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THESIS

 

”is LIBRAva *
Michigan State
University

 

 

 

 

This is to certify that the
dissertation entitled

Pure Azimuthal Shearing Deformations for an Extended Class
of Elastic Materials that are Characterized By a
Microstructurally Motivated lntemal Balance Principle

presented by

Hasan Demirkoparan

has been accepted towards fulfillment
of the requirements for the

Ph. D. degree in Mathematics

 

Wj Mum M

j r Professor’s Signature

' 06/13/2005

 

‘ Date

MSU is an Affirmative Action/Equal Opportunity Institution

 

PLACE IN RETURN Box to remove this checkout from your record.
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DATE DUE DATE DUE DATE DUE

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2/os'—"cm__—c:/crn momma-p.15"

 

fi.‘

Pure Azimuthal Shearing Deformations for an Extended Class
of Elastic Materials that are Characterized By a

Microstructurally Motivated Internal Balance Principle
By

Hasan Demirkoparan

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements

for the degree of
DOCTOR OF PHILOSOPHY
Department of Mathematics

2005

ABSTRACT

Pure Azimuthal Shearing Deformations for an Extended Class
of Elastic Materials that are Characterized By a

Microstructurally Motivated Internal Balance Principle
By

Hasan Demirkoparan

In the event of certain substructural reconfiguration in solids, it is sometimes
the case that the notion of simple deformation X —+ x must be broadened so as
to incorporate a more detailed kinematic description. This typically involves new
kinematic variables in addition to x = x(X). Each new kinematic variable is in turn
associated with an additional balance principle of the same tensor order as the new
kinematic field variable. Gurtin’s theory of configurational forces and the Ericksen
model of liquid crystals are examples of such theories.

Recently a new constitutive framework for treating finite deformations when con-
ventional elastic behavior is modified by the action of an additional microstructural
balance requirement was proposed by Pence and Tsai. This framework offers inter-
esting possibilities for the development of singular surfaces that can be interpreted
as locations of concentrated microstructural rearrangement. This thesis investigates

the solution of the pure azimuthal shear problem in this new framework.

To my parents and my wife,

iii

ACKNOWLEDGMENTS

I am very grateful to have known Prof. T. J. Pence both as a person and as a
mathematician. It has been a privilege to work with him, and this thesis would have
not been possible without his guidance. I would like to thank to the other thesis
committee members, Professors C. Y. Wang, C. Wei], H. Tsai and K. Promislov for

their advise and helpful discussions.

iv

TABLE OF CONTENTS

LIST OF TABLES ....................... vi
LIST OF FIGURES ............................................................................. vii
1 Introduction ................................................................................... l
2 .Pure Azimuthal Shear Problem for a Neo-Hookean material ........ 5
2.1 Introduction .................................. 5
2.2 Pure Azimuthal Shear Problem for a Neo-Hookean material ....... 5
3 The Internally Balanced Material ................................................. 14
3.1 Formulation of the Problem for the Internally Balanced Material ..... 14
3.2 Solution for Small k (6: << 0.") ....................... 21
3.3 Numerical Solution .............................. 27
3.4 Explicit Solution when d = a‘ (k = 1) ................... 39
3.5 Alternative Derivation from Energy ..................... 46
3.6 Direct Methods for Minimization when k = 1 ............... 56
4 A More Generalized Class of Material Models ............................. 74
4.1 Extended Material Model .......................... 74
4.2 Existence and Uniqueness for the Extended Model ............. 78
4.3 Numerical Solution for the Extended Model ................ 89
4.4 Regular Perturbation Solution for the Extended Model .......... 94 .
4.5 Boundary Layer Type Solution for the Extended Model .......... 100
4.6 Twist, Energy and Torque in the Extended Model ............. 114
5 Conclusions and Discussions ............ i ............................................. 118

A g(r) given by (3.115) is a strong local minimum of (3.168) if [till 5

‘Ilm .............................................................................................. 120
B Alternate Proof of Theorem 4.2.3 ................................................ 127
BIBLIOGRAPHY ............................................................................... 134

3.1

3.2

4.1

4.2

LIST OF TABLES

Normalized energies for A1 and Any for several 1/10 values when R.- = 1,

Ro=4andtb,-=O ............................ 61
Normalized minimum energies A4(nmin) for several 1,00 values when R.- =

1,Ro=4andt/J,-=0 .......................... 65
N umerically computed values of 1" and do for one term patching for several

t/Jo ando‘rvalues when R,=1, Ro=4 and 114:0 .......... 110
Numerically computed values of 7“, do and (II for two term patching for

several we and 61 values when R,- = 1, R0 = 4 and z/Ji = 0 ...... 111

vi

2.1

3.1

3.2

3.3

3.4

3.5

3.6

LIST OF FIGURES

Deformation of an originally vertical line segment for different amount
of twist we and w.- = 0. Specifically shown are we =
.75413, 1.50826, 3.0165, 5.49779 corresponding respectively to 43.20° ,
86.41°, 129.62°, 315° for a neo—Hookean material in standard nonlinear
elasticity theory. Note that R,- = 1 and Re 2 4 are used and multiples
of arcccs(Rf/Rg) = 1.50826 are selected in order to provide correspon-
dence with results of the more general theory that is the object of this
thesis ....................................

Comparison of normalized torque-twist relation when k = 0 and k = 0.02
for R..- = 1 and Re = 4. The dashed line shows the case k = 0.02.
The normalized torque is M / (47r h 61 R3123) ..............

Graphs of g(R) for five values of we: we 2 1, we = 2, we = 3, we = 5
and we=10 when R;=1, Re=4, w,=0 and k=0.1. For we=5
and we = 10 there is an apparent discontinuity at R,- -- 1 .......

Graphs of p(R) for five values of we: we =1, we = 2, we = 3, we = 5 and
we =10 when R,- =1, Re 2 4, 103': 0 and k = 0.1. The top curve
corresponds to we 2 1 and the middle one corresponds to we = 2. The
graphs of p(R) are almost identical for we = 3, we = 5 and we = 10
and so generates an essentially common lower curve ..........

Graphs of g(R) for five values of we: we = 1, we = 2, we = 3, we = 5
and we=10 when R,=1, Re=4, wi=0 and k=0.5. For we=2,
we = 3, we = 5 and we 2 10 there is an apparent discontinuity at
R, = ...................................

Graphs of p(R) for five values of we: we :..— 1, we 2 2, we = 3, we = 5 and
we = 10 when R.- =1, Re = 4, w, = O and k = 0.5. The top curve
corresponds to we 2 1. The graphs of p(R) are almost identical for
we = 2, we = 3, we = 5 and we = 10 and so generates an essentially
common lower curve ...........................

Graphs of g(R) for five values of we: we = 1, we = 2, we 2 3, we = 5
and we=10 when R,=1,Re=4,w,=0 and kzl. For we=2,
we = 3, we 2 5 and we = 10 there is an apparent discontinuity at
R,- = ...................................

vii

13

26

31

31

3.7 Graphs of p(R) for five values of we: we = 1, we = 2, we = 3, we = 5
and we =10 when R, =1, Re =4, w,- =0 and k = 1. The top curve
corresponds to we = 1. The graphs of p(R) are almost identical for
we = 2, we = 3, we = 5 and we = 10 and so generates an essentially
common lower curve ...........................

3.8 Graphs of g(R) for five values of we: we = 1, we = 2, we 2 3, we = 5
and we=10 when I?,-=1,Re=4,w,~=0 and k=2. For we=2,
we = 3, we = 5 and we 2 10 there is an apparent discontinuity at
R,- = 1 ...................................

3.9 Graphs of p(R) for five values of we: we = 1, we = 2, we = 3, we = 5
and we =10 when R, = 1, Re = 4, w,- = 0 and k = 2. The top curve
corresponds to we = 1. The graphs of p(R) are almost identical for
we = 2, we = 3, we = 5 and we 2 10 and so generates an essentially
common lower curve ...........................

3.10 The lower curve is the graph of \Ilext(k) and the upper curve is the nu-
merically estimated \IlemUc) when R,- = 1, Re = 4. Note that \IleritUc)
appears to take its minimum when k = 1 ................

3.11 Graphs of the normalized total energy, normalized F} and normalized E“
as afunction of k when R.- =1, Re=4 and we—w, =1 .......

3.12 The deformation of an initially vertical line segment for a material with
k = l for we = .75413 and we = 1.50826 = \Ilmee. These two values
of we were also used in Figure 2.1 for the neo—Hookean material. Here
Ri=1, Re=4 and 114:0 .......................

3.13 Comparison of the deformation of an initially vertical line segment for a
material with k = 1 to that of the neo—Hookean material (k = 0)

3.14 Graph of g(R) when k = 1 for the following ten values of we,
we = 0.150826, 0.30165, 0.452475, 0.6033, 0.754125, 0.90495, 1.055775,
1.2066,1.357425, 1.50826 = \Ilmee. Here R,- = 1, Re = 4 and w,- = 0 .

3.15 Graph of the family A3(n) with R,- = 1, Re = 4, 111i: 0 and we 2 4 for
n = 0 (the diagonal line), and 1,10, 100,1000,10000 ..........

3.16 Graph of the family A4(n) with R,- = 1, Re = 4, w,- = 0 and we = 4
for n = 0 (the diagonal line), and 1,10, 83.1,100, 1000, 10000. The
dashed line shows the case n = 83.15 which results the minimum
energy among the family A4(n) .....................

3.17 Graph of A1 and A4(83.15) when R,- = 1, Re = 4 and we = 4 ......

3.18 Normalized energy versus twist for R,- = 1, Re = 4, w,- = 0 and varying
we. Here \I/e,ex = 1.50826. The energies corresponding to A1, A2,
A3,ee, A4,ee and Any are normalized by dividing 27rhci. The nor-
malized energy associated with A4Inmie] is also indicated by stars for
we = 2, 2.5, 3, 3.5, 4 ............................

4.1 Numerical solution when R, = 1, Re = 4, w,- = 0 and we = 0.452478 for
three values of 61, d = 0.1,0.01,0.001. Here |we — 'lbz'l = 0.3\Ilmee, . . .

4.2 Numerical solution when R,- = 1, Re = 4, w,- = 0 and we = 1.5 for three
values of 07, (‘1 = 0.1,0.01,0.001. Here |we — 31M 2 0.995\Ilmee,

viii

34

35

43

45

45

63

64
66

66

91

92

4.3

4.4

4.5

4.6

4.7

4.8

4.9

4.10

4.11

4.12

4.13

4.14

4.15

Numerical solution when R,— — 1, Re —- 4, w,- — 0 and we- -— 3 for three
values of 01,0: 01, 0.01, 0001. Here |we— 1M“ — 1. 989‘11mex
Numerical solution when R,-- — 1, Re: 4 and t/Ji- — 0 and we— - 3 for three
values of a, a- —— 0.1, 0.01,0.001 and three values of we, we: 0.3\Ilmee,
0.995\Ilmee, 1.989‘Ilmee ..........................
One-term, two-term perturbation solutions and numeric solution with
w,- = 0 and we = 0.150826 when 51 = 0.1. Here R,- = 1, Re 2 4
and \I’max = 1.50826 so that we = 0.1‘Ilmee ...............
One-term, two-term perturbation solutions and numeric solution with
w. = 0 and we = 1.3 when a = 0.01. Here R,- =1, Re = 4 and
\Ilmee. = 1.50826 so that we 2 0.862\Ilm ................
One-term, two—term perturbation solutions and numeric solution with
1M: 0 and we = 0.97 when 6 = 0.1. Here R,- =1, Re = 4 and
‘I’max = 1.50826 so that we = 0643me ................
One-term, two-term perturbation solutions and numeric solution with
wi=0andwe=l.3whend=0.1. HereR,=l,Re=4and
\I’max = 1.50826 so that we = 0.862\Ilmee,. Since we > \Ilem, the two
term perturbation solution involves decrease of g(R) when R = R;
One—term outer solutions with R.- = 1, Re = 4, w,- = 0 and we = 3 for
the following six trial values Jo, 3., = 1,2.25,4,6.25,9,12.25. Here
do = 1 is on the far left and do = 12.25 is on the far right. The curves
order themselves sequentially in do. The envelope of vertical tangency
is shown as a dashed line .........................
One-term patched, one—term regular perturbation and numeric solutions
when R,- =1, Re = 4, w,- = 0, we = 1.5 < \Ilmee z1.50826 and
ft = 0.1. The patching point is at r = 1.220537 ............
One-term patched, two—term patched and numeric solutions when R,- = 1,
Re = 4, w,- = 0, we = 1.5 < \Ilmee z 1.50826 and (1 = 0.1 The
patching point is at r = 1.220537 for the one-term patched solution
and the patching point is at r = 1.114934699 for the two-term patched
solution ..................................
Two—term patched and numeric solution when R,- = 1, Re = 4, w.- = 0,
we = 3 > \Ilmee, z 1.50826 and 6: = 0.1. The patching point is at
r = 1.2336459 ...............................
One-term patched, two-term patched and numeric solution when R.- = 1,
Re = 4, w,- = 0, we = 10‘Ilmex and dz = 0.01. The patching point is
at r = 1.246533 for the one-term patched solution and the patching
point is at r = 1.245361 for the two-term patched solution ......
Normalized energy E / (27rhd) versus twist when R,- = 1, Re = 4 for the
four values of it, (t = 0,0.001,0.01, 0.1 are depicted as the curves
E1, E2, E3, E4 respectively ........................
Normalized torque M / (27rhd) versus twist when R, = 1, Re = 4 for the
four values of (3, d = 0,0.001,0.01, 0.1 are depicted as the curves
Ml, M2, M3, M4 respectively .......................

ix

93

93

97

98

99

100

104

111

113

113

115

116

CHAPTER 1

Introduction

In the event of substructural reconfiguration in solids, it is sometimes the case that
the notion of simple deformation X -+ x must be broadened so as to incorporate
a more detailed kinematic description. This typically involves new kinematic vari-
ables in addition to x = x(X). Each new kinematic variable is in turn associated
with an additional balance principle of the same tensor order as the new kinematic
field variable. The continuum mechanical theory known variously as “continua with
microstructure” [9] or “multi-field theory” [18] gives insight into the role of such addi-
tional balance equations. Gurtin’s theory of configurational forces provides a related
type of description [12]. A well-known example of such a theory is the Ericksen model
of liquid crystals [10]. In the liquid crystal theory, the additional field is a (unit) vector
that describes the liquid crystal orientation.

A new constitutive framework for treating finite deformations when conventional
elastic behavior is modified by substructural reconfiguration has recently been pro—
posed by Pence and Tsai [22]. Energy minimization then leads to an additional
balance principle that is associated with the richer kinematics. In this thesis, we
investigate the solution of the pure azimuthal shear problem in this new framework.
Azimuthal shear is most simply conceived as a twisting deformation of a circular

cylinder due to an imposed difference in twist \11 between in the inner and outer

radii. The constitutive model has, among other things, a natural parameter k 2 0
such that the neo—Hookean material in standard nonlinear elasticity is retrieved in
the limit I: -—> 0. In Chapter 2 the solution of the pure azimuthal shear problem
is summarized for a neo-Hookean material in standard nonlinear elasticity theory.
The methodology and results of the pure azimuthal shear problem for a neo-Hookean
material are well-known and will serve as a basis for comparison with the new results
described in this thesis. The new content of the thesis is divided into two parts as
presented in Chapters 3 and 4.

The first part of this thesis, Chapter 3, is concerned with a material model such
that the stored energy to be minimized is the sum of two terms. One term can
be viewed as penalizing macroscopic elastic deformation and the strength of this
penalization is described by a positive constant 6:. The second term can be viewed
as penalizing elastic substructural reconfiguration and its strength is described by a
positive constant a‘. The parameter k is the ratio 6/01“. In the first section of
Chapter 3 the pure azimuthal shear problem in this new framework is formulated.
It is confirmed that if the parameter k goes to zero, then the pure azimuthal shear
problem in the new framework reduces to the standard pure azimuthal shear problem
for a neo—Hookean material. Since the governing field equations are complicated for
a general k, it does not seem feasible to get a closed form solution for arbitrary
k. Hence, in Section 3.2, the pure azimuthal shear problem is studied for small
Ii: by a perturbation expansion about the base neo-Hookean solution. The torque-
twist relation is also presented for small k. An interesting aspect of the torque-twist
relation as determined by the perturbation analysis is that torque is not an increasing
function of twist for all values of twist if k 7e 0. The pure azimuthal shear problem
is then solved numerically for any k in Section 3.3. When k ¢ 0 the numerical

solution reveals evidence of a threshold value for twist at which smooth solutions

are no longer available. In Section 3.4 the pure azimuthal shear problem is solved
analytically in the special value k = 1 and a restriction in the form [‘11] S \Ilmex on
the imposed twist \I' is obtained for a classically smooth solution. With the exception
of k = 0 and k = 1 it does not seem likely that analytic solutions can be constructed
explicitly. In Section 3.5 the pure azimuthal shear problem is derived by minimizing
an appropriately reduced stored energy function which acknowledges the symmetries
inherent in the solution. In particular, we prove that the exact solution that is derived
in Section 3.4, is indeed an absolute minimum for the energy integral when k = 1.
In Section 3.6 direct methods of calculus of variations are used to investigate solution
possibilities for arbitrarily large twist ‘1! when k = 1. A minimizing curve is obtained
among a collection of test curves for the parametric energy integral when the twist is
arbitrarily large. Indeed we prove that this minimizing curve is an absolute minimum
of the parametric energy integral in the class of piecewise-smooth curves such that,
once restricted to the interior, their parametrization reduces to a function. The
importance of such a result is that if we seek minimizers of the nonparametric energy
integral for arbitrarily large twist \11 in the space of piecewise-smooth functions, then
the global minimum involves a discontinuity at the inner radius.

A more generalized material model is considered in the second part, Chapter 4.
The strain-energy function for the new material model augments the strain-energy
function of the first part by inclusion of an additional term. This additional term
can be viewed as penalizing the overall deformation, as opposed to penalizing specific
partial deformations. The strength of this penalization is described by a positive
constant (1. Thus there are now three terms to the stored energy with respective
strengths 6:, a” and a. The pure azimuthal shear problem subject to the same
boundary conditions is studied in this extended material model when d = a" (k = 1) .

Our interest is in the extent to which a > 0 gives different qualitative behavior from

the or = 0 treatment of Chapter 3.

In the first section of Chapter 4, the pure azimuthal shear problem is formulated for
the extended material model when d = of. Moreover the resulting boundary value
problem is expressed as an integral equation. In addition two modified boundary
value problems are also defined. These second boundary value problems for the pure
azimuthal shear problem are useful for obtaining existence and uniqueness results.
The principle of contracting mapping is then used to prove an existence and unique-
ness result for the pure azimuthal shear problem in Section 4.2. In Section 4.3 the
pure azimuthal shear problem is solved numerically in the extended material model.
In Section 4.4 a regular perturbation procedure is used to investigate the solution of
the pure azimuthal shear problem when the material parameter a/ci is small. It is
observed that if [‘11] > \Ilmex, then the regular perturbation cannot be used. Further
if [\IJI > \IImax or [\III = (1 — e)\II,e,ax with 0 < 6 << 1, then severe requirements are
placed on the smallness of a/d in order to obtain a useful approximation near the
inner radius. In Section 4.5 a boundary layer type analysis is developed in order to
get a useful approximation in such cases. In particular, inner and outer solutions
are obtained and a patching method is proposed for obtaining the full solution. This
patching method is found to give satisfactory agreement with the numerical solution.
The relation between torque, twist and energy in the extended material model is
studied in Section 4.6, which is then followed by a concluding discussion in Chapter

5.

