flat punch of arbitrary planform under the action of a normal centrally applied force The general method is applied here to the non-classical punch problem. A simple yet accurate relationship is established between the punch settlement and the applied force for an arbitrary flat punch. Specific formulae are derived for a punch whose planform has the shape of a polygon, a triangle, a rectangle, a rhombus, a circular sector and a circular segment. All the formulae are checked against the solutions known in the literature, and a good accuracy is confirmed. Theory. The presentation in this section will be made in terms of the elastic contact problem but one should keep in mind that all the results will be applicable in other branches of engineering science. Here, we outline the idea of the analytical treatment of the elastic contact problems which allows us to derive simple yet accurate formulae for various punch shapes. The governing integral equation is given by (5.1.8). The analytical approach is based on the integral representation for the reciprocal distance established in (1.1.27). Substitution of (1.1.27) into (5.1.8) gives, after changing the order of integration
337
5.4 Flat punch of arbitrary planform under a normal force
ρ
w (ρ,φ) =
2 ⌠ H π ⌡
2π
dx (ρ − x ) 2
2 1/2
a (φ0)
⌠ dφ ⌠ ⌡ 0 ⌡ 0 x
0
λ(
x2 , φ−φ0) ρρ0
(ρ20 − x 2)1/2
σ(ρ0,φ0)ρ0dρ0. (5.4.1)
Consider a flat-ended punch with a planform S whose boundary is defined in polar coordinates as ρ = a (φ).
(5.4.2)
Let the normal pressure distribution under the punch be σ =
ca (φ)
(5.4.3)
a 2(φ) − ρ2
1/2
where c is a constant which can be determined from the condition that the integral of σ over S should give the total force P . 2π
a (φ)
⌠ dφ ⌠ ⌡ ⌡ 0
0
ca (φ)
a 2(φ) − ρ21/2
2π
ρdρ = c ⌠ a 2(φ)dφ = 2 Ac = P ,
⌡
0
(5.4.4) where A is the area of S . It is noteworthy that the total force does not depend on the location of the coordinate system origin. This location can be determined from the condition that the stress distribution (5.4.3) should not produce any tilting moment about the origin, which leads to the two equations 2π
⌠ a 3(φ) cosφ dφ = 0 , ⌡ 0
2π
⌠ a 3(φ) sinφ dφ = 0. ⌡
0
The left hand sides of both equations are proportional to the x and y coordinates of the center of gravity which means that the origin of the system of polar coordinates should be located at the center of gravity of the domain of contact S . One gets immediately from (5.4.4) that
338
CHAPTER 5
P a (φ)
σ =
APPLICATION TO CONTACT PROBLEMS
.
(5.4.5)
2 A a 2(φ) − ρ2
1/2
For the case of a flat punch w = δ = const. Now substituting (5.4.5) in (5.4.1), we can verify how close to a constant will be the displacements, produced by the traction distribution (5.4.5). Integration with respect to ρ0 gives ∞
HP w (ρ,φ) = 2A 2π
×⌠ e
⌡
in(φ-φ0)
0
Σ n=-∞
F 1 −
ρ
xdx ⌠ (x )|n| 2 2 1/2 ⌡ ρ (ρ − x ) 0
|n| 1 x2 , ; 1; 1 − 2 dφ . 2 2 a (φ0) 0
(5.4.6)
Here F stands for the Gauss hypergeometric function. Further evaluation of the normal displacements can be carried out separately for each harmonic. The zero th harmonic has the form 2π
w0 =
HP π ⌠ a (φ)dφ. 2A 2 ⌡
(5.4.7)
0
It is important to note that the second harmonic is equal to zero for an arbitrary contour, and that all the odd harmonics will be zero if the expression for a (φ) does not contain odd harmonics. The expression for the fourth harmonic is 2π 4 i (φ-φ0)
HP 8 3⌠ w4 = ρ 2 A 35 ⌡ 0
e
dφ0
a 2(φ0)
.
(5.4.8)
The investigation of further harmonics shows that their amplitude decreases. Now consider in more detail the case of a square of side 2 l . The equation of the boundary in this case is a (φ)= l /cosφ for −π/4<φ<π/4, and the pattern is repeated outside this range. We can evaluate several non-zero harmonics: HP 4π l ln(1 + √2), w0 = 2A
HP 32ρ3cos4φ , w4 = 2A 105 l 2
(5.4.9)
339
5.4 Flat punch of arbitrary planform under a normal force
w8 = −
ρ HP 64 20 ρ 6 12 ρ 4 ρcos8φ( )2 + ( ) + ( ) , 13 l 13 l 2 A 3465 l
7 9 144 Γ( ) 6 Γ( ) 2 ρ4 2 ρ2 HP ρcos12φ ( ) + ( ) = l 19 2A Γ(17) l Γ( ) 2 2
w 12
11 7 5 7 3780 Γ( ) 1680 Γ( ) 1200 Γ( ) Γ( ) 2 2 ρ8 2 2 ρ6 ρ ( ) + ( ) + ( )10. + l l 25 l 23 1 21 Γ( ) Γ( ) Γ( ) Γ( ) 2 2 2 2 If we assume that the punch settlement δ≈ w 0 then the remaining harmonics may be called the solution error. Direct computations show that the error is less than 3% inside the circle ρ≤ l . The error is reasonably small outside the circle reaching 20% at the apex, and decreasing very rapidly with the distance from the apex. Taking into consideration that the error sign fluctuation will result in even smaller error in the total force value, we may assume (5.4.7) being the relationship between the punch settlement and the total force which can be rewritten in the form δ ≈
HP
,
g√A
(5.4.10)
where A is the punch base area, and g is a dimensionless coefficient depending on the punch geometry only. 2√ A g = 2 , (5.4.11) π ra where 2π
r
a
=
1 ⌠ a (φ) dφ 2π ⌡
(5.4.12)
0
can be called the average radius with respect to the center of gravity If the correct pressure distribution required to yield w =δ over the punch could be found then the total force P would be related to δ by a formula
340
CHAPTER 5
δ =
APPLICATION TO CONTACT PROBLEMS
HP . g 1√ A
Our basic assumption is that the g of equation (5.4.10), which we can calculate, is likely to be a good approximation to g 1. The problem now is to find the value of g for various punch planforms. One can compute the coefficient g for the square from (5.4.9) as 1 = 0.3611, g = π ln(1 + √2) which is very close to the approximate value 0.3607 given by Maxwell for the capacity of the square. Using the electrostatic analogy, one can easily deduce that our coefficient g is related to the capacity C of a flat lamina by g =
C . √A
(5.4.13)
Of course, closeness to the result of Maxwell does not mean that ours is so accurate. The value of g which seems to be accurate was obtained in (Noble, 1960) and (Solomon, 1964a), and is 0.367, so that our result is in error by 1.6% which is not bad. Now it seems reasonable to assume that formulae (5.4.10−5.4.12) are valid for an arbitrary flat punch, and we need to verify how good they really are for each specific case. We have found in the literature only one general formula of the type (5.4.10) suggested by Solomon (1964b). His result expressed through the coefficient g reads
g =
1/8 29/8 I 0
π11/8 A 1/4
.
(5.4.14)
where I 0 stands for the polar moment of inertia. One can easily verify that formula (5.4.14) is exact for a circle, so one should expect it to be sufficiently accurate for domains with the aspect ratio not far away from unity, but the error might be quite significant for oblong domains. For example, in the case of an ellipse with semi-axes a and b formula (5.4.14) gives g =
27/8 a b 1/8 3/2 ( b + a ) . π
The error of this formula can be quite significant for large aspect ratio ε= a / b . Our formulae (5.4.10−5.4.12) in the case of an ellipse are exact.
341
5.4 Flat punch of arbitrary planform under a normal force
Example 1: Polygon. Consider a flat punch, shaped as a polygon with n sides, with the only limitation that the function a (φ) describing its boundary be continuous and single-valued. The origin of the coordinate system is located at the center of gravity, as before. Let us number the polygon sides in a counter-clockwise direction from 1 to n , a k being the length of the k th side. The apex at which the sides a k and a k+1 intersect is numbered k +1. It is clear that the value of index equal to n +1 is understood as 1. We denote the distance from the center of gravity to the k th apex as b k. Let A k be the area of the triangle formed by a k, b k and b k+1, the total area A of the polygon being equal to the sum of A k. Then formulae (5.4.11) and (5.4.12) yield the following expression for the coefficient g 2√ A
g = n
Σ
π
k=1
Ak ak
ln
.
(5.4.15)
b k + b k+1 + a k b k + b k+1 − a k
The formula due to Solomon (5.4.14) in this case gives 29/8 g = 11/8 π
n
Σ k=1
2 A 3k
1 + 2 2 A ak
a 4k + 3( b 2k+1 − b 2k)2 1/8 48 A 2k
.
(5.4.16)
In the case of a regular polygon formulae (5.4.15) and (5.4.16) simplify to g =
4√tan(π/ n ) , 1 + sin(π/ n ) π√ n ln 1 − sin(π/ n )
g =
29/8 2cos2(π/ n ) + 1 π11/8 3 n sin(2π/ n )
(5.4.17)
and 1/8
(5.4.18)
respectively. Formulae (5.4.17) and (5.4.18), though looking different, give practically the same results in the whole range 3≤ n <∞. Consider several particular values of n . For an equilateral triangle ( n =3) formula (5.4.17) gives g = 0.3673. The value of g , which seems likely to be accurate, can be computed from (Solomon 1964a), and is equal to 0.3829, so that our result is in error by 4.1%. As we have seen earlier, the error of (5.4.17) for a square is 1.6%. Since formula (5.4.17) in the limiting case n →∞ gives the exact result for a circle g =2/π3/2=0.35917, we should expect that the error of (5.4.17) will decrease with n . For a regular pentagon g =0.3599. We did not find in the literature anything to compare with this result. The value of g for a regular hexagon is
342
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
0.3595 which is within the bounds given by Po´ lya and Szego¨ (1951) 0.35917< g <0.3635, and it is quite clear that the maximum possible error indeed decreases with n . It is noteworthy that the value of g does not change significantly in the whole range 3≤ n <∞. a2
Example 2: Triangle. In the case of a triangular punch with the sides a 1, and a 3, formula (5.4.15) simplifies as follows: b1 + b2 + a1 b2 + b3 + a2 1 6 1 + ln ln g = a2 b2 + b3 − a2 π√ A a 1 b 1 + b 2 − a 1 b 3 + b 1 + a 3 -1 1 . + ln a 3 b 3 + b 1 − a 3
(5.4.19)
The parameters in (5.4.19) can be determined from the well known geometric formulae A = [ p ( p − a 1)( p − a 2)( p − a 3)]1/2, b1 =
1 [2( a 21 + a 23) − a 22]1/2 , 3 b3 =
p = ( a 1 + a 2 + a 3)/2, b2 =
1 [2( a 22 + a 21) − a 23]1/2, 3
1 [2( a 23 + a 22) − a 21]1/2. 3
We are unaware of any report treating a triangle of general type but certain particular cases have been considered, so we can compare the results. When a 1= a 2= l , and the angle between these two sides is equal to α, the formula for the coefficient g can be rewritten in the form g =
6 2γ − α α α π γ -1 cot ) + ln tan( + ) , √tan(α/2) 2sin( )ln(cot 2 4 4 4 2 π (5.4.22)
where γ = tan (3tan(α/2)). -1
The isosceles triangle was considered by Rvachev et al (1977) who gave an approximate expression for the stress distribution but, for some reason, did not give the relationship between the total force and the punch settlement. They have though presented a graph depicting the location of the point of application of the force P as a function of the angle α. Their graph indicates a small variation (within 0.01 of the height of the triangle) of the coordinate about the center of gravity. We think that Rvachev et al did not realize that this fluctuation should be attributed to the approximate nature of their method, and
343
5.4 Flat punch of arbitrary planform under a normal force
that the exact location of the point is at the center of gravity. Of course, the previous statement should be understood as a conjecture since our method is also approximate. The case of α=π/2 was considered by Po´ lya and Szego¨ (1951) who gave the following bounds for g : 0.35917< g <0.4282. Okon and Harrington (1970) obtained g = 0.3867 as the most probable result. Our result is g = 0.374 which is well inside the admissible region and differs by 3.3% from the result of Okon and Harrington. The following bounds were established by Po´ lya and Szego¨ for a triangle with the sides a , a /2 and a √3 /2: 0.35917< g <0.517. Our result g = 0.3822 is within this interval. There seems to be no other source to compare with this result. Example 3: Rectangle. Consider a punch with a rectangular base, a and b being its semiaxes. Introduce the aspect ratio ε= a / b . Formula (5.4.15) in this case reduces to g =
2
.
(5.4.21)
1 1 sinh-1ε π √ε sinh-1( ) + ε √ε Howe (1920) suggested an approximate formula for the capacitance of a rectangle which in terms of the coefficient g reads 1
g =
.
(5.4.22)
1 1 1 (ε + 1) ε 2√ε sinh-1ε + sinh-1( )+ + 2 − 3 ε ε 3ε 3ε2 2
3/2
The result due to Solomon (5.4.16) in this case takes the form g =
29/8 ε 1 1/8 + . 12ε π11/8 12
(5.4.23)
We have found in the literature some numerical results which seem to be more or less accurate. Borodachev and Galin (1974) have considered the case of a narrow rectangular punch, and Noble (1960) investigated an equivalent problem of the electric charge distribution on a rectangular lamina. Their data, expressed in terms of the coefficient g , are presented below and compared with our result (5.4.21) and those due to Howe (5.4.22) and Solomon (5.4.23). The following relationship can be established between Borodachev-Galin’s coefficient γ and our g : γ=1/2π g √ε . Several useful conclusions can be drawn from the data presented. It seems logical to assume that the error of an approximate formula should change monotonically (or to have only one extremum) with respect to a certain
344
CHAPTER 5
ε = Borodachev and Galin Noble Formula (5.4.21) Howe (5.4.22) Solomon (5.4.23) Discrepancy (%) Formula (5.4.21) Howe (5.4.22) Solomon (5.4.23)
APPLICATION TO CONTACT PROBLEMS
0.020
0.050
0.100
0.125
0.150
0.200
0.250
0.500
1.000
0.7375 – 0.8031 0.6916 0.5402
0.5661 – 0.6072 0.5317 0.4819
0.4819 – 0.5037 0.4481 0.4423
– 0.4543 0.4771 0.4268 0.4304
0.4458 – 0.4576 0.4112 0.4211
0.4259 – 0.4306 0.3899 0.4071
– 0.4047 0.4128 0.3759 0.3969
– 0.3762 0.3742 0.3462 0.3715
– 0.3670 0.3612 0.3363 0.3613
-8.9 6.2 26.7
-7.3 6.1 14.9
-4.5 7.0 8.2
-5.0 6.1 5.3
-2.6 7.8 5.5
-1.1 8.5 4.4
-2.0 7.1 1.9
0.5 8.0 1.3
1.6 8.4 1.6
parameter. The fact that the discrepancy due to each formula jumps when moving from the data due to Noble to those by Borodachev and Galin, indicates that the results of at least one author are not exact. Our formula seems to perform better then the other two in a sufficiently wide range of aspect ratio. As expected, the formula due to Solomon performs well only when the aspect ratio is not far away from unity. There seems to be little change in the error of Howe’s formula (5.4.22). If this is really so, then its accuracy can be improved dramatically just by multiplication by a constant factor, say, 1.07. Example 4: Rhombus. Let α be the angle at one of the rhombus apices. Formula (5.4.15) in this case yields g =
2 cos(α/2) + sin(α/2) + 1 π√sinα ln cos(α/2) + sin(α/2) − 1
.
(5.4.24)
The same formula in terms of the rhombus semiaxes a and b and the aspect ratio ε = a / b has the form [2(ε + 1/ε)]1/2
g = π ln
.
(5.4.25)
1 + ε + (1 + ε ) 1 + ε − (1 + ε2)1/2
2 1/2
We did not find in mechanics literature any result related to a punch with a rhombus planform. In electrical sciences, the related problem of the capacity of a diamond was solved numerically by Okon and Harrington (1970). Their result, expressed in terms of the coefficient g , for a diamond with the aspect ratio a : b =0.7:1.65 is g =0.3855. Formula (5.4.25) gives g =0.3744 which differs by 3% from the result of Okon and Harrington. They also considered a rhombus with the aspect ratio 1:2. Their result, g =0.3705, almost coincides with ours which is
345
5.4 Flat punch of arbitrary planform under a normal force
g =0.3698. Example 5: Circular segment. Let the radius r and the angle 2α be the segment parameters. The location of its center of gravity is defined by x = kr, c where 2 sin3α . k = 1 3(α − sin2α) 2
(5.4.26)
The equation of the segment boundary with respect to its center of gravity takes the form a (φ) = r [− k cosφ + (1 − k 2sin2φ)1/2], and a (φ) = r
k − cosα cos(π − φ)
for 0≤φ≤π−γ or π+γ≤φ<2π;
for π−γ≤φ≤π+γ.
(5.4.27)
Substitution of (5.4.27) into (5.4.11−5.4.12) gives 2( α − g =
1 sin 2α)1/2 2
.
π γ π2E( k ) − E(γ, k ) − k sinγ + ( k − cosα) ln tan( + ) 4 2
(5.4.28)
where γ = tan-1(sinα/( k − cosα)). We have found only one numerical example to verify the accuracy of (5.4.28): Okon and Harrington (1970) have computed the capacity of a semicircle. Their result expressed through the coefficient g is 0.3724. Formula (5.4.28) gives 0.3714 with the discrepancy of 0.3%. Example 6: Circular sector. Repetition of the procedure, described in the previous paragraph, leads to the following result for a circular sector of subtended angle 2α: g =
2√α
.
πE(γ, k ) − k sinγ + k sinα ln[cot(α/2)cot((γ − α)/2)]
(5.4.29)
Here, k =2sinα/(3α), and γ=tan-1(sinα/(cosα− k )). Okon and Harrington (1970) obtained g =0.3668 for the case of a quadrant. Formula (5.4.29) for α=π/4 gives g =0.3639, with the discrepancy of 0.8%.
346
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
Discussion. We might enquire whether there exists any contour, other than an ellipse, for which an expression of the type (5.4.5) would be an exact solution to the integral equation (5.1.8). Expression (5.4.6) can provide the sufficient conditions: 2π
2 ⌠ ein(φ-φ0) F 1 − | n |, 1; 1; 1 − x dφ 2 2 ⌡ a 2(φ0) 0
(5.4.30)
0
should be equal to zero for n ≠0. The integral (5.4.30) will vanish for all odd In the case of even n , the n if a (φ) contains even harmonics only. hypergeometric function in (5.4.30) represents a finite polynomial in x / a (φ) of degree not greater than n −2 which means that the integral will vanish if ( a (φ))-2 contains harmonics not higher than the second, which corresponds to an ellipse. The question as to whether these conditions are also necessary requires an additional investigation. Solomon’s formula (5.4.14) can be considered as a particular case of a more general one, namely 2π
1 ⌠ ( a (φ))mdφ g = π√ A 2π ⌡ 2
0
2π
-1/2m
1 ⌠ ( a (φ))ndφ1/2n. 2π ⌡
(5.4.31)
0
for m =2 and n =4. One may ask now whether this choice of the parameters m and n is in any sense optimal. Direct computations show that for a regular polygon, m =1 and n =9 give much better accuracy than (5.4.18). In the case of a rectangle m =1 and n =7 are more accurate than (5.4.23). It is clear that the formula chosen by Solomon (5.4.14) is not the best particular case of (5.4.31). One can suggest many formulae of the type (5.4.31) but this would be just an exercise in curve-fitting, which is outside the scope of this book. Mossakovskii (1972) considered the case of a flat punch of nearly circular cross-section. He assumed a solution for the contact tractions in the form σ(ρ,φ) =
F 0(ρ,φ) + α F 1(ρ,φ) + α2 F 2(ρ,φ) + . . .
,
(5.4.32)
a 2(φ) − ρ2
1/2
where α is a small parameter. Expression (5.4.32) has two essential disadvantages as compared with (5.4.3): ( i )even in the case of an ellipse expression (5.4.32) gives an infinite series rather than the closed form exact solution; ( ii )the solution (5.4.32) at the point ρ=0 is a function of φ which is physically meaningless. Though Mossakovskii’s method can give reasonably
5.5 Inclined flat punch of general shape
accurate estimations for the relationship between the punch settlement and applied force, his claim of being able to evaluate the stress distribution exerted by the punch seems to be wrong because of incorrect assumption of a square root singularity at the punch boundary (for example, in the case of a square punch). The mathematically similar problem of sound penetration through an aperture of general shape in a rigid flat screen was considered in (Fabrikant, 1986b). The same method was used to find the electrical capacity of flat laminae (Fabrikant, 1986k).
1.
Exercise 5.4 Establish (5.4.6).
2.
Derive (5.4.7).
3.
Verify (5.4.15).
4.
Derive (5.4.19).
5. Generalize the theory for the case of a non-homogeneous half-space, with the modulus of elasticity being proportional to a power function of the depth.
5.5 Inclined flat punch of general shape An approximate analytical solution is given here to the contact problem of a flat inclined punch of arbitrary planform under the action of a normal non-centrally applied force. Some accurate relationships are established between the tilting moments and the angles of inclination of an arbitrary flat punch. Specific formulae are derived for a punch whose planform has the shape of a polygon, a triangle, a rectangle, a rhombus, a circular sector and a circular segment. All the formulae are checked against the solutions known in the literature, and their accuracy is confirmed. Theory. Consider a flat-ended punch with a planform S whose boundary is given in polar coordinates as ρ = a (φ). where the function a (φ) is bounded and single-valued. The punch is pressed against an elastic half-space by a normal force P applied at the point with cartesian coordinates x 0 and y 0. This loading is statically equivalent to a centrally applied force P and two tilting moments M x= Py 0 and M y=− Px 0. The case of a centrally applied force was considered in the previous section. It
347
348
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
remains here to consider the punch under the action of the tilting moments, and to superpose the results. Repeating the procedure of the section 5.4, we come to the same governing integral equation, namely, ρ
w (ρ,φ) =
2π
dx
a (φ0)
λ(
2 ⌠ ⌠ dφ ⌠ H π ⌡ (ρ2 − x 2)1/2 ⌡ 0 ⌡ 0
0
x2 , φ−φ0) ρρ0
(ρ20 − x 2)1/2
x
σ(ρ0,φ0)ρ0dρ0. (5.5.1)
Let the normal displacements under the punch be w = αx y − αy x ,
(5.5.2)
where αx and αy are the tilting angles about the axes Ox and Oy respectively. It is necessary to relate these angles to the tilting moments. Present the normal stress distribution under the punch as σ =
a (φ)ρ( p 1cosφ + p 2sinφ)
,
(5.5.3)
a 2(φ) − ρ21/2
where p 1 and p 2 are the as yet unknown constants. Make use of the condition that the integral of σ over S should be equal to zero. Since p 1 and p 2 are independent, this leads to the two equations 2π
⌠ ( a (φ))3cosφ dφ = 0 , ⌡ 0
2π
⌠ ( a (φ))3sinφ dφ = 0. ⌡
0
(5.5.4) It follows that (5.5.4) will be satisfied if and only if the origin of coordinates is located at the center of gravity of the domain of contact. The axis orientation will be discussed later. The relationships between the tilting moments and the parameters p 1 and p 2 can be established from the statics conditions M x = Py 0 = ⌠ ⌠ σ y dS ,
⌡⌡ S
which leads to
M y = − Px 0 = −⌠ ⌠ σ x dS,
⌡⌡ S
349
5.5 Inclined flat punch of general shape
8 ( p I + p 2 I x), 3 1 xy
Mx =
8 M y = − ( p 1 I y + p 2 I xy), 3
(5.5.5)
where I x, I y and I xy are the well known quantities of the moments of inertia and the product of inertia respectively. Now it is necessary to relate p 1 and p 2 to the angles αx and αy. This can be done by substitution of (5.5.3) into (5.5.1) which, after integration with respect to ρ0, yields ρ
∞
w (ρ, φ) = H
Σ
2 ⌠ (x )|n| 2 x d x 2 1/2 ⌡ ρ (ρ − x )
n=-∞
2π
×⌠ e
⌡ 0
in(φ-φ0)
0
3 − |n| 1 x2 , ; 1; 1 − 2 ( p cosφ0 + p 2sinφ0)dφ0. F 2 2 a (φ0) 1
(5.5.6)
Here F denotes the Gauss hypergeometric function. Further evaluation of the normal displacements can be carried out separately for each harmonic. Note that the zero th and all the even harmonics of w will be zero if a (φ) contains only even harmonics. The first harmonic will take the form 2π
w 1(ρ, φ) =
π ⌠ H ρ cos(φ − φ0)( p 1cosφ0 + p 2sinφ0) a (φ0) dφ0, 2 ⌡ 0
which can be simplified as w 1(ρ,φ) =
π H ρ[( p 1 J y + p 2 J xy)cosφ + ( p 1 J xy + p 2 J x)sinφ], 2
(5.5.7)
where the following quantities were introduced 2π
J x = ⌠ a (φ) sin2φ dφ ,
⌡
0
2π
J y = ⌠ a (φ) cos2φ dφ ,
⌡
0 2π
J xy = ⌠ a (φ) sinφ cosφ dφ
⌡
(5.5.8)
0
These quantities do not seem to have been used before in engineering practice so they do not have an accepted name. Since their tensor properties are similar to
350
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
those of the moments of inertia, we shall call J x and J y the linear moments of a two-dimensional domain about the axes Ox and Oy respectively, while J xy will be called the linear product of a two-dimensional domain about the axes Ox and Oy It is important to note that the third harmonic is equal to zero for an arbitrary contour. Here is the expression for the fifth harmonic 2π
128 4⌠ w 5(ρ,φ) = Hρ 315 ⌡ 0
cos5(φ − φ0) a 2(φ0)
( p 1cosφ0 + p 2sinφ0)dφ0,
which can be modified as w 5(ρ,φ) =
64 H ρ4{[( A + A ) p 1 + ( A − A ) p 2]cos5φ c6 c4 s6 s4 315
+ [( A
s6
+A ) p1 + ( A s4
− A c6) p 2]sin5φ}.
c4
(5.5.9)
Here, the following geometrical characteristics of the domain of contact were introduced 2π
A
c4
cos4φ dφ , = ⌠ ⌡ ( a (φ))2
2π
A
c6
0
cos6φ dφ , = ⌠ ⌡ ( a (φ))2 0 2π
2π
A
s4
sin4φ dφ = ⌠ , ⌡ ( a (φ))2 0
A
s6
sin6φ dφ = ⌠ . ⌡ ( a (φ))2
(5.5.10)
0
Investigation of further harmonics shows that their amplitude decreases. Now consider in more detail the case of a square of side 2 l . The equation of the boundary in this case is a (φ) = l /cosφ for −π/4<φ<π/4, and the pattern is repeated outside this range. We can evaluate the first two non-zero harmonics: w 1 = π H l ρln(1 + √2)( p 1cosφ + p 2sinφ), w5 =
128 H ρ4 ( p 1cos5φ + p 2sin5φ). 945 l 2
Since the amplitude of w 5 is significantly less than that of w 1, it seems natural to assume w ≈ w 1, and the remaining harmonics may be called the solution
351
5.5 Inclined flat punch of general shape
error. Direct computations show that the error is less than 3% inside the circle ρ≤ l . The error is reasonably small outside the circle, reaching about 20% at the apex and decreasing very rapidly with the distance from the apex. Taking into consideration that the error sign fluctuation will result in even smaller error in the integral characteristics sought, a direct comparison of (5.5.2) and (5.5.7) leads to π π (5.5.11) αx = H ( p 1 J xy + p 2 J x) , αy = − H ( p 1 J y + p 2 J xy). 2 2 The inversion of (5.5.11) gives p1 = −
2
J xyαx + J xαy
πH
J x J y − J xy2
,
p2 =
2
J yαx + J xyαy
πH
J x J y − J xy2
. (5.5.12)
Substitution of (5.5.12) in (5.5.5) finally gives the required relationship Mx =
16 ( m α + m 12αy) , 3π H 11 x
My =
16 ( m α + m 22αy) 3π H 21 x (5.5.13)
where m 11 = m 21 =
J y I x − J xy I xy J x J y − J xy2 J xy I y − J y I xy J x J y − J xy2
,
m 12 =
,
m 22 =
J xy I x − J x I xy J x J y − J xy2 J x I y − J xy I xy J x J y − J xy2
, .
