ALEXANDER MENDELSON National Aeronautics and Space Administration
MACMILLAN APPLIED F R E D
SERIE~
IN
MECHANICS
LANDIS,
EDITOR
The Analysis 01 Stress and Delormation by George W. Housner and Thad Vree1and, Jr. Analytical Methods in Vibrations by Leonard Meirovitch Continuum Mechanics by Walter Jaunzemis Plasticity: Theory and Application by Alexander Mendelson Statics 01 Delormable Bodies by Nils O. Myklestad
PLASTICITY: Theoryand Application
MACMILLAN
COMPANY,
COLLIERMACMILLAN
NEW
LIMITED,
YORK
LONDON
PREFACE
With the advent of the jet age followed closely by the space age, the theory of plasticity has been brought forcibly into the forefront of engineering application and design. Modern aircraft, missiles, and space vehicles must be designed on the basis of minimum weight, whieh invariably means designing into the plastic range to obtain maximum lo ad to weight ratios. Moreover, the facts of economie life have made the saving of material and more efficient design a necessity for even the more earthbound industriaI applications. This book is the result of the author's teaching for several years of a graduate course in plasticity for engineers at Case Institute of Technology. It was soon realized that although a number of excellent books on plasticity were available, none of them adequately met the requirements of the course. The available books were either too theoretieal and mathematieal for the average engineer and designer, or their main emphasis was placed on problems of large plastic deformations such as are encountered in metalforming processes. Very little has been published in textbook form on the most oimportant class of elastoplastic problems, in whieh the plastic strains are of the same order of magnitude as the elastic strains, which are of such prime Cor'~f'lrn to today's engineer. Furthermore, where such problems are treated, usual assumptions of perfect plasticity are used, no attempt being made take into account the strainhardening properties of real materials. A set of mimeographed notes was prepared whieh included the basie theory placed primary emphasis on the solution of elastoplastic problems for vii
viU
Prerace
materials with strain hardening. In particular, it was emphasized that with the present availability of highspeed computing facilities, many of the simplifying assumptions hitherto commonplace in plasticity calculations were no longer necessary. The present book is based on these notes. Following a brief introduction, Chapter 2 discusses some of the basic experiments concerning the elastoplastic behavior of metals. Chapters 3 and 4 describe the basic properties of the stress and strain tensors. Tensor notation is introduced and is frequent1y used together with the longhand notation, but a knowledge of tensor properties is not needed. Chapter 5 describes briefly the elastic stressstrain relations. Chapter 6 discusses the various yield criteri a and their experimental verification. In Chapter 7 the plasticity flow rules, or stressstrain relations, are derived and discussed, inc1uding a new set of equations which relate plastic strain increments to total strains rather than stresses. A series of practical problems for both ideally plastic and strainhardening materials is presented in Chapters 8 through 11. Chapter 8 deals with problems of spheres and cylinders. Chapter 9 is devoted entire1y to the powerful . method of successive elastic solutions, by means of which a large c1ass of otherwise intractable pro blems can be solved. First introduced by I1yushin some twenty years ago, this method has not yet gained wide acceptance in this country and, to the author's knowledge, is not even mentioned in any other current book in the English language. Chapter lO discusses plate problems, both for the pIane stress and pIane strain cases. Chapter Il gives the generaI solution to the elastoplastic torsion problem. The theory ofthe slipline field as appÌied to the pIane strain problem of plasticrigid materials is then presented in Chapter 12, and limit analysis of framed structures in Chapter 13. Chapter 14 discusses problems of creep at elevated temperatures and shows how the previously discussed plasticity methods can be applied to creep problems. It is realized that to treat these last three subjects adequately would require a book for each of them. It is hoped, however, that sufficient information is furnished herein to provi de the reader with a worthwhile introduction to, and basic understanding of, these subjects. In the author's experience the material inc1uded can be covered adequately in a onesemester graduate course. Chapters 3 through 5 may be omitted by those familiar with basic elasticity theory. Sections 6.3, 6.5, 7.6, 7.8, 12.6, 12.8, and 13.6 may also be omitted on a first reading or if time is short. It is hoped that this book will be found useful as a graduate text and as an aid to engineers and designers faced with the problem of designing into the plastic range. The author would like to acknowledge his appreciation to his colleagues
Prerace
ix
and coworkers at the Lewis Research Center of the National Aeronautics and Space Administration whose helpful discussions and combined efforts over the years are in no small measure responsible for much that is in this book. In particular, thanks are due to S. S. Manson, M. H. Hirschberg, and E. Roberts. A special debt of gratitude is owed to Professor R. H. Scanlan for his continuous encouragement and interest. But above all I offer humble thanks to Him who " ... favours man with knowledge and enables him to achieve understanding." 31"::l'?wm
Cleveland
A. M.
CONTENTS
Cbapter 1. Introduction
l
Cbapter 2. Basic Experiments 21 TENSILE TEsT 22 TRuE STRESSSTRAIN CURVE 23 COMPRESSION TEST AND THE BAUSCHINGER
4 4 7 EFFECT.
ANISOTROPY
24 25 26
INFLuENcE OF HYDROSTATIC PRESSURE. INCOMPRESSIBlLITY
27
EMPIRICAL EQUATIONS FOR STRESSSTRAIN CURVES
EFFECTS OF STRAIN RATE AND TEMPERATURE IDEALIZATION OF THE STRESSSTRAIN CURVE.
16 20
DYNAMIC AND KINEMATIC MODELS
,,~llIitJl'L"'C
13 15 16
3. Tbe Stress Tensor 31 TENsoR NOTATION 32 STRESS AT A POINT 33 PRINCIPAL STRESSES. STRESS INVARIANTS 3:"4 MAXIMUM AND OCTAHEDRAL SHEAR STRESSES 35 MOHR'S DIAGRAM 36 STRESS DEVIATOR TENsoR 37 PURE SHEAR
24 25 27 30 34 37 39 41 xi
Contents
xii Chapter
4. The Strain Tensor 41 STRAIN AT A POINT 42 PHYSICAL INTERPRETATION OF STRAIN COMPONENTS 43 FINITE DEFORMATIONS 44 PRINCIPAL STRAINS. STRAIN INVARIANTS 45 MAXIMUM AND OCTAHEDRAL SHEAR STRAINS 46 STRAIN DEVIATOR TENSOR 47 COMPATIBILITY OF STRAINS
Chapter
5. Elastic StressStrain Relations 51 EQUATIONS OF ELASTICITY 52 ELASTIC STRAIN ENERGY FUNCTIONS 53 SOLUTION OF ELASTIC PROBLEMS
Chapter
6.
Criteria for YieIding
61
EXAMPLES OF MULTIAXIAL STRESS
62 63
EXAMPLES OF YIELD CRITERIA
64
LODE'S STRESS PARAMETER. EXPERIMENTAL VERIFICATION OF
6.5
YIELD CRITERIA SUBSEQUENT YIELD SURFACES. LOADING AND UNLOADING
YIELD SURFACE. HAIGHWESTERGAARD STRESS SPACE
44 44 48 51 53 55 58 59
83 84 85 86 87
Chapter
67 68
Chapter
7. 71
Plastic StressStrain Relations
72 PRANDTLREUSS EQUATIONS 73 PLASTIC WORK. Two MEASURES OF WORK HARDENING 74 STRESSSTRAIN RELATIONS BASED ON TRESCA CRITERION 75 EXPERIMENTAL VERIFICATION OF PRANDTLREUSS EQUATIONS 76 GENERAL DERIVATION OF PLASTIC STRESSSTRAIN RELATIONS 77 INCREMENTAL AND DEFORMATION THEORIES 78 CONVEXITY OF YIELD SURFACE. SINGULAR POINTS 79 PLASTIC STRAINToTAL STRAIN PLASTICITY RELATIONS 710 COMPLETE STRESSSTRAIN RELATIONS. SUMMARY
Chapter
8.
81
82
EIastopIastic ProbIems of Spheres and Cylinders GENERAL RELATIONS THICK HOLLOW SPHERE WITH INTERNAL PRESSURE AND THERMAL LOADING
lO.
101 102 103
88 92
104
HOLLOW SPHERE. THERMAL LOADING ONLY HOLLOW SPHERE OF STRAINHARDENING MATERIAL PLASTIC FLOW IN THICKWALLED TUBES
98 98 100 104 108 109 110 119 121
141 145 148 150 156
164 164
PLATE
The PIane EIastopIastic ProbIem GENERAL RELATIONS ELASTOPLASTIC THERMAL PROBLEM FOR A FINITE PLATE
172 183 193 197 208
213 213 218
ELASTOPLASTIC PROBLEM OF THE INFINITE PLATE WITH A CRACK
DISTINCTION BETWEEN ELASTIC AND PLASTIC STRESSSTRAIN RELATIONS
HOLLOW SPHERE. RESIDUAL STRESSES. PRESSURE LOADING
9. The Method of Successive Elastic Solutions 91 GENERAL DESCRIPTION OF THE METHOD 92 THIN FLAT PLATE 93 THIN CrRCULAR SHELL 94 LONG SOLID CYLINDER 95 ROTATING DISK WITH TEMPERATURE GRADIENT 96 CIRCULAR HOLE IN UNIFORMLY STRESSED INFINITE
70 71 79
Chapter
Chapter
HOLLOW SPHERE. SPREAD OF PLASTIC ZONE. PRESSURE LOADING ONLY
64 64
70
xiii
Contents
STRAININVARIANCE PRINCIPLE
223 230
11. The Torsion ProbIem 111 TORSION OF PRISMATIC BAR. GENERAL RELATIONS 112 ELASTICITY SOLUTIONS 113 MEMBRANE ANALOGY 114 ELASTOPLASTIC TORSION. PERFECT PLASTICITY 115 ELASTOPLASTIC TORSION WITH STRAIN HARDENING 116 BAR WITH RECTANGULAR CROSS SECTION 117 BAR WITH CIRCULAR CROSS SECTION
234 234 240 245
12.
260
The SlipLine FieId
246 248 250 253
121
PLANE STRAIN PROBLEM OF A RIGID PERFECTLY PLASTIC
122 123 124 125 126 127 128
VELO CITY EQUATIONS
260 266
GEOMETRY OF THE SLIPLINE FIELD
268
SOME SIMPLE EXAMPLES
272 276 279 284 285
MATERIAL
NUMERICAL SOLUTIONS OF BOUNDARYVALUE PROBLEMS GEOMETRIC CONSTRUCTION OF SLIPLINE FIELDS COMPLETE SOLUTIONS. UPPER AND LOWER BOUNDS SLIP LINES AS CHARACTERISTICS
Contents xiv
Chapter 13. Limit Analysis 131 DESIGN OF STRUCTURES 132 SIMPLE TRUSS 133 PURE BENDING OF BEAMS 134 BEAMS AND FRAMES WITH CONCENTRATED LOADS 135 THEOREMS OF LIMIT ANALYSIS 136 METHOD OF SUPERPOSITION OF MECHANISMS 137 LIMIT DESIGN
Chapter 14.
141 142 143 144
Creep BASIC CONCEPTS MULTIDIMENSIONAL PROBLEMS UNIAXIAL CREEP IN INFINITE STRIP CREEP IN ROTATING DISKS
300 300 301 305 307 312 318 323
CHAPTER
1
327 327 331 333 335
INTRODUCTION Index
347
The history of plasticity as a science began in 1864 when Tresca [1] published his resuIts on punching and extrusion experiments and formulated his famous yield criterion. A few years later, using Tresca's resuIts, SaintVenant [2] and Lévy [3] laid some of the foundations of the modern theory of plasticity. For the next 75 years progress was slow and spotty, aIthough important contributions were made by von Mises [4], Hencky [5], PrandtI [6], and others. It is only since approximately 1945 that a unified theory began to emerge. Since that time, concentrated efforts by many researchers have produced a voluminous literature which is growing at a rapid rate. Brief but excellent historical sketches are furnished by Hill [7] and Westergaard [8]. The theories of plasticity fall into two categories: physical theories and mathematical theories. The physical theories seek to explain why metals flow plastically. Looking at materials from a microscopic viewpoint, an attempt made to determine what happens to the atoms, crystals, and grains of a when plastic flow occurs. The mathematical theories, on the other are phenomenological in nature and attempt to formalize and put into form the resuIts of macroscopic experiments, without probing very into their physical basis. The eventual hope, of course, is for a merger two approaches into one unified theory of plasticity which will both the material behavior and provide the engineer and scientist with the tools for practical application. The present treatise is concerned the second of these categories, i.e., the mathematical theories of plasticity l
2
Introduction [Ch. l
and their application, as distinct from the physicai theories. t'he latter beiong to the realm of the metai physicist or solidstate physicist. We start by defining roughly and intuitively what is meant by a metai flowing plastically. If one takes a thin strip of a metal such as aluminum and cIamps one end and applies a bending force to the other end, the end of the strip will deflect. Upon removal of this force, if this force is not too Iarge, the end of the strip will spring back to its originaI position, and there will be no apparent permanent deformation. If a sufficientIy Iarge Ioad is applied to the end, the end will not spring back all the way upon the removal of the Ioad but will remain permanent1y deformed, and we say that plastic flow has occurred. Our objective in this case will not be to determine why the permanent deformation took pIace but to describe what has happened in terms of stresses, strains, and Ioads. Solutions of this particular problem can be found, for exampIe, in references [9] and [lO]. In short, piasticity is the behavior of solid bodies in which they deform permanent1y under the action of externai Ioads, whereas eiasticity is the behavior of solid bo dies in which they return to their originaI shape when the externai forces are removed. Actually, however, the e1astic body is an idealization, because all bo dies exhibit more or less piastic behavior even at the smallest Ioads. For the socalled e1astic body, however, this permanent deformation is so small as to be practically not measurable, if the Ioads are sufficientIy small. Piasticity theory thus concerns itself with situations in which the Ioads are sufficient1y Iarge so that measurabie amounts of permanent deformation occur. It should·further be noted that piastic deformation is independent of the time under Ioad. Timedependent deformations are discussed briefly in Section 2.4 and in Chapter 14. The theory of piasticity can convenient1y be divided into two ranges. At one end are metalforming processes such as forging, extrusion, drawing, rolling, etc., which involve very Iarge piastic strains and deformations. For these types of problems the eiastic strains can usually be negIected and the materiai can be assumed to be perfectly plastico At the other end of the scale are a host of problems involving small piastic strains on the order of the eiastic strains. These types of problems are of prime importance to the structural and machine designer. With the great premium current1y piaced on the saving of weight in aircraft, missile, and space applications, the designer can no Ionger use Iarge factors of safety and "beef up" his design. He must design for maximum Ioad to weight ratio, and this inevitabIy means designing into the piastic range. Even in more prosaic industriaI applications the competitive market is forcing the application of more efficient design. In this book emphasis will be piaced primari1y on the second type ofprobIem, i.e., the elastoplastic problems, where the piastic strains are of the same
Ch. 1] Introduction
3
~rder o~ magnitude as the eiastic strains. Problems of Iarge piastic deformatlOns wI1I be treated only briefly, as will problems of creep and Iimit design. In Chapter 2 some simpie experiments to determine severai basic facts about the e1astoplastic behavior of metais will be discussed. References 1. 2.
3.
4. 5.
6. 7. 8.
H. Tresca, Sur l'ecoulement des corps solids soumis à de fortes pression Compt. Rend., 59, 1864, p. 754. ' ~. de SaintVenant, Memoire sur l'établissement des équations différenhelles des .mouvements intérieurs opérés dans les corps solides ductiles au delà des lImites où l'élasticité pourrait les ramener à leur premier état Compt. Rend., 70, 1870, pp. 473480. ' M. Lévy, Memoire sur les équations générales des mouvements intérieurs des corps solides ductiles au delà des limites où l'élasticité pourrait les ramener à leur premier état, Compt. Rend., 70, 1870, pp. 13231325. R. von Mises, Mechanik der festen Koerper im plastisch deformablen Zustant, Goettinger Nachr., Math.Phys. Kl., 1913, pp. 582592. H. He.ncky, Zur Theorie plastischer Deformationen und der hierdurch im Ma:enal hervorgerufenen Nebenspannungen, Proceedings of the 1st InternatlOnal.C:0ngress on Applied Mechanics, Delft, Technische Boekhandel en Druckenj, J. Waltman, Jr., 1925, pp. 312317. L. Prandt1, ~pannungsverteilung in plastischen Koerpern, Proceedings of the 1st In.ternatlOnal Congress on Applied Mechanics, Delft, 1924, pp. 4354. R. HI11, The Mathematical Theory of Plasticity, Oxford University Press London, 1950. ' H. M. Weste~gaard, Theory of Elasticity and Plasticity, Harvard University Press, Cambndge, 1952.
B..~. Sha~er and R. N. House, The ElasticPlastic Stress Distribution Wlthlll a Wlde Curved Bar Subjected to Pure Bending. J. Appl. Mech., 22, No.3, 1955, pp. 305310. lO. B. W. Shaffer and R. N. House, Displacements in a Wide Curved Bar J. Appl. Mech., 24, No.3, 1957, pp. 447452. ' 9.
Sec. 21]
CHAPTER
TensUe Test
5 p
2
BASIC EXPERIMENTS
P FIGURE
2.1.1
Tensile specimen.
and conventional strain by
l  lo lo
B=
In this chapter the results of some basic experiments on the behavior of metals is presented. The stressstra in curve in tension, one of the basic ingredients necessary in applying plasticity theory, is described in som~ detail. The effects of reverse loading, strain rate, temperature, and hydrostatlc pressure are briefly discussed. Idealizations of the stressstrain curve and various models of material behavior are described.
21
(2.1.2)
Initially the relation between stress and strain is essentially linear. This linear part of the curve extends up to the point A, which is called the proportional limito It is in this range that the linear theory of elasticity, using Hook's law is valido Upon further increase of the load, the strain no longer increase~ linear1y with stress, but the material still remains elastic; i.e., upon removal of the lo ad the specimen returns to its originallength. This condition will
TENSILE TEST
The simplest and most common experiment, as well as the most importa~t, is the standard tensile test. A cylindrical test specimen such as shown m Figure 2.1.1 is inserted into the tensile machine, the load is i.ncreased, and th.e readings of the load, the extension of the gage length inscnbed on the speclmen, and/or the decrease in diameter are recorded. A typicalload extension diagram is shown in Figure 2.1.2. The nominaI stress, defined as the lo ad divided by the originaI crosssectional area, is plotted against the conventional or engineering strain, defined as the increase in length per unit originallength. NominaI stress is represented by
p Ao
a =n
4
B
(2.1.1) FIGURE
2.1.2 Conventional stressstrain curve.
Basic Experiznents [Ch. Z
Sec. ZZ]
True StressStrain Curve
7
6
prevail until some point B, called the elastie limit, or yield potnt, is reached. In most materials there is very little difference between the proportionallimit A and the e1astic limit B. For our purpose, we shall consider them to be the same. Furthermore, the values of these points depend on the sensitivity of the measuring instruments. For some materials the yield point is so poor1y defined that it is arbitrarily taken to be at some fixed value of permanent strain, such as 0.2 per cento The stress at this point is usually called the offset yield strength, or the proof strength. Beyond the elastic limit, permanent deformation, called plastie deformation, takes pIace. The strain at the e1astic limit is of the order of magnitude of 0.001, or 0.1 per cento As the lo ad is increased beyond the elastic limit, the strain increases at a greater rate. However, the specimen will not deform further unless the load is increased. This condition is called work hardening, or strain hardening. The stress required for further plastic flow is calledfiow stress. Finally a point is reached, C, where the load is a maximum. Beyond this point, called the point of maximum load, or point of instability, the specimen "necks down" rapidly and fractures at D. Beyond C a complicated triaxial state of stress exists. The point C therefore represents the limit of the useful part of the tensile test as far as plasticity theory is concerned. The stress at the maximum load point Cis called the tensile strength, or ultimate stress. If at any point between the elastic limit Band the maximum load point C the load is removed, unloading will take pIace along aline parallel to the elastic line, as shown in the figure by B' C'. Part of the strain is thus recovered and part remains permanently. The total strain can therefore be considered as being made up of two parts, se, the elastic component, and sP, the plastic component:
(2.1.3)
stresses or to irregu!arities in. the specimen as well as to the rate of loadin . :ery httle plashc flow takes pIace at the upper yield point T;e lower yleld pomt should therefore always be used for d . . plastic flow calculations. eSlgn purposes and for
Furthe~more,
22
TRUE STRESSSTRAIN CURVE
W,e ha~e dis.cussed the plot of the nominaI stress versus the conventional . not t h e tme stress stram. . It . IS eVldent, however, that this nominaI st ress IS . . actmg m the specimen, since the crosssectionai area of th de creasi 'th 1 d A e speClmen IS f . ng Wl oa. t .stresses up to and near the yield, this distinction is no lmportance. At hlgher stress es and strains this difference becomes lmportant. The true stress can readily be obt' . l ame d f rom t h e nommal stress as foi oWS. If small changes in volume are neglected l' e the t ' . . ' . ., ma ertal IS assume d to b e mcompressible, then
?
Aolo = AI where Ao and lo are the originaI crosssectionai area and gage Iength and A and I are the current values. If P is the load , then the true st ress a IS . P
PI
A
Aolo
a==
The nominaI stress an is an = P/Ao and t he conventional t . . s = (l/Io) _ 1. Therefore , s ram IS
(2.2.1) Upon reloading, the unloading line, B'C', is retraced with very minor deviations. Actually a very thin hysteresis loop is formed, which is usually neglected. Plastic flow does not start again until the point B' is reached. With further loading, the stressstrain curve is continued along B'C as if no unloading had occurred. Point B' can thus be considered as a neW yield point for the strainhardened material. A few materials, such as annealed mild steel, exhibit a sharp drop in yield after the upper yield point B is reached, as shown by the dashed Une. The specimen will then extend at approximate1y constant lo ad to a strain of about lO times the initial yie1d before the lo ad will start increasing again as the material begins to work harden. The flat portion of the curve, called the lower yield, actually represents an average of a series of unstable jumps between the upper and lower yields caused by the propagation of Luder bands across the specimen. The upper yield point is very sensitive to small bendin~
I~
a
s?mewh~t
similar fashion, one recognizes that the conventionai or stram cannot be completely correct, since it is based on initial ength, where~s the Iength is continuously changing. A different definition ~as.therefore mtroduced by Ludwik [1] based on the changing Iength Thus e mcrement of strain for a given length is defined as . ~ngmeermg
de = di
(2.2.2)
l
the total strain in going from some initiallength lo to the length l is
_ t
s
di
l
= J/o 7 = In ~
(2.2.3)
Basic ExperiInents [Ch.2
Sec. 22]
e is called the natural, logarithmic, or true strain and it represents. a sort of average strain in going from the length lo to the length I. Its re1atlOn to the conventional strain is readily found, since l/Io = 1 + e:
dA
or
For small strains the two are practically identical, and for most problems considered the conventional strain will be used. The natural stra~n, howev~r, has several advantages. Natural strains are additive, but conventlOnal strams are noto Second, if a ductile material is tested in compression and in tension, the truestress versus truestrain curves are almost identical, whereas they are quite different if conventional strain is used. Finally, the incompressibility
condition to be used later becomes simply
(2.2.5)
dI
7
/[=
Hence
da = fj! = de a I
or
da de =
(2.2.4)
e = In (1 + e)
9
True StressStrain Curve
8
da de
a
a
(2.2.7)
= 1+e
On a plot of a versus e, the value of a at which the lo ad is a maximum occurs where the slope is equal to the stress; Le., one must draw a tangent to that point of the curve for which the subtangent is equal to 1, as shown in Figure 2.2.1. Discussions of the stressstrain curve and the strain distributions
whereas in terms of conventional strains it is
(2.2.6) which reduces to
__ __ __ ~
only in the case of small strains. If a plot is now made of true stress versus true strain for the tensil.e test previously described, the curve will be essentially the same up to a~d s~lghtly above the yield point. Beyond this point the two types of plots wl11 dlverge. The true stress will always increase until the rupture point and does not have a maximum at the point where the lo ad starts dropping. The true stress at the point of maximum lo ad can be found as follows. Since
p = aA
~
~~
__ __ ________ e
I I~
I ~I
I
FIGURE
True stressstrain curve.
2.2.1
in the neck of a tensile specimen after necking has started can be found in references [2] and [3]. Alternatively, the true stressstrain curve can be obtained by measuring the 'at,amemal strain rather than the longitudinal strain, provided the tensile specihas a circular section. Thus, if eD is the strain in the diametral direction,
eD
a
Aolo = AI
Also
AdI
+ IdA
= O
=
D  Do Do
(2.2.8)
Do is the initial diameter and D is the diameter at the true stress, a. The diametral strain is
da or
+ A da
~
I
at the point of maximum lo ad
dP = a dA
~
= O
(2.2.9)
Basic Experiments [Ch. 2
10
Sec. 22]
11
True StressStrain Curve
1,000,.
and from equation (2.2.5) the 10ngitudina1 true strain is (2.2.10)
e = 2eD = 21n.r;;
SAE 3140 240,000 psi
'" I
The true strain at any 10ad can therefore be determined by measuring the change in diameter of the specimen. From equation (2.2.10) it is seen that the true strain can a1so be written Ao e = 1n
(2.2.11)
A
~ )(
·iii
a. b" (/) (/)
Harddrawn copper
100
~ +(/) al ::l
~
Necking )( Fracture
o
The quantity on the right of equation (2.2.11) is called the true reduction in area. Equation (2.2.11) states that the true strain is equal to the true reduction in area. Figure 2.2.2 (from reference [4]) shows the true stressstrain curves for a
10L~L~~
0.01
Notural stra in FIGURE
e
1.0
2.2.3 True stressstrain curves on loglog coordinates.
variety of materials. The ends of the curves represent the points of fracture and the circ1e on each curve represents the maximum 10ad point or instability point for that curve. To show the complete curves to fracture, the abscissa is such that the elastic parts of the curves are too small to be seen. These curves are also shown replotted on 10glog coordinates in Figure 2.2.3. Note that most of the curves appear as straight lines on this 10g10g plot. This indicates that they can be represented by an equation of the form
320
280
'"ro
0.1
240
)(
·iii
a.
200
(J
b
= Aen
(2.2.12)
where A and n are material constants with n the slope of the curve when plotted on loglog coordinates. A is called the strength coefficient and n is called the strainhardening exponent. Il: folIows from equation (2.2.7) that for a material which behaves according equation (2.2.12), the true strain at the point ofmaximum load is given by e=n
o FIGURE
0.20
0.40 Natural Strain 
0.80
E
2.2.2 True stressstrain curves for several materials.
(2.2.13)
simple relation (2.2.13) has been found useful in fracture studies. It also a simple method for determining the instability point on the true curve. .l:tqluation (2.2.12) will, of course, not fit alI materials, nor will it be valid small strains or very 1arge strains. However, Marin [5] has studied 31
Basic Experiments [Ch. 2 . ,
12
'l d "ound that the average deviation between the theoretldifferent matena s an l l 2 r _ ' b equation (2.2.13) and the actual va ues was pe cal values of e as glVen y ce:. single quantity which represents the ability of a material to defor~ · l1y is the ductility of the material. The most commoI1 measur~ o plast lca 'l t' the per cent stram at ductilit is the per cent elongation in the tensl e tes ,1.e.,. . .. Y Th l'f l l'S the gage length at fracture and lo IS the mltlal gage r fracture. us, length, then the per cent elongation is
er
lr  lo
= ~
x lO
O
(2.2.14)
Together with the per cent e10ngation as given by equation (2.2.1.4),
o~~
t~e in~~al g~g~~:~~~~~ :!~~: ~!:t~~a~:tc:1:~:;~~~~y ~::~~so~:~~::g:n~n:eck~~; ~tarts, m~st of th\:eforma~i:~ a~c~~~~~~ ~~ un
must also specify
smallest cross section, with on1y a relatlVely sma amt th The longer the formation occurring throughout the rest of the g~ge e.~tbe' The ductility is a e length used, the smaller the per cent elongatlOn.wl . ;h:refore reported as the per cent e10ngation for a glVen g.age length. . ductility , however , is the true stram at fracture . A better measure "or l (2.2.15) . t' (2 2 15) can be written in terms of the reduction in area AlternatlVely, equa lOn ., at fracture. From equation (2.2.11) it follows that er
=
Ao In Ar
where Ao is the initia1 area and Ar is the area at fracture. As
menti~~ed ~;e~
vi~~::r~1 :~'!~;~: ~ :'~~::~~i;:7~:rt;u:'~:~~ :~~;r~:::,::~:~~Ymuoh
~nder load and will behave in a brittle fashion. un~xt~~~d~:~g:u~~~~t~ ~it 1 cause such a materia1 to fracture, whereas a matena V: d form under similar loads without fracturing. A cychc load above the w~ll cause a lowductility material to fail in relatively few cycles, whe:eas highductility materia1 will fail after a much larger number of Cyc:l~s (a for very low cycle fatigue). In meta1forming processes such as ro mg, ing, forging, etc., a sufficient amount of ductility is needed to prevent during the forming processo
Sec. 23]
23
Compression Test and the Bauschinger Effect
13
COMPRESSION TEST AND THE BAUSCHINGER EFFECT. ANISOTROPY
If instead of a tensile test one runs a compression test and plots nominaI stress against conventional strain, a different curve will be obtained than for the tensile test. However, if the true stress is plotted against the true strain, practically identical curves are usually obtained. The yield points in tension and compression will, for example, generally be the same. If, however, a metal is first deformed by uniform tension and the load is removed and the specimen is reloaded in compression, the yield point obtained in compression will be considerably less than the initial yield in tension. This has been explained as being the result of the residual stresses left in the material due to the tensile deformations [6]. A perhaps better explanation is based on the anisotropy of the dislocation field produced by loading [7]. This effect is called the Bauschinger effect, and is present whenever there is a reversal ofthe stress field. The Bauschinger effect is very important in cyclic plasticity studies. Unfortunately, however, it enormously complicates the problem and is therefore usually ignored. There are several simplified models used to describe the Bauschinger effect. These are illustrated in Figure 2.3.1 (from reference [8]). At one extreme it is assumed that the elastic unloading range will be double the initial yield stress. If the initial yield stress in tension is ao, then the specimen will yield in compression after being stressed in tension to a = al when
This is shown as path ABCDE in Figure 2.3.1. According to this theory, then, total elastic range of the material remains constant, the initial compressive being reduced by the same amount as the tensile yield is raised. At the other extreme there is isotropic hardening. This theory assumes the mechanism that produces hardening acts equally in tension and cornP1~eSSi1011. Thus compressive yielding will occur when
by the path ABCFG. This is the simplest ofthe theories to apply and the one most frequently used. lSetwl!en these theories there is a theory which assumes that the tensile and yields are independent of each other. The compressive yie1d is independent of the amount of tensile hardening and remains at a = ao
Basic Experiments {Ch. 2
Sec. 24]
Effects
or Strain Rate and Temperature
15
14
24 EFFECTS OF STR1UN RATE AND TEMPERATURE Tests on the effect of the rate of st rammg . . and of te . ertres of mild steel were carried out b M .. mperature on the propofincreasing the stra in rate l'S lYl a~Jome [9], among others. The effect genera y to mcrease th t '1 . in Figure 2.4.1. For materials with 1 . Id e enSI e yleld, as shown a ower yle , such as mild steel, the stress_    10 3 sec1 LA~
~_ 10 2
____________4~+~~€
~_
FIGURE
G FIGURE
2.3.1
Theories for Bauschinger effect.
as shown by AB CHI. Actually experiments indicate that the compressive yield stress usually lies between points H and D of Figure 2.3.1, such as at J. It should be noted that in this figure the curves after yielding are shown as a set of parallellines for simplicity. Actually areai stressstrain curve will show continuous curvature and varying slope after yielding when the lo ad is reversed. As an allied effect to the Bauschinger effect, any initial isotropy which is present is usually destroyed upon 10ading into the plastic range; i.e., if originally the tensile yield point was the same in all directions, it will no 10nger be so. Both the compressive and tensile yield values are changed in all directions by plastic yielding in one direction. Thus plastic deformation is anisotropie. For example, coldrolled sheet has markedly different properties in the thickness direction than in the pIane of the sheet, and usually somewhat different yield points in the rolling than in the transverse direction. We see that the material may have initial anisotropy due to the manufacturing process, and it may also develop anisotropy due to plastic yielding. For small plastic strains the second effect is probably not too important. As for the first effect, the material being used can be tested for anisotropy. If a large amount of anisotropy is found, a much more complicated Hnli~o1rr01DlC theory of plasticity may have to be used.
2.4.1
sec 1
_ _106 sec 1
Effect of strain ratl.
strain curve may approach that f Perti materia1s the reverse will be tr o da h ect.ly plastic material. For other '11 . ue, an t e stram harde' strain . some met mng with l tiWl mcrease . . rate [16]. These effect' s are lmportant m WhlCh are performed at very high t . a  ormmg processes . s ram rates These typ f not be dlscussed in this text. . es o processes will Temperature has a very important effect on me . d' tal propertles. At very 10w temperatures metals which 'l are very uctrle can bec b . l lustrated in Figure 2 4 2 (fr ti ome very nttle. This is .. om re erence [lO]) Th t Il d h '.. e emperature at which the ductility changes so rapid1y i s ca e t e transltlOn te temperature and strainrate ftì t mperature. Such strong 11' e ec s occur more .bodycenteredcubic structures. genera y In metals with
~
g
100 80
(;
.s 60 c
2u
40
.g 20
~ 200 I~~~..l.150 100 50 FIGURE
2.4.2
Effect of temperature.
16
Basic Experiments [Ch. 2
Sec. 26] Idealization 01 the StressStrain Curve
17
At the other end of the time and temperature scales is the phenomenon of creep. Creep is a continuous deformation with time under constant lo ad and occurs primarily at high temperatures, although some metals, e.g., lead, will creep at room temperature. Although it is questionable whether plasticity theories can be applied to the creep phenomenon, it is the usual practice to do so and, in Chapter 14, we shaU describe how this is done. A typical set of creep curves is shown in Figure 2.4.3.
~P (a)
Time FIGURE
2.4.3
Creep curves.
p
25 INrLUENCE or HYDROSTATIC PRESSURE. INCOMPRESSIBILITY (c)
Bridgman, in a series of classical experiments [11, 12] in which he carried out tensile tests under conditions of hydrostatic pressures up to 25,000 atm, showed that hydrostatic pressure has negligible effect on the yield point until extremely high pressures are reached. Furthermore, the shape of the stressstrain curve remains unaltered in the smaUstrain range. The major effect of hydrostatic pressure is to increase the ductility of the material and to permit much larger deformations prior to fracture. It has also been shown that the density, and consequent1y the volume, does not change even for very large plastic deformations. Thus, in the plastic range, a material can be considered as incompressible. These two experimental facts, i.e., the lack of influence of hydrostatic pressure and incompressibility, are very important in the development of plastic flow theories.
26
IDEALIZATION or THE STRESSSTRAIN CURVE. DYNAMIC AND KINEMATIC MODEL
Because of the complex nature of the stressstrain curve, it has customary to idealize this curve in various ways [l3]. Figure 2.6.1
p
(e)
Idealized stressstrain curves' ( perfect1y plastic' (c) rigid l' .' a) perfectly elastic, brittle' . .' , lllear stralll hardenin . (d) l ' ' () g, e astIC, perfectly e e1asttc, hnear strain hardening. 2.6.1
Basic Experiments [Ch. 2
d namic models which can be used idealized curves as well as cor~esP~;~~lgbe~hown subsequently that with the to describe the material.behavlO~: these idealizations are in many cases use of modern computmg mac mery, 18
Sec. 26} Idealization ol the Stl'essStl'ain Curve
.
not necessary. .. 26 1 are designated as dyn amie .models.d They 1 of The models shown m FIgure .: . by displacements. To devlse mo es replace stresses by forces and strams _ =op le = /ii OR
(j
O
~____~__~=========P=l.~~R====~______L________ T (o)
T
T~~:===========11~R==::P·===~LT (b)
T~____~__~==::::::::ll:R::::::::~_P_.__L_________ T (c)
IR
(d)
T
pII
IR
(e) T
19
this type to represent combined stresses acting in several directions would be extremely difficult. For this reason Prager [141 introduced ingenious kinematic models in which both the stresses and strains are represented by displacements. Figure 2.6.2 i11ustrates this type of model for the case of the rigid linearh"'dening plastie mat.,ial [81. The modeJ is taken to be a slotted bar, as shown. The bar is free to move aIong its length on the frictionless table T. But for the bar to move, tb, pin P must 'ngage tbe end of the b",. InitialJy pin p is at the center R of the slot and this point is marked as point O on the table. The distanee from p to either end ofthe slot is taken equal to tbe yie1d stress (lo of the rigid linearhardening materia1. The distance OP of the pin p from the fixed point O is taken equaI to the stress. Then the distance OR from the center ofthe slot R to the fixed point O is propor';onal to the strain, i .e., , ORjm, Wh.,e lan , m is tbe slope of the plastic Slressstrain curve. Thus plastie flow will take pIace when the pin 1S engaged at one end or the other ofthe slot. Figures 2.6.2 and 2.6.3 illustrate the different positions of the kinematic modeI and the corresponding stressstrain diagram. Note that for this particular model it has been assumed that the elastic unloading range EG i8 equaI to twice the initiaI yield, so that the yield point in compression G 1S Iess than the initial yield C.
~
T p8
T
E O I
(f)
IR
(g)
IR
T T
T B
T
A
F
IR
(h)
T
T
T~JL~=========r===IR==~T
G
(il
H
O
FIGURE 2.6.2 K inematic model. FIGURE 2.6.3 Stressstrain curve for model of Figure 2.6.2.
Basic Experiments [Ch.2
20
two stress variables, al and 0'2' a:e In a two_dimensional state of .stress, Th fore if the position of the plll specified instead of the single va:lab~s:be f::: to ~ove in two directions. In is to indicate the stat~ of stress, lt m o_dimensionaI frame rather than ~ slot. this case the rigid reglO~ must be ~:~il in Chapter 6, after the discuSS IOll of This will be discus sed III greater yield surfaces.
21
Ch. 2] Problems
Problerns
Show that natural strains are additive whereas conventional strains are not. Assume that a material behaves elastically up to the point of instability. Show that the natura l strain at this point is unity. 3. Derive equation (2.2.13). 4. Let the stressstrain curve of a material be given by a = Aen, where e is the conventional strain. Show that at the point of instability
1. 2.
n 1 n
e=
FOR STRESSSTR1UN 27 EMPIRICAL EQUATIONS
In a standard tensile test using a tin.diameter specimen with a lin. gage length, the following data were obtained. At a load of 10,000 lb, the conventional strain was 0.10, and at a lo ad of 12,000 lb, the conventional strain was 0.60. Find the true stresses and strains for these two conditions. Determine the strength coefficient A, the strainhardening exponent n, the change in gage length at the maximum load, and the maximum lo ad assuming equation (2.2.12) to hold. 6. A tensile lo ad is applied to a thinwalled hollow circular cyÙnder. Determine the change in wall thickness and in mean radius at the point of maximum load, if the stressstrain curve is given by a = Aen, where e is the conventional strain and a is the true stress. 7. Derive the incompressibility conditions (2.2.5) and (2.2.6). 8. The following data were obtained in a tensile test on a 0.505in.diameter specimen:
5.
CURVES
.
f
ivenmaterial
t the stressstralll curve o a g It is sometimes useful to rep:e~en b fitting the experimental data. E~uaby an equation obtained empl:lcallYhi~h will frequently fit most of a glVen tion (2.2.12) is such an equatlOn w. l mentioned, will not usually fit at . b t as was prevlOUS Y . One of the stressstralll curve, u, . d of the stressstralll curve. the lowstrain and highstralll en s d by Ludwik (1). It has the form .' l quations was propose (2.7.1) first such emplflca e 0"= 0'0
A frequently used form,
+ men
due to Ramberg and Osgood [151, is e =
(2.7.2)
i + k(ir
Diameter, in.
Some other forms that have been proposed are
a
= a + (b
0'=
c(a +
0'= 0'0
(2.7.3)
 a)(l  ene) e)n
. h ield strain, 0'0 the yield stress, turalloganthms, eo t e y ,. band c constants. where e is the base o na E the elastic modulus, and m, n, ,,~, , t f the stressstrain curve as 'bl t fit the plastlc par o It is also pOSSl e o . l f arbitrary degree, i.e., . d b polynomIa o Ya ately as deslre e :::; eo
a =
{ ao
m
e 2 + a1 e + a2 e + .. , + am
. h d . g all the
wh ere eo l'S the yield strain. beginning with a2 are zero.
6,750 9,250 10,400 10,900 11,100 11,200
Diameter, in.
Load,lb
0.419 0.402 0.375 0.361 0.354 0.326
11,000 10,800 10,200 9,700 9,500 8,950 Fracture
tanh~)
f
Ee
0.487 0.481 0.472 0.463 0.450 0.438
Load,lb
For linear stram ar enm
(a) Plot the true stressstrain curve. (b) Determine the strength coefficient A and the strainhardening exponent n. (c) Determine the maximum lo ad from the stressstrain curve and compare it with that obtained using equation (2.2.13). Consider a material whose stressstrain curve is given by a = 30,000 + 1.5 X 106 e, a > 30,000 psi. If a tensile specimen of this material is stretched to a strain of 0.004 in./in., at what stress will it yield in compression when the load is reversed, for each of the assumptions in Figure 2.3.1 ? For the dynamic models of Figure 2.6.1, show the relations between the constants and the parameters of the stressstrain curve. Denote the constants by k (k 1 and k 2 for the last model), the weight of the block
Basic Experiments [Ch. 2 22
by W, the friction coefficient by f.t, and the force by P. For example, for the first model, the equation of the stressstrain curve is a = Ee and the corresponding model equation is P = kx. Thus
Ch. 2]
15. 16.
General References
W. Ramberg and W . R . O sgoo d Des . f 23 T hree Parameters, NACA Techni;al M cnp 10n of StressStrain Curves by T. A. Trozera, O . D . Sher b y and J ote No. 902, July 1943 . E D Te~perature on the Plastic Deforma~"10n or~, Effect of Strain Rate and Cali[. (Berkeley) Tech. Repl S 2 of HIgh Purity Aluminum Un' , Dec. 1955. ., ero 2, lssue 44, Contract
N70NR2~~'
GeneraI References 11. For the kinematic model of Figure 2.6.2, show that e = OR/m. 12. Describe a kinematic model similar to that shown in Figure 2.6.2 for iso13.
tropic Sketchhardening. typical stressstrain curves that would be obtained using Ludwik's
Drucker, D . C ., St ressStram . Relations in h . Theory and Experiment ONR R t e Plashc RangeA Survey f ept. NRD41D32, 1950. o Goodier, J. N. and P J H 'd 1958.' . . o ge, Jr., Elasticity and Plasticity ' W'I . I ey, N ew York
expression for the following cases:
HIll, 1950. R., The Mathematical Th eory of Plasticity, Oxford U· mv. P ress London '
(a) n = 1. (b) ~ n < 1. (c) ao = 0, n = 0,
Johnson, W., and Plas . . for Mechanical Engineers' Van 'N os t rand,, London, 1962P.. B. Mellor ,t/city
°
t,
1.
References 1. P. Ludwik, Elemente der technologischen Mechanik, Springer, Berlin, 1909. 2. J. D. Lubahn and R. P. Felgar, Plasticity and Creep of Metals, Wiley, New 3. York,1961. G. E. Dieter, Jr., Mechanical Metallurgy, McGrawHill, New York, 1961. 4. H. Schwartzbart and W. F. Brown, Jr., NotchBar Tensile Properties of Various Materials and their Relation to the Unnotch Flow Curve and 5.
Notch Sharpness, Trans. ASM, 46, 998, 1954. J. Marin, Mechanical Behavior of Engineering Materials, PrenticeHall,
6.
Englew ood Cliffs, N.J., 1962. R. Hill, The Mathematical Theory of Plasticity, Oxford Univo Press, London,
7. 1950. D. Mclean, Mechanical Properties of Metals, Wiley, New York, 1962. 8. J. N. Goodier and P. G. Hodge, Jr., Elasticity and Plasticity, Wiley, New York,1958. M. J. Manjoine, Influence of Rate of Strain and Temperature on Stresses of Mild Steel, J. Appl. Mech., 11, A21l, 1944. lO. A. W. MagnusSOn and W. M. Baldwin, Low Temperature Brittleness, 9.
Mech. Phys. Solids, 5,172,1957. P. W. Bridgman, The Effect of Hydrostatic Pressure on the Fracture Brittle Substances, J. Appl. Phys., 18, 246, 1947. 12. P. W. Bridgman, Studies in Large Plastic Flow and Fracture with Emphasis on the Effects of Hydrostatic Pressure, McGrawHill, New
11.
13. 1952. W. Johnson and P. B. Mellor, Plasticity for Mechanical Engineers, 14.
Nostrand, Princeton, N.J., 1962. W. Prager, The Theory of PlasticityA Survey of Recent Proc. lnst. Mech. Engrs., London, 169, 41, 1955.
Sec. 31]
CHAPTBR
3
THE STRESS TENSOR
. familiar with the basic concepts of the theory It is assumed that the reader lS. . f tress and strain. However, to avoid .' . l d'ng the defimtlOns o s f h d to refresh the memory o t ose of elastlClty, mc u 1 fer to other texts an b . il having the student re k l'n elasticity we shall ne Y tly done any wor ' . readers who have not recen f these basic concepts, with partlcular . m . th e next three chapters some o . h are parreVlew d strain tensors WhlC . h erties of the stress an d t f plasticity theory. The rea er emphasls on t ose prop 6 . t t in the developmen o .' h may skip directly to Chapter . ticularly Impor an thoroughly familiar with elastlclty t eory d lthough we shall not con. ndorder tensors, an a Stress and stram are seco . t' s such it is important that . .h and thelr proper les a , cern ourselves Wlt tensors . t t tI' on known as tensor notatlOn. '1' 'th the subscnp no a . 'ting out long formulas or expresthe student be faml lar Wl . . t nly a time saver m wn f f f l' derivations and in the proo o This notatlOn lS no o . . l o extremely use u m sions, but lt lS a s . t f the past and present literature on theorems. Furthermore, a aJor pa~ ok owledge of this notation is ess.en1tìaf subject utilizes tensor notatlOn, aln . a npecific problems the usual lUlll'.l"'U~ . th literature In so vmg s .h in followmg e . d We shall therefore start Wlt a notation must, however, always be use . description of tensor notation.
n:
24
31
25
Tensor Notation
TENSOR NOTATION
A tensor is a system of numbers or functions which transform according to a certain law, when the independent variables undergo a linear transformation. We shall not concern ourselves here with the transformation laws for tensors but will merely record a few elementary properties, inc1uding the tensor notation. We shall accept the fact that stress and strain are tensors. A subscript notation is used which is really very simple. The coordinate axes are designated by the letter x, with a latin subscript. Thus XI does not meanjust one quantity, but three quantities, Xl> X2, and Xs, where Xl' X2' and Xs are used instead of X, y, and z (or r, B, z, etc.). Any other subscript, such asj, k, l, m, etc., can be used equally well. For twodimensional problems the subscript is understood to have a range of only two rather than three. A double subscript indicates a system of nine components if the range of each of the subscripts is three, or a system of four components if the range is two. For example, the stress tensor is designated by a" and stands for nine components: alj
==
a ll a2l
a12 a22
alS] a2S
aSl
aS2
ass
=
[ax 'T yX
(3.1.1)
[ 'T zx
e".
Similar1y, the nine components of the strain tensor are designated by Two subscripted quantities are said to be equal if their corresponding components are equal. Thus if A" = Bli> then All = B ll , A 12 = B 12 , etc. If two subscripted quantities are added, their corresponding components are added. Thus
+ Bll = A 12 + B 12 = A 2l + B 2l = All
Cll
C12 C2l
etc.
system is called a system of first order, a double subscript one of second order, etc. It is evident from the definitions of equality addition above, that these can apply only to systems of equal order. now come to the only "tricky" part of tensor notationthe summation ",vI>n."... " . Whenever a subscript is repeated, this indicates summation over of the subscript. Thus
Sec. 32] Stress at a Point
The Stress Tensor [Ch. 3
The simp1est secondorder symmetric tensor is the Kronecker delta or substitution tensor, defined by
26
Such a subscript is called a dummy subscript and it must be a letter not a number; i.e., all does not mean summation. A more complicated example is given by the increment of work per unit volume:
27
Off
oli
(3.1.2)
=O = l =
The advantage here becomes apparent, since we have written down one term instead of nine. Also the work increment stands out as the scalar product of ents the strain increment and the stress. Furthermore, 7"jf dSjf repres the work increment without being tied to any particular system ofaxes. It would thus include 7"11 dS + 7"22 dS + 7"33 ds 33 , where these are the principal stresses ll
i "# j i
=j
(3.1.3)
[~ ~]
It is called the substitution tensor because OffA f offA"c
= =
Af
(3.1.4)
Ai/C
22
andA strains. system having any number of subscripts is said to be symmetric in two of these subscripts if the components of the system are unaltered when the two subscripts are interchanged. Thus a secondorder system is called symmetric if
Finally, the convention l'S used to d . 1 erentiation by a eSIgnate partia1 d'fii comma. Thus
(3.1.5)
The stress and strain tensors are usually symmetric. A system is said to be skewsymmetric or antisymmetric if the interchange of the indices changes the signs of the components. Thus for a secondorder skewsymmetric system
32 STRESS AT A POINT Consider as show' . forces P 1 P 2a' "body n In FIgure 3.2.1 subjected t o a system of externa1 • Pa. Now consider l dIViding it into parts I and II h a p ane AB passing through the bod , as s own. If we consider part I , it l'S seen t hatY
This shows immediately that All = A ll = O
= A 22 = O A33 =  A 33 = O
A 22
Therefore a skewsymmetric tensor of second order is characterized by three quantities:
Pl
= A 32 =
P2 =
Al3
=
A 23  A 3l
P3 = A l2 = A 2l ~
It can readily be shown that every secondorder tensor Ajf may be
posed into the sum of a symmetric tensor ejf and a skewsymmetric tens
or
FIGURE
3.2.1
Loaded body.
The Stress Tensor [Ch. 3
28
Sec. 32} Stress at a Point
it is in equilibrium under the action of forees P 1 , P 2 , P 3 , P 4 , and the force P 12 that part II exerts on part I, P 12 being the resultant of the eontinuous distribution of forces on the pIane AB that part II exerts on part I. If a small area AA is taken in this pIane with a force AP acting on it, then the uni! stress acting at this point is defined as p
=
. AP 11m AA
29
(3.2.1)
t.A>O i l
The important thing to note here is that the unit stress, p, must be referred to a particular pIane. For any other pIane passing through the same point, it is obvious from consideration of Figure 3.2.1 that the force distribution on this pIane, and hence the unit stress, will be different. The unit stress, p, of course, need not be perpendicular to the pIane AB. In practice, therefore, the stress, p, is decomposed into two components, one normal to the piane of reference, called the normal stress, and one parallel to this pIane, called the shearing stress. The normai stress is taken as positive when it is tensile in nature and negative when it is compressive. To compIetely specify the stress at a point it is necessary to specify the stresses at that point on three mutually perpendicular pianes passing through the point. The stress on any arbitrary pIane through the point can then be determined in terms of the stress es on the three perpendicular planes, as will short1y be shown. Let the three mutually perpendicular planes be the planes perpendicular to the x, y, and z coordinate axes. Then the stresses acting on these pianes at their point of intersection are as designated in Figure 3.2.2. The stresses as shown are alI positive. The subscripts denote the direetion of the stress. The first subscript designates the normai to the pIane under consideration, and the second subscript designates the direetion of the stress. Thus 7" xy denotes a shearing stress acting on the face of the element that is perpendieuIar to the x axis, the stress aeting in the direction of the y axis. As mentioned previously, the normai stress is taken positive when it nt'r,r1llt'po tension and negative when it produees eompression. The positive G1reetlOtls of the components of shearing stress on any side of the cubie element taken as the positive direetions of the coordinate axes, if a tensile stress on same side would have the positive direction of the corresponding axis. It is seen from the figure that the complete specifieation of the stress at point is given by the nine quantities
FIGURE
3.2.2
Convention for stresses.
It is customary in engineering praetice to d Clx instead of Cl elete the second subseript on the quanhtles are designated' 1 xx, etc. In tensor notation th ' slmp y by Cl b e nme If one considers an infinite' l ti or, y some authors, 7"li' , , Slma rectangula 11 ' pO,mt m a body, then it readily ii Il r p~ra eleplped surrounding a st~Ì1c equilibrium of forees and ~ ows, as lS s~own in standard texts, pomt satisfy the following eqUation~~ents reqUIres that the stresses at
norm~l, stress and write
(3.2.2)
~ +
87"yZ
8x
Fj are the components of th
8y
8Cl
+ Biz =  Fz
e body forces per uni t volume. Also (3.2.3)
The Stress Tensor {Ch. 3
Sec. 33]
PromClpal . Stresses. Stress Invariants
31
z
30
OD..L ABC
.J = 00
N
In tensor notation these become simply
OA
(3.2.4)
m =
00
n
00
OC 08
The second line of(3.2.4) expresses the fact that the stress tensor is symmetric. There are therefore in generaI only six independent components of stress at a point rather than nine. (Note: There are some peculiar conditions for which the stress tenso will not be symmetric, as in the case when body moments r
c y
act (1).)
33 PRINCIP.AL STRESSES. STRESS INV.ARI.ANTS If we are given the six components of stress at a point with respect to some coordinate system (x, y, z), we can determine the stresses acting on any pIane through this point. This can be done by consideration ofthe static equilibrium of an infinitesimal tetrahedron formed by this pIane and the coordinate planes, as shown in Figure 3.3.1. In this figure we have shown the stresses acting on the three coordinate planes. These stresses are assumed to be known. We wish to find the stresses acting on the pIane ABC whose normal ON has direction cosines l, m, and n. Let the area of the infinitesimal triangle ABC be designated by AA. Then the areas ofthe faces AOB, COB, and AOC are equal to m AA, l AA, and n AA, respectively. Now let the stress vector acting on the face ABC be designated by Sand its x, y, and z components by Sx, Sy, and Sz as shown in Figure 3.3.1(b). Then for equilibrium of forces in the
x
c
y
x direction,
x (b) FIGURE
or
3.3.1
F
orces on infinitesimal tetrahedr ono
from (3.3.1) resuIts in Similarly,
Sy
= lr XY + mu y + n'1"Zy
Sz = lr xz
+ m'1"yZ + nuz
Sn
=
12ux
+ m 2 uy + n2 uz + 2(lm'1"xy + mn'1"yZ + nlr zx )
(3.3.3)
to obtain the resuItant shear stress Ss a Ct'lllg on thIS . pIane, To obtain the normal stress Sn acting on this pIane we project the
Sx, Sy, and Sz onto the normal ON, to get
S; = S2  S2n = S;,
+
S~
+
S2z  S2n
(3.3.4)
Sec. 33]
Principal Stresses. Stress Invariants
33
The Stress Tensor [Ch. 3
Expanding the determinant glves . . equation for S: a CUblC
32
Equations (3.3.1) give the x, y, and z components of the stress acting on this pIane, and equations (3.3.3) and (3.3.4) give the normal and shear stresses. Equations (3.3.1) can also be considered as the boundary conditions that have to be satisfied by the stress components 0!J at any point on the boundary of the body. Thus if the element of area ABC is considered to be an element of the boundary whose normal has the direction cosines l, m, and n, and Sx, Sy, and Sz are the components of the applied boundary forces at the point O, then equations (3.3.1) are precisely the boundary conditions that must be satisfied by the stress tensor. In tensor notation, if we replace l, m, and n by Il, 12 , and 13 , we can write (3.3.1) as
(3.3.5)
Suppose the pIane element ABC of Figure 3.3.1 is so oriented that the resultant stress S on this pIane element is normai to the pIane; Le., S = Sn and Ss = O. The pIane is then called a principal pIane at the point, its normal direction is called a principal direction, and the stress S = Sn is called a principal stress. At every point in a body there are at Ieast three principal directions. These principai stresses and principai directions can readily be found as follows. Assume the element ABC to lie in a principai pIane at point O so that S = Sn' Then S has the same direction cosines l, m, and n as the normal. The components of S in the x, y, and z directions are then Sx
= IS
Sy = mS
S3  11 S 2
12 S  13 = O