CHAPTER 2

Pure Azimuthal Shear Problem for

a Neo-Hookean material

2.1 Introduction

This section provides preliminary discussion from the well known conventional theory
of hyperelasticity in order to provide context for this thesis. Readers familiar with the
pure azimuthal shear problem for a neo—Hookean material can, if they wish, proceed

directly to Chapter 3.

2.2 Pure Azimuthal Shear Problem for a Neo-
Hookean material

In standard nonlinear continuum mechanics it is the case that the continuity equation,
the principles of linear and angular momentum, and the energy equation hold for any
material regardless of its constitution. However, unless the body can be regarded as
rigid, these equations are in general insufficient to determine the motion produced

by given boundary conditions and body forces. They need to be supplemented by

a further set of equations, known as constitutive equations, which characterize the
material composition of the body. Such a set of constitutive equations usually serves
to define an ideal material, and much of the work on modern continuum mechanics
has been concerned with the formulation of constitutive equations to model as closely
as possible the behavior of real materials.

We consider a continuous body which occupies a connected open subset of a three-
dimensional Euclidean point space, and we refer to such a subset as a configuration of
the body. We identify a specific configuration as a reference configuration and denote
this by B,.. Let points in B, be labeled by their position vectors X relative to some
chosen origin and let BB, denote the boundary of 8,. Now suppose that the body
is deformed quasi-statically from B, so that it occupies a new configuration, B with
boundary 88. We refer to B as the current or deformed configuration of the body.
The deformation is represented by the mapping X : B, —-> B which takes points X
in B, to points x in B, where x is the position vector corresponding to the point
X in B... The mapping x is called the deformation from B, to B.

In this section we review the well known treatment of this deformation in the
context of conventional continuum mechanics and hyperelasticity. In such a treatment

the equilibrium equations in the absence of body forces are given by
div 0' = 0, (2.1)

where a' is the Cauchy stress tensor and div is the divergence operator with respect
to current coordinates. The constitutive equation of an elastic material is given in

the form
a' = C(F), (2.2)

where G is a symmetric tensor-valued function defined on the space of deformation

gradients F. Here F = Gradx or in component form

6113

Ea — 8X07

 

(2.3)

where i and a 6 {1,2,3}. In general the form of G depends on the choice of
reference configuration and G is referred to as the response function of the material
relative to reference configuration 8,. A material whose constitutive law has the
form (2.2) is referred to as a Cauchy elastic material. From the point of view of both
theory and applications a more useful concept of elasticity, which is a special case of
Cauchy elasticity, is hyperelasticity (or Green elasticity). In this theory there exists
a stored energy function (or strain-energy function), denoted W = W(F), defined on

the space of deformation gradients such that the total stored energy is given by
E = [W(F) dV. (2.4)

Then for an unconstrained material

8W
_ _ —1 T
a — (3(F) — J 8F F , (2.5)

where J = detF. The Jacobian J of the deformation gradient F determines the
resulting volume change in the deformation. In this setting (2.1), (2.5) are the Euler-
Lagmnge equations associated with minimization of (2.4). For a volume preserving

(isochoric) deformation
J = 1. (2.6)

A material for which (2.6) is constrained to be true for all deformation gradients F is
said to be incompressible. The modification of (2.5) appropriate for incompressibility
is

_ 8W

where I is the identity tensor and the scalar p is an arbitrary hydrostatic pressure,

that is formally the Lagrange multiplier associated with the constraint (2.6).

7

A standard model for an incompressible material is given by the neo—Hookean

stored energy density W in the form

W = $1! (11 - 3): (2-8)

where u > 0 is the shear modulus of the material in the reference configuration,
11 = tr(B) and B = F FT, the left Cauchy-Green tensor of the deformation, and tr
denotes the trace. More detailed explanation of the above concepts, including restric-
tions upon W motivated by observer invariance and material symmetry requirements,
can be found in any standard nonlinear elasticity book [2], [3], [11], [21].

Consider a nonlinearly elastic thick-walled circular cylindrical tube whose natural

(unstressed) configuration is defined by
0<R53R§Re,038<27r,03Zgh, (2.9)
where (R, O , Z) are cylindrical polar coordinates. Consider the following deformation
r= R, 0=O+g(R), 2:: Z, (2.10)

where (r,6,z) are cylindrical polar coordinates associated with the deformed con—
figuration. The deformation (2.10) is called a pure azimuthal shear (it is also called
gyroscopic, rotational or circumferential shear in the finite elasticity literature [6],

[15], [24], [26]). The deformation gradient F is given by

P l
1 0 0
F: rg' 10 =I+TQ'€9®BR. (2-11)
0 0 1

 

 

where ’ denotes the derivative with respect to R. Since r = R we may also con-
sider that the derivative is with respect to r. Note that (2.6) is satisfied for this

deformation.

The left Cauchy-Green tensor B = F FT is

 

 

1 rg' 0 l
B = rg’ 1 + (rg’)2 0 (212)
0 O I

= I + rg' (e9 (83 e, + er (8) 89) + (rg')2 e9 (8) e9.
For a neo—Hcokean material the Cauchy stress tensor follows from (2.7), (2.8) as
a' = —-pI+uB. (2.13)

The equilibrium equations (2.1) in cylindrical coordinates are

60... 180.9 dare I _
6r 'i' '1: 80 + 62 + ‘7: (arr _000) _' 0: (2'14)

 

 

 

009,- 160.99 809,, 1 _
W+PW+W+F (0r0+00r)—0i (2'15)

802,. 16029 (9022 1 __
a. ; as a. flan—0' (2'16)

 

 

 

We will impose boundary conditions of prescribed twist

g(Ri) = 21).. g(Ro) = we. (2.17)
We also consider the boundary condition

0.430) = 09.. (2-18)

meaning that the outer surface R = Re supports a radial traction 02,. Replacing the

components of 0' in equilibrium equations (2.14)—(2.16) gives

9%?) + $00529” + %[—p(rg')2] : 0, (2.19)

 

3(wg’) 1501 + #(rg’f’ - 1?)

3r r ('90

 

1

501 - 19)

Bz : 0'

 

Since 9 is a function of r only, it follows from (2.19)—(2.21) that

8p

 

aT-+rwlg) 0.

5(1‘9’) 13p ,_
“ a. 755”“9—0’
0p_

52—0

Then equation (2.24) gives

19 = W, 9).

while equation (2.23) supplies

329 _ 0(7‘9’)

—_ 2 '.
as “T 0r + ”'9

 

Hence ap/BB is a function of r only and so p must be linear in 6; i.e.,

p = 60(7‘) 9 + 15(7‘).

On the other hand it is immediate from equation (2.22) that

8217
Br 86 — 0'

 

Hence co(r) is a constant function and p becomes
P=m0+flfl-

Because of the symmetry, for any 6
0,,(6) = or,(6 + 27r),

which gives

coEO.

10

(2.21)

(2.22)

(2.23)

(2.24)

(2.25)

(2.26)

(2.27)

(2.28)

(2.29)

(2.30)

(2.31)

Consequently

to = p(ri- (2.32)

Using the above result in (2.26) gives

5(7‘39’) _
6r _

 

0. (2.33)
After integration and replacing r by R
Cl
g(R) = 3:, + C2, (2.34)

where c1 and cz are integration constants. Using the above result in (2.22)

C2
p(R) = 52—271 + C3, (2.35)

where c3 is another integration constant. The boundary conditions (2.17) now give

(ll/Jo '— #103ng

 

 

 

CI = R12 _ R2 , (2.36)
(501' '- l/Jo R3

Cg = wi + W. (2.37)

Notice that |g’(R)| is decreasing in R with finite maximum value
I 2 (wt _ 1%)}?
- = 0. 2. 8

The boundary condition (2.18) supplies

. (libi _ wolzlfi1

c3 = (R? _ R3,)2 u - ,u + 02,. (2.39)

This completes the solution of pure azimuthal shear problem for a neo—Hookean
material. The methodology and results (2.9)—(2.39) are well-known and will serve as
a basis for comparison with the new results described in this paper. We will compute
some useful quantities for later use. The normal traction on the inner surface R = R,-
is

R§+R§
123 —R,?

 

a..(R.—) = -u (a — a)? + 09... (2.40)

11

Note that 0,,(R,) < 0,9,. The azimuthal component of the traction on the inner and

outer surfaces are given respectively by

2

R
UorfRi) = 2# RToR—Q (we — i/Ji), (2-41)

2

R.
09r(Ro) = 2l‘ R2 -—2 R2 (1pc — “(i)- (242)

 

Torques, M,- and [We acting on inner and outer surfaces are

2" 47r hR,-2 R:
M. = h f 42.? am.) d6 = ——R—‘:—_—-Re— (w. - a), (2.43)
0 o 1
2" 4n hR? R2

Note that M,- + Me = 0, which is expected from the general theory. Let M = Me =

—M,-. Twist ‘II is defined to be the difference between angular displacements at

R=Re and R=R,,:

‘1’ = (we — #2.)- (2-45)

The T argue-Twist relation now follows from (2.44), (2.45) as

47r,uhR,-2R2

A4: IP_R,°n QAQ

Note from (2.12) that, 11 == 3 + (rg’)2 which in turn gives W = u(rg’)2/2. The

energy E stored by this deformation follows from (2.4) and (2.34) as

2 7r u h R? R3

E = W \112. (2.47)
The work of the torque (2.44) is defined to be

- ~11

W = / M(w) all, (248)

o

where M = M (w) is given by (2.46) with w replacing \II. By direct calculation it is
found that

W=E, an)

12

 

 

 

 

.4-4 -2 0 2 4

X

 

Figure 2.1: Deformation of an originally vertical line segment for different amount of
twist we and w; = 0. Specifically shown are we :2 .75413, 1.50826, 3.0165, 5.49779
corresponding respectively to 43.20°, 86.41°, 129.62°, 315° for a neo—Hookean material
in standard nonlinear elasticity theory. Note that R.- = 1 and Re 2 4 are used and
multiples of arccos(R?/R§) = 1.50826 are selected in order to provide correspondence
with results of the more general theory that is the object of this thesis

which is also to be expected from the general theory.

We will conclude this section by considering the cross-section R, = 1, Re = 4 and
sketching the deformation of the line segment given by { (R, 8)|1 S R g 4, O = 7r/ 2}
in the reference configuration. Suppose w,- = 0 so that deformation is governed
by we whereupon g(R) = ((—16/15R2) + 16/15) we, g’(R) = (32/15R3) we.
Figure 2.1 shows the deformed location of the line segment for four values we:
we = 0.5 arccos(R?/R§) = 0.75413, we = arccos(R,-2/R,2,) = 1.50826, we =
1.5 arccos(R,2/R,2,) = 3.0165, we = (7/4)7r = 5.49779. Multiples of arccos(R,-2/R,2,) =
1.50826 are selected in order to provide correspondence with results of the more

general theory as later considered in Section 3.4.

13

CHAPTER 3

The Internally Balanced Material

3.1 Formulation of the Problem for the Internally

Balanced Material

The constitutive framework that we consider in this thesis is due to Pence and Tsai

[22] and is based on multiplicative decomposition of the deformation gradient
F=FF, an

where F and F” are respectively the conventional elastic part of the deformation and
a precursor internal deformation that we will term pre-elastic. A full discussion of the
rationale and motivation for this theory is given by Pence and Tsai [22]. In this thesis

we will consider materials that admit only volume preserving deformations; hence
det F = l. (3.2)
Further, the elastic part itself is also taken to be volume preserving and thus
det F = 1. (3.3)
Note that (3.2), (3.3) in turn give

det F" = 1, (3.4)

14

and that any two of (3.2), (3.3), (3.4) implies the third. Equilibrium in this theory is

based on minimizing the potential energy functional
E = / W(F,F*) dX, (3.5)

where W is a general storage energy function. The energy considered here is taken
to have the following form

0

of ,,
W—7(Il—3)+2

(i1 — 3)? (36)

where (1" 2 0 is the internal modulus, d 2 0 is the shear modulus and If = tr B’
and I“1 = trB with B‘ = F*(F*)T and 13 = info? Note that both a and B“ are

symmetric positive-definite with

det B = 1, (3.7)
and

det B’“ = 1. (3.8)

As discussed by Pence and Tsai [22] minimization of the potential energy functional
involves separate independent variations with respect to both x and F‘. Minimiza-

tion with respect to x at fixed F" gives

div 0’ = 0, (3.9)

a=aB—pr (am)

Here p is a Lagrange multiplier associated with the constraint (3.2). Minimization

with respect to F' at fixed x gives the internal balance

A2 A

aB +qB=a"B, (3.11)

where q is a Lagrange multiplier associated with the constraint (3.4). Equation (3.11)
is the additional internal balance principle that serves to generalize the present treat-

ment. It is a natural consequence of the energetic dependence on the decomposition

15

(3.1) which, as discussed Pence and Tsai, is motivated by the need to characterize
materials with complex microstructural features.

Pence and Tsai prove for arbitrary symmetric positive-definite tensor B obeying
detB = 1 that equations (3.7) and (3.11) can be solved uniquely for a symmetric
positive-definite tensor B and scalar q [22]. The proof given by Pence and Tsai is not
fully constructive, in that the analysis generates a sixth order polynomial equation
for q that may not have a closed form expression for the root of interest. The root
itself however is guaranteed to exist and selected by a bound requirement.

We now formulate the pure azimuthal shear problem for this material. This de-
formation is again assumed to be in the form (2.10) so that (2.11), (2.12) continue to
hold. This in turn ensures (3.2). We seek scalar functions g(r), p(r, 6, z), q(r,l9, z)
and a symmetric positive-definite tensor function B(r, 0, 2) such that equations (3.7),
(3.9), (3.10), (3.11) hold where B is given by (2.12). We again consider boundary
conditions (2.17), (2.18).

Since B is given by (2.12), equations (3.7) and (3.11) can in principal be solved
for B and q in terms of rg’ , implying, in particular, that q = q(r) and B = B(r).
Once B is known the components of 0 follow from (3.10) whereupon the equilibrium
equation (3.9) is anticipated to first require p = p(r) and second provide g(R) and
p(R). This is the strategy in what follows.

By using the argument given by Pence and Tsai, it can be shown that B and
q are independent of 9 and z by the following argument. Let b,- and b,- be the
eigenvectors of B and B, respectively, corresponding to the eigenvalues A? and A?
respectively. The internal balance (3.11) gives that B and B have common principal

frames whence b,- = b,. Thus by principal frame calculations

A? = (0z“)’1 (4 3? +q5lf), (3.12)

16

and

,\?

 

~ = -q+\/q2+4a*ci,\f

 

1

Since (3.7) gives 3‘? i3 A3 = 1, it follows from (3.13) that

fi(-q+\/q2+4da*x\f) 2863.

i=1

By using the definition of B

Xf=1,

1
A2:—
2 2

1
A2:—
3 2

and corresponding eigenvectors are

bl = 82)

b2:—

26:

 

 

(2 + (rg')2 — \/-4 + (2 + r2g’2)2 ),

 

(2 + (rg')2 + \/—4 + (2 + rzg’ 2)2 ),

 

(rg’)2 + \/—4 + (2 + rzg’ 2)2

 

2rg’

 

 

b3:—

09’)2 - \/-4 + (2 + rzg’ 2)2 e
2rg’

(3.13)

(3.14)

(3.15)

(3.16)

(3.17)

(3.18)

(3.19)

(3.20)

Note from (3.14)—(3.17) that q is independent of 6 and 2. Since b,- = b,, we can

write

B=A¥b1®b1+l§b2®b2+5l§b3®b3,

(3.21)

and conclude that B is independent of 9 and z with Bee = 0 and B9,, = 0. Thus B

can be written as

8..
B = Ere

 

0

L

Bra

Bee
0

ca.

DO

 

17

(3.22)

and by (3.7)

Bzz (Err B60 - 330) = 1

Now, using (3.22) in (3.10) the Cauchy stress tensor 0‘ can be written as

- q

 

 

313.. — p 31%.. 0
0 = 51539 5:300 — P 0
0 (1 dBu—p

The equilibrium equations (3.9) become

 

one... — p) 10(33..) 1 A . A . _
87' ‘l‘ T 60 + ;(OBT,— '- (1ng) — 0,

afdérol 13(5’390-19) 1 .~ _
a. +T—oe__+¥(2aB’9)‘0’

 

333.. — p)

Bz : 0'

Since B is a function of r only, the equations (3.25)—(3.27) reduce to

8B" (9p fr

 

 

a (97‘ *5;+;(Brr— ng)'—‘O,
,. aBrg 18p 2C} A
8r - FEB Bro 0’
0p
E — O.

(3.23)

(3.24)

(3.25)

(3.20)

(3.27)

(3.23)

(3.29)

(3.30)

From (3.30) it follows that p is a function of r and 9, and from (3.29) it follows

that 619/00 is a function of 1' only. Hence p must be linear in 6. By using similar

arguments as in the neo—Hookean case (Chapter 2), equations (3.24) and (3.28) lead

again to the conclusion that p is a function of r only. Then (3.29) gives

33. 2* -
i+—}B..=0,

 

a
r

18

(3.31)

and hence

~ d
Bro = :1), (3.32)

where all is an integration constant. Moreover (3.28) can be written as

dp Asia" 0 - .
07—0 Gr +‘;(Brr_800)‘ (3'33)

 

The internal balance equation (3.11) can be written in the following form
1:132 + p 13 = B, (3.34)

where we introduce the normalized parameter k and the normalized function p = p(r)

 

via
(1 q(r)
k = _ = . 3.35
a... p(r) a, ( )
Then the four nontrivial components of (3.34) generate
k (Bf, + BEG) + pBrr _1 = 0: (336)
k (13.. 13.. + 3.. ‘6.) + pr}... — rg’ = , (3.37)
k (Bfe + Bge) + p399 — (1 + (rg')2) = 0, (3.38)
k 133, + p8,. — 1 = 0. (3.39)

Therefore, in this new framework, the pure azimuthal shear problem reduces to
the following.

Find scalar functions p = p(r), p = p(r), g = g(r) , B" = 13.41") , Boo = Boo(r),
Bu = Bah) such that equations (3.23), {3.33), {3.36), (3.37), {3.38), (3.39) are
satisfied. Here Bro is given by (3.32) in terms of as yet undetermined constant (11.
This constant and any other integration constants will be determined from boundary

conditions (2.17) and (2.18).