It is clear that all these results can be rewritten in a matrix or a tensor form. One can verify that formulae (5.5.13) are invariant with respect to an arbitrary rotation of the axes. The same property holds for m 11+ m 22 and m 12− m 21. Strictly speaking, according to the reciprocal theorem, m 12 should be equal to m 21, so that formulae (5.5.13) generally do not satisfy this theorem. But we may state that this theorem is satisfied ’approximately’. We mean by this the following property which has been verified by several direct computations, namely, | m 12− m 21|/ m 11 <<1 and | m 12− m 21|/ m 22 <<1. This theorem will be satisfied exactly for any domain which has at least one axis of symmetry because in this case m 12= m 21=0, provided that the coordinate axes coincide with the central principal axes of the domain of contact. Since we have no numerical data for non-symmetrical domains which could be used to verify the accuracy of (5.5.13), we shall consider further only the case when the domain of contact has an axis of symmetry. In this case formulae (5.5.5), (5.5.11) and (5.5.13) simplify significantly, namely,
352
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
8 I p, 3 y 1
Mx =
8 I p, 3 x 2
My = −
αx =
π HJ p , 2 x 2
π α y = − H J y p 1, 2
Mx =
I 16 x α , 3π H J x x
My =
Returning back to the problem using the results of section 5.4, we traction distribution under the punch coordinates of its point of application σ =
P a (φ)
I 16 y α. 3π H J y y
(5.5.14)
(5.5.15)
(5.5.16)
of a non-centrally loaded flat punch and can write the following expression for the in terms of the applied force P and the x 0 and y 0 xx
yy
1 + 3 A 0 + 0, 4 Iy I x 1/2 2 2 2 A a (φ) − ρ
(5.5.17)
where A is the area of the domain S . An expression equivalent to (5.5.17) can be written in terms of the normal displacement δ and the tilting angles αx and αy σ =
2 a (φ) π H a 2(φ) − ρ2
1/2
αy αx δ + x − y , Jx Jy J0
(5.5.18)
where 2π
J 0 = ⌠ a (φ) dφ
⌡
0
The quantity J 0 may be called the polar linear moment due to the analogy with the moments of inertia and the property J 0= J x+ J y. One can note also that J 0 is proportional to the average polar radius. Expressions (5.5.17) and (5.5.18) are exact for an ellipse. We expect them to be reasonably accurate in the neighbourhood of the coordinate origin for an arbitrary punch planform with at least one axis of symmetry, while the error might become quite significant close to the boundary of the domain S . Let us rewrite formulae (5.5.16) in the form
353
5.5 Inclined flat punch of general shape
Mx =
A 3/2 hα , 2π H x x
My =
A 3/2 hα, 2π H y y
(5.5.19)
where hx =
32 I x 3 A 3/2 J x
,
hy =
32 I y 3 A 3/2 J y
.
(5.5.20)
We introduced the coefficients h x and h y for two reasons: since they are dimensionless they characterize the shape of S and do not depend on its size; both h x and h y are equal to the corresponding coefficients of magnetic polarizability which will simplify the comparison of our results with the numerical data available. There is an advantage of formulae (5.5.19) over the equivalent (5.5.16): the factors depending on the shape of S are separated from those depending on its size. One can draw from (5.5.19) an immediate conclusion that in the case when a domain S is magnified so that its linear dimensions double, its area quadruples, and the tilting moment should be multiplied by 8 in order to produce the same tilting angle. This conclusion is not so clear from (5.5.16). The remaining part of this section will be devoted to the evaluation of the coefficients h x and h y for various punch planforms. Several punch planforms are considered. Each configuration is related to its central principal axes and assumed to have at least one axis of symmetry coinciding with the axis Ox . The high degree of accuracy of formulae (5.5.20) is confirmed by comparison with available numerical solutions. Example 1: Polygon. Consider a polygon with n sides. The function a (φ) describing its boundary is bounded and single-valued. The origin of the coordinate system is located at the center of gravity, as before. Let us number the polygon sides in a counter-clockwise direction from 1 to n , a k being the length of the k th side. The apex at which the sides a k and a k+1 intersect is numbered k +1. It is clear that the value of index n +1 is to be understood as 1. We denote the distance from the center of gravity to the k th apex as b k; ψk stands for the angle between the axis Ox and the perpendicular to the side a k. Let A k be the area of the triangle formed by a k, b k and b k+1, the total area A of the polygon being equal to the sum of A k. The following expressions can be obtained for the moments of inertia n
Ix =
2 A 3k
Σ k=1
2 2 sin ψk + ak
b 2k+1 − b 2k 4Ak
sin2ψk +
a 4k + 3( b 2k+1 − b 2k)2 48 A 2k
cos2ψk,
(5.5.21)
354
CHAPTER 5
n
Iy =
2 A 3k
Σ k=1
2 2 cos ψk − ak
b 2k+1 − b 2k 4Ak
APPLICATION TO CONTACT PROBLEMS
sin2ψk +
a 4k + 3( b 2k+1 − b 2k)2 48 A 2k
sin2ψk.
(5.5.22) The linear moments can be computed in the form n
Jx =
Σ k=1
Ak a 2k
−( 1 + 1 )[ a 2 − ( b − b )2]cos2ψ k k+1 k b k+1 k bk
b k + b k+1 + a k 1 1 cos2ψk, + 4 A k( − ) sin2ψk + 2 a k ln b k + b k+1 − a k bk b k+1 (5.5.23) n
Jy =
Σ k=1
Ak a 2k
( 1 + 1 )[ a 2 − ( b − b )2]cos2ψ k k+1 k b k+1 k bk
b k + b k+1 + a k 1 1 sin2ψk. − 4 A k( − )sin2ψk +2 a k ln bk b k+1 b k + b k+1 − a k (5.5.24) Substitution of (5.5.21−5.5.24) into (5.5.20) gives the coefficients h x and h y for an arbitrary polygon. In the case of a regular polygon a k= a , b k= b = a /[2sin(π/ n )], Ak =[ a 2cot(π/n)]/4=[ b 2sin(2π/ n )]/2, ψk=2π( k -1)/ n , (5.5.21−5.5.24) simplify to Ix = Iy =
A = nA k,
and
formulae
1 2π 2π na 4 nb 4 π π cot cot2( ) + = sin 2 + cos , 64 3 24 n n n n (5.5.25)
Jx = Jy =
na nb π 1 + sin(π/ n ) π 1 + sin(π/ n ) cot ln = cos ln . 4 2 n 1 − sin(π/ n ) n 1 − sin(π/ n ) (5.5.26)
Substituting (5.5.25) and (5.5.26) in (5.5.20) leads to
hx = hy =
2π 16(2 + cos ) n
.
1 + sin(π/ n ) π π 9 n 3sin cos3( ) ln 1 − sin(π/ n ) n n 1/2
Consider several particular values of n .
(5.5.27) For an equilateral triangle ( n =3) formula
355
5.5 Inclined flat punch of general shape
(5.5.27) gives h x= h y=31/416/[27ln(2+√3 )]=0.5922. We did not find any numerical data to compare with this result. In the case of a square n =4, and h x= h y=4/[9ln(1+√2 )]=0.5043 which is inside the interval from 0.4973 to 0.5162 given by Okon and Harrington (1981) and within 3% from the result of de Smedt (1979) which is 0.5193. Since formula (5.5.27) gives the exact result for a circle in the limiting case of n →∞, namely, h x= h y=8/(3π3/2)=0.4789, we should expect that the error of (5.5.27) will decrease with n . The value of the 1/4 coefficients for a regular hexagon is h x= h y=40√2 /(3 81 ln3)=0.4830 which differs by 1.4% from the result 0.49 due to Okon and Harrington (1981), and it is quite clear that the maximum possible error indeed decreases with n . It is noteworthy that the values of the coefficients do not change significantly in the whole range 3≤ n <∞. In the case of a triangle with sides Example 2: Isosceles triangle. a 1= a 2= l , the angle between them being equal to α, formulae (5.5.20−5.5.23) give Ix =
1 4 l sinα sin2(α/2), 12
Jx =
2 α l cos sinα + sin(α + γ) − 2sinγ 3 2
Iy =
1 4 l sinα cos2(α/2), 36
2γ − α α α π γ + 2sin3( )ln(cot cot ) + ln tan( + ), 2 4 4 4 2 Jy =
2 α l cos −sinα − sin(α + γ) + 2sinγ 3 2 2γ − α α α + sinαcos ln(cot cot ), 2 4 4
with the result for the coefficients
h x = 8(tan(α/2))3/23sinα + sin(α + γ) − 2sinγ
+ 2sin3
2γ − α α α π γ -1 ln(cot cot ) + ln tan( + ) , 2 4 4 4 2
(5.5.28)
356
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
h y = 8√cot(α/2) 9−sinα − sin(α + γ) + 2sinγ
α α -1 2γ − α cot ) , + sinαcos ln(cot 4 4 2
where γ = tan-1(3tan(α/2)). The isosceles right triangle was considered by Okon and Harrington (1981) who gave the interval between 0.9829 and 1.021 for only one of the coefficients, which in our notation is h x. Our result for h x is 0.9255 which differs by less than 10% from theirs. The second formula (5.5.28) gives h y = 0.3995, and we found nothing in the literature to compare with this result. Example 3: Rectangle. Consider a punch with a rectangular base, a 1 and Introduce the aspect ratio ε= a 2/ a 1. Formulae a 2 being its semiaxes. (5.5.21−5.5.24) in this case reduce to 4 a a 3, 3 1 2 J x = 4 a 1sinh-1ε ,
4 a 3a , 3 1 2 J y = 4 a 2sinh-1(1/ε).
Ix =
Iy =
and formulae (5.5.20) yield hx =
4ε3/2 9sinh-1ε
,
hy =
4ε-3/2 9sinh-1(1/ε)
.
(5.5.29)
The coefficients of magnetic polarizability were computed by de Smedt (1979) for a rectangle with various aspect ratios ε. Here, we present his results along with those given by (5.5.29) Our formula (5.5.29) seems to perform satisfactorily in a sufficiently wide range of aspect ratio. The approximate expression for the stress distribution under the punch according to (5.5.17) takes the form σ =
P a (φ)
xx
yy
1 + 9 0 + 0. 4 a 21 a 22 1/2 2 2 8 a 1 a 2 a (φ) − ρ
(5.5.30)
Expression (5.5.30) can be used for analyzing the process of movement of the applied force P , say, along the axis Ox . This analysis can be done by
357
5.5 Inclined flat punch of general shape
ε= de Smedt h x= Formula (32) h x= de Smedt h y= Formula (32) h y= Discrepancy in h x (%) Discrepancy in h y (%)
0.1000 0.1287 0.1408 4.1070 4.6876 -9.4 -14.1
0.2000 0.1881 0.2001 2.0260 2.1488 -6.4 -6.1
0.3333 0.2531 0.2612 1.2600 1.2701 -3.2 -0.8
0.5000 0.3249 0.3265 0.8892 0.8708 -0.5 2.1
0.7500 0.4240 0.4165 0.6426 0.6228 1.8 3.1
0.8000 0.4436 0.4341 0.6130 0.5929 2.2 3.3
1.0000 0.5193 0.5043 0.5193 0.5043 2.9 2.9
requiring that the contact traction vanish at the edge. One can conclude from (5.5.17) that the boundary at which this occurs will always be a straight line. It is clear from (5.5.30) that the punch will be in contact with the half-space as long as x 0≤4 a 1/9, after which the punch will start separating from the half-space. Assuming that the new domain of contact is also a rectangle (of course, with a different aspect ratio), one can again apply the formulae of this section to analyze the process further. If we denote the width of the zone of separation by c , the following relationship holds c =
2 (9 x − 4 a 1) , 5 0
4 for x 0≥ a 1. 9
The last formula states, for example, that when the force P is applied at x 0=13 a 1/18 only a half of the punch will be in contact with the half-space. Unfortunately, there is no data to verify these relationships. Further analysis reveals that the core inside which the force can be applied without causing any separation is a rhombus with semiaxes 4 a 1/9 and 4 a 2/9 respectively. As one knows, in the case of a circular punch the core is a circle of radius equal to one third of the radius of the punch. The results due to (5.5.30) can be compared with the numerical data received in a personal communication from de Smedt. In order to make the comparison possible, we must set P =0, M x=0 in (5.5.30) and replace M y by (5.5.19), with the result σH =
9√ε a (φ) h y x
.
(5.5.31)
4 a 1 a 2(φ) − ρ2
1/2
Computations due to (5.5.31) were made for ε=0.5 along the axis Ox , the value h y was taken 0.8708 (see the preceding Table). Here are the results compared with those communicated by de Smedt Since the numerical method of De Smedt is also approximate, we are using the word discrepancy rather than the word error in the tables throughout this and adjacent sections. We can also compare
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x/a 1= 0.0833 0.2500 0.3333 0.5000 0.5833 0.6667 0.7500 0.8333 0.9167 de Smedt σ H = 0.1143 0.3501 0.4759 0.7523 0.9367 1.1460 1.4304 1.8303 2.8182 Formula (5.5.31) σ H = 0.1159 0.3577 0.4898 0.7999 0.9950 1.2392 1.5709 2.0886 3.1777 Discrepancy (%) -1.3 -2.2 -2.9 -6.3 -6.2 -8.1 -9.8 -14.1 -12.8
the same values along the axis Oy . One can use a formula similar to (5.5.31) replacing all x by y and interchanging a 1 and a 2, the value of h x was taken to be 0.3265. The results are: y/a 2= de Smedt σ H = our result σ H = Discrepancy (%)
0.1667 0.1756 0.1756 0.0
0.3333 0.3663 0.3673 -0.3
0.5000 0.6011 0.5998 0.2
0.6667 0.9014 0.9292 -3.1
0.8333 1.6413 1.5662 4.6
The agreement is satisfactory. Example 4: Rhombus. Let α be the angle at one of the rhombus apices, and l be its side. Formulae (5.5.21−5.5.24) in this case yield Ix =
14 α l sinα sin2( ), 6 2
Iy =
14 α l sinα cos2( ), 6 2
A = l 2sinα,
α α α cos(α/2) + sin(α/2) + 1 J x = 2 l sinαcos − sin + sin2( )ln , 2 2 2 cos(α/2) + sin(α/2) − 1 α α α cos(α/2) + sin(α/2) + 1 . J y = 2 l sinα−cos + sin + cos2( )ln 2 2 2 cos(α/2) + sin(α/2) − 1 The coefficients are determined as 8sin2 hx =
α 2
α α α cos(α/2) + sin(α/2) + 1 9(sinα)3/2cos − sin + sin2( )ln 2 2 2 cos(α/2) + sin(α/2) − 1 8cos2
hy =
,
α 2
.
α α α cos(α/2) + sin(α/2) + 1 9(sinα)3/2−cos + sin + cos2( )ln 2 2 2 cos(α/2) + sin(α/2) − 1 (5.5.32)
359
5.5 Inclined flat punch of general shape
The same formulae can be expressed in terms of the rhombus semiaxes a and b and the aspect ratio ε= b / a , giving hx =
2√2ε (1 + ε2) 91 − ε +
hy =
,
ε 1 + ε + (1 + ε ) 2 1/2 ln (1 + ε ) 1 + ε − (1 + ε2)1/2 2
2 1/2
2√2 (1 + ε2) 9ε3/2ε − 1 +
.
(5.5.33)
1 + ε + (1 + ε ) 1 + ε − (1 + ε2)1/2 2 1/2
1 (1 + ε )
2 1/2
ln
We did not find in mechanics literature any result related to a punch with a rhombus planform. In the electrical sciences, the mathematically equivalent problem of the coefficients of magnetic polarizability of a diamond was solved numerically by de Smedt (1979). Here, we present his results compared with those given by formula (5.5.33) ε= de Smedt h x= Formula (5.5.33) h x= de Smedt h y= Formula (5.5.33) h y= Discrepancy of h x (%) Discrepancy of h y (%)
0.1000 0.1181 0.1078 6.1820 4.5987 8.7 25.6
0.2000 0.1729 0.1627 2.7060 2.1982 5.9 18.8
0.3333 0.2341 0.2258 1.5240 1.3254 3.6 13.0
0.5000 0.3052 0.2986 0.9946 0.9095 2.2 8.6
0.7500 0.4101 0.4026 0.6703 0.6388 1.8 4.7
0.8000 0.4323 0.4230 0.6323 0.6052 2.1 4.3
1.0000 0.5193 0.5043 0.5193 0.5043 2.9 2.9
The deterioration of the accuracy of (5.5.33) for small values of ε can be attributed to the erroneous assumption of a square root singularity in (5.5.3) which is grossly incorrect for domains with sharp angles. The traction distribution under the punch can be expressed according to (5.5.17) as xx
yy
1 + 9 0 + 0. σ = 2 a 2 b 2 1/2 2 2 2 A a (φ) − ρ P a (φ)
Further analysis of the last expression reveals that the core inside which the force can be applied without causing any separation is a rectangle with semiaxes 2 a /9 and 2 b /9 respectively. In the case of ε=1 the rhombus transforms into a square, and all the results are in agreement with those of the Example 3.
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Example 5: Circular segment. Let the radius r and the angle 2α be the segment parameters. The location of its center of gravity is defined by x = kr, c
where k is given by (5.4.26). The equation of the segment boundary with respect to its center of gravity takes the form (5.4.27). Computation of the area and moments (5.5.21−5.5.24) yields A = r 2(α −
1 sin2α), 2
Ix =
1 2 Ar (1 − k cosα), 4
Iy =
1 2 Ar (1+3 k cosα−4 k 2), 4 (5.5.34)
Jx =
1 − k2 2 F(π−γ, k ) r − k sin3γ + (1 − k 2sin2γ)1/2sinγcosγ + 3 k2
+
Jy =
2k2 − 1 −sinγ + ln tan(π + γ ), E( , ) + 3( cos ) k k π − γ − α 4 2 k2
2 r sinγ k sin2γ − 3cosα − (1 − k 2sin2γ)1/2 cosγ 3
−
1 − k2 1 + k2 F( , ) + E( , ) k k π − γ π − γ , k2 k2
(5.5.35)
where γ = tan-1(sinα/( k − cosα)), and the functions F(⋅,⋅) and E(⋅,⋅) stand for the incomplete elliptic integrals of the first and the second kind respectively. Substituting in (5.5.20) leads to
hx =
+
4(1 − k cosα)
3 2 2 1/2 −k sin γ + (1 − k sin γ) sinγcosγ +
α − 1sin2α1/2 2
1− k 2 F(π−γ, k ) k2
2 k 2−1 −sinγ + ln tan(π + γ )-1, E( , ) + 3( cos ) k k π − γ − α 4 2 k2
hy =
4(1 + 3 k cosα − 4 k 2)
α − 1sin2α1/2 2
2 2 2 1/2 sinγk sin γ − 3cosα − (1 − k sin γ) cosγ
361
5.5 Inclined flat punch of general shape
−
-1 1 − k2 1 + k2 π − γ π − γ F( , k ) + E( , k ) . k2 k2
(5.5.36)
We can use this result to investigate the case of a circular punch under the action of a normal force P applied at x 0> r /3. From the classical theory we know that there should be a separation between the punch and the half-space. Assuming that the domain of contact after separation is a circular segment, one can get the following relationship between the coordinate x 0 and the size of the segment characterized by the angle α x0 = r
(1 − k )(1 + 4 k ) . 3( k − cosα)
(5.5.37)
The last expression is exact in two limiting cases: the complete circle α=π gives x 0= r /3, and α→0 results in x 0= r . The problem of an inclined circular punch was considered numerically in the book by Rvachev and Protsenko (1977). Here, we compare the results α (deg)= Rvachev et al. x 0= Formula (5.5.37) x 0= Discrepancy (%)
158.4 0.3583 0.3543 1.1
108.1 0.5833 0.5418 7.1
102.0 0.6250 0.5750 8.0
The agreement should be considered as surprisingly good, especially taken into consideration that Rvachev and Protsenko considered the domain of contact not in the form of a segment but having a more complicated shape. Consider a punch configuration obtained by the Example 6: Cross. orthogonal intersection of two equal rectangles with sides 2 a and 2 b . Introduce the aspect ratio as ε= b / a The area and the moments will take the form A = 4 a 2ε(2 − ε) ,
Ix = Iy =
J x = J y = 4 a ln[ε + (1 + ε2)1/2] + ε ln
4 4 a ε(1 + ε2 − ε3), 3 1 + (1 + ε2)1/2 . (1 + √2)ε (5.5.38))
The coefficients will be determined as hx = hy =
4ε(1 + ε2 − ε3) 1 + (1+ε2)1/2-1 2 1/2 ln[ + (1+ ) ] + ln . ε ε ε 9ε(2 − ε)3/2 (1 + √2)ε
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(5.5.39) The values of h = h x= h y were computed from (5.5.39) and compared with those given by de Smedt (1979). The results are presented below ε= de Smedt h = Formula (5.5.39) h = Discrepancy (%)
0.1000 0.2000 0.3333 0.4000 0.5000 0.6000 0.7500 0.8000 1.0000 1.5910 0.8720 0.6255 0.5725 0.5267 0.5069 0.4985 0.4997 0.5193 1.7382 0.8758 0.6006 0.5465 0.5049 0.4890 0.4893 0.4926 0.5043 -9.3 -0.4 4.0 4.5 4.1 3.5 1.9 1.4 2.9
Taking into consideration the shape complexity, we should consider the results agreement as surprisingly good, not only quantitatively but qualitatively as well: both sets of data display a relatively flat minimum around ε=0.75. Discussion. It is noteworthy that the change of the order of integration which led to (5.5.1) is valid inside the circle ρ≤ min { a (φ)} only. Nevertheless, one can obtain from (5.5.1) the exact solution for an ellipse and sufficiently accurate formulae for various punch planforms as it was demonstrated in the preceding Examples. The accuracy of formulae (5.5.20) can be improved by taking into consideration the fifth harmonic (5.5.9) in combination with the variational approach (Noble, 1960). The following functional assumes its maximum value at the exact solution of (5.1.8) I (σ) =
σ( N ) 2⌠ ⌠ σ( M ) w ( M )d S M − ⌠ ⌠ σ( M )⌠ ⌠ d S Nd S M. H⌡ ⌡ R ( M,N ) ⌡ ⌡ ⌡⌡ S
S
(5.5.40)
S
Taking σ( N ) d S ≈ w 1 + w 5, ⌡ ⌡ R ( M,N) N
H⌠ ⌠
(5.5.41)
S
and substituting (5.5.3), (5.5.7), (5.5.9), and (5.5.41) in (5.5.40), one gets, after integration with respect to ρ 2π
4 I = ⌠ ( a (φ))4( p 1cosφ + p 2sinφ) (αxsinφ − αycosφ) ⌡ 3 H 0
−
π π 4 ( p J + p 2 J xy)cosφ − ( p 1 J xy + p 2 J x)sinφ − ( a (φ))3([ p 1( A c6 + A c4) 63 3 1 y 3
363
5.5 Inclined flat punch of general shape
+ p 2( A s6 − A s4)]cos5φ + [ p 1( A s6 + A s4) + p 2( A c4 − A c6)]sin5φ)dφ. Considering now the functional I as a function of p 1 conditions ∂I = 0 , ∂p1
(5.5.42) and p 2, the extremum
∂I = 0, ∂p2
give two linear algebraic equations with respect to the unknowns p 1 and p 2. The complete solution is rather cumbersome. Here, we present the final result for the coefficients h x and h y which are valid only for domains having at least one axis of symmetry, and the central principal axes taken as the coordinate axes
hx =
32 I x 3 A 3/2 J x(1 + ηx)
,
hy =
32 I y 3 A 3/2 J y(1 + ηy)
,
(5.5.43)
where the correction terms ηx =
( B c4 − B c6)( A c4 − A c6) 42π I x J x
,
ηy =
( B c4 + B c6)( A c4 + A c6) 42π I y J y
, (5.5.44)
the parameters A c4 and A c6 are defined by (5.5.10), and 2π
2π
B c6 = ⌠ ( a (φ)) cos6φ dφ,
⌡
7
0
B c4 = ⌠ ( a (φ))7cos4φ dφ.