(3.3.9)
where
= 12 =
+ + Oz '1'2 2 + '1'zx 2 xy + '1'YZ
13 =
o x o yo z
Il
0x
0y

(OXOy
+ 2'1'Xy'1'YZ TZX

+ o y o z + o zo x )
(o x '1'2YZ
(3.3.10)
+ OyT 2ZX + Oz'1';y)
It can be proved. [2] that th e cub'IC equatlOn . (3 3 9) h h " .. as t ree real roots and consequent1y there are (at le t) h db as t ree prmclpal stre h' nate y 01, 02, and 03' Substituting a f h sses, w lch will be desig(3.3.6) enables one to solve for th nyo t es~ solutions back into equations 'f . e correspondmg di t' n, l m addition the identity f2 + m 2 + n2 = 1 . rec lOn cosines l, m, and 01, 02, and 03 are distinct the th . lS used. If the three roots b . ' ree correspondmg p' . 1 . e umque and orthogonal. If t f h nnClpa dlrections will 1 .. . wo o tese roots are e umque but the other two d' t' equa, one dlrectlOn will d' lrec lOns can be any t b o t e first. If aH three roots are equa, 1 t h ere are no uwo. lrections, . , orthogonal t h an any three directions can b h mque pnnclpal directions e c osen . This correspon d s to a state of hydrod . stress. stattc
Suppose instead of the axes x y and . were chosen at the point O Th' , h z, a dlfferent set ofaxes, x' y' and z' ", . en t e equat' f d stresses, (3.3.9) would be th lOn or etermining the principal esame, except that I l d . ' 1U terms of the stresses o' 0" • 1, 2, an 13 would be defined axes, Le., x, y, Oz, etc., wlth respect to the new coordinate
Sz = nS Il
and equations (3.3.1) give immediateIy
l 2 
+ m'1'yx + n'1'zx = O lr XY + m(Oy  S) + nrZy = O lr xz + m'1'yz + n(oz  S) = O
l(ox 
=
o'x '1"2 XY
+ o' + Oz' y
+ '1'YZ '2 + ...
etc.
S)
the principal str.esses are physical quantities . on the coordmate axes ch and obvlOusly do not I . ( osen. Hence the numb 2 'l''''V'H.''''' m 3.3.9) must be the same no ers 1, 1 , and 13 morder that they give the same' matter what coordinate axes are values for 01 , o 2, and 03' Th us
3
or in tensor notation (repiacing l, m, and n by Il, 12 , and 1 )
Il  o x
+
0y
+
o
z
=
o'x
+ U 'y + U z'
For equation (3.3.6) to have a nontriviai solution for l, m, and n, the minant of the coeftìcients must vanish, resulting in OX \OiJ 
8jj S\
=
'1'Xy \
'1'xz
S 0y 
'1'yz
S
'1'zx '1'Zy Oz 
\
S
=
O
. invariants for 12 and thtrd of 13' theIl' t12 , and l 3 are t h erefore caHed the first second s ress tensor re t' 1 " first and second invariants ' ~pec lve y. We will show later are partlcular1y important in plasticity
The stress Tenso r {Ch.3
34
If we choose the principal directions as the directions of the coordinate
Sec. 34] Max'unum and Octahedral Shear Stresses
respect and . these d ' . 35 . to l and m ,equatmg equat10ns are obtained for l and m: envatlves to zero, the following
axes, then the stress invariants take on the simple form Il = 1 = 2
 (~)Z 2
(3.5.4)
+ a3 2
. . (3 .5' 4) , this equation repres ents .a .' uatlOn . If the equaltty slgn lS used m eq C in Figure 3.5.1. The center lS circle in the '1"  a pIane, as shown by l T
B
which represents the region exterior to the circle C3 with center at (al + az)/2 and crossing the a axis at al and az. It follows therefore from (3.5.4), (3.5.5), and (3.5.6) that the admissib1e va1ues of'1" and a 1ie in the shaded region shown in Figure 3.5.1 bounded by the circles Cl' C z, and C3 • The maximum shearing stress, as is clear from the figure, is represented by the Iargest ordinate, AB, which is the radius of the circle Cz and is therefore equa1 to (al  a3)/2. To determine the orientation of the pIane that has this shearing stress, we make use of equation (3.5.2). The va1ue of u corresponding to the maximum shearing stress is equa1 to (al + a3)/2 (the center of the circle Cz). Substituting these values of'1" and a into equations (3.5.2), we get
We have thus obtained the same va1ues for the maximum shearing stress as in the previous section.
CT
36 STRESS DEVIATOR TENSOR FIGURE
3.5.1
Mohr's diagram.
. 2 and it crosses the a aX1S )/ obviously at O, (az + a3 , . (3 54) lies . d fi d by equatlOn ., Therefore the reglO n e n e . . t circles includes it as a boundary, smce 1t represen s
t  a and a = a3' a a  z.. d outside th1S c1rcle an with radii equai to or
greater than (az  a3)/2. . f (3 5 2) since az  a3 > O and 'd' the second equatlOn o .., Now conSl ermg b 1 ss than or equal to zero: < O, the numerator must e e az  al '1"z '1"z or
+ (a
+ (a
 a3)(a  al) ::; O
_ al
~ a3y ::;
(al; a
. t fami1y of circles with centers WhlCh represen s a
It is convenient in p1asticity theory to split the stress tensor into two parts, one called the spherical stress tensor and the other the stress deviator tensor. .The spherica1 stress tensor is the tensor whose elements are a m Oli. where U m is the mean stress, i.e.,
(3.6.1)
(3.6.2)
y
3
at (al
+
a3)/2 and
(3.6.2) it is apparent that a m is the same for all possib1e orientations axes; hence the name sphericai stress. A1so, since a m is the same in all
41
Sec. 37] Pure Shear
The StresS Tenso r [Ch. 3
where
40
directions, it can be considered to act as a hydrostatic stress. Now, it was ib1e shown in Section 2.5 that even very large hydrostatic pressure has a neglig effect on yielding and plastic fiow. Therefore, in plastic fiow considerations one can consider, if desired, the stress system obtained by subtracting the spheric state of stress from the actual state, rath" than working with the actual al state of stress. We therefore define a stress tensor called the stress
!(Ir + 312) = l'f(2Ir + 91112 + 2713 )
J2 = J3
(3.6.6)
fi One t" advantage of usin g t he stress deviator t . rs lllvanant of this tenso' l ensor lS now apparent Th h r " a ways zero Th' . e t e sum of the diagonal elements . . . lS can also be seen by taking The invariants J and J III equatlOn (3.6.3) or (3.6.4). 2 3 can, of course b . . components al}. For example, , e wntten III terms of the stress
deviator tensor SI} as follows:
(3.6.7)
(3.6.3) =
(3.6.9)
It is apparent that subtracting a constant normal stress in all directions will not change the principal directions. The principal directions are therefore the same for the deviator tensor as for the originaI stress tensor. In terms of r the principal stress es the deviator tenso is
Th'
.
J 2 
3 2 z'Toct
(3.6. IO)
between J d h u~ed1S relatlOnship to lend credence on p~:S7calt :
w111 be discussed later.
octahedral shear stress is sometimes g ounds to some plasticity theories, as
o PURE SHEAR
o o
o
r To obtain the invariants of the stress deviator tenso , replace S
S'
+ Vl in equation (3.3.9). This results in (3.
important state of stress is the one d . d~Slgnated as pure shear, or simple . If at a point in a body a set of = a _ O co or lllate axes can b f d y _ .a = : then the point is said t o ' e oun such that z state III a cyhndrical bar in to' . be III a state of pure shear. The i s the stress state rSlOn lS an example of pure shear. Another
(3.7.1)
The Stress Tensor [Ch. 3
Ch. 3]
14.
42
General References
Given the following stress tensor at a point:
It can readi1y be shown (4) that a necessary and sufficient condition for a
10,000 1,000 (  8,000
state of pure shear to exist is
(3.7.2)
and presented in this chapter are independent of the material properties and are therefore equally valid for bodies which behave elastically or plastically.
l=~ V3 15.
deu
16.
Given the tensor TI/eTici' (a) What is the order of this tensor? (b) lf i, j, and k range from 1 to 3, how many components does this tensor
2.
3.
have? (c) Write out the components. The generai Hooke's Iaw is given by
(a) Write out the equations this represents. (b) What is the order of this tensor? 4. Prove that every secondorder tensor au may in a unique way be decomposed into the sum of a symmetric tensor eu and a skewsymmetric tensor Pu· 5. Verify relations (3.4.4). 6. Verify equations (3.4.6), (3.4.7), (3.4.8), (3.4.10), and (3.4.11). 7. Discuss the Mohr diagram for the case a2 = a3 and determine the orientation of surface elements experiencing extreme shearing stresses. Consider also the case where al = a2 = a3' 8. Derive equations (3.6.7), (3.6.8), and (3.6.9). 9. Let ax = a = a = O, Txy = Ty" = Tzx = a. Calculate the octahedral shear z stress, J , Jy , and J , the principal stresses, the greatest shearing stress, and 1
2
3
the direction cosines for the greatest principal stress. 10. Show that subtracting a hydrostatic stress from a given stress state will not 11.
change the principal directions. Show that the second and third invariants of the stress deviator tensor can also be written
J2 J3
= !(S~ + S~ + S~) = t(Sr + S~ + S~)
12. Determine the invariants of the stress tensor and the octahedral stresses 13.
the case of uniaxial state of stress. The case of piane stress is partially characterized by a z = Txz = Tyz es = Determine the invariants of the stress tensor, the principal stress , maximum shear stresses, and the octahedral stresses for this case.
 8,000) 6,000 20,000
m 1
 V2
n
=
1
V
A circular cylinder 2 in. in diameteris sub' 6 a bending moment of 35,000 in lb an ~ecte~ t.o a tensile force of 30,000 lb Determine the stress invariants' the a moment of 50,000 in lb' stress, and the stress deviator prmclpal stresses, the maximum shea;
t~ns
Problems 1. Write out the expression for the work incremt:nt TU
1,000  6,000 6,000
Determine the stress invariants h :~esses, the maximum shear str~s~ ea~~~~ d~viator tensor, the principal ~ pane. upon which it acts and e normal and shear stresses actin~ on a pane wlth direction cosines
In conc1usion it should be emphasized that all the equations developed
dW =
43
~ ~wIstmg
~:._?Pil~ed . to. the two opposite sides of a
Show that if a uniform tension a square and a uniform compressio~ state of pure shear is obtained.
1
applied to the other two sides, a
References L
I. S. Sokolnikoff Mathemat' New York, 1956: p. 42. Ica l Theory oj Elasticity, 2nd ed., McGrawHill
2. 3. 4 .
Ibid., p. 17. ' O. Mohr, Abhandlungen aus dem G b' Wilhelm Ernst und Sohn B l' 1 e let del' technischen Mechanik 2nd d , er m, 914 p 192 ' e ., C E P . . p.earson, Theoretical ElastzClty, .. Harvard ,. . University Press, Cambn'd ge, 1959, 57.
GeneraI References Sokolnikoff, I . S., 1Y1athematical 'I Th New York, 1956. eory oj Elasticity, 2nd ed., McGrawHill ,
, S., and J. N. Goodier, Theory York, 1951.
01' 'J
Elastzclty, .. McGrawHill, New
... Sec. 41] Strain at a Point
CHAPTER
45
4 y
x
TRE STRA,lN TENSOR
FIGURE
4.1.1 Deformed body.
The components of displacement of the pomt . P o are
= Vo = Wo = Uo
When the relative position of any two points in a continuous body is changed the body is said to be deformed or strained. If the distance between every pair of points in a body remains constant during motions of the body, the body is called rigido The displacements of a rigid body consist of translations and rotations; translations and rotations are therefore ealled rigid body displacements. The analysis of strain eonsists of the study of the deformations of bodies whieh is essentially a geometrie problem and is unrelated to the material properties. The specifieation of strain at a point is therefore the same for elastie and for plastically deforming bodies.
41
x~ y~
Xo
 Yo
z~  Zo
The eomponents of displaeement of the pomt . P are
u = x'  x v = y'  y w = z'  z
We start the analysis of the strain at a point in a body by eonsidering body referred to an orthogonal set ofaxes as shown in Figure 4.1.1. vVl1"l'"'' two arbitrary neighboring points P o and P in the unstrained body. straining, the points move to P~ and P', respeetively. The eoordinates of and p' are xo, Yo, and Zo and of P, x, y, and Z. The eoordinates of x~, y~, and z~ and x', y', and z', respeetively. The veetor A deforms into veetor A'. The eomponents of A are A x , A y , and Az and the eomponents of
We assume that the displaeements are si l the eoordinates x y and z (A t Il n~ evalued eontinuous funetions " . e ua y as wlll b that alI their derivatives thr h . e apparent later, we have to therefore expand the d' 1 oug t e t.hlrd are also continuous.) We lSp aeements at P m a Taylor series about P o as :
= Uo
+ ox ou Ax + ou A OU oy y + oz Az
v = Vo
8v o + ox Ax + ~ A + 8v 8y Y oz Az
u
P~
w = Wo + ~w Ax + ow A + ow A x 8y Y oz z A~ = Ax 44
+ SAx
(4.1.2)
h
STRA.IN A. T A. POINT
are
(4.1.1)
(4.1.3)
The Strain Tenso r
[Ch. 4
neglected, since P is taken in the neighborhood
46
where higherorder terms are of P ' Making use of (4.1.1) and (4.1.2), o
8u A + 8u Az 8y y 8z
8u (x'  x)  (x~  xo) = 8x Ax
+
8v (y' _ y)  (y~  Yo) = 8x Ax
+ 8y y
(Z'  z)  (z~  zo)
8v A
+ 8v Az
8w
8w
(4.1.4)
8z
8w A
+= 8x A x + A 8y y 8z 
d
Now the change in the components of the vectors A an
z
is called the relative displacement tensor. In generaI, as can be seen, it is not symmetric. Nothing has been said so far as to whether the displacements of P o and P represent rigid body motions or not, or what part of them does. Since rigid body motions play no part in the analysis of strain, it is best to eliminate them from consideration at the start. This can be done as folIows. A rigid body motion, as mentioned earlier, is characterized by the fact that the length of aline joining any two points remains unchanged. Consider the vector A shown in Figure 4.1.1 and assume the vector A' to be obtained from .IL by rigid body displacements. Then
A2 = A ' 2 = (A x + 8Ax)2 + (A y + 8Ay)2 + (Az + 8Az)2 = A; + A~ + A; + 2(A x 8A x + A y 8A y + Az 8A.)
A' can be written
_ (x'  x)  (x~  xo)
XI 
) x') o  (x  Xo 
8A x = Ax  Ax I I _ ) = (y' _ y) (y~  Yo) A' A = (y  Yo)  (y Yo I SA y = Y Y ) _ (z'  z)  (zo  zo) SA = A~  Az = (Z'  z~)  (z  Zo z I