19

We conclude this section by showing that this boundary value problem reduces to
the pure azimuthal shear problem for the standard neo-Hookean material in the limit

k —+ 0. As It goes to zero it follows from (3.36) that

A

3., = i- + 0(1), (3.40)

from (3.37) that

13,9 = r—g— + 0(1), (3.41)

. 2 r2

Bog _ 1+; g +o(1), (3.42)
and from (3.39) (that

. 1

Bu 2 E + 0(1). (3.43)

Substituting from (340)-(343) into (3.23) and retaining only 0(1) terms gives

 

1 1 1 2 I2 2 I2
-(__:r_9__rg):1, (3,44)
9 p p p2
After simplifying, equation (3.44) reduces to p3 = 1 which gives
p = 1+ 0(1) (3.45)

independent of r. Therefore in the k —-+ 0 limit it follows to leading order that
8,, = 1, 8,9 = rg', Bag = 1+ r2g’2, Bu 2 1. (3.46)

Using the above results in (3.32) and in (3.33), give respectively to leading order that

r3 g’ = d1, (3.47)
and

d

d—: = —dr(g')2. (3-48)

20

Observe now that equation (3.47) is equivalent to equation (2.33) from the neo-
Hookean case. Similarly equation (3.48) retrieves equation (2.22) from the neo-
Hookean case. Hence as k —) 0 the results of the neo—Hookean analysis are obtained.
For future use we note that (3.45) implies p —> 1 in this limit, which by virtue of

(3.35) gives that q(r) ——> 01“.

3.2 Solution for Small k (3 << oz“)

At the end of the last section it was shown that the present theory with k = 0
reduces to the problem for the neo-Hookean type material of standard nonlinear
finite elasticity. In this section we will study the pure azimuthal shear problem for
small It by a perturbation expansion about the base neo—Hookean solution.

If k = 0, then the neo—Hookean solution applies. From equations (2.34) and (2.35)

this solution is given by

c
90(R) = 3% + 020, (349)
and
dcio
P0(R) = if + C30, (3.50)

where C10, C20, C30 are the constants computed from boundary conditions (2.17),
(2.18) and given by (2.36), (2.37), (2.39) respectively. Note that instead of c1, c2,
c3, the constants C10, c20, C30 are used respectively to emphasize that these are the
constants corresponding to zeroth order solution. The subscript o is used to denote
the zeroth order approximations for the unknown functions and tensor quantities.

Moreover from (3.46) the base neo—Hookean value of 13 is

 

 

1 rgo' 0
B0 = rgo’ 1+ (rgo’)2 0 , (3-51)
0 O 1

21

and from the remark after (3.48)
(10 = a‘. (3.52)

Assume that there is a small k solution in the form of following expansions for g(R),

p(R), q(R), B(R) for the given problem:

g(R) = 90(R) + 23.11 k" 974R), (3-53)
p(R) = p0(R) + 23:, k" MR), (3.54)
q(R) = <10 + 233:1 *5" (MR). (3.55)
B(R) _—. 130(3) + 23:, k" 13,,(12), (3.56)

where go(R), p0(R), go and 130(R) are as given above. Note that by using the

expansion (3.53) the left cauchy-green tensor, B, given by (2.12), can be written as:

 

 

 

 

 

 

- 1 r(90'+kgl’+...) Ol
B: r(go'+kgl’+...) 1+r2(go'+kgl’+...)2 0 , (3.57)
_ O 0 1 _
or equivalently,
' 1 rgo 0 - ' 0 rgl 0 -
B = rgo’ 1+ r2(go’)2 O + k rgl ’ 2r2go’gl’ 0 + . .. (3-58)
_ O 0 1 _ _ O 0 0 _
giving that
F 1 rgo’ 0 '
Bo = rgo’ l + r2(go ')2 0 a (3-59)
_ 0 O 1 _

 

 

and

" T

0 7‘91 I 0
B1

7°91, 27290,91, 0 - (3.60)

 

 

0 0 O

.. .J

Substituting from expansions (3.53)—(3.56) into the internal balance (3.34) gives

(03+kQ1'l'...)

k(I§0+kl§1+...)2+ (130+kf31+...)

 

at
—(Bo+kB1+...) =0, (3.61)
or equivalently,
* ‘ 2 ‘ Cll ‘ 2
(Bo — B0) + k(B0 - B1+ B1+ 5B0) + 0(k ) = 0. (3.62)
The 0(1) terms of (3.62) give
130 = BO, (3.63)

which was already evident from (3.51) and (3.59), while the 0(k) terms give

133 + g 130 + 131 — B1 = 0. (3.64)
Therefore

131 = B1 — 1‘33 — g 130. (3.65)

By using (3.59), (3.60) and (3.63)

 

 

-1-(ql/a*) - ”(9032 (904; 0
f3. = ... (13,)... 0 , (3-66)
_ ‘1 — (QI/O‘)
where
(Bar. = my — 27‘90’ — #6033 — g-mg (3.67)

23

(B1).» = 212903. ' — 1 — 37360)? — r4(go'>4 — £3; —- gear (3.68)

and the other components of 131 follow from the symmetry of 131. To find the
complete solution up to the order k , it is necessary to find ql, 91 and p1. If the com-
ponents of I30 and 131 , given by (3.59) and (3.66)—(3.68) respectively, are substituted

into (3.23), then it follows that

391

1p

 

1+ k( — 3 —— - r1190 )2) + 0(k2) = 1, (3.69)

from which it is immediate that

t

(3 + ”(90 ')2)~ (3-70)

 

(11:

If go ’ (r) is computed from (3.49) and used in (3.70), then ql is found as

4%

(11: _a. (1 _,_ W) (3.71)

where cm is the constant given by (2.36). By using the reduced equilibrium equations
(3.32) and (3.33) it follows that 91 and p1 are given by

83 20 —c
_ 10 + 10 11

— 9—73 2T2 + C21, (3-72)

91

and

 

= (1101120 — 3C10C11) & + 4C‘i'0é

pl 37‘4 T + C31. (3.73)

The constants cu, C21, C31 are determined by requiring gl(R,v) = 91(Ro) = 0 and
0,,(R0) = 0,9,. Hence

16 (R? + R3123 + R3) at.

 

 

 

011 = 2 C10 + 9 R? R3 1 (3-74)
8 R? + RS‘ c3
C2. = ( 9 R, R; 10, (3.75)
_4 ,3 _ , *
C31 :: C10[ ('10 + R303 C11 19C10)](10 (376)

312:;

24

The azimuthal component of the traction on the inner and outer surfaces are respec—

tively given by

A A

a (I
09,.(Ri) = QC“)? — [CCU-RT? + 0(k2), (3.77)
67 Cl
06r(Ro) = —2010§2' + ICC”? + 0(k'2). (3.78)

O 0

By using (3.78) the Torque can be computed from
27r
M = h / 123094123719. (3.79)
0

Then Torque- Twist relation is given by

47rhéR3R§
M: 123-123 11,-

 

4 7r h d R? R3 [9(R3 — R3)2 + 8(R;‘ + R? R3 + R3) \112]

1:
RE - R? 9 (R? - REV

\II + 00:2). (3.80)

 

Note from equation (3.80) that M (-\II) = —M(\IJ) by the terms thus far computed.

The torque can be normalized by dividing 4 7r h d R? R3. The normalized Torque-
Twist relation up to 0(k) as given by (3.80) is displayed in Figure 3.1. Here R,- = 1,
R0 = 4 and the solid line shows the neo—Hookean case k = 0. The dashed line shows
the case k = 0.02. Figure 3.1 shows that increasing k = d/a" results in a relative
softening.

Perhaps the most interesting aspect of (3.80) is that, to within the terms thus far
computed, M is not an increasing function of \I’ when k > 0 for all \II.

By using (3.80) it is immediate that

dM ”magma—1)

W = (R? — R2) (3'81)

 

 

+32rhc‘1RfRfi(R;-‘+R?R§+R3)k\ll2

3(R? _ R2)3 + 0(k2).

25

 

53.0.0.0
aroma

Normalized Torque
O

 

 

 

 

-0.1
-0.2
-0.3
l
0‘15 0 5
Amount of Twist

Figure 3.1: Comparison of normalized torque-twist relation when k = 0 and k = 0.02
for R, = l and R0 = 4. The dashed line shows the case 16 = 0.02. The normalized
torque is M/(4 7r h 67 R2123)

On the basis of the terms thus far computed one can solve dM/dIII = 0 for ‘11 and

hence M takes its extremum values at

\/§(R§ - REM/1 - k

 

 

 

 

\II = :t‘IJext(k), \Ilcxt(k) :2 fiJMR? + 12ng + R3). (3.82)
Also note from (3.82) that

lift 11....(k) = 00, (3.83)
and that

% _ V3023 ‘ R?) (3.84)

dk — _ J33 \/1'—""'E k (fun? + R3123 + 123,)
Restricting attention to small It, and in particular taking 0 < k < 1 it follows
that dWext/dk < 0. It will indeed be shown analytically in Section 3.4 that loss of

resistance occurs for k = 1 at sufficiently large ‘11. On the other hand, this thesis

26

will show that the subsequent decrease in M with \11 for \II > \Ilext is not found
analytically. Presumably this decrease is an artifact of the expansion procedure that
would be corrected by invoking a \11 dependent criterion for determination of the
number of terms necessary for a uniformly valid approximation. These issues will be

clarified in the following development.

3.3 Numerical Solution

In this section we present a numerical solution procedure for the pure azimuthal
shear problem which does not require an assumption of small k. The solution reveals
evidence of a threshold value for \11 at which smooth solutions no longer are available.

We begin with some additional reduction of the governing equations. Post-

multiplying (3.34) with 13—1 gives
H3 + p1 = B 13—1. (3.85)

Since the product of two symmetric matrices need not be symmetric, there are five

nontrivial components of (3.85) as follows

k B... + p = 3.41%.. — mm], (3.86)

k 3.. = Baby's... — 19.9], (3.87)

k 3.. = Bums... — (1 + 99933.9], (3.88)

4 Bee + p = B..[(1 + (r9933... — 193.9], (3.89)

k 13.. + p = .i. (3.90)
B...

27

The values 13,, and 1399 can now be obtained from (3.87) and (3.88) respectively in

terms of 13.9 and 13”, providing

 

 

 

. k )3.
B,, = [ , +1] f, (3.91)
82.2 7‘9
. k B 9
z A 1 I 2] __T_. .
Baa [Bzz + + (79 ) rg' (3 92)

Note, if B" as determined from (3.91) and Bag as determined from (3.92) are
both substituted into (3.86) and (3.89) respectively, then the resulting equations are
equivalent. Therefore only four of the five equations in (3.86)—(3.90) are independent.
In addition Bu can be obtained uniquely from (3.90) in terms of p by using the

positive-definiteness of B;

A _ 2
B2,: M V” “L4,“. (3.93)

2k
The constraint (3.23) yields

 

. . . 1
B3,, = B... B99 — B . (3.94)

Taken together (3.91)—-(3.94) indicates that Bra can be found as a function of g’ and

 

p. Indeed it is given by

Mr 9’
\/2k2(r2g’2 + 2) + (k3 — 1)p + (/4k + p2(1+ k3)
By using (3.93), (3.95) in (3.91), (3.92) the nontrivial components of 13 can be found

 

 

{3.0 = (3.95)

in terms of g’ and p. In principal (3.86), or equivalently (3.89), now supply an
equation relating p and g’ .
To get a numerical solution observe first from (3.32) and (3.95) that
2k1‘2g' 2 d?

2k2(7‘2g’2 + 2) + (k3 —1)p+ W(1 + k3) _ 57 _
Secondly upon substituting from (3.91), (3.92), (3.93), (395) into (3.86) and manip-

 

(3.96)

ulating the resulting equation it is found that
193p6 — (k3 —1)k?(1"2g'2 + 3)p5 + [1 — (r29'2 +1)ic:3 + l~:6](r2g'2 + 3)kp4+

28

(1 + 5k3 — 5k6 — k9)p3 + 209- 1) 218(ng ’2 + 3));2 + (k3 — 1)4 = 0. (3.97)

Let k be fixed and choose a trial value for (11. Then at a given radial location
7‘ = r, the equations (3.96) and (3.97) can be solved simultaneously for p, E p(r,)
and gj E g’(1'J-). Once g;- is known the tangent line approximation can be used
to estimate g(r) at a nearby point 1' provided that g(rj) is also known. This is the
strategy we will follow. At issue is the proper value of d1. Divide the interval [R4, R0]
into N equal subintervals and label the end points as R,- = r0, r1, . . . ,rN = R0. One
of the boundary conditions in (2.17) can be used as starting point, say g(ro) = 2,0,.

Then by the tangent line approximation

901) % 9(7‘0) + (9'00) AT), (398)

where Ar = (R0 — R,)/N . In general

is

g,- = g(r ~ g( m) + g’(rl) Ar), (3.99)

J
l=0(

for j = 1,2,... ,N. Since g(RO) = 1120, it is therefore necessary to choose d1 so that

= Z(g'(n) Ar). (3.100)
(=0

That is, d1 can be used as a shooting parameter. Using this numerical scheme it is
found that all and the overall solution curves are robust with respect to finer and
finer discretization (increase in N).

However it is found that the R = R, end point value of g’ ; that is, g’(R,-), is very
sensitive to change in N when 2110 is sufficiently large. For example take k = 0.1,
R,- = 1, R0 = 4, and «[2,- = 0. Then for 1P0 = 2 it is found that d1 = 2.84365 for
N = 300 and that d1 = 2.86951 for N = 3000. The associated numerical values of
g’(R) at R,- = 1 are found to be 9’ (1) = 8.54856 and g’ (1) = 9.05256 respectively.
On the other hand for ’l/Jo = 10 it is found that (11 = 3.1621814432 for N = 300 and
that (11 = 3.162276607355 for N = 3000. The associated numerical value of g’(R)

29

at R,- = 1 are found to be g’(1) = 736.36329 and g’ (1) = 7101.95319 respectively.
We also find for we = 2 that the numerical value of the derivatives at R = 1.01 are
given by g’(1.01) = 7.56906 and g’ (1.01) = 7.94672 respectively for N = 300 and
N = 3000. For we = 10 it is found that g’(1.01) = 23.48574 and g’(1.01) = 23.50641
respectively for N = 300 and N = 3000. Therefore only g’(R,-) is very sensitive to
change in N when we is sufficiently large. In general, if j 75 0, then g; values are
quite stable to increase in N even for large we values. In particular for sufficiently
large we , the numerically computed value of g’(R.,-) increases without apparent bound
as N gets larger. We speculate that this is numerical evidence for a lack of smooth
solutions at R = R, when we is sufficiently large.

For k = 0.1 Figure 3.2 shows the graph of g(R) for five values of we: we = 1,
we=2,we=3,we=5andwe=10whenRi=1,Re=4,w,-=0from
the numerical procedure outlined above. For we = 5 and we 2 10 the numerical
procedure does not converge to a finite value for g’(R,-) and the associated graphs for
g(R) show an apparent discontinuity at R = R,. Figure 3.3 shows the corresponding
p(R) for the same parameter values. The graphs of p(R) are ordered from top to
bottom for we = 1, we 2 2 and we = 3 and they are almost identical for we = 3,
we: 5 and we = 10. For we = 1 and we = 2 we find that p(Re) = 0.89962
and p(Re) = 0.89903 respectively. For we = 3,5, 10 we find that p(Re) = 0.89883.
However p(R1-) = 0.790, 0.490, 0064,0005, 0.001 for we = 1,2,3, 5,10 respectively.

For k = 0.5 Figure 3.4 shows the graph of g(R) for five values of we: we = 1,
we=2, we=3, we=5 and we=10 when R,=1, Re=4, w,=0. For we=2,
we 2 3, we = 5 and we = 10 the numerical procedure does not converge to a finite
value for g’(R,.-) and the associated graphs for g(R) show an apparent discontinuity
at R = R,. Figure 3.5 shows the corresponding p(R) for the same parameter values.

The top curve corresponds to we 2 1. The graphs of p(R) are almost identical for

30

 

 

 

 

 

 

 

 

 

 

 

Figure 3.2: Graphs of g(R) for five values of we: we = 1, we = 2, we = 3, we = 5
and we=10 when R,=1, Re=4, ¢i=0 and 19:01. For we=5 and we==10
there is an apparent discontinuity at R,- = 1

 

 

p(R)

 

 

 

 

Figure 3.3: Graphs of p(R) for five values of we: we = 1, we = 2, we = 3, we = 5
and we =10 when R,- =1, Re = 4, w,- = 0 and k = 0.1. The top curve corresponds
to we = 1 and the middle one corresponds to we = 2. The graphs of p(R) are almost
identical for we = 3, we = 5 and we = 10 and so generates an essentially common
lower curve

31

 

 

 

9(R)

 

 

 

 

 

 

 

Figure 3.4: Graphs of g(R) for five values of we: we 2 1, we 2 2, we = 3, we = 5
andwe=10whenR,-=1, Re=4,wi=0andk=0.5. For we:2, we=3,
we = 5 and we = 10 there is an apparent discontinuity at R,- = 1

we = 2, we = 3, we = 5 and we 2 10 and so generates an essentially common
lower curve. For we = 2,3,5,10 we find that p(Re) = 0.49934 and for we = 1 we
find that p(Re) = 0.49955. However p(R,-) = 0.3252,0.0042,0.0012,0.0005, 0.0002
for we = 1, 2, 3, 5, 10 respectively.

For k = 1 Figure 3.6 shows the graph of g(R) for five values of we: we = 1, we = 2,
we = 3, we = 5 and we =10 when R,- =1, Re 2 4, w,- = 0. Theassociated graphs for
g(R) again show an apparent discontinuity at R = R,- when we = 2, we = 3, we = 5
and we = 10. Figure 3.7 shows the corresponding p(R) for the same parameter values.
Note from the scale that this numerical solution gives 0 < p(R) << 1 for R > R,- = 1.
The top curve corresponds to we 2 1. The graphs of p(R) are almost identical for
we = 2, we = 3, we = 5 and we = 10 and so generates an essentially common lower
curve. For each we we find that p(Re) = 0.000422356. However p(R,) = 0.0001,
0.000001,0.0000003,0.0000001,0.000000056 for we 2 1, 2,3, 5, 10 respectively.

32

 

 

p(R)

 

0.1 -------- -------- -------- é -------- 1 -------- é ------- «

 

 

 

Figure 3.5: Graphs of p(R) for five values of we: we 2 1, we = 2, we 2 3, we = 5
and we = 10 when R, = 1, Re = 4, w,- = 0 and k = 0.5. The top curve corresponds
to we = 1. The graphs of p(R) are almost identical for we = 2, we = 3, we = 5 and
we = 10 and so generates an essentially common lower curve

 

 

1o . . , , .