⌡
0
Since expression (5.5.41) is approximate, there is no guarantee that (5.5.43) will be more accurate than (5.5.20). We performed the necessary computations for a rectangle. Here are the results compared with those of de Smedt (1979) ε= de Smedt h x= Formula (5.5.43) h x= de Smedt h y= Formula (5.5.43) h y= Discrepancy in h x (%) Discrepancy in h y (%)
0.1000 0.1287 0.1405 4.1070 4.5856 -9.2 -11.7
0.2000 0.1881 0.1988 2.0260 2.0985 -5.7 -3.6
0.3333 0.2531 0.2577 1.2600 1.2479 -1.8 1.0
0.5000 0.3249 0.3207 0.8892 0.8714 1.3 2.0
0.7500 0.4240 0.4165 0.6426 0.6463 1.8 -0.6
0.8000 0.4436 0.4376 0.6130 0.6190 1.3 -1.0
1.0000 0.5193 0.5331 0.5193 0.5331 -2.7 -2.7
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Comparison with similar data computed on the basis of formula (5.5.29) shows that the correction terms ηx and ηy in this particular case resulted in decreasing the value of discrepancy, positive as well as negative. We caution again that there is no guarantee that this will be valid for an arbitrary domain. For example, here are the data computed for a rhombus ε= de Smedt h x= Formula (5.5.43) h x= de Smedt h y= Formula (5.5.43) h y= Discrepancy of h x (%) Discrepancy of h y (%)
0.1000 0.1181 0.2268 6.1820 8.5600 -92.0 -38.5
0.2000 0.1729 0.1860 2.7060 2.5916 -7.6 4.2
0.3333 0.2341 0.2351 1.5240 1.4196 -0.4 6.8
0.5000 0.3052 0.3031 0.9946 0.9408 0.7 5.4
0.7500 0.4101 0.4058 0.6703 0.6490 1.0 3.2
0.8000 0.4323 0.4264 0.6323 0.6138 1.4 2.9
1.0000 0.5193 0.5091 0.5193 0.5091 2.0 2.0
Comparison with the data computed due to (5.5.33) indicates that the discrepancy decreased for ε≥0.2 while for ε=0.1 it has jumped in the opposite direction to −92%. The main reason for this is a jump in the value of the coefficients ηx and ηy when ε is very small. The following rule of thumb may be suggested for the user wishing to improve the accuracy: when the value of the correction coefficients ηx and ηy does not exceed small percentage of unity, their inclusion generally results in an improvement in accuracy, otherwise one should not use formulae (5.5.43). It is worthwhile to give the solution due to (5.5.42) for the case when the domain of contact has no axis of symmetry, and only the first harmonic of the displacements w 1 is taken into consideration. The result is
p1 =
p2 = where
αx( c 22 I xy − c 12 I x) + αy( c 12 I xy − c 22 I y) c 11 c 22 − c 122 αx( c 11 I x − c 12 I xy) + αy( c 12 I y − c 11 I xy)
(5.5.45)
c 11 c 22 − c 122
c 11 =
πH ( J y I y + J xy I xy) , 2
c 12 =
πH ( J xy( I x + I y) + I xy( J x + J y)) 4
c 22 =
πH ( J x I x + J xy I xy) , 2
Formulae (5.5.45) are different from the equivalent set (5.5.12) derived earlier. In the absence of any numerical data for a general domain, it is impossible to
365
5.5 Inclined flat punch of general shape
say whether formulae (5.5.45) are more accurate than (5.5.12), but they are definitely more complicated. It is noteworthy that in the case of a domain with an axis of symmetry both (5.5.45) and (5.5.12) simplify to the same equations (5.5.15). One can notice a certain similarity between the formulae derived and those related to the Saint-Venant theory of bending. This similarity will become more evident if, for example, we rewrite equation (5.5.17) in the form σ =
a (φ) 2 a 2(φ) − ρ2
1/2
M y M x P + 3 x − y . 4 I x I y A
We think that this similarity is not a pure coincidence since the method used in this section can also be called semi-inverse. The method can be developed further into a complete Saint-Venant type theory of elastic contact problems which could combine the simplicity and accuracy sufficient for a practical engineer. The mathematically identical problem of the magnetic polarizability of small apertures was solved in (Fabrikant, 1987j). Exercise 5.5 1. Find the moments of inertia and the J -moments for a circular sector characterized by the radius r and the polar angle 2α. Answer: 9α2 + 9αsinαcosα − 16sin2α , 36α
Ix =
1 4 1 r (α − sin2α), 4 2
Jx =
2 2 k 2−1 1− k 2 F(γ, k ) r 2 E(γ, k ) − k sin3γ − (1 − k 2sin2γ )1/2sinγcosγ + 3 k k2
I y = r4
α γ − α + 3 k sinαcosα + cos(α + γ) + sin2α ln(cot cot ) , 2 2
Jy =
2 1 − k2 F(γ, k ) r k sinγ(sin2γ − 3) + (1 − k 2sin2γ)1/2sinγcosγ − 3 k2
366
+
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
1+ k 2 cos2α ln(cotαcotγ−α) − cosα − cos(α+γ). E( , ) + 3 sin γ k k α 2 2 k2
Here k = 2sinα/(3α), and γ = tan-1(sinα/(cosα − k )). 2. Find the shape coefficients h x and h y for the circular sector described in the Exercise 1. Answer:
hx =
+
2(2α − sin2α) 1 − k 2 F(γ, k ) − k sin3γ − (1 − k 2sin2γ)1/2sinγcosγ 3/2 2 α k
2 k 2−1 cosα + cos(α+γ) + sin2α ln(cotαcotγ−α)-1, E( , ) + 3 sin k k γ α 2 2 k2
hy =
4(9α2 + 9αsinαcosα − 16sin2α) 2 k sinγ(sin γ − 3) 5/2 9α
+ (1 − k 2sin2γ)1/2sinγcosγ −
1 − k2 1 + k2 F( , ) + E(γ, k ) k γ k2 k2
α γ − α -1 + 3 k sinα−cosα − cos(α + γ) + cos2α ln(cot cot ) . 2 2
3.
Verify (5.5.4).
4.
Derive (5.5.7).
5.
Establish (5.5.14) −(5.5.16)).
367
5.6 Curved punch of general planform
5.6 Curved punch of general planform In this section, we analyze the elastic contact problem for a curved punch of non-elliptic planform under the action of a normal force. The punch base is assumed to be a quadratic surface. Some general relationships are established between the applied force and the punch settlement. Specific formulae are derived for a punch whose planform has the shape of a polygon, a rectangle, a rhombus and a cross. An example of a finite rigid cylinder lying on its generator and pressed against an elastic half-space is considered in detail. The method allows us to have singular tractions at the cylinder edges and zero tractions at the rest of the contact domain boundary. The last condition serves to determine the width of the domain of contact. All the formulae are checked against the previously published solutions, and a good accuracy is confirmed for a sufficiently wide range of values of aspect ratio. Theory. Consider a punch with a curved base and arbitrary planform. The punch is pressed against an elastic half-space by a normal force P . Let the boundary of the domain of contact S be given in the polar coordinates as ρ = a (φ), where the function a (φ) is bounded and single-valued. Here we assume that the domain of contact is prescribed. The case when the domain of contact is unknown (or partially unknown) will be discussed further. The punch base is assumed to be a quadratic surface. This limitation is not essential. The method can be applied to higher order surfaces as well. The transformed governing integral equation is given by (5.5.1). Let the normal displacements under the punch be w = g 0 + g x y 2 + g xy xy + g y x 2,
(5.6.1)
where g 0 is the punch penetration, and g x, g y and g xy are the known constants defined by the punch base geometry. Let the pressure distribution under the punch be σ(ρ,φ) =
a (φ)[α0 + ρ2(αxsin2φ + αxysinφ cosφ + αycos2φ)]
,
(5.6.2)
a 2(φ) − ρ2
1/2
where α0, αx, αy and αxy are the as yet unknown constants. We make use of the condition that the integral of σ over S should be equal to P . This leads to the expression
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CHAPTER 5
P = 2α0 A +
APPLICATION TO CONTACT PROBLEMS
8 (αx Ix + αxy Ixy + αy Iy ), 3
(5.6.3)
where A is the area of the domain of contact S ; and Ix , Iy and Ixy are the axial moments of inertia and the product of inertia respectively. The location of coordinate origin can be defined from the condition that the tilting moments produced by the tractions (5.6.2) should vanish. This condition yields the following set of equations: 2π
⌠ a 3(φ)[α + 3 a 2(φ)(α sin2φ + α sinφ cosφ + α cos2φ)]sinφdφ = 0, 0 x xy y 4 ⌡ 0
2π
⌠ a 3(φ)[α + 3 a 2(φ)(α sin2φ + α sinφ cosφ + α cos2φ)]cosφdφ = 0. 0 x xy y 4 ⌡
(5.6.4)
0
The point whose coordinates satisfy (5.6.4) will be called the punch centre. When the domain of contact has one axis of symmetry the punch centre is located on this axis. In the case of two axes of symmetry, the punch centre coincides with the centre of gravity. Now it is necessary to relate α0, αx, αy, and αxy to the parameters g 0, g x, g y, and g xy. This can be done by substitution of (5.6.2) into (5.5.1) which yields, after integration with respect to ρ0, ∞
w (ρ,φ) = H α0
Σ
ρ
2π
2 ⌠ ein(φ-φ0) F 2−| n |, 1; 1; 1 − x dφ ⌠ (x )|n| 2 2 1/2 2 2 a 2(φ0) 0 ⌡ ρ (ρ − x ) 0⌡ n=-∞
xdx
0
∞
+ H
Σ n=-∞
1−
ρ
2π
x 3d x ⌠ ein(φ-φ0) F (4−| n |, 1; 1; ⌠ (x )|n| 2 2 1/2 2 2 ⌡ ρ (ρ − x ) ⌡
0
0
x2 )(αxsin2φ0 + αxysinφ0cosφ0 + αycos2φ0)dφ0. 2 a (φ0)
Here F denotes the Gauss hypergeometric function. Further evaluation of the normal displacements can be carried out separately for each harmonic. Note that all the odd harmonics of w will be zero if a (φ) contains only even harmonics. The zero th harmonic will take the form
369
5.6 Curved punch of general planform
2π
π 1 2 w 0(ρ,φ) = H⌠ 2α0 + ( a 2(φ0) + ρ) 2 4 ⌡ 0
×(αxsin2φ0 + αxysinφ0cosφ0 + αycos2φ0) a (φ0)dφ0,
(5.6.5)
which can be simplified as π ρ2 H 2α0 J0 + αx B x + αxy B xy + αy B y + (αx Jx + αxy Jxy + αy Jy ), 4 2 (5.6.6) where the J -moments were defined in the previous section, and the following additional quantities were introduced: w 0(ρ,φ) =
2π
2π
B x = ⌠ a (φ)sin φ dφ , 3
⌡
B y = ⌠ a 3(φ)cos2φ dφ ,
2
⌡
0
0
2π
B xy = ⌠ a 3(φ)sinφ cosφ dφ.
(5.6.7)
⌡
0
Since the tensor properties of the B -moments are similar to those of the moments of inertia, we shall call B x and B y the cubic moments of a two-dimensional domain about the axes Ox and Oy respectively, B xy will be called the cubic product of a two-dimensional domain about the axes Ox and Oy . Here is the expression for the second harmonic 2π
w 2(ρ,φ) =
3 π H ρ2 ⌠ (αxsin2φ0 + αxysinφ0cosφ0 + αycos2φ0) a (φ0)cos2(φ−φ0)dφ0, 8 ⌡ 0
which can be modified as w 2(ρ,φ) = − C
xxyy
3 π H ρ2{−αx[( C − C ) cos2φ + 2 C sin2φ] + αy[( C xxxx xxyy xxxy yyyy 8
)cos2φ + 2 C
xyyy
sin2φ] + αxy[( C
xyyy
− C
xxxy
)cos2φ + 2 C
xxyy
sin2φ]}.
(5.6.8) Here, the following geometrical characteristics of the domain of contact have been
370
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
introduced: 2π
2π
Cxxxx = ⌠ a (φ)sin4φdφ,
Cxxxy = ⌠ a (φ)sin3φcosφdφ,
⌡
⌡
0
0
2π
2π
Cxxyy = ⌠ a (φ)sin2φcos2φdφ,
Cxyyy = ⌠ a (φ)sinφcos3φdφ,
⌡
⌡
0
0
2π
Cyyyy = ⌠ a (φ)cos4φdφ.
(5.6.9)
⌡ 0
The C -moments will be called the linear moments of the fourth order. Their relationships with the J -moments are easy to establish, for example, Jx = Cxxxx + Cxxyy , Jxy = Cxyyy + Cxxxy , etc .
It is important to note that the parameter α0 did not appear
in (5.6.8), and the parameters αx, αxy and αy will not appear in the expression for the fourth harmonic. The investigation of further harmonics shows that their amplitude decreases. In the case of an ellipse they vanish thus making the solution exact. It seems natural to assume w ≈ w 0+ w 2, and the remaining harmonics may be called the solution error. We should also take into consideration that the error sign fluctuation will result in even smaller error in the integral characteristics sought, like the total force P . Now we have an approximate expression for the displacements under the punch
w =
π H 2α J + αx B x + αxy B xy + αy B y + x 2−αx( Cxxxx − 2 Cxxyy ) 4 0 0
+ αxy(2 Cxyyy − Cxxxy ) + αy(2 Cyyyy − Cxxyy ) + y 2αx(2 Cxxxx − Cxxyy )
+ αxy(2 Cxxxy − Cxyyy ) − αy( Cyyyy − 2 Cxxyy ) + 6 xy αx Cxxxy + αxy Cxxyy + αy Cxyyy .
(5.6.10) Comparison of (5.6.1) and (5.6.10) leads to the following set of equations π H (2α0 J0 + αx B x + αxy B xy + αy B y) = g 0, 4
371
5.6 Curved punch of general planform
π H α (2 C − C ) + αxy(2 C − C ) − α y( C − 2 C ) = g x xxxx xxyy xxxy xyyy yyyy xxyy 4 x π H −α ( C − 2 C ) + αxy(2 C − C ) + α y(2 C − C ) = g y, xxyy xyyy xxxy yyyy xxyy 4 x xxxx 3π + αxy C +αy C = g xy. H αx C xxyy xyyy 2 xxxy (5.6.11) The last three equations of (5.6.11) can be solved with respect to αx, αy, and αxy. In the case of elastic contact problem when g 0 is prescribed, the value of α0 can be found from the first equation (5.6.11), after which the total force P is defined by (5.6.3), and the solution is completed. When the total force P is given, the value of α0 can be determined from (5.6.3) after which the punch penetration g 0 is given by the first equation (5.6.11). A significant simplification occurs when the domain of contact S has at least one axis of symmetry. In this case C = C = B xy=0. The last equation xxxy
xyyy
(5.6.11) becomes decoupled from the previous three. written explicitly αx =
− C
4[ g x(2 C
yyyy
xxyy
3π H ( C
xxxx
αy =
) + g y( C
yyyy
C
yyyy
xxxx
3π H ( C 2 g xy 3π H C
− 2C
xxyy
)]
− C 2xxyy)
− 2 Cxxyy ) + g y(2 Cxxxx − Cxxyy )]
4[ g x( C
xxxx
αxy =
The solutions can be then
C
yyyy
− C 2xxyy)
,
,
.
(5.6.12)
xxyy
Substitution of (5.6.12) into the first equation (5.6.11) gives α0 =
2 g − π H J0 0 3( C
g xβx + g yβy
xxxx
where
βx = B x(2 C
yyyy
− C
xxyy
C
yyyy
C 2xxyy)
−
) + B y( C
xxxx
,
(5.6.13)
− 2C
xxyy
),
372
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
βy = B x( Cyyyy − 2 Cxxyy ) + B y(2 Cxxxx − Cxxyy ). Expressions (5.6.2), (5.6.3), (5.6.12), and (5.6.13) give a complete and exact solution for an ellipse. We shall prove them to perform well for an arbitrary punch planform. We expect (5.6.2) to be reasonably accurate in the neighbourhood of the coordinate origin, while the error might become quite significant close to the boundary of the domain S . It
is
convenient
to
discuss
the
following
particular
cases:
g 0=2π√ A ,
g x= g y= g xy=0; g y=2π/√ A , g x= g xy= g 0=0; and the case g x=2π/√ A , g y= g xy= g 0=0. each case we compute the integral p =
In
H⌠ ⌠ σd S , A⌡ ⌡ S
which is proportional to the average value of σ, and is dimensionless, thus characterizing the shape of S and being independent of its size. We shall denote these parameters by p 0, p y and p x for each case respectively. There are two reasons for these definitions: i ) they correspond exactly to the parameters used in the theory of sound penetration through holes, so it would be easy to compare the numerical results; ii ) tabulation of these parameters for various shapes will simplify significantly the solution of any specific contact problem. Indeed, the relationship between the applied force P and the punch penetration g 0 can be rewritten as
P =
p A 0 g + √ A ( p y g y + p x g x). 2π H √ A 0
The last expression indicates that the knowledge of the shape coefficients and the punch area is sufficient for establishing the relationship between the applied force and the punch penetration. Formulae (5.6.3), (5.6.12) and (5.6.13) lead to the following expressions for the parameters p 0, p y and p x
p0
8√ A = , J0
px =
py =
8{8 J0 [ Ix ( Cyyyy − 2 Cxxyy ) + Iy (2 Cxxxx − Cxxyy )] − 3 A B y} 9 A 3/2 J0 ( Cxxxx Cyyyy − C 2xxyy)
8{8 J0 [ Ix (2 Cyyyy − Cxxyy ) + Iy ( Cxxxx − 2 Cxxyy )] − 3 A B x} 9 A 3/2 J0 ( Cxxxx Cyyyy − C 2xxyy)
.
.
(5.6.14)
373
5.6 Curved punch of general planform
Some further simplifications take place when the domain S possesses such a symmetry that all the moments about the axis Ox are equal to the similar moments about the axis Oy . In this case
py = px =
8(8 I 0 J0 − 3 A B 0) 3 A 3/2 J 20
,
(5.6.15)
where the moments with the subindex 0 indicate corresponding polar moments. Formula (5.6.13) also simplifies as follows: α0 =
B0 2 ( g + g y). g0 − J0 x π H J0
(5.6.16)
Formulae (5.6.2), (5.6.3), (5.6.12) and (5.6.13) are the main results of this section. The elastic contact problem for a wide variety of planforms can now be solved by a relatively simple computation of the geometrical characteristics (moments) of the domain of contact. Several examples are considered further. We present only the necessary computations of the moments involved. The domain of contact is assumed to be prescribed. The more complicated case when the domain of contact is partially unknown is discussed in the context of the problem of a rigid roller on an elastic half-space. Example 1: Polygon. Consider a polygon with n sides. The general notation of its parameters is the same as in the previous section, where the moments of inertia and the J -moments are computed. Here the remaining moments are presented. The C -moments can be computed by the following formulae n
C
xxxx
=
Σ−q cos2ψ k
k
− u kcos4ψk + v ksin4ψk + 4 s ksinψkcos3ψk + 2 t kcos4ψk,
k=1
n
C
xxxy
=
Σv cos4ψ k
k
+ u ksin4ψk + s kcos2ψk(1 − 4sin2ψk)
k=1
+
1 q ksin2ψk − 2 t ksinψkcos3ψk, 2
(5.6.17)
374
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
n
Cxxyy =
Σu cos4ψ k
− v ksin4ψk −
k
1 s sin4ψk + 2 t ksin2ψkcos2ψk, 2 k
(5.6.18)
k=1
n
Cxyyy =
Σ−v cos4ψ k
− u ksin4ψk − s ksin2ψk(1 − 4cos2ψk)
k
k=1
+
1 q sin2ψk − 2 t ksin3ψkcosψk, 2 k
(5.6.19)
n
Cyyyy =
Σq cos2ψ k
− u kcos4ψk + v ksin4ψk − 4 s ksin3ψkcosψk + 2 t ksin4ψk,
k
k=1
(5.6.20) where tk =
qk =
Ak ak
ln
Ak a 2k
(
b k + b k+1 + a k b k + b k+1 − a k
,
A 2k 1 1 − ), sk = 4 2 ( b b ak k k+1
1 1 + )[ a 2k + ( b k − b k+1)2]. bk b k+1
(5.6.21)
b 2k − b 2k+1 + a 2k 3 b 2k+1 − b 2k + a 2k3 , + uk = b k+1 bk 12 a 4k Ak
vk =
16 A 4k 3 a 4k
1 − 1 . b 3k+1 b 3k
The following formulae can be derived for the cubic moments n
Bx =
Σ−j cos2ψ k
k
+ r ksin2ψk + 2 f kcos2ψk,
k=1
n
By =
Σj cos2ψ k
k=1
k
− r ksin2ψk + 2 f ksin2ψk,
(5.6.22)
375
5.6 Curved punch of general planform
n
B xy =
Σ(j
k
− f k)sin2ψk + r kcos2ψk,
(5.6.23)
k=1
where 2Ak 2 ( b − b ), rk = k a k k+1
2 A k 3 b k + b k+1 + a k ln , jk = a k b k + b k+1 − a k
fk =
b k+1 − b k 2 1 + 1 j. ( b k + b k+1) A k1 + 4 k 4 ak
(5.6.24)
Substitution of (5.6.17−5.6.24) into (5.6.2, 5.6.3, 5.6.12 and 5.6.13) gives the complete solution for an arbitrary polygon. In the case of a regular polygon significant simplifications occur. The moments of inertia and the J -moments take the form (5.5.25) and (5.5.26), and the remaining moments are determined by 2π nb 3 π 1 + sin(π/ n ) sin + cos3( )ln , (5.6.25) B x =B y = 4 n n 1 − sin(π/ n ) C
= C
C
=
xxxx
xxyy
yyyy
3 nb π 1 + sin(π/ n ) cos ln 8 n 1 − sin(π/ n )
=
nb π 1 + sin(π/ n ) cos ln 8 n 1 − sin(π/ n )
(5.6.26)
Note that formulae (5.6.26) are valid for any regular polygon except the square, due to the fact that the trigonometric series summation n
Σ
2π sin ( k −1) = n 4
k=1
n
Σcos (k−1)2nπ 4
xxxx
3n , 8
(5.6.27)
k=1
is not valid for a square. expressed as C
=
= C
yyyy
The C -moments for a square of side 2 l can be
2√2 = l 4ln(1 + √2) − , 3
C
xxyy
=
2√2 l. 3 (5.6.28)
Formulae (5.6.3, 5.6.12, and 5.6.13) simplify for a regular polygon αx =
4(5 g x + g y) 3π H J0
,
αy =
4( g x + 5 g y) 3π H J0
,
αxy =
16 g xy 3π H J0
.
(5.6.29)
376
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
Again, one should note that formulae (5.6.29) are not valid for a square. formulae to follow are valid for any regular polygon including the square.
The
B0 2 ( g x + g y), g0 − α0 = J0 π H J0 4 8 P = Ag + I 0 − π H J0 0 3
A B0
( g + g ). y J0 x
(5.6.30)
The dimensionless coefficients p y and p x take the form 2π π 23+7cos sin n n . py = px = − 36 2π 1+sin(π/ n ) π 1+sin(π/ n ) ln ( n sin )1/2 n cos ln n n 1−sin(π/ n ) 1−sin(π/ n ) 8√2
(5.6.31) Consider several particular values of n . For an equilateral triangle ( n =3) formula (5.6.31) gives p y= p x=0.3782. We are not aware of any numerical data to compare with this result. In the case of a square n =4, and p y= p x=0.2697. The result due to De Smedt is 0.2645, with the discrepancy less than 2%. Since formula (5.6.31) in the limiting case n →∞ gives the exact result for a circle p y= p x=4/(3π3/2)=0.2394, we should expect that the error of (5.6.31) will decrease with n . The value of the coefficients for a regular hexagon is 0.2443, and again, we did not find in the literature anything to compare with this result. Example 2: Rectangle. Consider a punch with a rectangular base, a 1 and a 2 being its semiaxes along the axis Ox and Oy respectively. Introduce the aspect ratio ε= a 2/ a 1. Formulae (5.6.17−5.6.24) in this case reduce to 4 a 1 a 23, 3 Jx = 4 a 1sinh-1ε , Ix =
Iy =
C
= 4 a 1(sinh-1ε −
C
= 4 a 2(sinh-1
xxxx
yyyy
ε ), 3(1 + ε2)1/2
1 1 − ). ε 3(1 + ε2)1/2
4 3 a a , 3 1 2 Jy = 4 a 2sinh-1(1/ε). C
xxyy
= 4a1
(5.6.32)
ε , 3(1 + ε2)1/2 (5.6.33)
377
5.6 Curved punch of general planform
1 B x = 2 a 31ε(1 + ε2)1/2 − sinh-1ε + 2ε3sinh-1( ), ε 1 B y = 2 a 31ε(1 + ε2)1/2 − ε3sinh-1( ) + 2sinh-1 ε. ε
(5.6.34)
The coefficients p y and p x were computed by De Smedt (1979) for a rectangle with various aspect ratios ε. Here, we present his results along with those given by our method ε= De Smedt p y= our method p y= Discrepancy in p y (%) De Smedt p x= our method p x= Discrepancy in p x(%)
0.1000 2.9980 3.2809 -9.4 0.0376 0.0284 24.6
0.2000 1.3730 1.3959 -1.7 0.0639 0.0577 9.7
0.3330 0.7942 0.7782 2.0 0.0982 0.0963 1.9
0.5000 0.5229 0.5100 2.5 0.1399 0.1431 -2.3
0.7500 0.3491 0.3485 0.2 0.2022 0.2086 -3.2
1.0000 0.2645 0.2697 -2.0 0.2645 0.2697 -2.0
Our formulae seem to perform satisfactorily over a sufficiently wide range of aspect ratio. The traction distribution due to (5.6.3) can be compared with the numerical data received in a personal communication from De Smedt. Computations were made for ε=0.5, g y=2π/√ A , H =1, g 0= g x=0. Here are the results along the axis Ox , compared with those communicated by De Smedt x/a 1= De Smedt σ= our method σ= Discrepancy(%)
0.0000 0.2500 0.3333 0.5000 0.5833 0.6667 0.7500 0.8333 0.9167 -0.4715 -0.3933 -0.3249 -0.1238 0.0452 0.2515 0.5456 0.9462 2.0580 -0.4731 -0.3953 -0.3314 -0.1290 0.0232 0.2273 0.5141 0.9602 1.8556 -0.3 -0.5 -2.0 -4.2 48.8 9.6 5.8 -1.5 9.8
The relevant results along the axis Oy are given below y/a 2= De Smedt σ= our method σ= Discrepancy (%)
0.0000 -0.4715 -0.4731 -0.3
0.1667 -0.4765 -0.4744 0.5
0.3333 -0.4774 -0.4791 -0.3
.5000 -0.4837 -0.4907 -1.4
.6667 -0.5063 -0.5198 -2.7
.8333 -0.5311 -0.6138 -15.6
As we predicted, the discrepancy becomes quite significant close to the boundary. Example 3: Rhombus.