(
47
Sec. 41] Strain at a Point
= A2 (4.1.5)
+ 2(A x 8A x + A y 8Ay + Az 8Az)
where we have neglected higherorder terms in 8A i since we are considering only infinitesimal transformations. Therefore,
Substituting into (4.1.4) gives (4.1.8)
or (4.1.6)
But, from (4.1.6), (4.1.9) or
or, in tensor notation, simp1y
8A I = ul"A, . is called an affine transformation, ~nd if it The transformatlOn (4.1.6) d h' derivatives are small, then 1t be(;Oniles d that the UI an t e1r further assume ,r. Con (1) The tensor an injìnitesimal affine trans) orma l •
8Ul 8Ul 8Ul 8Xl 8X2 8xs UI,' =
8U2 8U2 8U2 8Xl 8X2 8xs 8us 8Xl
8us 8us 8X2 8xs
equation (4.1.9) must be true for alI values ofA x, Ay, and Az, a necessary sufficient conditiOll that the transformation (4.1.6) represent a rigid body is
8u 8x
8u 8y
8u 8z
8u = 8v = 8w = O 8x 8y 8z
8v 8x
8v 8y
8v 8z
8u 8v 8y = 8x
8w 8w 8w 8x 8y 8z
8u 8z
8w  8x
8v 8z
8w 8y
(4.1.10)
The Strain Tensor {Ch. 4
48
That is, for rigid body motion the tensor
UI,}
of equation (4.1.7) is skew
symmetric. Now every secondorder tensor can be decomposed into a symmetric tenso and a skewsymmetric tensor in one and only one way (see Problem 4, r 3). It follows therefore that if we decompose the tenso r UI.! into Chapter symmetric and skewsymmetric parts, the skewsymmetric part will represent ent rigid body motions, whereas the symmetric part will repres pure deforma
tion. Therefore, let UI.! = 1(U,.f
or
+ Uf.,) + 1(U'.f
U'.f =
where
Bu Bx Bli
=
.l.(BU 2 By .l.(BU 2 Bz
+ +
Bli
WIi
=
Bli
Bv By
l(BV BW) "2+Bz By
is called the strain tensor and
W!j
BW) Bx
+ BW) By
_~)
.l.(BU _ BW) 2 Bz Bx
Bx
(4.1.12) x
Bw Bz
FIGURE
4.2.1
Shear deformation .
axes, respectively. Hence if i ' dk and z directions, then A ~ l'A'an represent the unit vectors in the ve t d  l y and B = kB Af x, y, c ors are eformed into A' and 13' . h z· ter deformation, these Wlt components
.l.(BV _ BW) 2 Bz By
.l.(BW _ BV) 2 By Bz
1(BW _ BU) Bx Bz
+
.l.(BU 2 Bz
O
1(BV _ BU) Bx By
z
(4.1.11)
Bx
.l.(BU 2 By
O
To mterpret the meaning of the t . Figure 4.2.. 1 Consider two vectorss A ram andcomponent 13'11l1tlally .. BI!parallel = BZy, to use is made of the y and z
 Uf.,)
+ BV)
.l.(BV 2 Bz
BW) Bx
Thus Bx represents the extension h . 49 ~ vector originally parallel to th or c. ange m length per unit length of . '1ar mterpretations e x aXIs. By and Bz obviously have Slml . .
+ wli
.l.(BU 2 By BV) Bx
Sec. 42] PhySlca . l Interpretation of Strain Components
A' = i SA x _ B' = l. SBx
O
+ l'(A y + SA y ) + k SA + j SBy + k(Bz + SBz) z
Denote the angle between A' and 13' by product
(4.2.2)
e. Then from the definition ofthe
is called the rotation tensor. For cos
deformation therefore, equations (4.1.6) become
e =;:;; A' 13' =+ SA A .SB B x SBx + (A y + SAy) SB y + (B z + OBJ SA y
SA, = BliAf
y
z SA z
z
we have neglected the d (4.2.3) .. o e small. Therefore, pro ucts of the vect or mcrements, assuming
t b
42 PHYSICA.L INTERPRETA. TI0N
or
STRA.IN
_ A'· 13'
COMPONENTS The physical meaning of the components of the strain tensor (4.1.12) e1 readily be determined as follows. Assume the vector A to be parall14 to x axis so that A = Az = O and Ax = A. Then equation (4.1. ) y
immediately
SA x SA B11 = Bx = Ax = A 

A'B'
The Strain Tensor [Ch. 4 50
We have thus neglected all in~rements except draw the figure as shown in Figure 4.2.2. z
d an
z
and we can noW
8B
51
y I
1 R :r...:;.. 0
_"1
8By RI
8A
z
Sec. 43] Finite Deformations
0
1
  11 I 0,1 Il 
o
plll
y
FIGURE
k8Az
4.2.3 Shear deformation.
L::::::::::::~~~y
o FIGURE
4.2.2 Shear deformation.
4.3
N ow, from equation (4.1.14), 8A z
=
BZyAy
8B y
=
ByzBz
(4.2.5)
. t h e ng . ht angle between is the decrease m ex W ere ex h ) ( Aiso cos 8 = COS 7r /2 , .ii and B. Or cos 8 = sin ex ~ ex for small angles. From (4.2.4) and (4.2.5), a
= ByZ + BZy = 2ByZ
ase in the right angle between two vectors '1' y and z axes. AIso, from Figure Hence ByZ repres ents half the decre that were initially directed along the pOSi lVe 4.2.2 and equation (4.2.5),
FINITE DEFORMATIONS
In the previous discussion we assumed that the displacements and their derivatives were small; i.e., we considered only infinitesimal strains. If the strains become too large, it is obvious that the previous formulation can no longer be correct. In this case the strains are no longer linearly related to the derivatives ofthe displacement. Furthermore, the equilibrium equations must be satisfied in the deformed body and must therefore be written in terms of the deformed coordinates if these are considerably different from the undeformed ones. There are two methods of describing the deformation of a continuous body, when the deformations are large, the Lagrangian and the Eulerian. The Lagrangian method uses the initial coordinates of each particle to describe the deformation. The Eulerian method uses the coordinates ofthe particles in the deformed state to describe the deformation. We shall derive the elements of the Eulerian strain tensor using tensor notation for brevity. The power of this notation for derivation purposes, as was mentioned earlier, will be evident. Let the coordinates of two neighboring particles before deformation be al and al + dal' After deformation has taken pIace, let the coordinates of these particles be Xt and Xi + dXt. The initial distance between the particles is
8A z
L POP' ~ tan POP' = lI; =
BZy
8B y
L ROR ' ~ tan ROR' = JJ; =
Byz
(4.3.1)
the final distance between them is given by (4.3.2)
I I are e ual. If we noW rotate the q h . ' we obtain a figure such Thus angles POP and ROR l about t e ongm , R'OP'Q' through an ang e ByZ l'd' or shear ofthe elements . 4 2 3 This represents aSi mg that of Figure . . . of the sliding being proportional to the. H i s called a shear stram. to the xy pIane, the amount of the element from the xy pIane. ence ByZ derivations can be made for BXY and Bxz'
the Eulerian description we wish to describe everything in terms of the coordinates Xt. We therefore write
at = at(x!, X2, X3) dat = al,} dx} = ai,,, dx"
(4.3.3)
Sec. 44] Principal strains. Strain Invariants
53
The Strain Tensor [Ch.4
which reduces to El ~ Ex onIy if Ex « 1. Similarly, the decrease in the right angle between two elements finally directed aiong the y and zaxes is no Ionger 2EyZ but is given by
52
Therefore, We can also write Therefore,
(4.3.4)
Now the displacements are given by UI = XI  al al
=
XI  UI
or Hence ds 2
_
ds~
44
PRINCIPAL STRAINS. STRAIN INVARIANTS
= [Sjle  (Sii  UI,j)(SIIe  ul,le)1 dXj dXIe
'"
(I;>Djle _ Ule ,j  Uj ' le
= [Djle = (UIe,j
+ Uj,le
+ UI,jul,le)1 dXj dXIe
 UI,jUI,Ie) dXj dXIe
(4.3.5)
= 2Ejle dXj dXIe The tensor Ejle
= !(UIe,j + Uj,le
(4.3.6)
 UI,1UI,Ie)
. ical terms in engineering notation is called the Eulerian stram tensor. Some typ are EX
which again reduces to ex = 2EyZ onIy if Ey, Ez , and ex are small. In spite of the foregoing, however, it is possibie sometimes to treat probIems involving Iarge strains using the equations for infinitesimai strains. This is possibie ifwe do the probIem incrementally a small step at a time and after each step change the coordinates to correspond to the deformed body [4]. We are then solving, in essence, a successive series of small strain problems. We will elaborate on this concept more fully in the treatment of creep probIems.
=
~~ !((~~r + (:~r + (:f1 
OU
EXY =
! ( oy +
" ) uV OX
_
ou + OV OV + OW OW)
1. (OU __ 2
OX
(4.3.7)
oy
OX
oy
OX
uy
h t their products can be neglected, We see that ifthe derivatives are small, so t a . l obtained for the infinÌressions as prevlOuS y P these reduce to the same ex . l' t pretations given in Section 4.2 for tesimal strains. The simple physlca 111 elr. bI in this case Thus if El is the . h ver no longer app lca e . . l t finally parallel to the X axlS, small stra111S are, owe , change in length per unit length of an e emen it can be shown [21 that
In the study of the stress at a point we found that there exist at least three mutually orthogonal pianes which have no shear stress acting on them, Le., the principai planes. We now ask ourselves, do there exist in a similar fashion planes with no shoor strain? By this we mean pianes whose normals will not change orientation when the body is strained. Thus a vector A, which was originally normal to such a pIane, will either shorten or lengthen but will not change direction. The answer to this question is yes. Such pianes, as in the case of stress, are called principal planes, the normal directions to these pianes are the principal directions, and the corresponding strains are called principal strains. To find these directions and the corresponding strains we proceed as follows. Consider a vector A normai to the pIane ABC as shown in Figure 4.4.1. Upon straining it is assumed that A changes length by an amount SA but its direction remains the same; Le., SA is in the same direction as A. Since A and SA are parallel, the components of A and of SA are proportional; Le., (4.4.1) Now by definition the strain 8 in the direction of A is
SA
8=7
(4.4.2)
The Strain Tensor [Ch. 4
Sec. 45] Maximum. and Octahedral ShearStrains
55
The three homogeneous equations (4.4.5) will have a nonvanishing solution if, and only if, the determinant of the coefficients vanish. Thus
54
z
ByZ
=
O
(4.4.6)
~y
Equation (4.4.6) is of exactly the same form as (3.3.8) with stresses replaced by strains. All the remarks and derivations made there, therefore, apply here just as well. Thus (4.4.6) will have three real roots B1> B2' and B3, corresponding to the three principal strains. The invariants appearing in the cubie equation (4.4.7) X
FIGURE
4.4.1
Principal strain vector.
are I~
Hence from (4.4.1),
I~
(4.4.3)
= BxAx + BXyAy + BxzA z 8A y = BXyAX + ByA y + ByzAz 8A = BxzAx + ByzAy + BzAz z x
(4.4.4)
+ BxzA z = BXyAx + (By  B)A y + ByzAz = B  B)A = Bxz A x + Byz A Y + ( Z Z B)A x
+
BxyAy
and substituting into (4.1.14) gives immediately BAI = BliAj
which is the same as (4.4.5).
+
I{ =
BI
If =
(BIB2
B2
+ B3 + B2 B3 +
B3 Bl)
(4.4.9)
O O O
We note that if equations (4.4.5) are divided by A, then, since Ax/A = I, = m, and Az/A = n, A x , A y , and Az can be replaced in these equations by I, m, and n, respectively. By substituting into these equations the principal strains as obtained by solving the cubie (4.4.7), the principal directions can be obtained, just as was done for the stress tensor. Ay/A
(4.4.5)
. . th power of the tensor notation, equations (4.4.3) can To Illustrate agalll e be written
or
(4.4.8)
I~ = BIB2B3
Substituting (4.4.3) into (4.4.4) gives (BX 
+ By + Bz B;y + B~z + B;x  (BXBy + ByBZ + BzBx) BXByBZ + 2BxyBYZBZX  (BxB~z + ByB;X + BZB~y) Bx
or in terms of the principal strains
Writing out equation (4.1.14) we have 8A
I~
= = =
MAXIMUM AND OCTAHEDRAL SHEAR STRAINS Just as in the case of stress, there exists a direction at every point in a strained body for whieh the shear strain is a maximum. To find this direction we can proceed as follows. Let the coordinate axes be taken in the directions the principal shear strains and consider a vector A, as shown in Figure 4.5.1, having direction cosines I, m, and n with respect to these axes, whieh
Sec. 45] Maxirnum and Octahedral Shear Strains
57
The Strain Tensor [Ch. 4
Equation (4.5.4) gives the linear strain for a vector having direction cosines l, m, and n with respect to the principal strain directions, in terms of the principal strains. To determine the shear (J = RP/A, we have
56
3 Q
2
e
,lé==_2
FIGURE
4.5.1
Maximum shear vectors.
CI.' 11 strain so that the . . 1 2 d 3 Let the vector A SUller asma are deslgnated by , ,an ' . . com osed of two parts: a linear stram P h r of amount (J = RP/A, for point P moves to Q. The stram IS th~s _ RQ/A and a rotatlOn or s ea h of amount e  , small (J. To calculate the linear strain, e, we ave
+ 8A l )2 + (A 2 + 8A 2)2 + (A3 + 8A3)2 3 A 2 + 2A(RQ) ;;:; Ai + A~ + A~ + 2(A 1 8A l + A 2 8A 2 + A3 8A ) (A
+ RQ)2
= (Al
(4.5.1)
mall quantities have been neglected. Therefore, where squares of S
RQ
Al 8A l
A = e = A2 +
A 2 8A2 + A3 8A 3 2 A2 A
(4.5.2)
. t' (4114) which is given in expanded form in (4.4.4), From the baslc equa lOn ., , we have 8A l = elAl
(4.5.3)
8A 2 = e2 A 2 8A 3 = e3 A 3
+
(J2 = (PQ)2 = (8Al)2 A2 A
_ m and A3/A = n and sub,
(8A2)2 A
+
(8A3)2 A
or
(4.5.5)
or
(4.5.6)
Comparing equations (4.5.6) with (3.4.2), it is seen that they are of identical forms, with Ss replaced by (J and the principal stresses by the principal strains. The maximum shear strains and the corresponding directions can therefore be obtained in the exact same way as for the stresses. Thus designating the maximum shear strains by 'l'1> 'l'2, and 'l'3, we can write directly, by analogy to (3.4.4), 'l'l = ±!(e2  e3)
'l'2 = ±!(el  e3)
(4.5.7)
'l'3 = ±!(el  e2) and the same table of direction cosines is applicable as given on page 35. Thus the maximum shearing strain acts on a pIane bisecting the angle between the maximum and minimum principai strain directions and is equai to half the difference of these strains. If we consider the octahedral pianes for which l = m = n = ± 1/V3, we see from (4.5.6) that the shear strain on these planes, which we designate 'l'oct, is given by
+ e~)  i(el + e2 + e3)2 = M(el  e2)2 + (e2  e3)2 + (e3  el)2]
'l';ot = t(ei
+
e~
(4.5.8)
in exact analogy to the octahedral shear stress. In terms of the invariants of the strain tensor in analogy to (3.4.10) we have 'l';ct
Making use of the relations Al/A .= l, A 2/A stituting (4.5.3) into (4.5.2) results m
+
= %[I? +
3I~]
and in terms of nonprincipal strains this becomes
(4.5.9)
The Strain Tensor [Ch. 4 58
It is also apparent by the same analogy (comparing equations (4.5.4) and (4.5.5) with (3.5.1)) that a Mohr's diagram can be constructed for the strains
Sec. 47]
Compatibility of Strain
l . 59 neg ectIng products of the strains . The change In . vo lurne per unit volume is then
in an identical fashion to the stresses.
46 STR1UN DEVIATOR TENSOR As in the case of the stress tensor, the strain tensor can be separated into two parts, a spherical part ql} and deviator part eli' The spherical part is
The spherical strain tensor is thus ro o . deviator strain tensor then represe! p rtlOna~ to t~e volume change. The By analogy with the stress deviat~rat~:re .dl~tortlOna~ strain. invariants of the strain deviator tensor are SOl dlscussed In Section 3.6, the J{ = O
given by
J~
(4.6.1)
J~
= i(I? + 3I~) = i? (2I? + 9I{I~ + 27I~)
(4.6.4)
or where
e m
= !(81
+ e2 + e3)
is the mean strain. The deviator strain then
becomes
]
::: eyz 2e x 
ez 
ez 
ey
where
2e y 
Also it follows that
e xz
8 XY
3 =
em
ez 
e XY
3
e xz
e yz
ex 2e z 
ex 
ey
3
47
o
o
If we consider a rectangular parallelepiped with sides equal to a, b, and c, so that its initial volume is abe, then after straining its volume will be or
eli
ex = 
ey
ox
e XY
=
1
'2
(OU
OV)
oy + ox
STRAIN
defines six strain
OU
o o
o
or
COMPATIBILITY
The strain tensor placements; Le.,
or, in terms of the principal strains,
o
(4.6.6)
(4.6.2)
e yZ
_ l(OV
e yZ 
'2
OZ
components
OV
=
oy
OW)
+ oy
.
In
terms of three dis
ow oz
ez = 
O e xz
(4.7.1)
= t ( .!!. + OW) OZ
OX
It is obvious that if the displacement . of x, y, and z, then one can use . s are speclfied as continuous functions ponents uniquely. Let us n eq~datlOns (~.7.1) to compute the strain comd' ow consl er the Inverse pr bI . o em, 1.e., to calculate Isplacements U v and w in b d . several problems a:e ~ncounteread ~nYt~lv~n thelstrai~ c.omponents el}' Here . e rst p ace, It IS apparent that the
The Strain Tensor [Ch.4
60
solution will not be unique, for the strains represent pure deformation, whereas the displacements include rigid body motion which have no efIect on the strains. The problem can, however, be made unique by specifying the rigid body motion, i.e., specifying the displacement and rotation at some point in the body. However, a more difficult problem is encountered in calculating the displacements from the strains. Equations (4.7.1) are six equations for the three unknowns u, v, and w. It is evident, therefore, that these equations will not have a solution for any arbitrarily chosen strains, but that some restrictions must be placed on the strains in order that equations (4.7.1) have a solution. This can also be seen from the following physical considerations. Assume the body is divided into infinitesimal cubes. Let all these cubes be separated from each other and let each of the cubes be subjected to some arbitrary strains. It is obvious that if we noW try to put all the cubes together, we will, in generaI, no longer be able to fit all the cubes together the way they were before, to produce a continuous body. Between some of the cube boundaries there will be gaps; others will overlap. This shows that there must be some relationships between the strains at the different points of a body in order that the body remain continuous after straining, i.e., that the displacements be continuous functions of the coordinates. These relationships are called the compatibility relations, or, sometimes, the continuity relations. The compatibility equations are found in all standard texts on elasticity. They are derived, for example, in a particularly e1egant fashion in reference [3]. In engineering notation these equations are
Problems
Equations (4.7.2) are necessary and suffic' .. 61 components give singlevalued d' l lent condlt1Ons that the strain F' lSp acements for a ' l or a mult1ply connected regl'on h Slmp y connected region. , owever these d' . , c o n lt10ns are necessary but generalIy not sufficient. . Just as for the stress equations of Ch here that alI the relations present d ~Ptehr.3, lt should also be emphasized . l . e m t lS chapt pat 1'b'}' 1 lty relations (472) . d .. ,are m ependent of th er, mc . udmg the comtherefore hold for both elasticall . e mat~nal properties and y and plastlCally behavmg materi a1s.
Note on Shear Notation
The shear strains defined herein differ from . many authors by a factor of 1/2 T h ' the shear strams defined by shear strains by . us Tlmoshenko, for example, defines the
YXY
ou ov oy +OX
= 
ou
Yxz = 
OZ
ov YyZ
= OZ
ow
+OX ow
+ oy
The factor . lS . necessary for the te . 1/2 used h erem .. nsor defimt10n of strain as must b " e exerclsed m reading the e mt10n of shear strain is used.
by equation (4 1 12) C rexempltfied .. . are lterature to determine which d fi . . Problerns
1. Obtain the equat'lOns representing rigid body displ acements; l.e., . show that u
=
a
v = b W
=
C
+ WyZ
+ wzx + w",y
where a, b, and c are constants and
oe yZ oe zx ) o ( oeXY oz oz +OX +oy
f}2e z
= OXoy
These equations were first derived by SaintVenant.

wzy
 W"'z  wyx
The Strain Tensor [Ch. 4
62 2.
Let the rectangle of Figure 4.2.3 be a square. Show that the shear strain is equai to the extension of the diagonai of this square. 3. The dispiacement vector at a point in a body is given by il = C[i(lOx
+
3y)
+ j(3x +
2y)
+
References
63
GeneraI References Novozhilov, V. V., Theory oJ Elasticit P SokolnikofI, I. S., Mathematical Th ~,./r~am?~ Press, London, 1961. eO/yo E astlclty, McGrawHiII, New York, . 1956.
k(6z)]
Tlmoshenko, S., and J. N. Goodi er, Theory oJ Elasticity, McGrawHiII, New York, 1951. where C is a constant. Show that there is no rotation and compute the 4.
principai strains. Given the strain tensor
B'I
r=
0 005 .
l0.~04
0.004 0.001 O
o.L)
Determine: (a) The principai strains. (b) The direction cosines of the principai directions. (c) The Iargest shearing strain. (d) The octahedrai shear strain. 5. If a rectanguiar parallelepiped with initiai dimensions of 1 in. in the x direction, 2 in. in the y direction, and 3 in. in the z direction is strained to the condition of Problem 4, determine the finai dimensions. 6. Derive the Lagrangian strain tensor. 7. Show that in pIane polar coordinates the infinitesimai strains are given by
Br
Bu = Br
BrO
Bv = Br
1 Bv
BO
+
= r Be +
1 Bu
u
r
v
r Be  r
where u and v are the dispiacements in the radiaI and tangentiai directions, 8.
respectiveIy. Assuming axiai symmetry, derive the compatibility equation in pIane polar
coordinates. 9 Verify equations (4.5.9) and (4.5.10). 10. Verify equations (4.6.4) and (4.6.5).
References 1. L S. SokolnikofI, Mathematical Theory oJ Elasticity, McGrawHill, New York, 1956, p. 21. 2. Ibid., p. 31. 3. Ibid., p. 25. 4. W. Prager and P. G. Hodge, Theory oJ PerJectly Plastic Solids, Wiley, Nevi York, 1951, p. 118.
Sec. 51] Equations or Elasticity
CHAPTER
5
thermal expansion, T the temperature a b o v e ' 65 perature, G the shear modulus related to E some arbltrary reference temand ft by the wellknown relation
G=
E 2(1 + ft)
(5.1.2)
and El =
+ U y + Uz
Ux
= Il
(5.1.3)
Equation (5.1.1) can be solved for the st resses to glVe . (5.1.4)
where
ELASTIC STRESSSTRAIN RELATIONS
À
=
(5.1.5) ftE
(1 + ft)(l  2ft)
In engineering notation equations (5 .. 1 1) become 1
In the previous chapters the states of stress and of strain at a point in a body have been defìned and some of their properties discussed. Equations (3.2.2) give the conditions of equilibrium that must be satisfìed by the stresses at every point of a body, whereas equations (4.7.2) represent the compatibility conditions that must be satisfìed by the strains. The boundary conditions in terms of surface forces are given by equations (3.3.1). There remains to discuss the relations between the stresses and the strains at every point of the body. We shall discuss briefiy these re1ations for an elastic body, as a pre1ude to the generaI elastoplastic stressstrain re1ations to be subsequently
+ u z)] + exT

ft(u x
+ u z)] + exT
E [u z 
ft(u x
+ uy)] + exT
By
= E [uy
Bz
=
1 1
(5.1.6)
yz 
B_
zx 
51
ft(u y
= E [u x
B_
discussed.

Bx
1 2G
l+ft
'T'YZ
=
~ 'T'YZ
2G,'T'zx
=
~ 'T'zx
1
l+ft
EQUATIONS OF ELASTICITY
Hooke fìrst proposed a linear re1ation between stress and strain for a lo ad applied in one direction. The generalization of Hooke's law to three dimensions is given in Problem 3, Chapter 3. For an isotropic material this becomes, using the tensor subscript notation and inc1uding thermal strains,
It readi1y follows from (5.1.6) that
 2ft e = 1 EEl + 3exT
(5.1.7)
(5.1.8) where E is the elastic modulus, ft poisson's ratio, ex the coefficient of 64
Elastic stressstrain Relations [Ch. 5 66
where
Bm
and
are the mean strain and mean stress, respectively. Finally,
a m
Sec. 52]
El as t·lC Strain Energy Functions
where a is the coefficient Of l'mear thermal 67 . we O course have expanSlOn. From Hooke's law f
combining (5.1.8) and (5.1.1) results in (5.1.9)
where eli and Sii are the strain deviator and stress deviator tensors, respecrs tively. Thus the deviators of the stress and strain tenso are related to each other, in the elastic case, by the simple equation (5.1.9), whereas the spherical stress components are related to the spherical strain components by equation (5.1.8). It also follows that
J 1 = J1 = O
J~ =
(5.1.10)
and now maki ng use of assumption 3 th fìrst three of equations (5 1 6) h ' e total strains are as given by the the 1ast three of equations..(5 1,w 6) ereas N t Ho ok' es law and assumption 4 give to the extent that a may be a t'. o e that assumption 2 may be modifìed une lOn of temperature.
f'
4b2 J2 52
2
_
oct 
1
~ ~2
(5.1.11)
4G2 . oct
"
It should be noted that nothing in the foregoing discussion requires that B, G, /k, or a be constant throughout the body. They may, for example, be functions of temperature, so that if the bodyis not at a uniform temperature, these constants may have different values at different points in the body. It is probably worthwhile to brietly discuss the origin of the aT terms appearing in equations (5.1.1). We start by defìning the free thermal expansion of a material as that part of the expansion which is unintluenced by stress but is due to temperature rise alone. We also make the following four
assumptions: 1. The material is isotropic and, therefore, the free thermal expansion is the same in all directions. 2. The free thermal expansion is directly proportional to the ternulerature. 3. The principle of superposition of strains holds. 4. The thermal expansion does not intluence the shear strains.
Let us noW denote the strains due to temperature rise by primes and strains due to stress es by double primes. Then assumptions 1 and 2 us to write
B~
= B~ = B~ = aT
ELASTIC STRAIN ENERGY FUNCTIONS
If a body is deformed by the action of work on the body If the bod . . external forces, then the latter do . .' Y lS m equilibriu h g.oes ~nto kmetic energy, then this work i m so t .at none of this work tlOn m the body. The elastic str . s stored as stram energy of deforma am energy can be wntten . as follows [1]:
u = 1ajJBjJ
(5.2.1)
If we substitute into (52 .. 1) equatlOn . (5.1.4), there results U
=
GBIJBtJ
+ 2~ 82
_ 2G 2+ 3À aT8
(5.2.2)
Ma~ing use of the defìnitions of the in . devIator tensors, equation (522) b v~nants of the strain and strain .. can e wntten after some algebra1' c malllp. ulations as
U _ 2G + 3À /2 6 (Il  3aTI~)
+ 2GJ~
(5.2.3)
NowI'1 represents the spherical state of strain . . change, wh~reas J~ is an invariant of the ;h1~h lS pro~ortional to the a pure d1stortional strain E . eVlator stram tensor which as the sum of two energy ter~s q;~~~n (5.2.3) ~an therefore be con. rst term lS the energy involved
Elastic StressStrain Relations [Ch. 5
References
68
in changing the volume; the second term is the energy of distortion, which is designated by Ud and can be written Ud
= 2GJ~ = 3GY~ot
Ud
= 2G
and the piastic problem occurs in the stress" 69 occurs, the linear generalized H k' 1 straIn reIatlOns. If plastic flow 00 e s aw no Ion h Id " ger o sand the reiations between stress and strain become nonlInear In a . manner . Wh at happens to the reIatl'ons between stres dvery complIcated . occurs will be the subiect of Ch t s an straIn when plastic flow J ap ers 6 and 7.
or, from (5.1.10),
53 SOLUTION
or
1 J
3
2
2
= 4G T oot
(5.2.4)
Problerns Show how equation (5 1 4) can be ob . tamed from equation (5.1.1).
1. 2.
D~rive equation (5.1.9)'. .
3.
GlVen the following stress tensor at a point:
ELA.STIC PROBLEMS
To solve an elastic problem we have to find the stresses and the strains which will satisfy the previously derived equations. Thus the stresses must satisfy the equilibrium equations, (3.2.2) and (3.2.3), as well as the boundary conditions (3.3.5). The strains must satisfy the compatibility equations (4.7.2). Finally, the stresses must be related to the strains through the stressstrain relations (5.1.1) or their equivalent. The problem of finding a set of stresses and strains satisfying the above relations is known as thefirst boundaryvalue problem of elasticity. Alternatively, it is possible to reduce the above set of equations to three equations, called the Navier equations, involving only the displacement u! [2].
where V 2 is the Laplacian operator, À and Gare Lamé's constants previously defined, F! are the body forces per unit volume, and eis the invariant, defined in (5.1.5) and which can also be written
10,000 1,000  8,000
1,000  6,000 6,000
8,000 6,000 20,000
°
Assume the materiai behaves elasticall . 6 and Poisson's ratio of 3 D t . Y wlth an elastic modulus of 30 x 10 th .. e ermme the st . d . e mean strain assuming the tem . ram eVIator tensor. Calculate et =.7.5 X 10 6 in.fin.rF. perature nse at the point is 300 P and that 0
4. 5.
Venfy equations (5.1.10) and (5.1.11). . . Prove t . that the . axes of prmclpai stress coincid .h ,e Wlt the axes of principai s ram for an lsotropic materiai ob . 6. Prove that for the materiai of pey~g Hooke s Iaw. similar to Mohr's strain di ro em 5, Mohr's stress diagram will be 7 D . agram. • enve equation (5.2.3).
References 1. I. S. Sokolnikoff M th York, 1956, p. 84: a ematical Theory of Elasticity, McGrawHill 2. Ibid., p. 73. ' New
GeneraI References This last formulation is most convenient if the displacements are specified on the boundary rather than the forces. The problem is then known as the
second boundaryvalue problem of elasticity. From the previous derivations it is apparent that both the equilibrium equations and compatibility relations are independent of the properties of. the material under consideration. The equilibrium equations express the static equilibrium of an element of the body and the compatibility relations express the continuity of the body. These equations will therefore hold whether the body behaves elastically or whether plastic flow occurs. The same holds for the boundary conditions. The difference between the elastic problem
Hoffman, O ., and G . Sach s, Introduction to the Th McGrawHill, New York 1953 eory of Plasticity far Engineers Sokolnikoff I S M h .' . ' 1956. ' . ., at ematlcal Theory of Elasticity, McGrawHill, New York,
Sec. 62]
CHAPTER
6
EXAMPLES 01' MULTIAXIAL STRESS
62
.
. for a material in simpie tenslO n was In Chapter 2 the stressstram curve. . Id point at which the material that there eXlsts a yle . . h discussed and lt was s own. 11 In this case the stress is uniaxial and t~ls will begin to deform pIast1~a y. h t 'f there are several stresses actmg a point can readi1y be dete~mm.ed. Buwtw t 1 mbination of these stresses will t dlrectlOns? h a co . . d'a' at a point m lueren . 1 th t for a hydrostatic stress, 1.e., 1 . . ? W know for examp e, a cause Yleldmg. e.: ieldin does not occur even for very arge equal stresses in all dlrectlOns, Y . g 1 . test which can be performed th r example mvo vmg a . . values of stress. A s ano e . h' lled cylinder which lS bemg 'ffi 1t conslder a t mwa . without too much d 1 cu y: . . t d by a twisting moment T, and lS . 11 b a load P lS bemg tWlS e pu11ed axla y Y , . 6 1 1) by a pressure p. . pressurized interna11y (see FIgure '.' Il P d the torque T, it is posslble . h re p the aXlal pu , an 1 . By varymg t e pressu , h' h will also of course resu t m to get various combinations of stresses, w lC
p
FIGURE
70
6.1.1
Combined stresses in thinwalled cylinder.
or Yield Criteria
7l
different principal directions. The question here is: For what combinations of loads will the cylinder begin to yield plastically? Another simple example is the pIane stress problem of a thin rotating disk with or without temperature gradients. At every point of the disk (except possibly at the rim) there exists a state of biaxial stress. The question again arises: For which states ofbiaxial stress will the disk deform plastically? The criteria for deciding which combination of multiaxial stresses will cause yielding are called yield criteria. The first step of any plastic flow analysis is to decide on a yield criterion. The next step is to decide how to describe the behavior of the material after yielding has started. In this chapter we shall discuss the choice of a yie1d criterion.
CRITERIA FOR YIELDING
61
Examples
EXAMPLES 01' YIELD CRITERIA
Numerous criteri a have been proposed for the yielding of solids, going as far back as Coulomb in 1773. Many of these were originally suggested as criteria for faiIure of brittle materials and were later adopted as yield criteria for ductiIe materials. Some ofthe more common ones will be briefly discussed. Although some of these theories are no longer in use, they are inc1uded here both for their historic interest and to give the reader a feeling for the type of approach used in promulgating yield criteria.
Maximum Stress Theory, or Rankine Theory This theory assumes that yielding occurs when one of the principal stresses becomes equal to the yield stress in simple tension ao, or the yield stress in compression ao,c' Thus if al is the maximum principal stress and a2 is the minimum principal stress, yie1ding will occur in tension when al = ao and it will occur in compression when a2 = ao,c' For a material with the same yield in tension and compression, this criterion becomes
or
(6.2.1)
A simple plot illustrating this criterion for the case of biaxial stress with a3 = O is shown in Figure 6.2.1. The coordinates are the remaining principal stresses al and a2' Yielding occurs when the state of stress is on the boundary of the rectangle, for then one of the stresses is at the yield point in tension or compression. For example, consider a thinwalled cylinder subjected to an
Criteria for Yielding
[Ch. 6
Sec. 62]
Examples of Yield Criteria
72
73
CF2
CF2
CFo
2 1//
CF o,C
....1,/" CFo CF1
CFo,c
FIGURE FIGURE
6.2.1
6.2.2 Maximum strain theory.
Maximum stress theory.
increasing internaI pressure p. Let al be the circumferential stress and a2 the axial stress. Then al = 2a2' As the pressure is increased from zero, the stresses follow the dashed line of Figure 6.2.1, as shown, al always being equal to twice a2' At point 1 the cylinder is still elastic, neither stress having reached the value of ao. At point 2, al is equal to ao and yielding begins even though a2 is only !ao. This maximum stress criterion, however, shows very
Maximum Sbear Tbeory, or Tresca Criterio n
.This theory (sometimes called the Coulo w111 occur when the maximu h mb theory) assumes that yielding m s ear stress reaches th 1 s ear stress occurring und . l' e va ue of the maximum h. er Slmp e tenslOn Th . glVen by equation (3 4 4) d' . e maX1mum shear stress is . .. an 1S equal to h lf th d' maX1mum and minimum . . 1 a e 1fference between the . pnnC1pa stresses Fo . 1 Slllce a2 = a3 = the m . . r S1mp e tension, therefore . . ,aX1mum shear stress at . Id' 1 ' cntenon then asserts that yieldi '11 y1e 1S zao· The Tresca . d' . . ng Wl occur when a f SlX con 1tlOns lS reached: ny one o the following
°
poor agreement with experiment and is rarely used.
Maximum Strain Theory, or SaintVenant Theory
This theory assumes yielding will occur when the maximum value of the principal strain equals the value of the yield strain in simple tension (or compression), BO = ao/E. Thus if B1 is assumed to be the largest strain in absolute value, yielding will occur when
a2 
a3
a3 
al
= ± ao = ± ao
(6.2.4)
(6.2.2)
For the biaxial case with aO 3 ,we h ave or, for the biaxial case with
= al EB2 = a2 
EB1
a3 =
0,
= fLa1 =
fLa2
±ao ±ao
for for
la11;:;:: la21 la21;:;:: la11
A plot in the al a2 pIane showing the boundary at which yielding begins is shown in Figure 6.2.2. This theory also does not agree well with most experiments. It has, however, been used in the design of guns, since some experimental results on thickwalled cylinders are in agreement with this theory
al 
a2 = ao
if al > 0,
a2
if a2 >
al
>
if al >
a2
if al
a2 > a3' aO 2 , (6.2.10)
78
where
aO,e
. If . the yield stress in compresslO n . 1S
a
l
>
a3
>
a2' a2
=
° then ,
(6.2.11) If al > a 3' both negative, then a3
A plot in the
ala3
= ao,e
pIane is shown in Figure 6.2.5.
Sec.63]
Yield Surface. HaighWestergaard Stress Space
79
where atl and at2 are the tensile yields in the al and a2 directions, ael is the compressive yield in the al direction, and 'To is the shear yield strength.
63
YIELD SURFACE. HAIGHWESTERGAARD STRESS SPACE
In Section 6.2 we discussed several yield criteria and also plotted several twodimensional plots for biaxial stress cases, showing the curves at which yielding takes pIace. In the most generaI case, the yie1d criterion will depend on the complete state of stress at the point under consideration and will therefore be a function of the nine components of stress at the point. Since the stress tensor is symmetric, we can reduce this function to a function ofthe six independent components of the stress tensor. This yie1d criterion for a virgin material is then essentially the extension of the single yield point of the uniaxial tensile test to the sixcomponent stress tensor. For a materialloaded to the initial yield, it can be expressed by the relationship (6.3.1)
O"o,C
where K is a known function, or if desired, (6.3.2)
FIGURE
6.2, 5 InternaI friction theory.
ust the available list. t d by no means exha k The various yield cntena 1S e l' t d theories which attempt to ta e ther more comp 1ca e db l yield criterion propose y . Besides, there are o , into account anisotropy. For ~xhamd~ e, ant yields in tension and compresslOn . mat erials W1t 1uere [2] for anisotrop1C has the form (for biaxial stress)
. . r
ar + ef,
aelatl
'TO(ael
atl 2 (ael a
tl/ a t2) + a t2)1
'To
+ a~ + (ael
ala 2
Equation (6.3.1) represents a hypersurface in the sixdimensional stress space and any point on this surface represents a point at which yie1ding can begin. For example, for the simple tensile test at the yield point ao, the point ax = ao, a y = a z = T XY = T xz = 'TYZ = 0, must lie on this surface, and the point a x = a y = a z = T xz = 'TyZ = 0, T XY = 'To, which represents a thinwalled tube loaded in torsion to the torsional yield 'l'o, must also lie on this surface. The function appearing in (6.3.1) is called the yieldfunction and the surface described by (6.3.1) or (6.3.2) in the stress space is called the yield surface. Without specifying yet any particular form for this surface, the equation describing it can be simplified somewhat by making use of some of the previously discussed assumptions regarding the yielding of metals. If, as isotropy is assumed so that rotating the axes does not affect the yielding, can choose the principal axes for the coordinates, and then (6.3.1) can be

