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THEORY OF ELASTICITY

ENGINEERING SOCIETIES MONOGRAPHS Bakhmeteff: H ydraulics of Open Channels Bleich: Buckling Strength of Metal Structures Crandall: Engineering Analysis Elevatorski: Hydraulic Energy Dissipators Leontovich: Frames and Arches Nadai: Theory of Flow and Fracture of Solids Timoshenko and Gere: Theory of Elastic Stability Timoshenko and Goodier: Theory of Elasticity Timoshenko and Woinowsky-Krieger: Theory of Plates and Shells Five national engineering societies, the American Society of Civil Engineers, the American Institute of Mining, Metallurgical, and Petroleum Engineers, the American Society of Mechanical Engineers, the American Institute of Electrical Engineers, and the American Institute of Chemical Engineers, have an arrangement with the McGraw-Hill Book Company, Inc., for the production of a series of selected books adjudged to possess usefulness for engineers and industry. The purposes of this arrangement are: to provide monographs of high technical quality within the field of engineering; to rescue from obscurity important technical manuscripts which might not be published commercially because of too limited sale without special introduction; to develop manuscripts to fill gaps in existing literature; to collect into one volume scattered information of especial timeliness on a given subject. The societies assume no responsibility for any statements made in these books. Each book before publication has, however, been examined by one or more representatives of the societies competent to express an opinion on the merits of the manuscript.

THEORY

.-

1 ,

OF

ELASTICITY By S. TIMOSHENKO And J. N. GOODIER Professors of Engineering M echanics Stanford University

Ralph H. Phelps, CHAIRMAN Engineering Societies Library New York

BIBLIOTECA CENTRALÃ ONIVERSITATEA "POLITEHNICA•

ENGINEERING SOCIETIES MONOGRAPHS COMMITTEE

llllÕillH

A. S. C. E.

00128156

Howard T. Critchlow H. Alden Foster

A. I. M. E. Nathaniel Arbiter John F. Elliott A.S. M.E. Calvin S. Cronan Raymond D. Mindlin

A. I. E. E. F. Malcolm Farmer Royal W. Sorensen

A. I. Ch. E. Joseph F. Skelly Charles E. Reed

NEW YORK

To RÕNTOL'n°N'

N

McGRAW-HILL BOOK COMPANY, lNc. 1951

PREFACE TO THE SECOND EDITION

THEORY OF ELASTICITY Copyright, 1934, by the United Engineering Trustees, Inc. Copyright, 1951, by the McGrawHill Book Company, Inc. Printed in the United States of America. All rights reserved. This book, or parts thereof, may not be reproduced in any form without permission of the publishers. XIII

64719

The many developments and clarifications in the theory of elasticity and its applications which have occurred since the first edition was written are reflected in numerous additions and emendations in the present edition. The arrangement of the book remains the sarne for the most part. The treatments of the photoelastic method, two-dimensional problems in curvilinear coordinates, and thermal stress have been rewritten and enlarged into separate new chapters which present many methods and solutions not given in the former edition. An appendix on the method of finite differences and its applications, including the relaxation method, has been added. N ew articles and paragraphs incorporated in the other chapters deal with the theory of the strain gauge rosette, gravity stresses, Saint-Venant's principle, the components of rotation, the reciproca! theorem, general solutions, the approximate character of the plane stress solutions, center of twist and center of shear, torsional stress concentration at fillets, the approximate treatment of slender (e.g., solid airfoil) sections in torsion and bending, and the circular cylinder with a band of pressure. Problems for the student have been added covering the text as far as the end of the chapter on torsion. It is a pleasure to make grateful acknowledment of the many helpful suggestions which have been contributed by readers of the book.

S. TIMOSHENKO J. N. GoonIER PALO ALTO, CALIF.

February, 1951

THE MAPLE PRFSS COMP AN·t, YORK, P A.

V

PREFACE TO THE FIRST EDITION During recent years the theory of elasticity has found considerable application in the solution of engineering problems. There are many cases in which the elementary methods of strength of materiais are inadequate to furnish satisfactory information regarding stress distribution in engineering structures, and recourse must be made to the more powerful methods of the theory of elasticity. The elementary theory is insufficient to give information regarding local stresses near the loads and near the supports of beams. It fails also in the cases when the stress distribution in bodies, all the dimensions of which are of the sarne order, has to be investigated. The stresses in rollers and in balls of bearings can be found only by using the methods of the theory of elasticity. The elementary theory gives no means of investigating stresses in regions of sharp variation in cross section of beams or shafts. It is known that at reentrant corners a high stress concentration occurs and as a result of this cracks are likely to start at such corners, especially if the structure is submitted to a reversai of stresses. The majority of fractures of machine parts in service can be attributed to such cracks. During recent years considerable progress has been made in solving such practically important problems. ln cases where a rigorous solution cannot be readily obtained, approximate methods have been developed. ln some cases solutions have been obtained by using experimental methods. AB an example of this the photoelastic method of solving two-dimensional problems of elasticity may be mentioned. The photoelastic equipment may be found now at universities and also in many industrial research laboratories. The results of photoelastic experiments have proved especially useful in studying various cases of stress concentration at points of sharp variation of cross-sectional dimensions and at sharp fillets of reentrant corners. Without any doubt these results have considerably influenced the modero design of machine parts and helped in many cases to improve the construction by eliminating weak spots from which cracks may start. Another example of the successful application of experiments in the solution of elasticity problems is the soap-film method for determining stresses in torsion and bending of prismatical bars. The vii

viii

PREF ACE TO THE FIRST EDITION

PREFACE '1'0 THE FIRST FDITION

difficult problems of the solution of partial differential equations with given boundary conditions are replaced in this case by measurements of slopes and deflections of a properly stretched and loaded soap film. The experiments show that in this way not o~y a vis~al picture_ of the stress distribution but also the necessary mformat10n regardmg magnitude of stresses can be obtained with an accuracy sufficient for practical application. . . . . . Again, the electrical analogy wh1ch glv~s a means of mvest1gatmg torsional stresses in shafts of variable dlameter at the fillets and grooves is interesting. The analogy between the p.r?blem of bending of plates and the two-dimensio?al pr?blem of elast1~ity ~as also been successfully applied in the solut10n of 1mportant engmeermg problem~. ln the preparation of this book the intention was to give to eng1neers, in a simple forro, the neces.sary fundame~tal knowledge of. the theory of elasticity. It was also mtended to ~nn~ together solut10ns of special problems which may be of pract1cal 1mportance ~nd to describe approximate and experimental methods of the solut10n of elasticity problems. . . Having in mind practical applications of the the?ry of elast1c1ty, matters of more theoretical interest and those wh1ch have not at present any direct applications in engineering have be~n omitted in favor of the discussion of specific cases. Only by studymg such cases with all the details and by comparing the results of exact investigations with the approximate solutions usually given in the elementary books on strength of materiais can a designer acquire a thorough understanding of stress distribution in engineering structures, and lear.n to use, to his advantage, the more rigorous methods of stress analys1s. ln the discussion of special problems in most cases the method of direct determination of stresses and the use of the compatibility equations in terms of stress components has been. applied .. This method is more familiar to engineers who are usually mterested m the magnitude of str((sses. By a suitable introduc~ion o~ stress fu~ctions this method is also often simpler than that m which equat10ns of equilibrium in terms of displacements are used. ln many cases the energy method of solution of elas~icity pro_ble~s has been used. ln this way the integration of differential equat10ns 1s replaced by the investigation of minimum conditi?n~ of certain int~­ grals. Using Ritz's method this problem of vanat10nal calculu~ 1s reduced to a simple problem of finding a minimum o~ a f~ct1on. . ln this manner useful approximate solutions can be obtamed m many practically important cases.

1X

To simplify the presentation, the book begins with the discussion of two-dimensional problems and only la ter, when the reader has familiarized himself with the various methods used in the solution of problems of the theory of elasticity, are three-dimensional problems discussed. The portions of the book that, although of practical importance, are such that they can be omitted during the first reading are put in small type. The reader may return to the study of such problems after finishing with the most essential portions of the book. The mathematical derivations are put in an elementary forro and usually do not require more mathematical knowledge than is given in engineering schools. ln the cases of more complicated problems all necessary explanations and intermediate calculations are given so that the reader can follow without difficulty through all the derivations. Only in a few cases are final results given without complete derivations. Then the necessary references to the papers in which the derivations can be found are always given. ln numerous footnotes references to papers and books on the theory of elasticity which may be of practical importance are given. These references may be of interest to engineers who wish to study some special problems in more detail. They give also a picture of the modem development of the theory of elasticity and may be of some use to graduate students who are planning to take their work in this field. ln the preparation of the book the contents of a previous book ("Theory of Elasticity," vol. I, St. Petersburg, Russia, 1914) on the sarne subject, which represented a course of lectures on the theory of elasticity given in several Russian engineering schools, were used to a large extent. The author was assisted in his work by Dr. L. H. Donnell and Dr. J. N. Goodier, who read over the complete manuscript and to whom he is indebted for many corrections and suggestions. The author takes this opportunity to thank also Prof. G. H. MacCullough, Dr. E. E. Weibel, Prof. M. Sadowsky, and Mr. D. H. Young, who assisted in the final preparation of the book by reading some portions of the manuscript. He is indebted also to Mr. L. S. Veenstra for the preparation of drawings and to Mrs. E. D. W ebster for the typing of the manuscript.

S. UNIVERSITY OF MICHIGAN

December, 1933

TIMOSHENKO

CONTENTS V

f>REFACE TO THE SECOND EDITION.

vii

PREFACE TO THE FrnsT EDITION. .

. xvii

NoTATION . . CHAPTER 1. 1. 2. 3. 4. 5. 6.

INTRODUCTION 1

Elasticity . . . . . . . Stress. . . . . . . . . Notation for Forces and Stresses . . Components of Stress. Components of Strain. Hooke's Law. Problems . . . . . .

CHAPTER 2.

2 3 4 5 6 10

PLANE STRESS AND PLANE STRAIN

7. Plane Stress . . . 8. 9. 10. 11.

12. 13. 14. 15. 16.

Plane Strain . . . Stress at a Point . Strain at a Point . Measurement of Surface Strains Construction of Mohr Strain Circle for Strain Rosette. Differential Equations of Equilibrium. Boundary Conditions . . . . Compatibility Equations . . . Stress Function. . Problems . . . . . . . . .

CHAPTER 3. TWO-DIMENSIONAL COORDINATES 17. 18. 19. 20. 21.

22. 23. 24.

PROBLEMS

IN

11 11 13 17 19

21 21 22

23 26 27

RECTANGULAR

Solution by Polynomials. . . Saint-Venant's Principle. . . Determination of Displacements . Bending of a Cantilever Loaded at the End Bending of a Beam by Uniform Load . . . Other Cases of Continuously Loaded Beams . Solution of the Two-dimensional Problem in the Form of a Fourier Series. . . . . . . . . . . . . . . . . . . . . . . Other Applications of Fourier Series. Gravity Loading . . Problems . . . . . . . . . . . . . . . . . . . xi

29 33 34 35 39

44 46 53 53

xii

CONTENTS

CHAPTER 4.

TWO-DIMENSION AL PROBLEMS IN POLAR COORDINATES

General Equations in Polar Coordinates. . . . Stress Distribution Symmetrical about an Axis. Pure Bending of Curved Bars . . . . . . . . Strain Components in Polar Coordinates. . . . Displacements for Symmetrical Stress Distributions. Rotating Disks. . . . . . . . . . • • . . . . . Bending of a Curved Bar by a Force at the End . . The Effect of Circular Holes on Stress Distributions in Plates Concentrated Force at a Point of a Straight Boundary Any Vertical Loading of a Straight Boundary Force Acting on the End of a W edge . . . Concentrated Force Acting on a Beam. . . Stresses in a Circular Disk. . . . . . . . Force ata Point of an Infinite Plate. . . . General Solution of the Two-dimensional Problem in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . 40. Applications of the General Solution in Polar Coordinates . . 41. A Wedge Loaded along the Faces. Problems . . . . . . . . . . . . . . . . . . . . . . .

25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39.

CHAPTER 5. THE PHOTOELASTIC METHOD 42. Photoelastic Stress Measurement . . . . . 43. Circular Polariscope. . . . . . . . . . . 44. Examples of Photoelastic Stress Determination 45. Determination of the Principal Stresses . 46. Three-dimensional Photoelasticity . . . CHAPTER 6. 47. 48. 49. 50. 51. 52. 53.

CONTENTS

55 58 61 65 66 69 73 78 85 91 96 99 107 112 116 121 123 125

131 135 138 142 143

STRAIN ENERGY METHODS

Strain Energy . . . . . . Principle of Virtual Work . Castigliano's Theorem. . . Principle of Least Work. . Applications of the Principle of Least Work-Rectangular Plates. Effective Width of Wide Beam Flanges . Shear Lag . . . Problems . . . . . . . . . . . . . .

146 151 162 166 167 171 177 177

CHAPTER 7. TWO-DIMENSIONAL PROBLEMS IN CURVILINEAR COORDINATES 54. Functions ~f a Complex Variable. . . . . 55. Analytic Functions and Laplace's Equation Problems . . . . . . . . . . . . . . . 56. Stress Functions in Terms of Harmonic and Complex Functions . 57. Displacement Corresponding to a Given Stress Function. 58. Stress and Displacement in Terms of Complex Potentials 59. Resultant of Stress on a Curve. Boundary Conditions . 60. Curvilinear Coordinates. . . . . . . . . . . . . . .

179 181 182 183 186 187 190 192

61. Stress Components in Curvilinear Coordinates Problems . . . . . . . . . . . . . . . . 62. Solutions in Elliptic Coordinates . . . . . . 63. Elliptic Hole in a Plate under Simple Tension 64. Hyperbolic Boundaries. Notches . . 65. Bipolar Coordinates . . . . . . . 66. Solutions in Bipolar Coordinates Other Curvilinear Coordinates . .

CHAPTER 8. SIONS 67. 68. 69. 70. 71. 72. 73. 74. 75.

195 197 197

201 204 206 208 212

ANALYSIS OF STRESS AND STRAIN IN THREE DIMEN-

Specification of Stress at a Point . . . . . . Principal Stresses. . . . . . . . . . . . . Stress Ellipsoid and Stress-director Surface . . Determination of the Principal Stresses . . . Determination of the Maximum Shearing Stress Homogeneous Deformation . . . Strain at a Point . . . . Principal Axes of Strain. Rotation. Problem . . . . . . . .

CHAPTER 9.

xiii

213 214 215 217 218 219 221 224 225 227

GENERAL THEOREMS

Differential Equations of Equilibrium. Conditions of Compatibility . . . . . Determination of Displacements . . . Equations of Equilibrium in Terms of Displacements . . General Solution for the Displacements . The Principle of Superposition . 82. Uniqueness of Solution . . . . . . . . 83. The Reciproca! Theorem. . . . . . . . 84. Approximate Character of the Plane Stress Solutions . Problems . . . . . . . . . . . . . . .... 76. 77. 78. 79. 80. 81.

228 229 232 233 235 235 236 239 241 244

CHAPTER 10. ELEMENTARY PROBLEMS OF ELASTICITY IN THREE DIMENSIONS Uniform Stress . . . . . . . . . . . . . . . . . . Stretching of a Prismatical Bar by Its Own Weight . . Twist of Circular Shafts of Constant Cross Section . . Pure Bending of Prismatical Bars. Pure Bending of Plates . . . . . . . . . . .

245 246 249 250 255

ÜHAPTER 11. TORSION OF PRISMATICAL BARS 90. Torsion of Prismatical Bars . . . 91. Bars with Elliptical Cross Section. . 92. Other Elementary Solutions . . . . 93. Membrane Analogy. . . . . . . . . . 94. Torsion of a Bar of N arrow Rectangular Cross Section

258 263 265

85. 86. 87. 88. 89.

268 272

CONTENTS

CONTENTB

xiv 95. 96. 97. 98. 99. 100. 101. 102.

Torsion of Rectangular Bars. . . . . . . . . . . Additional Results . . . Solution of Torsional Probie~s·b~ En~r~· Meth~d ·. Torsion of Rolled Profile Sections The Use of Soap Filma in Solving·T~~i~n ·P~oble~~- ·. Hydrodynamical Analogies. . . . . . . . . . . . . . . Torsion of Hollow Shafts Torsion of Thin Tubes . . . . . . . . . . . . . . .

103. Torsion of a Bar in which ~e C~o~ Se~ti~n. R~~ains. Pia~e: 104. Torsion of Circular Shafts of Variable Diameter . Problems . . . . . . . . . . . . . . . . . . . . . . . CHAPTER 12. 105. 106. 107. 108. 109. 110. 111. 112. 113. 114. 115.

275 278 280 287 289 292 294 298 302 304 313

BENDING OF PRISMATICAL BARS

Bending of a Cantilever. Stress Function. . . . Circular Cross Section. . Elliptic Cross Section . . Rectangular Cross Section . . Additional Results . . . . . . . Nonsymmetrical Cross Sections. . Shear Center. . . . . . . . . . . . . The Solution of Bending Problems by the Displacements . . . . . . . . . . . . Further Investigations of Bending . . .

316 318 319

. . . . . . . . . Soap-film Method. . . . . . . . . . . . . . . . . . .

321 323 329 331 333 336 340 341

CHAPTER 13. AXIALLY SYMMETRICAL STRESS DISTRIBUTION IN A SOLID OF REVOLUTION 116. 117. 118. 119. 120. 121. 122. 123. 124. 125. 126. 127. 128. 129. 130. 131.

General Equations . . . . Solution by Polynomials. . Bending of a Circular Plate . . . . . • . . • • • The Rotating Disk as a Three-dimensional Problem. Force at a Point of an Indefinitely Extended Solid . . • . . . . Spherical Container under Internai or Externai Uniform Pressure. Local Stresses around a Spherical Cavity. • . . . . . . . . . . Force on Boundary of a Semi-infinite Body . . . . . . . . . . Load Distributed over a Part of the Boundary of a Semi-infinite Solid Pressure between Two Spherical Bodies in Contact . . . . . Pressure between Two Bodies in Contact. More General Case lmpact of Spheres . . . . . . . . . . . . . . . . Symmetrical Deformation of a Circular Cylinder . . . The Circular Cylinder with a Band of Pressure. Twist of a Circular Ring Sector. . . . . Pure Bending of a Circular Ring Sector

CHAPTER 14. THERMAL STRESS 132. The Simplest Cases of Thermal Stress Distribution . . . . . . 133. Some Problema of Plane Thermal Stress. . . . . . . . . . . 134. The Thin Circular Disk: Temperature Symmetrical about Center

343 347 349 352 354 356 359 362 366 372 377 383 384 388 391 395

898 404 406

135. 136. 137. 138. 139. 140.