Let a 1 and a 2 be its semiaxes along Ox and Oy
respectively. We denote its side l =( a 21+ a 22)1/2, and introduce the aspect ratio ε= a 2/ a 1. Formulae (5.6.17−5.6.24) in this case yield
378
CHAPTER 5
Ix =
l 4 ε3 3(1 + ε2)2
,
Iy =
l 4ε 3(1 + ε2)2
APPLICATION TO CONTACT PROBLEMS
,
A =
2 l 2ε (1 + ε2)
.
Jx =
4lε ε2 1 + ε + (1 + ε2)1/2 1 − ε + ln , (1 + ε2) (1 + ε2)1/2 (1 + ε2) 1 + ε − (1 + ε2)1/2
Jy =
4lε 1 1 + ε + (1 + ε2)1/2 − 1 − ε + ln . (1 + ε2) (1 + ε2)1/2 (1 + ε2) 1 + ε − (1 + ε2)1/2
Bx =
2l ε ε + 4ε − 3 + 2 − ε ln1 + ε + (1 + ε ) , 2 3 (1 + ε ) (1 + ε2)1/2 (1 + ε2) 1 + ε − (1 + ε2)1/2
By =
2 l 3ε 1 + 4ε2 − 3ε3 + ε2(2ε2 − 1)ln1 + ε + (1 + ε2)1/2. (1 + ε2) 1 + ε − (1 + ε2)1/2 (1 + ε2)3 (1 + ε2)1/2
(5.6.35) 3 3
3
2
2 1/2
(5.6.36) 1 + ε + (1 + ε ) 4lε ε 2 − ε + 5ε − 4ε + , 2 2 2 1/2 2 ln (1 + ε ) 3(1 + ε ) (1 + ε ) 1 + ε − (1 + ε2)1/2 2
3
4
2 1/2
C
=
C
1 + ε + (1 + ε2)1/2 1 4lε −4 + 5ε − ε2 + 2ε3 , ln + = (1 + ε2) 1 + ε − (1 + ε2)1/2 (1 + ε2)2 3(1 + ε2)1/2
xxxx
yyyy
1 + ε + (1 + ε2)1/2 4lε ε2 1 − 2ε − 2ε 2 + ε 3 + . ln xxyy (1 + ε2)2 3(1 + ε2)1/2 (1 + ε2) 1 + ε − (1 + ε2)1/2 (5.6.37) We did not find in the mechanics literature any result related to a punch with a rhombus planform. The mathematically equivalent problem of sound penetration through an aperture in the shape of a diamond was solved numerically by De Smedt (1979). Here, we present his results compared with those given by our method C
=
ε= De Smedt p y= our method p y= Discrepancy (%)
0.1000 4.6520 3.7425 19.6
0.2000 1.8890 1.6605 12.1
0.3333 0.9844 0.9192 6.6
0.5000 0.5933 0.5770 2.7
0.7500 0.3655 0.3661 -0.2
1.0000 0.2631 0.2697 -2.5
De Smedt p x= our method p x= Discrepancy (%)
0.0314 0.1944 -518.4
0.0549 0.1435 -161.3
0.0862 0.1345 -56.0
0.1270 0.1532 -20.6
0.1923 0.2050 -6.6
0.2631 0.2697 -2.5
Though our results are satisfactory for p y, they are unacceptable for p x when ε≤0.5. Despite the large relative errors of p y and p x for small ε, there is
379
5.6 Curved punch of general planform
reason to believe that the accuracy of solution of various contact problems will be quite satisfactory due to the following: different signs in discrepancy will partially compensate the total error; the term with p 0 generally dominates in real contact problems, and it will generally determine the total error of the solution. We shall see further that in the case of a rigid cylindrical roller our theory works well for the aspect ratios ε very far away from unity. An alternative approach which uses the variational principle and somewhat improves the accuracy, is discussed further on. Example 4: Cross. Consider a punch with a configuration obtained by the orthogonal intersection of two equal rectangles with sides 2 a and 2 b , ( a ≥ b ). Introduce the aspect ratio as ε= b / a . The area and some of the moments are given in the previous section. The remaining moments are
1 + (1+ε2)1/2 2 1/2 2 1/2 3 B x = B y = 2 a 2ε(1+ε ) + ln[ε + (1+ε ) ] + ε ln − √2. ε(1 + √2) 3
(5.6.38) The comparison between our results and those given by De Smedt (1979) are presented below ε= De Smedt p y= p x= our method p y= p x= Discrepancy(%)
0.1000 0.9675 1.6943 -75.1
0.2000 0.4854 0.6765 -39.4
0.3333 0.3271 0.3716 -13.6
0.5000 0.2671 0.2683 -0.5
0.7500 0.2523 0.2517 0.3
1.0000 0.2645 0.2697 -2.0
Taking into consideration the shape complexity, we should consider the results agreement as surprisingly good, not only quantitatively but qualitatively as well: both data display a relatively flat minimum around ε=0.75. The discrepancy becomes unacceptably large for ε≤0.3. It will be shown later on that the variational approach slightly improves the results. Example 5: Rigid roller. Consider a rigid right cylinder of length 2 l and radius r 0 lying on its generator and pressed against an elastic half-space by a normal centrally applied force P . This case corresponds to g x= g xy=0 and All the previously derived formulae are valid here. Formulae g y=−1/(2 r 0). (5.6.12) yield 2( C αx = −
yyyy
− 2 Cxxyy )
3π r 0 H ( C C − C 2xxyy) xxxx yyyy
,
380
CHAPTER 5
αy = −
2(2 Cxxxx − Cxxyy ) 3π r 0 H ( Cxxxx Cyyyy − C 2xxyy)
APPLICATION TO CONTACT PROBLEMS
,
αxy = 0.
(5.6.39)
Assuming that the contact region is close to a rectangle, we can use formulae (5.6.32−5.6.34) for all the moments involved. We denote the width of the contact region by 2 b . In the previously considered contact problems the contact region was prescribed. Here, we have the domain of contact partially unknown. Its length 2 l is prescribed, and the tractions should be singular at y =± l , while its width 2 b is as yet unknown and is to be found from the condition that the tractions vanish at x =± b . We define α0 from the condition α0 = −αy b 2.
(5.6.40)
Returning back to (5.6.2), one can see that the condition (5.6.40) will cause σ
Fig.
5.6.1.
The geometry pertinent to the rigid roller problem.]
to vanish along a part of an ellipse (see Fig. 2
x + b2
a (φ)pha/x y 2 αy b 2
= 1,
5.6.1)
(5.6.41)
381
5.6 Curved punch of general planform
while the stresses will remain singular at y =± l . Direct computations show that the ellipse semiaxis along Oy is at least several times greater then the other semiaxis, thus making the part of the ellipse (5.6.41) very close to a straight line and validating our assumption of a rectangular domain of contact. The ellipse semiaxis ratio is smallest for a square where it is equal to 2.4612. In this sense the case l = b is the least accurate and will be considered in more detail further. Formulae (5.6.3), (5.6.13) and (5.6.39) now give the following expressions for the total force P and the punch penetration g 0 P =
8 ( α x I x + α y I y ) − 2α y b 2 A , 3
(5.6.42)
g0 =
1 π H [αx B x + αy( B y − 2 J0 b 2)], 4
(5.6.43)
where αx and αy are defined by (5.6.39) and the moments by (5.6.32−5.6.34), and this completes the solution. We have found only one report (Kalker, 1972) where some formulae were presented for the case of a very narrow domain of contact with the aspect ratio ε= l / b >>1. These formulae in our notation take the form P =
l3 1 − ln2 1 + , 2 2ln(4ε) + 1 2ε r 0 H
(5.6.44)
g0 =
l2 2 2ln(4ε) + 1 . 4 r 0ε
(5.6.45)
We have received in private communication from Kalker some data generated by his numerical method. Although his computer program does not give the value of aspect ratio, we managed to compare the results by the following procedure. Assuming equality of his punch settlement with ours (5.6.43), we can find the value of aspect ratio ε which, being substituted into (5.6.42) allows us to compare the total forces. The results of comparison are given in Table 5.6.1. The agreement over a wide range of aspect ratios should be considered as very good, taking into consideration the approximate nature of our theory, and also the fact that Kalker’s program ignores the stress singularity at the roller edges. Such a good agreement allows us to claim that formulae (5.6.42−5.6.43) give a sufficiently accurate analytical solution to the problem of a rigid roller on an elastic half space. Surprisingly enough, our method seems to be the most accurate around ε=1, despite the fact that our assumption of a rectangular domain of contact is the least accurate in this case. We have compared the traction
382
CHAPTER 5
Table 5.6.1.
APPLICATION TO CONTACT PROBLEMS
Comparison of our method with the numerical results of Kalker
aspect ratio ε
punch settlement g *
23.31 10.77 6.799 4.879 3.757 2.511 1.843 1.282 0.9653 0.6741 0.4174 0.3018 0.2359 0.1933 0.1633 0.1411 0.1239 0.1101 0.9884E-01 0.8939E-01 0.6806E-01
0.2501E-03 0.1000E-02 0.2250E-02 0.4000E-02 0.6251E-02 0.1225E-01 0.2026E-01 0.3603E-01 0.5633E-01 0.1003 0.2263 0.4041 0.6351 0.9212 1.265 1.670 2.139 2.679 3.297 4.000 6.771
total force P* total force P* (Kalker) formula (5.6.42) 0.5379E-04 0.2590E-03 0.6618E-03 0.1298E-02 0.2200E-02 0.4934E-02 0.9110E-02 0.1853E-01 0.3240E-01 0.6748E-01 0.1925 0.4122 0.7499 1.231 1.882 2.734 3.817 5.173 6.847 8.890 18.40
0.6251E-04 0.2891E-03 0.7174E-03 0.1377E-02 0.2296E-02 0.5019E-02 0.9114E-02 0.1835E-01 0.3208E-01 0.6719E-01 0.1952 0.4216 0.7717 1.272 1.952 2.844 3.982 5.411 7.185 9.362 19.30
discrepancy (%) -16.20 -11.65 -8.409 -6.118 -4.389 -1.729 -0.4892E-01 0.9399 1.015 0.4326 -1.393 -2.279 -2.908 -3.335 -3.703 -4.043 -4.323 -4.612 -4.942 -5.306 -4.901
distribution due to (5.6.2) with the numerical results of Kalker. picture is presented below for r 0/ l =20, ε=0.9653, H =1, y/l =0.1. x/l = Kalker σ= our method σ= discrepancy(%)
0.1200 0.7967E-02 0.7960E-02 0.09
0.3600 0.7420E-02 0.7513E-02 -1.25
0.6000 0.6226E-02 0.6528E-02 -4.84
The typical
0.8400 0.3352E-02 0.4675E-02 -39.49
As predicted, the relative error becomes quite significant close to the boundary of the contact region. A similar behavior is observed along a line parallel to the axis Ox : y/l = Kalker σ= our method σ= discrepancy (%)
0.1000 0.7967E-02 0.7960E-02 0.09
0.3000 0.8146E-02 0.8177E-02 -0.38
0.5000 0.8623E-02 0.8762E-02 -1.61
0.7000 0.9498E-02 0.1018E-01 -7.17
0.9000 0.1819E-01 0.1570E-01 13.67
It is also of interest to compare the numerical results by Kalker with his approximate formulae (5.6.44−5.6.45). This comparison is given in Table 5.6.2. We expected Kalker’s formulae to be the most accurate for large ε with the
383
5.6 Curved punch of general planform
Table 5.6.2.
Comparison of formulae due to Kalker with his numerical results
aspect ratio ε
punch settlement g *
22.34 10.27 6.456 4.621 3.551 2.368 1.734 1.197 0.8848 0.5772
0.2501E-03 0.1000E-02 0.2250E-02 0.4000E-02 0.6251E-02 0.1225E-01 0.2026E-01 0.3603E-01 0.5633E-01 0.1003
total force P* total force P* Kalker formula (5.6.44) 0.5379E-04 0.2590E-03 0.6618E-03 0.1298E-02 0.2200E-02 0.4934E-02 0.9110E-02 0.1853E-01 0.3240E-01 0.6748E-01
discrepancy (%)
0.5163E-04 0.2459E-03 0.6243E-03 0.1223E-02 0.2079E-02 0.4706E-02 0.8838E-02 0.1873E-01 0.3471E-01 0.8364E-01
4.018 5.054 5.658 5.743 5.511 4.621 2.987 -1.103 -7.126 -23.95
accuracy decreasing with ε. This is not the case: the error is almost constant in the interval 2<ε<22, it decreases to zero around ε=1, after which the error changes sign and increases rapidly. This means either that the formulae of Kalker (5.6.44−5.6.45) are not exact asymptotically, or that his numerical procedure has an error of about 5%. Discussion. An alternative method can be suggested by using the variational approach (Noble 1960). We can use again the functional (5.5.40), which assumes its stationary value at the exact solution of (5.1.8). We take σ(N) dS ≈ w 0 + w 2, ⌡ ⌡ R(M,N) N
H⌠ ⌠
(5.6.46)
S
where σ is defined by (5.6.2) and w 0+ w 2 is given by (5.6.10). Substitution of (5.6.1), (5.6.2), (5.6.10) and (5.6.46) into (5.5.40) makes it possible to consider the functional I as a function of α0, αx, αy and αxy. The extremum conditions ∂I = 0, ∂α 0
∂I = 0, ∂α x
∂I = 0, ∂α y
∂I = 0, ∂αxy
give four linear algebraic equations with respect to the unknowns α0, αx, αy, and αxy. The complete solution is rather cumbersome. Here, we present the set of equations for the coefficients α0, αx, and αy which are valid only for domains having at least one axis of symmetry.
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c 11α0 + c 12αx + c 13αy = 4[ A g 0 +
APPLICATION TO CONTACT PROBLEMS
4 ( I g + Iy g y)], 3 x x
c 12α0 + c 22αx + c 23αy =
1 16 [ Ix g 0 + ( Dxxxx g x + Dxxyy g y)], 5 3
c 13α0 + c 23αx + c 33αy =
1 16 [ I g + ( Dxxyy g x + Dyyyy g y)]. 5 3 y 0
(5.6.47)
Here, c 11 = 2π H J0 A , c 12 =
1 4 4 πH[ B x A + Ix (2 J0 + 2 C − C ) − I (C − 2 C )], xxxx xxyy xxyy 2 3 3 y xxxx
c 13 =
1 4 4 πH[ B y A + I (2 J + 2 C − C ) − I (C − 2 C )], yyyy xxyy xxyy 2 3 y 0 3 x yyyy
c
4 π H [5 B x Ix + D (2 C − C ) − D (C − 2 C )], xxxx xxxx xxyy xxyy xxxx xxyy 15
22
=
c 23 =
2 π H [5( B x Iy + B y Ix ) − D ( C − 2C ) xxxx yyyy xxyy 15 + 2D
xxyy
c 33 =
(C
xxxx
+ C
yyyy
− Cxxyy ) − Dyyyy ( Cxxxx − 2 Cxxyy )],
4 π H [5 B y Iy + D (2 C − C ) − D (C − 2 C )]. yyyy yyyy xxyy xxyy yyyy xxyy 15
(5.6.48)
The D -moments are introduced similar to (5.6.9) as 2π
Dxxxx = ⌠ a 6(φ)sin4φdφ,
⌡ 0
2π
Dxxyy = ⌠ a 6(φ)sin2φcos2φdφ,
⌡
0
2π
D
yyyy
= ⌠ a 6(φ)cos4φdφ.
⌡
(5.6.49)
0
It is clear that the variational approach solution is more cumbersome than the one introduced earlier. It remains to be seen whether it will be more accurate. One advantage should be noted: the matrix of (5.6.48) is symmetric (as it is required by the reciprocal theorem) while the matrix of (5.6.11) is generally not
385
5.6 Curved punch of general planform
symmetric. all, the the the
Let us compare the results for several particular configurations. First of consider a regular polygon. In the tables hereafter the word simple refers to method introduced earlier, while the word variational refers to the solution of set of equations (5.6.47). In the case of a regular polygon we shall need polar D -moment only D0 =
8 2π 4 1 6 π π cos4( ). cos2( ) + b n sin 1 + 3 3 5 n n n
Here are the results of computations for a regular polygon with n sides n= simple p y= p x= variational p y= p x= Discrepancy (%)
3 0.3782 0.3409 9.9
4 0.2697 0.2612 3.2
5 0.2502 0.2472 1.2
6 0.2443 0.2429 0.6
7 0.2420 0.2412 0.3
9 0.2403 0.2401 0.1
∞ 0.2394 0.2394 0.0
Both methods seem to work well. If one considers the result by De Smedt for a square, 0.2645, as exact then this might be an indication that the variational approach is somewhat more accurate. In the limiting case of n →∞ both methods give the exact result for a circle. The D -moments for the rectangle will take the form Dxxxx =
24 a a 5, 5 1 2
Dyyyy =
24 5 a a, 5 1 2
Dxxyy =
8 3 3 a a. 3 1 2
Here are the numerical results computed for a rectangle ε= De Smedt variational Discrepancy De Smedt variational Discrepancy
p y= p y= (%) p x= p x= (%)
0.1000 2.9980 3.4239 -14.2 0.0376 0.0316 15.9
0.2000 1.3730 1.4523 -5.8 0.0639 0.0588 7.9
0.3330 0.7942 0.8023 -1.0 0.0982 0.0939 4.3
0.5000 0.5229 0.5166 1.2 0.1399 0.1370 2.1
0.7500 0.3491 0.3437 1.5 0.2022 0.1997 1.2
1.0000 0.2645 0.2612 1.2 0.2645 0.2612 1.2
Again, the general impression is that the variational approach is more accurate but not in all cases. For example, the discrepancy in p y for ε=0.1 increased as compared with the result from the simple method. It is up to the user to decide whether a somewhat better accuracy of the variational approach is worth more cumbersome computations. The mathematically equivalent problem of the analytical determination of the quadratic term in the low-frequency expansion, related to the sound transmission
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CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
through an aperture in a rigid screen, is considered in (Fabrikant, 1986d). Exercise 5.6 1. Find the D -moments for a rhombus, with semiaxes a 1 and a 2. Answer: Dxxxx =
4 a a 5, 5 1 2
Dyyyy =
4 5 a a, 5 1 2
Dxxyy =
2 3 3 a a. 15 1 2
2. Find the variational solution for a rhombus and compare it with the data of De Smedt, and with the simple solution. Answer: see the Table below ε= De Smedt variational Discrepancy De Smedt variational Discrepancy
0.1000 4.6520 -0.5952 112.8 0.0314 0.0090 71.5
p y= p y= (%) p x= p x= (%)
0.2000 1.8890 3.3549 -77.6 0.0549 0.1464 -166.7
0.3333 0.9844 0.9465 3.8 0.0862 0.1110 -28.8
0.5000 0.5933 0.5580 6.0 0.1270 0.1400 -10.2
0.7500 0.3655 0.3534 3.3 0.1923 0.1971 -2.5
1.0000 0.2631 0.2612 0.7 0.2631 0.2612 0.7
Conclusion: the discrepancy decreased for ε>0.33, but the results are unacceptable for ε<0.33. 3. Find the D -moments for a cross-shaped punch. Answer: D
xxxx
= D
yyyy
=
24 6 a ε(1 + ε4 − ε5), 5
D
xxyy
=
8 6 3 a ε (2 − ε3). 3
4. Find the variational solution for a cross and compare it with the data of De Smedt, and with the simple solution. Answer: see the Table below ε= De Smedt p y= p x= variational p y= p x= Discrepancy (%)
0.1000 0.9675 1.4346 -48.3
0.2000 0.4854 0.5822 -19.9
0.3333 0.3271 0.3397 -3.9
0.5000 0.2671 0.2606 2.4
0.7500 0.2523 0.2482 1.6
1.0000 0.2645 0.2612 1.2
Conclusion: Comparison of this table with the data from the simple solution leads to the same conclusion: the results become valid in a wider range of the aspect ratio ε, but the theory fails for very small ε.
387
5.7 Flat flexible punch of general planform
5.
Apply the theory above to the case of a roller, with the aspect ratio ε=1.
Answer: a (φ) l ln(1 + √2) − σ =
√2ln(1 + √2) − 1 y 2 x2 1 1 − 2 − l 2√2 2√2ln(1 + √2) − 1 l 2
π Hr0ln(1 + √2)[3ln(1 + √2) − √2] a 2(φ) − ρ2
1/2
≈
x2 y2 0.4869 a (φ) l 1 − 2 − l (2.4612 l )2 π Hr0 a 2(φ) − ρ2
1/2
2 l 3[12ln(1 + √2) − √2]
1.878 l 3 ≈ P = , π Hr0 9π Hr0 ln(1 + √2) [3ln(1 + √2) − √2]
g0 =
l 2[13ln2(1 + √2) − 6√2 ln(1 + √2) + 2] 4 r 0(1 + √2) [3ln(1 + √2) − √2]
≈
1.06558 l 2 r0
5.7 Flat flexible punch of general planform under shifting load The following mixed boundary value problem for a transversely isotropic elastic half-space is considered: a uniform tangential displacement is prescribed over a finite domain of general shape, while the remaining part of the boundary is traction-free. The problem may be interpreted as an external crack problem, with a remote shear loading, or as a contact problem, with a flexible punch, subjected to a tangential displacement. We use the term flexible to indicate a special kind of a punch which does not exert any normal pressure. A general relationship is established between the resultant shifting force and the displacement. A favorable comparison is made with the available numerical results.
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CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
Theory. Consider a transversely isotropic elastic half-space z ≥0, with the following mixed boundary condition prescribed on the plane z =0: σz = 0, for −∞<( x , y )<∞;
τ = 0, for ( x , y )∉ S ;
u = u ( x , y ), for ( x , y )∈ S .
(5.7.1)
Here S denotes the domain of contact in the elastic contact problem, or for the crossection of the connection between two elastic half-spaces in the external crack problem. Let the boundary of the domain S be expressed in polar coordinates as ρ = a (φ),
(5.7.2)
where a (φ) is a single-valued bounded function. can we rewritten from (2.6.2) as follows: 2π a (φ)0
G1
τ(ρ0,φ0)
⌠⌠ 2 ⌡⌡ 0
R
ρ0dρ0dφ0 +
2π a (φ)0
G2
⌠⌠ 2 ⌡⌡ 0
0
The governing integral equation
q τ(ρ0,φ0) q R
ρ0dρ0dφ0 = u (ρ,φ).
0
(5.7.3) Here the overbar indicates the complex conjugate value, and iφ
q = ρei φ − ρ0e 0,
R2 = q q.
(5.7.4)
The approach is based on the integral representation for the reciprocal distance established in (1.1.27). We also need an integral representation for the kernel in the second term of expression (5.7.3). It was established in (2.5.6). We consider further only the case where the right hand side in (5.7.2) u =const. Substitution of (1.1.27) and (2.5.6) into (5.7.2) gives, after changing the order of integration and retaining the zero harmonic only G1
ρ
2π
a (φ0)
τ(ρ ,φ )ρ dρ
0 0 0 0 ⌠ dφ ⌠ ⌠ 0 π ⌡ ⌡ (ρ2 − x 2)1/2 x⌡ (ρ20 − x 2)1/2 0
dx
0
2π
G2
2 i φ0
⌠ e + π ⌡ 0
ρ
dφ0 ⌠
a (φ0)
dx
⌡ (ρ − x )
0
2
2
1/2
⌠ ⌡ x
ρ20 − 2 x 2 ρ0(ρ20 − x 2)1/2
τ(ρ0,φ0)dρ0 = u . (5.7.5)
389
5.7 Flat flexible punch of general planform
We assume the shear stress distribution in the form τ =
c a (φ) , [ a (φ) − ρ2]1/2
(5.7.6)
2
where c is a complex constant. This loading is statically equivalent to a resultant force T = T x+i T y. Integration of (5.7.6) over S yields T = 2 Ac ,
(5.7.7)
where A is the area of the domain S . It is of interest to note that the relationship (5.7.7) does not depend on the location of origin of the system of coordinates. This location can be determined from the condition that the shear tractions should not produce any torque. This leads to two equations 2π
2π
⌠ ( a (φ))3cosφ dφ = 0, ⌡ 0
⌠ ( a (φ))3sinφ dφ = 0. ⌡
(5.7.8)
0
One can note that the left-hand side of each equation (5.7.8) is proportional to the x or y coordinates of the center of gravity. This means that the origin of the system of coordinates should be located at the center of gravity of the domain S . The axis orientation will be discussed later. It is now necessary to relate the tangential force T to the displacement u . This can be done by substitution of (5.7.6-5.7.7) into (5.7.5) which yields, after integration with respect to ρ0 2π
2π
2 iφ π u = G 1 T ⌠ a (φ0)dφ0 + G 2 T ⌠ e 0 a (φ0)dφ0. 8A ⌡ ⌡ 0
(5.7.9)
0
The complex expression (5.7.9) is equivalent to two real ones, namely, ux =
π { T x[ G 1 J 0 − G 2( J x − J y)] + 2 T y G 2 J xy}, 8A
uy =
π {2 T x G 2 J xy + T y[ G 1 J 0 − G 2( J y − J x)]}, 8A
(5.7.10)
The J -moments were introduced in (5.5.8), and some of them were computed in paragraph 5.6. One can now derive specific formulae for a variety of punch shapes. We leave this exercise to the reader.