atl)al
+
aelatl (~

a ) a2 t2
=
(6.3.3)
Criteria Cor Yielding
[Ch. 6
. . or compresdrostatlc tenslOn . . ' lways assume d that hy . t that only the stress devla ors Furthermore, smce 1t lS a. Id' we can assume • TI do es not influence y1e mg, . ::ter into the yield function and wnte
80
(6.3.4)
Sec.63] Yield SurCace. HaighWestergaard Stress Space
81
We can Iearn a great deai about the yieid surface defìned by equation (6.3.3) from simpie geometric considerations. We introduce the (al' a2, aa)
coordinate system, which represents a stress space called the Haigh Westergaard stress space [3]. Every point in this space having coordinates al> a2, and aa is a possibie stress state. Consider aline ON as shown in Figure 6.3.1
f1(Sl> S2' Sa) = K . nd S can be written in terms of the invariants a Alternatively, smce Sl' S2' a J1> J2 , and J a, where (6.3.5)
we can write f(J 2 ,Ja)
(6.3.6)
=K
ptions the yield criterion has been Subject therefore to the above two assum . 'riants of the stress deviator . f the two nonzero mva reduced to a functlOn o . tric in the principal stresses, tensor. We note in passing that!(J2 , J~) lS sy~mme aterial all of the principal d' m an lsotrop1C . . which is to be expecte smce . . Id' Thus whatever yield functlOn lS l the same role m yJe mg. stresses must p ay . . the rincipal stresses. chosen it must be symmetnc m . dP ' t ' the Tresca maximum shear , .d ly used y1el cn ena, 62 . .' discussed in Section ., are The two most Wl e h Mises y1eld cntenon, . l t criterion and t e von M' riterion is by far the Slmp es one s ecifìc cases of (6.3.6). The von. lses c p . ted with equatlOn (6.3.6), that can be assOC1a (6.3.7)
FIGURE
6.3.1 HaighWestergaard stress space.
passing through the origin, and having equal angies with the coordinate axes. Then for every point on this line the stress state is one for which (6.3.10)
Thus every point on this line corresponds to a hydrostatic or sphericai stress state, the deviatoric stresses Sl = (2a1  a2  aa)/3, etc., being equai to zero. Furthermorè, if we consider any pIane perpendicular to ON, the equation of this pIane will be (6.3.11)
where l 2 • t;;;M . 300 b = V2/3 a3  V2f3 a2 sin 30°  v 2/3 al Slll
a 
=
1/V6 (2a3
(6.3.14)
 a2  al) FIGURE
(6.3.15)
6.3.5 von Mises circ1e and Tresca hexagon.
then be a circular cylinder whose axis is the line ON equally inclined to the stress axes. The yield locus for the Tresca maximum shear criterion is a regular hexagon inscribed in the von Mises circ1e, as shown in Figure 6.3.5. This can be proved as follows. Consider the sector between the a2 and al axes in Figure 6.3.5. For any stress state in this sector,
The Tresca yield condition for this sector is then a2  al = ao. This will be represented by aline parallel to the a3 axis, since it is independent of a3, the distance of this line from the a3 axis being equal to 1/V2 (a2  al) = 1/'',12 ao [by equation (6.3.14)]. In a similar manner, a straight line is obtained for each sector, thus forming the hexagon shown in the figure. The corners of hexagon will touch the von Mises circ1e, as can be seen again from the sector, since the radius to the corner is equal to 1T
FIGURE
piane
of al on 7T pIane. 6•3•4 Pro;ection J
r
= (l/V2)ao = VI a cos 30°
3
o
Criteria for Yielding [Ch. 6 . Id · c1'rcle In the stress space the Tresca y1e Ml s e s · . . h d' ofthe von . l' d Wh1Ch lS t era lUS " ' b d' the von M1ses cy m er, as surface is a regular hexagonal cylmder mscn e m
86
shown in Figure 6.3.6.
Sec.63] Yield Surface. HaighWestergaard Stress Space
87
It can now readily be shown that if the yield locus is assumed to be convex and one circumscribes the von Mises circle by a regular hexagon, then alI possible yield lo ci must lie between the two regular hexagons inscribed in, and circumscribing, the von Mises circle. By the locus being convex, we mean simply that any straight line in the 7T pIane may pierce the locus in at most two points. It will be proved later that the yield surface must indeed be convex; for the present we shall merely assume it. Let the point A on the as axis in Figure 6.3.7 lie on the yield locus. By the symmetry conditions previously discussed, the points Band C must also lie on the yield locus. The
o
<TI
FIGURE
6.3.6 Tresca and von Mises cylinders.
. . 6 3 5 that the maximum difference between It is noW eV1dent from F1 g.ure. . . t eoD and is equal to
the von Mises and Tresca Cf1tena occurs a
V~ ao  (ljV2)ao
=

~
 1 = 0.156
V3
(ljV2)ao
as previously mentioned. h ar stress states. This follows The line at e = 00 corresponds to pure s e from (6.3.15), for
e = o:
FIGURE
and the mean stress is a
m
= i(al
+ az + as) = !(al + az)
he stress components al' az, . bt t the mean stress fro m t ) 1( _ ) O which corresponds to a Therefore, 1f we su rac t t 1( a  az al" and as, we get a stress s a e ~ a~~h: ~ ~lane itself, since al + az + as = O, state of pure shear (see p. 4 ). F other pIane parallel to the O nd a =  az· or any .. f it follows that as = a l . d' ated with the addltlOn o a have a case of pure shear, as 1ll 1C , we l 7T pane, ( ) . f t of amount ! al + a2 • sphencal state o s re~s d to a uniaxial stress, since The Hne at e = 30 correspon s 2a3  az  al = 1 az  al
as = az M
.
f
00 et a stress state al  az, , .
If we subtract a hydrostatlc stress o az. we g
6.3.7 Bounds on yield loci.
curve CAB is a convex piecewise smooth curve passing through CAB and having the proper symmetries. Any other curve through these points passing inside CAB will obviously not be convex. CAB is therefore a lower bound for the yield loci. Now draw a horizontalline symmetric about the as axis intersecting the adjacent axes of symmetry at C' and B'. Then it folIows that no piecewise smooth curve passing through C'AB' can lie outside C'AB', since then it would not be convex. Thus C'AB' represents an upper bound to the yield locus. The rest of the upper and lower bounds can be constructed from symmetry and are shown as the two regular hexagons in Figure 6.3.7. Thus ìt has been shown that all conceivable yield loci satisfying the conditions of isotropy, equal yield in tension and compression, independence of hydrostress, and convexity must lie between two regular hexagons as shown; convex curves are, of course, admissible. The usual Tresca yield locus is rep'res,enlted by the inner hexagon, and the von Mises circle circumscribes the
Criteria for Yielding [Ch. 6 88
inner hexagon and is circumscribed by the outer one. If the von Mises yield surface is taken as a reference, then the maximum deviation of any admissible yield surface is about 15.5 per cent.
Sec. 64] Lode's Stress Paramet ero Verification • of Yield C't
By means of (6 .42) . criterion can be written . , th e von MIses al 
64 LODE'S STRESS P.AR.AMETER. EXPERIMENT.AL VERIFIC.ATION OF YIELD CRITERI.A The first investigation of a yield criterion was performed by Tresca in 1864, in which he measured the loads required to extrude metals through dies of various shapes. On the basis of these experiments he arrived at the maximum shear stress criterion, previously discussed. It is evident that according to the Tresca criterion the intermediate stress has no effect on yielding. The von Mises criterion, on the other hand, gives equal weight to all three principal stresses. The simplest and most common type test specimen used to check these criteria is the thinwalled tube shown in Figure 6.1.1. The first such experiments were run by Lode [5], who tested tubes of steel, copper, and nickel under various combinations oflongitudinal tension and internaI hydrostatic pressure. Lode devised a very sensitive method of differentiating between the Tresca and von Mises criteria by determining the effect of the intermediate principal stress on yielding. According to the Tresca criterion, if al ~
a2
~
a3
.
Th . rl erla 89 us If (] varies from 0° to 30° ' 1 1 f l d' ' ft Wl vary from O to l AlI O oa mg from pure shear to sim l '  . . combinations range Oto 1. p e tenslOn are therefore mcluded in the
the yield criterio n is given by
2
a3
+ ft2)1/2
(3
ao
(6.4.4)
The . crit . . h b h difference between the Tresca an d von MIses y ow much the right si de of (644) d'ftì ena IS t en determined (6.4.4) agrees with (641) Th' :. I ers from 1. For ft = l, equation . . . ... IS IS the case of sim I t . cntena agree as expected T h ' P e enSlOn, and the two . . e maxlmum differ . h ence In t e range O to l o vlOusly occurs when Il. = O h' h . . bh . ;;; r , W IC IS the case of p h IS t en 2jv 3, which agrees again with th . ure s ear. The difference . e prevlOUS resuIts. Lode ran a series of tests S 30°). A ft of l correspond~ot:e~:f t~el range of ~ from l to l ( 30° S O in Figure 6.4.1. It is seen that th ~xla ~ompresslOn. The resuIts are shown e ata avor the von Mises criterion even 1.30
I
• Steel 1.2 5f ° Copper + Nickel
o
1.2O 0"1 0"3
~Ib 1.1 5
~7
1.10
To account for the influence of the intermediate stress in the von Mises criterion, Lode introduced the parameter ft, called Lode' s stress parameter
1.05
(not to be confused with poisson's ratio, ft):
1.00
2
.J.uQ =.)3+ J1 .f\
b'l o
VI
V
Il
o~
:f:\ J~
~ 00
oo
i ~
•
••
l< o
o
~
0"10"3
ì a:o=1?\
1.0 0.8 0.6 0.4 0.2
O
0.2 0.4 0.6
"q
0.8
+
1.0
J1
ft=
=
FIGURE
!(al !(al 
a2 
+ a3) a3)
ft is thus the ratio of the difference between the intermediate stress and
average of the largest and smallest stresses to half the difference the largest and smallest stresses, and is therefore a measure of the effect of intermediate stress. Comparing (6.4.2) to (6.3.15) for the case al ~ a2 ~ we see that ft =
V3 tan (]
6.4.1
Lode's test results.
appreciable deviation occurS. These d . . . . to lack of isotropy of the material S eVlatlOns were parttally attriFIgure 6.4.2 in the a l . . ome of the results are also plotted 1 a 2 pane, smce a is zero ~ th of the von Mises circular cyl' d 3 • h or ese tests. The interm ellipse, and the intersection of th T er Wlt. the ala2 piane is obviously e resca cylInder is a h ( as shown in Figures 6.2.3 and 6 2 4 ' exagon no longer plot between the hexagon and th 11:" FIgure 6.4.2 shows that the elIipse. e e Ipse, aIthough generally closer to
Criteria for Yielding [Ch. 6
Sec. 64] Lode's Stress Parame tero Verlfìcation . of Yield C't
The
90
M' von Ises criterion, on the other hand , b ecomes
1'1
. erla
91
1.01'NJ
(6.4.6) Both (6.~.5) and (6.4.6) piot as elIipses in the XY 6.4.3. It IS seen in the figure that th d UXT pIa?e, as shown in Figure abIy better than the Tresca c't .e ata fit the von Mlses criterion considern enon, aIthough . t he von Mises criterion do s t' appreclable deviations from ome lmes occur . Fi gure 6.4.4. shows a replot of 1.0
o
o r,,__, __~__~1.0
0'/0'0
FIGURE 6.4.2 Lode's test resultS. In 1931 Tay10r and Quinney [6] published their classica1 experiments, which were intended to settle this questiono They used copper and steel tubings which were very near1y isotropic and tested them very carefully in combined tension and torsion. They concluded that the deviations from the von Mises criterion were real and could not be explained on the basis of lack of experimental accuracy or isotropy. Their results are shown in Figure 6.4.3. 0.6
.~L
_ _ _ _ _ _ _ _
_ _ _ _ __
1.0      _ _..J
FIGURE 6.4.4 Results of Taylor and Qumney. .
0.5r0.4
some of these resuIts in the
m the literature. 0.2 0.1
o
I
and Figure 6 4 5 h . ?bt'ame d by Ros and Eichinger [7]pane, Other . ... s ows slmilar resuIts UlU2
.
tests of slmtlar nature can be found
• Copper
X Aluminum o Mild steel
0.1
0.2
o
0.3
0.4
0.5 0.6
0.7
0.8
0.9
1.0
O'x/O'o
FIGURE 6.4.3 Results of Taylor and Quinney. The theoretical curves are obtained as follows. Using the coordinate of Figure 6.1.1, for tension and torsion loads only, the stresses are T ' all the others being zero. The Tresca criterio n then becomes
(lx
XY
U~ 2 4"
J
or
+ T 2XY
=
(lo
1.0100 8alf li
This constitutes loading. (2) F = K
(6.5.2)
This is called neutralloading.
criterion. (3) F = K
65 SUBSEQUENT YIELD SURF ACES. LOADING AND UNLOADING So far we have discussed the initial yield surface at which a material will fust start yielding. For a perfectIy piastic material, this yield surface remains fixed, as is seen in the uniaxial tensile test, where the stress after yielding remains constant at the yie1d stress [Figure 2.6.1(d)]. However, for a material that strain hardens the yield surface must change for continued straining beyond the initiai yield. We know that for the uniaxiai case, if a material is strained beyond the yield point to some point such as B' in Figure 2.1.2, the load removed so that the stress state moves to C', and then the lo ad is in~ creased again, yielding will not take pIace until the point B' is reached again. Thus the yieid point has been raised in the workhardened material. In same way the yie1d surface in the case of multiaxial stress must "move or in some way, at least at the point where yie1ding initially took pIace. In equation (6.3.1) we defined a yield function by the relation
such that whenever the function F became equal to the constant K, yieiding would begin. K then represented an initial yie1d surface in the stress space. We can now generalize this type of relation to subsequent yie1d surfaces. Mter yie1ding has occurred, K takes on a new value (or values), der)endìng·, on the strainhardening properties of the material. If the material is UIUU~\"'U· and then loaded again, additional yie1ding will not occur until the new of K is reached. The function F can then be looked upon as a loading !ù..""fi,r>n or loading surface, which represents the lo ad being applied, and the tUllctllOn K is a yield function, or strainhardening function, and will depend on complete previous stress and strain history of the material and its
dF = 8F da < O 8ali
ti
This constitutes unloading, d' Geometrically the conditions (652) the stress state is on the yieid "'" d' are rea 11y visualized. F = K means surlace, F > Om tb out" from the yieid surface and pi t' fl . eans e stress state is "moving as lC ow lS occurri dF stress state is "moving in" fro th .e1 ng. < O means the taking pIace. dF = O corresPo:ds te Ytlh d surface and unioading is therefore o e case of the str t . . Id the yIe surface and is caIIed neu traIload'mg For t . ess h s ate movmg on no piastic flow occurs. If F < K th . ~ s ram ardening materiai . ' e stress state lS an ei t' as lC one. For periiect1y p1astlc materiais piastic fl ow occurs ""lor F=K
dF= O
(6.5.3)
The case dF > O does not exist. . The above discussion is illustrated in Figul'e 6.5.1. Pomt P represents the
FIGURE
6.5.1
Stress increment vector for loading.
Criteria for Yielding [Ch. 6
Sec. 65]
Sub sequent Yield Surfaces. Loading and Unloading
95
94
existing state of stress Iying on the yieid surface. If the stress state changes to Q, so that the vector da points "outward" from the cylinder, then Ioading is taking pIace. If Q lies on the surface, we have neutraIIoading, and if Q is inside the surface, we have unioading. As Ioading continues, the point representing the new existing state of stress will again lie on the surface, which will move and/or change its shape correspondingly. A similar picture can be drawn in the 7T pIane. Assume the materiai obeys the von Mises criterion, so that the initiai yieid Iocus is a circle of radius vi ao in the 7T pIane. Suppose that straining takes pIace to some point a~ > ao and the materiai is then unioaded. If we noW assume that the materiai remains isotropic, just as it was originally, then the new yieid Iocus is a circle of radius vi a~, which is Iarger than, but concentric with, the originaI yieid circle. We have thus assumed that the materiai strain hardens isotropically, as shown for the uniaxiai case by the curve ABCFG of Figure 2.3.1. For isotropic hardening, therefore, the yieid cyIinder will expand with stress and strain history but will retain the same shape as initially. For a Tresca material, the subsequent yieid Ioci will be a series of concentric
30
20
10
10
 20
1===::::::=:::::::
25~~~~LL~~~LJ FIGURE
6.5.3 Initiai and subsequent yield l OCI..
reguiar hexagons. Ihis is shown in Figure 6.5.2.
FIGURE
6.5.2 Subsequent yield loci.
The assumption of isotropic hardening is the simpiest one, mathematically, to use. However, it does not take into account a Bauschinger effect. The Bauschinger effect would tend to reduce the size of the focus on one side as that on the other si de is increased. The yieid surface would thus change shape as the yieiding progresses. Experiments to verify this effect are described in reference (9). Severai tests were carried out with aluminum alloy tubes various ratios or torsion and tension to obtain an initiai yieid surface. unioading and loading again in a specified manner, subsequent yieid were obtained. The results plotted in the Ta pIane are shown in Figure 6.5 Without going into a detailed discussion of this data, it is evident that initiai von Mises yieid ellipse does not just grow symmetrically but that definite Bauschinger effect exists. To account for the Bauschinger effect, Prager (reference (14) of Chapter
introduced . 2 the kinematic mode!' d'Iscusse d for the one d' . ec lOn .6 and illustrated for th t ..  ImenslOnal case in S dt 1 h a case m Figures 2 6 2 d 2 6 mo e t e totai eiastic range is m .nt' .. an .. 3. In this ame d yieid surface without deformo .talTh co~stant by transiating the initiai mg I. e modells repre t d b .. . avmg the shape of the yieid !'. sen e y a ngld frame surlace, as shown for th T hF' Igure 6.5.4. The state of stress b ef ' . e resca criterio n in position of a pin free to move 'th. orhe yleid occurs IS represented by the Wl m t e frame As th . o t e rame, yielding occurs Th f . . e pm contacts the side . e rame IS assumed to b . f h. f ro atlOn and to be perfectly h e constramed against smoot ,so that only fo t can be transmitted to it As th . rces normai to the frame transiate in a direction' norma~ Pt mthPushes against the frame it causes it to . o e surface at the . t f pom o contact At corners, If the motion of th . th d' e pm engages both side th f . e Irection of the motion f t h ' s, e rame transiates in described by the modei l' e ~'d e pm. Depending on the materiai being , .. , ngl perfectly pl t' .. elastic perfectly plastic the t t f as IC, ngld strain hardening or , s a e o stress and the st t f ' ' . m the model in different W F a e o stram are repreays. or example for a ri 'd t . ,>U O "I a"l
(7.6.4)
But the increment da~1 produces no plastic flow. Therefore,
strain history. 2. The relation between infinitesimals of stress and plastic strain is linear; (7.6.5)
i.e.,
(7.6.2) AIso, da%1 has been taken proportional to the gradient of j; therefore,
Although equation (7.6.2) seems very reasonable, it shouId be noted that there appears to be no theoretical justification for it. It is pureIy an assumption. Although the Cljlcl may be functions of stress, strain, and history of loading, (7.6.2) implies that they are independent of the da/cl' From assumption 1 it follows that for piastic deformation to take pIace
(7.6.6) where a is a scaler > O and from (7.6.4), (7.6.5), and (7.6.6) it follows that 8j da"l = 8j da" = .8f a 8j O 8a"l 8a"I "I 8a"l 8a"l >
(7.6.3) or
Hence
and from the linearity assumption 2 it follows that the superposition principie may be applied to the stress and strain increments. Thus if da;i and da7i are two increments producing piastic strain increments, dei; and deii', then an increment dali = daii + da7i will produce an increment,· dei; + deii'. Now assume that for a given state of stress a"l' an increment of stress da"l producing plastic flow is imposed. This increment da"l can be decomposed into two parts da~1 and da%1 such that da~1 produces no plastic flow and da~1 is proportional to the gradient of j(alj)' Geometrically this means that the vector da"l is decomposed into a component tangent to j and a component perpendicuiar to j, as shown in Figure 7.6.1.
(7.6.7) E~ation (?6.7) proves that the proposed decomposition is possible
by
;~~~nncnegd(7;6.2) adnd (7.6.7) [realizing that da"l in (7.6.2) can be r~placed
"I a"l pro uces no plasi' fl ] defJ must be proportional to a, or IC ow we see that every component of .;)
(7.6.8) or, combining with (7.6.7) gives
(7.6.9) w~re gli depends in generaI on stress, strain, and history of loading
ow the second of conditions (7.6.1) can be written
FIGURE
7.6.1 Decomposition of stress increment vector.
. (7.6.10)
3 Plastic StressStrain Relations [Ch. 7
114
But da;i produces no plastic flow, so that the increment dali = Cda;! : da;i for any value of C, positive or negative, will produ~e the sa~e plashc mcrement defJ. We can therefore write the strainhardemng condltlOn as
(Cda;i
+ da7i)defJ :2:
T
Sec. 76] Generai Derivation of Plastic StressStrain Relations
where \d'A = G dJ,\ and we recognize the PrandtlReuss equations. For the Tresca :Yield condition, assuming it is known which is the maximum principal stress al and minimum principal stress a3, we have
f
(7.6.11)
O
.1 I d p must vanish; otherwise C cou1d be chosen (a 1arge negative ua/i eli ' number) so as to violate (7.6.11). Therefore,
, BU t
115
= !(al 
a3)
of = O oa2
of oa3
Then
del
= !d'A
de~ =
O
deà = !d'A But Hence Comparing with (7.6.5) it is seen that
gli
of
(7.6.12)
= G c;vali
tuting (7.6.12) into (7.6.9) gives
li
(7.6.13)
(7.6.14)
or
which is the generaI stressstrain relation consistent with the originaI assumptions. . . Id f t' let unc lOn, Let us take some specific examples. For the von Mlses yle
f = J2 = ![(al  a2)2 of =
~{al
 !(a2
It is worth noting one ot,her importan,t fact from the previous derivation. ,I Since dari defJ = O and " dari is tangent to the yield surface, it follows that 11 defJ is normal to the yield surface, for the above equation merely represents a h,., . dot product oftwo vectors. This can also be seen from (7.6.14), since defJ is equa1 to the gradient of ftimes a scalar. It also follows from the above that the Prandt1Reuss equations imply the use of the von Mises criterion. .i To summarize: Starting with the definition of work hardening and postu , lating the existence of a loading function and linearity between increments of stress and increments of strain, we can arrive at the generaI flow rule (7.6.14) for a strainhardening material. It can also be shown that the plastic strain increment vector must always be normal to the yield surface. The scalar G, which depends in generaI on the stress, strain, and history, must be determined from experiment, and its derivation will be discussed short1y.
'{',l' ,
where G is a scalar which may depend on stress, strain, and history. Substi
oal
which are the same as equations (7.4.3). We note that the form of the flow rule or plastic stressstrain relations associated with the Tresca criterion is entire1y different than that for the von Mises; thus each yield condition has an associated flow rule, as was pointed out in Section 7.4. This is sometimes ignored and, for example, the Tresca criterion has been used with the von Mises flow rule. There is, however, no theoretical justification for that type of assumption.
+ (a2
+ a3)]
 a3)2
+ (a3
 al)2]
Perfectly Plastic Material For this case the work done by an external agency which slowly applies and removes a set of stresses is zero over the cycIe, or
Therefore,
dali de" = O
(7.6.15)
Plastic StressStrain Relations [Ch. 7
116
It should be remarked that this equation is not the same as the second of (7.6.1) with the equality sign. In (7.6.1) the equality sign is used only when
defJ = O. For ideaI plasticity it is also assumed thatf(atJ) exists and is a function of stress only, and that plastic flow takes pIace without limit when f(alj) = K and the material behaves elastically when f(al') < K. For plastic flow,
Sec. 76]
or
Generai Derivation of Plastic StressStrain Relations
ae =
(~r,n
=
{6~ [(al
and for the uniaxial tensile test
n=2
 a2)2
a e 
c
+ (a2
 a3)2
+ (a3
117
 a1 )2]f,n
!' al' TherelOre,
= 1/3
therefore, 8f df= dal} 8atJ
=O
(7.6.16)
~::~:;~~:splgent.erallYkused. O.ne defines the effective strain in~reme::t;=
Comparing (7.6.15) and (7.6.16) it is seen that defJ = dÀ 8f
which agree~ .with the previous definition in equation (7.2.9). The defimtlOn of effective plastic strain, ep , is not quite as simple Th as lC wor per umt volume; i.e.,
(7.6.17) (7.6.19)
8al}
where dÀ is a scalar.
and since
Determination or tbe Function G. Effective Stress and Effective Strain
(7.6.20)
For (7.6.14) to be of any practical use, it must be related somehow to the experimental uniaxial stressstrain curve. What we are looking for is some function of the stresses, which might be called the effective stress, and some function ofthe strains or strain history, which might be called effective strain, so that results obtained by different loading programs can all be correlated by means of a single curve of effective stress versus effective strain. This curve should preferably be the uniaxial tensile curve. The definition of effective stress can be arrived at rather simply; since it should reduce to the stress in the uniaxial tension test, it is a quantity which will determine whether plastic flow takes pIace or not, and it must be a positively increasing function of the stress es during plastic flow. Now the loading function f(atJ) also, by definition, determines whether additional plastic flow takes pIace. It is also a positively increasing function as long as plastic flow takes pIace and, if unloading takes pIace, plastic flow is not resumed until the highest previous value ofjis exceeded. The loading function f(alf) must therefore be some constant times the effective stress to some power; i.e.,
For example, if we assume again
then
For example, if f
= J 2 , it can readily be shown that (7.6.21)
and, if f = al  a3 with a1 > a2 > a3 as 10r !' the Tresca criterion, then (7.6.22)
Equation (7.6.21) expanded becomes de p =
vi [(de~)2 + (de~)2 + (def)2
(7.6.23)
and, in terms of principal strain increments , de p
(7.6.18)
+ 2(de~y)2 + 2(de~z)2 + 2(de~J2]1'2
=
Vi [(def)2 + (de~)2 + (de~)2]1/2
=
V3 [(de l)2 + (de~)2 + del de~]1'2
2
.
P P where A the incompressibility condition de 1 + de 2 + de3p  O h as been used find :e~ond. ~ethod for a.rriving at (7.6.21) is sort of intuitive. One seeks t~ efìmtlOn of effectlve plastic strain increment which when integrated
Plastic StressStrain Relations [Ch. 7
118
. f t' f only The simplest combination of plastic strain increments a unc lOn o ae . ". ." . which is positive increasing and has the correct d1menslOn lS
lS
de p = C VdefJ defJ
Sec. 77] Incrementai and Deformation Theories
or
119
(7.6.29)
where a~ = daelde p is the slope of the uniaxial stressplastic strain curve at the current value of a e . As an example, for 1= J 2 , equation (7.6.29) gives
To make this definition agree for simple tension we must have
de~ = de p
=
C V(de~)2
+ ì(de~? + ì(deD 2 = C yt de~ (7.6.30)
Therefore,
C= de p = and, for I
v't VidefJ defJ
= J2 ,
(7.6.24)
so that the integrated effective strain is a function of effective stress only; i.e., ep
=
f =f de p
H(ae) da e
(7.6.25)
It should be noted here that the definiti~n ~7.6.21) for de p has been derived  J nly Drucker has shown that 1t lS reasonably correct for almost f or I  2 o . . d' f ot ~(J J) The second intuitive approach for definmg ep lS, o course, n
any J. 2, 3' based on any specific loading function. . ow in a position to determine the functlOn G. It should first be We are n . ial tensile r d that for the previous formulation to agree W1t. h t he umax . rea 1ze d Id ust be the slope of that curve (in the plastic range). Substlcurve, a e ep m tuting the basic equation defJ = G 881 di alJ
into (7.6.21) gives
\ Equations (7.6.30) constitute the flow rule (or plastic stressstrain relations) I CJssociated with the von Mises yield criterion. They are the wellknown Prandt1Reuss relations we obtained previously. If we replace the plastic strain increments in the above equations by total strain increments, the LévyMises relations are obtained which are valid only if the plastic strains are so large that the elastic strains can be neglected. As a final note, a generaI flow law such as (7.6.14) can also be obtained on the basis of a hypothesis that there exists a plastic potential (similar to the strain energy density function) which is a scalar function of stress, g(atj), from which the plastic strain increments can be obtained by partial differentiation with respect to the stresses. Thus
(7.6.31) where df3 is a nonnegative constant. The plastic potential g(ali) was first introduced by Melan [14]. By comparison with (7.6.14), it would appear that the plastic potential should play the same role as the yield function, and indeed Bland [15] has proved that they must be the same function, so that g in (7.6.31) can be replaced by I; (7.6.31) and (7.6.14) are then the same.
(7.6.26)
(7.6.27) or and the generaI plastic stressstrain relation becomes yt(8118atj) de p defJ = v(811 8amn)(8118amn)
(7.6.28)
77
INCREMENTAL AND DEFORMATION THEORIES
Equations such as (7.6.30) are called incrementai stressstrain relations because they relate the increments of plastic strain to the stress. To obtain the total plastic strain components, one must integrate these equations over the whole history ofloading. Hencky [16] proposed total stressstra in relations
Plastic StressStrain Relations [Ch. 7
120
whereby the total strain components are related to the current stress. Thus, instead of (7.6.30), one would have (7.7.1) The plastic strains then are functions of the current state of stress and are independent of the history of loading. Such theories are called t~!al or cfeformation theories in contrast to the incrementai or flow theories previously d~~cribed. This type of assumption great1y simplifies the problem; however, as was previously shown, the piastic strains cannot in generaI be independent of the loading path and deformation theories cannot generalIy be correct. There has often been a tendency therefore to ignore alI deformation theory as of little value. It can easiIy be shown, however, that for the case of proportional or radiaI loading, Le., if all the stresses are increasing in ratio, the incrementaI theory reduces to the deformation theory. For if atj = Ka?h where a?i is an arbitrary reference state of stress (nonzero) and K is a monotonically increasing function of time, then Sti = KS?l and a e = Ka~ and (7.6.30) becomes p
deiJ
Sec. 78]
Convexity of Yield Surface. Singular Points
ProbIems of plastic flow in thermally stressed disks and cylinders have been handied in this way and good results obtained using deformation theory . .On the other hand, it will subsequent1y be shown that with the present ';ldespread availability of highspeed computors, many simplifying assumptlOns heretofore made, inc1uding the use of deformation theories under doubtful conditions, are often unnecessary.
78
CONVEXITY OF YIELD SURFACE. SINGULAR POINTS
In Section 6.3 the statement was made that the yield surface was convex. A proof, as given in reference [13], will now be presented. Consider some state of stress a~ inside the loading surface, as shown in Figure 7.8.1. Let some
l,
er' jl
I
I
3de p SO
= 2 a oe
121
~,~ ....... 
tl
*.
eri!
which can be immediateIy integrated to give (7.7.2) FIGURE
so the plastic strain is a function only of the current state of stress and is independent of the loading path. Furthermore, it has been proposed by Budiansky [17] that there are ranges of loading paths other than proportionalioading for which the basic postulates of piasticity theory are satisfied by deformation theories. Budiansky's theory postulates the occurrence of corners or singuiar points on the successive yield surfaces and, although the existence of such singular points has as yet not been established experimentalIy, one cannot rule out the possibility of loading paths other than proportionalloading for which total plasticity theories may give satisfactory answers. From a practical viewpoint, there are a great many engineering problems where the loading path is not far from proportionalloading, provided one is careful when unloading occurs to separate the problem into separate parts, the loading parts, and the unloading parts.
7.8.1 Stress path produced by external agency.
externai agency add stresses along some arbitrary path inside the surface until a state of stress ai} is reached which is on the yield surface. Only eiastic changes have taken pIace so far. Now suppose the external agency to add a very smalI outward pointing stress increment dati which produces smalI plastic strain increments defJ, as welI as eiastic increments. The external agency then releases the dati and the state of stress is returned to a* aiong an I . ti e astlc path. The work done by the external agency over the cyc1e is (7.8.1) If the piastic strain coordinates are superimposed on the stress coordinates as in Figure 7.8.2, oW may be interpreted as the scalar product of the vecto;
Plastic StressStrain Relations (Ch. 7
122
Sec. 79] Plastic StrainTotal Strain Plasticity Relations
123
convex. On the other hand, if the surface is not convex, there exist some poln:ts ali and ajj such that the vector alJ  ajj forms an obtuse angle with the vector dali> as shown in Figure 7.8.4. This completes the convexity proof.
FIGURE
7.8.2 Stress and plastic strain increment vectors.
* and the vector defJ plus the scalar product of dalJ and defJ· Now, ali  ali .. . ' (7 6 l) from the strainhardening defimtlOn equatlOn ., , dali defJ ;:::: O or
IdaljlldefJl cos () ;:::: O
(7.8.2) FIGURE
7.8.4 Surface not convex, obtuse angles possible.
or That is the vectors dali and delj ma ke an acute angle with each other. In a similar'fashion, since the magnitude of ali  ajj can always be made larger than the magnitude of dali' it folIows that p
(ali  ajj)defJ ;:::: O or
laii  ajjlldefJl
CoS '" ;::::
Equation (7.6.14) iinplies that the yield surface has a unique gradient. It may happen, however, that the yieid surface has vertices or corners where the gradient is not defined. For exampIe, the Tresca hexagon has no unique normai at the corners, where two of the stress es are equa!. Such points are called singular points or singular yield conditions. Such points can be treated by introducing an auxiliary parameter as described in reference [18].
O
Hence (7.8.3) Thus t h e vect or ali  ali* makes an acute angle with the. vector defJ l for all choices of ajj. Therefore, alI points .ajj must lie on one SI de of ~ P ane pe~endicular to defJ, and, since defJ is normal to the yield sur~ace, thlS pIane :V111 ~e tangent to the yield surface. This must be true for alI pomts alj ~n the Yl~ld surface so that no vector alj  alj* can pass outside the surface mtersectmg b the surface twice, as shown in Figure 7.8.3. The surface must therefore ~
79
PLASTIC STRAINTOTAL STRAIN PLASTICITY RE LA TIONS
The PrandtIReuss equations relate the piastic strain increments to the stresses. We shall now derive a similar set of equations involving onIy strains. These equations enabie one to compute the piastic strain increments from the totai strains without recourse to the stresses. In effe et , they provide a simpie method for separating the totai strains into their eiastic and piastic components. The advantage of this formulation will become evident Iater when certain iterative methods for solving piasticity probIems are discussed. Assume some loading path to a given state of stress and totai piastic strains efJ· Let the Ioad be increased by a small amount, producing additionai piastic strains LlefJ. The totai strains can now be written (7.9.1)
FIGURE 7.8, 3 Convex surface, only acute angles possible.
where efj is the elastic component of the totai strain, efJ is the accumulated piastic strain up to (but not including) the current increment of Ioad, and
Plastic StressStrain Relations {Ch. 7 124
p
Sec. 79]
Plastic StrainTotal Strain Plasticity Relations
125
•
.' lastic strain due to the increment of load. eij 1S 1S the mcrement of: P' t be computed. Define modified total strains presumed to be known, ei} 1S o as follows: (7.9.2)
~efJ
or, in expanded form, A P ue x
~ep (2 ex' = 3

ey, 
e z')
~ep 3 (2 ey'

e z, 
,) ex
~ep (2 ez' = 3

, ex
ey')
eet
A P uey =
Then (7.9.3)
A P ue z
eet
eet
Subtracting the mean strain from the diagonal components of both sides of
(7.9.10)
equation (7.9.3) results in (7.9.4) .' d ' is the modified strain . here ee is the elastic stram dev1ator tensor an eij W. ti F H ke's law and the PrandtlReuss relatlOns, devlator tensor. rom 00 eeli