 

g(R)

 

 

 

 

 

 

 

 

 

 

Figure 3.6: Graphs of g(R) for five values of we: we = 1, we = 2, we = 3, we = 5
and we=10 when R.-=1, Re=4, w,=0 and 1:21. For we=2, we=3, we=5
and we 2 10 there is an apparent discontinuity at R,- = 1

33

x10

 

 

 

p(R)

 

 

 

 

Figure 3.7: Graphs of p(R) for five values of we: we = l, we = 2, we = 3, we = 5
and we 2 10 when R,- = 1, Re = 4, w, = 0 and k = 1. The top curve corresponds
to we 2 1. The graphs of p(R) are almost identical for we = 2, we = 3, we = 5 and
lite = 10 and so generates an essentially common lower curve

For k. = 2 Figure 3.8 shows the graph of g(R) for five values of we: we 2 1, we 2 2,
Lie 2 3, we 2 5 and we 2 10 when R,- = 1, Re = 4, w,- = O. The associated graphs for
g(R) again show an apparent discontinuity at R = R,- when we = 2, we = 3, we = 5
and we = 10. Figure 3.9 shows the corresponding p(R) for the same parameter values.
The lower curve corresponds to we 2 1 and graphs of p(R) are almost identical for
we 2 2, we = 3. we 2 5 and 'ci've = 10 and so generates an essentially common top
curve. For are = 2.3.5.10 we find that p(Re) R3 —0.9987 and p(Re) = -—0.9991
when we 2 l. H(;)we\'er p(R) = —0.6505,—0.0080.—0.0025,—0.00ll,—0.00IO for
Le = l. 2. .3. 5 10 respectively:

It is tempting to seek a correlation between the issue of an apparent discontinuity
at R = R, for sufficiently large L-‘e and the previously obtained \I/ex,(k) in Section

32 This gives rise to the notion of a smooth solution region in (we. A‘)-space for

.34

 

 

g(R)

 

 

 

 

 

 

 

 

Figure 3.8: Graphs of g(R) for five values of we: we 2 1, we = 2, we = 3, we = 5
and we=10 when R,=1, Re=4, ¢i=0 and k=2. For we=2, we=3, we=5
and we = 10 there is an apparent discontinuity at R,- = 1

 

p(R)

 

 

 

 

Figure 3.9: Graphs of p(R) for five values of we: we = 1, we = 2, we = 3, we 2 5
and we = 10 when R.- = 1, Re = 4, w, = 0 and k = 2. The top curve corresponds
to we = 1. The graphs of p(R) are almost identical for we = 2, we = 3, we = 5 and
we = 10 and so generates an essentially common lower curve

35

fixed R,, Re, w,. Using R,- = 1 and Re = 4 in (3.82) gives

q, (k)._15\/3\/1-—k
ext -— fim .

Let w, = 0. Then the curve (3.101) can be used to estimate the boundary for

 

(3.101)

the proposed smooth solution region in (we,k)—space. On the basis of following the
argument it is found that smooth solution region is in fact somewhat larger than that
predicted by (3.101). Based on the numerical solutions presented above it can be
assumed for a given k that there is an \Ilem(k) for which g’(R,-) —> oo. Equivalently
there is a critical value of shooting parameter d1, say de,,t(k), for which g’(R,-) —* 00.

Indeed it can be seen from (3.96) that

R?
dcrit (k) = «I; -

Then for a given Re and N equation (3.96) with (11 = dem in conjunction with

 

(3.102)

and (3.97) can be solved simultaneously for pj E p(r,) and g]. E g’(rJ-) as before.
In principle \Ilerit(k) can now be approximated by (3.100). However the sensitivity
of g’(R.-) to N gives difficulty. Since p,- (j = 0,...,N) has no such sensitivity,
an alternative is to find a good fitting polynomial for p(r). Once a closed form

approximating function for p(r) is available, equation (3.96) can be solved for g’ (7‘)

giving

 

 

\/4 1.23% + (k3 — 1) d? p(r) + (1 + k3) d; ./4"7c + p(_r)2

I :2}:
gm ,/2kr6—2k2d§r2

(3.103)

 

Note that since we > w,- it is the positive root of (3.103) that is of interest. If the
positive root of (3.103) is integrated numerically from R,- to Re with d1 = cm, one
then obtains an estimate for \IlemUc).

By using the outlined method, the maximum amount of twist versus 1: is depicted
in Figure 3.10. Note that \IIemUc) apparently takes its minimum when k = 1. The

graph of \Ilext(k) is also depicted in the same figure. Note that \IlextUc) is defined if

36

 

010)

  
 

Maximum Twist
A

N 0’
I

 

 

 

 

 

Figure 3.10: The lower curve is the graph of \IJextUc) and the upper curve is the
numerically estimated \Ilerit(k) when R,- = 1, Re = 4. Note that \IlemUc) appears to
take its minimum when k = 1

k E (0, 1] and that it gives a reasonable estimate for the critical amount of twist only
if k is sufficiently small.

For k = 0.1, R,- = 1, Re = 4, w,- = 0, we find that \Ilerit(0.1) = 3.030996. Note
in comparison that ‘Ilext(0.1) = 1.667811 < ‘Ilem(0.1). While (3.101) could likely be
improved by computing more terms in the perturbation expansion, the central issue
of an apparent threshold angle we remains. Namely, both the perturbation expansion
and the numerical solution procedure give evidence that there is a maximum amount
of twist above which there is no smooth solution for the pure azimuthal shear problem
under consideration.

With respect to the later development of Section 6, it is useful to remark that the
numerical estimate for \Ilem(1) is 1.50826 when R = 1, Re = 4.

Note that the energy density (3.6) is separable in i1 and If. Hence the potential

37

 

2.5

 

 

 

 

2 5
51.5
9, a
g 3
1 I
E :
O 1
Z :
0.5 ;
. l 1

0o 05 1 15 2

1:

Figure 3.11: Graphs of the normalized total energy, normalized E and normalized

E“ asafunctionofkwhenR,=1, Re=4 and we—w,=1

energy functional (3.5) can be decomposed into two parts
E=E+E2

where

A

E=/%(i1—3)dX
is the energy corresponding to elastic part of the deformation and

E‘ =/“7(1;—3) dX

is the energy corresponding to pre-elastic part of the deformation.

(3.104)

(3.105)

(3.106)

Let we — w,- be fixed. For given Ri, Re and k the numerical solution presented in

this section can be computed numerically to find (3.105) and (3.106). This numerical

computation reveals the following relation between E and E‘ for a fixed amount of

twist. If 0 < k < 1 then R > E" and if k > 1 then E < E‘. Moreover 1.3} = E“

when k = 1 meaning that the energies corresponding to elastic and pre-elastic part

of the deformation are equal.

38

For we — w,- = 1, Figure 3.11 shows the numerically estimated graphs of the
normalized total energy, normalized E and normalized E“ as a function of I: when

R,- = 1, Re = 4. Energies are normalized by dividing whd.

3.4 Explicit Solution when (31 = a* (k = 1)

In this section we will show for k = 1 that the governing field equations for azimuthal
shear can be solved explicitly up to integration constants. Then using the boundary
conditions we will show that the pure azimuthal shear problem for k = 1 has a unique
solution provided that the twist \II is not too large thereby verifying the notion of a
threshold twist as described above.

Recall from Section 3.3 that (3.97) supplies an equation relating p and g’ . However
it is only for k = 1 that ensuing algebraic simplifications as given next will take place.
That k = 1 is special in this regard could have been anticipated by the discussions
in Pence and Tsai.

If (3.97) is used with k = 1, then p is a root of the equation
64 p4 [102 - (019')4 + 20932 - 3)] = 0- (3107)

Hence

 

p = 0, :l:\/(rg’)4 + 2(rg’)2 — 3. (3.108)

However if components of B are computed from (3.91), (3.92), (3.93), (3.95) by using

 

either p = i\/(rg’)4 + 2(1‘g’)2 — 3, then one finds that the component equations
(3.36), (3.37), (3.38) of the internal balance equation (3.34) are not satisfied. Thus
thase roots are spurious and p = 0 is the only solution possibility for the internal

balance equation (3.34). This is consistent with the uniqueness result for the solution

of internal balance. For p = 0, it follows from (3.91), (3.92), (3.93), (3.95) that

I
B — ——T—g——— (3.109)

To _ 4 + (rg’)2’

39

B... = ——-——., (3.110)
4 + (79))
A I 2
oo = m) (3.111)
W
B... =1, (3.112)

and it is verified that (3.34) holds. Now equations (3.109) and (3.32) give

 

I
d
——’3—— = --§, (3.113)
4 + (79’)2 7"
which in turn provides
2d1
' = —————. 3.114
9 T T, _ d‘i ( )
After integration and replacing r by R it is found that
4 - d2
g(R) = arctan R d2 1 + d2, (3.115)
1

where d2 is another constant. By applying the boundary conditions (2.17) and elim-

inating d2 it follows that

R4—d2 R‘.‘—d2
° 1—arctan ' 1

d1

 

= (w. - 10.). (3-115)

 

arctan

5.3

Equation (3.116) can be solved for (11 if and only if the twist \II as given by (2.44)

obeys
R2
|\IJ| S \Ilmex, \Ilmex = arccos(F’2). (3.117)

By restricting the boundary values as indicated in equation (3.117) the constants d1
and d2 are found to be

R-2R2 sin‘II
d = ' ° , 3.118
1 V/R;1 + Rf, — 2R,.2R§ cos ‘11 ( )

 

 

124—3?

d? (3.119)

 

d2 = w,- — arctan

40

Note that cos \I'mex = Rf/Rg implies sin \Ilmex = «Rf, — Rf/Rg whereupon the asso-

ciated values of d1 and (12 are
(d1)ma.r : R321 (d2)max : 1192'- (3.120)

In particular (3.120)1 shows that (d1)mee = dem(1) as given by (3.102). In addition
\Ilmex = arccos(1/16) = 1.50826 when R,- = 1, Re = 4 which is consistent with
\Ilem(1) as given at the end of Section 3.3.

Substituting for B", R99, 9’ from (3.110), (3.111), (3.114), into the equilibrium

equation (3.33) gives

d . a d —(r ’)2
211: = 05(71271‘9/7) + ;(W). (3.121)

This equation simplifies to

2’2_

dr _ 0. (3.122)

Using (3.122) and replacing r by R gives
p(R) = d3, (3.123)

where d3 is another constant. This constant can be computed from the boundary

condition (2.18) with the help of (3.24):

d (31

3 = fi 123— ct? — 00 (3.124)

rr?

where d1 is the constant given by (3.118). Since p = q/a“ and p = 0 it is immediate

that
q = 0- (3.125)

Thus it has been shown that if k = 1 and \Il obeys (3.117) then B is given from
(3.22), (3.109)-(3.112), g is given by (3.115), p is given by (3.123) and q is given by
(3.125), where the constants d1, d2, d3 follow from (3.118), (3.119), (3.124). Note

that (3.125) is consistent with the k = 1 numerical solution given in Section 3.3.

41

We remark that (3.114), (3119) give

 

 

 

 

2R2R2 ' \II
g'(R) = = 08‘“ . (3.126)
R\/R4(R‘,-‘ + Rf, — 2133123 cos\II) — RfRf, sin2 \1'
Thus |g’(R)| is decreasing in R with
, 2R3, sin \II
, 2R22 sin \II
9 (Re) — RARE _ Re cos\II)’ (3.128)
Observe also that
9,031) —* 00 as ‘1’ -+ ‘I’max- (3.129)

The normal traction on the inner surface R = R,- is

& 61
Urr(R’i) : 'fii V R? _ (if _ fl V R: _ df + 0197-1 (3°130)
and we observe again that an.(R.-) < 09,. By using (3.24) and (3.32) the shear

traction on the inner and outer surfaces are
0,0(Re) = 61—. (3.131)

The Torque-Twist relations follows from (2.44), (3.118), (3.131) and is given by
M = Me = —M.- where

_ 27rhc‘erRg sin\II
\/R§+R3—2R,?R§cos\ll’

 

M g 111...... (3.132)

 

Note that

g—Agfiwm) = 0. (3-133)

Thus \Ilmex is similar to Wen as discussed in Section 3.2. More generally, the notion
of a threshold value for \II has been verified.

To compute the total stored energy E first note from (3.22) that,

i1 = 3.. + Bag + 13... (3.134)

42

 

 

 

 

 

 

.3- 4
44 -2 o 2 4
X

Figure 3.12: The deformation of an initially vertical line segment for a material with
k = 1 for we = .75413 and we = 1.50826 = \Ilme... These two values of we were also
used in Figure 2.1 for the neo—Hookean material. Here R,- = 1, Re = 4 and w.- = 0

By using If = tr B“ = tr C“ , where C" 2' (F‘)T F" is the right Cauchy-Green tensor

of the pre—elastic part of the deformation and F’ = F4 F, one finds
(3* = [fi‘1 F]T [13“1 F] = FT 3'1 F. (3.135)

Then by (2.11) and (3.22) it follows that

I 2 A _ ’ A A
(1 + (rg )A)B,:, 27:9 Br9 + 390 + .1 . (3.136)
Brr899 — 830 B 2‘

At this point an interesting observation following from (3.109)—(3.112), (3.134),

1;:

 

 

(3.136) is that the values of i1 and If are equal for k = 1. This common value

is given by

1“1 = I; = ,/4 + (rg’)2 +1. (3.137)

The total stored energy E for this deformation follows from (3.5), (3.6), (3.136) as
R0
E: nh(d+a’)/ (1/4-11- (rg’)2 —2)rd7‘. (3.138)
R.

43

Evaluating this integral with the aid of (3.114), (3.118) and using a“ = (it (since

k = ci/a‘ = 1) gives

 

E=27rhd(\/R§+Rg—2R3Rgcos\II+Rf—R§). (3.139)

Note as a result of (3.137) that the energies corresponding to elastic and pre—elastic

part of the deformation are equal for k = 1; i.e.,

/%(If —3)dX = [SJ—(1} —3)dX '1‘? (3140)

The work of the torque M follows from (2.48) and (3.133) as

 

v

W=27rhd(\/R3+Rg—2R§R§cos\II+Rf—Rg). (3.141)
By direct comparison of (3.139) and (3.141) it is again found that
W = E. (3.142)

We conclude this section by sketching three graphs for the deformation when
k = 1. The first is analogous to Figure 2.1 which we recall corresponds to the
neo—Hookean material (k = 0). Consider again the deformation of a line segment
occupying {(R,9)|1 S R g 4, 6 = 7r/2} in the cross-section R,- = 1, Re = 4 in the
reference configuration. It now follows from (3.117) that the solution (3.115) holds
only if |\IJ| S ‘Ilmex = arccos(ilg) 2 1.50826. Suppose w,- = 0 so that deformation
is governed by we. Figure 3.12 shows the deformed location of the line segment for
two values we: we 2 0.5‘Ilmex = 075413 and we = \Ilmee. = 1.50826. Observe that
9’(R.-) = 00 for 111.. = \Ilmax.

For we = .75413 and we = 1.50826 = \I’mex Figure 3.13 compares the defor-
mation of an initially vertical line segment for a material with k = 1 to that of
the neo-Hookean material (19 = 0). Since it is possible to twist the neo—Hookean
material beyond \Ilmex , the deformation of the initially vertical line segment for the

neo—Hookean material is also depicted for two more values of we > \Ilmex.

44

 

 

 

 

 

 

Figure 3.13: Comparison of the deformation of an initially vertical line segment for a
material with k = 1 to that of the neo-Hookean material (k = 0)

 

 

 

 

 

 

 

 

 

 

1.6 . . e. . .
1.4 --------- r -------- r ...... : ........ . ........ . ....... -
1.2 ........................................ i
1 ........ .............. 1 ........... 1 ....... _
5.50.8 - -- E- ----------------- : ------- -
0.6» - .........
0.4, - L ------ .1--- ------- i -------- 1----—..
0.2 ----- ........ ........ . ....... q
0 " 1 1 i 1 1
1 15 2 2: 3 35 4

Figure 3.14: Graph of g(R) when k = 1 for the following ten val-
ues of we, we = 0.150826,0.30165,0.452475,0.6033,0.754125,0.90495,1.055775,
1.2066,1.357425, 1.50826 = \Pm. Here R,- = 1, Re = 4 and w.- = 0

45

Figure 3.14 shows the graph of g(R) for ten values of we: we 2 \I/mex = 1.50826,
we = 0.9\IImax = 1.357425, we = 0.8\Ilmax = 1.2066,. . ., we 2 0.1\Ilmax = 0.150826. It

is again observed that g’(R,-) -—1 00 as we —+ \I/mex.

3.5 Alternative Derivation from Energy

In this section it will be shown that the formulation for the pure azimuthal shear
problem can be obtained directly by minimizing an appropriate energy functional
that acknowledges the a—priori symmetries of azimuthal shear. This will serve as a
basis for later developments and in particular will clarify the interpretation of the
results of the previous section for the case w > \Ilmex. Consider the deformation
given by (2.10) for a material whose energy function is given by (3.6) and subject
to constraint (3.7). Building the constraint into the minimization with Lagrange
multiplier 'r gives the functional

H = H(k) = 0*] g(r; -— 3) +311 — 3) — T(det 1‘3 — 1) dV. (3.143)
9 .

A A A A

Upon use of the symmetries g = g(r), 1' = T(r), B" = Br,(r), Bra = 8,9(7‘),

Egg = 1399(7‘), E... = E... (r), the functional H becomes

. 3° 1 . k 4 .
H = 27rha / [—(I1 — 3) + —(11 — 3) — 7(det B -— 1)] rdr, (3.144)
R, 2 2

with f1 and I f given by (3.134) and (3.136) respectively. Thus (3.144) may be written

as
H03 3m 81:01 Bee, 3221 73919,) =
Re A A A A
7rho£t A A(r, B", Bro, B99, B..,T,g,g’)dr, (3.145)
where

A : [k(Brr + BOO + Bzz _' 3) _ 2’l—(BzzBrr BOO — 3221330 —' 1)

46

 

 

1 '28..—2 '8. 8 1
+(( +(1‘9)) TAQ 9+ 90+ _3)],._

 

 

. . . 3.146
BrrBBQ — 372-9 822 ( )
The Euler-Lagrange equations for minimization of (3.145) are
6A _ 6A _ 6A _ 6A _ 0
aBrr 7 837-9 , 8890 ’ 6322 ,
8A d 8A 8A
_=0,___.._=0, 3.147
()7 dr 89’) 8g ( )

The first five equations in (3.147) give
k + 332(27'9’31—9390 - Ego _ 839 - (T9,)2B39 — 2TBZZBQQ : 0, (3.148)

A

8.. [((rg')2 + 1)8.. 8.. + 8.. 8.. — r g' (83. + 8.. 8.9)] (3.149)
+ 2 T 8.9 = 0,

k + 834898.83 — BE. — BE. — 099283.) — 2TB..8.. = 0. (3.150)
I: 83, —- 278.. — 1 = 0, (3.151)

8.. (8.. 89., — 83.) — 1 = 0. (3.152)

The last equation in (3.147) admits an immediate first integral by virtue of 6/1/89 =

0, the result being
‘ 3 I ‘ 2 ‘ J
B..(r g B... — r Bra) = E, (3.153)

where d is an integration constant. Note that equation (3.152) is equivalent in form to
(3.7). Moreover, if p = —2r, then equations (3.39) and (3.151) are also equivalent. It
remains to show that (3.148), (3.149), (3.150), (3.153) provides the same completions
as (3.36), (3.37), (3.38), (3.39) for suitably chosen d1 and (I. Although we do not

47

have the proof of equivalency of these two systems for general [6, such equivalency
can be demonstrated for both 19 = 0 and k = 1.
Let us first consider the neo-Hookean limit case k = 0. By taking k = 0 and

p = —27', equations (3.148), (3.149), (3.150) reduce to
832(27‘9'31—0300 — B30 — BEG — (T9,)2830) + P322369 = 0, (3-154)

A

Bzz[((7'g’)2 + ”En-Bra + 3993,49 — Tg’(éfg + Brréggfl - pBrg = 0, (3.155)

8342.48-89 — BE. — Bf. — (rg’12BE.)+ p8..8.. = o, (3.156)
and (3.153) reduces to

r3g'Err — r2E.g = 0. (3.157)
Moreover (3.151) gives

8.. = —. (3.158)

Now substituting from (3.158) into (3.154)1-(3.156) gives after some manipulation that

p2399 = [Tg’Bro — 39912 + 839, (3.159)
P2316 = [(0932 + 1)BrrBr9 + BOOBrO — T9'(Bfo + BwBoall, (3-160)
pZErr = [rg’Err —- Erg]2 + 133,. (3.161)

Now, (3.159) and (3161) are quadratic equations for Egg and B... respectively and
using the positive—definiteness of B can be solved uniquely in terms of Br6 and p.