The Long Circular Cylinder . The Sphere . . . . General Equations . . . . . Initial Stresses . . . . . . . Two-dimensional Problema with Steady Heat Flow . . Solutions of the General Equations . . . . . . . . . .

CHAPTER 15.

XV

408 416 421 425 427 433

THEPROPAGATION OFWAVESINELASTIC SOLIDMEDIA

141. . . . . . . . . . . . . . . . . . . 142. Longitudinal Waves in Prismatical Bars. 143. Longitudinal lmpact of Bars. . . . . . 144. Waves of Dilatation and Waves of Distortion in Isotropic Elastic Media. . . . . . . . . . . . . . . . . . . . 145. Plane Waves. . . . . . . . . . . . . . . . . . . . . . . 146. Propagation of Waves over the Surface of an Elastic Solid Body.

438 438 444 452 454 456

THE APPLICATION OF FINITE DIFFERENCE EQUATIONS IN ELASTICITY 461 1. Derivation of Finite Difference Equations 465 2. Methods of Successive Approximation. 468 3. Relaxation Method. . . . . . . . . 473 4. Triangular and Hexagonal N ets . . . . 477 5. Block and Group Relaxation. . . . . 479 6. Torsion of Bars with Multiply-connected Cross Sections. 480 7. Points Near the Boundary. . . . . . . . . . 483 8. Biharmonic Equation . . . . . . . . . . . . 490 9. Torsion of Circular Shafts of Variable Diameter 495 AuTHOR !NDEX. 499 SuBJECT lNDEX.

APPENDIX.

NOTATION x, y, z Rectangular coordinates. r, O Polar coordinates. ~' '1

Orthogonal curvilinear coordinates; sometimes rectangular coordinates. R, >f;, o Spherical coordinates. N Outward normal to the surface of a body. l,m,n Direction cosines of the outward normal. A Cross-sectional area. Moments of inertia of a cross section with respect to x- and y-axes. I v Polar moment of inertia of a cross section. g Gravitational acceleration. p Density. q Intensity of a continuously distributed load. p Pressure. X, Y,Z Components of a body force per unit volume. X, :Y,Z Components of a distributed surface force per unit area. M Bending moment. Mt Torque. . ln each case all the equations of elasticity are satisfied, but the solutions are exact only if the surface forces are distributed in the manner given. ln the case of pure bending, for instance (Fig. 22), the bending moment must be produced by tensions and compressions on the ends, these tensions and compressions being proportional to the distance from the neutral axis. The fastening of the end, if any, must be such as not to interfere with distortion of the plane of the end. If the above conditions are not fulfilled, i.e., the bending moment is applied in some different manner or the constraint is such that it imposes other forces on the end section, the solution given in Art. 17 is no longer an exact solution of the problem. The practical utility of the solution however is not limited to such a specialized case. It can be applied with sufficient accuracy to cases of bending in which the conditions at the ends are not rigorously satisfied. Such an extension in the application of the solution is usually based on the so-called principle of Saint-Venant. This principle states that if the forces acting on a small portion of the surface of an elastic body are replaced by another statically equivalent system of forces acting on the sarne portion of the surface, this redistribution of loading produces substantial changes in the stresses locally but has a negligible effect on the stresses at distances which are large in comparison with the linear dimensions of the surface on which the forces are changed. For instance, in the case of pure bending of a rectangular strip (Fig. 22) the cross-sectional dimensions of which are small in comparison with its length, the manner of application of the externa} bending moment affects the stress distribution only in the vicinity of the ends and is of no consequence for distant cross sections, at which the stress distribution will be practically as given by the solution to which Fig. 22 refers. The sarne is true in the case of axial tension. Only near the loaded end does the stress distribution depend on the manner of applying the tensile force, and in cross sections at a distance from the end the stresses are practically uniformly distributed. Some examples illustrating this statement and showing how rapidly the stress distribution becomés practically uniform will be discussed later (see page 52). 1

This principie was stated in the famous memoir on torsion in Mém. savants étrangers, vol. 14, 1855. Its relation to the principie of conservation of energy is di:icussed later (see p. 150).

34

THEORY OF ELASTICITY

TWO-DIMENSIONAL PROBLEMS

19. Determination of Displacements. When the components of stress are found from the previous equations, the components of strain can be obtained by using Hooke's law, Eqs. (3) and (6). Then the displacements u and v can be obtained from the equations

20. Bending of a Cantilever Loaded at the End. Consider a cantilever having a narrow rectangular cross section of unit width bent by a force P applied at the end (Fig. 26). The upper and lower edges are free from load, and shearing forces, having a resultant P, are distributed along the end x = O. These conditions can be satisfied by a proper combination of pure shear, with the stresses (e) of Art. 17 reprel sented in Fig. 24. Superposing the ;, pure shear r zv = - b2 on the stresses X (e), we find p

ôu ÔX

av

=E,,,

fJy

ÔU

=

oy

Ey,

ÔV

+ ÔX

(a)

= 'YX11

The integration of these equations in each particular case does not present any difficulty, and we shall have severa! examples of their application. It may be seen at once that the strain components (a) remain unchanged if we add to u and v the linear functions U1

=a+ by,

v1

= e - bx

1

= E (u,, - vuy),

Ey

=

E1 (uy

Ex

= E [u,,

Ey

=

E [o-11

'YX11 =

GTX'll

1

- v(uy

+ u,)]

= E [(1 - v )u,, - v(l

v(u.,

+ u.)]

=

1

-

1

b2 - 2d4

E [(1 -

2

v2 )u11

-

v(l

;e

~~

y FIG. 26.

To have the longitudinal sides y (rX'll)ll=±c

=

±e free from forces we must have

= -b2 -

from which

4

d c2

2

=o

To satisfy the condition on the loaded end the sum of the shearing forces distributed over this end must be equal to P. Hencei

f from which

e Tzv ·

-e

dy =

f

e

-e

(b2 - ~e y

2)

dy = P

~ !:.

4 e

Substituting these values of d4 and b2 in Eqs. (a) we find

+ v)o+

:e

(a)

y2

b2 =

and in the case of plane strain the strain components are:

1

=O

u11

- vu,,),

: 1 1

= d4xy,

TzY = -

(b)

in which a, b, and e are constants. This means that the displacements are not entirely determined by the stresses and strains. On the displacements due to the interna! strains a displacement like that of a rigid body can be superposed. The constants a and e in Eqs. (b) represent a translatory motion of the body and the constant b is a small angle of rotation of the rigid body about the z-axis. It has been shown (see page 25) that in the case of constant body forces the stress distribution is the sarne for plane stress distribution or plane strain. The displacements however are different for these two problems, since in the case of plane stress distribution the components of strain, entering into Eqs. (a), are given by equations E,,

O"z

i

35

3P u,,=-2caxy, 11]

Txy

=

3P (1 -

-

4c

v)u,,]

1

It is easily verified that these equations can be obtained from the preceding set for plane stress by replacing E in the latter by E/(1 - 11 2), and 11 by 11/(l - 11). These substitutions leaveG, which is E/2(1 + 11), unchanged. The integration of Eqs. (a) will be shown later in discussing particular problems.

O"y=Ü

y2) c 2

N oting that ic 3 is the moment of inertia I of the cross section of the cantilever, we have Pxy -

-,

1

_ !:_ ! (c2 _ 12

1

O"y

=o

y2)

(b)

The minus sign before the integral follows from the rule for the sign of shearing stresses. Stress Tz~ on the end x = O is positive if it is upward (see p. 3).

THEORY OF ELASTICITY

36

TW" f

X

r-... ,Neutralaxis(I=l44J)

4

.

assumed that longitudinal fibers of the bent bar are in simple tension or compression. From the first of Eqs. (48) it can be shown that the stress ur is always positive for the direction of bending shown in Fig. 42. The same can be concluded at once from the direction of stresses ue acting on the elements n - n in Fig. 42. The corresponding tangential forces give resultants in the radial direction tending to separate longitudinal fibers and producing tensile stress in the radial direction. This stress increases toward the neutral surface and becomes a maximum near this surface. This maximum is always much smaller than (ue)max.· For instance, for b/a = 1.3, (ur)max. = 0.060(ue)ma:.:.; for b/a = 2, (ur)max. = 0.138(ue)max.; for b/a = 3, (ur)max. = 0.193(ue)max.. ln Fig. 43 the distribution of ue andurfor b/a = 2 is given. From this figure we see that the point of maximum stress ur is somewhat displaced from the neutral axis in the direction of the center of curvature. 28. Strain Components in Polar Coordinates. ln considering the displacement in polar coordinates let us denote by u and v the compo-

..............

~1~1.2

65

Er

(49)

= ar

As for the strain in the tangential direction it shpuld be ebserved that it .depends not only on the displacement v b~t ~l~o- on the ~adÍ~l cfi;_ placement u. ·A.ssumjpg, ior instaf!ce, th!!t .:the pointsa ·and doUIJ,e element abcd. (Fig. 44) have only the radiál displacement u, the new le'ngth bf the a.rc adis (r + u) dO and the tangential strain is therefore (r

+ u) d O r dO

r dO

u

= -

r

The difference in the tangential displacement of the sides ab a:nd cd of

66

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

THEORY OF ELASTICITY

the element abcd is (av/ao) do, and the tangential strain dueto the displacement V is accordingly av/r ao. The total tangential strain is thus 1

u

EB

=

av

au

av ar

1

= E (u, -

Eg

=

'Yr9 =

1

E ( can be combined, and ~e then obtain the stress function for the two equal and opposite forces applied at O ànd Oi, in tµe forro -cf>(x,y +a)

4>2 = 4>1 - (c/>1

1 a) + âc/> ây

=

2Ma

-1rT

l'ubstituting (a) in ~q. (b), and noting (see page 57) thaij

ª"' . ,;. ª"' am..

· itfJ "" fJr

8

+ ao oosr 8·

(69)

ây

cos 3 8

(70)

X

fa)

X

Fm. 56.

Having the distribution of stresses, the corresponding displacements can be obtained in the usual way by applying Eqs. (49) to (51). For a force normal to the straight boundary (Fig. 52) we have .

au

Er

'YrO

=

2P cos 8

=ar = - ?rE_r_ r

When a il!I very i;mall, this app:roaches the value ()y

+ ain 8 coa 8)

a3 4>1

-

Eo = '.!±

(b)

(8

If the directions of the couples are changed it is only necessary to change the sign of the function (70). A series of stress functions obtained by successive differentiation .has been employed to :solve the problem of stress concentration due to a semicircular notch in a semi-infinite plate in tension parallel to the edge. 1 The maximum tensile stress is slightly greater than three times the undisturbed tensile stress away from the notch. The strip with a semicircular notch in each edge has also been investigated. 2 The stress-concentration factor (ratio of maximum to mean stress at mini~um section) falls below three, approachmg unity as the notches are made larger.

+ cf>(x,y)

c/>i =,-a i>cf>

= -M11'.

in which M is the moment of the applied couple. . Reasoning in .the sarne manner, we find that by differentiation of 't'h .i. we obtain the stress f unct10n c/>2 for the case when two equal and opposite couples M are acting at two points O and 0 1 a very small distance apart (Fig. 56b). We thus P . find that

= -

ur = ...,...

8 coa 8)

1("

This resultant balances the externai force P, and, as the stress components Tro and u 9 at the straight edge are zero, solution (66') satisfies the boundary conditions. Having the solutions for vertical 11nd horizontal concentrated forces, solutions for inclined forces are obtained by superposition. Resolving the inclined force P into two components, P cos a .;ertically and P sin

$9

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

+

àu r ii{J

àv =

r ao

àv

+ àr -

V

2P cos 8

?rE

r

(e)

rv = 0

t F. G. Maunsell, Phil. Mag., vol 21 p 765 1936 2 C B · · . ' . ' . H p ·. · Ling, J. Ap?lied Mechanics (Trans. A.S.M.E.), vol. 14, p. A-275 1947; · oritsky, H. D. Smvely, and C. R. Wylie, ibid., vol. 6, p. A-63, 1939. '

straight boundary of the plate. The horizontal displacements are obtained by putting 8 = ±r/2 in the first of Eqs. (g). We find

Integrating the first of these equations, we find .. 2P u = - rE cos fJ log -r

where f(O) is a function of O only. (e) and integrating it, we obtain

v=

= - (l ~;)P 8 sin

8

+A

(d)

(u)

Substituting in the second of Eqs.

in which F(r) is a function of r only. third of Eqs. (e), we conclude that j(O)

+ f(fJ)

J

~: sin fJ + =~ log r sin fJ -

sin 8

f(fJ) d8

+ F(r)

(e)

Substituting (d) and (e) in the

+B

cos O,

F(r) = Cr

(f)

where A, B, and C are constants of integration which are to be determined from the conditions of constraint. The expressions for the displacements, from Eqs. (d) and (e), are u = - 2p cos

v=

rE 2vP - sin 8 rE

olog r

-

(l -Ev)P fJ sin 8 r

2P

.

+ r-E log r sm 8 + (l

+A

(1 - v)P r

~Ev)P sin

8

E

+A

sin 8

+B

cos 8

O cos 8

= -

2P rE log r

(g)

cos O - B sin 8 + Cr

+B

(h)

To find the constant B let us assume that a point of the x-axis ata distance d from the origin does not move vertically. Then from Eq. (h) we find

B

,.. = _ (1 - v)P, 2E

2P

= rE log d

Having the values of all the constants of integration, the displacements of any point of the semi-infinite plate can be calculated from Eqs. (g). Let us consider, for instance, the displacements of points on the

(u)

9 =2

,, = _ (1 - v)P 2E

9 = -2

(7 l)

The straight boundary on each side of the origin thus has a constant displacement (71), at all points, directed toward the origin. We may regard such a displacement as a physical possibility, if we rémember that around the point of application of the load P we removed the portion of material bounded by a cylindrical surface of a small radius (Fig. 52b) within which portion the equations of elasticity do not hold. Actually of course this material is plastically deformed and permits di8placement (71) along the straight boundary. The vertical displacements on the straight boundary are obtained from the second of Eqs. (g). Remembering that vis positive if the displacement is in the direction of increasing 8, and that the deformation is symmetrical with respect to the x-axis, we find for the vertical displacements in the downward direction ata distance r from the origin (v)

Assume that the constraint of the semi-infinite plate (Fig. 52) is such that the points on the x-axis have no lateral displacement. Then v = O, for 8 = O, and we find from the second of Eqs. (g) that A = O, C = O. With these values of the constants of integration the vertical displacements of points on the x-axis are (u)e=o

91

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

THEORY OF ELASTICITY

90

9=

,.. = - (v)

-2

.. = 2p log ~ - (l

9 =2

rE

r

+

v)P rE

~72)

"

At the origin this equation gives an infinitely large displacement. To remove this difficulty we must assume as before that a portion of material around the point of application of the load is cut out by a cylindrical surface of small radius. For other points of the boundary, Eq. (72) gives finite displacements. 34. Any Vertical Loading of a Straight Boundary. The curves for (j"' and -rZll of the preceding article (Fig. 53) can be used as influence lines. We assume that these curves represent the stresses for P equal to a unit force, say 1 lb. Then for any other value of the force P the stress (j"' at any point H of the plane mn is obtained by multiplying the ordinate HK by P. If several vertical forces P, P 1, P 2, • • • , act on the horizontal straight boundary AB of the semi-infinite plate, the stresses ori the horizontal plane mn are obtained by superposing the stresses produced by each of these forces. For each of them, the (j"' and -rZ!I curves are obtained by shifting the (]',, and -rZll curves, constructed for P, to the new origins 01, 02, . • . . From this it follows that the stress(]',, produced, for instance, by the force P 1 on the plane mn at the point D is obtained by multiplying the ordinate H 1K 1 by P 1. ln the sarne manner the (j"' stress at D produced by P 2 is H 2 K 2 • P 2, and so on. The total normal

TWO-DIMENSIONAL PROBLEMS IN POLAR COORDINATES

THEORY OF ELASTICITY

92

stress at Don the plane mn produced by P, P1, P2, ""' = DD1 ·

P

+ H1K1

· P1

+ H2K2

is

· P2

---

another manner by means of a stress function in the form (a)

+ ··

Hence the ... ~ 1/ ~

~

~

dy J cf> dx _ _!!___ X = ay2 . ds + ax ay. ds - iJy dy

(ª")

-

y

=

-

a'q, dx

a'q, dy ax 2 ds - axayds

=

dy ds

('") dx +~ ax ay ds

=

!!:_ ('") ds

By

F

B A

-

d ds

f

~

[ª"ay - i axA aq,]B ~ [ª" + ª"]º

+ iF, ~ F,,

(ª") [ª"]B ay ay (ª") ['•]B

d Ads B

y ::

- i

ÔX

i iJy

A

or, using Eq. (e) of Art. 58

The components of the resultant force on the are AB are therefore

f

J: - [X::+ J:

It will be evident from Eqs. (e) that if the curve AB represents an unloaded boundary, so that X and Y are zero, avn if rp and âvhere n is a positive or negative integer, F,,, Fy, and Af are zero, since thc functions 1

Equations (d) and (e) serve to establish an analogy betv•een plane stress and the slow motion of a viscous fluid in two düncnsious. See J. N. Gvodier, Phil. Mag., series 7, vol. 17, pp. 554 and 800, 1934. 1 These boundary eonditions lead to an analogy with the transverse deflections of elastic plates. An aecount oí this analogy, with references, is given by R. D. Mindlin, Quart. Applied Math., vol. 4, p. 279, 1946.

1ii.

192

PROBJ,f','JIIS l!{ CURVILINEAR C00RDI,VAT1~8

THEORY OF ELASTICITY

in brackcts return to their initial valucs \Vhcn thc circuit is completed. These functions by themsclves could not represent stress due to loads applicd at the origin. The function log z = log r ie does not return to it.1:1 initial value on completing a circuit round the origin, since 8 increases by 27r. Thus if 'if;(z) = (! log z, or x(z) = Dz log z, '\vhere C and D are (complcx} constants, Eq. (103) \Vill yield a non-zero value for Fr iF11 • Similarly x(z) = D log z "\vill yield a non-zero value of Jlf if D is imaginary, but a zero value if D is real. 60. Curvilinear Coordinates. l>uJar coordinates r, 8 (Fig. 124) may hc regarded ai'! spccifying thc position of a point as thc int.ersection of a circle (of radius r) and u radial line (at thc anglc Ofrom the initial line). A changc from Cartesian lo plar coordinates is effected by means of the equations

+

and it is usually most convenient to begin '"ith these. example, the two equations x = e cosh

\vhere e is a constant.