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APPLICATION TO CONTACT PROBLEMS
Note that the change of the order of integration in (5.7.5) is valid inside the circle ρ=min( a (φ)) only, nevertheless, the solution given by (5.7.9) is exact in the case of an ellipse. This is easy to explain. It is known (Willis, 1970) that a traction distribution over an ellipse, given by a polynomial multiplied by a (φ)/[ a 2(φ) − ρ2]1/2, produces a polynomial displacement. In our case the displacement is constant. Since expression (5.7.9) gives the exact value of the integrals in (5.7.2) for ρ=0, the result is exact all over the ellipse. Indeed, formula (5.7.10) yields for an ellipse with semiaxes a and b ( a ≥ b ) ux =
(2 − k 2)K − 2E 1 G 1K + G 2 T, 2a x k2
uy =
1 (2 − k 2)K − 2E G 1K − G 2 T, 2a y k2
where K and E are complete elliptic integrals of the first and second kind, with the argument k =(1−( b / a )2)1/2. In the isotropic case G 1=(2−ν)/(2πµ), G 2=ν/(2πµ), and the last result coincides with Mindlin’s (1949). We can always choose the coordinate axes orientation so as to make J xy vanish. In this case formulae (5.7.9) simplify as follows π [ G 1 J 0 − G 2( J y − J x)] T y. 8A (5.7.11) Since expressions (5.7.10) and (5.7.11) are exact for an ellipse, we may expect them to be reasonably accurate for an arbitrary domain S . This assumption was justified in previous paragraphs when compared with various numerical results available in the literature. ux =
π [ G 1 J 0 − G 2( J x − J y)] T x, 8A
uy =
In order to verify the accuracy of our method, some numerical computations were performed by Kalker. Equations (5.7.11) may be rewritten as T x = C x u x,
T y = C y u y,
with 8A . π[ G 1( J x + J y) + G 2( J x − J y)] (5.7.12) The values of C x and C y, due to formulae (5.7.12), are presented in the table below, and compared with the numerical results of Kalker. Computations were made for an isotropic body, with the shear modulus µ=0.5, and the Poisson coefficient ν=0.3. The accuracy of the numerical procedure was Cx =
8A , π[ G 1( J x + J y) − G 2( J x − J y)]
Cy =
391
5.7 Flat flexible punch of general planform
ε= C x our method C x Kalker Discrepancy (%)
0.1 5.16 4.644 10.
0.2 6.12 5.650 7.7
0.3 6.87 6.445 6.2
0.5 8.1 7.766 4.1
0.7 9.19 8.904 3.1
1.0 10.67 10.43 2.3
C y our method C y Kalker Discrepancy (%)
4.32 3.991 7.6
5.32 5.044 5.2
6.13 5.904 3.7
7.56 7.373 2.5
8.85 8.667 2.1
10.67 10.43 2.3
assessed by computation of C x and C y for an ellipse, for which the exact solution is well known. The exact solution was about 4% above the numerical result by Kalker. All our results are also above the numerical ones. If we assume that error pattern for a rectangle is the same as for an ellipse (this means, for example, that our 10% discrepancy translates into 10−4=6(%) error), then our formulae must be considered as surprisingly accurate over a wide range of aspect ratios. We expect the error of our method to increase monotonically with decreasing ε, since the assumed traction distribution (5.7.6) is less realistic for a narrow rectangle than for a square. The fact that the discrepancy of C y with the numerical results does not change monotonically, indicates some flaws in the Kalker’s numerical procedure. We have also compared the shear traction distribution due to (5.7.6), (5.7.7) and (5.7.12) for a square with a 1=4, along the line y =0.5. The tangential displacement u x was assumed equal to unity, and u y=0. The comparison is given in the table below. Taking into consideration the approximate nature of both methods, the agreement
x= τ our method τ Kalker Discrepancy (%)
0.5 0.084 0.0863 −2.6
1.5 0.090 0.0923 −2.5
2.5 0.107 0.105 1.7
3.5 0.172 0.222 −29.
should be considered surprisingly good, except for the point close to the boundary, where neither method may claim to be accurate. Kalker has made computation for a cross, with a =4 and ε=0.5. The remaining parameters were taken the same as for rectangle. His result C x= C y=9.033, our result is 9.26, with the discrepancy of 2.4%. We have also compared the shear stress distribution along the line y =0.5. The tangential displacement u x was assumed equal to unity, and u y=0. The comparison is given in the table below.
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CHAPTER 5
τ our method τ Kalker Discrepancy (%)
0.0996 0.0997 −0.08
0.104 0.106 −1.9
APPLICATION TO CONTACT PROBLEMS
0.124 0.120 2.9
0.199 0.254 −27.
The agreement is good, except for the point close to the boundary. Discussion. Some qualitative analysis of the general solution is given here. us rewrite equations (5.7.11) as follows
ux =
πG1 J0 8A
[1 − κ] T x,
uy =
πG1 J0 8A
[1 + κ] T y,
κ =
G 2( J x − J y) G1J0
Let
.
(5.7.13) First of all, consider the case when J x= J y. We have from (5.7.13) that κ=0, and hence the elastic compliance will be the same in any direction. Due to the fact that a circle has the greatest value of J 0 of all domains with the same area A , one may come to the conclusion that a circular domain is the easiest to move. A numerical comparison with a square shows that the ratio of the linear polar moments of a square to that of a circle is equal to (2/√π )ln(1+√2 )=0.9945 which is very close to unity. Taking into consideration that our theory is approximate, it is surprising that the difference can still be observed. Comparison of a cross with a circle shows that its compliance may become arbitrarily small as ε→0. Now let us deform the domain S so that J x is no longer equal to J y while keeping its area constant. Formulae (5.7.13) show that it will become easier to move the domain in the oblong direction, and a greater force will be required for a displacement in the perpendicular direction. In the extreme case of a needle, the limiting value for κ is G 2/ G 1; in the case of isotropy ( G 2/ G 1)=ν/(2-ν)≤1/3. It is noteworthy that for G 2=0, the elastic compliance does not depend on the direction of the shift. In the case of isotropy, this corresponds to the Poisson ratio being equal to zero. Exercise 5.7 1. Verify the invariance of formulae (5.7.10) with respect to the rotation of axes. 2. Find the relationship between the shifting force and displacement of an elliptical punch, when its semiaxis a < b .
Answer: u x
(2 − k 2)K( k 1) − 2E( k 1) 1 T , G K( k 1) − G 2 = 2 2a 1 x k1
the
translational
393
5.8 Reissner-Sagoci problem for general domains
uy
(2 − k 2)K( k 1) − 2E( k 1) 1 T . = G 1K( k 1) + G 2 2 2a y k1
3. Find the tangential compliance of a cross-shaped punch, with the longer side 2 a and the aspect ratio ε. Hint : use (5.7.12) and (5.5.38) 4. Find the tangential compliance in the direction of the axes of symmetry for a rhombus-shaped punch.
5.8 Reissner-Sagoci problem for general domains The problem of torsion of a transversely isotropic elastic half-space by a punch of general planform is considered. An approximate analytical solution is obtained by the general method. A general relationship is established between the torque and the torsion angle. Some specific formulae are derived for punches having the planform of a polygon, a rectangle, and a cross. Theory. Consider a transversely isotropic elastic half-space z ≥0. A flexible punch of general planform S is attached to the half-space surface, and a torque M z is applied, producing the torsion angle ω. We need to relate the torsion angle to the torque. The mathematical formulation of the problem leads to the following mixed boundary conditions on the plane z =0: σz = 0, for −∞<( x , y )<∞; u x = −ω y
and
τzx = 0
uy = ωx,
and
τyz = 0, for ( x , y )∉ S ;
for ( x , y )∈ S .
(5.8.1)
Here S denotes the domain subjected to torsion, and ω is the torsion angle. Introduce the complex tangential displacements u = u x+i u y, and the complex shear tractions τ=τzx+iτyz. Let the boundary of the domain S be expressed in polar coordinates as ρ = a (φ),
(5.8.2)
where a (φ) is a single-valued bounded function. The governing integral equation is given by (5.7.3). The approach is based on the integral representations established in (1.1.27) and (2.5.6). Substitution of (1.1.27) and (2.5.6) into
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APPLICATION TO CONTACT PROBLEMS
(5.7.3) gives, after changing the order of integration and retaining the first harmonic only ρ
2 u = G ⌠ πρ 1 ⌡ 0
a (φ0)
2π
x 2d x (ρ2 − x 2)1/2
⌠ cos(φ−φ ) dφ ⌠ 0 0 ⌡ ⌡ 0 x
τ(ρ0, φ0) dρ0 (ρ20 − x 2)1/2
2π ρ 0 2 τ(ρ0, φ0) dρ0 d x x i φ0 1 i φ⌠ ⌠ ⌠ + e d φ G 2 −e 0 πρ ⌡ (ρ20 − x 2)1/2 ⌡ (ρ2 − x 2)1/2 0⌡ x 0 a (φ )
ρ
+ e ⌠ -i φ
⌠e ⌡
3 i φ0
⌡ (ρ2 − x 2)1/2 0
a (φ0)
2π
x 2d x
dφ0 ⌠
3ρ20 − 4 x 2
τ(ρ0,φ0)dρ0. ⌡ ρ20(ρ20 − x 2)1/2
0
x
(5.8.3) Note that the change of the order of integration in (5.8.3) is valid only inside the circle ρ=min[ a (φ)], and the fact that we have ignored all the harmonics but the first one. Nevertheless, it will be shown further that the results are exact for an ellipse, and seem to be reasonable for a wide variety of nonelliptical shapes. Let the shear traction distribution under the punch be τ = τzx + iτyz =
a (φ) ρ(i q ycosφ − q xsinφ) [ a 2(φ) − ρ2]1/2
,
(5.8.4)
where q y and q x are the as yet unknown constants. Make use of the condition that in the case of pure torsion the resulting force should be equal to zero, which means that the integral of τ over S should vanish. Since q y and q x are independent, this leads to two equations, namely, 2π
⌠ ( a (φ))3cosφ dφ = 0, ⌡ 0
2π
⌠ ( a (φ))3sinφ dφ = 0 . ⌡
0
(5.8.5) One can note that the left-hand side of each equation (5.8.5) is proportional to the x or y coordinates of the center of gravity. This means that the origin of the system of coordinates should be located at the center of gravity of the domain of contact. The axis orientation will be discussed later. The relationships between the torque M z and the parameters q y and q x can be established from the statics conditions
395
5.8 Reissner-Sagoci problem for general domains
M z = ⌠ ⌠ τyz x dS − ⌠ ⌠ τzx y dS,
⌡⌡
⌡⌡
S
S
which leads to Mz =
8 ( q I + q x I x). 3 y y
(5.8.6)
where I x and I y are the moments of inertia. Next, it is necessary to relate q y and q x to the torsion angle ω. This can be done by substitution of (5.8.4) in (5.8.3), which yields, after integration with respect to ρ0 2π
π u = G ρ⌠ cos(φ−φ0)(i q ycosφ0 − q xsinφ0 )dφ0 4 1 ⌡ 0
2π
−
iφ π G 2ρei φ⌠ a (φ0)(−i q ycosφ0 − q xsinφ0)e 0 dφ0 8 ⌡ 0
2π
+
3iφ 3π φ G 2ρe-i ⌠ a (φ0)(−i q ycosφ0 − q xsinφ0)e 0 dφ0. 8 ⌡
(5.8.7)
0
The overbar everywhere denotes the complex conjugate value. can also be rewritten as follows: u =
Expression (5.8.7)
π G ρ[(i q y J y − q x J xy)cosφ + (i q y J xy − q x J x)sinφ] 4 1
−
π φ G 2ρei [− q y(i J y − J xy) − q x( J xy + i J x)] 8
+
3π φ G 2ρe-i {−i q y[4 C − 3 J y + i(3 J xy − 4 C xxxy)] yyyy 8
− q x[4 C xyyy − 3 J xy + i(3 J x − 4 C xxxx)]}
(5.8.8)
The J -moments were introduced by (5.5.8), and the C -moments are defined by (5.6.9). Substitution of the last two conditions (5.8.1) in (5.8.8) yields
396
CHAPTER 5
ω(− y + i x ) =
APPLICATION TO CONTACT PROBLEMS
π G [(i q y J y − q x J xy) x + (i q y J xy − q x J x) y ] 4 1
−
π G ( x + i y )[− q y(i J y − J xy) − q x( J xy + i J x)] 8 2
+
3π G 2( x − i y ){−i q y[4 C yyyy − 3 J y + i(3 J xy − 4 C xxxy)] 8
− q x[4 C xyyy − 3 J xy + i(3 J x − 4 C xxxx)]}
(5.8.9)
Separation in (5.8.9) of the terms related to x and y will lead to two linear algebraic equations for the unknowns q y and q x. In the general case, the parameters q y and q x are complex, and one has to solve four linear algebraic equations. These equations may be simplified by the assumption that J xy=0. One can always achieve this by appropriately choosing the coordinate axis directions. In the case of symmetry, these axes will coincide with the principal axes of inertia. For the sake of simplicity, we assume also that C = C =0, xyyy
xxxy
which will always be the case if domain S has at least one axis of symmetry. Under these assumptions, expression (5.8.9) yields just two equations with real coefficients, namely, 5 3 3 π G q J + π G 2[ q y( J y − C yyyy) + q x( C xxxx − J x)] = ω, 4 1 y y 4 2 2 3 5 3 π G 1 q x J x + π G 2[ q y( C yyyy − J y) + q x( J x − C xxxx)] = ω. 4 2 4 2 (5.8.10) The solution of (5.8.10) is
qy =
qx =
4ω[ G 1 + 3 G 2(3 − 4 c x)] π( G 1 + G 2) J y[ G 1 + 3 G 2(3 − 2 c x − 2 c y)] 4ω[ G 1 + 3 G 2(3 − 4 c y)] π( G 1 + G 2) J x[ G 1 + 3 G 2(3 − 2 c x − 2 c y)]
,
. (5.8.11)
Here the parameters were introduced c x = C xxxx/ J x,
c y = C yyyy/ J y.
(5.8.12)
By substituting (5.8.11) in (5.8.6), one can find the relationship between the
397
5.8 Reissner-Sagoci problem for general domains
torque and the angle of torsion in the form 32ω[( G 1 + 9 G 2)( I y/ J y + I x/ J x) − 12 G 2( c x I y/ J y + c y I x/ J x]
Mz =
3π( G 1 + G 2)[ G 1 + 3 G 2(3 − 2 c x − 2 c y)]
.
(5.8.13) One can verify that the solution (5.8.11) and (5.8.13) is exact for an ellipse. Indeed, consider an ellipse with semiaxes a and b ( a ≥ b ). The necessary geometrical characteristics are J x = 4 b [E( k ) − (1 − k 2)K( k )]/ k 2,
J y = 4 b [K( k ) − E( k )]/ k 2,
C xxxx = 4 b [2(2 k 2 − 1)E( k ) + (1 − k 2)(2 − 3 k 2)K( k )]/(3 k 4), k = [1−( b / a )2]1/2.
C yyyy = 4 b [(2 + k 2)K( k ) − 2(1 + k 2)E( k )]/(3 k 4),
(5.8.14) Here K( k ) and E( k ) are the complete elliptic integrals of the first and the second kind respectively. Substitution of (5.8.14) in (5.8.11) and (5.8.13) yields
qy =
qx =
ω{ G 1 k 2[E − (1− k 2)K] + G 2[(8−7 k 2)E − (1− k 2)(8−3 k 2)K]}
,
π b ( G 1+ G 2){ G 1(K−E)[E −(1− k 2)K] + G 2[ k 2K(K+E)−(K−E)(K+3E)]} ω{( G 1 k 2 − 8 G 2)[K − E] + G 2 k 2[5K − E]
,
π b ( G 1+ G 2){ G 1(K−E)[E −(1− k 2)K] + G 2[ k 2K(K+E)−(K−E)(K+3E)]}
(5.8.15) 2ω a { G 1 k E − G 2[8(1− k )(2− k )K − ( k −16 k +16)E]} 3
Mz =
2
2
2
4
2
3( G 1+ G 2){ G 1(K−E)[E −(1− k 2)K] + G 2[ k 2K(K+E)−(K−E)(K+3E)]}
.
(5.8.16) The abbreviations E and K in (5.8.15) and (5.8.16) stand for E( k ) and K( k ) respectively. The same results can be obtained by direct substitution of (5.8.4) in (5.7.3), with an exact computation of the integrals involved, by using relevant formulae from Appendix A5.1. In the case of isotropy, G 1=(2−ν)/(2πµ), G 2=ν/(2πµ), and formulae (5.8.15) and (5.8.16) simplify as follows: qy =
qx =
µω{ k 2[E − (1− k 2)K] + 2ν(1− k 2)[2E − (2− k 2)K]} b {[K − E][E − (1− k 2)K] + νE[2E − (2− k 2)K]} µω{ k 2[K − E] + 2ν[2E − (2 − k 2)K]} b {[K − E][E − (1− k 2)K] + νE[2E − (2− k 2)K]}
,
,
(5.8.17)
398
Mz =
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
2πµω a 3{ k 4E + 4ν(1 − k 2)[2E − (2 − k 2)K]} 3{[K − E][E − (1− k 2)K] + νE[2E − (2− k 2)K]}
.
(5.8.18)
Formulae (5.8.17) and (5.8.18) are in agreement with the corresponding results of Mindlin (1949). Some misprints are noticed in the relevant formula of (Kassir and Sih, 1968). The result by Willis (1970) seems to be in error, since it indicates that the parameters q y and q x do not depend on elastic constants, which is incorrect. In the case when J x= J y and c x= c y, formulae (5.8.11) and (5.8.13) simplify significantly, namely,
qy = qx Formulae exact for accurate particular
8ω = , π J 0( G 1 + G 2)
Mz =
64 I 0ω 3π J 0( G 1 + G 2)
(5.8.19)
(5.8.11) and (5.8.13) are the main results of this section. They are an ellipse, and there is reason to believe that they will be sufficiently for a general shape. The derivation of specific formulae for any shape is left to the reader.
Example 1: Regular polygon. Consider a punch shaped as a regular polygon with n sides. The necessary moments were computed in paragraph 5.6. We can use expressions (5.8.19), with the result qy = qx =
Mz =
8ω π 1 + sin(π/ n ) π nb ( G 1 + G 2)cos ln n 1 − sin(π/ n )
32ω b 3sin(π/ n )[2 + cos(2π/ n )] 9π( G 1 + G 2) ln
1 + sin(π/ n ) 1 − sin(π/ n )
,
.
(5.8.20)
In the limiting case of n →∞, formulae (5.8.20) give the results for a circular punch qy = qx =
4ω , 2 π b ( G 1 + G 2)
Introduce the stiffness parameter
Mz =
16ω b 3 3π( G 1 + G 2)
(5.8.21)
399
5.8 Reissner-Sagoci problem for general domains
D = M z( G 1 + G 2)/ω = 64 I 0/(3π J 0),
(5.8.22)
which is a geometric characteristic of domain S . If we take the ratio of the stiffness of a regular polygon D to the stiffness D of a circle, having the p c same area, the result is: Dp Dc
=
2sin(π/ n )[2 + cos(2π/ n )] 1 + sin(π/ n ) 3ln 1 − sin(π/ n )
.
(5.8.23)
Elementary analysis of (5.8.23) shows that an equilateral triangle has stiffness 1.24 times greater than that of a circle having an equal area. The corresponding ratio for a square is 1.05, and further increase in n makes the polygon stiffness practically indistinguishable from that of a circle. One may prove a theorem stating that of all simply connected domains, having the same area, the circle has the lowest stiffness. This is opposite to the relevant theorem in the Saint-Venant theory of torsion of rods, where a circle has the greatest stiffness. This is easy to explain since in the torsion of a half-space, one has to twist not only an imaginary rod, but its surroundings as well. Example 2: Rectangle. Consider a punch with a rectangular base, a 1 and a 2 being its semiaxes along the axis Ox and Oy respectively. Introduce the aspect ratio ε= a 2/ a 1. The area and the necessary moments were computed in paragraph 5.6. Formulae (5.8.11) and (5.8.13) for a rectangle take the form
qy =
qx =
ω[( G 1 − 3 G 2)(1 + ε2)1/2sinh-1ε + 4 G 2ε] π a 1ε( G 1+ G 2){( G 1−3 G 2)(1+ε2)1/2sinh-1εsinh-1(1/ε)+2 G 2[εsinh-1(1/ε)+sinh-1ε]} ω[( G 1 − 3 G 2)(1 + ε2)1/2sinh-1(1/ε) + 4 G 2] π a 1( G 1+ G 2){( G 1−3 G 2)(1+ε2)1/2sinh-1εsinh-1(1/ε)+2 G 2[εsinh-1(1/ε)+sinh-1ε]}
,
,
(5.8.24) − 3 G 2)(1 + ε ) [ε sinh (1/ε) + sinh ε] + 4 G 2ε(1 + ε )} Mz = . 9( G 1+ G 2){( G 1−3 G 2)(1+ε2)1/2sinh-1ε sinh-1(1/ε)+2 G 2[εsinh-1(1/ε)+sinh-1ε]} (5.8.25) In order to verify the accuracy of our theory, some computations were made by Kalker for a rectangle with a 1=4 and various aspect ratios. The half-space was assumed to be isotropic, with the shear modulus µ=0.5, and the Poisson coefficient ν=0.3. The quantity C z= M z/ω was computed by using the universal software developed by Kalker. Our results due to (5.8.25), compared with numerical results by Kalker, are presented in the table below. 32ω a 31{( G 1
2 1/2
3
-1
-1
2
400
CHAPTER 5
ε= C z our method C z Kalker Discrepancy (%)
0.1 48.3 36.74 23.9
0.2 62.9 51.29 17.5
APPLICATION TO CONTACT PROBLEMS
0.3 77.9 66.47 14.7
0.5 113.7 102.4 9.9
0.7 160.6 148.4 7.6
1.0 258. 241. 6.6
The accuracy of the numerical procedure was assessed by computation of C z for an ellipse, where the exact solution is well known. The exact solution was about 7% above the numerical result by Kalker. All our results are also above the numerical ones. If we assume that error pattern for a rectangle is the same, as for an ellipse, then our formulae must be considered as surprisingly accurate over a wide range of aspect ratio. We have also compared the shear stress distribution due to (5.8.4), and (5.8.24) for a rectangle, with a 1=4 and ε=0.3, along the line x =0.5. The torsion angle was assumed equal to unity. The shear traction is presented in polar form. The comparison for the modulus |τ| and its argument arg (τ) in degrees is given in the table below y= |τ| our method |τ| Kalker Discrepancy (%)
0.15 0.139 0.145 −1.9
0.45 0.219 0.228 −4.2
0.75 0.375 0.366 2.3
1.05 0.806 1.11 −37.7
arg (τ) our method arg (τ) Kalker
112.57 112.14
141.27 140.72
154.3 152.94
161.03 161.15
The agreement of the argument of τ is very close. The agreement in modulus is satisfactory, except for the point close to the boundary where neither method may claim to be accurate. The fact that the discrepancy does not change monotonically suggests some flaws in the Kalker’s numerical procedure.
Example 3: Cross. Consider a punch with a configuration obtained by the orthogonal intersection of two equal rectangles with sides 2 a and 2 b ( a ≥ b ). Introduce the aspect ratio as ε = b / a . The area and the moments are given in paragraph 5.5. Formulae (5.8.11) and (5.8.13) in this case yield qy = qx =
ω π a ( G 1+ G 2) ln(ε + (1+ε2)1/2) + εln
, 1 + (1+ε ) (1 + √2)ε
2 1/2
401
5.9 Interaction between punches subjected to normal pressure
Mz =
64ω a 3ε(1 + ε2 − ε3) 9π( G 1 + G 2)ln[ε + (1+ε2)1/2] + εln
. 1 + (1+ε ) (1 + √2)ε 2 1/2
Kalker has made computation for a cross, with a =4 and ε=0.5. The remaining numerical data was taken the same as for rectangle. His result C z=154.7, our result is 167.9, with the discrepancy of 7.9%. Taking into consideration the assumed error of Kalker’s software being 7%, our result might be considered very accurate. We have also compared the shear traction distribution along the line y =0.5. The torsion angle ω was assumed equal to unity. The comparison is given in the table below. x= |τ| our method |τ| Kalker Discrepancy (%)
0.5 0.120 0.118 1.5
1.5 0.280 0.287 −2.5
2.5 0.535 0.512 4.4
3.5 1.12 1.47 −23.
arg (τ) our method arg (τ) Kalker
135. 135.
108.4 108.0
101.3 101.1
98.13 98.72
The agreement is good, except for the point close to the boundary.
1.
Exercise 5.8 Establish (5.8.3).
2.
Derive (5.8.8).
3.
Verify (5.8.11) and (5.8.13).
4.
Consider the torsion of a half-space by a rhomboidal punch.
5. Compare the torsional rigidity of a rectangular punch with the torsional rigidity of a rhomboidal one, having the same area and the same aspect ratio.