After substituting Egg and B... in (3.160), the resulting equation is

 

(2 Bro — TQ’PZ) \/p4 + 431'!) (T 9,192 — BT59)
2-1-21'2(g’)2

 

— (3.162)

48

Tg’lp4 + 43w (7.9, P2 '— BN9”
2 + 2 7'2 (g’)2

 

:0,

which can be solved for Bra in terms of p. By using the resulting expressions for

A

B", Bro, B99 as functions of p in (3.152) one obtains

p3 +Px/P4 (1 +2139”)
2 + 2139’?

 

= 1, (3.163)
which has root
p = 1, (3.164)

and so retrieves the neo—Hookean limit (viz.(3.45).) If other roots of (3.163) are used
to find the nontrivial components of B, it can be shown that the internal balance
equation (3.34) is not satisfied. This is consistent with the uniqueness result for the
solution of internal balance equation (3.34). It then follows that (3.148)—-(3.l53) again
lead to (3.46). If (3.46) is used in (3.157), then (3.157) reduces to an identity. If (3.46)
is used in (3.145) with k = 0, then H is a functional of 9’ only. Indeed it is the
same energy integral (2.4) as given in the standard neo—Hookean case. By solving
the Euler-Lagrange equations associated with this integral, one obtains g(R) given
by (2.34). This shows that the solution of the Euler-Lagrange equations associated
with vanishing first variation to the functional (3.144) for k = O indeed retrieves the
k = 0 neo—Hookean result.

Next we will consider the case k = 1 which corresponds to a = 01*. By taking

k = 1 and p = —27', equations (3.148), (3.149), (3.150) reduce to

1 + 332(27‘9'Br9396 — 5’39 - 339 - (T9,)2BEB) + szzBBG = 0, (3.165)

B..[((rg')2 + Diana, + 6998., — rg'aéfg + 6.309)] — p119 = o, (3.166)

1+ B:2(27"9’37'1‘891'9 _ BEG — Bgr _ (T9,)283r) + péZVZBTT : 0’ (3'167)

49

and (3.153) does not change. An explicitly constructive solution procedure for obtain-
ing p from (3.165)—(3.167) is not obvious. However it follows by direct verification
that a solution of (3.165)—(3.167) is given by p = 0 in conjunction with components
of B given by (3.109)—(3.112). If (3.109)—(3.112) are substituted into (3.153), then
resulting expression reduces to an identity provided that cl = 0. This shows that the
Euler-Lagrange equations associated with the functional (3.144) for k = 1 retrieves
the k = 1 result for the internally balanced elastic material model.

By using 16 = 1 with components of B given by (3.109)—(3.112) in (3.145) we
obtain

12..
H1 2 27rhé /R.- (rm — 27‘) (11". (3.168)
Here the subscript 1 in H1 is notation to indicate that k = 1. It therefore follows
for k = 1 that if |\II| S \Ilmax then g(r) as given by (3.115) is an extremal of (3.168)
obeying (2.17). Indeed (3.113) is the first integral of the Euler-Lagrange equation for
(3.168).

The formal theory of calculus of variations can now be used to prove that g(r)
given by (3.115) is in fact a strong local minimum of (3.168) if |\II| S \Ilmax (See
Appendix A). Thus minimization of (3.168) is central to the k = 1 theory, suggesting
that (3.168) can be used to investigate solution possibilities for |\II! > \Ilmax. This will
be the focus of the next section using direct methods of the calculus of variations.

We will close this section by proving that g(r) given by (3.115) is indeed an
absolute minimum of (3.168) if Nil 3 \Ilmax. We proceed on the basis of a convexity
argument using standard procedures of the calculus .of variations as reviewed next.

Consider the basic problem of calculus of variations: Find a function y(:r) on the

interval (a, b] that minimizes the definite integral

b
lel = / F($,y($),y’(r)) dx, (3.169)

50

subject to
y(a) = A, y(b) = B. (3.170)

In general, the minimum value of J [y] depends on the eligible functions specified by

the problem. Hence, the basic problem is stated more precisely as to find
1;};ng | W) = A, W?) = B}, (3-171)

where S is the collection of eligible functions specified by the problem. Any
g(x) E S will be called as a test function. Given that y’ appears in the integrand
F (m,y(:c),y’ (123)) of J [y], it is rather natural to use S = Cl, that is the space of
smooth test functions. However for many problems, it is more useful to work with
continuous and piecewise-smooth test functions, denoted by D1. The following are

well-known from the calculus of variations [1], [28].

Definition 3.5.1 A test function is called an admissible test function if it also sat-

isfies all the prescribed constraints on y(2:) such as the end conditions.

Let g(a‘) be an admissible test function for minimizing J [y] and introduce the nota-

tion {W(z) := F(:I:, g(x), 37(3)), 131,,(x) := 6F(:c, g(x), 3)’(a:))/0y, etc.

Theorem 3.5.2 A C1 solution 3) that minimizes J [y] must be a solution of the

Euler-Lagrange equation

Ry - —-(F.y') = 0- (3.172)

Proof: See [28], page 13. D

51

Definition 3.5.3 A smooth test function which. satisfies (3.172) is called a smooth

extremal for the basic problem (3.169)-(3.170).

Definition 3.5.4 An admissible test function y(:z:) is said to be a weak local min-
imum for the basic problem {3.169)~(3.170) if J[y(:c)] Z J[3)(a:)] for all admis-
sible functions of the type y($) 2 10(3) + €h(:r) with |h(:1:)[ S 1, |h’(:r)[ S 1,

h(a) = h(b) = 0 and e > 0 sufiiciently small.

Definition 3.5.5 An admissible test function 32(23) is said to be a strong local min-
imum if 3 e > 0 such that J[y($)] 2 J[y(2:)] for any admissible y satisfying

lit/(~73) - 3?($)| < 6 in [a,b]-

Weak and strong local minima can also be defined by using weak and strong neigh-

borhoods of y($) as described next

Definition 3.5.6 For a given 6 > 0, the set of all piecewise-smooth test functions

satisfying
[31(1?) - 9($)l < 6 (3-173)

for each at E [a, b], is called a strong e-neighborhood of g(x) and denoted by

Us(6,i]($)) .

Definition 3.5.7 For a given 6 > 0, the set of all piecewise-smooth test functions

satisfying the inequality (3.173) and
|y’(:r) - i?’(x)| < 6 (3.174)

for each :1: E [a, b], is called a weak e-neighborhood of g(x) and denoted by
UW(€ag($))

52

Definition 3.5.8 An admissible test function y(:r) is said to be a strong local mini-

mum if 3 e > 0 such that J[y(a:)] 2 J[y(:r)] for all y E U5(e,y(:r)).

Definition 3.5.9 An admissible test function y(:r) is said to be a weak local mini-

mum if El 6 > 0 such that J[y(a:)] Z J[y(:r)] for all y E Uw-(e,y(:c)).

If S 2 D1; i.e., the set of all piecewise-smooth functions in [a, b], then the Euler-
Lagrange equation (3.172) alone is not sufficient to characterize the minimization.

The characterization needs to be modified as follows.

Theorem 3.5.10 In order for the admissible piecewise-smooth test function g of the
basic problem (3.169)—(3.170) to render the integral J [y] a weak local minimum, it is

necessary that

A

EN) = c + / Eye) dt, (3.175)

for some constant c.

Proof: See [28], page 38. D

Equation (3.175) is called the integral form of the Euler-Lagrange equation.

Corollary 3.5.11 At any point :1: where g' is continuous, it is necessary that 9

satisfies the Euler-Lagrange equation (3.172).

Definition 3.5.12 A piecewise-smooth test function g(x) that satisfies (3.175) is

called an extremal for the basic problem (3.169)-—(3.170).

53

It is more effective and convenient to work with the Euler-Lagrange equation (3.172).
Hence, it is important to know whether the extremals for a basic problem are smooth
and the locations of discontinuities of their derivative when they are not smooth.
These discontinuities are known as corners of extremals. Information on discontinu-

ities can be obtained from the so—called Erdmann’s corner conditions.

Definition 3.5.13 A corner point of a test function y(:1:) is a point :r where the left
derivative and right derivative of y(:r) exist at :z: (and are denoted by y’_ and y’+

respectively) but they are not equal.

Theorem 3.5.14 If y(:1:) is an extremal, then 13],,(1‘) is continuous at a corner point.

Proof: See [28], page 45. [:1

Theorem 3.5.15 If g(z) is an extremal, then the quantity (F—g'fiw is continuous

at a corner point.

Proof: See [28], page 46. [:1

Theorem 3.5.16 (Hilbert ’3 Theorem)Suppose that g is an extremal, 2:0 is not a

corner point, and 13‘ y.y,(a:0) 51$ 0. Then 7) is C2 and

(man (Riot/1L (F... — F.) = 0. (3.176)

holds in some neighborhood of .130.

Proof: See [28], page 48. [:1
Equation (3.176) is sometimes called the ultradifferentiated form of the Euler-

Lagrange equation.

54

Definition 3.5.17 A scalar valued function f : R" ——> R is said to be convex if
f(/\w+ (1 — /\)2) S Af(w)+(1— /\)f(z) (3.177)

Vw,z€ R", and VA 6 [0,1].

Theorem 3.5.18 If F (x,y, y’) is differentiable in its arguments and convex in y
and y’ for each x, then any admissible extremal of the basic problem (3.169)-— (3.170)

renders J [y] a global minimum and conversely.

Proof: See [28], page 108. CI

Theorem 3.5.19 A scalar valued function of one variable f (y) is convex if and only

if f” is everywhere non-negative.

Proof: See [28], page 105. E]
This completes the quick review of the appropriate machinery from the calculus
of variations. We may now on this basis rapidly show that g(r) as given by (3.115)

is indeed an absolute minimum. For (3.168) the Lagrangian F is

F(r,g,g') = r\/4 + (rg’)2 — 2r. (3.178)

From (3.178) it follows that

(92F 4r3
8—9’5 = (4 + (T9,)2)3/2 > 0. (3.179)

 

Hence F (r,g,g’ ) = F (r,g’ ) is convex in g’ for each fixed r by Theorem 3.5.19. If
[‘11] S \Ilmax, then it follows by the Theorem 3.5.18 that g(r) given by (3.115), (3.118),

(3.119) is an absolute minimum of (3.168) obeying (2.17).

55

3.6 Direct Methods for Minimization when k = 1

At the end of the last section it was shown that if [‘1'] is restricted by [W] S \I/max
where \Ilma,‘ is given by (3.117), then g(r) given by (3.115) provides an absolute
minimum for (3.168) among functions obeying (2.17). The significance of this result
is that minimization of (3.168) subject to (2.17) governs the k = 1 problem. In this
section we will investigate the minimization of (3.168) subject to (2.17) for arbitrarily
large twist \II. Since g’(R,-) -> 00 as W —> \Ilmax, it is conjectured that minimizers
might involve vertical line segments. Since such a vertical line segment cannot be
considered as a function of R, it is expedient to use the parametric form of the
energy integral (3.168) for further investigation.

Consider the basic problem of calculus of variations in parametric form: Find a
piecewise-smooth curve C = {(x,y) = (x(t),y(t)) : t1 S t S t2} that minimizes the

definite integral

t—2
J0: / f(t,x,y,x,y)dt,= / f(t,x,y,x,y)dt, (3.180)
C t—l

subject to

WHILE/(’51)) = (331,91): ($(tzlay(t2)) = ($2ay2)' (3-181)

The dots indicate derivatives with respect to the parameter and f is a given function,
continuous in all its arguments when (x(t),y(t)) lies in a given region G; x, 3]
have arbitrary values and are not zero simultaneously. The function f cannot be
completely arbitrary because the integral under consideration must depend only on
the curve C and not on any particular parametric representation of the curve. By

the discussions given in [1], p 39—40 this independence of parametric representation

will hold if and only if:
(i1) t does not appear explicitly in the integrand function f; that is,
f(ta$.y.i.3)) = f($.y.i3.3'l)- (3-182)

56

(i2) The integrand function f (x, y, x, 3)) must be positively homogeneous and of the

first degree relative to the second pair of arguments; that is,

f (x. y. mi. me) = m f (:v. y, i. 3)) Vm > 0- (3-183)

As a result of homogeneity

If the common ratio in (3.184) is denoted by K = K (x, y, x, y), then
fix = 92 K, fey = ~33? K, fyg = 1132 K. (3.185)
Suppose an admissible curve
C: x = at). y = @(t) (t. s t 3 6) (3186)

gives the functional JC a weak relative minimum. Then functions (13(t), (p(t) must

satisfy the Euler-Lagrange equations in parametric form

f. = A + f ME) d5. f. = B + f me as (3.187)

in which A and B are constants.

If arc length 3 is used as parameter, then

d
:i: = i = cosw, y = "g 2: sinw, (3.188)

ds ds

where w is the angle between the tangent to the curve and the x-axis. Along each

smooth arc of a curve

at d

_ . __ = __ ... = 3.1 9

dsf. f. o, d812, f. o, ( 8 )
and at a corner point 3 = so

fi(3(i) " I'd-(36) 2' 0) fy(5(i) " 131(36): 0' (3.190)

57

(Here 53 and 3; denote the one-sided limits lime—.3; and limsfisa respectively.) A
smooth solution of (3.189) is called an extremal and (3.190) express the Weierstrass-
Erdmann corner conditions. Note that (3.190)1 is equivalent to corner condition given
by Theorem 3.5.15 and (3190);; is equivalent to corner condition given by Theorem
3.5.14 for the nonparametric case.

If parametric forms of r = x(t) and g = y(t) are used, then the energy integral

(3. 168) becomes

Hp[x(t),y(t)] = 27rhoz [t2 E W — 2x] 36 dt, (3.191)

t1

where the subscript p in Hp indicates that the parametric form of the energy is used,

if : dIII/dt, 3] : dy/dt, ($(t1),y(t1))=(R1,2,/Ji) and ($02), 3102)) = (R0, 2110)’

Note that the Lagrangian of (3.191) can be written as
f(x, y, x, y) = x 43':2 + x2y2 — 22:31. (3.192)

Homogeneity of f in the last two variables is immediate since

 

f(x, y, mi, my) = x \/4(m:i:)2 + x2(my)2 — 2x(mx) (3.193)

 

= x \/4m2x2 + x2m2y'2 — 2xmx

= m[x\/4:i:2 + x2 '2 — 2x33]

: mf($7y3j:7y.)'

Thus condition (i2) is satisfied. Moreover, condition (i1) is satisfied since t does not
appear explicitly in the Lagrangian (3.192). Therefore the value of the energy integral
(3.191) depends only on the curve C not on any particular parametric representation

of the curve.

58

The Euler-Lagrange equations (3.187) are

 

[ 4x1i:

t
and

x3y
\/4i2 + $2312

where 51 and 52 are constants. Note that (3.195) is equivalent to (3.113) and by

 

= 52, (3.195)

taking the derivative with respect to t in (3.194) the first Euler— Lagrange equation

 

 

becomes
d [ 423x . x2y2
— — 2x] = 4x2 + 2:2 '2 + — 2a. 3.196
dt 43';2 + 33292 y /4:i:2 + $292 ( )

From (3.195) and (3.196) it follows that y(t) = constant and curves described by
(3.115) are extremals whereas x(t) = constant are not extremals. An extremal
consisting of an arc of (3.115) and a straight line segment y(t) = constant will not
satisfy the Erdmann’s Corner Conditions (3.190).

For [\IJI S ‘Ilmax, we consider the composite curve consisting of the curve obtained
from (3.115) by vertical translation (with d1 = R? and d2 = 1b,») in conjunction with
the line segment x = R,- between y = w,- and y = «to — \Ilmax. This composite
curve will be called A1. Thus A1 consists of a vertical translation of the \P = \Ilmax
extremal so as to meet the x = R0 end condition y(Ro) = 1110, along with an x = R,-
vertical segment so as to connect to the x = R,- end condition y(R,) = 212,-. Since
this vertical segment is confined to the boundary, it need not satisfy Euler-Lagrange
condition (3.195) and (3.196).

We begin with some direct comparisons. In particular, it will now be shown that
Al as described above provides a minimum to (3.191) among a collection test curves:

AnH, A2, A3(n), A4(n) that we now define.

59

In the remainder of this section we will take 11), = 0 and 1/10 > 0. For (b0 S \Ilmax,

we note that A1 can be parameterized as follows

mt): R‘ if t6 [O’Ri] (3.197)

t if te [R.,Ro],

(7pc— \Ilmax) t/R1 if t E [0,R1]
(wo — \I’max) + arctan \/(t4/R§) — 1 if t E [R,-, R0],

 

x(t) =

Then the corresponding energy can be computed from (3.191) as
leAI] = Ed + E3 (3.198)

where

Ed -_— when? (a, — rm], E, = 277hci [, /Rg — 123+ R3 — R3]. (3.199)

Note that first term Ed is the energy corresponding to vertical segment x = R,- and
second term E, is the energy corresponding to the curve obtained from (3.115) by
vertical translation. The associate matching requires that d1 = R? and d2 = 0 in
(3.115). Since \Ilmax = arccos(Rf/Rg), the energy expression (3.198) may also be

written as

Hp[A1] = 2nhd [RE (the — arccos(fi—‘ZD + VR: — R? + (R,2 — 123)] (3.200)

As we —> \IJmm = arccos(Rf/Rg) the energy associated with the discontinuity Ed
goes to zero as expected.
Let Any be the curve corresponding to neo—Hookean solution (2.34), (2.36) and

(2.37) with w, = 0. Then it has the parametrization
an(t) =t ift E [R,-,Ro], (3.201)

_ '"RgRg'wo + $0 R3
_ (R?) — R?) V (R?) - Rf)

 

ynHU) 1ft 6 [Ri,Ro].