X

(a)

The first, V.'hen r is givcn various constant values, represents the family of circlc::i. The seC'ond, \vhen Ois givcn various constant values, reprcsenüi the fumily of radial lincs. Equati.ons (a) are a special case of equations of the forro

F i(x,y)

=

t,

F 2(x,y)

= 1J

(b)

Giving defini.te constant values to ~ and 1J, these eqiiations wHl represent t-..,vo {'urves \Vhich v.ill interscct, V.'hen F1 (:.r,y), F2 (x,y) are suitable functions. Different >'alues of t and 1J v.·ill yield different curves anda different point of intersection. Thus each point in the xy-plane v.•ill be charactcrized by definite values of t and ?J-the values '''hich make the t\VO curves given by Eqs. (b) pass through it-and t, 1J may be regardcd a::;" coordinates" of a point. Since given values of t, 1J define +.he point by n1eans of t""'º intersecting curves, they are called curvi~ linear coordinaU:s. 1 Polar coordinates, with the associated stress cornponents, proved vcry useful in Chap. 4 for problems of concentric circular boundaries. 'fhe stress and displacernent on such a bo11ndary become functions of O only, sincc ris constant. lf the boundaries consist of other curves, for instance ellipses, it is advantugeous to use curvilinear coordinates one of \vhich is constant on each boundary cur,·e. If Eqs. (b) are solvcd for x and y, \Ve shall have t\VO equations of the form (') 1 The general theory of carvílincar coordinates \Vas lnt on the ellipae. -

-----·-

l'iS 1HUTG'l POLITEH!lll'. 11tv\\)0/l.,,RA. \

ft\St \GTE.CA. CEKTitALA ~

194

THEORY OF ELASTICITY

tion of

PROBLEMS IN Cú'RfllLINEAR COORDlA'ATES

z, and thcrefore satisfy the Cauchy-Riemann equations (e) of

Art. 55, also therefore the Laplace equations (f) and (g) of Art. 55. The curvilinear coordinates to be used in this chapter will all be derived from equations of the form (g), and as a consequence will possess furthcr special propcrties. The point x, y having the curvilinear coordinates ~' 71, a neighboring point X dx, y dy will he.ve curvilinear coordinates ~ d~, 11 d11, and since there will be two equations of the type (e) V·te may \Vrite

+

dx

ax

= a~ d~

+

+

+

(h)

lf only ~is varied, the increments dx, dy correspond to an elcment of are ds; along a curve 1J = constant, and

dx

ox

(i)

= a~ d~,

Since

z

+

(dx)'

~

(dy)'

[

ox

GiY +(:~)'](d()'

dx = -d11

ª" '

Procceding as above we shall find

that

ox

Ô1/ =

and that ds,

where

Y

ª"

? increasf'ng

Jd11, and

=

dy/dx

= -

Fia. 127.

cota

Comparing this last result with Eq. (n), we see that the curves t = constant, 1J = constant, intersect at right angles, the direction 11-increasing making an angle (ir /2) a with the x-axis (Fig. 127). Consider for instance the elliptic coordinates defined by Eq. (f). Wc have

(j)

f'(.I) = e sinh .1 =e sinh ~ cos 1/

= f(t) we ha vc

âz

ôy =Jcosa

-J sin a,

+

'fhus

(a,,1· ~

7J = constant, in the direction ~-increasing, and the x-axis (Fig. 127). ln the sarne \Vay, if only 1/ is varied, the incrcments dx and dy of Eqs. (h) corrcspond to an element of are ds~ alonga curve~ = constant, and instead of Eqs. (i) v.·e havc

+

âx 011 d11,

ax

.ay

ar

d

,

ª' ~ ª' +' ª' ~ arfi'ATb'S

THEORY OF ELASTICITY

200

2~o

The greatest value, occurrin~ .at the ends of the major axis, \vhere 11 = Oand11",andcos211 = l,1s 28 sinh 2tc (o-.),,,,... = cosh 2tc - 1

When a

=

b, so that the cllipse becomes a circle, both (o- 0 )max. and

(o-~)mt~. reduce to 28, in agreemcnt '\\-ith the value for the circular hole

under uniform all-round tension found on page 81. The displacements can be evaluated from Eqs. (n) and (111) or (99). 1'hcy are of course continuous, bcing represented by single-valued continuous functions. The problem of un-iform pressure 8 \Vithin an elliptical holc, 1 and zero stress at infinity, is easily obtained by combining the above solut.ion "'ith the state of uniform stress (Te = (T" = -8, derivable from the complex potcntial Y.,(z) = -Sz/2. 63. Elliptic Role in a Plate under Simple Tension. As a second problem, consider t-he infinite platc in a state of simple tcnsile stress S in a direction at an angh: f3 below thc positive x-axis (Fig. 129), dist.urbed by an elliptical hole, '\\-·ith its major axis along the x-axis, as in X O a thc preceding problem. The ellipp b tical bole \Vith major axis perpenx' y' dicular or parallel to the tension 2 s y iR a 8pecial case. The more general problem is, however, no more 1'10. 12\l, difficult by the prcscnt method. From its solution we can find the cffect of the elliptical bole on any statc of uniform plane stress, spccified by principal stresses at infinity in any oricntation with rCBpect to the hole. Let- Ox', Oy' be Cartesian axes obtained by rotating Ox through the angle f3 soas to bring it parallel to the tension S. Then by F.qs. (105), (lOB)

It is easily sho,vn from Eqs. (e) that c2 = a2 - b2,

sinh 2to

2ab

,,

cosh 2tc

~-,

a2

+ b2

= --,,-

(T,,

and with these v.·c find that (o-.)mo a constant if the quantity in parentheses is made to vanish.

Thu::1 B

=

-A

co;; 2 'lo

(e)

'fo find thc rcsultant force t.ransmittcd 11·e may apply Eq. (103) of \rt. 59 to thn narro'v secLion EOB, Fig. 130, more prc('iscly to the Io,ver part of t.he limiting ellipse ~ = O bet\veen thc hyperbolas = and _ . . . _ 7J '17íl 7J - 11" - 1Jo· 0 n t 1us elhpse !; becomes i7J, 1 becomes -i7/, and \t·e have from Eq. (103), (e) and (d)

1

+ R) cot '17]~.:;;:-~·

F~ - i"Fu = i"[A11 - (A = i[A(:ir - 2'170

+ 2 cot 7Jo) + 2B cot 7Jo]

'Th·Is pro bl ()til (also the case of shear loadin") \Yll.S solvcd b~c A A G 'ffitb "·J·.1"11, TechRA · ept. rronau( Research Comm. (Great Britain) 1027-1928 vol II 668· andH:-;;bz "h , ,.,p., h · - eu er, . o.ngew. 1rial . Mech., vol. 13 p. 439 1933· or "Kcrba ,e re," p. ;)5, Berlin, IV38. ' , ' pannungs-

THEORY OF ELASTICITY

206

PROBLEMS IN CURVILINEAR COORDINATES

Since A and B were taken as real, F,. is zero and, using Eq. (e)

Fu

-A(11" - 211 0

=

+ sin 21'1o)

which determines A 'vhen the total tension Fv is assigned. The stress and displacement components are easily found from Eqs. (109), (110), ( 111). The first gives 4A cosh I; sin 11 u~+u~= cosh 21; - cos 217

- -e-

The value of 11 1 along the hyperbolic boundary is found by setting 11 = 110 in this expression. It has a maximum, -2A/c sin 1)0, at the waist ,vhcre I; = O. Ncuber 1 has exprcssed this as a function of the radius of curvature of thc hyperbola at the "''aist. He has solved, by anothcr mcthod, the problems of bcnding and shear of the plate as well as tension. 66. Bipolar Coordinates. Problems involving two nonconcentric circular boundaries, including thc spccial case of a circular hole in a semi-infinite plate, usually require the use of thc bipolar coordinates t, 11, defined by (a) z = ia coth -!r r = t i11

207

It may be seen from Fig. 131 that 61 - 62 is the angle between the two l~nes joinin~ the "poles" -ia, ia to the typical point z, when this point hes to the nght of the y-axis, and is m.inus this angle when the point lies to the left. It follo\\'S that a curve 1/ = constant is an are of a circle passing through the poles. Several such circles are drawn in Fig. 131. From Eqs. (e) it is clear that a curve ~ = constant will be a curve for ,~·hich r1/r2 = constant. Such a curve is also a circle. It surrounds the pole ia if r1/r2 exceeds unity, t.hat is, if tis positive. It surrounds the other pole -ia if ~is negative. Several such circles are drawn in

+

+

Replacing coth ir by (eit e-!t)/(etr - e-Ir) and solving the first equation for et, it is easily shown that this is equivalent to

r=

y

z +ia log--.

(b)

z - ia

+

The quantity z ia is represented by the line joining the point -ia to the point z in the xy-plane, in the sense that its projections on the axes give the real and imaginary parts. The sarne quantity may be represented by r 1e1º1 ,vhere r 1 is the length of the linc, and 01 the angle it makes ,vith the x-axis (Fig. 131). Similarly z - ia is the line joining the point ia to the point z, and may be represented by r28' 9• (Fig. 131). Then Eq. (b) becomes

~ + i 11

= log

(~ eie•e-'º')

= log

~ + i(01 -

r,

= log-,

"

Fig. 131. They form a family of coaxal circles with the two poles as limiting points. The coordinate '1 changes from 1r to -'Ir on crossing the segmcnt of the y-axis joining ~he poles, its range for the ''rhole plane being -'Ir to 'Ir. Stresses and d1splaccments -.,vill be continuous across this segment jf they are r~presented by periodic functions of 11 \vith period 2'1r. Separat1on of real and imaginary parts in Eq. (a) leads t 0 t X=

92)

a sin 11 ' cosh ~ cos 1)

y

=

a sinh ~ cosh ~ C08 '1

(d)

Differentiation of Eq. (a) yields

so that ~

Fio. 131.

1 . h 2_.\ 1 --iacosec 2 2

(,)

1 Loc. cit. For a cornparison of Neubcr's rcsults v;ith photoelastic and fatigue teats of notched platcs and grooved shafts sce R. E. Peterson and A. M. Wahl, J. Applied Mechanics, vol. 3, p. 15, 193fi, or S. Timoahenko, "Strength of MaterialB," 2d ed., vol. 2, p. 340. Sec also M. "'.\f. }'rocht, "Photoelasticity ," voL 2.

and 2 ia

e

_ dz/dt - dZ/df

=

-

'nh211 - .i cosech2 - .i 2 2

s1

'Soo the deriva.tion of Eq. (e) in Art. 54.

(:11' ~ ,.,.1>:11" 1 • • should be zero. The two states of strain t/ . . , "1>:11' • • • , and

238

THEORY OF ELASTICITY

E/ 1 .. , 'Y:rv" , and consequently the t\vO states oi stl'eSS u/ . . . , r""' . . • , and u:r" . . . , r:rv" • . . , are thercfore identical. 'Ihat is, the equations can yicld only one solution corresponding to given loads.1 The proof of uniqueness of solution was bascd on the assumption that the strain energy, and hence stresses, in a body disappear when it is freed of external forces. Howevcr there are cases \Vhen initial stresses may exist in a body \Vhile externai forces are absent. An example of this kind ,~·as encountered in studying the circular ring (see ltrt. 39). lf a portion of the ring between two adjacent cross sections

is cut out, and the ends of the ring are joincd again by welding or other mcans, a ring with initial stresses is obtained. 2 Severa! examples of this kind wcre discusscd in considering two-dimensional problems. We can also have initial stresses in a simply connectcd body dueto some nonelastic deformations during the process of forroing the body. \Ve may have, for instance, considerablc initial stresses in large forgings due to nonuniform cooling and also in rolled metallic bars due to the plastic flow produced by cold v:ork. For determining these initial stresses the equations of elasticity are not sufficient, and additional information regarding the process of forming thc body is necessary. It should be notcd that in all cases in which thc principlc of superposition can be used the deformations and stresses produced by externai forces are not affected by initial stresses and can be calculated in exactly the sarne manner as if there wcre no initial stresses. Then the total stresses are obtained by superposing the stresses produced by externa} forces on the initial stresses. ln cases when the principle of supcrposition is not applicable, the stresses produced by external loads cannot be detcrmined -..vithout knowing the initial stresses. We cannot, for instance, calculate bending stresses produccd by lateral loads in a thin bar, if the bar has an initial axial tension or compression, -..vithout kno"ving the magnitude of this initial stress. 1 This thcorem is dueto G. Kirchhoff. See his Vorle:rungen über Math. Phys., Mechanik. • The ring represents the simplest example of multiply-connected bodies. ln the case of such bodies general equations of clasticity, cxpressed in terms of stress components, are not sufficicnt for determining stresses, and to get a complete solution an additional invcstigation of displacementa :is necessary. The :6rst investigations of this kind were made by J. H. Michell, Proc. London Math. Soe., vol. 31, p. 103, 1899. See also L. N. G. Filon, Brit. Assoe. Advancement Sei. Rept., 1921, p. 305, and V. Volterra, Sur l'équilibre des corps élastiques multiplement connexés, Ann. école norm., Paris, series 3, vol. 24, pp. 401-517, 1007. Furtber references on initial stresses are given in the papcr by P. Neményi, Z. angfl'W. Mat/i. Mech., vol. 11, p. 59, 1931. ·

GENERAL THEOREMS

239

83. The Reciproca! Theorem. Limiting ourselves to the t\.vodimensional case let us consider the plate under two different loading conditions, and denote by X,, Y 1, X 1, and Y1 the components of the boundary and the volume forces in the first case' and by X 2, Y2, X 2, • and Y2 1n the second case. For the displacements, thc strain componcnts, and stress components in the two cases we use the notation ,,,,,,d,,,,,, Ui, Vi, Ez, E~, 'Yr'll, u,,, 1z must ali:!O be zero. So also must k in Eq. (a). Then (i) reduces to 4> = o - -1 -' - e,zt 2 l ~

(J)

+

However \b and 0 0 are related hy (e) in v.·hich we can now take k substituting (j) in (e) and using (d), vre have

=o

O.

(k)

=

o

(j)

ª'•) ay• ª' (60 + ''•) az•

ay• "•'4> + az•

(h)

=a+bx+cy

Using (a) and thc first of (b),

=

where Eq. (e) has been used in the last step. .-\lso, on aecount of (d), vre can replace /iY.c' in Ul by -a•e0 /ay•. Then Ul becomes

Q 20 0

a•e, ª' (e, + ª'•) (1 + v) ay• àz' ---

ª"

~,.1

..

+ 1 +.,0 0

-21 1 +' "0,z'

Also, since u, is zero, and us and ""are given by the first t11.'o of Eqs. (b), we can 'vrite V1'4> = 0, and thercfore, using (a)

(1

ª' (ª'• +1 +' .. 0 •) = 0

ax ay az•

Thesc, with (g), show that all three second derivatives with respect to :t and y of the function (of x, y, and z) in brackets vanish. Thus this function 1nust be linear in x and y, and V.'e can v.·rite

\b = (b)

)

This equation may be used in plaee of the :first of (130). and Jast can be rcplaccd by

where a, b, ande are arbitrary functions-of z. respeet to z, V.'e find

which are satisfied, as before, by = _,

'

(g)

àz'

+ -ay- o , ax

'1s

(ª'•

ª' az• +1+ .. 0 • =O ay•

(JT,y

-

243

GENERAL THEOREMS

-·

and therefore, from (d), (1)

The remaining equations of (130) are satisficd on account of Eq. (a) and the vanishin.g of "" Tn, Tv•· ~Ve can now obtain a stress distribution by choosing a functi.on \bo of z and y wh1ch satisfios Eq. (!), fiuding 0 0 from Eq. (k), and \b from Eq. (j). Thc strc.'IBes are then found bythe formulas (b). Each will consist of two parts, the first dcrived from \bo in Eq. (j), the seeond from the term -

~ 1 ~.,

e,z•.

ln view of Eq. (1),

the first part is exactly like the plane stress components detennined in Chaps. 3 to

1,I

244

'l'HEORY OF l!.'LASTICITY

7. The second part, being proportional to z•, may be made as email as we plea.ee compared with the first by restricting ourselves to plates ·which are sufficiently thin. Hence the conclusk'!l. that our solutions in Chaps. 3 to 7, which do not satisfy ali the compatibility conditioilll, are nevertheless good approximations for thin plates. The "exact" solutions, represented by stress functions oi the form (j), will requfre that the stresses at thc boundary, as cLscwhere, havc a parabolic variation ovcr the thickness. However any change from this distribution, so iong as it does not alter the intensity of force per unit length of boundary curve, will only alter the stress in the immediate neighborhood of tbe edge, by Saint-Venant's principie (page 33). The type of solution considcrcd abovc will alwaya represent the actual stress, and the components "'''Tu, T,, will in fact bc llero, cxccpt close to the edges. Problems 1. Show that •• = •• =

'Y··

k(y' =

+ z'),

l'••

=

'Y••

=

k'xyz

o

where k, k' are small constants, is nota possible state of strain. 2. A solid is heated nonuniformly to temperature T, a function of x, y, and z. If it is snpposed that each element has unrcstraincd thcrinal cxpansion, the strain components v;ill be

•·=•.=•.=aT,

1'zu = 1'•• = 1'•• =

O

where a is thc const:i.nt cocfficient of thcrmal cxpansion. Prove that this can only occur when Tis a linear function of x, y, and z. (The stress and consequent further strain arising 'vhen Tis not linear are discussed in Chap. 14.) 3. A disk or cylinder of the shape shown in Fig. 137a is compressed by forces P at C and D, along CD, causing extension of AB. lt is then compressed by forces P along AB (Fig. 137a) causing extension of CD. Show that these extensions are equal. 4. ln the general solution of Art. 80 v•hat choice of the functions o, 1, ,, , will give the general solution for plane ~train (w =O)?