1 S =
2G
li
~ ~elPi 2G~À with
Hence
,  (1 + 2G ~)~ePI'
elJ 
~À
eet
given by (7.9.7) or alternatively by
(7.9.5)
1
(7.9.6)
(7.9.11) and the primed quantities are the modified total strains as given by equation (7.9.2). Equations (7.9.10) are equivalent to the Prandt1Reuss equations, The stresses do not appear in these equations and the increments of plastic strain can be computed from the total strains. Note that since they have been derived by use of the Prandt1Reuss equations, they implicit1y make use ofthe von Mises yield function. It should also be emphasized that the equivalent total strain defined by (7.9.7) is a purely mathematically defined quantity without any direct physical meaning, even in the uniaxial case. However, it can be related to the uniaxial stressstrain curve as follows: From equation (7.2.12),
We noW define an equivalent modified total strain by
so that, from (7.9.6),
and, from (7.9.5), (7.9
(7.9.12)
Plastic StressStrain Relations [Ch. 7
126
Sec. 710]
Complete StressStrain Relations. Summary
127
where higherorder terms in L\Bp have been neglected. Substituting into (7.9.14) and solving for L\Bp gives
Substituting this value for L\.:\ into equation (7.9.8) gives
(7.9.16) or
= L\B p +
2(1
+ IL)
3E
(7.9.13)
ae
Referring to the uniaxial stressstrain curve as shown in Figure 7.9.1, let
For linear strain hardening, equations (7.9.15) and (7.9.16) are obviouslyj exact. Equation (7.9.16) shows how Bet is related to L\Bp through the geometryl ofthe uniaxial stressstrain curve. We shall use this relationship subsequent1y together with equations (7.9.10) and (7.9.11) to solve specific problems. For want of a better name we will refer to equations (7.9.10) as the plastic straintotal strain equations. If one desires to use the total or deformation theory of plasticity, it can be shown [19] that it is only necessary to remove the primes and increment symbols from equations (7.9.9) through (7.9.11); Le.,
where
B
(7.9.17) FIGURE
7.9.1
Relation between Beh a e, and D.Bp •
L\a e be the increment in stress to which corresponds a plastic stra.in increment A d let e be the stress at the end of the incremento Then Bet lS the sum of
an a . 1 . l' d b y the plastic stra in increment plus the total elasti~ stram mu hp le 2/3(1 + IL)' Solving equation (7.9.13) for L\Bp results m llBp
(7.9.14) now readi1y be eliminated from equations (7.9.13) or (7.9.14). as .. 1 d b a le. Let the stress precedmg the mcrement of oa e e,ll,.' ' lO OWS. . b t glVes A eo Then expanding a a e == ae,! 1 + UU e in a Taylor senes a ou a e ,i1 approximately n e a ca ~ Il
a = a ._ + (dadB e) 1
e
e,I
p
il
L\B p
+ ...
(7.9.15)
710
COMPLETE STRESSSTRAIN RELATIONS. SUMMARY
In the previous sections the relations between the increments of plastic strain and the stress es at any instant were discussed in some detail. The fundamental problem in applying plasticity theory is to determine the total plastic strain as a function of the history of loading or history of stress. Suppose a body is loaded along some specified load path to some finalload condition. To calculate the plastic strains at this finalload condition it is theoretically necessary, in generaI, to integrate the infinitesimal plastic strain increments over the actualloading path. Although this can be done in relatively simple cases, it is usually more expeditious to assume the lo ad applied in small finite increments and calculate the finite increments of plastic strain
~"""""~f~,~Plastic StressStrain Relations [Ch. 7
128
for each of the load increments. An these increments of plastic strain are then added to give the total plastic strain. The integration is thus replaced by a
lI
Sec. 710]
Complete StressStrain Relations. Summary
129
Tresca yield criteria, the following relations previously derived will be used For the von Mises criterio n : .
summation. Let the totalloading path be divided into N increments of load. Assume that the plastic strains have been computed for the first i  l increments of lo ad and we now wish to compute them for the ith increment of load. The total strains at the end of the ith increment can be written with thermal
ae =
v'3:l; =
J2
Toct
= V!SljS!/
strains included, as (7.10.1)
(7.10.3)
llep where 0 = al! = a x + a y + a z. The first two terms on the right side of equation (7.10.1) represent the elastic part of the total strain, the third term is the thermal strain, the fourth term is the plastic strain accumulated in the first i  1 increments of load, and the fifth term is the plastic strain due to the ith increment of load. In expanded form these equations are
=
VtllefJ llefJ
= V:t[(lle~)2 + (lle~)2 + (lle~)2 + 2(lle~y)2 + 2(lle~z)2 + 2(lle:x)2]1/2 =
~3 [(lle~)2 + (lle~)2 + (lle~y)2 + (lle~z)2 + (lle:x)2 + lle~ lle~]1/2
v2
= 3 [(Il exp  llef)2 +
+ 6(lle~y)2 +
(lle~
 lleD2
6(lle~z)2
+ (Ile:
 lleD2
+ 6(lle:x)2]1/2
(7.10.4)
or (7.10.2) ll
eXY =
1 + f.L TXY + L: lle~y,1c + lle~y,! e 1c=1
eyZ =
1 + f.L e Tyz + L: lle~z,1c + lle~z,! 1c=1
ll
ezx =
1
+ f.L Tzx
e
(7.10.5)
ll
L: lle~z,1c + lle:x,! + 1c=1
In the above equations the sums are known and the problem is to calculate the plastic strain increments for the current or ith increment of load, and the corresponding stresses. To do this it is necessary to use one or another of the plastic stressstrain relations discussed in previous sections. A yield criterion must be chosen and the associated flow rule as given by equation (7.6.14). In particuIar, since we shall concern ourselves only with the von Mises and
.1tnd Ile; is .reiated to a e through the uniaxiai tensile stressstrain curve as shown In Flgures 7.3.2 or 7.9.1.
2 P1astic StressStrain Re1ations [Ch. 7
130
ll
L LlefJ.k =
eli  0l1em
k=l ll
eli = eli em
: eet
L LlefJ.k
k=l
= jeli = j(e x + ey + ez) = viieljeli =
~2 [(e~
_
e~)2 + (e~

e~)2 + (e~
+ 6(e~y)2 + 6(e~z)2 +
Complete StressStrain Re1ations. Summary
131
where ue.tl is the va1ue of the equiva1ent stress at the end of the (i  l)st increment of lo ad and (due/dep)t_l is the slope of the uniaxiai tensile curve repiotted as true stress versus true plastic strain. Equation (7.10.9) is exact for)inellJ:_strain hardening. The above reiations'are shown graphical1y in Figure 7.9.1. If the deformation or totai theory of plasticity is used, aH the above relations are valid if the Ll's are removed from all the previous equations and the primes are removed from equations (7.10.6) through (7.10.7). Equation (7.10.8) becomes
A1ternatively we define eli = eli 
Sec. 710]

e~)2
6(e~x)2]lf2
(7.10.10) (7.10.6)
Then
and by the use of (7.10.10), the uniaxial stressstrain curve can be repiotted as a curve of ep versus eet as shown in Figure 7.10.1. This curve can then be used instead of the originaI stressstrain curve. 24 ~1O3
....
or
~ 20
c
'o'+VI
/
16 /
+C
Q)
(3
> '3
cr
Q)
(7.10.7)
(3
/
12 /
8
.L
+
~
4
V 4 8 12 16 20 24x10 3 Equivolent plostic stroin, ep
O
FIGURE
7.10.1
Equivalent total strainequivalent plastic strain curve.
For the Tresca criterion,assume and
Ae e Ll. p, eh
al
>
a2
>
as.
Then
and u e are related to each other by (7.10.8)
(7.10.11) or, alternatively,
Furthermore, for smaH increments (7.10.9)
and the relation between
aT
and Llep is taken from the uniaxiai tensile curve.
2 References Plastic StressStrain Relations [Ch. 7
132
he stressstrain relations discussed in this chapter are just one of fo~r T of relations that must generally be satisfied in solving an The other three sets of relations are the same as for any elastIclty
elastopl~s~lC
~er~blem.
13.
14.
where SIj is the stress deviator tensor. Show that equations (7.9.7) and (7.9.11) are equivalent.
11. 12.
problem. These are
1 The equations of equilibrium of stresses. . The straindisplacement or compatibility relatlOns.
2:
133
Show that the PrandtlReuss relations imply that the principal axes of stress and of plastic strain illcrement coincide. Derive equations (7.2.12) and (7.2.13) using tensor notation only. Determine the equivalent stress aT for the Tresca criterion by means of equation (7.6.18). Assume al > a2 > a3. Aiso determine the effective plastic strain increment by the two methods described by equation (7.6.20) and what follows. Prove that
lO.
3. The boundary conditions. To obtain a complete solution we must find a set of stresses and s~rai~s which satisfy these four sets of relations. In the next. several chapters lt wlll be shown how these relations are adapted to speclfic problems an~ h~; solutions to these problems can be obtained. In all that foll?ws, as. m e . . ,lS assume d that the material is homogeneous, lsotroplc, and precedmg, lt strain hardens isotropically.
References 1.
Problems Show that the equivalent stress a e defi ne d by equation (7.2.9) can also be written a e = VJS/jSjj = VJ(Si + S~ + S5
1.
2.
Show that the equivalent plastic strain increment de p defined by equation (7.2.10) can also be written de p =
3. 4. 5. 6.
7.
V j.delJ
delJ
= V j.[(del? + (de~)2 + (de~)2]
3. 4.
5.
f (7 3 5) is valid for the PrandtlReuss relations. Show t h at equa lOn ., f (7 3 2) Show that equation (7.3.3) follows directly from equ~ lon .. , . 'th the . rel af lOns (7 ., 4 2) are mconslstent Wl Show that the stressstram
6.
defi~ition (7.~.1).
7.
Show that the following expressions for the effectIve p as lC are equivalent:
8. 9.
(7621) and (7622) making use of equations (7.6.20). Denve equatlOns .. . . . l f strain increment
de p = =
vi [(de~)2 + (de~)2 + (de~)2 + 2(de~y)2 + 2(de~z)2 + 2(de~x)2]1/2
lO.
~3 [(de~)2
11.
= V2 [(de~ 3
_
+ (den 2 + de~ de~ + (de~y)2 + (de~z)2 + (de~J2]1/2 de~)2 + (de~
(de~  deD 2 + 6(de~y)2 + 6(de~z)2 + 6(de~J2]1/2
 den 2 +
From the fact that the plastic strain increment :ector. is ~or~~l surf ace, prove that the PrandtlReuss equatlOns lmp y von Mises yield criterion. 9. Derive equation (7.6.30) from equation (7.6.28).
8.
2.
12.
13.
~s:h~r:~~ 14.
B. SaintVenant, Mémoire sur l'établissement des équations differentielles des mouvements intérieurs opérés dans les corps solides ductiles au dela des limites où l'élasticité pourrait les ramener à leur premier état, Compt. Rend., 70, 1870, pp. 473480. M. Lévy, Mémoire sur les équations géneralés des mouvements intérieurs des corps solides ductile au dela limites où l'élasticité pourrait les ramener à leur premier état, Compt. Rend., 70, 1870, pp. 13231325. R. von Mises, Mechanik del' festen Koerper in Plastisch deformablem Zustand, Goettinger Nachr. Math. Phys., Kl., 1913, pp. 582592. L. Prandtl, Spannungsverteilung in plastischen Koerpern, Proceedings 01 the 1st International Congress on Applied Mechanics, Dellt, Technische Boekhandel en Druckerij, J. Waltman, Jr., 1925, pp. 4354. E. Reuss, Beruecksichtigung der elastischen Formaenderungen in der Plastizitaetstheorie, Z. Angew. Math. Mech., lO, 1930, pp. 266274. R. Hill, The Mathematical Theory 01 Plasticity, Oxford Univo Press, London, 1950, p. 25. D. R. Bland, The Two Measures of WorkHardening, 9th International Congress 01 Applied Mechanics, Univo de Bruxelles, 1957, pp. 4550. H. Ford, Advanced Mechanics 01 Materials, Wiley, New York, 1963, p. 416. A. M. Wahl, Effect of Transient Period in Evaluating Rotating Disk Tests Under Creep Conditions, J. Basic Eng., 85, 1963, pp. 6670. W. Prager, Strain Hardening Under Combined Stress, J. Appl. Phys., 16, 1945, pp. 837840. I. Todhunter and K. Pearson, A History 01 the Theory 01 Elasticity and Strength 01 Materia/s, VoI. II, Part 1, Cambridge Univo Press, 1893, p. 166. D. C. Drucker, Some Implications of Work Hardening and IdeaI Plasticity, Quart. Appl. Math., 7, 1950, pp. 411418. D. C. Drucker, A More Fundamental Approach to Plastic StressStrain Relations, 1st U.S. Congress 01 Applied Mechanics, ASME, New York, 1952, pp. 487491. E. Melan, Zur Plastizitaet des raeumlichen Kontinuums, Ingr.Arch., 9, 1938, pp. 116126.
Plastic StressStrain Relations [Ch. 7
134
15. 16.
D. R. Bland, The Assoeiated Flow Rule of Plasticity, J. Mech. Phys. Solids, 6, 1957, pp. ZI~~r Theorie Plastiseher Deformationen und der hierdureh H. Z. Hene y, Z Angew Math Mech., im Material hervorgerufenen Naehspannungen,. . .
CHAPTER
8
4, 1924, pp. 323334. t of Deformation Theories of Plasticity, B. Budiansky, A Reassessmen J A I Mech 26 1959, pp. 259264. . . . pp Koiter "Str~ssStrain Relations, Uniqueness, and VanatlOnal Theo18. ~~~'for Ela~tiePlastie Materials with a Singular YieId Surfaee, Quart.
17.
19.
Appl. Mdat1h., 11, ~~\P~:~~~5~~aetieal Solution of Plastie Deformation A. Men esonan . . , h R R281959 Problems in ElastiePlastie Range, NASA Tec. ept.  , .
ELASTOPLASTIC PROBLEMS OF SPHERES AND CYLINDERS
GeneraI References k
D C StressStrain Relations in the Plastie Range, a Survey of Theory
Druea:~' Ex~eri~ent, Office ofNaval Research, Contract N7 onr358, NR041o32, Hi1l,D~~. ~~;OMathematical Theory of Plasticity, Oxford Univo Press, London,
1950·W ., an d p .B. Mellor , Plasticity for Mechanical Engineers, Van Nostrand, Johnson, l t"t Offiee Prineeton N.J., 1962. N hd' P M' StressStrain Relations in Plastieity and Thermop as ICI y, ag of\~";vaI'Researeh, Contraet Nonr222 (69), Tech. Rept. No.9, 1960.
81
GENERAL RELATIONS
Spheres and cylinders are widely used as pressure vessels, in the chemical industry, for example, as well as many other places. The loads involve high pressures and sometimes high temperatures and high temperature gradients. The elastic stress and strain distributions are relatively simple to obtain, particularly since the loading is usually reasonably symmetric. The solutions in the elastoplastic range, however, become complicated, and so simplifying assumptions of various types are made. These usually involve assuming the material to be incompressible in both the elastic and plastic ranges, and assuming it to be perfectly plastic in the plastic range. With these assumptions closedform solutions can be obtained. We shall first present some of these classical solutions. Subsequently it will be shown how these problems can be solved without the usual simplifying assumptions. For later use we record here the equilibrium, compatibility, straindisplacement, and stressstrain relations in spherical coordinates and polar coordinates assuming spherical and axial symmetry, respectively.
SphericaI Coordinates
80
The stresses are designated by a r and ao = = 8. The equilibrium equations reduce to
a
and the strains by
8
r
and
(8.1.1) 135
Sec. 81]
GeneraI Relations
Elastoplastic Problems of Spheres and Cylinders [Ch. 8
136
where Fr is the body force per unit volume. The strains are related to the displacements by
du
Br
= dr
Be
=
r = Bq,
U
where sgn stands for "the sign of." We note that if the plastic strains vary monotonically with the applied load, equation (8.1.9) can be integrated to give (8.1.10)
(8.1.2)
where u is the radiaI displacement. Combining both of equations (8.1.2) gives
137
Note also that the Tresca yie1d criterio n in this case coincides with the von Mises criterion.
the compatibility equation dee Be  Br _ O dr + r 
(8.1.3)
Because of symmetry the shear stresses and shear strains are zero as well as the tangential displacements. The stressstrain relations are
= 1 [(1 E
We assume axial symmetry and either pIane strain or pIane stress. The equilibrium equations then become (8.1.11)
(8.1.4) Be
PoIar Coordinafes
",)ae  ",a r]
where Fr is the body force per unit volume, The straindispIacement reIations and corresponding compatibility equation are
+ T + ee <X
p
(8.1.12)
wh ere epr and eeP are the total plastic strains. From the incompressibility condition it follows that (8.1.5)
which are the same as equations (8.1.2) and (8.1.3). The stressstrain reIations are given by
For the von Mises yield criterio n, the equivalent stress becomes (8.1.6) (8.1.13)
so that the yield criterion is (8.1.7) and the equivalent plastic strain increment is
(8.1.8) The Prandt1Reuss relations thereupon reduce to (8.1.9)
For the case of pIane stress, a z = O, and for the case of pIane strain Bz = O or ez = constant for generalized pIane strain. In both cases the shear stresses and strains are zero. The von Mises and Tresca criteria do not coincide in this case as they do for the case of spherical symmetry. The yield criteria and corresponding plasticity relations wilI be described subsequent1y as they are used. SeveraI examples wiIl now be discussed beginning with the case of a thick hollow sphere.
Elastoplastic Problems of Spheres and Cylinders [Ch. 8
138
82
Consider a sphere with inner radius a and outer radius b, subjected to an internaI pressure p and a radiaI temperature distribution T(r). It is obvious that complete symmetry about the center will exist so that the ra~ial and any two tangenti al directions will be principal direct~ons. E.quatlOns .(8.1:1) through (8.1.10) apply. We start by finding the elastlc solutlOn. Sub~tltutmg the stressstrain relations (8.1.4) (with the plastic strains set to zero) mto the compatibility equation (8.1.3) and making use of the equilibrium equation (8.1.1), the following solution for the stresses can readi1y be obtained:
r
2E = "31 sr aT r2 dr + 32 ( 1 lfLr a
a r
3
3"
)
Cl
2
C + r3
(8.2.1)
(8.2.2)
arCb) = O
resulting in C2
=
_pa 3
3E 1 Cl =   b3 lfL a3
Sb aT r
2
a
3
dr
+ 23 b3pa_ a3
From equations (8.2.5) the yield condition becomes
a
EaT T=='(1 
fL)ao
r
As a specific example, assume a temperature distribution resulting from an outward flow of heat due to an inner surface temperature of T o and outer surface temperature of zero. This steadystate temperature distribution will be given by
(8.2.3) T
For convenience the following dimensionless quantities are now introduced: b {3==.
139
We note that in the case of pressure loads only, the stress distribution is independent of Poisson's ratio [see equations (8.2.5) and (8.2.4)]. The assumption that is often made that the material is incompressible in the elastic range (fL = t), as weII as the plastic range, therefore leads to no error in the elastic stress distribution. In the case oftemperature loads, however, assuming fL = t instead of 0.3, for example, results in approximately a 30 per cent error in the elastic thermal stresses. The strains are not independent of Poisson's ratio even for the case of pressure loading. In what follows, the effect of Poisson's ratio is always taken into account. The conditions for the onset of yielding in the sphere can now be investigated. In terms of the dimensionless stresses defined in (8.2.4), the yield criterion (8.1.7) is written
where Cl and C2 are integration constants. Note that E and a have been assumed constant in obtaining the above solution. The constants Cl and C2 can be obtained using the boundary conditions arCa) = p
Hollow Sphere with Internai Pressure and Thermal Loading
The strains can be computed from (8.1.4) and the displacement from (8.1.2).
THICK HOLLOW SPHERE WITH INTERNAL PRESSURE AND THERMAL LOADING
a
Sec. 82]
or
T
1) = {3~(~ 1)  1 =
Toa
(~
b a r
p
(8.2.7)
p==. p
p=='
a
ao
Sr = a r ao
(8.2.4)
where
To
EaTo
= '(1;fL""""');ao
Se =  ae ao
where ao is the yield stress in uniaxial tension. Equation (8.2.1) can now be written in the dimensionless form as
(8.2.5)
Evaluating the integrals and substituting into the yield condition (8.2.6) results in
Consider first the case of pressure only. Then the yield condition becomes (8.2.9)
140
Elastoplastic Problems or Spheres and Cylinders
[Ch. 8
Yielding wiIl first occur at the smallest value of p, i.e., p = 1, and the di~en sionless pressure necessary to first cause yielding, the criticai pressure, wlll be P or1t =
2(P  l) 3{33
(8.2.10)
Sec. 83)
Hollow Sphere. Spread or Plastic Zone. Pressure Loading Only
so that assuming E = 31 X 10 6 , ao = 31,000 psi, IL = 0.3, and IX = 7.5 X lO  6 per °F, the temperature difference between the inner and outer surfaces for yielding to start is 130°F. For both pressure and temperature acting, equation (8.2.8) gives, for {3 = 2, 112P
A plot of the ratio of this criticaI pressure as a function of .the ratio of the outer to the inner radii {3 is shown in Figure 8.2.1. For a .glVen value o~ {3, yielding wiIl start at the inner surface at a pressu:e as glVe.n by equatlOn (8.2.10) or Figure 8.2.1. As P is increased, the plashc zone w1l1 spread from
+ (7p2
==========
0.4
FIGURE
8.2.1
= 0.75
compared to 0.583 for pressure alone. The effect of the temperature gradient in this case has been to retard the onset of yielding. A complete discussion of the effects of temperature and pressure on yielding is given in reference [1].
2/3 ~_ _ _ _ 0.6
1.5
 12)7'01 = 7p3
For a value of 7'0 = 0.4, which corresponds to only a 37°F temperature difference T o, yielding will first occur at p = 1 for a value of
P or1t Perit 0.8
141
2.0
f3 Variation of criticaI pressure with {3, pressure loading only.
the inner surface toward the outer surface. Note that as {3 ~pproaches i.nfi~ity 2. 3' so that if the pressure is equal to t the yleld stress, yleldmg is sure to take pIace, no matter what the dimensions of the sph~re. Considering the case of temperature only, equation (8.2.8) glVes
So far, only the start of yieiding has been considered. The spread of the piastic zone through the sphere is investigated next. The pressure problem and the temperature problem will be discussed separately in Sections 8.3, 8.4, and 8.5 under the assumption that the material is perfectly plastico The generaI solution for strainhardening materials un del' combined pressure and thermal gradient is presented in Section 8.6.
Porlt approaches
(8.2.11) For this case yielding will also first star t at the inner surface. Howeve~, if both pressure and temperature are present, yielding may start at any radms, depending on the relative values of P, 7'0, and {3.  2. Then for the case of internaI pressure alone, A s an example l et {3 equation (8.2.10) gives 7 Por1t = 12 = 0.583 For temperature alone, equation (8.2.11) gives 7'O,orlt = 1.4
83 HOLLOW SPHERE. SPREAD or PLASTIC ZONE. PRESSURE LOADING ONLY When only internaI pressure is acting, yieiding will begin at the inner surface at a pressure given by equation (8.2.10); Le.,
(8.3.1) As the pressure increases, the plastic zone will spread outward toward the outer surface. Let the radius to the end ofthe plastic zone be re' Since the material is assumed perfectly plastic, at every point in the plastic region the equivalent stress is equal to the yield stress and since for this case ao > a" (8.3.2)
142
Elastoplastic Problems or Spheres and Cylinders [Ch. 8
Sec. 83] Hol1ow Sphere. Spread or Plastic Zone. Pressure Loading Only
143
p
in the plastic region. Substituting into the equilibrium equation (8.1.1) gives
1.6
dSr
dp
2
=P
Sr = 2ln p
or
+C
Butat
Sr = p
p= 1
Therefore,
0.4
C =P
and
Sr
= 2ln p
So
=
 P
(2 In P
+ 1) 
}
P
p
< Pc 
o~~~~
(8.3.3)
1.0
Pc
Equations (8.3.3) give the stresses in the plastic region. Note that no stressstrain re1ation was needed to obtain these stresses. The problem is therefore called statically determinate. At the plastic zone boundary, i.e., at P = Pc, the radiaI stress is
FIGURE
8.3.1
Plastic zone radius versus applied pressure,
We can now consider the elastic part of the sphere as a new sphere with inner radius re and outer radius b, with an internaI pressure given by equation (8.3.4). Since at this new inner radius the sphere is just at the yield point, equation (8.3.1) must apply with f3 replaced by f3e = b/re, and Pcrit replaced
r
2In Pc + i (1  ~~) = 2ln ~ + ~ (1 _ r~) a 3 b
p =
3
(8.3.5)
Equation (8.3.5) gives the pressure required to cause the plastic zone to reach a radius re or, alternative1y, for a given internaI pressure p, equation (8.3.5) could be solved for the plastic zone radius re' A pIo t of the pressure versus the plastic zone radius is given in Figure 8.3.1 for f3 = 2. When re becomes equal to b, the sphere is completely plastico This will occur at a pressure [from (8.3.5)]
p = 2ln f3
(8.3.6)
2.
(8.3.7)
= 2ln b So = 1 + 21n!. b
by Sr.e. Hence
or
=
Sr = 2 In P  2 In f3
= 21n.e f3
2 f3e 3  1 2lnpe  P = 3~
f3
From (8.3.3) and (8.3.6) the stresses will be
(8.3.4)
Sr.e = 2ln Pc  P
__2.2 ~__
As a check, note that So  Sr = 1 for all r. So far th~ st.ress~s i~ the plastic part of the sphere have been computed. The str~ss d~stnbutlOn m the elastic part ofthe sphere can be readily obtained by ~onsldenng the elastic portion of the sphere to be a new sphere with inner radlU~ ~e, outer radius b, and with the pressure at the inner radius equal to the cntical . ( pressure for a sphere with these dimensions. The elastic solu t'lOn, equatlOns 8.~.5) (without temperature terms), can be used for this fictitious sphere replacmg a by re and f3 by f3e' Thus
or (8.3.8)
4
144
Elastoplastic Problems or Spheres and Cylinders [Ch.8
Equations (8.3.3) and (8.3.8) give the complete stress distribution in the sphere for given ratio of applied internaI pressure to yield strength, with the plastic zone boundary Pc given by equation (8.3.5). To calculate the strains and displacements in the sphere, the stressstrain relations and straindisplacement relations are used. For convenience we define, as was done for stresses in equations (8.2.4), "dimensionless" strains and displacements as folIows:
Sec.84] Hollow Sphere. Residual Stresses. Pressure Loading
145
acting at the inner radius re. From (8.3.11) the displacement at r> r is obtained by replacing f3 by f3e and P by rlre, resulting in  c U=
2[(1  2p,)p + __ 1+ p, f33] 
3
2
3f3e
(8.3.13)
p2
When the plastic zone reaches the outer radius b , P
= f3 ' c f3
 l ,an d
(8.3.14) (8.3.9)
To compute the strains before yielding begins, equations (8.2.5) (with temperature terms deleted) are substituted into the stressstrain relations (8.1.4), resulting in
The error in assuming p, = 1 is about 30 per cent , for p, = O.. 3 In alI the previous computations it was assumed that the dimensions of the s~here do not change as the pressure increases. This is, of course, not true, smce t.he sphere grows with increase in internaI pressure, the inner radius becommg a + ua and the outer radius b + Ub' A rigorous analysis would the:efore have to take into account the change of dimensions of the shell. ThlS can become particularly important for large strains.
(8.3.10)
84 HOLLOW SPHERE. RESIDUAL STRESSES. PRESSURE LOADING
and from the straindisplacement relation (8.1.2), (8.3.11)
When yielding starts P is equal P crit given by equation (8.2.10), and the displacement at the inner surface, P = 1, is
If ~he pressure is removed from the sphere discussed in Section 8.3 after plastIc flow has occurred over part of the sphere, residual stresses will result. To find the residual stresses it is necessary to superpose on the stress system due to the internaI pressure p and temperature T a completely elastic stress system due to a pressure p and temperature T. This will be correct as long as yieiding in reverse does not occur; i.e., the residual stresses are not lar~e ~nough to pro.duce yielding. To see this, consider two stress systems satIsfymg the folIowmg two sets of equations: , cf,a r
Note that if incompressibility had been assumed in the elastic region, p, = ·h the first term on the right of equation (8.3.12) disappears. For a Poisson's ratio of 0.3 and f3 of 2, the error in the displacement of the inner surface at the beginning of yieid would be about 7 per cento At the outer surface the error is 38 per cento As yielding progresses to some radius re, we can consider as before a sphere with inner radius re, outer radius b, and criticaI pressure 2 f3~  1
P cr1t
="3r
d r
+
2
,
,
ar 
ae
r
=0
(8.4.1 )
a;(a) = p a;(b) = O
p Elastoplastic Problems of Spheres and Cylinders
146
[Ch. 8
and 1/
dar
dr
Br"
+2
"
r
Sec. 84]
Hollow Sphere. Residual Stresses. Pressure Loading
147
Adding to the stresses given by (8.3.3) and by (8.3.8) gives for the residual stresses
Il
a r  ao = O
r
1 (" =E a r  2,."ao")  ~T ~
(8.4.2) or
s; = ~3
[3 In
p 
~ P orit
(1  ~)]
b+ 3
2 3
s~ = "3 a;(a) = p
p
In P  P
}
p
crit
1
(1 + 2 3)]
p :$ Pc
(8.4.5)
p
a;(b) = O
.
d
tem corresponds to the system of stresses in the sphere with The system correp . phere with temperature T and InternaI pressuref onds to the stress es In a s . s~P. If the two systems are added together, th~re is obt.aIne~ a system o stresses a r  a r + a "n etc ., satisfying the followmg equatlOns.
~:; ~:=:re s;sand internaI pressure P. 
doubleprime~
I
dar
+ 2 ar 
dr dBo
ao
S~
=
1~
P or1t
80
dr =  r Br
= ~ (a r 
BO
1 [(1 =E
area)
The superscript r is used in the above equations to indicate residual stresses. When P = l (at the inner surface), S; = O, as expected, and
=O
r Br 
(8.4.6)
2,."ao)
(8.4.3)
+ Br
 ,.,,)ao  ,."ar]
+ BOP
=O
arCb) = O
Thus the resultant system corresponds to the unloaded sphere with.per~anent plastic strains due to the fint system. If plastic flow occurs durmg t e u~ l d' the elastic doubleprimed system can no longer be added to t e O~~g:::i system, but it is necessary to solve another plastic flow problem for the new plastic strains. For the case of pressure loading only, the elastic stresses due to a pressure equal to p are, from (8.2.5), So" 
(8.4.4)
(8.4.7)
an~ since p ~ P or1t, a residual compressive stress resuIts. Upon reapplication of a pressure Iess than or equai to the originaI maximum, onIy eiastic strains will occur. The shell has thus been strengthened by the initial pressurization. If the material work hardens, an even greater strengthening can be achieved. In the above derivation it has been assumed that no piastic flow takes pIace during the unioading; i.e., there is no yieiding in compression due to the residual stresses. If such yieiding occurs, then not only is our assumption that the unioading is eiastic vioiated, but the situation may be dangerous with regard to the safety of the sphere. The maximum value of applied pressure P such that if the sphere is unioaded there will be no reversed plastic fiow is called the shakedown pressure, Ps' This pressure can be found as follows. For reversed yieIding the yieId criterion can be written S; 
S~ =
l
(8.4.8)
The maximum residuai stress will occur at P = l, where S; = O. From (8.4.7) it therefore follows that P s = 2Por1t
(8.4.9)
Elastoplastic Problems of Spheres and Cylinders
148
[Ch. 8
Ps
Sec. 85] Hollow Sphere. Thermal Loading Only
Yielding will first occur at r = a, (p = 1), and the criticaI temperature difference at which yielding will first start is given by
1.6
4/31::=:::::::=====
1.2
=
T
0.8
O.orit
1.5
2.5
2.0
3.0
3.5
f3 FIGURE 8.4.1 Variation or shakedown pressure with thickness ratio for hollow sphere with internaI pressure.
As long as the applied pressure is less than twice the criticaI pressure, the residual stresses will be elastico Making use of equation (8.3.1) the shakedown pressure can be written direct1y as a function of the thickness ratio f3:
2(f33  l) _ 2(f32 + f3 + l) f3(2f32  f3  l) f3(2f3 + l)
S8  Sr
p
Sr
= 0,
C
= O.
= 21n p + C
Hence
Sr = 21np } p::::: Pc S8 = 121n p
HOLLOW SPHERE. THERMAL LOADING ONLY
2.0 1.6 +
1.2
~
~~
0,8 0.4
2345678 FIGURE
8.5.1
TO.orlt
f3 as a function of f3.
(8.5.4)
Note that the stresses in the plastic region are independent ofthe temperature. The radius of the plastic zone, re, of course, depends on the temperature.
(8.5.1)
'ti.
(8.5.3)
2
and since Sr(1)
For the case of a temperature gradient as given by equation (8.2.7), yielding will occur as given by equation (8.2.11). Then
= 1
The equilibrium equation now becomes
or Figure 8.4.1 shows the shakedown pressure as a function of f3.
(8.5.2)
A plot of TO.crit versus f3 is shown in Figure 8.5.1. If TO exceeds TO.crit, the plastic zone will spread outward to some radius r . With~n th~s zo~e, i.e:, for r ::::: re, the yield criterion IS8  Srl = l will appl;. But SInce In thlS reglOn the tangential stress will be a large compressive stress and the radiaI stress will be a small compressive stress, the yield criterion can be written
(8.4.10)
85
149
FIGURE
8.5.2 Two plastic zones due to temperature gradient.
150
Elastoplastic Problems of Spheres and Cylinders
[Ch. 8
As the temperature is further increased, a second plastic zone unconnect~d to the first may start at a new radius, depending on the value of f3 as shown m Figure 8.5.2. This is due to the fact that since there are no exter~al forces acting on the sphere, the resultant force acting on any cr.oss sectlOn .must vanish. Thus the tangenti al stress vari es from compreSSlOn at the mner surface to tension at the outer surface. The inner surface will begin flowing plastically in compression, but if the temperature gradient is high eno~gh, the outer surface will start flowing plastically in tension, thus produclllg two plastic zones, one in tension and one in compression. A detailed discussion is given in reference [l].
Sec. 86]
Hollow Sphere of StrainHardening Material
151
(8.6.3)
(8.6.4) where (8.6.5) The boundary conditions used in deriving the above equations were Sr(1) = p
86 1I0LLOW SPIIERE MATERIAL
or
We now consider the generaI case of a hollow sphere of strainhardening material with both pressure and thermal loads. Equations (8.1.1) through (8.1.9) apply and in addition the assumption is made that the plastic str~i~s are varying monotonically, so that equation (8.1.10) may be used. If thIS IS not the case, equation (8.1.9) is used instead, and the calculation performed in steps or increments as described subsequently for more generaI types of problems. In addition, we use the dimensionless quantities defined ~y equations (8.2.4) and (8.3.9). The equilibrium compatibility and stressstralll relations are now written
(8.6.6)
Sr(f3) = O
STRAINHARDENING
For the elastic case equations (8.6.2) reduce to (8.2.5). For the case of a perfectly plastic material, the solution was given in the previous sections. We shall consider here only the case of a strainhardening materiaI. To obtain a complete solution to the problem, it is necessary to determine the plastic strain distribution €f through the sphere. This wiII, of course, depend on the stressstrain curve of the materiaI. The plastic strain distribution can be obtained in the following manner. The equivalent stress is related to the equivalent plastic strain through the stressstrain curve of the materiaI. Thus or, for this case,
dEe +EepEr _ dp Er Ee
O
= Sr  2/hSe + (1  /h)r + €f = (l  /h)Se  /hSr + (l  /h)r +
(8.6.1)
where f is a known function representing the stressstrain curve. It therefore follows that
ISI2
€~
ISI
Substituting the last two of equations (8.6.1) into the second, combining with the first, and integrating results in the following equations:
l
(8.6.7)
:s; l
Also, from equation (8.6.3),
(8.6.8) (8.6.2) A complete solution can now be obtained by an iterative or successive approximation method. One chooses a distribution of €f (say zero). S is computed using equation (8.6.3), merely to determine hs sign at the different
Elastoplastic Problems oi Spheres and Cylinders [Ch.8
152
radiaI positions. A first approximation to S is then obtained from equation (8.6.7), and a first approximation to the plastic strain distribution can be calculated from (8.6.8). A better value of S can then be computed from (8.6.7) and the next approximation for E~ obtained from (8.6.8). If the process converges, we will thus obtain the proper values of E~ and S such that (8.6.7) and (8.6.8) are satisfied simultaneously. The individuaI stresses Sr and So can then readily be computed. Thus a complete solution is obtained in both the elastic and plastic regions. There is no need to treat the two regions separately as was done for the perfect1y plastic material in previous sections. It should be noted, however, that equations (8.6.7) and (8.6.8) apply only for ISI > 1; for ISI:::; 1, E~ is set equal to zero. The integrands Efj p appearing in the previous equations are therefore generally zero over part of the integration range. A similar technique is described in reference [9]. As a specific example, consider a sphere made of a material whose stressstrain curve is given by the following equation: ae = 30,000
+
l36,000e~/2
ae ;::: 30,000
Sec. 86]
Hollow Sphere oi StrainHardening Material
153
1.8
2.0
P FIGURE
8.6.2 Variation of plastic stral'n wl'th t'a d'lUS fol' various pressures: 13 = 2, To = O.
was used to perform the integrations. The cases shown are for illustrative ~urposes only. Any combi~ation ~f geometry, loading, and material proper
Equation (8.6.7) now becomes S
= I~I
(1
+ 0.1434Iefll/2)
Results of calculations performed by the iterative procedure described are shown in Figures 8.6.1 and 8.6.2 for 13 = 2 and temperature distribution given by equation (8.2.7). In performing the calculations, the thickness of the sphere was divided into 40 equally spaced intervals and Simpson's rule
tI es can be used .and a rapI? solutlOn obtained. The time required to obtain a complete. solutlOn for a glven loading condition, using a highspeed digitaI computer, lS on the order of a few seconds. This .type of successive approximation method will be discussed at greater lengt~ m. Chapter 9, wher~ several nu~erical examples will be given. Right now 1t wl11 be shown that lf the matenal strain hardens linearly the solution can for some cases be obtained in closed formo For linear strain hardening it follows from Figure 8.6.3 that 1 m
Ep
= ,y; CI SI
 1)
(8.6.9)
2.0
Pc
where t~e strainh.ardening parameter m is defined as the ratio of the slope of the stramhardenmg part of the stressstra in curve to the elastic modulus Then, from (8.1.10), .
1.5
(8.6.10) 1.0 L~:::::1~:::"L~7~~:;': 0.5 P FIGURE 8.6.1 Variation of plastic zone radius with applied pressure for different temperature gradients: 13 = 2, T = ToCf3/p  1)/(13  1), ae = 30,000 + 136,000 ep l/2 •
C?nsider the case of pressure loading only. ao will always be positive and > O. Therefore, S/I SI = +1 and
ar wl11 always be negative, so that ao  a r p Er
1 m
= f i l (l  S)
(8.6.11)
154
Elastoplastic Problems of Spheres and Cylinders
[Ch. 8
"e
Sec. 86]
Hollow Sphere of StrainHardening Material
155
which is the same result previously obtained for the perfectly plastic material. Obviously the onset ofyield depends onIy on the yie1d stress. As P is increased, the plastic zone spreads to Pc and Sr.c can be considered to be the criticaI pressure acting on a sphere with inner radius Pc and outer radius (3c = (3/ Pc' Thus (8.6.18)
Hence 2 3_ (33. [(  (33  1 1
'3 Pc
1  m) 1  m I  m (3~  1]  p,)m P  (1 _ p,)m In Pc  3(1  p,)m ~
+ 2(1
(8.6.19)
or e
~
FIGURE
P
Stressstrain curve for linear strain hardening.
8.6.3
= 4(1  p,)m[((33  1)/(33] p~ + 2(1  m) In Pc + t(1  m)((3~  1)/(3~ 2m(1  p,)
+ (l 
m)
If the plastic zone extends to p = Pc, then, from (8.6.11), making use of the first of equations (8.6.1), it follows that
(8.6.20)
which relates the pressure P to the plastic zone radius Pc' Note that if m = O, l  m 1 m   I n P   2  (Sr
Il
+ P)
m m ...!..dp = {1  ml P _ 1  m (S m n Pc 2m r.c
P"P
+ P)
P
~
Pc
(3/~
1
[p  (/ ~ p,~m (In Pc 
l  m S = 2(1 _ p,)m (1  S)
!Sr.c 
!P)]
3
+ 2p3 Cl
(8.6.13)
(8.6.14)
which is the value previously obtained for the perfectly plastic material [equation (8.3.5)]. As an example, for (3 = 2, m = 0.1, and p, = 0.3, the pressure required for yielding of the complete sphere, Pc = (3, is 1.83, compared to l.39 for a perfectly plastic material. It thus takes a 32 per cent higher pressure for the strainhardening sphere to yield completely as compared to the perfectly plastic sphere. To obtain the stresses we substitute into equations (8.6.2). Thus
and, since S = 1 when P = Pc, we have (8.6.15)
Substituting into (8.6.13) gives t p~
= (33 (33_
2 (3~  1
= 2lnpc + '3~
P;::: Pc
where S r.c is the value of Sr at P = Pc' Substituting into (8.6.3) and (8.6.4) resu1ts in Cl =
P
(8.6.12)
[ 1 m ] 1 P  2(1 _ p,)m (2 In Pc  Sr.c  P)
Sr
=
So
=
(l  :3) Cl + l ~ p, [l ~ m In Pc  l ;."mm (Sr.c + P)] } P + (l + _1 ) Cl + _1_ [1  m lnp _ 1  m(S + P)] 2p3 p, m 2m r.c P
+
1 
p;::: Pc
c
(8.6.21) (8.6.16) Sr
=
P
+
2(1  m) In P lm + 2(1  p,)m
So
=
P
+
2(1  m) In P 1  m + 2(1  p,)m
At the onset of yield when Pc = 1, we have Sr.c = P and therefore (8.6.17)
 p,)m(p~/p3)(p3  1) 1  m + 2(1  p,)m
+ 4(1
+ .2. (l 3
}
 p,)m(p~/p3)(2p3 + 1) 1  m + 2(1  p,)m
P < Pc 
(8.6.22)
156
Elastoplastic Problems ol Spheres and Cylinders [Ch. 8
Note that if m = O these reduce to the previously obtained values for the perfect1y plastic material. Thus to obtain the complete stress distribution, Pc (or P) is obtained from (8.6.20), Sr.c from (8.6.18), and then the stresses from (8.6.21) and (8.6.22).
Sec.87] Plastic Flow in ThickWalled Tubes
157
Equations (8.1.11), (8.1.12), and (8.1.13) are now written dSr So  Sr dp =  p 
(8.7.2)
87
PLASTIC FLOW IN THICKWALLED TUBES
A considerable amount of work has been done on the problem of plastic flow in a thickwalled tube under internaI pressure with and without temperature gradients. Solutions have been obtained, for example, in references [1] through [8]. These solutions differ in the yield criteria used and in the plastic stressstrain relations. Some solutions use the von Mises yield criterio n and the associated flow rule [3]. Others use the Tresca criterio n and its flow rule [7]. Reference [4] uses the Hencky total strain relations. In other papers complete incompressibility is assumed in both the elastic and piastic regions. Of the references cited, only [8] takes into account strain hardening of the material. There are three cases that can be treated: (1) pIane strain, e z = O; (2) generalized pIane strain, e z = constant =I O; and (3) tube with open ends, P = O. We shall present a generaI solution for a strainhardening material, inc1uding radiaI temperature gradients, which can take into account any of these cases. The Tresca criterion and its associated flow rule will be used, since in this case it offers some simplifications. For this purpose it will be assumed that ao > az > ar' It is shown by Koiter [7] that this is true for a large range of conditions. We introduce the same dimensionless quantities as in the problem of the sphere; Le.,
EaT T
= (1  {L)ao
EaTo
y
= (1 + {L)ao
r
p=
a
b (3=a
Er EO
Ez
+ Sz) + (1  {L) T + = So  {L(Sr + Sz) + (1  {L) T + = Sz  {L(Sr + So) + (1  {L) T + = Sr  {L(So
= So  Sr
where a and bare the internaI and externai radii, ao is the yield stress, and eo is the yield strain.
E:
(8.7.3)
E:
If the Tresca criterion and its associated flow rule are used, then, assuming So > Sz > S"
(8.7.4) and
S
=
So  Sr
=
1
at yielding
For boundary conditions it is assumed that S,(a) = P S,(b)
=
O
(8.7.5)
and the conditions at the end of the tube are determined by case l, 2, or 3 above. For pIane strain Ez = O. For generalized pIane strain, Ez is a constant which can be determined from the end loads on the tube. Thus let the axial force acting on the tube be F. Define
Then it readily follows from the condition F*
S
Er
=
J:
Szp dp
and the third of equations (8.7.3) that
(8.7.6)
Elastoplastic Problems or Spheres and Cylinders
158
[Ch.8
z
+ 2(1
= _1_ [(1  2p,)P f321
 f1)
2(1  f1)
In any case
€z
(8.7.7)
To reIate €r to the definition
JIi
(8.7.8)
= [32 _ 1 1 Tp dp
€p, the two methods indicated in Section 7.6 may be used. If
is used, then from (8.7.4), (8.7.15)
is a knownconstant. From the Iast of equations (8.7.3), (8.7.9)
Substituting this reIation into the first two of equations (8.7.3), making use ofthe equilibrium and compatibility equations (8.7.2), and integrating resu1ts after some aIgebraic manipuIations in the following soIution: P  1 8 = r p2 p2
JP Tp dp + 11 JP€r dp + (1 1  f11 P 
2
86 = P  T+! p2 p2
1) Cl 2"
P
JP1 Tp dp +  1 1 f1 2 (€r + fP1 €r dp) + P
(8.7.10)
f32 = f32 1_ 1 (J13 P + 1 Tp dp  1 _ f12
On the other hand, if the definition (7.6.20) is used, then it follows that (8.7.16) The two definitions differ by the familiar constant 2/V3 and either one can be used. Since the definition based on the pIastic work increment appears to be more consistent with the Tresca criterion, we shall use it, and assume for the case under consideration that (8.7.14) may be written
(1 + p\)Cl
2P 2JP Tp dp + 11 p 2 8 =  T+2 €r + 2 Cl p2 P 1 f1 2 P Cl
159
(8.7.14)
JP1 Tp dp]
For a tube with open ends, P = O and €z
Plastic Flow in Thick·Walled Tubes
Let the stressstrain curve be given by a reIation of the form
If the axiaI force is due to internaI pressure onIy, then F* = P/2 and €
Sec.87]
fii P €r dp ) 1
(8.7.11)
181 ;: : 181 : .:;
1
(8.7.17)
1
To use the successive approximation method, it is preferabIe, as was done the for case of the sphere, to rewrite equations (8.7.11) and (8.7.17) as follows:
(8.7.12) (8.7.18)
YieIding will begin at p = 1 when 8 = 1, so that the criticaI pressure will be 8 = f(l€fI) sgn 8
€r = O
and, from (8.7.12),
Therefore, f321( 1 Pcr1t = ~
+T
f32 2_ 1 J131 Tp d) P
(8.7.13)
For P < Pcrlt, we have the eIastic soIution, which agrees with the classicaI eIastic soIution. For P ;:::: P cr1t a pIastic zone wiII spread out to some radius Pc' The soIution for generaI strain hardening can be obtained by an iterative or successive approximation method, as indicated for the sphere.
181;:::: 181::.:;
l} 1
(8.7.19)
An initiaI distribution of €r (such as zero) is assumed. The signs of 8 through. out the cross section of the tube are then determined from (8.7.11), and the actuaI vaIues of 8 are calcuIated from (8.7.19). A better approximation can now be obtained for the €r using equation (8.7.18). The process is repeated unti! convergence is obtained. For the cases ofthe sphere and the tube heretofore discussed, the successive approximation method has been found to converge fairly rapidIy using the techniques described. However, this may not aIways be true. A generaI dis. cussion of the convergence of the successive approximation method is given in Chapter 9.
Elastoplastic Problems of Spheres and Cylinders
160
[Ch.8
For the case of linear strain hardening, a solution can be obtained in closed form for the above problem. As for the sphere problem, equation (8.6.11) is used: 1 m ef = (1 m
Sec.87]
Plastic Flow in ThickWalled Tubes
161
Equation (8.7.26) gives the relationship between the piastic zone radius Pc and the applied pressure P, for a given temperature distribution T. For a perfect1y piastic materiai this reduces to
 S)
P
= In Pc
 Sr.c
= In Pc
+
(8.7.20)
f
P€p
1
...!.. dp
P
lm
=   (In Pc  Sr.c  P) m
f3~  1 [
2f3~
1
+ T(Pc)