60

 

 

[ we [I N. Energy for A1 ] N. Energy for Any ]

 

 

 

 

 

 

 

 

 

 

2 1.4605 1.7056
2.5 1.9605 2.4835
3 2.4605 3.3440
3.5 2.9605 4.2720
4 3.4605 5.2570

 

 

 

Table 3.1: Normalized energies for A1 and An” for several we values when R.- = 1,
Re 2 4 and w, = 0

Then the energy for the present case k = 1 using the k = 0 neo—Hookean deformation

can be computed from (3.191) as

 

33103

leAnHl =27Thé{Ri2—RZ+R3 \/1+ (R?_R2)2 _

 

 

 

R4 (t2 12.2sz R2 «p R4 w?
2 o o i o 0 o 0 o o
R*‘\/1 + (162,.2 — 123;)2 +(123— R3)210g [123— 123+ 1+ (12,?- Rg)2l

 

-- .22
(fi—%) x—e :—er “0)

R3R3¢o2 16g [31L + \/1+ (Raft/23 ] }

Note that He[A,,H] z 5.2570(2nhd) when Ri = 1, Re 2 4 and we = 4. In
the following table, Table 3.1, the normalized minimum energies for A1 and An” are
presented for several we values when R,- = 1, Re = 4 and w,- = 0. The energy in this
table is normalized via division by 2 77 h a. It is consistently verified that A1 < An”.

We now construct a broader class of comparison functions. Let A2 be the straight

line connecting (Re 0) to (Re, we). It is given by the parametrization

x2(t) =t ift e [Rene], (3.203)
920) = ”(t—1,25% ift e [R,-,Re].

61

The corresponding energy is computed from (3.191) to be

He[A2] = 2 77 ha [R3 — 173]. (3.204)

 

{4 (R. — R.)2 + RS 12313” — [4 (R. — R.)2 + R? 212213/2]
“Ra-302413 .

Let A3(n) be the family of curves given by the parametrization:

+2nhd]

3:3,..(t) =t ift e [Rene], (3.205)

y3,,,(t) = we [1+ (arctan —T%:—:—f§-)-)/arctan n] ift E [Iti,Re],

with n 2 0 and not generally an integer.
Figure 3.15 shows 6 curves of this family for the case R,- = 1 and Re = 4 (implying
\Ilmex 2 1.50826) and we = 4, namely n = 0,1, 10,100, 1000, 10000. Note that in the

limit as n —+ 0 these rectifiable curves tend to A2, and as n —> oo tend to the step

function
t if t e R,,Re
x3,ee(t) = [ 1 (3.206)
Re if t e [Re,Re+we],
- 0 if t e [ReRe]
93,000) =
(t—Re) if t e [Re,Re+we].

By numerical computation it is found that Hp[A3(n)] increases with n. The corre-

sponding energy as n —> 00 can be computed from (3.191) as
”11.520 Hp[A3(n)] = 2 77 hd R3 we. (3.207)
Let A4(n) be the family of curves given by the parametrization:
x4,,,(t) = t if t 6 [Re Re], (3.208)

2%))/arctann] ift E [RaRo].

y4,n(t) = we [(arctan R0

62

 

G(R)

 

 

 

 

 

 

 

Figure 3.15: Graph of the family A3(n) with R.- = 1, Re = 4, w,- = 0 and we = 4
for n = 0 (the diagonal line), and 1, 10, 100,1000, 10000

with n 2 0. Figure 3.16 shows 7 curves of this family for the case R,- = 1 and Re = 4
and we = 4, namely n = 0,1,10,83.1,100,1000, 10000. The dashed line shows the
case n = 83.1, the significance of which will shortly be made clear. In the limit as

n —-+ 0 these rectifiable curves tend to A2, and as n —> oo tend to the step function

. “f 0, .
$4,000): R' 1 t6[ R] (3.209)

t if t e [R,-,Re],

«at if t6 [0.1%]
we if t e [ReRe].

314,00“) =

In this case the corresponding energy as n —> 00 can be computed from (3.191) as
lim He[A4(n)] = 2 7r 66. R? we. (3.210)

Numerical calculation shows for this family of curves that the corresponding energy

He[A4(n)] initially decreases with n after which He[A4(n)] subsequently increases

63

 

 

 

.5
0|

 

 

 

 

 

Figure 3.16: Graph of the family A4(n) with R,- = 1, Re = 4, w,- = 0 and we = 4
for n = 0 (the diagonal line), and 1, 10, 83.1, 100, 1000, 10000. The dashed line shows
the case n = 83.15 which results the minimum energy among the family A4(n)

with n. Let nmie = nmie(R,-, Re, we) denote the transition value of n between these
two behaviors. We find for example nmin(1,4, 4) z 83.15 and the corresponding
minimum energy among A4(n) is Hp[A4(83.15)] z 3.70242 (2rhé). In Table 3.2
the transition value of n and corresponding normalized minimum energies among
A4 (71min) are presented for several we values when R,- = 1, Re = 4 and w,- = 0. Once
again the energy in this table is normalized via division by 2 77 h 6.

Taken together the family of curves A3(n) with n decreasing, the single curve A;
and the family of curves A4(n) with n increasing form a larger continuous family of
curves with minimum He at A4[nmie(R,-,Re,we)].

We find on the basis of extensive numerical calculation that the energy He at
A4[nmie(R,-, Re, we)] is still greater than the energy corresponding to A1 for the same
values of R,- , Re and we. In particular comparison of Tables 3.1 and 3.2 shows this

for a few selected cases. Figure 3.17 for R, = 1, Re = 4 and we 2 4 compares the

64

 

 

[ 212.. [[ n [Normalized Energy]

 

 

 

 

 

 

 

 

 

 

2 18.08 1.54436
2.5 28.14 2.10070
3 42.52 2.64498
3.5 61.00 3.17782
4 83.15 3.70242

 

 

 

Table 3.2: Normalized minimum energies A4(nmie) for several we values when R,‘ = 1,
R0 = 4 and $1 = 0

graph of A1 with the graph of A4(83.l5) which gives the minimum energy in this
extended family.

Consequently, for a given Re and fixed we 2 ‘Ilmex the minimum among the
computed energies remains Hp[A1].

Formally, A1 and hence He[A1] is defined only if we 2 \Ilmex. However it is natural
to extend A1 to 0 S we S \Ilmex via (3.115) and this is understood in what follows.
The graph in Figure 3.18 shows normalized energies E1 , E2, E3, E4 and E,” as a
function of the amount of twist we corresponding to A1, A2, A300. 114,00 and An”,
respectively, when R,- = 1 and Re = 4. The normalized energy associated with
A4['nmin] as given in Table 3.2 is also indicated by stars for we = 2, 2.5, 3, 3.5, 4.

If we E [0, \Ilmex] then the torque-twist relation is given by (3.132). If we 2 \Ilmex
then we note from (2.48), (2.49) that it is enough to compute BW/Bwe which can be

computed from (3.200). Hence the torque-twist relation for A1 is given by

 

 

27rhéR§R§sinwe if 0322,3me
M = JR? +R;‘,—-2R,2 R3 coswe (3-211)
27th§6 if we 2 am

The conspicuous feature of (3.211) is that the twisting moment associated with
this extended deformation A1 is monotonically increasing with we on 0 S we S \Ilme,x

after which the twisting moment remains constant. The physical interpretation is

65

 

 

G(R)

 

 

 

 

 

 

 

 

1 ' ' ' I
0-5 --------------- J" - — (dashed line) A4(83,1) _
—— (solid line) A1
0 ..... 4 """"" 1 ______________________________

Figure 3.17: Graph of A1 and A4(83.15) when R,- = 1, Re = 4 and we = 4

 

0|

A

Normalized Energy
to (D

 

 

 

 

Amount of Twist

Figure 3.18: Normalized energy versus twist for R,- = 1, Re 2 4, w,- = 0 and varying
we. Here \I/max = 1.50826. The energies corresponding to A1, Ag, A3,ee, A4,ee
and Any are normalized by dividing 2 77 h (‘1. The normalized energy associated with

A4[nmie] is also indicated by stars for we = 2, 2.5, 3, 3.5, 4

66

that conventional macroelastic resistance to twist occurs until we 2 \Ilmex after which
substructural rearrangement concentrates at the singular surface R = R,- so as to
relieve any further macroelastic resistance. This novel effect is not present in the
conventional theory of hyperelasticity, although it is reminiscent of other phenomena
in the conventional theory such as cavity formation and shear banding at critical load
values [5], [14], [19], [23], [27].

It will now be proven that Al given by (3197) provides a weak local minimum
for the parametric energy integral (3.191) subject to (x(tl),y(t1)) = (R,,0) and
($(t2), y(t2)) = (Re, we) in the family of piecewise-smooth curves lying in the closed
region a = {(x,y)|R.- s a: 3 120,0 3 y s 7).}-

To this end we begin with a technical detail. Namely we note that Al given
by (3.197) is a differentiable curve only if t 75 1. However the location of non-
differentiability at t = l is an artifact of the parametrization. If A1 is reparameterized
in terms of arc length, then it is also differentiable at the location associated with
the original t = 1. To this end let 5 be the arc length parameter. Then A1 can be

parameterized by

R,- if s E [0,we— \I'mex]
‘3 h(s) if s E [we—\Ilmex,se],

(3.212)

 

s ifs E [0,we — \Ilmex]
we — \Ilmex + arctan \/(h4(s)/Rf) —— l ifs 6 [we - ‘Ilmex, se],

 

where se is the total arc length and

h(s)
0

By using (3.197), it follows that ( 3 213) )can be written as

’1(3) 4
s— — we — \Ilmex +/ \/1-l~::7 4R dt. (3.214)

 

-—R4 t2

The integral in (3.214) does not have a readily explicit antiderivative hence there is no
readily explicit form for h(s). However using (3.214) it can be seen that parametriza-

tion of A1 given by (3.212) is a C1 parametrization.

Proposition 3.6.1 The A1 given by {3.197) is a weak local minimum of {3.191) sub-

ject to (x(t)),y(t1)) = (Re-,0) and (x(tg),y(t2)) = (Re,we) in the class of piecewise-

smooth. curves lying in the closed region G = {(x.y)]R,,- S x S Re, 0 S y S we}.

Proof: Let 6 2 0 be a given small nonnegative number. Consider the curve defined

by

5(7) = 51(7) + e ((7), (3.215)

W) = 910) + 6 720).
where ((0) 2 77(0) 2 0, C(Re) = 77(Re) = 0 and ((t) Z 0 if 0 S t S Rf. Here x1(t)
and y1(t) are given by (3.197); i.e., (x(t), y(t)) is a small perturbation of (x1(t), y1(t))
that is confined to lie in the closed region G .

Then the variation in H,e as a result of this perturbation is defined by
AH. = Hp[x1(t) + . ((03.0) + e 1200] - H.930, 91(1)], (3216)

or equivalently

R0
AHP=(27rhd) {f[x1+6C,y1+677,x'1+eC,y°1+67'7]— (3.217)
0

f[xla 111,151: 91]} dt)

where “dot” denotes d/dt. By using the Taylor Series expansion for the first term

in the integrand, (3.217) becomes

R0
AHP = (27fhd) {f[$1,y1,£61,?j1]+ (3.218)
0

68

of of of . of. . .
€[8_xl-C+ Can-*- 5.13—1—(+— Cyln 7”] +6"°2[l + — flzlaylixlvyllldt’

Hence (3.218) can be written as
2
AHP=(2rh&)[661+%621+...], (3.219)

where 61 is the first variation given by

3° 3f 3f (9-f (9f
61 2 —~ + — +— +-——— dt. 3.220
/0 [Bxlc (93/177 6231 8y1 7}] ( )

Since A] has a C1 parametrization, ('3 f /0x1 and a f /0y‘) are differentiable with

respect to t. Hence integrating by parts in (3.220) is permissible, which in turn gives

‘51: logici'ayfl l

R0 at (8f 8f 6f
/o “of 6231 dt(8x1)lC+ leg—g a—gh)]n}dt.

The boundary terms in (3.221) vanishes by virtue of ((0) 2 77(0) = 0, ((Re) 2

 

+ (3.221)

0

q(Re) = 0 so (3.221) reduces to

61 = /0R0{[§—xfl -— Big—bl (+ [8%]; — 52(6‘3—i)] 1]} dt. (3.222)

Note that the integral in (3.222) is naturally written as a sum of the two integrals

associated with the different parts of the composite curve A1:

32' .
61=/0 {[gxil—%(g—1{1)]c+ [g—yfl—ac-i; 62%)]77} dt. (3.223)

Ro I
+fR, {[§£*%(%)]C+[§i—% %)]n]dt.

The second integral in (3.223) is identically zero as the portion of A1 restricted to
[R.-,Re] is the solution of the corresponding Euler-Lagrange equation for (3.168).
Moreover since the Lagrangian (3.192) is independent of y, it follows that (3.223)

reduces to

61 2 Am { [g1- — %(66_1f1)] C— [1(3le 7)} dt. (3.224)

69

Since x1, y1 are given by (3.197), it follows that the integral in (3.224) simplifies to
Rt
61 = / 2w. — 91......) ((7) dt. (3.225)
0
In view of the restriction C (t) 2 0, it is immediate that
61 2 0. (3.226)

Consequently for a sufficiently small 6 2 0, the sign of AH? is determined by the

sign of 61 which, in turn, is positive by virtue of (3.226). Thus
AH. = lesvl(t) + e C(t).y1(t) + . 0(01 —- lercl(t).y1(t)l 2 o, (3.227)

which shows A1 is a weak local minimum of (3.191). This completes the proof of
Proposition 3.6.1. C]

We conclude this section by proving that A1 is indeed an absolute minimum of
(3.191) in the class of piecewise-smooth curves such that, once restricted to the interior
of G , their parametrization reduces to a function y = f (x) The importance of such
a result is that if we seek minimizers of (3.168) when we > \Ilmax in the space of
piecewise-smooth functions g(R) then the global minimum involves a discontinuity
at R = R,- of magnitude [[9]] = we — \Ile,ex such that g(R) on R,- < R S Re is given
by a vertical translate of the we = \Ilmex solution. The energy Ed associated with the
discontinuity is given in (3.199)1 and the energy Ee associated with the translation

is given in (3.199)2.

Proposition 3.6.2 Let y(x) be any given function E D1([Re-,Re]) with y(R.) = 12:“

and y(Re) = we. Let y(x) be the function given by

g(x) = (a. — 9,...) + arctan (TE-)4 — 1 x e [17,, Re]. (3.228)
Then
HIE/(513)] _ H1 [g(xll 2 R22 (271' lid!) ("/90 _ ‘I’max — 121'), (3229)

70

where H1[y(x)] is given by (3.168) (x and y(x) are used respectively instead of r
and g(r)) and \IImax is given by (3.117).

Proof: Since the Lagrangian of (3.168), F (x, y’) = x 4 + (xy’)2 — 2x, is convex in
y’ for a fixed x E [R,-, Re] it follows that

F(Ivyl) _ F(SE, 9’) Z 1231’ (y, _ ill)? (3230)
where

F.y’ = Ey’

 

(3.231)

y=f1

Integrating (3.230) from Re- to Re gives

R0 R0
/ F(x.y’> —F(x.7)dx 2 f F... (y'—0’>dz. (3232)
Hi Ri
whereupon
R0 A
fills/(13)] — Haw] 2 (2.7...) f _ 13./(y’— :7) dz. (3233)

By integrating by parts from (3.233)

R0
— (3.234)

Ri

H.000] — H.399] 2 omega, 0 — 27>]

 

[201,372.10 — .0) dz}.

Ri
Since y(x) is the solution of the Euler—Lagrange equation for (3.168), d[13‘ y,]/ dx = 0.
Hence equation (3.234) reduces to

R0

 

H.090] — H.129» 2 (2 «h a) [Ry] (y — 27)] (3235)

Rt.

Moreover y(Re) = y(Re) = we, so (3.235) reduces to

H1[y(:r)l—H1[39(:v)l2-(27129) [F37 mas—3(a)». (3.236)

x=R1

 

71

Now F.y’ == x3y’/(/4+ (73y’)2 implies F y, = RE. Using this result in (3.236) in
conjunction with the boundary values y(E) : w, and y(R.) = (1110 - \I/mex), it

follows that (3.236) becomes
H1[y(:v)l - H1 [31(7)] 2 R? (2 77 h a) (we — Wm... — 17,-). (3.237)

This completes the proof of Proposition 3.6.2. El

Theorem 3.6.3 Let A1 be the curve given by (3.197). Let C be any curve given by

the following parametrization

x(t) = Rd 2ft 6 [O’R‘] (3.238)

t z‘ft e [R.,R..].

(1/21 tl/Ri 2f t 5 [0,52]
900) 2'f t E [Rifle].

y(t) =

where cp(t) is a piecewise-smooth function with go(R,-) = w,- and (p(Re) = we. Then

le$(t).y(t)l — lefliiU). 3110)] Z 0- (3-239)

Proof: Note that

R0

le$(t).y(t)l=27rhé 0 f[$(t).y(t).i¢(t).i/(t)ldt, (3-240)
.. Rt . . Ha I
=2rha( 0 fl:v(t).y(t),x(t).y(t)ldt+fm le.y($).y($)ld$).

where F [x, y(x), y’ (x)] = x(/4 + (xy’)2 — 2x is the nonparametric Lagrangian. Simi-

larly
R i

le$1(t).yl(t)l = ma 0 711230330330,7.0)] dt (3.241)

72

R0
+277hd/ F[x,y1(x),y’1(x)] dx.
Ri

Therefore

lel‘U). y(t)] - lex1(t). 910)] =

Ri
27Thé {flxayaxayl —f[$11ylaaflay.1]}dt+
0

R0
27% R. {F [$.y($),y’($)l-F [audits/{(006133-

By Proposition 3.6.2, equation (3.242) gives

le$(t).y(t)l - le$1(t).yl(t)l 2

Ri

(3.242)

(3.243)

277hé f[x, y, 6:, g] — f[x1,y1,x'1,y'1] dt + R?277h&(we — wee. — 773,-).
0

However after computing the integral in (3.243) it is found that

le$(t)a y(t)] — le$1(t). 3110)] Z 2M5! R? [1171'— (w. — ‘I’maxfl

+Ri2 2717161 [(wo "' \I’max — 127)],
or simply

lem(t)ay(t)l — HPl$1(t)v y1(t)l Z 0

This completes the proof of Theorem 3.6.3. Cl

73

(3.244)

(3.245)

CHAPTER 4

A More Generalized Class of
Material Models

4.1 Extended Material Model

We now extend the model in Chapter 3 along lines suggested by Pence and Tsai [22].
In particular, we consider an extended material model whose strain-energy function

is given by

a a"
W—§(11—3)+3

(If - 3) + (i1 - 3). (4-1)

MIQ>

where a Z 0, 07" Z 0, d 2 0. The form (4.1) augments the previous form (3.6) by
inclusion of the additional 11 -— 3 term. Our interest is in the extent to which a > 0
gives different qualitative behavior from the a = 0 treatment of Chapter 3. The
boundary value problem for pure azimuthal shear with boundary conditions (2.17) is
again the focus of study.