CHAPTER 10 ELEMENTARY PROBLEMS OF ELASTICITY IN THREE DIMENSIONS 85. Uniform Stress. ln discussing the equations of equilibrium (127) and thc boundary conditions (128), it \Vas stated that the true solution of a problem must satisfy not only Eqs. (127) and (128) but also the compatibility conditions (sec Art. 77). These latter conditions contain, if no body forces are acting, or if the hody forces are constant, only second derivatives of the stress components. lf, therefore, Eqs. (127) and conditions (128) can be satisfied by taking the stress components cither as constants oras linear functions of the coordinates, Fru. 138. the compatibility conditions are sat-.. isfied identically and these stresses are the correct solution of the problem. AB a very simple example we may take tension of a prismatical bar in the axial direetion (Fig. 138). Body forces are neglected. The equations of equilibrium are satisfied by taking u,, = eonstant,

(a)

It is evident that boundary conditions (128) for the lateral surface of the bar, \vhich is free of externa} forces, are satisfied, bflcause all stress components, except cr"', are zero. The boundary conditions for the ends reduce to u,

=X

(b)

i.e., we have a uniform distribution of tensile stresses over cross sections of a prismatical bar if the tensile stresses are uniformly distributed over the ends. ln this case solution (a) satisfies Eqs. (127) and (128) and is the correct solution of the problem because the compatibility conditions (130) are identically satisfied. lf the tensile stresses are not uniformly distributed over the ends, solution (a) is no longer the eorrect solution beeause it does not satisfy the boundary conditions at the ends. Thc true solution becomes more complicated because the stresses on a cross section are no longer uni245

246

THEORY OF ELASTICITY

PROBLEMS OF ELASTICITY IN THREE DIMENSIONS

formly distributed. Examples of such nonuniform distribution occurred in the discussion of tv;o-dimensional problcms (see pages 51 and 167). As a second example consider the case of a uniform hydrostatic compression with no body forces. The equations of equilibrium (127) are satisfied by taking

The displacements u, v, w can now be found by integrating Eqs. (e), (d), and (e). Integration of Eq. (e) gives

fl~

=

Uu

= 'hich can be replaced with sufficient accuracy by an are of a circle of radius R/v, when the

THEORY OF ELASTICITY

PROBLEMS OF ELAST!CITY IN THREE DIME1\'SJO,\'S

deformation is small. ln considering the upper or lower sides of the bar it is evident that "\.Vhile the curvature of these sides after bending is convex down in thc lengthv;rise dircction, the curvature in the crossv,ise direction is convex up\vard. Contour lines for this anticlastic surlace will be as sho"\\'Il in Fig. 142a. By taking x and u constant in the first of Eqs. (h) we find that the equation for the contour lines is z2 - vy2 = constant

paths of the two rays is equal to an uneven number of half wave lengths of the light. The picture shov.·n in Fig. 142b, rcpresenting the hyperbolic contour lines, V.'as obtained hy this means. 89. Pure Bending of Plates. The result of the previous article can be applied in discussing the bending of plates of uniform thickness. If stresses 11% = Ez/R are distributed over thc edges of the plate parallel to the y-axis (Fig. 143), the surfacc of the plate v.·ill hecome 1 an A-,~:::=-==~==-=-=====-=-~«11-x _;,,.,.. 1 "''-;7 anticlastic surface, thc curvature of >vhich in planes parallel to the xz-plane is 1 / R and in the perpendicular direction is - v/ R. FIG. 143. If h denotes the thickness of thc platc, .L1f1 the bending moment per unit length on the edges parallel to the y-axis and

254

They are therefore hyperbolas with the asymptotes z2-vy2=0

FIG. 142a.

From this equation the angle a (Fig. 142a) is found from 1 2 tan a=-

'

This equation has been used for detcrmining Poisson's ratio 11. 1 If the upper surfacc of the beam is polished and a glass plate put over it, there will be, after bending, an air gap of variable thickness between the glass plate and the curved surface of the beam. This variable thickness can be measured optically. A beam of monochromatic

the moment of inertia per unit length, the relation between from the previous article, is 1

II

_\ _____ _ FIG. 142b.

light, say yello"-' sodium light, perpendicular to the glass plate, will be reflected partially by the plate and partially by the surface of the beam. The two refleeted rays of light interfere ,vith each other at points where the thickness of the air gap is such that the difference between the 1 A. Cornu, Compt. rend., vol. 69, p. 333, 1869. See also R. Straubel, Wied. Ann., vol. 68, p. 369, 1899.

Mi Elu

I2M1 =

Ehª

.Af 1

255

and R,

(a)

When V.'e have bending momcnts in t"-'O perpendicular directions (Fig. 144), the curvatures of the deftection surface may be obtained by superposition. Let 1/R 1 and X l/R2 be the curvatures of the deflection surface in planes parallel to the coordinate planes zx and zy, respectively; and lct frf 1 and M 2 be the bending z moments per unit length on the FIG. 144. edgcs parallel to the y- and xaxes, respcctivcly. Then, using Eq. (a) and applying the principie of superposition, we find 1 12 Ri= Ehs (M1 - vM2) 1

12

Ri= Ehª (.A.-12 - vM 1) 1

(b)

It is assumed that dellectioirn are small in comparison with the thickness of the plate.

257

THEORY OF ELASTICITY

PROBLEMS OF ELASTICITY IN THREE DIMEA'SfO,\'S

The moments are considered poRitive if they produce a deflection of the plate which is convex do\\'Il. Solving Eqs. (b) for M 1 and M 2, we find

The formulas (136) are used in the theory of plates when the bending moments are not uniform, and are accompanied by shear forces and surface pressures. For these circumstances they can be deduced from the general equations of Chap. 9 as approximations valid when the plate is thin. The elementary theory of bending of bars can bc related to the general equations in a similar manner. 1

256

M,

=

Eh' ( 1 12(1 - v 2) R1

Eh' M 2 = 12(1 -

For small deflections

\Ve

v 2)

1+ 1)

(

R2

R,

(o)

v R1

1

can use thc approximations

à'w - ay2

1

à'w

1

1) + v R2

- ax 2 '

R,

'l'hen, ""'·riting (135) v.·e find

=-D(::~+ v!~) M2 = -D(:~~+ v ::~) M1

(136)

The constant D is callcd the jlexural rigidity of a plate. ln the particular case when thc plate is bent to a cylindrical surface vrith generators parallel to the y-axis we have a2w/ay 2 =O, and, from Eqs. (136),

M, (137)

For the particular case in which M1 = M2 = M, we have 1

1

1

R,

R,

R

'l'he plate is bent to a sphcrical surface and the relation between the curvature and the bending moment is, from Eq. (e), Eh' M ~ 12(1 - v)

1

ºR

D(l + ,) =--li-

We shall have use for thcse rcsults later.

(138)

J. N. Goodicr, Tran8. Roy. Soe. Can., 3d ser., sec. III, vol. 32, p. 65, 1938.

259

TORSION

CHAPTER 11 TORSION 90. Torsion of Prismatical Bars. It has already been shown (Art. 87) that the exact solution of the torsional problem for a circular shaft is obtained if "'e assume that the cross sections of the bar remain plane and rotate 'vithout any distortion during t\'.'ist. This theory, developed by Coulomb,1 was applied latcr by Navier 2 to prismatical bars of noncircular cross scctions. Making the above assumption he arrived at the erroneous conclusions that, for a given torque, the angle of tvrist of bars is inversely proportional to the centroidal polar moment of inertia of the cross section, and that the maximum shearing stress occurs at the points most remate from the centroid of the cross section.ª It is easy to see that the above assumption is in contradiction with the boundary conditions. Take, for instance, a bar of rectangular cross section (Fig. 145). From Navicr's assumption it follows that at any point A on the boundary the shearing stress should act in the direction perpendicular to FIG. 145. the radius 0.4.. Resolving this stress into two components T~· and -r~,, it is evident that thcrc should be a complementary shearing stress, equal to T 11 ,, on thc clement of the lateral surface of the bar at the point A (see page 4), \Vhich is in contradiction \Vith the assumption that the lateral surface of the bar is free from external forccR, the twist being produccd by couples applied at the ends. A simple experiment >vith a rectangular bar, represented in Fig. 146, sho>\'S that the cross sections of the bar do not remain plane during torsion, and that the distortions of rectangular elements on the surface of the bar are greatest at the middles of the sides, i.e., at the points which are nearest to the axis of the bar.

The correct solution of the problem of torsion of prismatical bars by couples applied at thc ends was given by Saint-Venant. 1 He used the so-callcd semi-inverse method. That is, at the start he made ccrtain assumptions as to the deformation of the twisted bar and showed that \Vith these assumptions he could satisfy the equations of cquilibrium (127) and the boundary conditions (128). Then from the uniqueness of solutions of the elasticity equations (Art. 82) it follows that the assumptions made at thc start are correct and the solution obtained is thc exact solution of thc torsion problem. Consider a prismatical bar of any cross section t>visted by couples applied at the ends, Fig. 147. Guided by the solution for a circular shaft (page 249), Saint-Venant assumes that the deformation of the twisted shaft consists (a) of rotations of cross sections of the shaft as in thc case of a circular shaft and (b) of warping of the cross sections ,,,.hich F1G. 146. is the sarne for ali cross sections. Taking the origin of coordinatcs in an end cross section (Fig. 147) we find that the displacements corresponding to rotation of cross sections are

,.,

=

-8zy,

V

= 8z:x

(a)

where 8z is the angle of rotation of the cross section at a distance z from the origin. The warping of cross sections is defined by a function

w

z

~

e;;(x,y)

(b)

F1G. 147.

With the assumed displacements (a) and (b) we calculate the components of strain from Eqs. (2), which give E:

= Eu = E, = 'Yt11 =

i!w 7 •• -_ iJ:x

iJw 'Yu·=oy

1

"Histoire de l'académie," 1784, pp. 229-269, Paris, 1787. 1 Navier, "Résumé des leçons sur l'applieation de la mécanique,'' 3d ed., Paris, 1864, edited by Saint-Venant. • These conclusions are correct for a thin elastie layer, corresponding to a alice of the bar between two eross sections, attached to rigid pia.te~. See J, N, Qoodi.er, J, Appli~d Phyil., vol. 13, p. 167, 1942.

u

Y'

1

0

(º"' ) +az=IJ (º"' iJy+x) i!u + iJz

= 8 iJx - Y

(e)

i!c

M tm. savants étrangers, vol. 14, 1855. See also Saint-Venant's note to Navier's book, loc. cil., and 1. Todhunter and K. Pearson, "History of the Theory of Elaaticity," vol. 2.

The corresponding components of stress, from Eqs. (3) and (6), are do;

=

CTu

=

ífz

=

T"ll

=

and Eq. (e) bccomcs

)dy -(ª"'ay +x)dx ~O

( ay, ax y ds

O

=ao(:~ - y) Tuz = G8 (~: + x)

r::

(d)

It can be seen that with the assumptions (a) and (b) regarding the deformation, thcrc \Vill be no normal stresses acting bet\\·een the longitudinal fibcrs of the shaft or in the longitudinal direction of those fibers. There also ,,,jll be no distortion in the planes of cross sections, since E,,, Eu, 1'Xll vanish. V!i'c have at each point pure shear, defined by the components Txz anel Tuz· The function >Jt(x,y), defining warping of cross section, must now be determined in such a way that equations of equilibrium (127) will be O,__ _ _ __,__,x satisfied. Substituting expres~iions (d)

in these equations and

neglecting body forces we find that the function Y, must satisfy the equation

ª"' + ª"' ~ o ax2

iJy2

(139)

Consider nowthe boundary conditions (128). For the lateral surface of the bar, which is free from externai forces and has normais perpendicular to the z-axis, \Ve have X = Y = Z = Oand cos (N z) = n = O. The first two of Eqs. (128) are identically satisfied and the third gives (•)

which means that the resultant shearing stress at the boundary is directed along the tangent to the boundary, Fig. 148. It \Vas shown before (see page 258) that this condition must be satisficd if the lateral surfacc of the bar is free from externai forces. Considering an infinitesimal element abc at the boundary and assuming that s is increasing in the direction from e to a, we have

d'

m

=

dx

cos (Ny) = - ds

(140)

ds

1'hus each problcm of torsion is reduced to the problcm of finding a function Y, satisfying Eq. (139) and thc boundary condition (140). An alternativc procedure, which has the advantagc of leading to a simpl('r boundary condition, is as follows. In view of the vanishing of ff,,, ffy, ff,, Tri1 [Eqs. (d)], the equations of equilibrium (127) reduce to iJT:, = Ü

ª'

ªTy, = Ü

,

ª'

iJrn

ax

'

+ iJTy, ay

=O

The first two are alrcady satisfied sincc r'"" and r 11., as given by Eqs. (d), are independent of z. The third means that we can express r,.. and Ty, as T,..

ª"'ay

ª"'ax

=-1

(141)

v;•here 4> is a function of x and y, callcd the stressfunction. 1 From Eqs. (141) and (d) \Ve havc

ª'ay ~ ª' (ª"'ax - )·

-ªº~ao(ª"'+x) ax iJy

Y

(f)

Eliminating Y, by differentiating the first with respect to y, the second with respect to x, and subtracting from the first, we find that the stress function must satisfy the differential equation

Fio. 148.

l = cos (Nx) = dy,

261

TORSION

THEORY OF ELASTICITF

260

o2q, a2q, ax2+ay2=F

(142)

/ which satisfies Eq. (142) and is zero at the boundary. Several applications of this general theory to particular shapes of cross sections will be shown later. Let us consider now the conditions at the ends of the twisted bar. The normals to the end cross sections are parallel to the z-axis. Hence l = m = O, n = ± 1 and Eqs. (128) become

X=

(g)

±Txz,

in which the + sign should be taken for the end of the bar for which the external normal has the direction of the positive z-axis, as for the lower end of the bar in Fig. 147. We see that over the ends the shearing forces are distributed in the sarne manner as the shearing stresses over the cross sections of the bar. It is easy to prove that these forces give us a torque. Substituting in Eqs. (g) from (141) and observing that cf> at the boundary is zero, we find

263

forces, and sets up at the ends the torque given by Eq. (145). The compatibility conditions (130) need not be considered since the stress has been derived from the displacements (a) and (b). Thus all the equations of elasticity are satisfied and the solution obtained in this manner is the exact solution of the torsion problem. It was pointed out that the solution requires that the forces at the ends of the bar should be distributed in a definite manner. But the practical application of the solution is not limited to such cases. From Saint-Venant's principle it follows that in a long twisted bar, ata sufficient distance from the ends, the stresses depend only on the magnitude of the torque Mt and are practically independent of the manner in which the tractions are distributed over the ends. 91. Bars with Elliptical Cross Section. Let the boundary of the cross section (Fig. 149) be given by the equation x2 y2 a2 b2 - 1 = O (a)

+

FIG. 149. Then Eq. (142) and the boundary condition (144) are satisfied by taking the stress function in the form

Thus the resultant of the forces distributed over the ends of the bar is zero, and these forces represent a couple the magnitude of which is M1 =

ff

(Yx - Xy) dxdy = -

f f ~:xdxdy

-JJ~:

Integrating this by parts, and observing that cf> we find Mt = 2ff ct>dxdy

+ b2y2 -

x2 e/>= m ( a2 in which m is a constant.

)

1

(b)

Substituting (b) into Eq. (142), we find a2b2 m = 2(a2 + b2) F

Hence y dx dy

(h)

_

a

e/> - 2(a2

= O at the boundary, (145)

bF

+ b2)

(x + 2

a2

y2

)

(e)

b2 - 1

The magnitude of the constant F will now be determined from Eq. (145). Substituting in this equation from (e), we find

~t = a~ ! ~ 2 2 2

each of the integrals in the last member of Eqs. (h) contributing one half of this torque. Thus we find that half the torque is due to the stress component Txz and the other half to Tyz· We see that by assuming the displacements (a) and (b), and determining the stress components T.,., Tyz from Eqs. (141), (142), and (144), we obtain a stress distribution which satisfies the equations of equilibrium (127), leaves the lateral surface of the bar free from externa!

2 2

Smce

!! ~2d ~

X

( :2

ff

2

x dxdy

dy = I 'V= 7rbaª 4'

+:

2

ff

2

y dxdy -

ff

dxdy)

(d)

THEORY OF ELASTICITY

265

TORSION

in which

we find, from (d),

A = 7rab, from which (e)

F= Then, from (e),

(f)

Substituting in Eqs. (141), the stress components are 7

"''

2M1Y = - 7rab 3 '

(146)

The ratio of the stress components is proportional to the ratio y/x and hence is constant along any radius such as OA (Fig. 149). This means that the resultant shearing stress along any radius OA has a constant direction which evidently coincides with the direction of the tangent to the boundary at the point A. Along the vertical axis OB the stress component 'Tyz is zero, and the resultant stress is equal to T.,., Along the horizontal axis OD the resultant shearing stress is equal to Tyz· It is evident that the maximum stress is at the boundary, and it can easily be proved that this maximum occurs at the ends of the minor axis of the ellipse. Substituting y = b in the first of Eqs. (146), we find that the absolute value of this maximum is

are the area and centroidal moment of inertia of the cross section. Having the stress components (146) we can easily obtain the displacements. The components u and vare given by Eqs. (a) of Art. 90. The displacement w is found from Eqs. (d) and (b) of Art. 90. Substituting from Eqs. (146) and (148) and integrating, we find x

w

= M 1 (b2 - a2)xy (150) 7ra 3b3G

-L-

This shows that the contour lines lôrque Frn. 150. for the warped cross section are hyperbolas having the principal axes of the ellipse as asymptotes (Fig. 150). 92. Other Elementary Solutions. ln studying the torsional problem, SaintVenant discussed several solutions of Eq. (142) in the form of polynomials. To solve the problem let us represent the stress function in the form

+ F4 (x2 + y2)

q, = 1

(a)

Then, from Eq. (142),

a2 1 ax2

a21

+ a:J1

(b)

=O

and along the boundary, from Eq. (144), 'T max.

2Mi = 7rab2

(147) 1

For a = b this formula coincides with the well-known formula for a circular cross section. Substituting (e) in Eq. (143) we find the expression for the angle of twist a2 + b2 (148) 8 = Mt . 7ra 3b3G The factor by which we divide torque to obtain the twist per unit length is called the torsional rigidity. Denoting it by C, its value for the elliptic cross section, from (148), is (149)

;

~Ili,; 1

'

+ F4 (x + y 2

2

)

=

constant

(e)

Thus the torsional problem is reduced to obtaining solutions of Eq. (b) satisfying the boundary condition (e). To get solutions in the form of polynomials we take the function of the complex variable (x

+ iy)n

(d)

The real and the imaginary parts of this expression are each solutions of Eq. (b) (see page 182). Taking, for instance, n = 2 we obtain the solutions x2 - y2 and 2xy. With n = 3 we obtain solutions xª - 3xy2 and 3x2y - yª. With n = 4, we arrive at solutions in the form of homogeneous functions of the fourth degree, and so on. Combining such solutions we can obtain various solutions in the form of polynomials. Taking, for instance,

q,

= F 4- (x2

+ y2) + 1

= F -

2

[12 (x2 + y2) - 2a-1 -

(xª -

3xy2)

+ b]

(e)

TORSION

THEORY OF ELASTICITY

266

· l tºon of Eq· (142) in the form of a polynomial of the degree weobtamasou1 . · third · l · with constants a and b which will be adjusted later. Th1s p_o~ynomial is~ so _ut10n of the torsional problem if it satisfies the boundary cond1t1~n (144), i.e., if the boundary of the cross section of the bar is given by the equation

!