2
f3~ _ l
(Pc Pc
J
TP dp
]
(8.7.27)
Therefore, l Cl = f32 _ 1
[f/J1 Tp dp + P
f32(1  m)
 m(l _ /h2) (In Pc  Sr.c  P)
]
(8.7.21)
To obtain the stresses we now substitute into equations (8.7.10). Thus, since
f f p€f p €p
...!.. dp
1
Also, from (8.7.11),
S
= 
T
2fP Tp dp + "22 (P + P 1 P
+"2
Cl)
+
m
P
/h
S;
When P = Pc, S = 1. Therefore, from (8.7.22),
2
Pc
fPC Tp dp ] _ P
1
Pc
l  m
=   (In P  Sr  P) m
p:O;
Pc
l  m
p
lm (1 _ 2) (l  S)
(8.7.22)
2 Cl = p~ [ 1 + T(Pc)  "2
P
dp
= ,:n (In Pc  Sr.c  P)
p;::: Pc
then, from (8.7.10),
(8.7.23)
1
Substituting into the expression for Cl' (8.7.21), we get
1[1 + =
f3/~
T(Pc) 
l
fa
:~
P :o; Pc Tp dp]
[p + ;2 J: Tp dp 
m(~ =
:2)
(In Pc  Sr.c  P)] (8.7.24)
As a check, at the onset ofyield, Pc = l, Sr.c = P, and (8.7.24) reduces to (8.7.13) for the criticaI pressure. At P = Pc we can consider a new tube with inner radius Pc and outer radius f3c with Sr.c equal to Perito Thus Sr.c
= 
f3~  l [ 2f3~ l
+
2
T(Pc)  f3~ _ 1
J/Jc Tp dp]
and
(8.7.28) Sr
lfP Tp dp + (1 lm2) (In Pc 1 /h m
= "2 P
'
 Sr.c  P)
(8.7.25)
Pc
p;::: Pc
Solving (8.7.24) for P gives P
= f32  1 (1  /h2)m [p~ (l + T(Pc)  22 f32
1  /h2m
2
Pc
fPC Tp dp)
+
1
l
 f3+r 
f/J Tp dp] 1
lm 2 (In Pc  Sr.c) /hm
(8.7.26)
To obtain the complete stress distribution, we compute P or Pc from (8.7.26), Cl from (8.7.23), Sr.c from (8.7.25), and the stress es from (8.7.28).
Elastoplastic Problems of Spheres and Cylinders
162
[Ch. 8
GeneraI References
15.
ProbIerns 1.
Show that for the sphere with radiaI symmetry, the von Mises yield criterion becomes
16. 17.
163
Plot the plastic zone as a function of the applied pressure for a sphere with linear strain hardening. Assume f3 = 2, m = 0.1, and jJ, = 0.3. Compare the resu1ts with those for a perfectly plastic materia!. Repeat Problem 15 for a tube. Perform complete numerical analysis of the problem of the sphere, a :o; r:O; b, T(a) = T o, and T(b) = O. Assume E = 30 X 10 6 , jJ, = 0.3, et = 10 5 , a e = 30,000 + 136,000 (e p + 10 4 ) for a e > 30,000, and E, et, and jJ, are independent of temperature.
and the PrandtlReuss equations reduce to
de;
=
 2de~ =
de p sgn (a r  ae)
Explain why one would expect the Tresca and von Mises yield criteri a to coincide for the case of a sphere with radiaI symmetry. 3. Derive equations (8.2.1) and (8.2.3). 4. Obtain the equations for all the strains and displacements in the ·sphere before yielding begins for pressure loading only, for thermalloading only, and for the case when both thermal and pressure loading exist. 5. Show that the steadystate temperature distribution in a sphere of inner radius a and outer radius b is equal to
References 1.
2.
T
=
Toa (~b  a r
1)
if the inner and outer surfaces are kept at temperatures of T o and zero, res pectively. . 6. Using equations (8.3.3) and (8.3.8), show that the stresses are contmuous across the elastoplastic boundary. 7. Compute the displacements and strains in a sphere with pressure loading only, for r :o; re. Assume a perfectly plastic material and that the. dimensi~ns remai n fixed. Determine the errar in the displacements of the mner radlUs for the fully plastic case if jJ, is assumed to equal 0.5 instead of 0.3. 8. Show that for a hollow sphere with a temperature distribution given by equation (8.2.7), the tangenti al stress is compressive and the radiaI stress is near zero in the regio n adjacent to the inner circumference, so that the yield criterion in this regio n can be written
Se  Sr = 1
9. Starting with equations (8.6.1), derive equations (8.6.2) and (8.6.4) using boundary conditions (8.6.6). Derive equation (8.6.9). 11. Derive equations (8.6.12). 12. Derive equation (8.7.6). 13. Derive equations (8.7.10) through (8.7.12). 14. Show that the definition (7.6.20) for the equivalent plastic strain increment Ieads to equation (8.7.16) for the case of a tube with the Tresca criterion and associated f10w rule, if ae > az > aro lO.
2. 3.
4.
5.
6.
7.
8.
9.
W. Johnson and P. B. Mellor, Plasticity for Mechanical Engineers, Van Nostrand, Princeton, N.J., 1962. R. Hill, The Mathematical Theory of Plasticity, Oxford Univo Press, London, 1950. P. G. Hodge and G. N. White, A Quantitative Comparison of Flow and Deformation Theories of Plasticity, J. Appl. Mech., 17, 1950, pp. 180184. D. N. de G. Allen and D. G. Sopwith, The Stresses and Strains in a Partially Plastic Thick Tube Under InternaI Pressure and EndLoad, Proc. Roy. Soc. (London), A205, 1951, pp. 6983. M. C. Steele, Partially Plastic ThickWalled Cylinder Theory, J. Appl. Mech., 19, 1952, pp. 133140. R. Hill, E. H. Lee, and S. J. Tupper, The Theory of Combined Plastic and Elastic Deformation with Particular Reference to a Thick Tube Under InternaI Pressure, Proc. Roy. Soc. (London), A19l, 1947, pp. 278303. W. T. Koiter, On Partially Plastic ThickWalled Tubes, Biezeno Anniversary Volume in Applied Mechanics, N. V. De Technische Uitgeverij H. Stam, Haarlem, 1953, pp. 232251. D. R. Bland, Elastoplastic ThickWalled Tubes of WorkHardening Material Subject to InternaI and External Pressures and to Temperature Gradients, J. Mech. Phys. Solids, 4, pp. 209229. I. S. Tuba, ElasticPlastic Analysis for Hollow Spherical Media Under Uniform RadiaI Loading, J. Franklin Inst., 280, 1965, pp. 343355.
GeneraI References Hill, R., The Mathematical Theory of Plasticity, Oxford Univo Press, London, 1950. Hotfman, O., and G. Sachs, Introduction to tlle Theory of Plasticity for Engineers, McGrawHill, New York, 1953. Johnson, W., and P. M. Mellor, Plasticity for Mechanical Engineers, Van Nostrand, Princeton, N.J., 1962.
p Sec. 91]
CHAPTER
GeneraI Description of the Method
165
The solution is known to be y = eX. To find the solution by Picard's method, we proceed as follows. Integrate (9.1.1) to give
9
(9.1.2) Assume as a first approximation for y,
il) =
Substitute this value for y on the right side of (9.1.2) and calculate a second approximation for y:
THE METHOD OF SUCCESSIVE ELASTIC SOLUTIONS
+
y(2) = 1
f: il)
dx = 1
+x
Substitute the second approximation for y and calculate the third approximation:
i 91
1
GENERAL DESCRIPTION OF THE METHOD
In Chapter 8 it was indicated how the sphere and tube problems can be solved for arbitrary strain hardening by a successiveapproximation method. This method is nothing more than the extension of Picard's method (see reference [l]) of successive approximations to nonlinear equations. The method was apparent1y first used in plastic flow problems by Ilyushin [2] in his treatment of a thin shell. Ilyushin refers to it as the method oJ successive elastic solutions, since each iteration involves essentially the solution of an elastic problem. Before the advent of modern highspeed computing machinery, this method could be used only for relatively simple problems. However, with current widespread availability and use of digitaI computers it now becomes possible to solve simply and quickly many problems whose elastic solution can be obtained by numerical methods.
3
)
= 1+
i
x
o
i
2
2) dx = 1 + x + ~
2
Continue in this way to get
i
in +
)
= l +
1)
= l +
4
i i in) x
i o x
O
3
X2
)
dx = 1 + x + 
2!
X2
dx = l + x + 
2!
x3
+ 
3!
xn + ... + n!
As n gets larger and larger, it is seen that the infinite series for eX is approached. The exact solution can thus. be approached as closely as desired by taking more and more approximations. This technique can be direct1y extended to the generaI elastoplastic problem in the following manner. For convenience the pertinent equations given in previous chapters will be repeated here. The equilibrium and compatibility equations are independent of the plasticity relations and are given by equations (3.2.2) and (4.7.2); i.e.,
Before proceeding to describe this method for the generaI elastoplastic problem, we shall first give an illustration of the method of successive approximations for a simple differential equation [3]. Consider the equation dy y = O dx

164
(9.1.3)
y(O) = 1
(9.1.1)
The Method of Successive Elastic Solutions
166
[Ch. 9
and
Sec. 91]
GeneraI Description of the Method
167
the ~ssociated fiow rule. For definiteness we shall consider the Prandt1Reuss relatlOns, but any other set of relations can be used equally wel1. Thus
(9.1.4) (9.1.6)
~ (_ oSzx oy
+
OSXy oz
+
oSyz) = 02Sy ox oz ox
+
OSYZ OX
+
8szx) 8y
oy
~ (_ OSXy 8z
OZ
=
02sz OX oy
where The stressstrain relations depend on the plasticity theory used and we can write (9.1.7)
Alternati:,el y, the plastic strain increments can be related to the modified total strams as described in Section 7.9; i.e., (9.1.5)
2~ TXY + S~y + tls~y
SXy
=
SYZ
= 2G TyZ +
SYZ
_ 1 2G
Szx
1
Szx 
Tzx
+
p
p
A
I
+
uS yZ
+
USzx
uSx 
A P _
tls ( ' , , 3 2sx  S  S ) Set y z
A P _
tls ( ' , , 3 2s y  s z  s x ) s
US y 
p
p
et
A P
where s~, s~, etc., are the total accumulated plastic strains up to,but not including, the current increment of loading, tls~, tls~, tls~, etc., are the plastic strain increments due to the current increment of loading. The plastic strain increments are related to the stresses through the yield criterion and
(9.1.8)
168
The Method or Successive Elastic Solutions
[Ch. 9
where etc. (9.1.9)
Sec. 91]

Generai Description or the Method
169
I p
6~j,n
r
6ep
r
~
x~~""';""":""":""":""":""":""":NN
differentiai equation is integrated to give 1  p,2 pw 2 r 2 eo =   E  8
1 + p, +  r2 
fr
o aTrdr
1  p,
+ 2
fr er o
r
e~
dr
1 + p, fr ( p P) d' Cl C2 + 2r2 o er + eo l' l + T + f2
....
(954)
.'
"10If'lOIf'lOIf'lOIf'lO oo'
~
'"
~,
~~
l''' ~,
~
O
'Vi o
5,2 xlO 3
~ 1''
~
\ "~ t
40
\
vi" \Il
~ +
(J)
80
~
4.0
.~ 4.6
\
\
\.
\
(J)
\\ l"
\
4.4
\
4.0
5.0
'b ~ O
As an example, the solution obtained in this manner for a lOin.diameter parallelsided disk is shown in Figure 9.5.2. The value of pw 2 was taken as 1,500 and a temperature was assumed as shown in Figure 9.5.1. The strainstrain curve of Figure 9.3.1 was used. Again little difference was found between deformation theory and incrementaI theory. The computations for one iteration are shown in Table 9.5.1. As seen from Figure 9.5.2, a straight
0.04
l'q
0.08
Se
( b)
9.5.2 (continued)
\
l\}
4.2
Radius, in.
FIGURE
\
(Te
+
\ 3.0
O ee
1\
(Tr
4.8
'"""
\
140
2.0
\,
5.0
r.
"Ì'
\
1.0
o .er
~\
"
100
180 O
203
FIGURE
9.5.3
Variation of strain with change of strain.
drawn as shown and the intercept at zero Oe is obtained. This furnishes a new starting estimate. Three or four more successive approximations are carried out, and another, similar, extrapoiation is made. This technique reduced the number of successive approximations for this problem from ~O to 12, However, it shouId be noted that if a highspeed digitaI computor ~s used, the time per iteration is on the order of 0.1 sec and the number of lterations required is of secondary importance.
i
I
I
! Il
Sec. 95]
Rotating Disk with Temperature Gradient
205
The Method or Successive Elastic Solutions [Ch. 9
204
In this manner equations (9.5.1) and (9.5.11) can be written
FiniteDifference Formulation
Clar.l 
If it is desired to take into account variations of E,
or h along the disk, then it is necessary to use the finitedifference method [8]. This formulation is very generaI and can take into account not only initial v~riation~ in the disk thickness, but also the changes in thickness and radms durmg the loading process, if the disk should grow and change shape. In addition, we shall also consider the case of a disk with a centraI hole. We start again with the equilibrium and compatibility equations (9.5.1) and (9.5.2). It is convenient this time to solve the problem in terms of ~tress~s rather than strains, since the boundary conditions are generalIy glven m terms of stresses. The stressstrain relations are written as er =
~ (ar 
P,ae)
+ exT + ef + ~ef
ee =
i
(ae  p,ar)
+ exT + e~ + ~e~
+ Gl a e.I1
Dlae.l = Flar.ll
C;ar.l  D;ae.l = F/a r.I1  G;ae.I1
p"

Hl
+ H(
 P!
(9.5.13)
where
 P,11 b Pl,  l El 
1
2
Hl =
~ (rl  rl 1)(Pihlrt + Pl1hl1rt_1)
(9.5.10)
(9.5.14) where ef and e~ represent the total plastic strains up to the current increment of loading, and ~ef and ~8~ are the increments of plastic strain due to the current increment of loading. Substituting (9.5.10) into the compatibility equation (9.5.2) gives the compatibility equation in terms of stresses:
Equations (9.5.1) and (9.5.11) are two equations for the two stresses ar and ae' We proceed by putting them into finitedifference form as folIows. Let the disk radius be divided into N intervals (not necessarily equal). There are thus N + 1 stations, the first station being at the center for a solid disk: or at the inner radius for a hollow disk. The last station is at the outer radms. Equations (9.5.1) and (9.5.11) are written in finitedifference form at the midpoints of theseintervals. Thus at the midpoint of the (i  l)st interval,
It is to be noted that if the disk dimensions do not change appreciably with loading, alI these coefficients except P( depend only on the initial geometry, material properties, and operating conditions of the disk and are evaluated once and for alI. Only P; is a function of the plastic flow. Aiso for a solid disk, rl 1 can equal zero and so some of the primed coefficients be come infinite. This can be avoided by assuming for the first radiaI station some small nonzero number, rather than zero. Equations (9.5.13) give the stresses at the ith station in terms ofthe stresses at the (i  l)st station. Solving for the stresses at the ith station gives
+ 112.1 ae.I1 + m1.1
ar.l = 11l.lar.I1 ae.l = 121.1 ar.I1
+ 122.1 ae.I1 + m2.1
(9.5.15)
where l
_ D;Gl + DjG; CjD;  C;D j
12.j 
l 21.j
C:Fì = CD' l
j
ClF; C'D j i
l
_ C;Gj + CjG; CjD;  C;D j
22.j 
(9.5.12) m2.j =
CjP:  HjC;  H!Cj Cj D'1  C'D j j
(9.5.16)
The Method or Successive Elastic Solutions [Ch. 9
206
Sec. 95] Rotating Disk with Temperature Gradient
or, in matrix notation,
207
Therefore,
[::::J
[~:::: ~::::J [:::: ~:] + [::::]
=
or
al
=
+ MI
Llal_l
(9.5.17)
where a l' all> L lo and MI are the indicated matrices. Equation (9.5.17) represents a linear recurrence re1ation.between the stresses at the ith station and the stresses at the (i  l)st station. Obvious1y by successive application of (9.5.17), the stresses at the ith station can be linear1y related to the stresses at the first station. Let this linear relation be written
For a solid disk
ar.l
=
aO.l;
a r •l
=
therefore, aO.l
=
arCR) all.N + 1
bl .N+l
+ a12.N + 1
(9.5.23)
For a disk with a centraI hole of radius R o, with inner prescribed pressure we get
ar.l = ar(R o),
(9.5.18) (9.5.24)
where AI and BI are as yet unknown and al are the radiaI and tangential stresses at the first station. Substituting (9.5.18) into (9.5.17) gives Alal
+ BI
+ BIl) + MI
= LI(AI_lal
(9.5.19)
or
Now al will depend on the boundary conditions and is complete1y arbitrary, whereas (9.5.19) must be true for all values of al' It therefore follows that both sides of the equation must vanish identically. Hence AI = AI_1LI
(9.5.20)
Also, for the second station, equation (9.5.18) gives
Thus al is now known. The stress es at every station can then be direct1y computed by means of (9.5.18). To summarize then, the LI and MI matrices are computed from (9.5.16) and (9.5.14), the At and BI matrices from (9.5.20) and (9.5.21), and then the stress es are computed from (9.5.18), using either (9.5.23) or (9.5.24). It is to be noted that this straightforward procedure takes into account with equai ease variations of E, h, p, or even fl, along the radius of the disk. Furthermore, if the dimensions of the disk are changing during the plastic flow process, this can readily be taken into account. For if l' is the current radius to a given point P, and l" was the radius to the point P before plastic flow took pIace, then approximate1y l'
= 1"(1 + e&)
(9.5.25)
and equation (9.5.17) gives a2 = L 2a l A2 = L2
Hence
+M2
and
(9.5.21)
B2 = M 2
and, similar1y, if H was the underformed thickness at the point P and h is actual thickness after deformation, then approximate1y I
Beginning therefore with A 2 and B 2 as given by (9.5.21), all the other A's and B's can be computed successively by the recurrence relations (9.5.20). For the Iast station [the (N + l)st], l'N+1 equals R, the disk radius, and equation (9.5.18) becomes or [
a r ' N + l ] = [all.N+l
a12.N+l] [ar.l]
aO.N+l
a22.N+l
a2l.N+l
aO.l
+
[b
l N l ' + ]
b2•N+ l
(9.5.22)
I
h = (1
+
H ef)(1
I
+ e&)
(9.5.26)
Thus at any stage of the plastic flow process the values of l'I and hl appearing in equations (9.5.14) can be corrected by means of (9.5.25) and (9.5.26). The finitedifference formulation presented will of course give direct1y and quickly the elastic solution for a disk of arbitrary profile with variable properties, if P; is set equal to zero in equations (9.5.16). For the plastic problem P; is not zero, and its values can be determined by successive approximations, as thoroughly described in the previous examples. We shall
I:
I
l! .1
i i
I
l'
Sec. 96] 208
The Method of Successive Elastic Solutions [Ch. 9
Circular Hole in Uniforrnly Stressed Infinite Plate
209
Integrating resuIts in
return to this formulation at a Iater time in discussing the creep of a rotating disk. Solution by this method of the disk problem of the previous example gave almost identical answers.
A. 'f'
Ar2
+ BInr +
C 
Jr :1 Jr r Jr g(r)drdrdr a l
a
(9.6.6)
a
!he tri~le integraI can be somewhat simplified by making use of (9.6.5) and
96
CIRCULAR HOLE IN UNIFORMLY STRESSED INFINITE PLATE
As a final example of a probIem involving rotational symmetry, consider the case of a thin infinite plate uniformly stress ed containing a circuiar hoIe of radius a. Solutions to this probIem by iterative methods similar to that discussed herein are given in references [9] and [lO]. The present solution is taken from reference [lO]. The equilibrium and compatibility equations are given by (9.4.1) and (9.4.2); i.e.,
llltegrat1~~ by parts. The constants A and Bare determined from the bound
ary condlt1ons (9.6.7) where . hGeo is . the applied uniform stress at infinity. Note that t e yleid stress. Then
Geo
'S, assummg the nght sldes are known. These equations can now be solved by of the numerous methods of solving large sets of simultaneous linear
~.~
The PIane Elastoplastic Problem [Ch. lO
220
Sec. 102] Eiastopiastic Thermai Problem for a Finite Piate 10
y VV V V V
algebraic equations. Once these are solved, the stresses can be computed O. 9
from the relations
O'y,l1
=
TXy,!J
=
4>1+ l,i
+ 4>1 l,i
4>11,J+1 
82