402
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
5.9 Interaction between punches subjected to normal pressure A general theorem is established which relates the resultant forces, acting on a set of arbitrary punches, with their generalized displacements through a system of linear algebraic equations. The theorem is applied to the case of arbitrarily located elliptical punches. Several specific examples are considered. Theory. Consider a set of N arbitrary punches penetrating an elastic half-space z ≥0. Let Sn be the domain of contact for the n th punch, and P n be the normal force acting on the n th punch. The friction forces between the punches and the half-space are neglected. The problem is to find the relationships between the generalized displacements of the punches and the acting forces. The boundary conditions for the problem are for M ∈ Sn , for M ∉ Sn ,
w = w n( M ) σz( M ) = 0
n =1,2,..., N ,
(5.9.1)
where w denotes the normal displacement of a point at the boundary z =0, and σ stands for the pressure distribution. The prescribed function w n is defined by the punch face. By using the known solution of the Boussinesq problem and the principle of superposition, we can write N
w( Q) = H
σ (T )
Σ n=1
⌠ ⌠ n n dS . ⌡ ⌡ R ( T n, Q ) n
(5.9.2)
Sn
Substitution of the boundary conditions (5.9.1) in (5.9.2) integral equations. The exact solution of these equations present time, even for the case of several circles. Here, we do not really need to know these solutions if we integral characteristics only. We can single out, without first punch, and consider the related integral equation σ (T )
leads to a set of N is not known at the we are to show that are interested in the loss of generality, the
σ (T )
N
Σ
1 1 n n w 1( Q 1) = H ⌠ ⌠ dS1 + H ⌠ ⌠ dS . ⌡ ⌡ R ( T 1, Q 1) ⌡ ⌡ R ( T n, Q 1) n n=2 S1
(5.9.3)
Sn
Suppose that the functions σ0, σx and σy are known, satisfying respectively the following integral equations inside S 1
403
5.9 Interaction between punches subjected to normal pressure
σ (Q )
⌠ ⌠ 0 1 d S = 1, ⌡ ⌡ R ( T 1, Q 1) 1
(5.9.4)
S1
σ (Q )
⌠ ⌠ x 1 dS = x, ⌡ ⌡ R ( T 1, Q 1) 1
(5.9.5)
S1
σ (Q )
⌠ ⌠ y 1 dS = y. ⌡ ⌡ R ( T 1, Q 1) 1
(5.9.6)
S1
Multiplication of both sides of (5.9.3) by σ0( Q 1) and integration over the area S 1 yields σ (T )
⌠ ⌠ σ ( Q ) w ( Q ) d S = H ⌠ ⌠ σ ( Q )d S ⌠ ⌠ 1 1 d S + 1 ⌡⌡ 0 1 1 1 ⌡ ⌡ 0 1 1 ⌡ ⌡ R ( T 1, Q 1) 1 S1
S1
N
+H
Σ n=2
S1
σ (T )
⌠ ⌠ σ ( Q )d S ⌠ ⌠ n n d S . ⌡ ⌡ 0 1 1 ⌡ ⌡ R ( T n, Q 1) n S1
(5.9.7)
Sn
By interchanging the order of integration in (5.9.7) and taking into consideration the fact that σ0 satisfies (5.9.4), the following result can be obtained:
⌠ ⌠ σ ( Q ) w ( Q )d S = H P + 1 ⌡⌡ 0 1 1 1 1 S1
N
Σ⌠⌡ ⌠⌡ w n=2
1n
( T n)σn( T n)d Sn ,
Sn
(5.9.8) where P 1 is the total force acting on the first punch, and σ (Q )
0 1 dS , w 1n( T n) = ⌠ ⌠ ⌡ ⌡ R ( T n, Q 1) 1
(5.9.9)
S1
which is proportional to the normal displacement in the domain Sn due to a flat punch in S 1 under the action of a unit force. By invoking the mean value theorem, which is valid as long as σn does not change sign, we obtain the linear algebraic equation
404
CHAPTER 5
⌠ ⌠ σ ( Q ) w ( Q )d S = H P + 1 ⌡⌡ 0 1 1 1 1 S1
APPLICATION TO CONTACT PROBLEMS
N
Σw
1n
( C n) P n
(5.9.10)
n=2
The exact location of the point C n is not known but the fact that C n∈ Sn allows only a limited variation within Sn , and in many cases provides sufficiently close upper and lower bounds for the parameters sought. By using the same procedure, N −1 additional linear algebraic equations can be derived for the remaining punches. This set of equations provides the necessary relationships between the normal displacements of the punches and the applied forces. Let us derive similar relationships for the angular displacements. Multiplication of both sides of (5.9.3) by σx( Q 1) and integration over the area S 1 yields σ (T )
⌠ ⌠ σ ( Q )w ( Q )d S = H⌠ ⌠ σ ( Q )d S ⌠ ⌠ 1 1 d S + ⌡⌡ x 1 1 1 1 ⌡ ⌡ x 1 1 ⌡ ⌡ R ( T 1, Q 1) 1 S1
S1
N
+ H
Σ n=2
σ (T )
S1
⌠ ⌠ σ ( Q )d S ⌠ ⌠ n n d S . ⌡ ⌡ x 1 1 ⌡ ⌡ R ( T n, Q 1) n S1
(5.9.11)
Sn
By changing the order of integration in (5.9.11) and taking into consideration that σx satisfies (5.9.5), the following result can be obtained
⌠ ⌠ σ ( Q ) w ( Q )d S = H − M + 1y ⌡⌡ x 1 1 1 1 S1
N
Σ n=2
⌠ ⌠ x α ( T )σ ( T )d S , ⌡ ⌡ 1n n n n n Sn
(5.9.12) where M 1y is the tilting moment acting on the first punch about the axis Oy , and α1n( T n) =
σ (Q ) 1⌠ ⌠ x 1 d S 1. x ⌡ ⌡ R ( T n, Q 1)
(5.9.13)
S1
If σn does not change sign we can use again the mean value theorem to yield
⌠ ⌠ σ ( Q ) w ( Q )d S = H −M + 1y ⌡⌡ x 1 1 1 1 S1
N
Σα n=2
1n
( A n) x n P n
(5.9.14)
405
5.9 Interaction between punches subjected to normal pressure
where An(xn,yn) ∈ Sn . The additional N −1 equations relating the punch rotations about the axis Oy with the corresponding tilting moments can be obtained in a similar way. Multiply both sides of (5.9.3) by σy( Q 1) and integrate over S 1. procedure leads to the equation N
Σ
⌠ ⌠ σ ( Q ) w ( Q )d S = H M + 1x ⌡⌡ y 1 1 1 1 S1
where M
1x
The
⌠ ⌠ y β ( T )σ ( T )d S , ⌡ ⌡ 1n n n n n Sn
n=2
(5.9.15) is the tilting moment about the axis Ox acting on the first punch,
and β1n( T n) =
σ (Q ) 1⌠ ⌠ y 1 d S 1. y ⌡ ⌡ R ( T n, Q 1)
(5.9.16)
S1
As before, application of the mean value theorem gives
⌠ ⌠ σ ( Q ) w ( Q )d S = H M + ⌡⌡ y 1 1 1 1 1x
N
Σβ
S1
1n
n=2
( B n) y n P n.
Three sets of linear algebraic equations of the type (5.9.10), (5.9.14) and are the main results of this paragraph. It is clear that each equation interpreted in terms of the reciprocal work. In order to use them, one know the normal displacements outside every punch in the system due types of loading which, at the moment, is available for the elliptical only. This particular case is considered below.
(5.9.17) (5.9.17) can be need to to three punches
Application to elliptical punches. Consider the interaction of a set of N flat elliptical punches arbitrarily located on a transversely isotropic elastic half-space. Let a n and b n be the major and the minor semiaxes of the n th ellipse; X n and Y n define its center, and θn be the angle between the axis Ox and the major semiaxis a n; and P n be the normal force acting upon the n th punch. The functions σ0, σx and σy have the form (Lur’e, 1955): σ0 =
1 1 − x 2 − y 2 -1/2 2π b 1K( k 1) a 21 b 21
406
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
x2 y 2 -1/2 σx = 1 − 2 − 2 2π b 1D( k 1) a1 b 1 x
σy =
1 − x 2 − y 2 -1/2 2π b 1B( k 1) a 21 b 21 y
The boundary conditions (5.9.1) in this case will take the form w n = δn − αn x + βn y ,
for n =1,2,..., N .
(5.9.18)
Substitution of (5.9.18) into (5.9.10), (5.9.14) and (5.9.17) yields respectively a1
δ = HP 1 + K( k 1) 1
N
Σ
α = HM − 3D( k 1) 1 1y
3 B (k1) where α1n =
β
1n
=
K( k 1)
n=2
a 31
a 1 b 21
F (φ1n, k 1)
β1 = H M + 1x
P n,
(5.9.19)
,
(5.9.20)
,
(5.9.21)
N
Σα
x P 1n n n
n=2
N
Σβ
y P 1n n n
n=2
F (φ1n, k 1) − E (φ1n, k 1) K( k 1) − E( k 1)
,
(5.9.22)
E (φ1n, k 1) − (1 − k 21) F (φ1n, k 1) − k 21(ρ21n − 1)1/2/ρ1n(ρ21n − k 21)1/2 E( k 1) − (1 − k 21)K( k 1)
, (5.9.23)
B( k 1) =
D( k 1) =
E( k 1) − (1 − k 21) K( k 1) k 21 K( k 1) − E( k 1) k 21
,
,
(5.9.24)
(5.9.25)
407
5.9 Interaction between punches subjected to normal pressure
K( k 1), F (φ1n, k 1), E( k 1), E (φ1n, k 1) stand for the complete and incomplete elliptic
integrals of the first and second kind respectively; and ρ1n, φ1n are defined by φ1n = sin-1(
L =
1 ), ρ1n
ρ1n = [ L + ( L 2 − k 21 x 2n/ a 21)1/2]1/2,
1 2 [ k + ( x 2n + y 2n)/ a 21], 2 1
where x n and y n are coordinates of a certain point inside S n, and k 1 is the eccentricity of the first ellipse k 1 = [1 − b 21/ a 21]1/2. Each of the equations (5.9.19), (5.9.20) and (5.9.21) represent the first of a set of N equations. When the acting forces are known, the three sets of equations define the normal and the angular displacements of the punches. In the case where the displacements are known, the three sets of linear algebraic equations have to be solved for P n, M nx and M ny. It is also important to notice that each equation in the set is valid in the system of coordinates located at the center of the ellipse. Example 1: Two equal elliptical punches. Consider the case where N =2, a 1= a 2= a , b 1= b 2= b , X 1= Y 1=0, X 2= l , Y 2=0, θ1=θ2=0. If we also have P 1= P 2= P then, due to the symmetry of the system, the set of equations, equivalent to (5.9.19), reduces to just one equation, namely, a F (φ, k ) P , δ = HP + K( k ) K( k ) with the immediate result P0
P = 1 +
F (φ, k ) K( k )
,
where P 0=δ a / H K( k ) denotes the force which produces a δ when acting on a solitary punch. Equation (5.9.27) between the punches decreases the value of the force required settlement. The upper and lower bounds for (5.9.27) by taking
(5.9.26)
(5.9.27)
punch settlement equal to shows that the interaction necessary to produce the P can be obtained from
408
CHAPTER 5
a , φ = sin-1 l − a respectively.
and
APPLICATION TO CONTACT PROBLEMS
a . φ = sin-1 l + a
(5.9.28)
We shall also consider the central estimation for P defined by
a φ = sin-1( ). l
(5.9.29)
Figure 5.9.1 plots the ratio P/P0 versus l/a for a =2, b =1. The solid line gives the upper bound, the dashed line gives the lower one and the circles represent the central estimation. As one can see, the maximum possible error of the
Fig.
5.9.1.
Interaction between two elliptical punches.
central estimation is less than 9% for l/a >3.5, it is less than 5% for l/a >5, it is less than 2% for l/a >8, and it is less than 1% for l/a >12. Since there is no accurate solution available for this case, it is difficult to say how great is the real error of the central estimation, but there is a reason to believe that it is much less than that indicated above. This belief is founded in a comparison of the central estimation for two equal circular punches with the numerical solution of Kobayashi (1939). The values of the ratio P/P0 for various d = l/a are given in Table 5.9.1. If one takes Kobayashi’s solution as exact then the maximum error of the central estimation does not exceed 0.4% in the whole range of 2≤ d <∞. Even if one assumes the accuracy in the case of two elliptic punches to be ten times
409
5.9 Interaction between punches subjected to normal pressure
Table 5.9.1 Comparison of two solutions
d
upper bound
lower bound
central estimation
result of Kobayashi
error (%)
2.0 2.2 2.4 2.6 2.8 3.0 3.5 4.0 5.0 7.0 10.0 ∞
0.82213 0.83172 0.84030 0.84804 0.85505 0.86143 0.87515 0.88638 0.90367 0.92611 0.94522 1.
0.50000 0.61457 0.66379 0.69940 0.72728 0.75000 0.79241 0.82213 0.86143 0.90367 0.93381 1.
0.75000 0.76900 0.78517 0.79915 0.81136 0.82213 0.84427 0.86143 0.88638 0.91637 0.94005 1.
0.75272 0.77014 0.78545 0.79898 0.81096 0.82162 0.84370 0.86093 0.88602 0.91619 0.93999 1.
0.36 0.15 0.04 -0.02 -0.05 -0.06 -0.07 -0.06 -0.04 -0.02 -0.007 0.0
worse than the accuracy of the central estimation for two circular punches, this would still give the maximum error of 4% which is not bad. Bearing this in mind, we shall evaluate the central estimation only in the examples to follow. If the normal forces are applied centrally then the angles of inclination of the punches will be defined by (5.9.20) and (5.9.21) as α1 = −α2 = −
3D( k ) H α12 lP , a 31
β1 = β2 =0,
(5.9.30)
where
F (φ, k ) − E(φ, k ) , K( k ) − E( k ) and φ is defined by (5.9.29). α12 =
k = (1 − b 2/ a 2)1/2,
(5.9.31)
In the case α1=α2=0 there should be tilting moments applied to the punches whose value can be determined from (5.9.20) as M
1y
= −M
2y
= α12 lP .
(5.9.32)
Example 2: Four equal elliptical punches. Consider the configuration shown in Fig. 5.9.2. Let equal vertical forces P be applied to each punch. The punches are numbered in the clockwise direction starting from the one at the coordinate system origin. Due to the symmetry of the system, it is sufficient to consider just one equation of each of the sets (5.9.19−5.9.21). The result is
410
CHAPTER 5
Fig.
5.9.2.
a δ = HP 1 + K( k )
APPLICATION TO CONTACT PROBLEMS
Geometry of four punches interaction
F (φ12, k ) K( k )
+
F (φ13, k ) K( k )
+
F (φ , k ) 14
, K( k )
(5.9.33)
a3 α = H M − P l (α13 + α ), 14 3D( k ) 1 1y
(5.9.34)
ab 2 β = H M1x + Pc (β12 + β13), 3B( k ) 1
(5.9.35)
where α1n and β1n are defined by (5.9.22) and (5.9.23) respectively, and φ1n = sin-1(
1 ), ρ1n
for
n =2,3,4;
ρ12 = 1 + ( c 2 − b 2)/ a 21/2,
ρ14 =
1/2 ρ13 = L + ( L 2 − k 2 l 2/ a 2)1/2 ,
(5.9.36)
L =
l a 1 2 k + ( l 2 + c 2)/ a 2, 2
(5.9.37) When the forces and the tilting moments are known, equations (5.9.33−5.9.35) define the normal and the angular displacements of the punches.
411
5.9 Interaction between punches subjected to normal pressure
The mathematically equivalent problem of diffusion membranes was solved in (Fabrikant, 1985a, 1987k).
through
perforated
Exercise 5.9 1. Consider the interaction between the four elliptical punches depicted in Fig. 5.9.2. Find the tilting moments if the punches are not allowed to tilt. Answer: M = − Pc (α12 + α13) , M = Pl (α13 + α ). 1x
1y
14
2. Find the limiting case of formulae (5.9.22−5.9.25) for the case of circular punches. a a1 a 21 1/2 2 -1 1 1 − 2 , Answer: α →β → sin − 1n 1n r π r r 1n
1n
1n
F (φ , k 1)
a1 2 -1 , B( k 1)→D( k 1)→π/4 , sin → K( k 1) r π 1n and the equations (5.9.19−5.9.21) will take the form 1n
N
Σ
2 2 a δ = HP 1 + π 1 1 π 4 a 31 3π 4 a 31
α1 = H M
n=2
a
1 P nsin , r1n -1
N
1y
−
β = H M 1x + 3π 1
Σα
xP 1n n n
n=2
N
Σβ
yP 1n n n
n=2
,
,
where a 1 is the radius of punch one and r
1n
is the distance between the first
punch centre and a point inside S n. 3. Find the generalized displacements of an elliptical punch due to the action of several concentrated loads, applied normally outside the punch. The Hint : Consider the procedure of shrinking of the areas Sn , n =2,3,..., N . accuracy of equations (5.9.19−5.9.21) will increase. In the limiting case Sn →0 formulae (5.9.19−5.9.21) give an exact solution to the problem of several concentrated forces acting outside an elliptical punch. 4.
Consider the interaction of several punches on a non-homogeneous elastic
412
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
half-space, with the modulus of elasticity being proportional to a power function of the depth.
5.10 Interaction between flexible punches under shifting loading The mixed boundary value problem for a transversely isotropic elastic half-space is considered for the case where uniform tangential displacements are prescribed over several domains of arbitrary shape and the rest of the half-space boundary is stress free. The problem can be interpreted either as an interaction between two elastic half-spaces interconnected through several areas and subjected to remote shear loading, or as a contact problem of several flexible punches, connected to the half-space, with different tangential displacements prescribed. A general theorem is established which relates the resulting tangential forces, acting on each domain, with their generalized displacements through a system of linear algebraic equations. The theorem is applied to the case of arbitrarily located elliptical domains subjected to uniform tangential displacements. Several specific examples are considered. Theory. Consider a transversely isotropic elastic half-space z ≥0. Let the tangential displacements be prescribed over several simply connected domains S n while the rest of the half-space boundary is stress free. The mathematical formulation of the boundary conditions on the plane z =0 is u x = u x( x , y ),
u y = u y( x , y ),
σz = τzx = τyz = 0,
for ( x , y )∈ S n,
for ( x , y )∉ S n,
n =1,2, ..., N .
(5.10.1)
The boundary value problem (5.10.1) can be interpreted either as an interaction between several flexible punches S n bonded to the half-space and subjected to tangential displacements, or as the case of two elastic half-spaces connected in S n and subjected to a remote shear loading, the first interpretation being more general since the tangential displacements each punch can be treated as independent. Hereafter we shall call S n the domain of contact for the n th punch. The punches are assumed to be flexible, so that they do not exert any normal tractions. Let T and T be the components of the resulting tangential nx
ny
force applied to the n -th punch. The problem is to find the relationships between the generalized displacements of the punches and the acting forces. Introduce the complex tangential displacements u = u x+i u y, and the complex tangential tractions τ=τzx+iτyz. In view of the principle of superposition, the governing integral equation can we rewritten from (5.7.5) as follows:
413
5.10 Interaction between flexible punches under shifting loading
G G 1 ⌠ τ(ξ,η) dξdη + 2 ⌠ ⌠ q τ(ξ,η) dξdη = u ( x , y ). 2 ⌠ 2 ⌡⌡ ⌡⌡ R q R n=1 Sn Sn N
Σ
(5.10.2) Here q = x − ξ + i( y − η),
q = x − ξ − i( y − η),
R 2 = q(5.10.3) q.
Substitution of the boundary conditions (5.10.1) in (5.10.2) leads to a set of N integral equations. The exact solution of these equations is not known at present, even for the case of several circles. Here it will be shown that we do not really need to know these solutions if we are interested only in the relationship between the applied forces and the punch displacements. We can single out, without loss of generality, the first punch, and consider the related integral equation τ (ξ,η)
G1
⌠⌠ 1 u 1( x 1, y 1) = 2 ⌡ ⌡ R 11 S1
G2
⌠⌠ dξdη + 2 ⌡⌡ S1
q 11 τ1(ξ,η) q11 R11
dξdη
q τn(ξ,η) G2 τn(ξ,η) G 1 1n ⌠ ⌠ ⌠ ⌠ + dξdη + dξdη. 2 2 ⌡⌡ ⌡ ⌡ R 1n q R 1n 1n n=2 Sn Sn N
Σ
(5.10.4)
Here u n stands for tangential displacement of the n -th punch, τn is the tangential tractions exerted by the punch, the point ( x n, y n)∈ S n, and q
kn
R
kn
= x k − ξn + i( y k − ηn),
q
kn
= x k − ξn − i( y k − ηn),
(ξn,ηn)∈ S n.
= ( q q )1/2, kn kn
(5.10.5)
We have dropped for simplicity the subscripts of ξn and ηn in (5.10.4) and thereafter. It should not produce any confusion for the reader. Equation (5.10.4) can be rewritten as G1
G
2 u 1( x 1, y 1) = ∆⌠ ⌠ R 11τ1(ξ,η) dξdη − Λ2⌠ ⌠ R 11τ1(ξ,η) dξdη 2 2 ⌡⌡ ⌡⌡ S1
S1
414
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
G G 1 ⌠ R τ (ξ,η) dξdη − 2 Λ2⌠ ⌠ R τ (ξ,η) dξdη. 2 ∆⌠ 2 ⌡ ⌡ 1n n ⌡ ⌡ 1n n n=2 Sn Sn N
+
Σ
(5.10.6) Suppose that the function τ0, is known, satisfying the following integral equation inside S 1
⌠ ⌠ τ (ξ,η) R dξdη = a x 2 + a y 2 + a xy + a x + a y + a , 1 2 3 4 5 6 ⌡⌡ 0 S1
(5.10.7) where a 1, ...
, a 6 are complex constants.
One can verify that
∆ ⌠ ⌠ τ0(ξ,η) R 11 dξdη = 2 a 1 + 2 a 2 = d 1 = const,
⌡⌡ S1
Λ2⌠ ⌠ τ0(ξ,η) R 11 dξdη = 2 a 1 − 2 a 2 + 2i a 3 = - c 1 = const,
⌡⌡
(5.10.8)
S1
which means that the function τ0 renders constant the first two integrals in expression (5.10.4). This important property will be used in the derivation to follow. Multiplication of both sides of (5.10.4) by τ0 and integration over the area S 1 yields G
⌠ ⌠ u (x ,y ) τ (x ,y ) dx dy = 1 ⌠ ⌠ τ (x ,y ) dx dy ⌠ ⌠ 1 1 1 1 2 ⌡⌡ 0 1 1 ⌡⌡ 1 1 1 0 1 1 ⌡⌡ S1
S1
+
G2
⌠ ⌠ τ (x ,y ) dx dy ⌠ ⌠ 1 1 2 ⌡⌡ 0 1 1 ⌡⌡ S1
q 11 τ1(ξ,η) q11 R11
dξdη
τn(ξ,η) G 1 ⌠ ⌠ ⌠ ⌠ τ ( x , y ) d x d y dξdη 2 ⌡⌡ 0 1 1 1 1 ⌡ ⌡ R 1n n=2 S1 Sn N
+
S1
S1
Σ
τ1(ξ,η) R 11
dξdη
415
5.10 Interaction between flexible punches under shifting loading
G2
⌠ ⌠ τ (x ,y ) dx dy ⌠ ⌠ + 1 1 2 ⌡⌡ 0 1 1 ⌡⌡ S1
q 1n τn(ξ,η) q1n R1n
Sn
dξdη.
(5.10.9)
By changing the order of integration in (5.10.9) and taking into consideration the property (5.10.8), the following result can be obtained G
G
⌠ ⌠ u (x ,y ) τ (x ,y ) dx dy = 1 d T + 2 c T 1 1 2 1 1 2 1 1 ⌡⌡ 1 1 1 0 1 1 S1
τ0 ( x 1 , y 1 ) G 1 ⌠ ⌠ ⌠ ⌠ ( , ) d d τ ξ η ξ η d x 1d y 1 2 n R 1n ⌡ ⌡ ⌡ ⌡ n=2 Sn S1 N
+
Σ G2
⌠ ⌠ τ (ξ,η) dξdη ⌠ ⌠ + 2 ⌡⌡ n ⌡⌡ Sn
S1
q 1n τ0( x 1, y 1) q1n R1n
d x 1 d y 1 .
(5.10.10)
where T 1 is the complex representation for the total tangential force acting on the first punch. Introduce the notation Φ(ξ,η) = ⌠ ⌠ ⌡⌡ S1
τ0 ( x 1 , y 1 ) R 1n
d x 1d y 1,
Ψ(ξ,η) = ⌠ ⌠ ⌡⌡ S1
q 1n τ0( x 1, y 1) q1n R1n
d x 1d y 1 . (5.10.11)
Now expression (5.10.10) can be rewritten as G
G
⌠ ⌠ u (x ,y ) τ (x ,y ) dx dy = 1 d T + 2 c T 1 1 2 1 1 2 1 1 ⌡⌡ 1 1 1 0 1 1 S1
G2 G 1 2 ⌠ ⌠ Φ(ξ,η) τn(ξ,η) dξdη + 2 ⌠ ⌠ Ψ(ξ,η) τn(ξ,η) dξdη. ⌡⌡ ⌡⌡ n=2 Sn Sn N
+
Σ
(5.10.12) By evoking the mean value theorem, we come to the linear algebraic equation, for the forces T n applied to each punch.
416
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
G
G
⌠ ⌠ u (x ,y ) τ (x ,y ) dx dy = 1 d T + 2 c T 1 1 2 1 1 2 1 1 ⌡⌡ 1 1 1 0 1 1 S1
G2 G 1 2 Φ( x n, y n) T n + 2 Ψ(ξn,ηn) T n . n=2 N
+
Σ
(5.10.13)
The exact location of the points ( x n, y n) and (ξn,ηn) is not known but the fact that they belong to S n allows only a limited variation, and in many cases provides sufficiently close upper and lower bounds for the parameters sought. By a similar argument, N −1 additional linear algebraic equations can be derived for the remaining punches. This set of equations provides the necessary relationships between the tangential displacements of the punches and the applied forces, and represents the main result of this paragraph. It is clear that each equation can be interpreted in terms of the reciprocal work. In order to use equations (5.10.13), one needs to know the explicit expression for τ0 which, at present, is available for the elliptical punch only. Application to elliptical punches. Consider the interaction of a set of N flexible elliptical punches arbitrarily located on an elastic half-space, with uniform tangential displacements u n=const prescribed in S n. Let a n and b n be the major and the minor semiaxes of the n th ellipse; X n and Y n define its center, θn be the angle between the axis Ox and the major semiaxis a n, and T n be the tangential force acting upon the n th punch. The function τ0 for the first punch is x2 y 2 -1/2 τ0( x , y ) = 1 − 2 − 2 . a1 b1
(5.10.14)
Substitution of (5.10.14) in (5.10.10) yields, after computation of the integrals involved (see Appendix A5.1 for details) 2 a 1 u 1 = G 1K( k 1) T 1 + G 2[(2 − k 21)K( k 1) − 2 E( k 1)] T 1/ k 21 N
+
Σ
1 2
n=2
2 ⌠⌠ 2 ⌠⌠ ⌠ ⌠ G D τ d S + ( G / k ) L τ d S − i( G / k ) M τ d S 1⌡ ⌡ 1n n n 2 1 2 1 n , 1n n ⌡ ⌡ 1n n n ⌡⌡ Sn Sn Sn (5.10.15)
where
417
5.10 Interaction between flexible punches under shifting loading
D 1n = [ F (φ1n, k 1) − F (ψ1n, k 1)], L 1n = {(2 − k 21)[ F (φ1n, k 1) − F (ψ1n, k 1)] − 2[E(φ1n, k 1) − E(ψ1n, k 1)]}, M 1n = [(1 − k 21sin2φ1n)1/2 − (1 − k 21sin2ψ1n)1/2].