For the extended material (4.1) the minimization procedure again yields (39)

where now the Cauchy stress (3.10) is given by

a=aB+dB—pI. (4.2)

74

In polar coordinates (3.9) again reduces to (2.14)—(2.16). The interstitial balance is

still given by (3.11) namely
a‘B—ciB —qB=0. (4.3)

Here p is the hydrostatic pressure and q is the Lagrange multiplier associated with
the constraint (3.4) and B is still the left Cauchy-Green tensor given by (2.12). When
61 = 01* the pure azimuthal shear problem reduces to the following first order nonlinear

ordinary differential equation

- , d
arg’+—-arg “— (4'4)

where d is an integration constant.

This result can be obtained as follows. Note from Section 6 that if (i = oz“ (k = 1)
then p = 0 or equivalently q = 0. Therefore the nontrivial components of B are
still given by (3.109)—(3.112). The Cauchy stress tensor (4.2) can thus be expressed
in terms of r and g’ (r) One now finds from (2.14)—(2.16) that p = p(R) whereupon

(2.16) is satisfied automatically and (2.14) yields (4.4). The remaining equation (2.15)

then gives
dp . d 1 I 2 (37(902
—=2a— —— —ar —-———, 4.5
d7 d7( 4+ (r992) (‘9 ) \/4+ 0.7)? ( )

from which p = p(r) can be computed once g(r) is known.

Attention shall henceforth be restricted to this case 6. = a“. Note that (4.4) with
a 2: 0 and d1 = d/c‘r retrieves (3.113) of the previous development.

Equation (4.4) when written as a polynomial in g’ is quartic and hence can be
solved for g’ . It will be shown that if a > 0 then equation (4.4) subject to boundary
conditions (2.17) has a unique solution. Hence it is necessary to pick the correct root
of the quartic polynomial. Even though it is possible to distinguish the correct root

from extraneous ones, it does not seem feasible to integrate this root to get an explicit

75

form of g(r). Therefore the solution of (4.4) subject to (2.17) will be investigated by
alternate means.
Note that equation (4.4) can be written as

9': .1. [i _ 5‘9, ] (4.6)
a 7‘3 4+ (79’)2

 

or equivalently

_ ,1 '1 3 " 9’(€)
g(r)—g(awafeu 0Lfl+(€g,(€»,de (4.7)

By integrating the first integral in (4.7) and using the first boundary condition in

 

 

(2.17), g(Re) = w,, equation (4.7) can be written as

 

 

d r2—R.2 61 '
T' = i ——2——2t -— F , I d, 4.8
g(> Mam”, a]... (mom ( )
where
F , I : g’(€) . 49
(69(5)) \/4+(£g'(o)2 < )

By using the other boundary condition in (2.17), g(Re) = we, the constant (1 can be

computed as

F(€,9'(€)) 616- (4-10)

2aRzR2 2737R2R2 3"
d=(w..—w.) * ° 7 f

63—6.? + fez—R? ...-
Therefore boundary value problem (4.4), (2.17) is equivalent to the following integral

equation

90‘) = 1111' + (100 — 1Pi)P(7‘)+

r (3 RO
(Pm—1) f F(r,g'<s))ds+;P<r) / F(r,g'evde (4.11)

R,-

QIQ)

where

_ R: (.2 - R?)

P”) “ (Ra—R?)

 

(4.12)

76

By introducing

g(r‘€)_ P(r)—1 13.359377. (413)
’ F(r) 12.935312... '

the integral equation (4.11) can be written in a more compact form

.. R0

g(r) = 4. + (4.. - 21;.) P0) + 3 Hag) F(eg'e» 44. (4.14)

Q Hi
The boundary value problem, (4.4), (2.17) is usually called the first boundary value
problem (the value of the unknown function is prescribed at each end of the interval).
The second boundary value problem is a boundary value problem for which the
value of the unknown function is prescribed at one end point and the slope is pre-
scribed at the other end point. For latter purposes we will define the following second

boundary value problems: Find a solution of (4.4) either subject to

g(R’i) = wi) gl(R0) : m0, (415)

or else subject to
g’(lii) = mi, g(Ro) = 1/10- (4.16)

Integral equation (4.8) also follows for the second boundary value problem (4.4),
(4.15). However the constant d in (4.8) is now determined from the latter boundary
condition in (4.15). By setting 9’ (Re) = mg in (4.6) and solving for d it is found that

time R3
(/4+ R3 m3-

Consequently the second boundary value problem (4.4), (4.15) is equivalent to the

dzameR3+

 

(4.17)

following integral equation

dme R2 r2 — R?

a(/4+R§m34 QRET‘2

 

g(r) = a. + [me 123+ (4.18)

A

f f . negro) 44.

R

77

where F (E ,g’ (6)) is still given by (4.9). Alternatively this second boundary value

problem is equivalent to the initial value problem for

 

A I l
arg’+_fl_g—: (

_ o , 4.19
,F_—4+(rg)2 72 ° 4+R3m3) ( )

with initial condition
g(Ri) = 7.91- (4.20)

By a similar development the second boundary value problem (4.4), (4.16) gives

 

 

d—am-R3+ ém‘Ri (421)
- 1 z \/4 + R? m? .
and integral equation
_ 3 (i m,- R? R2 — 7'2
g(T‘) — wo "' [mi Ri + a m] 2 R3 7‘2 (4.22)

A

0 R0
+; / F(r,gro) 46

The corresponding initial value problem is

" ’ 1 rim-R3
ar ’+————a—r—“(-’——-—=—(am,R?+ z ' ), 4.23
9 4+(.~¢)2 .2 W ( ’

subject to the initial condition

 

g(Ro) : 410- (4.24)

4.2 Existence and Uniqueness for the Extended

Model

In this section it will be proven that the first order boundary value problem (4.4),
(2.17) has a unique smooth solution on [Re-, Re] for every 0 < R, < Re < oo whenever

a>0.

78

Since it is more convenient to deal with integrals than derivatives, we will use the
integral equation form (4.8) of the first boundary value problem (4.4), (2.17). Then
the principle of contraction mapping will be used to prove existence and uniqueness.

Let S be a normed linear space with the norm denoted by ||..|| A sequence of
elements yn of S is a Cauchy sequence if for every 5 > 0, there exist N such that
Hyn — ymll < 6 whenever n, m > N. S is a complete space if every Cauchy sequence
in S converges to a limit element in S.

An operator T mapping a normed space S into itself is said to be a contraction

mapping on the space if there is a number A, 0 < /\ < 1, such that for all 11:, y E S
IlTx - Tyll S A llx - yll. (4-25)

The following theorem is well-known.

Theorem 4.2.1 Principle of Contraction Mappings: Every contraction mapping T
defined on a complete normed linear space S has one and only one fired point (i.e.,
y = Ty has exactly one solution). For any point yo 6 S, the sequence of iterates

yn = Tyn—i = = T"y0 converges to the fixed point y and

n

A
Ilyn - 31” S 1_ ,\ llyr - ml (426)

 

Proof: See [4], page 26. Cl

Lemma 4.2.2 If

A _ . 2
2,3 (12" RR‘) } < 1, (4.27)

max {go—l (R0 — R.)

then the first boundary value problem (4],), {2.17) has one and only one solution.

79

Proof: Let So be the space of continuously differentiable functions on [R4, R0] with

the norm

 

||g(r)||=maX{&I;1gRo lg(r)| 1.2%. '33)“) (4.28)

where z(r) is the positive weight function given by

2(r)— Rg— 21R. (4.29)

The space So is complete with this norm, and convergence of a sequence gn(r) in this
norm to g(r) implies that g(r) is continuously differentiable and that gn(r) -—> g(r)
and g;(r) —-> g’ (r) uniformly.

Let T be the operator defined by the right side of (4.14). Observe that

3TgIr)=(zl.— w.)P’(r)+ 3]R° HI (re F(t’égI )) d4 (4.30)

where Hr(r,€) = BIT/8r. Since P’ (r), g’ (r) and Hr(r,{) are continuous functions
in r, the derivative of Tg(r) with respect to r as given in (4.30) is continuous in 1'.
Thus T maps So into So.

To see when T is a contraction mapping consider

T91()-T92( =3] HIro Hégi(€))—F(€,g’2(€)))d€. (431)

Hence

a Re _
|T91(r)-Tg2(r)IS-] lH(r,€)l |F(€,gi(£))-F(€,9§(€))ld€. (4-32)

0‘ R.-

Since F (5 , g’ (£ )) is Lipshitz in g’ with Lipshitz constant 1 / 2, inequality (4.32) gives

‘ ’ d R0 ‘ 1 I I
ITgi(r)-ng(r)| .<_ 3 ] IHIr,oI 5 lam—92mm, (4.33)
Rf
which implies that
|T91(r)-Tg2(r)l< 55-min —g.I r)” )]R°z( €)|H(r€)|d€ (4.34)

80

Since 0 < P(r) < 1 It IS clear that |H(r,{)| S 1 for every r E [R2 R ] Therefore

 

 

 

 

 

 

 

inequality (4.34) provides
A R0
IT’91( )— Tg2I )I< :2— ]3z zI4) 44 ”91(7‘) — 92(1')||- (435)
Computing the integral in equation (4 35) gives
ITg2I ) Tg2I )I s 3 (R2 — a)? Il91( ) 92( )II (4.36)
Similarly it can be shown that
1 d - —
mlng91(T)"T92( )} <
1 a “0 -
3 ) —3] zI4) IH.Ir.4)I 44 ”91(7") g2( )II (4.37)
and by using (4.13) it is clear that
2 R2
I 2r,I 4)I< _ M22122) (4-38)
for every r E [R.-,Ro]. Thus
1 d — —
Tl 379919) — T92(T)}l S
1 52 R3 “0
33-3 2.3 (12% R2) ] zI4) 44 “91(7‘) —g2(r)ll (439)
Computing the integral in (4 39) gives
d - 1 d R§(Ro— R2-—)2
Idr{ g2I )— T22 r()<}l 3m 3 23022 R2r)ug2I )— 92(T)ll- (440)
After Simplification equation (4 40) becomes
“ 1 & R42)(Ro — R4)
3(1)—{T91(d)- Tg2Ir)}| 3 $5 3 ,3 ( 122+ R.- l|91( )— 92(r)ll-(4-41)
Note that
1 R3 — R?
22213322135} = R3 (4'42)
81

Consequently (4.28), (4.36), (4.41) and (4.42) together give

IITQIU‘) - T920)” S (4.43)
‘ 6 RO— - 2
max {3304, — 1).-)2. 3%} H.910") — g2(r)ll-
Hence, if
- a _ _ 2
M=maX{§93(R2—R2)2,3E°R1—B‘)—}<1 (4.44)

then T is a contraction mapping on So. This completes the proof of Lemma 4.2.2.
C]

The inequality (4.27) provides a restriction that can be interpreted in various ways.
For example if d > 0, R0 > R..- > O are given then (4.27) requires sufficiently large
positive (1. Alternatively if a > 0, d > O, R,- > 0 are given then (4.27) requires that
R0 is sufficiently close to R2. It can also be shown upon modifying the norm (4.28)
with a different positive weight function that it is possible to weaken the inequality
(4.27) thereby making the result less restrictive. However such modification does not
seem to give an existence and uniqueness result that applies to arbitrary a > 0,
d > 0, R0 > R.- > 0. As we now show, such a result is obtainable by bootstrapping
the result with a development that first obtains an existence and uniqueness result

for the second boundary value problem.

Theorem 4.2.3 The Second boundary value problem, (4.4), (4.15) has a unique so-
lution on [R,-, R0] for every R, > R,- > 0.

Proof: Recall that the second boundary value problem (4.4), (4.15) is equivalent to
initial value problem (4.19), (4.20). Let (r = 04/43 and d: d/éz. Then (4.19) is

(4.45)

 

82

where d follows from (4.17).
Let

2 _3 , 9’ _ i
H(r29(7‘)2g(7‘))— 9+ 43339,), 33- (446)

Then initial value problem (4.19), (4.20) is equivalent to following initial value prob-

lem
H (r2g(r)2g'(r)) = 02 g(Ri) = 4/22. (4-47)

Let U = (0,00) x (—oo,oo) x (—oo,oo). Clearly U is an open convex region
of the three dimensional Euclidean-space and H is defined and continuous on
U. Moreover 0H/6g and 8H/Bg’ exist and are continuous on U. Additionally
H(R.-,g(R2-),g’(R,-)) = 0 has a solution in U and the Jacobian J = BH/ag’ is not
equal to zero at this point. Hence by the existence and uniqueness theorem for implic-
itly defined initial value problems (for example [20] pages 44-47) there is an interval
containing R,- such that the initial value problem (4.47) has a unique solution on this
interval.

Let the maximal solution domain for g(r) be the interval (a, b). Clearly R,- E
(a, b). It will now be established that b 2 R, by showing that the alternative as-
sumption b < R, leads to a contradiction.

Note that (4.45) can be written as

2107 9’

 

 

: — —- —— ——-———-—— . 4.48
g 3 33 4+ (33),) ( )
Integrating (4.48) from R,- to r provides
1 ’ J 9’(8)
= - + -- - - (18. 4.49
g(r) g(R2) 3 ] (33 4+ (8938”,) ( )
Let R1 and R2 be any two numbers with a < R,- < R1 < R2 < b. Then
1 R” J g’(8)
(R —(R 2:] —— __ds 4.50
9 2) 9. 1) a 121 (33 4+ (sg’(s))2) ( l

83

Since

 

 

 

 

__ 9’0") _ 9’(T) i _1_ '
|,.2. «“3933”,! S |T3l+l\/4+(rg’(r))2| S R§+R2 (4.51)
equation (4.50) implies that
|g(R2) — g(R1)I < I3 + i)1IR2 — HI. (4162)
_ R? R.- _

Assume that b is finite. Since it > 0, values of g(r) for r near b and less than b
satisfy a Cauchy condition and thus there exists lim,._..b g(r). Call this limit 9;, and

let gncw(r) be a solution of the initial value problem
H(T, g(T), 9,0.» = 0) g(b) = gb' (453)

Since H is independent of g(r), it satisfies a global Lipschitz condition in 9. There-
fore the solution gnew(r) is an extension of g(r). Since the domain of this extension
contains b, the domain (a, b) is not maximal. This contradiction comes from the
assumption b < R0. Hence the maximal solution domain for g(r) can be extended to
any R, < 00. Of course this means that the second boundary value problem, (4.4),
(4.15) has a unique solution on [R, R0] for every R0 > R2- ) 0. This completes the
proof of Theorem 4.2.3. D

Note that an alternative proof of the above theorem is presented in Appendix B.

Similarly it can be shown that

Theorem 4.2.4 The Second boundary value problem, {4.4), (4.16) has a unique so-
lution on [R2, R0] for every R, > R,- > 0.

To get the main result of this section, the following standard result from the theory

of ordinary differential equations will be now utilized.

84

Theorem 4.2.5 Uniqueness and Continuability of Solutions of Initial Value Prob-

lems: Consider the following initial value problem

y”(t) + f (t2 y(t), v’(t)) = 02 (4-54)
900) = 3102 (4.55)
3/(150) = 316- (456)

If f (t,y(t),y' (t)) is continuous and satisfies a Lipschitz condition in a domain D
which contains the point (to,y0,y6) then there is only one solution of (4.54), (4.55),
(4.56). Furthermore, this solution can be uniquely continued arbitrarily close to the

boundary of D .

Proof: See [4], page 10. D

Theorem 4.2.6 The second order nonlinear boundary value problem (4.4), {2.17)

has a unique solution on [R2, R0] for every 0 < R.- < R, < 00.

Proof: According to Lemma 4.2.2 there is some finite length A such that all first
boundary value problems for (4.4) on any interval of length less than A has a unique
solution. Let n be an integer such that (R,- — Ro)/2" = A < A. Divide the interval
[R2, R0] into 2" equal subintervals. Then by Lemma 4.2.2 on each subinterval all
first boundary value problems for (4.4) have unique solutions. Now if it can be shown
that first boundary value problem for (4.4) has a unique solution on an interval whose
length is 2A then the same argument can be repeated for intervals whose length are
4A, 8A, etc. until 2" A = (R.- — R0). This in turn would establish that the first

boundary value problem (4.4) has a unique solution on the interval [R2, R0]. Without

85

loss of generality we may assume that n = 1. Let Rm be the midpoint of the interval

[Rb R2,]. Then by Theorem 4.2.3 the second boundary value problem (4.4) subject to

90%) = Ila, 9’01".) = m (4.57)

has a unique solution. Moreover by Theorem 4.2.4 the second boundary value problem

(4.4) subject to

9’(Rm) = m 90%) = 2122 (4.58)

also has a unique solution. To continue the proof of Theorem 4.2.6 we now consider
the following claim
Claim 4.2.7 The solution gl(r, m) of (4.4), (4.57) and its derivative g,’(r, m) =

agz/8r(r, m) are each strictly increasing in m. Namely, if 1712 > m1, then

91(r2m2) > 9le m1) (4.59)
on [122,122.22] and

91(73 m2) > 9103 ml) (4.60)
on (R2,le-

Proof of Claim 4.2.7: Suppose m2 > m1. Clearly (4.59) is true for those
r 6 (R2, Rm] and sufficiently close to Rm. If

9i(Rn2m2) = 95(Rmm1) (4-61)

for some R, E (R2, R22], then the second boundary value problem (4.4) subject to

g(Ri) = 1% g,(Rn) = 9l(Rn2m2) (4-62)

has a unique solution by Theorem 4.2.3. This implies that g,(r,m1) E g,(r,m2)
for every r E [R.-,R,2]. Moreover since f (r, g’ ) is Lipshitz, then every initial value

problem has a unique solution by the Theorem 4.2.5. Therefore g,(r, m1) E g,(r, mg)

86

for every r E [Rm Rm]. In other words, branching of the solution at r = Rn is not
possible. Hence gz(r,m1) E g,(r,m2) for every 1' E [R2,Rm]. But this implies that
m2 = gf(r,m2) = gf(r,m1) 2 m1 contrary to the assumption m2 > m1. Thus g,’ (r, m)
is a strictly increasing function in m. Moreover since g1(R2-,m1) = g)(R.-,m2) = 2]),-
and gf(r,m2) > g; (r, m1) on [R2, Rm], we have g,(r, m) is also an increasing function
in m for every r E (Ri, Rm]. This completes the proof of Claim 4.2.7. [I]

I The proof of Theorem 4.2.6 now continues. It similarly can be shown that the
solution g,(r, m) of (4.4), (4.58) and its derivative g;(r, m) are respectively decreasing
and increasing functions of m.

In particular g1(Rm,m) is an increasing function of m and g,.(Rm,m) is a de—
creasing function of m. Since g,(r, m) and g,(r, m) are monotone functions of m
without jump discontinuities, they are continuous in m on [R2,RNJ and [erRo]

respectively. Therefore every first boundary value problem (4.4) subject to

and every first boundary value problem (4.4) subject to

g(Rm) : 2Ibma g(Ro) = $0 (4.64)

have continuous solutions in m. To continue the proof of Theorem 4.2.6 we now
establish the following claim

Claim 4.2.8 If every first boundary value problem (4.4), (4.63) and every first
boundary problem (4.4), (4.64) have continuous solutions in m, then these solutions
both map the real line R onto itself.