2

(x2

+

y2) - _!_ (x• - 3xy2)

2a

+b =

(f)

O

By changing the constant b in this equation, we obtain various shapes of the cross · l section. . Taking b = - /1 a• we arrive at the solution for the eqmlatera1tnang e. tion (f) in this case can be presented in the forro (x - y13 y - ja)(x

+ V3 y

E

qua-

267

By changing a, Saint-Venant obtained the family of cross sections shown in Fig. I52a. Combining solutions in the form of polynomials of the fourth and eighth degrees, Saint-Venant arrived at the cross section shown in Fig. 152b.

On the basis of his investigations, Saint-Venant drew certain general conclusions of practical interest. He showed that, in the case of singly connected boundaries and for a given cross-sectional area, the torsional rigidity increases, if the polar moment of inertia of the cross section decreases. Thus for a given amount of material the circular shaft gives the largest torsional rigidity. Similar conclusions can be drawn regarding the maximum shearing stress. For a given torque and cross-

- ja)(x +ia) =O

which is the product of the three equations of the sides of the triangle shown in Fig. 151. Observing that F = -2GIJ and substituting

q,

= -GIJ

U

(x 2

+ y 2)

-

ia (x 3

-

3xy 2 )

-

;

2 1a ]

(g)

into Eqs. (141), we obtain the stress components Txz and Tu•· Along the x-axis, Txz = O, from symmetry, and we find, from (g),

a 3

y

T

Frn. 151.

(b)

= 3GIJ (2ax _ x 2)

2a

yz

(h)

3

GIJa

(k)

=2

At the corners of the triangle the shearing stress is zero (see Fig. 151). Substituting (g) into Eq. (145), we find G1Ja 4

Mt = · - - 15 v3

3 IJGlp 5

(l)

= -

Taking a solution of Eq. (142) in the form of a polynomial of th~ fourth degree containing only even powers of x and y, we obtain the stress funct10n

= -GIJ

[~ (x• + y•)

_

~ (x4

_ 6z•y•

+ y4) + ~ (a

- 1)

J

The boundary condition (144) is satisfied if the boundary of the cross section is given by the equation

x'

;l

!.li. IH.

+ y'

- a(x' - 6x y

2 2

+y +a 4

)

1 = O

y

Frn. 152.

The largest stress is found at the middle of the sides of the triangle, where, from (h), T'max.

/J \ '

sectional area the maximum stress is the smallest for the cross section with the smallest polar moment of inertia. Comparing various cross sections with singly connected boundaries, Saint-Venant found that the torsional rigidity can be calculated approximately by using Eq. (149), i.e., by replacing the given shaft by the shaft of an elliptic cross section having the sarne cross-sectional area and the sarne polar moment of inertia as the given shaft has. The maximum stress in all cases discussed by Saint-Venant was obtained at the boundary at the points which are the nearest to the centroid of the cross section. A more detailed investigation of this question by Filon 1 showed that there are cases where the points of maximum stress, although always at the boundary, are not the nearest points to the centroid of the cross section. 1

L. N. G. Filon, Trans. Roy. Soe. (London), series A, vol. 193, 1900. Paper by G. Polya, Z. angew. Math. Mech., vol. 10, p. 353, 1930.

See also the

268

THEORY OF ELASTICITY

TORSION

Taking n = 1 and n = -1 in expression (d), and using polar coordinates r and y,, we obtain the following solutions of Eq. (b):

bar, subjected to a uniform tension at the edges anda uniform lateral pressure. li q is the pressure per unit area of the membrane and S is the uniform tension per unit length of its boundary, the tensile forces acting on the sides ad and bc of an infinitesimal element abcd (Fig. 154) give, in the case of small deflections of the membrane, a resultant in the upward direction -S(a 2z/ax 2 ) dx dy. ln the sarne manner the tensile

1

q, 1 = r cos Y,,

"'' =-;;. cos"'

Then the stress function (a) can be taken in the form

F 4

q, = - (x 2

+

y 2)

-

in which a and b are constants.

Fa r cos if; 2

-

+ Fb• - 2 a-r cos if;

(r 2

FIG. 153.

(m)

- - b2

It will satisfy the boundary condition (144) if at the boundary of the cross section we have a, converges very rapidly and there is no difficulty in calculating Tmax. with sufficient accuracy for any particular value of the ratio b/a. For instance, in the case of a

~

l _ cosh (n7ry/2a)] c s n7rx} d d _ 32G8(2a)3(2b) 0 cosh (n7rb/2a) 2a x Y 7r4

.,

16G8a = 2G8a - - -2 -

0.267 0.282 0.291 0.312 0.333

Let us calculate now the torque M 1 as a function of the twist o. Using Eq. (145) for this purpose, we find

[

we have

k2

0.263 . 0.281 0.291 0.312 0.333

or, observing that

Tmax.

Severa!

TABLE OF CONSTANTS FOR TORSION OF A RECTANGULAR BAR

The stress components are now obtained from Eqs. (141) by differentiation. For instance,

T

(159)

= k2G8a

_!_5 tanh n7rb n

2a

we have

The ~unction.

= ao(x 2 - a 2) (y2 - a2)

(d)

Substituting this in (165) we find from the minimum condition that a minimum. If we substitute in this integral 2GO for q/S, we arrive at the integral (165) above.

ªº

M1 = 2Jf e/> dx dy = ~-jGOa 4 = 0.1388(2a)4GO Comparing this with the correct solution (162) we see that the error in the torque is about li per cent. To get a closer approximation we take the three first terms in the series (e). Then, by using the condition of symmetry, we obtain

(a)

au

ªªº

=0

'

au âa2

e/>

(b)

Thus we obtain a system of linear equations from which the coefficients a 0, a 1, a 2, . . . can be determined. By increasing the number of terms in the series (a) we increase the accuracy of our approximate solution, and by using infinite series we may arrive at an exact solution of the torsional problem. 1 1 The condition of convergency of this method of solution was investigated by Ritz, loc. cit. See also E. Trefftz, "Handbuch der Physik," vol. 6, p. 130, 1928.

= (x2 - a2)(y2 - a2)[ao

+ a1(x2 + y2)]

(e)

Substituting this in (165) and using Eqs. (b), we find

ªº

5 259 GO

=

8 · 277 ·

a2'

ª

5 3

1

=

35

GO

8 · 2 · 277 · a4

Substituting in expression (145) for the torque, we obtain Mi

=0 . . . '

8 a2

The magnitude of the torque, from Eq. (145), is then

ln the approximate solution of torsional problems we replace the above problem of variational calculus by a simple problem of finding a minimum of a function. W e take the stress function in the form of a series

in which cf>o, cp 1, cp 2, . . . are functions satisfying the boundary condition, i.e., vanishing at the boundary. ln choosing these functions we should be guided by the membrane analogy and take them in a form suitable for representing the function cp. The quantities ao, a 1, a 2, . . . are numerical factors to be determined from the minimum condition of the integral (165). Substituting the series (a) in this integral we obtain, after integration, a function of the second degree in a 0, a 1, a 2, . . . , and the minimum condition of this function is

5GO

=

= VCi~~

+i

· ! · fü)GOa 4 = 0.1404G0(2a)4

This value is only 0.15 per cent less than the correct value. A much larger error is found in the magnitude of the maximum stress Substituting (e) into expressions (141) for the stress components w~ find that the error in the maximum stress is about 4 per cent, and to get ª better accuracy more terms of the series (e) must be taken. e It ~an be seen from the membrane analogy that in proceeding as xplamed above we generally get smaller values for the torque than the correct value. A perfectly fiexible membrane, uniformly stretched at a

1

See S. Timoshenko, Bull. lnst. Ways of Gommunication, St. Petersburg, 1913 nd Proc. London Math. Soe., series 2, vol. 20, p. 389, 1921. '

284

THEORY OF ELASTICITY

cf>

=

00

00

2:

2:

m11"X

n7r'Y

cos 2a cos '2b

amn

285

TORSION

tho boundary and uniformly loaded, is a system with an infinite number of degrees of freedom. Limiting ourselves to a few terms of the series (e) is equivalent to introducing into the system certain constraints, which reduce it to a system with a few degrees of freedom only. Such constraints can only reduce the flexibility of the system and diminish the volume bounded by the deflected membrane. Hence the torque, obtained from this volume, will generally be smaller than its true value. E. Trefftz suggested 1 another method of approximate determination of the stress function cf>. With this method the approximate magnitude of the torque is larger than its true value. Hence by using the Ritz and the Trefftz methods together the limits of error of the approximate solution can be established. ln using Ritz's method we are not limited to polynomials (e). We can take the functions cf>o, cf> 1, c/>2, ••• of the series (a) in other forms suitable for the representation of the stress function cf>. Taking, for instance, trigonometric functions, and observing the conditions of symmetry (Fig. 159), we obtain (f)

where a = b/a. Substituting in (f) we obtain the exact solution of the problem in the forro of an infinite trigonometric series. The torque will then be

00

00

~

~

128G8b 2

32ab

Lt Lt 11'4mn(m2a2 + n2) . mn11"2 m=l,3, ... n=l,3, ...

(g)

This expression is brought into coincidence with expression (160) given before if we observe that oo

1 m2

~

Lt

n=l,3,5, ..•

n 2 (m 2a 2

tanh ma11" - ma11" 2 2 = 96m 2 -i(ma1r/2) 3 11' 4

l

+n

2 )

As another example, in the case of a narrow rectangle, when b is very large in comparison with a (Fig. 159), we may take, as a first approximation, cf> = GO(a 2 - x 2) (h)

n=l,3,5, ... m=l,3,5, •..

Substituting in (165) and performing the integration, we find that

2: 00

u = 1r~b

~ Lt

a2mn

(m2 n2) (i2 + b2

cf> = Ge(a 2

m=l,3,5, ... n=l,3,5, .•.

2: m=l,3,5, ...

2:

n = 1,3,5, ...

amn

x 2)[1

fj=!

a

Equations (b) become

?r2ab 4

-

(m2a2 + n2) - 2G8. mn11'2 I6ab ( -1) mtn-1 = O b2

and we find 128G8b 2 ( -1) 11" 4mn(m 2a 2

m+n_ 1 2

+n

-

e-ll(b-yl]

(k)

and choose the quantity /j in such a manner as to make the integral (165) a minimum. ln this way we find

00

-2G8

which coincides with the solution discussed before (Art. 94). To get a better approximation satisfying the boundary condition at the short sides of the rectangle, we may take

2

)

1 E. Trefftz, Proc. Second Intern. Congr. Applied Mech., Zürich, 1926, p. 131. See also N. M. Basu, Phil. Mag., vol. 10, p. 886, 1930.

@_

'\}2

Dueto the exponential term in the brackets of expression (k) we obtain a stress distribution which practically coincides with that of the solution (h) at all points a considerable distance from the short sides of the rectangle. Near these sides the stress function (k) satisfies the boundary condition (144). Substituting (k) into equation (145) for the torque, we find

287

THEORY OF ELASTICITY

TORSION

which is in very good agreement with Eq. (161) obtained before by using infinite series. A polynomial expression for the stress function, analogous to expression (e) taken above for a rectangle, can be used successfully in all cases of cross sections bounded by a convex polygon. If

ln the particular case when m = !, p = q = 1, a = a 1, we have y = ±aif;(x/b) = ± v'x/b[l - (x/b)J, and we obtain

286

are the equations of the sides of the polygon, the stress function can be taken in the forro

cp

= ( a1X

+ biy + C1) (a2X + b2y + C2)

· · · (anX

+ b,.y + Cn) "1; "1;amnXnym

and the first few terms of the series are usually sufficient to get a satisfactory accuracy. The energy method is also useful when the boundary of the cross section (Fig. 161) is given by two curves 1 Y

= aif; (~)

A

= -

Go 11 a 2' 1 + 13 b2

f

c

a

e

where

a~

Frn. 161.

+ a1if;)

Substituting into the integral (165) we find, from the equation dl/dA =O, GO A where a=

Io1 i/;

3

fo

Mi = -A b(a

8= 1

+ a1)ª (1 if;ª dt 3

}o

i Such problema were discussed by L. S. Leibenson. See his book "Variational Methods for Solving Problema of the Theory of Elasticity," Moscow, 1943. See also W. J. Duncan, Phil. Mag., series 7, vol. 25, p. 634, 1938.

(cJ

3Mt (b1c1 3 + 2b2c2 3)G

(a)

LuftJahrt-forsch., vol. 20, 1944, tranlated as N.A.C.A. Tech. Mem. 1182, 1948. A more elaborate formula, taking account of the increased stiffness resulting fron;- the junctions of the rectangles, was developed on the basis of soap film and tors10n tests by G. W. Trayer and H. W. March, Natl. Advisory Comm. Aeronaut., Rept. 334, 1930. 3 Comparison of torsional rigidities obtained in this manner with those obtained by · t s is · g1ven · . ex?enmen for severa! types of rolled sections and for various dimensions m the paper by A. Fõppl, Sitzber. Bayer. Akad. Wiss., p. 295, München, 1921. See also Bauingenieur, series 5, vol. 3, p. 42, 1922. 2

From Eq. (145) we find the torque

bL

is obtained with sufficient accuracy from Eq. (155) by putting, instead of b, in this equation the developed length of the center line, 2 namely, b = 2a - e. ln the case of a channel section (Fig. 162b) a rough approximation for the angle of twist is obtained by taking for the fianges an average thickness c2, subdividing the cross section into the three rectangles, and substituting in Eq. (155), b1c1 3 + 2b 2c23 instead of bc 3, i.e., assuming that the torsional rigidity of the channel is equal to the sum of the torsional rigidities of the three rectangles. a Then

3

i/; dt

G8ba 3 11 ª 2 1 + 13 b2

Frn. 162.

(#/dt) 2 dt 1

DJ (b)

(a)

cp = A(y - aif;)(y

= 0.0736

An approximate solution, and a comparison with tests, for sections bounded by a circle and a chord has been given by A. W eigand. 1 Numerical methods are discussed in the Appendix. 98. Torsion of Rolled Profi.le Sections. ln investi~ating the torsion of rolled sections such as angles, channels, and 1-beams, the formulas derived for narrow rectangular bars (Art. 94) can be used. If the cross section is of constant thickness, as in Fig. 162a, the angle of twist

and

The boundary conditions will be satisfied if we take for the stress function an approximate expression

Mi

'

ll

l1u,1.

288

To calculate the stress at the boundary at points a considerable distance from the corners of the cross section we can use once more the equation for a narrow rectangle and take T

narrow rectangle

T1

= G8c. Using this, we obtain from (d) dT dr

= _

2T1

r

(d')

e

A T1r T=---

r

(b)

The sarne approximate equations can be used for an I-beam (Fig. 162c). At reentrant corners there is a considerable stress concentration, the magnitude of which depends on the radius of the fillets. A rough approximation for the maximum stress at 1 these fillets can be obtained from the 1 membrane analogy. Let us consider a 1 cross section in the forra of an angle of 1 1 constant thickness e (Fig. 163) and with 1 radius a of the fillet of the reentrant comer. r / Assuming that the surface of the membrane e _______ ....-v1'at the bisecting line 00 1 of the fillet is L " approximately a surface of revolution, with FIG. 163. axis perpendicular to the plane of the figure at O, and using polar coordinates, the Eq. (151) of the de:flection surface of the membrane becomes (see page 57) 1

ºª

+ !T

from which, by integration,

= e(}(]

Then, from Eq. (a), we obtain for the :flanges of the channel

r

289

TORSION

THEORY OF ELASTICITY

(f)

e

where A is a constant of integration. For the determination of this constant, let us assume that the shearing stress becomes zero at point 01 at a distance c/2 from the 3.Sr-r--.----,.--~-boundary (Fig. 163). Then, from (f),

A _ a+ (c/2)

T1[a

+ (c/2)] e

3.0 t-t--+---+---+------l

_

- 0 2.5

and

t---t---+---+----1-----1

"l:mux

'1

2.0 t - - - \ - - + - - - 1 - - - - - 1 - - - - 1

1

/,-(

2

dz dr 2

+ ! dz

= _

r dr

!1 S

(e)

Remembering that the slope of the membrane dz/dr gives the shearing stress T when q/S is replaced by 2G8, we find from (e) the following equation for the shearing stress:

dT 1 dr+ -;:r = -2G8

(d)

The corresponding equation in the arms of the angle at a considerable distance from the corners, where the membrane has a nearly cylindrical surface, is dr

-

dn

= -2G8

(e)

in which n is the normal to the boundary. Denoting by T 1 the stress at the boundary we find from (e) the previously found solution for a

Substituting in (f) and taking r =a, we find Tmax. =

T1

(1

+ 4: )

(g)

1.0 ~-_,__--.1-_---l._

o

0.5

1.0

J.5

__J

2.0

For a = !e, as in the Fig. 163, we a/e have Tmax. = l.5T1. For a very Frn. 164· small radius of fillet the maximum stress becomes very high. Taking, for instance, a = O.lc we find Tmax. = 3.5T1. More accurate and complete results can be obtained by numerical calculations based on the method of finite differences (see Appendix). A curve of Tmax./r1 as a function of a/e obtained by this methodI is shown in ~ig. 164 (curve A), together with the curve representing Eq. (g). It w1ll be seen that this simple formula gives good resultswhen a/ e is less than 0.3. 99. The Use of Soap Films in Solving Torsion Problems. We have ~een that the membrane analogy is very useful in enabling l,lS to visualize the stress distribution over the cross section of a twisted bar. T~ B~ J. H. Ruth, J. Applíed Mechan_ícs (~rans. A.S.M.E.), vol. 17, p. 388, 1950. e nse of the curve towards the nght is required by the limiting case as the fillet radius is increased in relation to the leg thickness. References to earlier attempts to solve this problem including soap-film measurements are given by I. Lyse and B. G. Johnston 1 Proc, A.S.Ç,E. 1 19351 p. 4691 and in. the above paper.

290

THEORY OF RLASTICITY

Membranes in the form of soap films have also been used for direct measurements of stresses. 1•2 The films were formed on holes cut to the required shapes in flat plates. To make possible the direct determina·· tion of stresses, it was found necessary to have in the sarne plate a circular hole to represent a circular section for comparison. Submitting

FIG. 165.

both films to the sarne pressure, we have the sarne values of q/S, 3 which correspond to the sarne values of G8 for the two bars under twist. Hence, by measuring the slopes of the two soap films we can compare the stresses in the bar of the given cross section with those in a circular 1 See papers by Taylor and Griffi.th, loc. cit.; also the paper by Trayer · an~ March, loc. cit. . . . . 2 A survey of this and other analogies for torsion, with references, is g1ven by T. J. Higgins, Experimental Stress Analysis, vol. 2, no. 2, p. 17, 1945. a It is assumed that the surface tension is the sarne in both films. This was proved with sufficient accuracy by the tests.