(5.10.16)
Here k 1=(1− b 21/ a 21)1/2, K( k 1), F(φ1n, k 1), E( k 1), E(φ1n, k 1) denote the complete and
incomplete elliptic integrals of the first and second kind respectively; ψ1n and φ1n
are defined according to formulae (A5.1.20−A5.1.21) from Appendix A5.1, namely, φ1n = tan
-1
x n y n + ( a 21 y 2n + b 21 x 2n − a 21 b 21 )1/2 y 2n − b 21
ψ1n = tan
-1
,
x n y n − ( a 21 y 2n + b 21 x 2n − a 21 b 21 )1/2 y 2n − b 21
.
(5.10.17)
In the case when φ <ψ , it should be replaced by π+φ . 1n 1n 1n point ( x n, y n)∈ S n. transforms into
We recall that the
If the mean value theorem is applicable, equation (5.10.15)
2 a 1 u 1 = G 1K( k 1) T 1 + G 2[(2 − k 21)K( k 1) − 2E( k 1)] T 1/ k 21 N
1 + 2
Σ n=2
2 2 G 1[ F (φ1n, k 1) − F (ψ1n, k 1)] T n + ( G 2/ k 1)[(2 − k 1)( F (φ1n, k 1)
− F (ψ1n, k 1)) − 2(E(φ1n, k 1) − E(ψ1n, k 1))] T n
− i( G 2/ k 21)[(1 − k 21sin2φ1n)1/2 − (1 − k 21sin2ψ1n)1/2] T n.
(5.10.18)
It is important to note that in the case of symmetric configuration, some of the terms in (5.10.18) vanish; the parameters φ and ψ do not necessarily 1n 1n correspond to the same point ( x n, y n); quantities which can assume any value (5.10.18) represents the first of the set of coefficients, which can be obtained in a
they should be interpreted as fuzzy corresponding to ( x n, y n)∈ S n. Equation N linear algebraic equations with fuzzy similar manner. The solution can also
418
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
be interpreted as a fuzzy domain. Analysis of this domain will allow us to find the upper and lower bounds for the quantities of interest. As we shall see further, in some cases the variation between the upper and the lower bounds becomes so small that allows us to obtain a reasonably accurate solution to the problem. When the acting forces are known, the set of equations define the tangential displacements of the punches. In the case when the displacements are known, the set of linear algebraic equations has to be solved for the forces T n. It is also important to notice that each equation in the set is valid in a system of coordinates located at the center of the ellipse. Example: Two equal elliptical punches. Consider a 1= a 2= a , b 1= b 2= b , X 1= Y 1=0, X 2= l , Y 2=0, θ1=θ2=0. displacements be prescribed as u 1= u 2= u where u is a real symmetry of the system, we may also assume T 1= T 2= T , resulting force is directed along the axis Ox ). Taking properties
the case where N =2, Let the tangential quantity. Due to the and ℑ[ T ]=0 (i.e. the into consideration the
D 1n( x , y ) = D 1n(− x , y ) = D 1n( x ,− y ) = D 1n(− x ,− y ), L 1n( x , y ) = L 1n(− x , y ) = L 1n( x ,− y ) = L 1n(− x ,− y ), M 1n( x , y ) = − M 1n(− x , y ) = − M 1n( x ,− y ) = M 1n(− x ,− y ),
(5.10.19)
⌠ ⌠ M τd S = 0, ⌡ ⌡ 1n n
(5.10.20)
we have
Sn
and the set of equations, equivalent to (5.10.15), reduces to just one equation, namely, 2 au = 2{ G 1K( k ) T + ( G 2/ k 2)[(2 − k 2)K( k ) − 2E( k )] T } − { G 1F( k ,φ) T + ( G 2/ k 2)[(2 − k 2)F( k ,φ) − 2E( k ,φ)] T },
(5.10.21)
where, due to symmetry, φ is defined as b φ = cos-1 2 2 2 1/2 ( x − a + b )
(5.10.22)
419
5.10 Interaction between flexible punches under shifting loading
Introduce the notation T0 =
2 au G 1K( k ) + ( G 2/ k )[(2 − k 2)K( k ) − 2E( k )] 2
(5.10.23)
where T 0 can be interpreted as the force needed to produce the displacement u when acting on an isolated punch. Equation (5.10.21) can be rewritten as follows:
T =
T0 2 − B
,
B =
G 1F( k ,φ) + ( G 2/ k 2)[(2 − k 2)F( k ,φ) − 2E( k ,φ)] G 1K( k ) + ( G 2/ k 2)[(2 − k 2)K( k ) − 2E( k )] (5.10.24)
Since B ≥0, one may conclude that the punch interaction reduces the force needed to produce the displacement u , as compared to an isolated punch. The upper and lower bounds for T can be obtained from (5.10.24) by taking
φ = cos-1
b , [( l + a )2 − a 2 + b 2]1/2
and
φ = cos-1
respectively.
b . [( l − a )2 − a 2 + b 2]1/2
(5.10.25)
We shall also consider the central estimation for T defined by
. ( l 2 − a 2 + b 2)1/2 b
φ = cos-1
We can consider in a similar manner the case when the displacement is prescribed in the Oy direction by the formal substitution of u by i u , and T by i T in expression (5.10.21). The result is
T =
T0 2 − C
,
C =
G 1F( k ,φ) − ( G 2/ k 2)[(2 − k 2)F( k ,φ) − 2E( k ,φ)] G 1K( k ) − ( G 2/ k 2)[(2 − k 2)K( k ) − 2E( k )]
. (5.10.26)
Figure 5.10.1 plots the ratio T/T0 versus l/a for a =2, b =1, ( G 2/ G 1)=1/3, with the
420
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
tangential displacements being directed along the axis Ox . The upper and lower bounds are given by circles, while the central estimation is given by a solid line.
Fig.
5.10.1. Interaction between two elliptical punches, with shifting load in the Ox direction
As one can see, the maximum possible error of the central estimation is less than 15% for l/a ≥2.5, it is less than 10% for l/a ≥3, it is less than 7% for l/a >3.5, and it is less than 5% for l/a ≥4. Since there is no accurate solution available for this case, it is difficult to say how great is the real error of the central estimation, but there is a reason to believe that it is much less than indicated above. This belief is supported by the same argument as in section 5.9. Figure 5.10.2 gives the same plot for the case when the tangential displacements are prescribed in the Oy direction. The respective errors are 12%, 7%, 5%, and 3.5%. One can see that the accuracy of the central estimation is significantly better than in the first case. This should be expected since the interaction in the case of displacements prescribed along the axis Ox is stronger then in the second case.
Discussion. It is appropriate to consider certain limiting cases. In the case of a circular punch the eccentricity k1→0, and formula (5.10.15) will simplify
421
5.10 Interaction between flexible punches under shifting loading
Fig.
5.10.2. Interaction between two elliptical punches, with shifting load in the Oy direction
π = G T + 2 1 1
2a1u1
+ G 2⌠ ⌠
⌡⌡ Sn
a1 -1 τ ( x, y) dxdy ⌠ ⌠ sin G 1 ⌡ ⌡ ( x 2 + y 2)1/2 n n=2 Sn N
Σ
a 1( x 2 + y 2 − a 21)1/2 ( x − i y )2
τn ( x , y ) d x d y .
(5.10.27)
Here we used the integrals (A5.1.22−A5.1.23). The interaction of two equal circular punches subjected to loading T in the Ox direction leads to the following expression for the central estimation 2 au =
π a a ( l 2 − a 2)1/2 G 1 T + G 1 T sin-1( ) + G 2 T . 2 l l2
Introducing again the notation T0 =
4 au , πG1
(5.10.28)
422
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
the following result can be obtained from (5.10.28) T0
T =
(5.10.29)
G 2 a ( l 2 − a 2)1/2 2 -1 a 1 + sin ( ) + 2 l G1 π l If the tangential displacements are prescribed in the Oy direction, we obtain T0
T =
(5.10.30)
G 2 a ( l 2 − a 2)1/2 2 -1 a 1 + sin ( ) − 2 l G1 π l All the results are valid for the case of an isotropic body, provided that G 1=(2−ν)/(2πµ) and G 2=ν/(2πµ), where µ is the shear modulus and ν is the Poisson coefficient.
1.
Exercise 5.10 Verify (5.10.6).
2.
Derive (5.10.12).
3. Consider the interaction of two equal elliptical punches, when their major semiaxes are orthogonal to each other. 4. Find the generalized displacements of an elliptical punch due to the action of several concentrated loads, applied tangentially outside the punch. Hint : Consider the procedure of shrinking the areas S n, n =2,3,..., N , to a point. The accuracy of equation (5.10.18) will increase. In the limiting case S n→0 formula (5.10.18) gives an exact solution to the problem of several concentrated forces acting outside an elliptical punch.
5.11 Contact problem for a rough punch The term rough punch is used here as the opposite to the term smooth punch, indicating a punch which does produce shear traction at its base. Consider a circular punch, with general base, penetrating transversely isotropic elastic half-space. The punch is subjected to the action of an axial force P and a tangential force T = kP , acting in the Ox direction, with k being the friction
423
5.11 Contact problem for a rough punch
coefficient. traction τ
zx
Let the domain of contact be a circle of radius a . The shear is assumed to be proportional to the pressure σ, namely, τ = k σ. zx
The governing integral equation will take the form (due to (2.2.13)): 2π a
⌠⌠ ⌡⌡ 0
2π a
σ(ρ0,φ0)ρ0dρ0dφ0
w (ρ,φ) = − α kℜ⌠ ⌠ H ⌡⌡
R
0
0
0
Here w (ρ,φ) is the punch settlement. form Nσ =
σ(ρ0,φ0)ρ0dρ0dφ0 iφ
ρe
i φ0
.
(5.11.1)
− ρ0e
Let us rewrite (5.11.1) in the operator
w − α k Mσ, H
(5.11.2)
where N and M are the integral operators in the left- and the right-hand sides of (5.11.1) respectively. The values of α and k for real bodies are less that unity, and hence their product is convenient to use as a small parameter. Let the following expansion hold ∞
σ(ρ,φ) =
Σ(αk) σ (ρ,φ). n
(5.11.3)
n
n=0
By substitution of (5.11.3) in (5.11.2) and equating the terms with equal powers of the small parameter, the following infinite system of integral equations can be obtained: Nσ0 =
w , H
Nσ1 = −Mσ0, ...,
Nσn = −Mσn-1, ...
(5.11.4)
The operator N −1, inverse to N is known from (1.4.10), so we can write the formal solution to the problem as follows: ∞
σ =
Σ n=0
w (α k )n(−N −1 M)nN −1 . H
(5.11.5)
Example. Consider the case of a flat circular punch. Let ω be the settlement of its centre, and δ be its tilting angle about the Oy axis. Then w (ρ,φ) = ω + δρcosφ. Substitution of (5.11.6) in the first equation of (5.11.4) yields
(5.11.6)
424
CHAPTER 5
σ0(ρ,φ) =
APPLICATION TO CONTACT PROBLEMS
1 ω + 2δρcosφ , π H ( a 2 − ρ2)1/2
(5.11.7)
2
which is, in fact, the solution for an inclined smooth punch. Substitution of (5.11.7) in the second equation of (5.11.4) yields the second term of the expansion 2 ω a ln[( a 2 − ρ2)1/2/ a ] x 2d x 1 d ⌠ a + x 1/2 + σ1(ρ,φ) = 3 cos φ + δ ln ρ dρ⌡ ( x 2 − ρ2)1/2 a − x πH ρ( a 2 − ρ2)1/2 a
ρ
a a a 3ln[( a 2 − ρ2)1/2/ a ] + cos2φ. + ρ2( a 2 − ρ2)1/2 ( a 2 − ρ2)1/2 2( a 2 − ρ2)1/2
(5.11.8)
The procedure may be continued, and further terms of the expansion may be obtained. It is left to the interested reader. The resultant force P and the tilting moment M can be obtained by integration of the sum of (5.11.7) and (5.11.8) as follows: P =
2a αk ω + a δ, πH π
M =
2 a 22 αk ω . aδ − π H 3 π
(5.11.9)
If the point, where the normal load P is applied, is shifted along the axis Ox by the distance b from the punch centre, then M = Pb , and the following relationship may be established between the punch settlement ω and the tilting angle δ: δ =
ω b + (α k /π) a . (2/3) a − (α k /π) b a
Formula (5.11.10) shows that the punch tilts even loading. A similar effect can be observed in the (Chapter 3) and in two-dimensional contact problems order to eliminate tilting, the normal load should x =−α ka /π.
1.
Exercise 5.11 Verify (5.11.5).
2.
Derive (5.11.7) −(5.11.8).
3.
Establish (5.11.9).
(5.11.10) in the case of a central case of a bonded punch (Muskhelishvili, 1946). In be applied at the point
425
Appendix A5.1
4.
Find the corrective term in the expression for P (5.11.9) due to σ2.
Answer: P =
2a 1 αk a δ. ω 1 − α 2 k 2 + 6 π πH
5. Find the normal displacement w outside the punch, taking into consideration the first two terms in the solution expansion. 2 a a a Answer: w = ωsin-1( ) + δρsin-1( ) − (ρ2 − a 2)1/2cosφ π ρ ρ ρ
+
4α k a -1 a (ρ2 − a 2)1/2 ρ cosφ cos + ln 2 ω ( ) 2 ρ ρ ρ (ρ − a 2)1/2 π
+ δ(ρ2 − a 2)1/2ln
ρ δ 2 3 -1 a a cos ( ) 2 1/2 + ρ (ρ − a ) ρ23 2
a 2 ρ2 + 2 a 2 ρ + (ρ2 − a 2)1/2 ln 2 cos2 − φ . 3 6 (ρ − a 2)1/2
6.
Try to find an exact closed form solution to the problem.
Appendix A5.1 Here the method of computation of several integrals over an elliptical domain is given. We introduce the notation R = [( x − x 0)2 + ( y − y 0)2]1/2 Z 0 = 1 −
x 20 a
2
−
y 20
1/2 b 2
The following integral is to be computed I = ⌠⌠ ⌡⌡ S
f ( x 0, y 0)d x 0d y 0 RZ 0
,
(A5.1.1)
where S is an ellipse with semiaxes a and b ( a ≥ b ), and f is a general function. We use the method of Rostovtsev (1961), slightly modified in order to simplify
426
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
the computations. Introduce a new system of polar coordinates ( r ,θ), with the origin at the point ( x , y ), namely, x 0 = x + r cosθ,
y 0 = y + r sinθ.
(A5.1.2)
First of all, consider the case when ( x , y )∈ S . (A5.1.1) yields 2π
r1
ab dθ ⌠ ⌡ ( b cos θ + a 2sin2θ)1/2 ⌡
I = ⌠
2
2
0
Substitution of (A5.1.2)
in
f ( x + r cosθ, y + r sinθ) d r , [( r 1 − r )( r + r 2)]1/2
0
(A5.1.3)
where r 1 and − r 2 are the roots of algebraic equation ( x + r cosθ)2 ( y + r sinθ)2 + − 1 = 0, a2 b2 namely, r 1,2
( Q 2 + PZ )1/2 = P
P =
cos2θ sin2θ + , a2 b2
Q
,
Q =
Z = 1 −
x cosθ y sinθ + , 2 a b2
x2 y2 . − a2 b2
(A5.1.4)
Now, we may split the integral (A5.1.3) in two, according to the scheme 2π
π
r1
2π
r1
r1
⌠ dθ ⌠ d r = ⌠ dθ ⌠ d r + ⌠ dθ ⌠ d r. ⌡ ⌡ ⌡ ⌡ ⌡ ⌡ 0 π 0 0
0
(A5.1.5)
0
Make a formal replacement of θ by π+θ in the second integral of (A5.1.5). Taking into consideration that r 1(θ)= r 2(π+θ), transformation of (A5.1.5) may proceed as follows 2π
r1
π
r1
π
r2
π
r1
π
0
⌠ dθ ⌠ d r = ⌠ dθ ⌠ d r + ⌠ dθ ⌠ d r = ⌠ dθ ⌠ d r + ⌠ dθ ⌠ d r. ⌡ ⌡ ⌡ ⌡ ⌡ ⌡ ⌡ ⌡ ⌡ ⌡ 0 -r 0 0 0 0 0
0
0
0
2
(A5.1.6) In the last term of (A5.1.6) we have made a formal replacement of r by − r .
427
Appendix A5.1
Taking into consideration that the expression f ( x + r cosθ, y + r sinθ) remains unchanged with the replacements of θ by π+θ and r by − r , the two integrals in (A5.1.6) can be combined, and the final result will read π
r1
ab dθ ⌠ 2 2 1/2 ⌡ ( b cos θ + a sin θ) ⌡
I = ⌠
2
f ( x + r cosθ, y + r sinθ)d r , [( r 1 − r )( r + r 2)]1/2
2
-r 2
0
(A5.1.7)
We recall the integral r1
1 1 − k ) Γ( + k ) r 2 2 − 2k . Γ( n + 1 − k ) Γ( k + 1) r 1 Γ( n +
n
Σ
n
r dr n ⌠ 1/2 = r 1 ⌡ [( r1 − r)( r + r2)] k=0 -r 2
(A5.1.8) Now various integrals can be computed fairly easily. are used in the main text of this paper
⌠⌠ ⌡⌡
d x 0d y 0 RZ0
S
⌠⌠ ⌡⌡ S
q d x 0d y 0 q RZ0
Here are the results which
π
ab dθ 2 2 1/2 = 2π b K( k ), ⌡ ( b cos θ + a sin θ)
= π⌠
2
2
(A5.1.9)
0
π
ab e2iθ dθ 2 2 2 2 2 2 1/2 = 2π b [(2 − k )K( k ) − 2E( k )]/ k . ( b cos θ + a sin θ ) ⌡
= π⌠ 0
(A5.1.10)
Now consider the case when the point ( x , y )∉ S . of (A5.1.2) in (A5.1.1) yields α2
In this case substitution
r1
ab dθ θ) d r ⌠ f ( x + rcosθ, y + rsin1/2 , 2 2 1/2 [( r 1 − r )( r + r 2)] ⌡ ( b cos θ + a sin θ) ⌡
I = ⌠
α1
2
2
-r 2
(A5.1.11) where α1 and α2 are the angles between the axis Ox and the two lines passing through the point ( x , y ) tangentially to the ellipse. We can find these angles by using classical formulae of differential geometry. The parametric equation of the ellipse can be written
428
CHAPTER 5
X = a cost ,
APPLICATION TO CONTACT PROBLEMS
Y = b sin t .
(A5.1.12)
The equation of the lines passing through the point ( x , y ) tangentially to the ellipse may be written as y − b sin t x − a cost . = b cost − a sin t
(A5.1.13)
Equation (A5.1.13) can be solved for sin t , and the two solutions are x ( x 2 b 2 + y 2 a 2 − a 2 b 2)1/2]
b[a2y
(sin t )1,2 =
x2b2 + y2a2
.
(A5.1.14)
.
(A5.1.15)
This gives the following two expressions for cost a [ b 2 x ± y ( x 2 b 2 + y 2 a 2 − a 2 b 2)1/2]
(cos t )1,2 =
x2b2 + y2a2
Now we can write tanα = 1,2
y − b sin t . x − a cost
(A5.1.16)
Substitution of (A5.1.14−A5.1.15) in (A5.1.16) yields tanα1,2 =
( a 2 y 2 + b 2 x 2 − a 2 b 2)1/2
xy
x2 − a2
.
(A5.1.17)
We can now compute the integrals
⌠⌠ ⌡⌡ S
d x 0d y 0 RZ0
α2
ab dθ = π b [F(φ, k ) − F(ψ, k )], ⌡ ( b cos θ + a 2sin2θ)1/2
= π⌠
2
2
α1
⌠⌠ ⌡⌡ S
q d x 0d y 0 q RZ0
α2
2iθ dθ ab e 2 2 ⌡ ( b cos θ + a 2sin2θ)1/2
= π⌠
α1
= π b {(2 − k 2)[F(φ, k ) − F(ψ, k )] − 2[E(φ, k ) − E(ψ, k )]
(A5.1.18)
429
Appendix A5.1
− 2i[(1 − k 2sin2φ)1/2 − (1 − k 2sin2ψ)1/2}/ k 2.
(A5.1.19)
Here xy + ( a 2 y 2 + b 2 x 2 − a 2 b 2)1/2
tanφ = cotα1 =
y2 − b2
tanψ = cotα2 =
.
xy − ( a 2 y 2 + b 2 x 2 − a 2 b 2)1/2 y2 − b2
.