24>1,1
4>11,J1 
48 2
(10.2.2) 4>1+1,1+1
+ 4>1+1.11
The strains are computed from (10.1.3) and the plastic strain increments from either (10.1.4) or (10.1.5), together with the stressstrain curve. The function f1g is now changed and the solution obtained again. The process is continued until convergence is obtained. It is to be noted that only the right sides of the set of n x m simultaneous equations change from iteration to iteration. It should also be noted that although equation (10.2.1) has been written for equal spacing between stations, it is possible to write the finitedifference formulas for unequal spacing. Once the calculation has converged for a given increment of load (in this case, thermalload) and f1g determined, f1g is added to g, the lo ad is incremented, and a new calculation started to determine the value of the plastic strain increments and stresses due to the new increment of load. It is to be remembered that at any station for which O'e is less than the yield stress, the plastic strain increments are set equal to zero. As an example, such a solution was obtained for a square plate with a parabolic temperature distribution given by T = TO(y2  t), the constant To being raised in increments untillarge zones of plastic fiow occurred. Linear strain hardening was assumed with the strainhardening parameter m taken to be 0.1. Some of the results are shown in Figures 10.2.3 and 10.2.4. Only one quadrant of the plate is shown, the other three quadrants being identical because of symmetry. In these computations 20 stations were taken in the x direction and 20 in the y direction, resulting in a set of 400 simultaneous equations to solve. Figure 10.2.3 shows the rate of growth of the regions of incipient fiow The curves are the loci of alI the points of incipient plastic fiow for a . value of loading parameter TO = ToErx/ao. Plastic fiow starts first at centers of the four sides of the plate and moves rapidly inward. Plastic does not start at the center of the plate unti! it is well developed at the Once plastic fiow has started at the center, however, the rate of growth this zone is greater than at the sides. Figure 10.2.4 shows the plastic strain trajectories for the maximum
221
V
l/ / / O. 8 Ta /1/' l/ / I V ..... V / O. 71 f ....,V /" ~ ~ V i> <J>
L
"'" " '\
~
+<J>
223
Since the solution obtained was an incrementaI one using finite increments of load, it was thought worthwhile to run an experiment to determine the effect ofincrement size f1To on the final solution. It was found that f1To = 2.5 ,,:as appro~imately the largest increment size that produced no appreciable dlfference In the final stress and strain distribution at 'To = 20. Calculations were also performed using the two iterative techniques described in Section 9.1, that is, using equations (10.1.4) to calculate the plastic strain increments and using equations (10.1.5). As expected for very large load increments the first method (using the PrandtlReuss relations) diverged, whereas the se~ond method, using the plastic straintotal strain equations, always converged. For those cases where both methods converged they gave identical answers. The solution to this problem can be found discussed in greater detail in reference [2]. !he above calcul~:ions were performed for a very thin plate. For a very thlCk plate, a condltlOn of pIane strain would exist at planes far from the surfaces. To obtain a solution to this problem only minor modifications of previous calculation method are necessary. Thus g and f1g must be computed using equations, (10.1.14) instead of (10.1.13). Also it is now necessary to compute U z • ThlS can be done as for the case of the cylinder in Section 9.4. From the third of equations (10.1.8),
2
1.2
'vi'"
aT uolE
B
Sec. 103] Elastoplastic Problem or the Infinite Plate with a Crack
103 ELASTOPLASTIC PROBLEM OF TUE INFINITE PLATE Wl TU A CRACK T~e elastic solution of the infinite plate with a crack was first obtained by InglIs [3] ~nd was used by Griffith [4] in his theory ofbrittle crack propagation. It was pOInted out by Orowan [5] that, for ductile materials, the plastic strain energy was a major factor in the energy baI ance of the system and could not be ignored in any analysis of crack stability. Since no solution was available
224
The PIane Elastoplastic Problem [Ch. IO
for the plastic strain field, various assumptions were introduced to take into account the plastic zone at the tip of a crack, such as Irwin's "equivalent crack length" [6] and Neuber's "plastic particle" [7]. In the present section it will be shown how the previous method can be used to obtain at least approximately the elastoplastic strain field in the vicinity of a crack, for a strainhardening material. Since the solution can be obtained only numericalIy, the presence of the mathematical singularity at the tip of the crack makes accurate answers very difficult to obtain. The solution to be presented is therefore not intended to provide accurate quantitative results, nor is it necessarily, or even probably, the best method of solving this problem. It is intended primarily to provide qualitative information on the effect of strain hardening and on the differences that might be expected between pIane stress and pIane strain solutions. Consider the case of an infinite plate with a centraI crack 2 units long, with a uniaxial tensile lo ad at infinity perpendicular to the pIane of the crack. As before, alI the stresses are made dimensionless by dividing by the yield stress and alI the strains are divided by the yield strain. The tensile stress at infinity is also divided by the yield stress. Since half the crack length is taken as unity, the x and y coordinates are dimensionless in terms of the halfcrack length. We shalI attempt a solution using finite differences, as in the case of the rectangular plate of Section 10.2. We are faced, however, with two problems in applying the finitedifference methods previously described. In the first pIace, it is obviously impossible to cover an infinite region with a finitedifference grido Second, the crack tip is a singular point of the stress field and, as pointed out in reference [8], the error in the finitedifference formulation in the vicinity of the singularity spreads to other points. The best procedure in this type of problem is to subtract out the singularity, if possible, giving a new problem with different boundary conditions, but which will be welI behaved. In the present problem we attempt to minimize both the above difficulties by subtracting out the elastic solution from the problem. The elastic solution contains a singularity at the crack tip and also satisfies the boundary conditions. We are thus Ieft with a wellbehaved problem with homogeneous boundary conditions. Furthermore, it has been shown experimentally by Dixon [9] that for most materials (mild steel, which has a Iower yield point, is one exception), the strain field outside the plastic zone is the same as the eiastic strain field; Le., the elastic solution prevails outside the plastic zone. The assumption is therefore made that the infinite plate can be replaced by finite rectangle with an edge crack upon whose boundaries the elastic field acts. Because of symmetry, only one quadrant of the plate need considered. We pro cee d, therefore, in the following manner.
Sec. 103]
Elastoplastic Problem or the Infinite Plate with a Crack
225
A rectangular section is chosen as shown in Figure lO .3.1. On the upper, . left, and nght boundaries of this rectangle, the elastic stress field for the y
I
I I I
I I
I I
I I I 1
FIGURE
10.3.1
I
J __
x
Rectangular boundary for finitedifference grido
infini~e plate acts. The lower boundary is a Hne of symmetry. The differential ~quatlOn to be satisfied is equation (10.1.16) without the temperature term;
1.e., '\144> = g(x, y) 
~g(x, y)
Let
(10.3.1) (10.3.2)
where.4>e is the elastic solution to the problem. Then 4>e satisfies the differential equatlOn '\144>e
=O
(10.3.3)
and the boundary conditions. 4>p then satisfies '\144>p
=  g(x, y) 
~g(x, y)
(10.3.4)
~omogen~ous boundary conditions. If it is assumed that the elastic pr.evatls on the upper, left, and right boundaries of the rectangle 10 FIgure 10.3.1, then we areleft with the problem offinding a function
~OllltlC1,n.
The Piane Elastoplastic Problem [Ch. lO
226
By
r/>
(B8~z _ B8f z) By Bx
(11.1.11)
(11.1.12)
such that
_ Br/> OX
(11.1.13)
The Torsion Problem [Ch. 11
238
Then the equilibrium equation (11.1.9) is identically satisfied and the compatibility equation (11.1.11) becomes \j
2
2 _ 8 4> 4> = 8X2
82 4> _
+ 8y2 
2Ga  g(X, y)
(11.1.14)
For the elastic problem g(x, y) is equal to zero. The boundary conditions for the problem can be obtained direct1y from equations (3.3.1). The first two of these are identically satisfied and the third reduces to
Sec. IlI]
239
It should be noted that equation (11.1.15) is equivalent to the statement that the resultant shear stress at the boundary is tangent to the boundary. This must always be true if the Iaterai surface of the bar is forcefree. It aiso follows, since equation (11.1.17) and consequent1y equation (11.1.15) hold for any line 4> = constant in the cross section, that the resultant shear stress at any point is tangential to the 4> = constant line passing through that point. The lines 4> = constant are called the stress trajectories. Furthermore, from (11.1.13) it folIows that the resultant shear stress T at any point is given by
(11.1.15)
From Figure 11.1.3 it can be seen that
Torsion of Prismatic Bar. Generai Relations
T = VT;z or
T
=
+
T~z = J(:~r + (:~r
(11.1.19)
Igrad 4>1
The resultant shear stress is thus equai to the gradient of the stress function 4>. Let us now calculate the resultant forces and moments acting on any cross section. The force in the x direction is given by
Qx =
JJTxz dx dy = JJ:~ dx dy
= J[4>(x, A) FIGURE
11.1.3 Geometry of boundary.
dy
l = cos (n, x) = ds
(11.1.16) m
= cos (n, y) = 
dx ds
M
= JJ(TyZX =
4>=0
on the boundary
=O
(11.1.20)
where the double integraI is taken over the area of the cross section and 4>(x, A) and 4>(x, B) are the values of 4> at two opposite points ofthe boundary at a given value of x, and are consequent1y equai to zero because of the boundary condition (11.1.18). In a similar fashion, it folIows that the y component of the resultant stress is zero. Finally, the torque acting on the section is computed from
and (11.1.15) becomes, upon combining with (11.1.13),
or 4> equals a constant along the boundary. In the case of simply connected boundaries, e.g., solid bars, this constant can be chosen arbitrari1y, since we are interested only in the derivatives of 4>. It is, therefore, for convenience chosen to be zero, so that the boundary condition becomes
 4>(x, B)]dx
JJ(:~
 Txzy)dx dy x
+ :~Y)dX dy
(11.1.21)
Integrating by parts and making use again of the boundary condition (11.1.18) results in M = 2
JJ4> dx dy
(11.1.22)
240
The Torsion Problem [Ch. 11
If a solution is obtained for equation (11.1.14) subject to the boundary condition (11.1.18), then the stress es can be computed from (11.1.13) and the strains from (11.1.1 O). (The calculation of the plastic strains will be discussed subsequent1y.) The equilibrium and compatibility equations, the stressstrain relations, and the boundary conditions will all be satisfied. The assumptiòn (11.1.1), or equivalently (11.1.4), thus leads to the correct solution of the torsion problem. It should be noted, however, that the solution requires the same stress distributions to act on every cross section, including the end sections. However, by SaintVenant's principle, if the beam is sufficient1y long, the solution will be valid for all cross sections far enough away from the ends for any stress distribution acting on the ends, provided the resultant force is zero and the resultant moment is given by (11.1.22). Equation (11.1.14) is the wellknown Poisson's equation, which is encountered so frequent1y in mathematical physics. Its solution can be obtained by many different techniques. For the elasticity problem, the right si de is a constant, and solutions, at least for simple shapes, can readily be obtained. For the plasticity problem the right side is a function of the plastic strains and is therefore unknown until the solution is obtained. (The problem is nonlinear.) For materials with or without strain hardening we can use the method of successive elastic solutions described in Chapter 9. For perfect1y plastic materials, a simpler, more specialized approach, to be described subsequent1y, can sometimes be used. In either case the elasticity solution is a prerequisite to the plasticity solution, and we shall therefore first present some solutions to the elasticity problem.
Sec. 112] Elasticity Solutions
241
the ~ou~dary condi~ion (11.1.18), g, = 0, will obviously be satisfied, and substItUtlllg (11.2.2) lllto (11.1.14) results in (11.2.3) Hence (11.2.4) is the solution for the elastic case. T~e torque req~ir~d to produce the angle of twist per uni t length a is obtallled by subStItUtlllg the solution (11.2 .4) into (11 .l . 22), g1Vlllg .. (11.2.5) The consta~t GJ is called the torsional rigidity, or torsional stiffness, of the bar, for ObVlOUS reasons. To obtain the shear stresses, the solution (11.2.4) is differentiated to give T xz
T yZ
= =
og,
a2
oy =  2 a2
og, ox
+ b 2 Gay
b2
(11.2.6)
= 2 a2 + b 2 Gax
and the resultant shear stress is
112 ELASTICITY SOLUTIONS Equation (11.1.14), with g(x, y) equal to zero, can sometimes be solved by guessing a solution, if the boundary of the cross section is of a simple shape. For example, if the bar has an elliptic cross section, the equation of the boundary curve being
(11.2.7)
T~e maximum stress will occur on the boundary at the point closest to the aXlS of the bar at x = 0, y = b. Thus (11.2.8) T(10 obtain the warping function w(x, y, a), we first compute the strains from 1.1.10) :
then choosing for a stress function (11.2.9)
The Torsion Problem [Ch. 11
242
Sec. 112] Elasticity Solutions 243
and then, from equation (11.1.5),
y 2
2
8x = bb + a ay
8w

2
l
a2
(11.2.10)
o
Therefore, b2
w = b2
_
a2
+ a2 axy + constant FIGURE
and since at the origin w must vanish, the constant must equal zero, and
w(x, y, a) = bb
2 _
2
a2
+a
2
axy
EquilateraI triangle.
11.2.1
If we therefore choose as a stress function (11.2.11)
The lines of constant W are therefore equilateral hyperbolas. If the cross section is a circ1e, then bis equal to a in aH the previous formulas, resulting in
~ = c(x  :3)(y  ~  ja)(y +
:3
+
ja)
. ~ will be zero on the boundary. Substitutinginto (11.1.14) [withg(x y) _ O] glVes for c: ' 
v'3
C = Ga 2a
Hence T XZ
= Gay
T yZ
=
T T
Gax
(11.2.12)
= Gar
max =
~ = ~: (V3 x a)(y  :3  ;a)(y + :3 + ;a)
(11.2.14)
is the solution to the problem. The stresses the tor ue and . q, the warpmg function can be computed as before. '
Gaa
Fi As a :~~1 ;xample,. consider a bar of rectangular cross section, as shown in gure . . . For thIS case the solution is not as simple as for the previous
w=o Next consider a bar whose cross section is an equilateral triangle as shown in Figure Il.2.1. The origin of x and y axes is taken at the centroid. Then the
y
equation of the straight lines representing the three sides are
b
a x=
v'3
o
x
(11.2.13)
FIGURE
11.2.2 Bar of rectangular cross section.
244
The Torsion Problem [Ch. 11
cases, since it is no longer possible to guess a solution. Equation (11.1.14) must be solved by one of the available methods. For example, the method of separation of variables is used in reference [2], resulting in an infiniteseries solution, whereas in reference [3] the Green's function is used, resulting in a double infinite series. Here the finitedifference method will be used, since it is relatively simple and can be direct1y extended to elastoplastic problems. We note first that because of symmetry, only one quadrant of the section need be considered. For a square cross section the diagonals are also lines of symmetry and only one octant is used. The quadrant is divided into a grid of n x m points as shown in Figure Il.2.3. At each of the grid points,
• • • • • • • • • • 2
3
• • • • • • • • • • • • • • • • •
;1
245
and right boundaries of the quadrant For a . th d' . square sectlOn only the poi t on e lagonal and to the right of th d' le ' ns resulting in ~n(n _ l) . e lagona or left) need be considered, 2 equatlOns. Along the lower boundar b y, ecause of symmetry, equation (11.2.15) becomes
4>1+1.1
+ 4>11.1 + 24>1.2
= 2Gah2
(11.2.16b)
and along the Ieft bùundary,
4>1.11
+ 4>1.1+1 + 24>2.1
 44>1,}
(11.2.16c)
Along the diagonal of a square,
j }1
24>1+1.1
+ 24>1.11
 44>1.1 = 2Gah2
2
n
 44>1.1 = 2Gah 2
(11.2.16d)
.~UCh ~ soIut~on :vas obtained for a square cross section using an Il x Il gn . as s own In FIgure 11.2.3. The 55 resulting equations were solved b stralghtforward GaussSeidel process esee reference [9]) Th . Ya st· h' h . e maXlmum shear less, w IC occurs at the centers of the sides and th t below with the values from reference [2]: ' e orque are compared
3
such as the point designated by (i, j), equation (11.1.14) is written in finitedifference form:
+ 4>11.1 + 4>1.11 + 4>1.1+1
(11.2. 16a)
44>2.1  44>1.1 = 2Gah 2
FIGURE 11.2.3 Finitedifference grid for rectangle.
4>1+1.1
 44>1.1 = 2Gah2
and at the center,
m
• • • • • • • • • • • • • • • (,;;) • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
Sec. 113] Membrane Analogy
(11.2.15)
where h is the grid spacing assumed constant and is the same in both the x and y directions. An equation such as (11.2.15) can be written for each of the n x m grid points, resulting in a set of n x m simultaneous linear equations for the unknown values of 4> at each of the points. Once the ,s are determined as the solution of this set of equations, the shear stress can be obtàined from equations (11.1.13) by numerical differentiation and the torque for a given angle of twist per unit length a from equation (11.1.22) by numerical integration. Actually the number of equations to be solved is (m  1) x (n  1) rather than m x n, since the boundary conditions require 4> to be zero at the
Tmax/Gaa M/Gaa 4 4>1.1/Gaa 2
Re/. [2]
Finite Difference
1.351 2.250
1.343 2.244
0.589
0.589
It is seen that the solution with this many grid points is sufficientIy accurate.
113 MEMBRANE ANALOGY Solutions of the elastic torsion r bI mental1y by means of th P o em can also be obtained experisid e membrane analogy suggested by PrandtI [4] C er a membrane such as a soap film havin h . onsection ofthe bar being twisted Ifth d g t. esame shape as the cross . e e ges of thlS membrane are fixed and a
The Torsion Problem [Ch. , 11
246
, appl'l d toe one si, de the membrane will deflect an amount glven pressure lS [2] by 269)' , , of the following equations (reference , p, by the so1u t lOn
Sec. 114] Elastoplastic Torsion. Perfect Plasticity
247
in the plastic region, whereas in the elastic part of the bar \12
(11.3.1) h is the deflection of a point of the membrane, p the applied pr~ssuref' w ere z "1 h in the membrane. Companson o and ~ the constant tenslO~ ~~~4un[~i;:~(X, y) equal to zero for the elastic equatlOns (11.3.1) and ( '. ) d' t 1 The deflection z of the membrane the analogy lmme la e y. 1 ] h prob em s ows ' n d if 2Gex is equal to p/S, z is equal to . to stre7 at any point is proportional to the The maXlmum s ope o . t and the volume under the membrane resultant shear stress at thde pom _ O pIane) is proportional to the torque (b , 1f the membrane an t h e z nit len th. The membrane analogy lS usefu or ~; complicated shapes and has been used producmg a tW1St of ex per u, determining the stress fun~tlO~ °lastic torsion for perfect1y plastic materials very successfully. Hs extenslOn o p will be discussed in Section 11.4.
correspo~ds t~e
:~::~~b;ane
etwe~n,
114 ELASTOPLASTIC TORSION.PERFECT PLASTICITY , 1arge, p 1as t'lC flow (will occur. the If the applied torque is sufficlently P bI Since 2 and '11 1 s occur at the boundary see ro em maximum stress Wl . a way will start at some point on the boundary and reference [5]), a pla~tlC z,one h is increased, Additional plastic d d the mtenor as t e torque sprea towar other oints in the cross section, For the SUbS~qUent1Yblstar\~! yield ~riteria of von Mises and of Tresca zones may case of the torslOn pro em, both reduce to T 2xz
+ T yz2
k2
(11.4.1)
, von Mises ., . . h ield stress in slmple shear. Accor d'm g to the . k lS t e y . ;d d' t the Tresca cntenon lt lS where ,.. v 3 an accor mg o , , cntenon, k lS equal to Go/ : d . , pIe tension If the matenalls , . /2 h e G is the ywl stress m Slm equal to Go , W er , (11:. 4 l) mus . t hold everywhere in the plashc perfect1y plastic, then oequatlOn region. In terms of the stress functlOn
(11.4.3)
and = Oon the boundary ofthe bar. The eIastopIastic boundary is unknown and is determined from the conditions that and its first derivatives (the shear stresses) are continuous across this boundary and that the resultant shear stress is Iess than or equaI to k inside the eIastic region. Equation (11.4.2) can aIso be written /grad / = k
(11.4.4)
inside the pIastic region, In other words, the sIope of the surface is a constant, equal to k, in the pIastic region, it is not greater than k in the elastic region, and the height and sIope of the surface are continuous across the eIastopIastic boundary, The above conditions on for a perfectly pIastic materiaI suggest the extension of the membrane analogy to a partially pIastic bar [6, 7]. A roof of constant slope, proportionaI to k, is erected with the membrane as its base. As the membrane is pressurized and deflects, it wiII approach the roof, The region of the membrane corresponding to the region of the bar flowing pIasticalIy wiII be pressed against the roof and wiII have the same sIope as the roof. The rest of the membrane, corresponding to the elastic region, wiII not be touching the roof and wiII have a smaller sIope. The membraneroo! analogy furnishes a simple physicaI and intuitive picture ofthe growth ofthe pIastic zones as the torque is increased. To obtain quantitative resuIts wiII usualIy entail a considerabIe amount of Iabor, For the case of complete pIastic yieIding, the soIution becomes much simpIer, In this case the membrane wiII be in contact with the whole roof, and it is no Ionger necessary to use a membrane, Instead, one constructs a roof of the proper sIope, This can be done by simpIy heaping dry sand onto a pIate whose shape is simiIar to the cross section of the bar. Since the torque is equaI to twice the volume of sand (see Problem 5), the torque required to produce complete yieIding can readiIy be determined. Thus for a circIe of radius a, the volume of the sand hilI (in this case, a eone) is
where h is the height of the heap. Since the sIope of the sand hilI corresponds to the shear yieId stress k,
,
(:~r + (:~r = k
= 2Ga
2
h
k=a
The Torsion Problem [Ch. 11
248
and the torque is given by
(11.4.5)
Sec. 1151 Elastoplastic Torsion with Strain Hardening
249
where eo and ao are the yield strain and yield stress, respectively, related to each other by ao = Eeo and a is a characteristic linear dimension of the cross section. The system of equations to be solved for a simply connected cross section can now be written
Similarly, for a rectangle with dimensions a x b [8], 2 M p = 112 a (3b  a)k
(11.4.6) (11.5.2)
Il .. 4 1 shows the sandhill analogies for the above two cases. · F 19ure
u=o
on boundary
(11.5.3)
BU
 Bg foE
FIGURE
11.4.1
b
(11.5.4) (11.5.5)
)oj
Sandhl'll analogies for circular and rectangular cross sections.
2
Et
'/22
= V3 v
Ey
+ Ex
(11.5.6)
115 ELASTOPLASTIC TORSION WITH STRAIN HARDENING
(11.5.7)
. blem for strainhardening materials has received relativ~ly The torslOn pro. . , t '11 be shown that the method of succeSSIVe little attention. In thlS sectl.on l Wl . i ns can readily be adapted to the elastic solutions or succeSSIve ap~ro;'lmat o then becomes a simple limiting
(11.5.8)
torsion problem. The pe;fe::~bie~s ~~ ~::ar strain hardening. The plastic case of the mo~e genetr~ will be used and for convenience the following straintotal stram equa lOns , dimensionless quantities are introduced.
u=_rP
 2Geoa
7'
7' xz
= x  2Geo
l + P, = 7' xz ao
7'
g = x/a
'T}
= Y 
=
7'z y
2Geo
y/a
1+p,
= ao
7'YZ
The relationship expressed in equation (11.5.7) is obtained from the uniaxial stressstrain curve and, as is evident, relations in terms of plastic strainstotal strains are being used. The successive approximation method proceeds in the usual manner. The plastic strainsare assumed to be zero everywhere. Equations (11.5.3) are solved by any available method. The stresses, the total strains, and equivalent total strain are computed by means of equations (11.5.4) through (11.5.7) with the help of the stressstrain curve. If at any point in the cross section the equivalent plastic strain as computed from (11.5.7) is negative, this point is in the elastic region, and the plastic strains at this point are set equal to zero. Otherwise, new approximations to the plastic strains are calculated by means of equations (11.5.8). One then returns to equation (11.5.2) and process repeated until convergence is obtained. The method wilI be illustrated for bars of rectangular and circular cross sections. For a circular cross with linear strain hardening the solution can be obtained in closed
The Torsion Problem [Ch. 11
250
116
BAR WITH RECTANGULAR CROSS SECTION
The eiastic solution for a bar with a rectanguiar cross section by means of finite differences was presented in Section 11.2. To obtain the eiastopiastic solution the function g(x, y) is subtracted from the right side of equation (11.2.15), which in terms of the dimensioniess quantities defined in (11.5.1) becomes
Sec. 116] Bar with Rectangular Cross Section
251
The results of such a calcuiation for a square cross section with Il x Il grid points as shown in Figure 11.2.3 are shown in Figures 11.6.1 through 11.6.4. In these calcuiations linear strain hardening was assumed. Equation (11.5.7), reiating the equivalent piastic strain to the equivalent totai strain, can then be written (see Section 7.9)
Ep
= 1 + j{l
+ ft)(mf1
(11.6.3)
 m)
where gl,j
1 (P = 2H Ex,l,!+l

P Ex,l,jl 
P ElI,Hl,j
+ ElI,ll,! P )
4.0
(11.6.2) 3,2
and H is the grid spacing divided by a. The onIy difference between equations (11.6.1) and (11.2.15) is in the numbers appearing on the right side, which now depend on the piastic strains and change from iteration to iteration. Equations (11.6.1) can tlierefore be solved in the same way as equations (11.2.15) for the eiastic case. Once the values of U are determined at all the grid points, corresponding values of piastic strains are computed by means of equations (11.5.4) through (11.5.8), as fully described in Section 11.5. The gl,! are then recomputed and equations (11.6.1) solved again, the process being repeated unti! convergence, is obtained.
2.4
x
o Q.E
'"
1.6
0.8
o FIGURE
11.6.2 Variation or E:;'ax with f3.
m 1.0 m
1.0 4,0
2.0
3.2
1.6
M* )(
1.2
o E
...
0.8
QQ5 ~ _
2.4
O
o FIGURE
6
11.6.1 Variation or T max with f3.
FIGURE
11.6.3 Variation or M* with f3.
The Torsion Problem [Ch. Il
252
where the strainhardening parameter m is the ratio of th~ slope of the linear hardening curve to the slope of the elastic curve, as pr~vlOusly ~efined. For the perfect1y plastic case m is equal to 0, and for the elastlC case m lS equal to 1.
Sec. 117] Bar with Circular Cross Section
253
The figures show the effects of the strainhardening parameter and the angle of twist on the maximum stress, the maximum plastic strain, the size of the plastic zone, and the torque. The results are also summarized in Table Il.6.1. Although the calcuiations are described using deformation theory, a similar calcuiation, increasing et in steps, gave aimost identical results. This is in agreement with similar calcuiations in reference [11]. As shown in reference [12], incrementaI and deformation theories give identical results for a perfectly piastic material of any cross section or a strainhardening materiai of circuiar cross seption. For strainhardening materiais of noncircuiar cross sections they will yie1d different results. It appears, however, that the differences will in generaI be slight.
117 BAR WITH CIRCULAR CROSS SECTION For a bar with a circular cross section the solution is greatly simplified. In particular, for the case of linear strain hardening, a c1osedform solution can be obtained. In polar coordinates the displacements are Ur =
m
f3
M*
T max
€~ax
o
2 3 4 5
1.786 1.918 1.955 1.977
0.751 0.751 0.751 0.751
0.820 1.824 2.851 3.959
2 3 4 5 6
1.813 1.997 2.094 2.166 2.228
0.785 0.825 0.862 0.899 0.934
0.758 1.623 2.434 3.240 4.003
1.838 2.073 2.223 2.347 2.465
0.818 0.893 0.963 1.032 1.099
0.701 1.478 2.209 2.919 3.618
1.890 2.220 2.471 2.717 2.966
0.881 1.022 1.156 1.290 1.426
0.600 1.250 1.870 2.488 3.116
0.05
0.10
2
3 4
5 6 0.20
2 3 4
5 6
O
Ue
= etrz
(11.7.1)
and the only nonzero strain is (11. 7.2) The stressstrain relation can therefore be written
(11. 7.3) The von Mises equivalent stress reduces to (11. 7.4) and the equivalent plastic strain is
(11.7.5)
(11.7.6)
t,
, ì
The Torsion Problem [Ch. 11
254
Sec. 117] Bar with Circular Cross Section
255
and the criticaI angle of twist per unit length will be
Let
r (i==P
(Je  S 2Gso = e
(11.7.7)
(11.7.13) or
where a is the radius of the bar. Then equation (11.7.6) can be written in dimensionless form as
(11.7.8)
The stressstrain curve can be written in dimensionless form as
(11.7.9)
To summarize, the strainhardening solution is found as follows. The elastoplastic boundary Pc is first determined from equation (11.7.11). The stress and strain in the elastic region for P :::; Pc are then computed from equations (11.7.2) and (11.7.3) with s:z set equal to zero. In the plastic region P > Pc, equation (11.7.10) is solved, usually by an iterative method. s:z can then b~ computed from equation (11. 7.5) and the shear stress from (11. 7.3). Once the shear stress is known throughout the section, the torque can be computed by integration. Let us now consider the case of linear strain hardening. Equation (11. 7.9) for the stressstrain curve can be written
and combining with (11.7.8) results in
(11. 7.14) Hence
equatio~
(11.7.10) becomes
which can be solved iteratively for €p' Equation (11.7.10) is valid only in the plastic region. Let this regio n exten.d  l • To determine the position of the e1astoplastrc bet ween P = Pc and P boundary, i.e., Pc, let Sp = O when (Je = (Jo or when €p
Hence from (11. 7.8)
+ p,) V3 (3
2(1 Pc =
3
V3 (3p  2(1 + p,) + 2(1 + p,)m/{l  m)
il
M 3 = 27T' M * = 2G TOP 2dP soa o substituting
(3c
 2(1 
v3+ p,)
P
~
Pc
(11.7.15)
Note that the criticaI value of P is obtained when the numerator of (11. 7.15) vanishes, which resuIts again in equation (11.7.11). Once the equivalent plastic strain is known from equation (11.7.15), the shear strain and the stress are computed from (11.7.5) and (11.7.3). have thus obtained a complete solution in closed formo To compute the define 1'0 == Toz/2Geo. Then
which depends only on Poisson's ratio and the yield str~in but no~ on stressstrain curve. The value of (3 at which plastic flow )ust starts 1S from equation (11.7.11) by setting Pc equal to 1. Thus the criticaI value will be
=
_{tf3P 1(3
1'0 
"2 P 
P
€o
P :::; Pc P ~ Pc
(11. 7.16)
The Torsion Problem [Ch. 11
256
results in
7rt 
M* = where
p~) + tB(l
v'37r[tA(1 
 p2)]
v'3f3
A = 3
B=
+ 2(1 + p,)m/(l 2(1 + p,) 3 + 2(1 + p,)m/{l
m)
(11.7.17)
Sec. 117] Bar with Circular Cross Section
257
ResuIts of eomputations using the above formulas are shown in Figures 11.7.1 through 11.7.3. Figure 11.7.1 shows the elastoplastie boundary as a funetion of f3. Figure 11.7.2 shows the effeet of the strainhardening parameter on the shear stress for f3 = 5.0, and Figure 11.7.3 shows the effeet of these parameters on the torque.
 m)
Note that for Pc = 1 (no plastie flow) the torque reduees to ~he elastic torque as given in equation (11.2.12). For a perfeet1y plastie matenal, m
and or
=
°
A =
~3
B
= i(1 + p,)
11111111!!~~~§~§~===:0.10 g.05
0.8
M* = 2'1r(1; p,) (1  !pn 3 3 M =
m
0.20 0.15
1.0
2;~io [1  12J3a3 (~:r]
and we reeover the classical solution as given in referenee [lO].
Te
,...
0.6
004
(11.7.18)
o
0.2
004
0.6
0.8
1.0
P
20 FIGURE
11.7.2 Variation of TO with P for various m: f3
18
= 5.0.
16
14
m 0.20
4
12 3
1310
M*
8
2 6 4
2
o
o
0.2
11.7.1
Variation of pc with f3.
8
12
16
20
13
Pc FIGURE
4
FIGURE
11.7.3
I
Variation of M* with f3 for various m.
I l!
I
l;
258
The Torsion Problem [Ch. 11
9.
Problems 1.
2.
3. 4. 5.
6. 7. 8.
Show that the Ioading is radiaI for the torsion problem of a bar with a circular cross section, so that totai plasticity theories may be used, as long as there is no unloading. Show that the maximum shear stress for a solid bar of elliptic cross section under torsion occurs on the boundary at the point c10sest to the axis of the bar. Determine the stresses, the torque, and the warping function for the triangular cross section bar of Figure 11.2.1. Show that if the Iaterai surface of the bar is stressfree, the resultant shear stress must be tangent to the boundary. Show that if p/S is equai to 2Ga, the applied torque acting on a bar is equal to twice the volume between a membrane of the same shape as the cross section of the bar and the z = O pIane. Determine the torque acting on a bar of circular cross section by calculating the volume under a membrane of the same shape. Show that for the torsion problem the yield criteria of von Mises and Tresca both reduce to equation (11.4.1). Using the SaintVenant assumptions (11.1.4), show that for a bar of circular cross section the radiaI and tangential displacements become Il,
9.
Generai References
=O
IlO
= arz
and consequently the only nonzero strain is eoz = tra. Calculate the torques required to produce a twist of 0.004 rad/in. in a 2in.diameter shaft if the material is perfectly plastic, and if the material strain hardens with m = 0.1. Assume E = 30 X 106, P, = 0.3, and Uo = 30 X 103•
References 1. B. SaintVenant, Mémoire sur la torsion des prismes, Mem. Acad. Sci. Math. Phys., 14, 1856, pp. 233560. 2. S. Timoshenko and T. N. Goodier, Theory of Elasticity, McGrawHill, New York, 1951, p. 275. 3. V. Kantorovich and V. I. Krylov, Approximate Methods of Higher Analysis, P. Noordhoff, Groningen, 1958, p. 70. 4. L. Prandtl, Zur Torsion von prismatischen Staeben, Physik. Z., 4, 1 pp. 758759. 5. I. S. Sokolnikoff, Mathematical Theory of Elasticity, McGrawHill, New York, 1956, p. 117. 6. A. Nadai, Der Beginn des Fliessvorganges in einem tortierten Angew. Math. Mech., 3, 1923, p. 442454. 7. A. Nadai, Theory of Flow and Fracture of Solids, VoI. 1, M(~GJrav{E[jll, New York, 1950. 8. W. Johnson and P. B. Mellor, Plasticity for Mechanical Engineers, Nostrand, Princeton, N.J., 1962, p. 132.
~:J~'l~~~ga,
259
Matrix Iterative Analysis, PrenticeHalI, Englewood Cliffs,
lO.
W. Prager and P G Hodg J Th New York, 1951, ·P. 72. e, r., eory of Perfectly Plastic Solids, Wiley,
11. 12.
J. H. Huth, A Note on Plastic Torsion J. A I M h W. Prager, An Introduction to the' Mafte~at~C ~~ 1955, pp. 432:~34. J. Appl. Phys., 18, 1947, pp. 375383. lca eory of PlastIclty,
i'
GeneraI Reference Johnson, W. and P. B. Mellor Plasticity for M h . Princeton, N.J., 1962.' ec amcal Engineers, Van Nostrand, ,l' Prager, W., and P G Hodge J Th York, 1951. '. ,r., eory OJ Perfectly Plastic Solids, Wiley New
Sec. 121] PIane Strain Proble m of a R19ldPerfect1y Plastic Material O
CHAPTER
12
o
Bx = Bx(X, y)
By = By (X, y)
a x = a x ( X, y)
a y = ay(x, y)
a z = az(X, y)
'TXY
261
(12.1.1)
= 'TXY(x, y)
Since 'T z = 'T = O it ~ 11 h . and a i: a pri~cipaI' stres~. ows t at the z dIfection is a principaI direction z
. For a rigidpIastie materiaI th l ' totaI strains and strain incre~ et e astle stralllS are neglected, so that the l' . en s are equa! to the corresp d' strallls and strain increments, and the Lé _ . . on lllg astle vy Mlses relatlOns result III [see equation (7.2.5)]
?
(12.1.2)
TRE SLIPLINE FIELD and since
121
PLANE STRAIN PROBLEM OF A RIGIDPERFECTLY PLASTIC MATERIAL
In the previous chapters it was shown how the successiveapproximation method can be applied to a variety of probIems, inc1uding pIane strain and pIane stress eIastopIastic probIems. In these probIems the constrain,ts imposed by the eIastic parts of the materiaI prevented unrestrained pIastic flow. In many metaIforming processes, such as rolling, drawing, forging, etc., Iarge unrestricted pIastic flows occur exeept for very smaH eIastic zones. For sueh probIems it may not be unreasonabIe to negIect the eIastic strains and assume the materiaI to be l'igidpel'fect/y plastic, as shown in Figure 2.6.1(b). Any eIastic part of the body is then assumed to act as a rigid inc1usion and the pIastic parts can flow freeIy at constant equivaIent stress. A great dea! of work has been done on soIutions of this type of probIem under conditions of pIain strain, using the theory of slip lines. This will be devoted to a brief discussion of this theory. We begin by writing equations of pIane strain for a rigidpIastic body. By pIane strain is the condition wherein the dispIacements aH occur in paraHeI pIanes in body, say, pIanes paralleI to the xy pIane, and aH stresses and strains independent of z; i.e., BZ = Bxz = ByZ = 'T xz = 'TyZ = O
260
dBz
= O, the last equation gives (12.1.3)
and also the mean stress is (12.1.4)
The von Mises yield criterion for this case becomes (12.1.5)
where k is yield stress in simple shear and h b . inste d f Th ". 'T as een wntten for brevity e eqU1hbnum equations to be satisfied are a o 'T xv·
8ay 8y
+
8'T 8x = O
(12.1.6)
Equations (12 .1.6) and (12 .. 1 5) represent three equations in the three unkno y of stre wns a x, a , and. 'T. If the boundary conditions are given only in terms any re;:::~~~~: t~qU~tlOns are .sufficie?t to give the stress distribution without W e ~ ressstralll relatlOns. Such problems are ealled statically . ·s : pOlllted out a similar situation in discussing the sphere In ectlOn 8.3. However, if displacements or velocities are specified
262
The SlipLine Field [Ch. 12
.
then thestressstrain relations must be used to f the b oundary over par t o , h comrelate the stresses to the strains and the problem becomes muc more of and k plicated. . . t The principal stresses in the plastic field can be wntten m erms G m as follows:
Sec. 121] Piane Strain Problem of a RigidPerfectly Plastic Material
263
Having determined the principai stresses and directions, the maximum and minimum shearing stresses and directions can readily be determined. These shearing stresses act on the pianes bisecting the principai directions as described in Section 3.4. Their values are given by
(12.1.10) (12.1.7)
or, from (12.1.4) and (12:1.5),
The maximum shear directions will be designated by the ex and ,8 directions. ex, called thefirst shear direction, is taken 45° clockwise from the first principai direction, as shown in Figure 12.1.1, and,8, the second shear direction, is 900 counterclockwise from the first shear direction or 45° counterc1ockwise from the first principai direction, as shown. Let B be the angie which the first shear direction makes with the x axis (measured counterc1ockwise). Then B = e/>  45°
(12.1.8)
1 tan 2B =  tan 2e/>
t step is to find the prindpal directions. We define thefirst principal direc~:ne:s the direction of the maximum principal stress. L~t e/> ~e the angle between the first principal direction and the x axis as shown m FIgure 12.1.1. Th
and, from (12.1.9), tan 2B
=
G
y