'i:.

l i:'I:,

l,I

TORSION

291

shaft under the condition that they have the sarne angle of twist O per unit length and the sarne G. The corresponding ratio of the torques is determined by the ratio of the volumes between the soap films and the plane of the plate. For obtaining the contour lines of the films the apparatus shown in Fig. 165 was used. 1 The aluminum plate with the holes is clamped between the two halves of the cast-iron box A. The lower part of the box, having the form of a shallow tray, is supported on leveling screws. The mapping of contour lines is done by using the screw B passing through a hole in a sheet of plate glass sufficiently large to cover the box in any possible position. The lower end of the screw carries a hard steel point whose distance from the 6rfllrfl glass plate is adjustable by the ---;-_-_-_-_ _ 1,_s.s_.• s ffl__lfrfl~~r11l~r11lr11_,, screw. The point is made to - . .· " /•" ( ( , - -:. -=..:r:...-1-j..--;:- -"-._ ·1rfl /"" approach the film by moving the ', ' 1 ,. ,....;.t-, ,, 't or11 ,.. '-~,', ~ I j \ l I / / glass pia te until the distortion of "\~\'-....:,~'/"~..{.:--' the image in the film shows that '~~~--/~~/." ,,,/,{' contact has occurred. The record is \ 1 11 made on a sheet of paper attached to /\/ 1 v, the board E, which can swing about 1 1 a horizontal axis at the sarne height : 1 1 1 as the steel recording point D. To mark any position of the screw, it is Fra. 166. only necessary to prick a dot on the paper by swinging it down on the recording point. After the point has been made to touch the film at a number of places, the dots recorded on the paper are used for drawing a contour line. By adjusting the screw B this can be repeated for as many contour lines as may be required. When these lines have been mapped, the volume and the corresponding torque can be obtained by summation. The slopes and the corresponding stresses are obtained by measuring the distances between neighboring contour lines. The slope can be obtained optically with much more accuracy by projecting a beam of light on to the surface of the film and measuring the angle of ~he_ reflected ray. The normal to the film is then half way between the mc1dent and the reflected rays. A special instrument was constructed for this purpose by Griffith and Taylor. Figure 166 represents an e~ample of contour lines obtained for a portion of an 1-beam (wooden ~mg spar of an airplane). From the close grouping of the contour hnes at the fillets of the reentrant corners and at the middle of the Upper face, it follows that the shearing stresses are high at these places. The projecting parts of the flange are very lightly stressed. The 1 See the paper by Taylor and Griffi.th, loc. cit.

292

THEORY -OF ELASTICITY

TORSION

rnaxirnurn stress in the middle portion of the web is practically constant along the side of the web and equal to that in a narrow rectangle for the sarne angle of twist. The application of soap-filrn rneasurernents to such cross sections as ellipses and rectangles, for which exact solutions are known, shows that the rnaxirnurn stress and the torque can be rneasured with an accuracy of 1 or 2 per cent. At the points of great stress concentration, as in the case of fillets of very srnall radii, an accuracy of the sarne order is not readily obtained. 1 100. Hydrodynamical Analogies. There are several analogies between the torsional problern and the hydrodynamical problern of the rnotion of fluid in a tube. Lord Kelvin 2 pointed out 0.-----t--X that the function [c/> 1 see Eq. (a), Art. 92] which is sornetirnes used in the solution of torsional problerns is identical with the stream function of a certain irrotational rnotion of y "ideal fluid" contained in a vessel of the sarne FIG. 167. cross section as the twisted bar. Another analogy was indicated by J. Boussinesq. 3 He showed that the di:fferential equation and the boundary condition for determining the stress function e/> (see Eqs. 142 and 144, Art. 90) are identical with those for determining velocities in a laminar motion of viscous fluid alonga tube of the sarne cross section as the twisted bar. 4 Greenhill showed that the stress function e/> is mathematically identical with the strearn function of a motion of ideal fluid circulating with uniform vorticity, 5 in a tube of the sarne cross section as the twisted bar.6 Let u and v be the components of the velocity of the circulating fluid ata point A (Fig. 167). Then from the condition of incompressibility of the ideal fluid we have

õu+õv =O õx

õy

(a)

i See the paper by C. B. Biezeno and J. M. Rademaker, De Ingenieur, No. 52, 1931. See also papers by P. A. Cushman, Trans. A.S.M.E., 1932, H. Quest, . Ingenieur-Archiv., vol. 4, p. 510, 1933, and J. H. Huth, Zoe. cit. 2 Kelvin and Tait, "Natural Philosophy," pt. 2, p. 242. a J. Boussinesq, J. math. pure et appl., series 2, vol. 16, 1871. 4 This analogy was used by M. Paschoud, Compt. rend., vol. 179, p. 451, 1924. See also Bull. tech. Suisse Rom. (Lausanne), November, 1925. · 6 The analytical expression for vorticity is the sarne as for rotation "'• discussed on p. 225, provided u and v denote the components of the velocity of ~he fl~id. . .· G A. G. Greenhill, Hydromechanics, an article in the Encyclopaedia Bntanmca, llth ed., 1910.

''l

irr lllr!

293

The condition of uniform vorticity is

õv

õu

-õx - -õy = constant

(b)

By taking U

= Õc/> ay'

V= -

Õc/>

õx

(e)

we satisfy Eq. (a), and from Eq. (b) we find

a2c1>

a2q,

2

2

ax + oy

= constant

(d)

which coincides with Eq. (142) for the stress function in torsion. At the boundary the velocity of the cir2

ax

+ ààytf>2 + df) dy

Substituting in Eq. (a), à

2 w,) = 1 -G az C

V

Py

+ vT + e

319

If the x-axis is an axis of symmetry of the cross section, bending by a force P in this axis will result in a symmetrical pattern of rotation w. of elements of the cross section (corresponding to anticlastic curvature), with a mean value of zero for the whole cross section. The mean value of àw,/àz will then also be zero, and this requires that e in Eq. (b) be taken as zero. If the cross section is not symmetrical we can define 1 bending without torsion by means of the zero mean value of aw,/az, again of course requiring the zero value for e. Then Eq. (b) shows that aw,/ àz vanishes for the elements of cross sections at the centroids-that is, these elements along the axis have zero relative rotation, and if one is fixed the others have no rotation-about the axis. With e zero Eq. (a) beeomes a2q, + a2q, = _v_ Py _ df (172) ax2 oy 2 1+ v I dy Substituting (172) in the boundary condition (d) of the previous article we find 2 aq, dy + aq, dx = aq, = [Px _ f( ) ] dy (173) ày ds àx ds as 21 y ds From this equation the values of the function q, along the boundary of the cross section can be calculated if the function f(y) is chosen. Equation (172), together with the boundary condition (173), determines the stress function q,. ln the problems which will be discussed later we shall take function f(y) in such a manner as to make the right side of Eq. (173) equal to zero. 2 q, is then constant along the boundary. Taking this constant equal to zero, we reduce the bending problem to the solution of the differential equation (172) with the condition q, = O at the boundary. This problem is analogous to that of the deflection of a membrane uniformly stretched, having the sarne boundary as the cross section of the bent bar and loaded by a 1.Jontinuous load given by the right side of Eq. (172). Severa! applications of this analogy will now be shown. 107. Circular Cross Section. Let the boundary of the cross section be given by the equation x2 + y2 = r2 (a) 1 J. N. Goodier, J. Aeronaut. Sei., vol. 11, p. 273, 1944. A different definition Was proposed by E. Trefftz, Z. angew. Math. Mech., vol. 15, p. 220, 1935. 2 See S. Timoshenko, Bult. Inst. Engineers of Ways of Communications, St. Petersburg, 1913. See also Proc London Math. Soe., series 2, vol. 20, p. 398, 1922.

THEORY OF ELASTICITY

BENDING OF PRISMATICAL BARS

The right side of the boundary condition (173) becomes zero if we take

The maximum shearing stress is obtained at the center (y = O), where

P (r2 _ y2) f (y) = 21

(3 + 2v)Pr2 (r..,.)max. = 8(1 + 11)!

320

(b)

Substituting this into Eq. (172), the stress function cjJ is then determined by the equation a2q, a2q, 1 + 2v Py ( ) àx 2 + ày 2 = 1 + v T e and the condition that q, = O at the boundary. Thus the stress function is given by the deflections of a membrane with circular boundary of radius r, uniformly stretched and loaded by a transverse load of intensity proportional to 1 + 2vPy

-T+vT

It is easy to see that Eq. (e) and the boundary condition are satisfied in this case by taking (d) e/> = m(x2 + y2 - r2)y where m is a constant factor. This function is zero at the boundary (a) and satisfies Eq. (e) if we take (1

m

+ 2v)P

= 8(1

+

v)I

Equation (d) then becomes

q,

= (1

+ 2v)P (x2 + y2 _

8(1

+ v)I

r2)y

(3 + 2v)P ( 2 _ 8(1 + v)I r (1 + 2v)Pxy " 11• = 4(1 + v)I

'Tzz

=

X

2 _ 1 - 2v 2) 3 + 2v Y

(r

(1 + 2v)Pr 2 (-rzz)ll=±r = 4(1 v)J

+

= ±r) is (h)

It will be seen that the magnitude of the shearing stresses depends on the magnitude of Poisson's ratio. Taking v = 0.3, (g) and (h) beco me p p (k) (r,,,) 11=±r = 1.23 A (-r,,,)max. = 1.38 A' where A is the cross-sectional area of the bar. The elementary beam theory, based on the assumption that the shearing stress r,,, is uniformly distributed along the horizontal diameter of the cross section, gives

4P

Tzz

= 3A

The error of the elementary solution for the maximum stress is thus in this case about 4 per cent. 108. Elliptic Cross Section. The method of the previous article can also be used in the case of an elliptic cross section. Let x2

a2 + b2y2 -

f(y) = -

(174)

(a)

1 =O

be the boundary of the cross section. vanish if we take

The vertical shearing-stress componen:, -r,,, is an even function of x and y, and the horizontal component -r11, is an odd functi.on of the sarne variables. Hence the distribution of stresses (17 4) g1ves a resultant along the vertical diameter of the circular cross section. Along the horizontal diameter of the cross section, x = O; and we find, from (174), (3 + 2v)P 2 _ 1 - 2v y2), .,11, = O (j) 'Tzz = 8(1 + v)J 3 + 2v

(g)

The shearing stress at the ends of the horizontal diameter (y

(e)

The stress components are now obtained from Eqs. (171):

321

The right side of Eq. (173) will

~(~:y2 - a2)

(b)

Substituting into Eq. (172), we find

a2q, ax2

a2q,

Py (ª2 b2 + 1

+ ay2 = T

" )

(e)

+v

This equation together with the condition q, = O at the boundary determines the stress function q,. The boundary condition and Eq. (e) are satisfied by taking (1

q, = 2(1

+ v)a + vb 2 2 2

+

v)(3a

2

P(

+ b )·I

x

2

a2

+ b2 y

2

-

ª

2)

y

(d)

.------··-·_.....,.;...;..;_..,,

f

1NST1TUTUl flüUTEMll'C i

\

11!~1~;~c~~~~R~U.

~

322

THEORY OF ELASTJCITY

BENDING OF PRISMATJCAL BARS

When a = b, this solution coincides with solution (e) of the previous article. Substituting (b) and (d) in Eqs. (171), we find the stress components

109. Rectangular Cross Section. The equation for the boundary line in the case of the rectangle shown in Fig. 188 is

T:u

T11•

+ +

+ +

2(1 v)a 2 b2 P [ 2 = (1 v)(3a2 b2). 2I a (1 + v)a 2 + vb 2 Pxy = - (1 + v)(3a 2 + b2 ) l

_

x2 _ (1 - 2v)a 2 y 2] 2(1 v)a2 b2

+

+

(175)

T,,, Tyz

= Ü

+

+

b2 P [a 2 _ (1 - 2v)a 2 y 2] b2) 21 2(1 v)a 2 b2

+

+

(Tzz)max. =

Pa

2

2J [

1 -

a

+3a2vb + /(1 + v)] b2

2

2

If b is very small in comparison with a, we can neglect the terms containing b2 /a 2, in which case {Tzz)max. =

If we substitute into Eq. (173) the consta.nt Pa2 /2J for f(y), the expression Px 2 /21 - Pa 2 /21 becomes zero along the sides x = ±a of the rectangle. Along the vertical sides y = ± b the derivative dy/ds is zero. Thus the right a side of Eq. (173) is zero· along the boundary line and we can take

b -.,

11_0. Additional ~esul~s. Let us consider a cross section the boundary of which cons1sts of two vertical s1des y = ±a (Fig. 189) and two hyperbolas1 (1

+ v)x2

- vy2 = a2

(a)

It is easy to show that this makes the right side of Eq. (173) on page 319 zero at the boundary if we take f(y) = : ;

The shearing stresses, calculated from (k), are

Pa2 (T zz) z=0, 11=0 = 'if 2

(Tzz)x=o,

y=b

= ~~

+ ma 2b2 - 2a2b2(m

(1 ~ py2 + 1~ p)

Substituting into Eq. (172), we find éJ2

+ nb 2)

(l)

The approximate values of the shearing stresses given on the second lines of the table (see page 326) were calculated by using these formulas. It will be seen that the approximate formulas (Z) give satisfactory accuracy in this range of values of a/b. If the width of the rectangle is large in comparison with the depth maximum stresses much larger than the value 3P /2A of the elementary theory are found. Moreover ü b/a exceeds 15 the maximum stress is no longer the component Tzz at x =O, y = ±b, the mid-points of the vertical sides. It is the horizontal component 7'11, at points x = a, y = ± TJ on the top and bottom edges near the corners. V alues of these stresses are given in the table 2 on page 329. The values of TJ are 1 See Timoshenko, loc. cit. 'E. Reissner and G. B. Thomas, J. Math. Phys., vol. 25, p. 241, 1946.

ax2

X

éJ2

+ ay2

Frn. 189.

=O

This equation and the boundary condition (173) are satisfied by taking Then the shearing-stress components, from Eq. (171), are

r.,.

= 2pl ( - z2

T11•

= 0

=

O .

+ _v_ y2 + ~) l+v

l+v

At each point of the cross section the shearing stress is vertical. The maximum of this stress is at the middle of the vertical sides of the cross section and is equal to Ttll3x.

=

Pa 2

2f

The problem can also be easily solved if the boundary of the cross section is given by the equation 1

n~

~),

(± = ( 1 a >X > -a (b) 1 This problem was discussed by F. Grashof, "Elastizitãt und Festigkeit "p 246 1878. ' . ,

For

li

=

Then an approximate expression for the stress function is

i, this cross-section curve has the shape shown in Fig. 190.

By taking

q,

1

f(y) =

Pa2 [ 1 'iI - (

± by);]

2

+

a2q, _ _ 11_Py + Pa ay2 - 1 + 11 I - 2bl 11

!-1 (± 1!)• b

. equat•ion and the boundary condition are satisfied by Th1s taking

1

Pa 2 11 :Z:.: _ 1 q, = 2(1 11)! [ y

(a•

+

+

)-

b (

± 11 ) -+1] v b

(y - a)[x

~

FIG. 190.

Txz

= 2(1

p

+ 11)! (a

2 -

x

'

Tyz

(1~11)Ixy

= -

(e)

y

dx = dy

/

I

FIG. 191.

th d (Art 109) we may arrive By using the energy me . o .. th s Let us consider for instance, the · t 1 t"onmmanyo ercase · ' . b at an app:oxima e so. u ~- 191 The vertical sides of the boundary are g1ven y cross sectt~on sh~wn+mb a~~· the ~ther two sides are ares of the circl~. the equa 10n Y - - • x2

+ y2

- r• =O

The right side of Eq. (173) vanishes if we take

f(y) =

{z. (r• -

y2)

- (2a

+ y)

2

+ By + · · ·) 3

+ y) tan a]

=

O

tan 2 a

P - (2a I

tan 2 a

1

Substituting from (e) and integrating, we arrive at the equation of the boundary, x 2) '

r 2 )(Ay

+ y) tan

l-a X

2

a

(a)

FIG. 192.

An approximate solution may be obtained by using the energy method. particular case when

r

I

p (2a 21

a2q, a2q, -ax - 2 = 1-+.,,-v Py - 2 + ay I

Tyz

-

+ y• -

Equation (172) for determining the stress function q, then becomes

. ( ) for stress components may be derived. The equafrom which the express10ns e b f df the condition that at the boundary tion of the boundary can now e oun ~om the direction of shearing stress coincides w1th the tangent / ' to the boundary. Hence I \

y = b(a 2

+ (2a + y) tan a][x f(y) =

. d·ifferent way In discussing stresses in a · t the sarne result m a · . h W e can arnve a . which is large in comparison w1th the dept ' we rectangular beam t_he w1dthl ot~ f r the stress function [Eq. (g), Art. 109] the used as an approximate so u ion o expression - - " - Py (x2 - a•) - 1 +V 2[

Txz

b2)(x2

The right side of Eq. (173) is zero if we take

Substituting in Eqs. (171) we find 2)

= (y• -

in which the coefficients A, B, . . . are to be calculated from the condition of minimum energy. Solutions for many shapes of cross section have been obtained by using polar and other curvilinear coordinates, and functions of the complex variable. These include sections bounded by two circles, concentric 1 or nonconcentric,2 a circle with radial slits, 3 a cardioid, 4 a limaçon, 5 an elliptic limaçon,• two confocal ellipses, 7 an ellipse and confocal hyperbolas, 8 triangles and polygons 9 including a rectangle with slits, 10 anda sector of a circular ring. 11 111. Nonsymmetrical Cross Sections. As a first example let us consider the case of an isosceles triangle (Fig. 192). The boundary of the cross section is given by the equation

· · (173) · h s i e q, must be constant th left side of the boundary cond1t1on vams e ' .. , e along the boundary. Equation (172) becomes

a2q, ax2

331

BENDING OF PRISMATICAL BARS

THEORY OF ELASTICITY

330

'

(d) '

= 1-"+V = 3~

In the (b)

1 A solution is given in A. E. H. Love's "Mathematical Theory of Elasticity," 4th ed. p. 335, and in I. S. Sokolnikoff's "Mathematical Theory of Elasticity," p. 253. 2 B. R. Seth, Proe. Irulian Acad. Sei., vol. 4, sec. A, p. 531, 1936, and vol. 5, p. 23, 1937. 3 W. M. Shepherd, Proe. Roy. Soe. (London), series A, vol. 138, p. 607, 1932; L. A. Wigglesworth, Proe. London Math. Soe., series 2, vol. 47, p. 20, 1940, and Proe. Roy. Soe. (London), series A, vol. 170, p. 365, 1939. ' W. M. Shepherd, Proe. Roy. Soe. (London), series A, vol. 154, p. 500, 1936. •D. L. Holl and D. H. Rock, Z. angew. Math. Meeh., vol. 19, p. 141, 1939. 6 A. C. Stevenson, Proe. London Math. Soe., series 2, vol. 45, p. 126, 1939. 7 A. E. H. Love, "Mathematical Theory of Elasticity," 4th ed., p. 336. 8 B. G. Galerkin, Bull. I nst. Engineers of W ays of Communieation, St. Petersburg, vol. 96, 1927. See also S. Ghosh, Bull. Calcutta Math. Soe., vol. 27, p. 7, 1935. 9 B. R. Seth, Phil. Mag., vol. 22, p. 582, 1936, and vol. 23, p. 745, 1937. 'º D. F. Gunder, Physies, vol. 6, p. 38, 1935. 11 M. Seegar and K. Pearson, Proe. Roy. Soe. (London), series A, vol. 96, p. 211,

1920.