(A5.1.20)
(A5.1.21)
In the limiting case of a circle, b → a , k →0, and formulae (A5.1.18−A5.1.19) will take the form
⌠⌠ ⌡⌡
d x 0d y 0
⌠⌠ ⌡⌡
q d x 0d y 0
S
S
RZ0
q RZ0
a , ( x + y 2)1/2
(A5.1.22)
( x 2 + y 2 − a 2)1/2 . ( x − i y )2
(A5.1.23)
= 2π a sin-1
= 2π a 2
2
The more complicated integrals can be computed in the same manner. Here are some additional integrals, where x and y are inside the domain S . 2 ⌠ ⌠ x 0d x 0d y 0 = π b 3 1 1 E − K + π b x 2 1 2K − (2 + k 2)E + π b y 2 1 2E − (2 − k 2)K ⌡ ⌡ RZ 0 k 2 1 − k 2 k4 k4 S
2 2 ⌠ ⌠ x 0 y 0d x 0d y 0 = 2πb xy1 − k 2 − k E − 2K RZ 0 ⌡⌡ k 4 1 − k 2 S
2 2 ⌠ ⌠ y 0d x 0d y 0 = π b 3 1 K − E + π b x 2 1 − k 2E − (2 − k 2)K ⌡ ⌡ RZ 0 k2 k4 S
+ π b y2
1 2 2 2 4 2(1 − k ) K − (2 − 3 k )E k
3 2 ⌠ ⌠ x 0d x 0d y 0 = π b 3 x 1 2 − k E − 2K + π b xy2 1 ( − 8 + 5 k 2)K + (8 − k 2)E ⌡ ⌡ RZ 0 k 4 1 − k 2 k6 S
430
CHAPTER 5
+ π b x3
APPLICATION TO CONTACT PROBLEMS
1 2 4 2 4 6 (8 − 3 k + k )K − (8 + k + 2 k )E 3k
2 ⌠ ⌠ x 0 y 0d x 0d y 0 = π b 3 y 1 (2 − k 2)K − 2E 4 RZ 0 ⌡⌡ k S
1 + π b x 2 y 6 ( − 8 + 9 k 2 − k 4)K + (8 − 5 k 2 − k 4)E k + π b y3
1 2 4 2 6 (8 − 11 k + 3 k )K − (8 − 7 k )E 3k
2 ⌠ ⌠ x 0 y 0d x 0d y 0 = π b 3 x 1 (2 − k 2)K − 2E 4 RZ 0 ⌡⌡ k S
+ π b xy 2
+ π b x3
1 2 4 2 4 6 (8 − 15 k + 7 k )K + ( − 8 + 11 k − 2 k )E k
1 2 4 2 4 6 (8 − 9 k + k )E − (8 − 13 k + 5 k )K 3k
3 ⌠ ⌠ y 0d x 0d y 0 = π b 3 y 1 (2 − k 2)E − 2(1 − k 2)K 4 ⌡ ⌡ RZ 0 k S
+ π b y3
6 1 2 4 2 4 6 (8 − 17 k + 11 k )E + ( − 8 + 21 k − 19 k + 6 k )K 3k
+ π b x2y
1 2 4 6 2 4 6 (8 − 19 k + 14 k − 3 k )K − (8 − 15 k + 7 k )E k
4
⌠ ⌠ x 0d x 0d y 0 = πbx 4 1 ( − 48 + 8 k 2 − 4 k 4 − 6 k 6)E + (48 − 32 k 2 + 5 k 4 + 3 k 6)K 8 ⌡ ⌡ RZ 0 12 k S
+ π bx 2 y 2
1 2 4 2 4 8 (48 − 16 k − k )E − (48 − 40 k + 4 k )K 2k
431
Appendix A5.1
+ π by 4
1 2 4 2 8 (16 − 16 k + 3 k )K − (16 − 8 k )E 4k
+ πb3 x2
+ πb5
1 8 − 5 k 2 − k 4 2 + πb 3 y 2 1 (8 − 3 k 2)K − 8 − 7 k 2E − − k E (8 )K 6 2 6 2 2k 1 − k 2k 1− k
1 − 2 + 4 k 2E + (2 − 3 k 2)K 4 2 2 4 k (1 − k ) 1 − k
3
⌠ ⌠ x 0 y 0d x 0d y 0 = πbx 3 y 1 (48 − 40 k 2 − 2 k 6)E − (1 − k 2)(48 − 16 k 2 + k 4)K 8 ⌡ ⌡ RZ 0 3k S
1 + π bx y 3 8 (1 − k 2)(16 − 8 k 2)K − (16 − 16 k 2 + k 4)E k 1 + π b 3 xy 6 (8 − 5 k 2)K − (8 − k 2)E k 2 2
⌠ ⌠ x 0 y 0d x 0d y 0 = πbx 4(1 − k ) (48 − 16 k 2 − k 4)E − (48 − 40 k 2 + 4 k 4)K ⌡ ⌡ RZ 0 12 k 8 2
S
+ π bx 2 y 2
+ π by 4
1 ( − 48 + 72 k 2 − 20 k 4 − 2 k 6)E + (1 − k 2)(48 − 48 k 2 + 5 k 4)K 2k8
1 2 4 2 2 4 8 (48 − 80 k + 31 k )E − (1 − k )(48 − 56 k + 12 k )K 12 k
+ πb3 x2
1 2 4 2 6 (8 − 7 k + k )K − (8 − 3 k )E 2k
+ πb3 y2
1 2 2 4 + πb 5 1 2 − k 2E − 2K (8 5 )E (8 9 + 2 )K − k − − k k 2k6 4 k 4 1 − k 2
3
2 ⌠ ⌠ x 0 y 0d x 0d y 0 = πbx 3 y (1 − k ) (1 − k 2)(16 − 8 k 2)K − (16 − 16 k 2 + k 4)E ⌡ ⌡ RZ 0 k8 S
432
CHAPTER 5
+ π bx y 3
APPLICATION TO CONTACT PROBLEMS
1 2 4 6 2 2 4 8 (48 − 104 k + 64 k − 6 k )E − (1 − k )(48 − 80 k + 33 k )K 3k
1 + π b 3 xy 6 (8 − 7 k 2)E − (1 − k 2)(8 − 3 k 2)K k 4
2 4 ⌠ ⌠ y 0d x 0d y 0 = πbx 4(1 − 2 k + k ) (16 − 16 k 2 + 3 k 4)K − (16 − 8 k 2)E 8 ⌡ ⌡ RZ 0 4k S
+ π bx 2 y 2
+ π by 4
(1 − k 2) (48 − 80 k 2 + 31 k 4)E − (1 − k 2)(48 − 56 k 2 + 12 k 4)K 8 2k
1 2 4 6 2 2 4 6 8 ( − 48 + 136 k − 132 k + 50 k )E + (1 − k )(48 − 112 k + 85 k − 24 k )K 12 k
+ πb3 x2
(1 − k 2) (8 − k 2)E − (8 − 5 k 2)K 6 2k
+ πb3 y2
1 2 4 2 2 5 1 2 2 6 ( − 8 + 11 k − 2 k )E + (1 − k )(8 − 7 k )K + π b 4 (2 + k )K − (2 + 2 k )E 4k 2k
The case where x and y are outside the domain S :
⌠ ⌠ x 0d x 0d y 0 = πbx 1 [F(φ) − F(ψ)] − [E(φ) − E(ψ)] ⌡ ⌡ RZ 0 k2 S
+ π bx
sin φ cos φ sin ψ cos ψ 1 1 1 + π by 2 − − ∆(ψ) ∆(φ) k ∆(ψ) ∆(φ)
⌠ ⌠ y 0d x 0d y 0 = πby 1 [E(φ) − E(ψ)] − (1 − k 2)[F(φ) − F(ψ)] ⌡ ⌡ RZ 0 k2 S
− π by
sin φ cos φ sin ψ cos ψ 1 − k2 1 1 + π bx 2 − − ∆(ψ) ∆(φ) k ∆(ψ) ∆(φ)
Appendix A5.1
2 ⌠ ⌠ x 0d x 0d y 0 = πbx 2 1 − (2 + k 2)[E(φ) − E(ψ)] + 2[F(φ) − F(ψ)] ⌡ ⌡ RZ 0 2k4 S
+ π bx 2
sin φ cos φ sin ψ cos ψ 1 sin ψ cos ψ 2 sin φ cos φ − − − (1 − k 2) 2 (2 + k ) ∆(ψ) ∆(φ) ∆3(φ) 2k ∆3(ψ)
+ π bxy
1 − k 2 1 1 3 − k 2 1 1 − − 4 3 3 2 ∆(φ) − ∆(ψ) k ∆ (φ) ∆ (ψ) 1 − k
+ π by 2
1 2 4 2[E( φ) − E(ψ)] − (2 − k )[F( φ) − F(ψ)] 2k
+ π by 2
1 sin φ cos φ sin ψ cos ψ sin φ cos φ sin ψ cos ψ + −2 − − ∆(ψ) ∆3(ψ) 2 k 2 ∆(φ) ∆3(φ)
− πb3
1 sin φ cos φ − sin ψ cos ψ 2 ∆(ψ) 2(1 − k ) ∆(φ)
2 ⌠ ⌠ x 0 y 0d x 0d y 0 = πbx 2 1 − k − 3 1 − 1 + (1 − k 2) 1 − 1 RZ 0 ⌡⌡ ∆3(φ) ∆3(ψ) 2 k 4 ∆(φ) ∆(ψ) S
+ π bxy
1 − k 2 k2 2 + 4 2 [E( φ) − E(ψ)] − 2[F(φ) − F(ψ)] k 1−k
+ π bxy
1 − k 2 sin φ cos φ sin ψ cos ψ 2 − k 2 sin φ cos φ sin ψ cos ψ − − − ∆(ψ) k 2 ∆3(φ) 1 − k 2 ∆(φ) ∆3(ψ)
+ π by 2
1 − k2 1 1 1 1 1 1 1 3 − − 3 + 3 + πb3 2 − 4 2 k ∆(φ) ∆(ψ) ∆ (φ) ∆ (ψ) 2 k ∆(φ) ∆(ψ)
2 2 ⌠ ⌠ y 0d x 0d y 0 = πbx 2 1 − k 2[E(φ) − E(ψ)] − (2 − k 2)[F(φ) − F(ψ)] ⌡ ⌡ RZ 0 2k4 S
+ π bx 2
1 − k 2 sin φ cos φ sin ψ cos ψ sin φ cos φ sin ψ cos ψ + (1 − k 2) −2 − − 2 ∆ ( φ ) ∆ ( ψ ) 2k ∆3(φ) ∆3(ψ)
433
434
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
+ π bxy
1 − k2 2 1 1 1 1 − ( k − 1)( 3 − 3 ) + (3 − 2 k 2) 4 ∆ φ ∆ ψ) ( ) ( k ∆ (φ) ∆ (φ)
+ π by 2
1 2 2 2 4 (3 k − 2)[E( φ) − E(ψ)] + 2(1 − k ) [F(φ) − F(ψ)] 2k
+ π by 2
sin φ cos φ sin ψ cos ψ 1 sin ψ cos ψ 2 sin φ cos φ − − − (1 − k 2) 2 (2 − 3 k ) ∆ φ ∆ ψ ( ) ( ) ∆3(φ) ∆3(ψ) 2k
1 sin φ cos φ sin ψ cos ψ − πb3 − ∆(ψ) 2 ∆(φ) 3
⌠ ⌠ x 0d x 0d y 0 = πbx 3 1 (8 − 3 k 2 + k 4)[F(φ) − F(ψ)] − (8 + k 2 + 2 k 4)[E(φ) − E(ψ)] + ⌡ ⌡ RZ 0 6k6 S
+ π bx 3
1 2 4 sin φ cos φ sin ψcos ψ − + 4 (8 + k + 2 k ) ∆(ψ) ∆(φ) 6k
sin φcos φ sin ψcos ψ sin φcos φ sin ψcos ψ − − + 3(1 − k 2)2 5 +(− 7 + 6 k 2 + k 4) 3 3 ∆ (φ) ∆ (φ) ∆5(ψ) ∆ (ψ) + π bx 2 y
1 1 2 1 6 (9 k − 15) ∆(φ) − ∆(ψ) + 2k
1 1 1 1 − 5 +(1 − k 2)(10 − 3 k 2) 3 − 3 − 3(1 − k 2)2 5 ∆ (φ) ∆ (ψ) ∆ (φ) ∆ (ψ) + π bxy 2
1 2 2 6 ( − 8 + 5 k )[F( φ) − F(ψ)] − (8 − k )[E( φ) − E(ψ)] 2k
+ π bxy 2
1 2 4 sin φcos φ sin ψcos ψ − + 2 ( − 8 + 3k + 2k ) ∆(ψ) ∆(φ) 2 k (1 − k ) 4
sin φcos φ sin ψcos ψ sin φcos φ sin ψcos ψ +(7 − k 2) 3 − − 3(1 − k 2) 5 − + 3 ∆ (φ) ∆ (φ) ∆ (ψ) ∆5(ψ) + π by 3
1 1 1 1 1 2 1 2 1 − 3 + 3(1 − k 2) 5 − 5 6 3(5 − 3 k ) ∆(φ) − ∆(ψ) + (10 − 8 k ) 3 ∆ (φ) ∆ (ψ) ∆ (φ) ∆ (ψ) 6k
435
Appendix A5.1
+ π xb 3
1 − 2[F(φ) − F(ψ)] + (2 − k 2)[E(φ) − E(ψ)] 2 2 k (1 − k )
+ π xb 3
sin φcos φ sin ψcos ψ 1 2 sin φcos φ sin ψcos ψ − − + (1 − k 2) 3 2 (−2+k ) ∆(ψ) ∆(φ) ∆ (φ) ∆3(ψ) 2 k (1 − k )
+ π yb 3
1 1 1 1 1 − 3 4 3 ∆(φ) − ∆(ψ) − 3 ∆ (φ) ∆ (ψ) 2k
4
2
2
⌠ ⌠ x 0 y 0d x 0d y 0 = πbx 3 1 − 3(1 − k 2)(5 − k 2) 1 − 1 + ∆(φ) ∆(ψ) ⌡ ⌡ RZ 0 6k6 S
1 1 1 1 − 3 − 3(1 − k 2)3 5 − 5 +(1 − k 2)2(10 − k 2) 3 ∆ (φ) ∆ (ψ) ∆ (φ) ∆ (ψ) +π bx 2 y
1 2 2 2 4 6 −(8 − k )(1 − k )[F( φ) − F(ψ)] + (8 − 5 k − k )[E( φ) − E(ψ)] 2k
+ π bx 2 y
sin ψcos ψ 1 2 4 sin φcos φ − + 4 (8 − 9 k + 3 k ) ∆(ψ) ∆(φ) 2k
sin φcos φ sin ψcos ψ sin φcos φ sin ψcos ψ − − 3(1 − k 2)2 5 − +(7 − k 2)(1 − k 2) 3 3 ∆ (φ) ∆ (φ) ∆ (ψ) ∆5(ψ) + π bxy 2
1 − k2 1 1 1 1 − − (10 − 6 k 2) 3 − 3 + (15 − 3 k 2) 6 ∆ φ ∆ ψ ( ) ( ) 2k ∆ (φ) ∆ (ψ)
1 1 + 3(1 − k 2) 5 − 5 ∆ (φ) ∆ (ψ) +π by 3
1 2 2 2 6 (8 − 3 k )(1 − k )[F( φ) − F(ψ)] − (8 − 7 k )[E( φ) − E(ψ)] 6k
+ π by 3
1 sin ψcos ψ sin φcos φ sin ψcos ψ 2 sin φcos φ − − 7(1 − k 2) 3 − + 4 (8 − 7 k ) ∆ ( φ ) ∆ ( ψ ) 6k ∆ (φ) ∆3(ψ)
sin φcos φ sin ψcos ψ + 3(1 − k 2) 5 − ∆ (φ) ∆5(ψ)
436
CHAPTER 5
APPLICATION TO CONTACT PROBLEMS
+ π xb 3
1 1 1 2 1 2 1 − 3 4 (3 − k ) ∆(φ) − ∆(ψ) − (1 − k ) 3 2k ∆ (φ) ∆ (ψ)
+ π yb 3
1 2 4 (2 − k )[F( φ) − F(ψ)] − 2[E( φ) − E(ψ)] 2k
+ π yb 3
1 sin φcos φ sin ψcos ψ sin φcos φ sin ψcos ψ 2 − − − ∆(ψ) ∆3(φ) 2 k 2 ∆(φ) ∆3(ψ)
2
⌠ ⌠ x 0 y 0d x 0d y 0 = πbx 31 − k − (8 − 5 k 2)[F(φ) − F(ψ)] + (8 − k 2)[E(φ) − E(ψ)] ⌡ ⌡ RZ 0 6k6 2
S
+ π bx 3
−
sin φcos φ sin ψcos ψ sin φcos φ 1 − k2 + 7(1 − k 2) 3 − − (8 − k 2) − 4 ∆(ψ) ∆(φ) ∆ (φ) 6k
sin φcos φ sin ψcos ψ sin ψcos ψ − − 3(1 − k 2)2 5 3 ∆ (φ) ∆5(ψ) ∆ (ψ)
+ π bx 2 y
−
1 1 1 1 − k2 (15 − 12 k 2) − − (1 − k 2)(10 − 4 k 2) 3 − ∆(φ) ∆(ψ) ∆ (φ) 2k6
1 1 1 + 3(1 − k 2)2 5 − 5 ∆ (φ) ∆ (ψ) ∆ (ψ) 3
+ π bxy 2
1 2 2 2 4 6 (8 − 7 k )(1 − k )[F( φ) − F(ψ)] + ( − 8 + 11 k − 2 k )[E( φ) − E(ψ)] 2k
+ π bxy 2
1 2 4 sin φcos φ sin ψcos ψ − − 4 (8 − 11 k + 2 k ) ∆(ψ) ∆(φ) 2k
sin φcos φ sin ψcos ψ sin φcos φ sin ψcos ψ + 3(1 − k 2)2 5 − (7 − 5 k 2)(1 − k 2) 3 − − 3 ∆ (φ) ∆ (φ) ∆ (ψ) ∆5(ψ) + π by 3
1 − k2 1 1 1 1 1 1 + (10 − 9 k 2) 3 − 3 − 3(1 − k 2) 5 − 5 − 3(5 − 4 k 2) − 6 ( ) ( ) ∆ φ ∆ ψ 6k ∆ (φ) ∆ (ψ) ∆ (φ) ∆ (ψ)
+ π xb 3
1 2 4 (2 − k )[F( φ) − F(ψ)] − 2[E( φ) − E(ψ)] 2k
437
Appendix A5.1
+ π xb 3
sin φcos φ sin ψcos ψ 1 sin φcos φ sin ψcos ψ − (1 − k 2) 3 − − 2 2 ∆ ψ ∆ φ ( ) ( ) 2k ∆ (φ) ∆3(ψ)
+ π yb 3
1 1 1 2 1 2 1 − 3 4 − (3 − k ) ∆(φ) − ∆(ψ) + (1 − k ) 3 ∆ (φ) ∆ (ψ) 2k
3
⌠ ⌠ y 0d x 0d y 0 = πbx 3(1 − k ) 3(5 − 2 k 2) 1 − 1 − (1 − k 2)(10 − 2 k 2) 1 − ∆(φ) ∆(ψ) ∆3(φ) ⌡ ⌡ RZ 0 6k6 2 2
S
−
1 1 1 + 3(1 − k 2)2 5 − 5 ∆ (ψ) ∆ (φ) ∆ (ψ) 3
+ π bx 2 y
1 − k2 (8 − k 2)(1 − k 2)[F(φ) − F(ψ)] − (8 − 7 k 2)[E( φ) − E(ψ)] 6 2k
sin φcos φ sin ψcos ψ sin φcos φ sin ψcos ψ 1 − k2 − − − (7 − 3 k 2)(1 − k 2) 3 (8 − 7 k 2) + π bx y 4 ∆ φ ∆ ψ ( ) ( ) ∆ (φ) 2k ∆3(ψ) 2
sin φcos φ sin ψcos ψ − + 3(1 − k 2)2 5 ∆ (φ) ∆5(ψ) + π bxy 2
+ π by 3
1 1 1 1 1 1 (1 − k 2)2 − (15 − 6 k 2) − + (10 − 7 k 2) 3 − 3 − 3(1 − k 2) 5 − 5 6 ∆(φ) ∆(ψ) ∆ (φ) ∆ (ψ) ∆ (φ) ∆ (ψ) 2k
1 2 4 2 2 4 6 − (8 − 13 k + 6 k )(1 − k )[F( φ) − F(ψ)] + (8 − 17 k + 11 k )[E( φ) − E(ψ)] 6k
+ π by 3
1 sin φcos φ sin ψcos ψ 2 4 sin φcos φ sin ψcos ψ + (1 − k 2)(7 − 8 k 2) 3 − − − 4 (− 8 + 17 k − 11 k ) ∆(ψ) ∆(φ) ∆ (φ) 6k ∆3(ψ)
sin φcos φ sin ψcos ψ − 3(1 − k 2)2 5 − ∆ (φ) ∆5(ψ) + π xb 3
1 − k2 1 1 1 1 + (1 − k 2) 3 − 3 −3 − 4 ( ) ( ) ∆ φ ∆ ψ 2k ∆ (φ) ∆ (ψ)
+ π yb 3
1 2 2 4 − 2(1 − k )[F( φ) − F(ψ)] + (2 − k )[E( φ) − E(ψ)] 2k
438
CHAPTER 5
+ π yb 3
APPLICATION TO CONTACT PROBLEMS
sin φcos φ sin ψcos ψ 1 2 sin φcos φ sin ψcos ψ − + (1 − k 2) 3 − 2 − (2 − k ) ∆ ( φ ) ∆ ( ψ ) ∆3(ψ) ∆ (φ) 2k
In the integrals above, the abbreviations F(φ), E(φ), F(ψ), E(ψ) were used to denote F(φ, k ), E(φ, k ), F(ψ, k ), E(ψ, k ) respectively. The notation ∆(φ) stands for (1 − k 2sin2φ)1/2 The case of a circle of radius a and x, y inside the domain S : 2 ⌠ ⌠ x 0d x 0d y 0 = π2 a 1 4 a 2 + 5 x 2 − y 2 16 ⌡ ⌡ RZ 0 S
⌠ ⌠ x 0 y 0d x 0d y 0 = π2 a 3 xy RZ 0 8 ⌡⌡ S
2 ⌠ ⌠ y 0d x 0d y 0 = π2 a 1 4 a 2 − x 2 + 5 y 2 16 ⌡ ⌡ RZ 0 S
3 ⌠ ⌠ x 0d x 0d y 0 = π2 ax 1 6 a 2 + 7 x 2 − 3 y 2 32 ⌡ ⌡ RZ 0 S
2 ⌠ ⌠ x 0 y 0d x 0d y 0 = π2 ay 1 2 a 2 + 9 x 2 − y 2 RZ 0 32 ⌡⌡ S
2 ⌠ ⌠ x 0 y 0d x 0d y 0 = π2 ax 1 2 a 2 − x 2 + 9 y 2 RZ 0 32 ⌡⌡ S
3 ⌠ ⌠ y 0d x 0d y 0 = π2 ay 1 6 a 2 − 3 x 2 + 7 y 2 32 ⌡ ⌡ RZ 0 S
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447
Subject Index
Subject Index
The page numbering in this Index is approximate because present edition of the book does not conserve the pages from original publication. Abel operator, 27-30, 42, 61, 65, 82, 92, 94, 183, 287 Abel integral, 168 Abel generalized integral equation, 182-187 Annulus, uniform pressure distributed over, 90 Approach, 2, 5, 6, 36, 182, 268, 280, 303, 350, 379, 402, 407, 413 Axisymmetric pressure, 5 problems internal mixed-mixed, 148-176 external mixed-mixed, 176-182 crack, 227, 239 Bessel functions, 4,5 integrals of, 253-256, 348 349 Bonded contact, 7, 145, 148-176 asymmetric punch problem, 187-193 concentrated load, influence of, 200-206 flexible punches, 434 non-homogeneous half-space, 186 two-dimensional problems, 447 Boundary conditions, 3, 27, 39, 44, 50-59, 80, 91, 105, 111, 131, 145, 188, 239, 337, 406, 412, 422, 433
Boundary value problems, see also Axisymmetric problems internal mixed of type I, 80-91 internal mixed of type II, 105-111 external mixed of type I, 91-101 external mixed of type II, 111-131 Boussinesq problem, 422 see also Concentrated load, point force solutions Circular crack, 7, 83, 85, 89, 90, 109, 237, 285-319 disk, 1, 3-4, 38 punch, 2, 7, 26, 88, 145, 148, 163, 172-176, 186-193, 341-350, 418, 432, 445-448 sector, 277, 360, 382-383 segment, 276, 359, 376 Complex conjugate, 38 constants, 114 functions, 107 stress intensity factor, 120 stresses, 74 tangential displacements, 72 Compliance, 411 Concentrated load outside a crack, 237-239 outside a punch, 175, 200, 211, 349, 433 point force solutions, 75-80 Contact, adhesive see Bonded contact Contact problems see also Boundary value problems
448
for a circular punch, 341-350 for a flat punch of general shape, 350-383 for a flexible punch under shifting load, 406-412 for a general curved punch, 383-406 for a rough punch, 445-448 for a smooth punch, 337-341 interaction between punches, 422-445 Reissner-Sagoci problem, 412-422 Contact, torsional see Contact problems, Reissner-Sagoci problem Contour, 269, 279, 352, 360 Crack problems, see also Boundary value problems, asymptotic behavior of stresses and displacements near the crack rim, 262-268 circular sector shape, 277 circular segment shape, 276 close interaction of pressurized coplanar circular cracks, 285-304 close interaction of coplanar circular cracks under shear loading, 304-319 concentrated load outside a circular crack, 237-239 cross-shaped, 278 flat crack of general shape, 268-284 flat crack under arbitrary normal loading, 227-234 general crack under uniform shear, 284-285 hexagon-shaped, 273 penny-shaped crack under uniform pressure, 252-257 penny-shaped crack under uniform shear loading, 257-262 plane crack under arbitrary
Subject Index
shear loading, 239-252 point force loading of penny-shaped crack, 234-237 rectangular, 274, 281, 285 rhomboidal, 275
a
Deformation, energy of, 318 Dirac delta-function, 109 Dirichlet conditions, 7 problem, 17 Displacement complex tangential, 72 normal, 2, 7 outside a penny-shaped crack, 255, 258-260 polynomial, 3 under a bonded punch, 148-176 under a circular punch, 341-346 under a point load, 75-80 Dual integral equations, 3-4 Elastic constants, 71, 73, 75, 241, 319, 418 Elastic half-space, 2, 7, 8, 26 see also Elastic constants non-homogeneous, 81 transversely isotropic, 71, 77 Electrical capacity, 362 Electrical charge distribution, 358 Electrical polarizability, 272 Electrified disk, 1 Elliptic integrals, 134, 377, 409, 417, 439 Elliptic punch, see Punch elliptic Equilibrium conditions, 191 equations, 72 Force see also Concentrated load; complex tangential, 76 friction, 8, 191 normal, 148
449
Subject Index
resultant, 83, 134, 153 shearing, 187, 191 unit, 305 Fourier series, 7, 9, 17, 37, 81, 105, 112, 193, 284 Fracture, see Crack problems Fredholm integral equations, 304 Friction coefficient, 445 forces 8, 191 Function see also Bessel functions complex stress, 152-172 Green’s, 1, 6, 234, 245, 340 hypergeometric, 36 potential, 5, 26, 232, 339 Green’s Green’s
functions
see
functions,
Half-space see also Elastic half-space; Axisymmetric problem; Bonded contact; Boundary value problems; Contact problems Boussinesq problem, 75-80 loading over annulus, 173-174 non-homogeneous, 186 torsion of, 412-422 transversely isotropic, 71 Harmonic function, 3, 7, 30, 41, 44, 73, 228, 239 Inclusion problem, 131-144, 302 Indentation, see Bonded contact; Boundary value problems; Contact problems Integral equations see also Dual integral equations approximate solutions for general domains, 341-411 for a bonded punch problem, 193, 206 for a half-space under shear loading, 105, 111 for a rough punch, 446
for a smooth punch problem, 338 for mixed boundary value problems of potential theory, 27, 40 for non-homogeneous half-space, 80, 91 set of, 406-444 Integral representation for the reciprocal of the distance between two points, 9-17 involving inverse cube of the distance, 19-25, 101-105 of more general type, 60-70 Integro-differential equation, 229, 241, 257 Isotropic medium, 238, 255, 259-260, 344-345 Jacobi polynomial expansion, 182 Kernel of the integral equation, 302, 407 singular part of, 198, 209 degenerate part of, 202,211 Klein-Gordon equation, 17 Laplace equation, 29 operator, 3, 41, 134, 286 L-operator, 17-19 Loading, axisymmetric, 5 distributed over annulus, 90 point force, 200-206 shear, 239-252 uniform, 252-257 Magnetic polarizability, 369-383 Mehler-Fok transform, 200 Moment complex, 135 inertia, 366,382 linear, 365
450
Subject Index
linear of fourth order, 387 tilting, 83, 94, 187-193, 367 Neumann boundary conditions, 7, 32 Normal stress, 71 Penetration, see Indentation Poisson coefficient, 75, 166, 171, 318 operator, 17 Polynomial displacements, 3, 408 kernel, 198, 209 loading, 6, 231, 232, 268, 307 profile, 8 Punch see also Contact problems arbitrary planform, 350-421 bonded, 148-176, 187-193 circular, 341-350 circular sector base, 360, 382-383 circular segment base, 359, 376 cross-shaped, 378, 397, 405, 421 curved, 383-406 elliptical, 409, 417, 426-433, 438-445 inclined, 346-350, 362-383 interaction between, 422-445 polygonal base, 355, 369, 391, 418 rectangular, 358, 372, 380, 394, 403, 410, 419 rigid roller, 398-402 rhomboidal, 359, 374, 380, 395 rough, 445-448 settlement, 89, 165, 172, 341, 350, 357, 400 smooth, 337-341 triangular base, 356, 371 Rectangle, see Punch, rectangular; Crack problems, rectangular Reissner-Sagoci problem, 412-422
Rhombus, see Punch, rhomboidal; Crack problems, rhomboidal Saint-Venant theory of bending, 382 of torsion, 419 Separation in contact problems, 373, 377 Shear modulus, 238 Shear stress, 136, 304, 407, 420 Singular integral, 153 Singular integral equation, 152, 198 Sphere, 15, 39 Spherical bowl, 1 Spherical coordinates, 262 Spheroidal inclusions, 302 Strain energy, 318 Stress complex, 74 components of in cylindrical polar coordinates, 345-346 function, 152-172 intensity factor, 86, 90, 97, 120, 143, 237, 251, 263-267, 289, 309 intensity function, 86, 91, 97 normal, 71 Stress-strain relations, 71 Surface displacements, 165 quadratic, 383 tractions, 164 Toroidal coordinates, 1, 15 Traction, see Stress Transforms, see under particular name Transversely isotropic material, 71-74 Uniform electrical charge distribution, 38 pressure, 90, 180, 252, 268, 290 shear tractions, 130, 257, 284, 307
451
Subject Index
tangential 433
displacements,
406,
Variational approach, 280, 379, 402
Work, 85, 95, 426, 438