2'T
G
x
(12.1.11)
It follows therefore that y
FIGURE
12.1.1 Principal directions and ex and,8 lines.
· (3 .. 3 6) for the prineipal directions it follows that Then from equat lOn
tan 2e/> =
2'T Gx
_
G y
.. WhICh glVes two va1ues of 't'I. differing by 90°. The second principal direction is taken 90° counterc1ockwise from the first.
cos 2B =
ii
. 2B sm =
Gy 
(12.1.12) 2k
Gx
At every point in the piastic fieId, the angle which the maximum shear direction makes with the x axis is determined by equations (12.1.11) or (12.1.12). If curves are now drawn in the xy pIane such that at every point of each curve the tangent coincides with one of maximum shear directions, then two families of curves called shear lines, or slip lines, will be obtained. Obviously, since the maximum and minimum shear directions at a point are orthogonal to each other, the two families of slip lines will form an orthogonai set. These two families of curves will be called the ex lines and ,8 lines, respectiveIy.
It shouId be carefully noted that aiong an ex line ex is varying and ,8 is constant, and aiong a ,8 line ,8 is varying and ex is constant. ex and ,8 are mereIy parameters or curvilinear coordinates used to designate the point under
The Slip.Line Field [Ch. 12 ,
264
the point P shown m ( Thus f.i ) con SI'der at1'on , J'ust as x and y designate the point. 12.. 1 2 can be designated P (Xl> Yl) or p 0:3, {"2 • ' F 19ure f3 lines
~
Sec. 121] Piane Strain Problem of a Rigid.Perfect1y Plastic Materia!
265
From (12.1.14) it is seen that the state of stress can be determined in terms of two independent quantities, a m and e. The equilibrium equations can be written in terms of these quantities by substituting (12.1.14) into (12.1.6). Thus
oa ox
E! 
oe. oe) 2k (cos2eox + s1ll2eoy
oa ( oe. oe) E! + 2k cos 2e   Slll 2e oy oy ox
= O
(12.1.15) = O
or defining x
FIGURE 12.1.2 Families of et and (3 lines. we can write , n the maximum shear planes equals the t' (3 4 5) Thus the The normal stresses actmg o . . l tresses as shown by equa 10n .. . 'f l t the o: and (3 lines are given by average of the pnnClpa s stresses acting normal and tangen la o
ox _ cos 2e oe _ sin 2e oe = ox ox oy
O O
(12.1.13)
oX _ sin 2e oe + cos 2e oe = oy ox oy
(12.1.16)
Now the choice of the x and y axes is arbitrary. If we choose the x and y axes at a given point to coincide with the o: and (3 directions at this point, then e = Oand
as illustrated in Figure 12.1.3. y
o o ox = 00:
o oy
=
o 0(3
and equations (12.1.16) become
ox oe 0(3 + 0(3 =
L~x
FIGURE 12.1.3 Stresses normal and tangential to Finally, we can express a x , a y, an d
7'
et
and (3 lines.
in terms of a m and
e, as follows:
ax =am ksin2e ay = a m + k sin 2e = k cos 2e
(12.1.17) O
Equations (12.1.17) are called compatibility equations (not to be confused with the strain compatibility equations). Each equation contains derivatives in only one direction. Integrating, along the o: curve along the (3 curve
(12.1.18)
7'
. d by use of Mohr's diagram. An these results can a1so be ob tame
where Cl and C2 are constants. These equations were first derived by Hencky in 1923, [14].
266
The SlipLine Field [Ch. 12
Sec. 122] Velo city Equations
From equations (12.1.18) it is apparent that if X and 8 are prescribed on the boundary, it should be possible to proceed along constant a and fllines to determine X = a m/2k and 8. Ifthe displacements or ve10cities are prescribed over part of the boundary, as is frequently the case, these equations are not sufficient to obtain a solution and the velo city equations following must also be used.
122
267
~~~;ename. are no extensions,
only shearing flows in the slip directionshence
Now consider the velocities in the slip directions. From Figure 12.2.1, y
VELO CITY EQUATIONS
The LévyMises relation (12.1.2) can also be written q
d8 X  d8 y d8 XY
_ 
ax 
ay
(12.2.1)
7'XY
~~~~x FIGURE
In addition, the incompressibility condition with d8 z = O is
12.2.1 Velocities in a and fl directions.
(12.. 2.2)
It is convenient to divide the strains by dt, the increment of time, and write
these equations in terms of velocities. Of course, these equations remain homogeneous in t, which acts merely as a scaling parameter. Then
=
Va
cos 8  vp sin 8
v =
Va
sin 8
Vx y
8 = O, t1he x ~xis will coincide with the a direction and the condition that e norma stralll rates be zero can be written
.
where Vx is the velocity in the x direction; i.e., Vx = du/dt and similarly vy = dv/dt. Equations (12.2.1) and (12.2.2) become ax 
(12.2.5)
:!
8a
(8v x /8x)  (8v y /8y) _ (8v x /8y) + (8v y /8x) 
+ vp cos 8
= (8V  x)
= O
8x
0=0
. = (8V 8y
0=0
y)
8p
=O
or, from (12.2.5),
ay
27' 8va 88 8x  vp 8x = O
(12.2.3)
Now since the principal axes of stress and of plastic strain increment coincide (see Section 7.2), it follows that the maximum shear stress lines and maximum shear velocity lines coincide, or that the stress slip lines are the same as the velo city slip lines. Also the strain rates normal to the a and p directions are equal to the mean strain rates [see equation (4.5.4)]. Therefore,
88 va 8y
+
8vp
8y =
(12.2.6)
O
~hre' since the hX dire~tion. is the same as the a direction and the fl direction is same as t e y dlrectlOn,
, I ,
88
8va p 8 = O a  v 8a 8vp 8fl +
Va
88 8fl = O
(12.2.7)
,
268
The SlipLine Field
[Ch. 12
Sec. 123J Geometry of the SlipLine Field 269
lf f3 is kept constant in the first equation and a in the second equation, we can write along an aline dv«  Vp de = (12.2.8) along a f3 line dvp + v« de =
1. He~cky's first ~heorem states that the angle between two slip lines of ?ne famIly at the pomts where they are cut by a slip line of the other family IS constant along their lengths. This is shown in Figure 12 31th I e1 and e2 being equa!. . ., e ang es
° °
These are the compatibility equations for the velocities first derived by Geiringer in 1930, [15]. lf the problem is statically determined, the slip line field and the stresses can be found from equations (12.1.18) (or their equivalent) and the stress boundary conditions. The velocities can then be computed from (12.2.8) (or their equivalent) using the velocity boundary conditions, since de will now be known from the stress solution. lf, however, the problem is not statically determined, which means that the stress boundary conditions are insufficient to obtain a unique slipline field, then equations (12.1.18) must be solved simultaneously with (12.2.8) using both the stress boundary conditions and the velo city boundary conditions. This is an extremely difficult problem and must usually be done by trial and error. A slipline field satisfying all stress conditions is assumed. The velocities are then computed and a check made to see if the velo city boundary conditions are satisfied. lf not, the slipline field is modified and the procedure repeated as often as necessary. This is obviously a very laborious process, since the construction of just one slipline field is a lengthy task. It is worthwhile to note some of the differences between Hencky's stress equations (12.1.18) and Geiringer's velocity equations (12.2.8). 1. Hencky's equations relate two unknowns, X and e, by two equations. Geiringer's equations relate three unknowns, v x , V y , and de, by two equations. 2. Hencky's equations give the stress state all along a known slip line ifthe stress state is known at one point on the slip line. Geiringer's equations will not give the velocities along a known slip line, if they are known at one point. 3. Hencky's equations force certain restrictions on the geometry of the slipline field, as we will shortly see. Geiringer's equations pIace no restriction on the geometry ofthe slipline field, except through the boundary conditions.
123
GEOMETRY OF TUE SLIPLINE FIELD
Hencky's equations, as mentioned above, impose some rather severe restrictions on the geometry of the slipline field, which are a great aid computation. We willlist a number of these but prove only a few to the method of attack.
FIGURE 12.3.1 Demonstration of Hencky's first theorem. 21: A) II a Iines (f3lines) turn through the same angle in going from one f3 line (a me to another.
~.
lf on~ a line (f3 line) is straight between two f3 Iines (a lines) then alI a lmes (f3 !mes) are straight between these two f3 Iines (a lines). Furthermore these straIght segments have the same length. ' 4. lf both the a and f3 Iines t' h . . in the . .are . s raIg t m a certam regI'on , alI the stresses reglOn are ~onstant. ThIS IS called a field of uniform stress state. 5. Along a straIght shear line the state of stress is constant 6. lf the. state of stress is constant along a curve, then eith~r the curve is e~be~~ed I~.~ field of constant stress or else the curve is a straight shear line Ii' e ~a II of curvat~re ~f the a Iines (f3 Iines) where they intersect a give~ f3 ~~ (ad~me). decrease m duect proportion to the distance traveled in the pOSIIve IrectlOn~: the f3 line (a line). Therefore, if the plastic zone extends ;;.r el~oUgh, the radlI of curvature eventuaIIy become zero, so that neighboring ~ne~run together and the solution ends at the envelope of the sIip lines IS IS uencky's second theorem. .
T:
tur~ ~:::e sroc~ed along a given sIip .line of one fami1y, the centers of curvap lmes of the other famIly form an involute of this sIi l' Of~h Thl~ el~velope of the slip lines of one fami1y is the locus of t~e I~:~PS e s IP mes of the other family.
270 The SlipLine Field [Ch. 12 Sec. 123}
lO. The envelope of the slip lines of one family is a limiting line across which the shear lines of the other family cannot be continued. Il. If the radius of curvature of an aline (ft line) jumps discontinuously as it crosses a f3line (a line), al1 a lines (f3 lines) crossing the f3 Ii ne (a line) wiII suffer the same jump in radius of curvature. This also means that the derivatives ofthe stresses are discontinuous across the slip line.
Geolnetry of the SliPLine Field
271
There are many similar theorems, but they are not of practical interest. Hencky's first theorem can easily be proved as foIIows. Referring to Figure 12.3.1, along the a line AD, the first of Hencky's equations (12.1.18) gives
c
D
I ~ R/3c,q cr
and along the f3 line CD, the second equation gives
> (
~ R/3 FIGURE
Therefore, (12.3.1)
Also along AB, and along BC
12.3.2 Proof of Hencky's second theorem.
~p,
of curva'ure R, ha, decreased by an amoun' and a' 'he poin' D, Rs has decrea,ed by an amo un, 'o firs' order of 'mali quan'i'ie,. In tbc limi,
~',
XA + BA = XB + BB Xa  Ba = XB  BB
Therefore, (12.3.5) (12.3.2) or, comparing with (12.3.1),
or, more conveniently for computational purposes, since Aa == R"AB and Af3 == RpAB, equations (12.3.5) can be written (12.3.3)
AR"  Rp AB == O ARp + R" AB == O
which proves Theorem 1. Theorem 2 is a direct corol1ary, since from equation (12.3.3),
along a f31ine along an a line
(12.3.6)
Equations (12.3.5) are the mathematical statement of Theorem 7 and hence Theorem 7 is proved. , Theorem 3 is also a direct corol1ary, since if one of the lines is straight, say AD, then BD  BA = O and therefore Ba  BB = O, so the other line is also straight. Similatly Theorems 4 through 6 foIIow directIy from Hencky's equations (12.1.18). Theorems 7 through lO are based on the theory of pIane curves. Thus let the radii of curvature of the a and f3 lines at the point A in Figure 12.3.2 be designated by R" and R p, respectively. At the point B the radius
The proof, .of the o'ber ,heorem, are given in Prager aud Hodge [lJ and "'11 not be gwen here. Figure 12.3.3 illu,tra'" 'ome of 'he" prope"i". ABCD and A'B'C'D' are neighboring a shear lines. AA' BB' CC' and DD' are.infinitesimal arcs ofthe f3lines. The center of curvat;re or'these'arcs form an Involute PQRS of the slip line ABCD. (An involute is the curve obtained by unwinding a fiexible "ring originally lY;ng on 'he curve, tba' 'he "ring 'angen, 'o 'he curve. The original curve is 'he evolu'e of the involute aUd" tbe locu, of the radii of curvature of the involute.) At tbc point T where the lOvolu'e PQRST mee', the line ABCDT, the di,tance be'ween tbe
"a1~ay,
'0
~ip
!!r1Mi.'W.,y? The SlipLine Field [Ch. 12
272
Sec. 124] Some Simple Examples
273
the straight lines be the (X lines and the circular arcs the f3 lines. Then from Hencky's first equation, since e is constant along an (X line, X must also be constant along an (X line, and, from the second equation, since evaries linear1y with distance aiong a f3 line, X must vary Iinearly with distance aiong a f3line. Thus the mean stress is constant in the radiaI direction and varies Iinear1y with the angie measured from the x axis. To find the stress components we then make use of equations (12.1.14). This type of slipline fieid is called a centered fan. Note that the center of the fan is a singuiar point of the stress field, since it can have any one of an infinity of values.
FIGURE
,
.
12.3.3 Involute and limiting line. (Reference [1].)
. lines becomes zero, and so does the radius of curvature of
~:~g;~i~:~~r~:h T, as stated in ~heorem 7, It is a point on the envelope of the
(X
lines and is a cusp of the
f3 Ime.
Indentation by a Punch We now consider a probIem which combine s, or "patches" together, the state of uniform stress and the centered fanthe indentation of a semiinfinite body by a flat rigid punch in the form of an infinite strip. Figure 12.4.2 shows a typicai pIane [2]. It is assumed that the surface of the punch
124 SOME SIMPLE EXllMPLES
y
qlines
/3 lines
State of Uniform Stress lf the stress is constant throughout the fie1d, the slip lines form two ,sets of orthogonal straight lines. This follows direct1y fro~ ~encky's eqUatlOn~ (12.1.18), for if X is constant, then e is constant. ThlS IS the converse o Theorem 4 of Section 12.3.
" '''Centered Fan
FIGURE
'd SII'pline field composed of a set of radiallines originating from C onSl er a 'F' 12 4 l Let a point and a set of concentric circular arcs as shown m Igure . . . C(
FIGURE
12.4.1
lines
Centered fan.
12.4.2 Slip lines under lubricated flat punch.
and body are perfectly lubricated, so that there is no friction between them. It is also assumed that there is a constant pressure over the face of the punch. For the boundary conditions we have that over the segment AB there is a uniform pressure kp, and the rest of the boundary is stressfree. We consider only the case of incipient plastic flow, since once plastic flow progresses, the shape of the boundary GABC changes considerably and it is necessary to satisfy the boundary conditions on the deformed boundary. Assume now that plastic flow occurs over a segment AG of the free boundary as shown. The length of this segment is as yet not known. From the boundary condition on this segment
onAG
274
The SlipLine Field [Ch. 12
Sec. 124J Some Simple Examples 275
From the yie1d condition,
are at 135° (or 45°). As before, region AEB is a constant stress region, the slip lines being straight lines with 8 = in. From the boundary condition,
X=
it folIows that Ux
Intuitively we would expect
Ux
= ±2k
8
to be compressive and we tentatively assume onAG
= in
X=
t
X=
t}
71"
8=4
y
l 2
P}
inAEB
(12.4.5)
to
X
lp
=~
alongAE
Similar results hold in the regions GBD and BDC. The pressure p exerted by the punch to produce this state can readily be determined. The li ne AF is an IX line and line HIJK is a f3 line. The compatibiIity relation to be satisfied along the f3 line (Hencky's second equation) is
+8=
constant
AlongHI, inAGF
(12.4.2)
Now consider the boundary AB. Since it has been assumed that there is no friction, T = O along this boundary, so that U
+ Uy =
4k
along AF
X
it folIows that
x
Now AF and AE are straight IX slip lines and it follows from Theorem 3 of Section 12.3 that alI the shear lines in between these two are straight, or region F AE is a centered fan. The stresses are then constant along any radiaI line from A to the arc FE and vary linearly along any arc such as IJ from the value
(12.4.1)
Since the shear stress is zero, AG is a principal direction and the slip lines must be at ± 45° with AG. This also follows from the last of equations (12.1.14). The IX lines make 45° angles with AG and the f3lines 135° (or 45°), as shown in Figure 12.4.2. Consider the triangular region AGFformed by AG and the slip lines AFand GF. By Theorem 6 of Section 12.3, this is a constant stress regipn. The slip lines are straight lines with 8 = 71"14. The mean stress X is a constant and must satisfy equations (12.1.18) throughout this region. Since on the boundary AG
U
=  kP} along AB
1'=0
X
= !
8
=
i
X
= t(1
 p)
and
8
= in
Hence
! + (12.4.3)
and
AlongJK,
i = te1  p) + in
or (12.4.6)
Therefore,
(U x~ or
Ux
=
Uy
The velocity distribution is readily determined from the Geiringer equations (12.2.8). Ifthe punch is moving with a velocity Uo in the negative y direction, then region ABE moves as rigid body attached to the punch with the same
uyf = k 2 ± 2k = k(2  p)
(12.4.4)
where the plus sign has been chosen (see Problem 7). It folIows then, just as for the segment AG, that AB is a principal direction and the slip lines make ± 45° angles with AB. This time the f3lines are at 45° with AB and the IX lines
velocity. In region AEFG, v'" equals zero and vp equals Uo/VI Region AEF thus moves out with velocity Uo/V2 and region AGF moves in the direction FG with the same velocity. The above solution was obtained by Prandtl [3J. An alternative solution given by Hill [4J, assumes the rigidplastic boundary to be HIJKLMN instead
276
The SlipLine Field [Ch. 12
of GFEDC. An analysis similar to the previous one shows that regions AHI and AJK are constant state regions and AIJ is a centered fan. The stresses in these regions are the same as previously obtained but the velocities are different, the outward flow velocity being twice that of the previous solution. Actually an infinity of solutions can be obtained between the two limiting solutions discussed above. This illustrates one of the difficulties of the pIane strain solution for a rigidplastic material. More than one solution (or no solution) may be obtained for a given problem, and the "correct" solution may be impossible to ascertain. The only truly satisfactory method is to solve the complete elastoplastic problem using the Prandt1Reuss relations. This, of course, will in generaI be extremely difficult. It is often possible, however, to determine the most probable solution, and sometimes a minimum force criterion may be used. In addition, we note the nonuniqueness of the boundary values due to the quadratic yield conditions. Thus in equation (12.4.1) the negative sign was chosen for G x on the basis of intuition. If the plus sign had been chosen, the pressure exerted by the punch, equation (12.4.6), would have come out negative, which is impossible. So we know that the negative choice was correct. However, the correct choice of sign was not really known a priori, and this will often be the case.
125
Sec. 125] Numerical Solutions ol BoundaryValue Problems
.x FIGURE
12.5.1
NumericaI soIution of first boundaryvaIue probIem.
w~ich is a rough approximation to the true intersection point of the a~d f3 h~es. (If the IX and f3 lines happened to be straight lines, the inter
(1, 2), IX
sectlOn pomt would be exact.) The points (2 3) (3 4) (4 5) d t . d' , , , , , ,etc., can be e ermme approxlmately the same way From H k' . . enc y s equatlOns we have
NUMERICAL SOLUTIONS OF BOUNDARYVALUE PROBLEMS We can therefore solve for
In the above example of the punch indentation, the slip lines and the solution were obtained completely in closed form from the boundary conditions. In generaI, however, numerical or graphical methods will be necessary. In this section a brief discussion of the simplest numerical methods will be presented. For this purpose we must distinguish among three types of boundaries, as shown in Figures 12.5.1 through 12.5.3. Figure 12.5.1 shows a case where the boundary curve Co is not a slip line. The values of X and 8 are given on this boundary and it is desired to construct the slipline fieH We choose a number of stations on the arc Co and try to construct the lines passing through these points. The various shear lines are """'~5~'" 1,2, 3, etc., and a grid point at the intersection of the ith IX line with f3 line is designated i,j. Now consider the points (I, l) and (2, 2) on the curve Co. At the (1, 1) draw a straight line with angle 81 1> representing the IX line through point. At (2, 2) draw aline with angle 822 + 71/2. The two lines intersect
277
y
812

X22
+ 822
X12
and 812 , giving us
812 = X22
In the same way we find
811 = X12
X11

= X12
X11

+ 812
+ 822 + 811
(12.5.1)
(12.5.2)
2
We ~an now ?ro.ceed to find X and 8 at (1,3), (2, 4), (3, 5), and (4,6). We thus obtam the sh?hne .field and the stresses in the entire region bounded by C :~d t~he termmal shp lin~s AP and QP. A little reflection indicates that th: u 10n cannot. be carned beyond region APQ without some additional . ThlS leads us to the following theorem Given an arc C whi h ~~ not a slip lin~ a~d all the stresses acting at every ;oint along the a:c, th:n e complete shphne field and the corresponding stresses can be determined
l',
Il
The SlipLine Field [Ch. 12
278
within the region bounded by Co and the intersecting .termi~aI slip lines as shown in Figure 12.5.1. Region APQ is called the reglOn oJ mfluence of the . h d' b' l are Co· If the are Co is itself a slip line, then the prevlOUS met o 1S o VlOUS. y inapplicable. If a second slip line, intersecting the. first one,. as ~hown 1ll 12.. 52 , is also given , a solution can be obtamed. For 1f e 1S known · F 19ure
See. 126]
Geometrie Construetion of SlipLine Fields
279
similar to those previously outlined. Details of the solution techniques for all three types of boundaryvalue probIems, as well as methods for improving the accuracy, can be found in references [1], [4], and [5].
126
GEOMETRIC CONSTRUCTION OF SLIPLINE FIELDS
y
A geometrie construction for the stress and velo city fieIds, which is frequent1y very useful and Ieads to a better insight into the principles underlying slipline theory, has been suggested by Prager [6]. For this purpose we make use of two planes, called the stress pIane and the physical pIane, as shown in Figure 12.6.1. Consider a point p undergoing plastic flow. The stress vector A
y
c L~x FIGURE
12.5.2 Second boundaryvalue probiem.
along both slip lines, then e can be determined ~t t~e adjoining net points by use of Hencky's first theorem. The complete shpIme field can then ~e constructed within the quadrilateral shown in Figure 12.5.2. !o determme the stress es it is necessary to know the value of X at just one pomt on the boundary sli; line, for by use of Hencky's equations (12.1.18) X can then be computed throughout the region. . . Alternatively, ifthe curve Co is a slip line, a solutio~ can ~e obtamed 1f o~ a second intersecting curve, not a slip line, either X or e1S specIfìe~. The sol~tlOn can then be obtained in the region indicated in Figure 12.5.3, usmg techmques
~
rrx (a)
TI FIGURE
FIGURE
12.5.3 Third boundaryvalue probiem.
(b)
12.6.1 Stress pIane (a) and physicai pIane (b).
acting at the point p will depend on the orientation of the area element through the point p upon which it acts. This is shown in the physical pIane ofFigure 12.6.1(b). The figure shows the traces ofseveral area elements whose UVLLH"L~ lie in the xy pIane. These area elements actually contain the point p are here shown separated for clarity. The shaded side of a given trace ret,res:ents material, and the stresses shown are those transmitted from the ummal:teCl side to the shaded side. Instead of identifying an area element by direction of its normaI, it is convenient to identify it by the direction of
280
The SlipLine Field [Ch. 12
the trace of the element on the xy pIane. Thus angles will be measured counterelockwise from the negative y axis instead of the x axis. On the stress pIane, Figure 12.6.1(a), a Mohr cirele is plotted for the stress state at the point P of the physicai pIane. The Mohr's cirele is constructed using the following convention. A shear stress which will cause the element to rotate in a elockwise sense is considered as positive, counterelockwise as negative. Thus the stress state (a x,  T xy) on pIane P A is shown as point A of the stress pIane. On pIane PB, whose normai stress is a y, the shear stress is positive (elockwise) and (ay, T XY ) gives the point B in the stress pIane. We note that the angle
and exceeds the yieid limit. The multiplier À corresponding to such a admissibie stress fieid is called a statically admissible multiplier. The
À+
=
DI
De
(13.5.4)
The upper bound theorem is then stated as follows' Th' . smallest kinematically admissible multiplier' i e f < .À+ e safety factor IS the be summarized by the reiation ' .. , . Both theorems can (13.5.5)
Th~ above theorems furnish upper and Iower bounds for . the Ioad. The questlOn of the uniqueness of the l t' .d d . so u lOn must stIlI be umqueness of the safety factor follows from equation (13 5c05n)slItere . TIhe be shown [4 12 13] . '" can a so at the start ~f piasti~ ~s111g. the ~rincipie of virtual work, that the stress field
fo~~:; which faH on th~~::a~g~~~::~::t !~ ;7~~if!~t~~~:sa:~~ea~::rse:: mi~:t: :!:!I:h:x;:f~eF~!:!e l~s~. . ~f. ;~:s:ntdheAor1.embs"lct~nSidedr the indeters 111 an the end B is Ul
(b) FIGURE
13.5.1
CoIlapse of bea m with distributed load.
Limit Analysis [Ch. 13
316
simply supported and the distributed load is of intensity P. The beam will collapse when yield hinges are formed at A and at some as yet unknown point designated by x = g. The correct solution can readily be obtained. Let R be the reaction at the point B. Then the moment at any x is given by M(x) = R(L  x)  tP(L  X)2
(13.5.6)
M(O) = RL  tPL2 = MQ
(13.5.7)
M(g) = R(L  g)  tP(L  g? = MQ
AIso, since at x =
317
Equating the rates of work done, by the upper bound theorem PL2 T=3MQ
or
p+
= PL2 = MQ
(13.5.11) 12
:omdPari~g ,,:ith equation (11.5.9), it is seen that an upper bound has bee loun Wh1Ch lS reasonably dose. n
g,
Since yield hinges form at x = O and x =
Sec. 135J Theorems of Limit Analysis
. A .lo,,:er bound to P is obtained as follows. Calculatin the gl' moment d1stnbutlOn due to the load p [assuming M(O) _ M] Q resu ts III
g, M must be a maximum,
(13.5.12)
M'W = R + P(L  g) = O Solving equations (13.5.7) and (13.5.8) for R, RL
+ 2v.r2 =
PL2
+ 4V2
r = MQ = 2
p = MQ = 6

g = Lg =

(13.5.8)
g, and P, 4.82843
No,,: t.his m~me~t ~istribution is not an admissible one (not stati calI adm1ss1ble), Slllce 1t glVes values greater than M Th t h ' y Q. us e maX1mum value f th '" . o m o.ccurs at x/L = i'z and is equal to 25/24 If " . we erelore mult1ply alI the moments and l b Th . oad by 24/25, an adm1ss1ble moment distribution which ' y eorem 2 lS a lower bound, will be obtained. Hence p = 24/25 x 12 = 11.52
= 11.65685
(13.5.9)
.r
(13.5.13)
is a lower bound for the collapse load. The collapse load is thus bounded by
2  v 2 = 0.58579 11.52 :s; p :s; 12
Equations (15.5.9) give the exact answer to the problem. However, for more complicated problems the exact solution may become very difficult. An approximate solution can be obtained by finding upper and lower bounds. We note first that if some arbitrary value g were chosen for the hinge, an upper bound would result, for the correct value of g will cause the yield limit to be reached at a lower load. Let us then arbitrarily choose the midpoint g = t as the hinge point. Then if the hinge A rotates through an angle e, g will rotate through an angle 2e and the total internaI work will be
or we can write p = 11.76
± 0.24
which is sufficiently accurate for engineering purpose~. F~r greater accuracy one could now take the value x/L = 2_ h h prevlOUS moment distribution w a s ' ' . 12, W ere t e
~~;o~~n:en~:S~tion o:tainda new an: :;~:~r:~;~ra~:u~~n:n~h~~~:c;~~~~f:~ ower oun . The collapse load is theQ found to be 11.65674 :s; p :s; 11.65714
The externalload P moves an a verage distance of iLe and the external work We = PL(iLe) = iPL2 e
(13.5.14)
(13.5.15)
l bForbthe above simple example there is little difference in the amount of a or etween the exact solution and the method of upper d l b For l an ower ounds is fa;:~;e~~o~P ex problems, however, the method ofupper and lower bound~
318
Limit Analysis
[Ch. 13
136 METHOD OF SUPERPOSITION OF MECHANISMS In the simple examples previously discussed, it was possible to list alI collapse mechanisms and to determine which of these was the correct one. For complicated structures this is no longer possible and more systematic methods are needed to detennine the limit loads. One such generaI method will be briefly discussed in this section. The interested reader is referred to references [2, 14, 15, 16] for more comprehensive treatments. The method oJ superposition oJ meehanisms due to Symonds and Neal [17, 18] basically involves combining or superposing various elementary meehanisms to obtain the collapse mechanism, the true collapse mechanism being distinguished by the lowest collapse load. The method will be described by means of an illustrative problem taken from reference [17]. Consider the frame shown in Figure 13.6.1, loaded as indicated. Let the fulIy plastic moment in the legs be Mo and in the horizontal beams 2Mo. There are lO possible sections where plastic hinges can occur, as indicated by the numbers in the figure. The hinge in the beam with the distributed load may occur at some unknown distance x as shown. The first step is to determine the number of linearly independent meehanisms for the structure. It can be shown that this will equal the linear1y independent static equations of equilibrium, which in tum is equal to the number of bending moments necessary to specify the bending moment diagram completely minus the degree of redundancy of the structure; Le.,
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