332

~

BENDlNG OF PRlSMATICAL BARS

an exact solution of Eq. (a) is obtained by taking for the stress function the expresision

The caf'e shown in Fig. 194 can be treated in a similar manner. Assume, for example, that the cross section is a parabolic segment and that the equation of the parabola is x 2 = A(y +a) Then we take p j(y) = 21 . A(y +a) y---+--r---1

=

p 61

[x

2 -

!

3

(2a

+ y)

2

J(y -

a)

The stress components are then obtained from Eqs. (171): âc/>

Tz•

= ây -

âc/>

Px 2

2T

Tu• = - âx =

P + 61 (2a + y)

2

=

2 y'ãp 2 27 4 [-x

ª

+ a(2a + y)]

2 v3P

(e)

27a4 x(a - y)

Along the y-axis, x = O, and the resultant shearing stress is vertical and is represented by the linear function (Tzz)z-o

=

2v3P

27a 3 (2a

+ y)

The maximum value of this stress, at the middle of the vertical side of the cross section, is (d)

By calculating the moment with respect to the z-axis of the shearing forces given by the stresses (e), it can be shown that in this case the resultant shearing force passes through the centroid C of the cross section. Let us consider next the more general case of a cross section with a horizontal axis of symmetry (Fig. 193), the lower and upper portions of the boundary being given by the equations = ..p(y) X = -..p(y) X

for for

X

>O

X

does not depend on 8, so that the third term in (a) gives zero when applied to cf>. W e now transform the compatibility equations (130) (see page 232) to cylindrical coordinates. Denoting by 8 the angle between r and the x-axis we have [see Eq. (13)]

""' = . Consider now the boundary conditions. The resultant shearing stress at the boundary (Fig. 218) must be in the direction of the tangent to the boundary, hence 'TrB

cos (NJ;) -

"B•

cos (Nr) = o

or, by using Eqs. (e),

This shows that cJ> must be constant at the boundary, and we satisfy this condition hy taking solutions of Eqs. (h) such that c/>o, c/>1, e/>., ••• are zero at the boundary. Having obtained c/>o, cj> 1, ••• the successive approximations for the stress components are now obtained from Eqs. (e). Introducing the new variables J; and r, these equations can be represented in the following form: (i)

f m which we conclude that the expression in the parentheses must be a constant. ~:noting this constant by -2c, the equation for determining the stress function e/> is

U sing now the expansion

(d)

= O

and the series (g), we find as the first approximation

We introduce now, instead of coordinates r and z (Fig. 218), new coordinates J; and

r.

J; = R

-r,

a2cJ> a1;2

+

a2c/> ar2

(Tez)o = G

r=z

and Eq. (d) becomes

i;) ª"' +

3

+ R ( 1 -n,

ai;

(e)

1

J;

--J; = 1

1

1-R

1;2

+ R + R2

.

we shall now solve Eq. (e) by successive approximations. e/> = c/>o

and determine c/>o, c/>1, c/>2,

(1 + 2R1;) ª"'º + ª"' ar ar 2 G [ (1 + 1;) ª"'º + ª"' R

1

(,,.,e)i = G [ (,,.e.)i

Considering J;/R as a small quantity, and using the expansion

!:

~~o

For the second approximation we find from Eqs. (i) 2c = O

I' 1

393

(f)

Assume

+ c/>1 + c/>2 + · · ·

(g)

. in such a manner as to satisfy the equations

=

(k)

1

ai;

]

ai;

]

For the third approximation, (,,.,e)2 = G [ ( 1

+ ~ +~:)ªa"'/ + ( 1 + ~) ~t1 + ªa~2 ]

(,,.e.). = G [ ( 1

+ ~ + ~:) ª~º + ( 1 + ~) ªa~' + ªa~2 J

(l)

We apply this general discussion to the particular case of a ring of circular cross section of radius a. The equation of the boundary (Fig. 218) is

(h)

1; 2

+r

2

-

a•=

o

(m)

and the solution of the first of Eqs. (h), satisfying the boundary condition, is

c/>o = 1.

''

'li

~ '

e

2 (!;2 +

r2 - a•)

l~Sil Tv Púl\T~ f----··---·---= ,., ' T l\JI..

'

1~-11 ~O

~i

\

R /4..

'

~i~~\OTECI\ Cf.N'f'RÀ~_.

AXIALLY SYMMETRICAL STRESS DISTRIBUTION

THEORY OF ELASTICITY

394

horizontal diameter of the cross section of the ring (Fig. 21S) from the second of the Eqs. (q), we find

The first approximation for the stress components, from Eqs. (j), is he)o

= -cGr,

(n)

(re,)o = -cG~

This is the sarne stress distribution as for a circular shaft. value of the torque is M1 = - Jf Crrer + -r.e~) d~ dr

(-rezh = -cG (t

The corresponding (o)

For the inner point i, Ç

Substituting from Eqs. (n),

+ ~ f. + 8R

e= 2(M,)o G7ra 4

T

' re

=

r

O, and

2 + a2ç) SR 4R2

13 _E+ 3a

16 R 2

= a, and we have

(-re,); = -cGa (1

dha 4 (M1)0 = - -, 2

r= o

395

+ 4R ~ !: + 17 ~) 16R 2

For the outer point O, t = -a, and

To get the second approximation we use the second equation of (h). tuting for o the expression above we find

Substi-

(-ro,)o = cGa (1 -

~!: + 17 ª) l6R2 2

4R

Using Eq. (r), the values of these stresses become The solution of this equation, satisfying the condition that 1 vanishes at the boundary, is 1 = 3- e~ - (~2 + r2 - a2) SR Substituting this in Eqs. (k) we find the second approximation for the stress components

-cG

(r +Hf)

-cG

1e a 2 [ + -SR- - -SR (r t

a2)

J

. (TOz, )

! j: l

+

a22 ar2

+

~ .!:__ º of the function e/>.. Thus the problem of calculating the correct1ons 1[t reduces to ~he solu_t10n of a system of equations similar to Eqs. (5) of the precedmg art1cle, and these can be treated by the iteration method. . . 3. Relaxation Method. A useful method f?r treatmg d1fference equations, such as Eqs. (8') in the p~eceding art1cl~, was developed by R. v. Southwell and was called by h1m the relaxati~n m_ethod. Southwell begins with Prandtl's membrane analogy, 1 wh1ch is based on the fact that the differential equation (4) for torsional problems has the sarne form as the equation

aw 2

ax2

aw 2

+ ay2

q

= -

S

(9)

for the deflection of a uniformly stretched and laterally _lo_a~ed me~n this equation w denotes deflection from the m1tially honb rane · l · · f th d" zontal plane surface of the membrane, q is the mtens.1ty o e istributed load, and S is the constant tensile force per umt length ~f the boundary of the membrane. The problem is to find the de~ect10n w as a function of x and y which satisfies Eq. (9) at every pomt of the membrane and which vanishes at the boundary.. Let us derive now the corre~ sponding finite-difference equa... ~ ~ Uj tion. For this purpose we replace ° 1 the membrane by the squar~ net 3 o of uniformly stretched strmgs, qô2 Fig. 1. Considering point O and Fw. 6. denoting by Sô the tensile force in the strings, we see that the st~ings. 0-1 and O-~ exert on the node O, Fig. 6, a force in the upward direct10n, equal to

Só ( Wo

~ W1 + Wo ~ Ws)

(10)

A similar expression can be written for the forc~ exerted by t~e two other strings, 0-2 and 0-4. Replacing the contmuous load actmg on i

2

See p. 268. We consider the deflections as very small.

469

APPENDIX

THEORY OF ELASTICITY

468

the membrane by concentrated forces qô 2 applied at nodal points, we can now write the equation of equilibrium of a nodal point O as follows: (11) This is the finite-difference equation, corresponding to the differential equation (9). To solve the problem, we have to find such a set of values of the deflections w that equation (11) will be satisfied at every nodal point. We begin with some starting values woº, w1º, w2º, w3°, w4°, . . . of the deflection. Substituting them into Eqs. (11) we shall usually find that the conditions of equilibrium are not satisfied and that, to maintain the assumed deflections of the net, we need to introduce supports at the nodal points. The quantities such as

Ro = qô 2 + S(w1º

+ w2º + W3° + W4° -

4woº)

(12)

will then represent the portions of the load transmitted to the supports. W e call these forces residual forces, or residuals. Imagine now that the supports are of the screw-jack type, so that a controlled displacement may be imposed at any desired nodal point. Then by proper displacements of the supports we can ultimately make all residual forces (12) vanish. Such displacements will then represent the corrections which must be added to the initially assumed deflections woº, w1º, w 2°, ... to get the true values of w. The procedure which Southwell follows in manipulating the displacements of the supports is similar to that developed by Calisev 1 in handling highly statically indeterminate frames. We first displace one of the supports, say support O, Fig. 6, keeping the other supports fixed. From such equations as (11) we can see that to a downward displacement w0' will correspond a vertical force - 4Swo' acting on the nodal point O. The minus sign indicates that the force acts upward. Adjusting the displacement so that , Ro that is, (13) Ro - 4Swo' = O, Wo = 4S we make the residual force (12) vanish and there will no longer be a pressure transmitted to the support O, but at the sarne time pressures Swo' will be transferred to the adjacent supports and their residual forces will be increased by this amount. Proceeding in the sarne way ! K. A. Calisev, Tehnicki List, 1922 and 1923, Zagreb. German translation in Pubs. lntern. Assoe. Bridge and Structural Engineering, vol. 4, p. 199, 1936. A similar method was developed in this country by Hardy Cross.

471

THEORY OF ELASTICITY

APPENDIX

with all the other supports and repeating the procedure several times, we can reduce all residual forces to small quantities, which can be neglected. The total displacements of the supports, accumulated in this procedure, represent then the corrections which must b~ added with the proper signs to the starting values woº, w1°, w2°, ••• morder to obtain the true deflections of the stretched square net. To simplify the calculations required by the procedure described, we first put Eq. (11) in nondimensional form by substituting

The denominator 1,000 is introduced into Eq. (17) for the purpose of making the iJ;'s such large numbers that half a unit of the last figure can be neglected. Thus we have to deal with integer numbers only. To make our example as simple as possible we will start with the coarse net of Fig. 2. Then we have to find values of iJ; only for three internal points for which we already have the correct answers (see page 463). W e make our square net to a large scale to have enough space to put on the sketch the results of all intermediate calculations (Fig. 7). The calculation starts with assumed initial values of i/;, which we write to

470

qó2

(14)

w=sl/I ln this way we obtain

-~

1 + (ih+ 1/12

+ 1/13 + 1/14 -

(15)

41/;o) =O

= 1

+ (i/; 1º + 1/12º + i/;3º + 1/14º -

41/;oº)

(16)

which in this case are pure numbers. Our problem now is to add to the assumed values i/;o 0, 1/11°, 1/12°, . such corrections as to annul all residuals. Adding to i/;oº a correct1on 1/lo' we add to the residual To the quantity -41/;o', and to the residuals of the adjacent nodal points the quantities i/;o'. Taking i/;o' = To/4 we shall annul the residual at the nodal point O and shall somewhat change the residuals at the adjacent nodal points. Proceeding in the sarne way with all nodal points and re?eating the proced~~e many times we shall in due course reduce the residual forces to neghg1ble values and so obtain the values of iJ; with sufficient accuracy. The corresponding values of w will then be obtained from Eq. (14). ~ To illustrate the procedure let us consider the problem of torsion of a square bar, already discussed in Art. 1. ln ~his c~se we have the differential equat;on (4). To bring it to nond1mens10nal form let us put

2

(17)

q, = 2G8ó iJ; 1,000

The finite-difference equation (5) then becomes 1,000

+ (i/; 1 + 1/12 + i/;a + 1/14 -

41/;o)

-2 -J -6

700

where i/;o, iJ; 1, • . . are pure_ numbers. The problem then reduces to finding such a set of values of 1/1 that Eq. (15) will be satisfied at all inner points of the net. At the bo~nd­ ary iJ; is zero. To get the solution we proceed in the manner descr1b~d above and take some starting values i/;oº, i/;1°, 1/12°, • . • • They w1ll not satisfy the equilibrium equations (15) and we shall have residuals To

o

=

O

(18)

-6

-12

900

:}$ o

-12 -z -J

-6 700

!l =i

:-

-2 -J -6

-12

900

-~ =~ -X -:~

=~

-'.

-1

~i

-1 -);

-2 -3

o

-)( -:~

+i

-2

o

~

-~

-:;: -2

: s-!'}

-J

-6 700

so -~

-12

!~ --fi -~

=i -z :5 --1

-:~ -J -~ -6 +~

-:~

-6 -12

~

1100

900

_,

-~ -:~ -~ -~

-2 -J

-~

-~ - -~ -~

-~ -~ -~

--;~ j

+i 7õ~ =~

-~ +~

o

--1-~ -12 -i -6 :li:;~ :5 o

-12

900

l r

ti=-

a Fm. 7.

the left above each nodal point. The values 700 900 and 1 100 are ' ' ' intentionally taken somewhat different from the previously calculated correct values. Substituting these values together with the zero values at the boundary into the left-hand side of Eq. (18) we calculate the residual forces for all nodal points. These forces are written above each nodal point to the right. The largest residual force, equal to 200, occurs at the center of the net, and we start our relaxation process from this point. Adding to the assumed value 1,100 a correction 50, which is written in the sketch above the number 1,100, we eliminate entirely the residual force at the center. Thus we cross out the number 200 in the sketch and put zero instead. Now we have to change the residuals in the adjacent nodes. W e add 50 to each of those residuals and write the new values -50 of the residuals above the original values as shown in the figurP,. This finishes the operation with the central point of th~

472

THEORY OF ELASTICITY

APPENDIX

11et. We have now four symmetrically located nodal points with residuais -50 and it is of advantage to make corrections to all of them simultaneously. Let us take for all these points the sarne correction, equal to -12. 1 These corrections are written in the sketch above the initial values, 900. With these corrections the values 12 X 4 = 48 must be added to the previous residuais, equal to -50, and we will obtain residuais equal to -2, as shown in the sketch. At the sarne time the forces -12 will be added to the residuais in the adjacent points. Thus, as it is easy to see, -12 X 4 = -48 must be added to the residual at the center and -12 X 2 = -24 must be added at the points closest to the corners of the figure. This finishes the first round of our calculations. The second round we again begin with the point at the center and make the correction -12, which eliminates the residual at this point and adds -12 to the residuais of the adjacent points. Taking now the points near the corners and introducing corrections -6, we eliminate the residuais at these points and make the residuais equal to - 26 at the four symmetrically located points. To finish the second round we introduce corrections -6 at these points. The sketch shows three more rounds of calculations which result in further 'reduction of the residuais. · The required values of if; will be obtained by adding to the starting values all the corrections introduced. Thus we obtain

get a better approximation we must advance to a finer net. Usmg the method illustrated in Fig. 5, we get starting values of y,, for a square net with mesh side õ = -§-a. Applying to these values the standard relaxation process the values of if; for the finer net can be obtained and a more accurate value of the maximum stress can be calculated. With the two values of maximum stress found for 0 = ia and õ = ia a better approximation can be found by extrapolation as explained in Art. 1 (see page 465). ' 4. Triangular and Hexagonal N ets. ln our previous discussion a square net was used,. but sometimes it is preferable to use a triangular or hexagonal net, F1gs. 8a and 8b. Considering the triangular net,

700 - 6 - 3 -

-!

= 688.5,

900 - 12 - 6 - 3 - 2 - 1 1,100

+ 50 -

:g0 oeo 5

2

=

i.311oeo 2 =

12 - 6 - 3 - 2 = 1,127

o.08610ea 2

2

= l.752G1Jõ 2 = 0.1095GIJa 2

5~~7 G1Jõ

2

= 2.254G1Jõ 2 = 0.1409G1Ja 2

which are in very good agreement with the results previously obtained (see page 463). It is scen that Southwell's method gives usa physical picture of the iteration process of solving Eqs. (15) which may be helpful in selecting the proper order in which the nodes of the net should be manipulated. • We take the correction -12, instead of work with·integer numbers.

-!ii1-

(aJ

(6) Fw. 8.

~~~ GIJõ 1

!º

= 876,

Equation (17) then gives for cf> the values 6

473

= -12.5, since it is preferable to

Fig. 8a, .we se~ that the distributed load within the hexagon shown by dotted lmes will be transferred to the nodal point O. If 0 denotes the mesh side, the side of the above hexagon will be equal to õ/-\!'3 and thc area of the hexagon is V3 õ2/2, so that the load transferred to each nodal ~oint will be V3 o2q/2. This load will be balanced by forces in the strm~s 0-1, 0-2, . . . , 0-6. To make the string net correspond to the uniformly stretched membrane, the tensile force in each string must .be equal to the tensile force in the membrane transmitted through one s1de of the hexagon, i.e., equal to So/V3. Proceeding now as in the preceding article, we obtain for the nodal point O the following equation of equilibrium:

,.

or w1

w e introduce

+ w + · · · + Ws 2

3 qó 2

-

6wo

+2S

=O

(19)

a nondimensional function 1f; defined by the equation 3 qó 2

w

+ 1/; + . . · + 1f; 2

6 -

61/;o

+1

= O

(21)

tion methods. . . d th ln the case of a hexagonal net, Fig. 8b, the load d1str:bute over e equilateral triangle, shown m the figure by dotted lines, will be transferred to the nodal point O. Denoting by ó the length of the mesh side we see that the side of the 2 triangle will b~ ó .y3 and its area 3 V3 ó / 4. 2 The corresponding load is 3 V3 qó / 4. This load will be balanced by the tensile forces in the three strings, 0-1, 0-2, 0-3. To make the string net correspond to the ~ a .\ stretched membrane we take the tensile Fm. 9. forces in the strings equal to Só .y3. The equation of equilibrium will then be w 1 + W2 + wa - 3wo. Só .y3 + 3 or W1

+ W2 + W3

-

3wo

6cf>o = 3G8ó 2 = GBa 3

2

and

Such an equation can be ·written for each internal n~dal p.oint, and for the solution of these equations we can, as before, use iterat10n orrelaxa-

ó

there will be only one internai point O of the net, and the values of the required stress function cf> are zero at all adjacent nodal points 1, 2, ••. , 6 since these points are on the boundary. The finite-difference equation for point O is then obtained from Eq. (19) by substituting cf>o for w 0 and 2G8 for q/ S, which gives

(20)

=2s1/I

and the finite-difference equation becomes 1/;i

475

APPENDIX

THEORY OF ELASTICITY

474

4

-- o

(22)

To get the finite-difference equations for torsional problems we have to substitute in Eqs. (19) and (22) 2G8 instead of q/ S. . As an example let us consider torsion of a bar the cross. sect10n ~f which is an equilateral triangle, i Fig. 9. The rigorous solut10n for th1s . case is given on page 266. Using the relaxation method it is natural to select for th1s c~se a triangular net. Starting with a coarse net we take th~ mesh s1de ô equal to one-third of the length a of the side of the triangle. Then This example is discussed in detail in Southwell's book, referred to above.

G8a 2 18

(23)

= --

Let us now advance to a finer net. To get some starting values for such a net let us consider point a - the centroid of the triangle 1-2-0. Assume that this point is connected to the nodal points O, 1, and 2 by the three strings a-0, a-1, a-2 of length ó/y'3. Considering the point a as a nodal point of a hexagonal net, Fig. 8b, substituting into Eq. (22) o/0 for ó, 2G8 for q/S, and taking W1 = W2 = o, W3 = o, Wo = cf>a, we obtain cf>

ª

=

!3 (cf>o + G8ó 2

2 )

= G8a 27

2

(24)

The sarne values of the stress function may be taken also for the points b, e, d, e, and f in Fig. 9. To get the values of the stress function at points k, l, m, we use again Eq. (22) and, observing that in this case W1 = w2 = wa = O, we find k

2

V3 qó =o

+ 34 sq õ2

cf>o

=

cf>m

=

1

G8a 2

= 54

(25)

ln this way we find the values of cf> at ali nodal points marked by small circles in Fig. 10. It is seen that at each of the nodal points a, e, and e there are six strings ~·----a 10 as required in a triangular net, Fig. 8a. Fm. · But at the remaining nodes the number of strings is smaller than six. To satisfy the conditions of a triangular net at all internai points we proceed as indicated by dotted lines in the upper portion of Fig.10. ln this way the cross section will be divided into equilateral triangles with sides õ = a/9. From symmetry we conclude that it is sufficient to consider only one-sixth of the cross section, which is shown in Fig. 1la. The values of cf> at the nodal points O, a, b, and k are already determined. The values at the points 1, 2, and 3 will now be determined, as before,

l

476

. E (22) and the values of cp at three adjacent points. by usmg q. point 1, for example, we will get 3 cf>o + c/>b + cf>a - 3c/J1 + 4 . 2GO g - O

For

(ª)2. -

and substituting for obtain

,i..ª'

,i..b,

,i..o

'I'

'I'

'I'

cf> 1

(26)

= "''\ · G8a 2

left-hand side of Eq. (28), we find the corresponding residuais

Ro = 1/11º

the previously calculated values, we

t d Ali these values are written ln a similar way c/>2 and c/>a are ca1c~1a e . . . . lla 1 They down to the left of the corresponding nodal pomts ~n Fig. . O will now be taken as starting values in the relaxat10n process.

1

+

c/> 2

+ ... + cp 6 -

6cf>o

ª2 + 3G8 81

=O

To bring it into purely numerical form we introduce the notation cf>

and obtain

Goa2i/I* = 486

1/1 = 486cf>

Goa2

or

+ 1/16

(27)

= O (28) E (27) are written to the The starting values of 1{;, calcu1ate d f rom q. ' 1 . t the . t ·n F1º rr 1lb Substituting these va ues m o left of t h e nodal pom s 1 ,.... · r

1/11 + 1/12 + . . .

The constant factor Goa2 is

- 61/;o + 18

omitt~d in_thfe figu;e.

* The number 486 is introduced m th1s ormu a so integers only.

61/;oº

+ 18

(29)

+/

.+/.

+/

+!

+/

+/

+!

+/

-11

N

+!

-f/

+!

+!

+/

-f/

FIG. 12.

ln the case of torsion, Eq. (19) will be replaced by the equation c/>

+!

I

(e)

fbJ Fra. 11.

+ •hº + · · · + 1/;6° -

The residuais, calculated in this way, are written to the right of each nodal point in Fig. llb. The liquidation of these residuals is begun with point a. Giving to this point a displacement i/;a' = -2 we add [see Eq. (29)] +12 to the residual ata and -2 to the residuals at all adjacent points. Thus the residual at a is liquidated and a residual -2 appears at point b. We are not concerned with residuals at the boundary, since there we have permanent supports. Considering now the point e and introducing there a displacement +2 we bring to zero the residual there and add +2 to the residuals at b, d, and e. All the remaining residuals will now be brought to zero by imposition of a dis-

1

(aJ

477

APPENDIX

THEORY OF ELASTICITY

that we may work with

placement - 2 at point f. Adding to the starting values of if; all recorded corrections, we obtain the required values of 1{;, and from Eq. (27) we obtain the values of cp. These values, divided by G8a 2, are shown in Fig. llc. They coincide with the values which can be obtained from the rigorous solution (g) on page 266. 5. Block and Group Relaxation. The operation used up to now in liquidating the residuais consisted in manipulation of single nodal points, considering the rest of the points as fixed. Sometimes it is better to move a group of nodal points simultaneously. Assume, for example, that Fig. 12 represents a portion of a square net and that we give to all points within the shaded area a displacement equal to unity while the rest of the nodal points remain fixed. W e can imagine that all nodal points of the shaded area are attached to an absolutely rigid weightless plate and that this plate is given a unit displacement, perpendicular to the plate. From considerations of equilibrium (Fig. 6),

I' 478

we conclude that the displacement described will produce changes of residuais at the end points of the strings attaching the shaded plate to the remaining portion of the net. If O and 1 denote the nodes at the ends of one string, the contributions to the residuals due to displacement Wo and W1 are Ro = -Sõ Wo õ

and

W1

Ri= Sõ Wo

1

-

W1

õ

If now we keep point 1 fixed and give to point O an additional displacement .6wo, we get the increments of the residual forces .6Ro = - S .6wo,

Introducing dimensionless quantities according to our previous notation, qõ2 R qõ2 = r,

we find .6ro = -

w=si/I

.61/lo,

and the liquidation of the residuals by subsequent point relaxation will proceed more ra pidly. Instead of giving the fictitious plate a displacement perpendicular to the plate and constant for all the nodal points attached to the plate, we ~an r~tate the plate about an axis lying in its plane. The correspondmg d1splacements of nodal points and changes of residuals can be readily calculated. So we can liquidate not only the resultant residual force sustained by the fictitious plate but also the resultant moment about any axis chosen in the plane of the plate. We can also discard the notion of the fictitious plate and assign to a group of points arbitrary selected displacements. If we have some i~ea of the shape o~ the deflection surface of the net we can select group d1splacements wh1ch may result in acceleration of the liquidation process. 6. Torsion of Bars with Multiply-connected Cross Sections. It was shown 1 that in the case of bars with multiply-connected cross sections the stress function q, must not only satisfy Eq. (4), but along the boundary of each hole we must have

- Jan

aq, ds = 2GOA

We see that unit increment in 1{10 produces changes in the residuais equal to .6ro

= -1,

M1 =

479

APPENDIX

THEORY OF ELASTICITY

+1

These changes are shown in the figure. The residuais of the rest of the nodal points of the net remain unchanged. If n denotes the number of strings attaching the shaded plate to the rest of the net, the unit displacement of the plate results in diminishing by n the resultant of the residual forces of the shaded portion of ~v~lL the net. Choosing the displacement so that the resultant vanishes we get residual forces which are self-equilibrat" _ .......~-'-'-'-'-'-lf-='/./ ing and as such lend themselves more t. .2 -4.8 readily to liquidation by subsequent FIG. 13• point relaxation of the normal kind. ln practical applications it is advantageous to alternate sequences of block displacement with sequences of point relaxation. Assume, for example, that the shaded area in Fig. 13 represents a portion of the triangular net. The number n of strings attaching this portion to the rest of the net is 16 and the resultant of the residuais shown in the figure is 8.8. Consequently an appropriate block displacement in this case will be 8.8/16 = 0.55. After such a displacement the resultant of the residual forces, acting on the shaded portion of the net, vanishes

(30)

where A denotes the area of the hole. ln using the membrane analogy the corresponding equation is

-S

aw

Jan

-ds = qA

(31)

which means that the load uniformly distributed over the area of the hole 2 is balanced by the tensile forces in the membrane. N ow applying finite-difference equations and considering a square net, we put Sõ for the tension in the strings, Wo for the deflection of the boundary of the hole, and Wi for the deflection of a nodal point i adjacent to the hole. Instead of Eq. (31) we then have

Sõ

2:

(wi

~ Wo) + qA =

O

or (32) See p. 296. The hole is represented by a weightless absolutely rigid plate which can move perpendicularly to the initial plane of the stretched membrane. 1

2

where n is the number of strings attaching the area of the hole to the rest of the net. The equilibrium equation (11) is only a particular case of Eq. (32) in which n = 4. W e can write as many equations (32) as there are holes in the cross section. These equations together with Eqs. (11) written for each nodal point of the square net are sufficient for determining the deflections of all nodal points of the net, and of all the boundaries of the holes. Consider as an example the case of a square tube, the cross section of which is represented in Fig. 14. Taking the coarse square net, shown in the figure, and considering the conditions of symmetry, we observe that it is necessary in this case to calculate only five values, a, b, e, d, and e, of the stress function. The necessaryequationswill be obtained by using Eq. (32) and the four Eqs. (11) written for the nodal e d e b a points a, b, e, d. Substituting 2GO FIG. 14. for q/S and observing that n = 20

/

ye

/

and A

=

16ó 2 we write these equations as follows:

relaxation procedure is used for all the points. Very often the points cl_ose to ,th~ boundary are connected with it by shorter strings. Due to d1fference m lengths of strings some changes in equilibrium equations (11) _and (19)_ should be introduced. The necessary changes will now be d~scusse~ m co~n~ction with the example shown in Fig. 15. A flat specimen with semicircular grooves is submitted to the action of tensile forces uni~or~y distributed over the ends. Suppose that the difference of prmc1pal stresses at each point has been determined by the photoelastic method, as explained in Chap. 5, and that we have to determine the sum of the principal stresses, which, as we have seen Frn. 15. (page 465), must satisfy the differential equation (6). For the points at the boundary one of the ~wo principal stresses is known, and using the results of the photoelastic tests, the second principal stress can be calculated, so that the sum of these two stresses along the boundary is known. Thus we have to solve the differential equation (6) the values of e/> along the boundary being known. ln using the finit~-difference method and taking a squar~ net we conclude, from symmetry, that only one-quarter of the spec1men should be considered. This portion 4/00

20e - 8b - Se - 4d = 2b - 4a = a - 4b +e+ e = b - 4c d e = 2c - 4d e =

+ + +

16 · 2GOó -2GOó 2 2 -2GOó 2 -2GOó - 2GOó 2

481

APPENDIX

THEORY OF ELASTICITY

480

4100

4JOO

460(}

S400

6700

8SOfJ

/OSfJO

1!41JO

2

These equations can be readily solved and in this way we obtain

4100

4100

S/(J(J

54/JO

8400

!!lfJO

!2JOO

A IJ600

4100

4100

4()00

J900

J900

4200

J800

JSOO

.J.SOO

2600

2

e = 1,170. 2Goó 2

488

and also the values a, b, e, and d. These values obtained with a coarse net, do not give us the stresses ' . . with sufficient accuracy and an advance to a finer net is necessary. The results of such finer calculations, made by the relaxation method, 1 can be found in Southwell's book. 7. Points Near the Boundary. ln our previous examples the nodP.l points of the net fall exactly on the boundary and the sarne standard 1

R. V. Southwell, "Relaxation Methods in Theoretical Physics,'' P· 60, Oxford

University Press, New York, 1946.

4100

J800

3200

2100

1000 Frn. 16.

with _the boundary values of cf> is shown in Fig. 16. Considering point A of this figure, we see that three strings at that point have standard length ó ~hile the fourth is shorter, say of length mó (m :::::: 0.4 in this case). Th1s n:11:1st .be taken ~nto consideration in the derivation of the equation of eqmhbrmm of pomt A. This equation must be written as follows: Só (cf>a - c/>1 ó

+ cf>a -ó

c/>2

+ cf>a

- c/>a 13

+ cf>a m/3 - cf> 4)

=

O

482

THEORY OF ELASTICITY

APPENDIX

or cf>1

+ cf>2 + "'ª + _!_m 4 -

(3

+ !) m "'ª

=

o

ln applying to point A the standard relaxation process and giving to cf>a an increment equal to unity, we will introduce the changes in residuals shown in Fig. 17a. This pattern must be used in liquidating residuals at point A. ln considering point B we see that there are two shorter strings. Denoting their lengths by mô and nõ we find that in the liquidation of residuais at B the pattern shown in Fig. 17b should be used. Introducing these changes at the points near the boundary and using the standard relaxation process at all other points the vall!es of , shown in Fig. 16, will be obtained. 1

+1 o----o - 41 - 4q, 3 t

+

q, 5

+ cpg)

We consider here only simply connected regions.

485

APPENDIX THEORY OF ELASTICITY

484 Similarly we find

4 ô ~ _!_ (6o - 4 2 - 44 + 1 + n) éJy4 õ4 ~ ~ .!. (4o - 2(1 + 2 +a+ 4) + 6 + s + 10 + 12] ax2 éJy2 õ4 . . . t Eq (34) we obtain the required finite-difference Substitutmg m o · equation

20 - 8(1 + 2 + a+ 4) + 2(6 + s + 10 + 12) - O (36) o + q, 5 + q, 1 + 9 + n -

statisfi;~ ~~~v:~: ;~~:~~~n~a~!::~t~ew:t~~~

This equation mfusht be the boundary o t e P1ª e. function we integrate Eqs. (35). dy l =cosa= ds

,.

· 23) th t Observing (Fig. 20, p. a m = sin

and

we write Eqs. (35) in the following forro: 2 dy a2 q, dx a = !!:__ ds ay2 + ds ax ay ds ay dx a2 q, dy a _ - ds éJx2 - ds ax ay 2

and by integration we obtain _ aq, ax

.

Eq. (39) will introduce a third constant, say C, so that the final expression for will contain a linear function Ax + By + C. Since the stress components are represented by the second derivatives of , this linear function will not affect the stress distribution and the constants A, B, C can be taken arbitrarily. From the boundary values of and its A e first derivatives we can calculate the approximate values of at the nodal points of the net adjacent to the boundary, sucli Frn. 19. as points A, C, E in Fig. 19. Having, for example, at point B the values B and (aq,/ax)B, we obtain

=

:: =

f f

_

A = B -

= -dx/ds

a

(ª"') x (ª"') = y =

(37)

!!:__

ds ax

Yds

(38)

Xds

To find we use the equation aq, _ aq,dx as - ax ds

+ aq,dy ay ds

(ª"') O ÔX

B

Similar formulas can be written also for point E. A better approximation for these quantities can be obtained later when, by further calculation, the shape of the surface representing the stress function q, becomes approximately known. Having found the approximate values of at the nodal points adjacent to the boundary, and writing for the remaining nodal point within the boundary the equations of the forro (36), we shall have a system of linear equations suffi.cient for calculating all the nodal values of . The second differences of q, can then be used for approximate calculation of stresses. The system of Eqs. (36) may be solved directly, or we can find an approximate solution by one of the processes already described. The various methods of solution will now be illustrated by the simple example of a square plate loaded as shown 1 in Fig. 20. Taking coordinate axes as shown in the figure, 2 we calculate the boundary values of starting from the origin. From x = O to x = 0.4a we have no forces applied to the boundary, hence

which, after integration by parts, gives

=

aq, X ÔX

aq,

+ y éJy

-

J( asax + a2q,

X

y

a2q, ) ds éJy

(39)

ÔS

Substituting in this equation the values of thedderivat;ves g;v;n bÍt E (37) d (38) we can calculate the boun ary va ues o . sh~~ld be ::ted th~t in calculating the first derivatives \38), twt~ co~­ stants of integration, say A and B' will appear and the mtegra wn m

1 This is one of many cases discussed by P. M. Varvak, "Collection of Papers on Structural Mechanics,'' Kiev Structural Institute, vol. 3, p. 143, 1936 (Russian). A solution, also numerical, of a similar problem is given in the book by K. Beyer, "Die Statik im Eisenbetonbau,'' 2d ed., vol. 2, p. 733, Berlin, 1934. 2 The system is obtained by rotating clockwise by 'Ir the axes used in Fig. 20,

p. 23.

The constants of integration will be calculated from the conditions that for the point x = 0.4a, the common point of the two parts of the boundary, the values of q, and aq,/ax must have the sarne values for both parts. Hence

and integration gives aq, -=A, ax

q,

=

+ B,

Ax

h ·s which as mentioned A B C are constants along t e x-axi ' ' O H ere ' ' b. ·1 We assume A = B = C = . before can be chosen ar itrari y. .d f the Then ~ vanishes along the unloaded portion of the bottom s1 e o J

y

~1

2

0.8a'L-~-1-~-+-i, 1

11111

11

111

lllll!P

i-0.728 'tf/.óB +J.IB +J.óB 2

1 -t----+---.x '

i,

9. Torsion of Circular Shafts of Variable Diameter. In this case, as we have seen (page 307), it is necessary to find a stress function wLlch satisfies the differential equation

a2 q, ôr 2 -

-

;r~ (f/>1

4f/>o -

- f/>3)

=

O

(41)

n::

i

FIG. 23.

1,

+ '1>2 + f/>3 + 'Í>4

3 aq,

a2 q,

r ôr + ÔZ 2 = o

(40)

at every point of the axial section of the shaft, Fig. 24, and is constant 1J.long the boundary of that section. Only in a few simple cases have

The problem then is to find such a set of 1 be satisfied at every nodal point of the ::~ of '1>.that Eq. (41) will assumed constant value at the bound . will be equal to the either by direct solution of Eqs. (41) :;~·y oTnhe1sfptrhob.ltem ct~n be treated A o e l era 10n methods . s an exa~ple, let us consider the case shown in Fig 25 l th . ::;:;:d~!t~ap1~ change in the .