Proof of Claim 4.2.8: Let g;(Rm,m) and g,(R,,,,m) be the solutions of first
boundary value problems (4.4), (4.63) and (4.4), (4.64) respectively. Consider the
map h : R —-> R defined by m —) gz(Rm,m). Let 1]) be any real number. To show

surjectivity we need to show that there exists m E R such that h(m) = gl(R2,2, m) =

87

w. Since the first boundary value problem (4.4) subject to g(Ri) = 112,-, g(Rm) = 1,!)
has a unique solution let m be the value of the derivative at Rm. For such an m it
is clear that h(m) = 91(Rm, m) = 1b. Therefore h(m) = g,(R,,,,m) is onto. Similarly
g,(Rm, m) is onto. This completes the proof of Claim 4.2.8. CI

The proof of Theorem 4.2.6 now continues. As a result of Claim 4.2.8, there must

be exactly one number mo for which

91(Rm, mo) = gr(Rm, mo)- (465)

By definition

91'(Rm, mo) = mo = g£(Rm, mo). (4.66)
For m = mo, the function g(r) defined by

g(r) = g(r’m) r E [R4, Rm] (4.67)
gr(7'2m) T E [RmaRol

has a continuous derivative at r = Rm. Hence it is the only solution of the nonlinear
boundary value problem (4.4), (2.17). This completes the proof of Theorem 4.2.6.

El

Theorem 4.2.6 establishes the existence of a unique classically smooth solution for
the pure azimuthal shear problem in the extended material model (a > 0, d = a" >
0) for all R4, R0, 1b., 1110. This contrasts with the lack of a classically smooth solution

in the precursor model (a = 0, d = (1* > 0) whenever
R2

I212. — WI > ‘I’max = arccos(fi’g). (4.68)

Recall that in this case we found that a portion of the deformation localized at
R = R.-. Thus the expectation for the extended theory is that a —+ 0 should in
some sense retrieve the localized deformation studied in Section 3.4, suggesting a
boundary layer type phenomena for sufficiently small a > 0 whenever we — 1M

becomes sufficiently large.

88

4.3 Numerical Solution for the Extended Model

In this section we present a numerical solution procedure for the pure azimuthal
shear problem in the extended material model. Note that for a fixed or > 0 the pure
azimuthal shear problem is governed by (4.4) and boundary conditions are still given

by (2.17). Note also that for a fixed (3: > O (4.4) can be written as

3 T3 9' J (4 69)
g 4 + (rg’)2
where
a = a/a, J = d/a. (4.70)

Differentiating (4.69) to eliminate J gives

 

g”(7‘) + f (7", 9’0"), 6) = 0, (471)
where
I - — g_l 27.2 9’2
f(r’g’a)- r [3+4+a[4+r2g’2]3/2l' (4'72)

For a given it the numerical solution of (4.71) subject to (2.17) can be found by
various “shooting”, “parallel shooting”, “collocation” or “finite-difference” methods
[16], [17]. Here we employ the finite—difference method to numerically solve (4.71)
subject to (2.17).

Let R,- = 7'0 < r1 < - -- < rn_1 < 7“,. = R0 be an equally spaced partition of the
interval [Rb R0]. The derivativas in (4.71) are replaced by suitably chosen difference
quotients defined in terms of the points 1“,. A typical choice is to use the centered

difierence approximations

 

 

9'01)“ g(rj+1)2';zg(Tj—1), (4.73)
9.0,], g g(r)“) - 29g» + 903-1). (4.74)

89

When these approximations are used in (4.71), we obtain a system of equations of

the form

g(r)-+1) -29(rj) +9634) = f(r,- g(r)-+1) _9(7‘j-1)).

 

 

4.
h2 2h ( 75)
for j =1,2,...,n — 1 where g(ro) = 11),, 903,) = 2,00.

In (4.75) the quantities g(rl), g(rg), . . . , g(rn_1) are unknown since any solution

g(r) of (4.71) will probably not satisfy (4.75) exactly. We think of the system (4.75)
as an approximation to the equation in (4.71) and hope that for a sufficiently small
spacing, h = (R0 — R;)/n, the solutions 90",) of (4.75) are good approximations to
the exact values of g(rj).

Since the function f (r, g’ , o?) is nonlinear in g’ , we are faced with solving a non-
linear system of n — 1 equations in n -— 1 unknowns. Newton’s method for several
variables can be used to determine g(rl),g(rz), . . . ,g(rn..1). Note that Newton’s

method applied to (4.75) takes the form
Gk+1 = Gk — J(Gk)—1P(Gk) (476)

where Gk = [g(k)(r1),g(k)(rg),...,g(")(rn-1)]T and where the 1"" component of the

vector function P(G) is given by

 

1).-(G) = g(r...) — 29m) + you-..) — I22 f(r,, 9""J+1)2';f("j-1)). (4.77)

Since BPj/ag(n) is zero except for Z: j — 1,j,j +1, the (n — 1) x (n — 1) Jacobian
Matrix J (Gk) is tridiagonal; and it is relatively easy to apply Newton’s method to
(4.75) even for large n. The starting vector Go could have components obtained by

a linear interpolation of the boundary conditions:

(0) .=I1_Z_Rl_' RO‘TJ'
g (’3) R.-R.“’°+R.-R.-

There are several programs to solve boundary value problems based on finite-

 

z/a. 133' s n - 1. (4.78)

difference method. MATLAB’s bvp4c is one such program and it is well explained in

90

 

 

 

 

 

 

 

 

 

0.5
0.4-
0.3-
E
‘5
0.2-
/ .’
0-1 ' ----- (dotted line) a=0.1
—- (dashed line) a=o.o1
— (solid line) $0001
0 x r

 

 

 

 

1 1.5 2 2.5 3 3.5 4
R

Figure 4.1: Numerical solution when R,- = 1, R0 = 4, (D,- = 0 and” 1120 = 0.452478 for
three values of 67, d =2 0.1, 0.01,0.001. Here [1/20 — 4),] = 03me

[25]. MATLAB’s bvp4c is utilized to solve (4.75) subject to (2.17) for several values of
5:. The specific problem that we consider involves R,- = 1, R0 = 4, 212,- = 0 whereupon
we consider the effect of increasing 1110 for various 67 > 0. The limit (31 —> 0 should
then retrieve the results of Chapter 3 wherein we recall that \Ilmam z 1.50826 for a
geometry with R0 = 4R,-.

In the first case we investigate the solution of (4.75) when R, = 1 , R0 = 4, «[2,- = 0
and 1P0 = 0.452478. Values it = 0.1, Er = 0.01 and (I = 0.001 are used in Figure 4.1
and corresponding graphs are depicted by dotted, dashed and solid lines respectively.
Note that 2/10 — w, = 0.452478 << \Ilma,x = 1.50826. In this case the three graphs are
almost identical.

In the second case we investigate the solution of (4.75) when R.- = 1, R0 = 4,
w,- = 0 and 2/20 = 1.5. Note that \IIm,x 2 1.50826 when R,- = 1 and R0 = 4. Hence
for this case 2/20 x \Ilmx. 57 = 0.1, (r = 0.01 and 67 = 0.001 are used in Figure 4.2 and

the corresponding graphs are depicted by dotted, dashed and solid lines respectively.

91

 

g(R)

 

 

: ----- (dotted line) 22:0.1
[’ — — (dashed line) a=0.01
— (solid line) a=0.001

1 1.5 2 2.5 3 3.5 4
R

 

 

 

 

 

 

Figure 4.2: Numerical solution when R,- = 1, R0 = 4, w,- = 0 and 7110 = 1.5 for three
values of a, a = 0.1, 0.01,0.001. Here [2,00 — 7,0,] = 0995me

As 6 tends to zero we observe that graph of the g(R) moves upward meaning that
smaller 67 gives rise to more angular displacement on the interior R,- < R < R0.

In the third case we investigate the solution of (4.75) when R,- = 1, R0 = 4,
21),- = 0 and we = 3. Values c1 = 0.1, d = 0.01 and (r = 0.001 are used in Figure 4.3
and corresponding graphs are depicted by dotted, dashed and solid lines respectively.
Note that 2120 — 1/2, = 3 > \Ilmax = 1.50826. Hence for this case 7,00 > ‘Ilmax. As
suggested by the previous material model, 67 = 0, the graph of g(R) converges to a
vertical line segment near 7‘ = R,- as Er -—> 0.

In Figure 4.4 the previous three graphs are superimposed. For 7,120 > \Ilmax, as
suggested by previous material model, (i = 0, the graph of g(R) converges to a

vertical line segment near 7‘ = R,- as 67 -—> 0.

92

 

 

 

 

 

0 5 3' ----- (dotted line) a=0.1 .
- ' — — (dashed line) G=0.01
— (solid line) a=0.001

1 .5 2 2.5 3 3.5 4
R

 

 

 

 

‘

Figure 4.3: Numerical solution when R,- = 1, R0 = 4, w, = 0 and 7,120 = 3 for three
values of Er, G- = 0.1,0.01,0.001. Here [1170 — 42,] = 1.989‘1/max

 

 

   

 

----- (dotted line) a=0.1
— — (dashed line) a=0.01
— (solid line) a=0.001

 

 

 

 

 

 

 

Figure 4.4: Numerical solution when R, = 1, R0 = 4 and 212,- : 0 and 1110 = 3
for three values of (i, 6: = 0.1,0.01,0.001 and three values of 1120, $0 = 0.3\Ilmax,
0.995\Ilm, 1.989‘I’max

93

4.4 Regular Perturbation Solution for the Ex-

tended Model

In this section it will be shown for small 57 > O that a regular perturbation can be

used to get an approximate solution of (4.4) subject to (2.17) for the case [2110 —- 1b,] <

\Ilmax. If [1,00 — 70,-] > \IJmax the technique presented here does not apply. Further if

[1170 — 4),] = (1 — e)\II.m,x with 0 < 6 << 1, i.e [1110 — 2b,] close to but less than \Ilmax,

then severe requirements are placed on the smallness of Cr in order to obtain a useful

regular perturbation solution approximation near R = R4.

Let

and

Then equation (4.69) becomes

FWMfl+mMfl+~J

 

6773(gf,(r) +5zg’1(r) +...) +

 

\/4+T2(96(7‘) +c'i9’1(7‘)+---)2

By expanding in 6: equation (4.81) provides

 

 

T3 96 3 r I 491
+ a r + { , }] + O a?
4 + r2(g(’,)2 [9° (4 + r2607)” ( )

=4+a$+omw

The 0(1) equation is

r396 _ -
4 + 73(96)? 0'

 

94

mm)

MW)

(4.81)

aw)

aw)

Note that by renaming the function and constant in the equation (4.83) it can be

seen that (4.83) is identical with (3.113). Hence the solution is

4__ ’2
7‘ d0

J02

 

go(r) = arctan + 60. (4.84)

Since attention is restricted to [1/20 — 112,] < \I/nm,x = arccos(R? R3) , it follows that do
and 50 can be computed uniquely from the boundary conditions (2.17) as in Section

3.4. Indeed do and 50 are given by (viz. (3.118) and (3.119))

 

 

 

— R?R2 sin ‘11
d = . ° , 4.85
0 «R? + Rf, -- 2R3R3 cos ‘11 ( )
R4 — d2
50 = 21;,- — arctan 1 dz 0 . (4.86)

0

At this point an important observation is that if [we — 4),] > \I/mM then there is
no 90(7') for which boundary conditions (2.17) are met. Hence the technique under

consideration does not apply. Moreover from equation (4.84)

2 do
I 7. = _ _ , 4.87
90( ) 1" mg ( )
and using (4.85) in equation (4.87) gives
, 2R3, sin‘II
90(Ri) = (4-88)

 

R,(R§ cos \IJ — RE)’
where ‘11 = [(00 — 112,]. Therefore 96(R4') —> 00 as [2120 -—¢,] —> ‘Ilmax. Hence for fixed (r
if We — 71),] = (1 — e)\IIImam with 0 < 6 << 1 then the first term, 5173 g’ (r) , in equation
(4.69) becomes large as r —-> 17,-. Thus the smallness of a for a useful approximation
must correlate with the closeness of [1110 — $1] to \Ilmax, as discussed at the end of this
section.

Deferring these issues for the present, the 0(a) equation associated with (4.82) is

47.39!

3 I 1

T90+ 2 123/2
[4+r(go)l

 

95

which can be solved for g]

, _ (J, - 7‘3 96) l4 + 7‘2(94’1)2l3/2

 

 

 

_ 4.
91 4 T3 ( 90)
Substituting from (4.87) into (4.90) gives after some simplification
J1 T3 2 (1.05 T'
, = 2 [_.___ _ ____] 4.91
91 [7'4 _ ng/Q [7'4 —d2]2 ( )
Therefore
1r2 d ,/ 4 — 2 —J 2 ,
91(7‘) = - 108 2+ ———-—°]— d1 T ”—2 or +51. (4.92)
2 —d0 7'4 —
By using boundary conditions
91(R’i) = 0? 91(R0) = 07 (493)

constants (11 and 51 can be determined uniquely:

 

 

 

 

 

_ R3 + Jo 133+ 30 ‘10 R3 J" R?
dl— (0.5 log RQ—do —0.5 log 123—620] 113—113 _R4_ :202)
1 1
__T_____=______ , 4.94
/(,/Rg—d3 «Rf-d3) ( )
and
_ — 2 2
51: d1 _ d017,, _110g R +d_o] (4.95)

\/Rg-d§ Rg-dd ng do
Consequently the perturbation expansion solution of (4.4) subject to (2.17) with

We — 111.] < \Ilmax is given by
g(r) = 90(1) + 5491(1) + 0(642) (4.96)

where 90(7), 91(7‘) are given by respectively (4.84) and (4.92). This procedure can
be continued to any order Er" by further expansion of (4.82) to obtain a first order
ODE for 9,, containing dn. Each such ODE for n > 1 is to be solved subject to

gn(R,) = 9,,(Ro) = 0. Thus truncation of the perturbation expansion at order n

96

 

0.16 . . . . -
0.14- /
0.12- -
0.1 . 1

A / "I
030.08- /
0.06 - /

0.04 -

 

 

 

 

 

—- — (dashed line) one-berm
0.02 ~ ----- (dotted line) two-term
— (solid line) numeric

1 1.5 2 2.5 3 3.5 4
R

 

 

 

 

 

 

Figure 4.5: One-term, two-term perturbation solutions and numeric solution with
w,- = 0 and 1P0 = 0.150826 when 64 = 0.1. Here R,- = 1, R0 = 4 and \Ilm =1.50826
so that 11),, = 0.1\I'max

results in an expansion that exactly satisfies the boundary conditions (2.17), but only
solves the ODE (4.4) to 0(51").

Figure 4.5 shows the graph of one-term, two-term perturbation solutions and nu-
meric solution with w,- = 0 and 1110 = 0.150826 2 0.1\II,,W when 61 = 0.1, R,- = 1,
R0 = 4. These three graphs look like almost identical. However if any part of
the graph is enlarged then it becomes apparent that two-term perturbation solution,
shown by the dotted line, is better than one-term perturbation solution, shown by
the dashed line, as expected from general perturbation theory.

Figure 4.6 shows the graph of the one-term, the two-term perturbation solutions,
and the numeric solution with 1b,- = 0 and do = 1.3 = 0.862‘1’max when 6 = 0.01,
R,- = 1, R0 = 4. These three graphs again look like almost identical. However
if any part of the graph is enlarged then it again becomes apparent that two-term
perturbation solution, shown by the dotted line, is better than one-term perturbation

solution, shown by the dashed line, as expected from general perturbation theory.

97

 

 
    

 

/ /
l A;

— —— (dashed line) one-term
0 - ----- (dotted line) two-term
-— (solid line) numeric

1 1.5 2 2.5 3 3.5 4
R

 

 

 

 

 

 

 

 

 

 

Figure 4.6: One—term, two-term perturbation solutions and numeric solution with
$4: 0 and 11),, =1.3 when 6: = 0.01. Here R,- =1, R0 = 4 and \II,,m =1.50826 so
that 1110 = 0-862\I”ma.:1:

As mentioned earlier the smallness of 67 for a useful approximation is expected to
correlate with the closeness of [190 — 11),] to \Ilmx. Numerical investigation of (4.96)
gives useful insight about this correlation. In particular we note that the correction
gl(R) involves 91 ’(R) < 0. Thus the two term expansion for g’(R) will involve
g’ (R) < 0 at some R if 54 is sufficiently large. This breakdown first occurs at
R = R.-. For fixed R5, R0, 1/2, and 57 there is a critical value of 1&0, say \Ilcm, such

that if 1120 2 \Ilcrit then
90,014) + €791,014) S 0, (4.97)

which is contrary to the physical expectation that g(R,-) is an increasing function
of 1%. Hence (4.97) provides an analytic expression to relate smallness of 64 with
1,00. Indeed by solving (4.97) numerically it has been found that \IJm, = 1.0887 2
0.722‘11max when R,- = 1, R0 = 4, 11’),- = 0, d = 0.1 and that ‘I’Cm = 1.301 =
0.862‘1’max when R,=1, R0 = 4, w,- = 0, d = 0.01.

98

 

0.8 -

0.6 ' ' i

g(R)

0.4 -

 

02* — — (dashed line) one-term ~

----- (dotted line) two-term

— (solid line) numeric

1 1.5 2 2.5 3 3.5 4
R

 

 

 

 

 

 

Figure 4.7: One-term, two-term perturbation solutions and numeric solution with
w,- = 0 and we = 0.97 when 64 = 0.1. Here R,- =1, Re = 4 and \Ilem =1.50826 so
that we = 0.643\Ilmee

Since \Ilem = 1.0887 when R, = 1, Re 2 4, w,- = 0 and 57 = 0.1 the values
of we = 0.97 and we = 1.3 are selected in the following graphs to emphasize the
change in the nature of perturbation expansion. It can be seen from Figure 4.7 that
the two-term perturbation solution, shown by the dotted line, is better than the
one-term perturbation solution, shown by the dashed line, as expected from general
perturbation theory. However as we increases through \I’em the value of the derivative
g’(R,-) in the two term expansion becomes negative. For we = 1.3 this gives a graph
for the two term expansion involving initially decreasing g(R). (see Figure 4.8). As
a result the two—term , perturbation solution becomes a worse approximation to the
numeric solution (shown by the solid line) than the one-term perturbation solution.
On the other hand recall from Figure 4.6 that two—term perturbation solution is better

than one-term perturbation solution when (r = 0.01 and we 2 1.3.

99

1.5

 

 

 

 

 

 

 

'. — — (dashed line) one-term
i: ----- (dotted line) two-berm
— (solid line) numeric

 

 

 

 

 

 

1.5 2 2.5 3 3.5 4
R

Figure 4.8: One-term, two-term perturbation solutions and numeric solution with
w.- = 0 and we = 1.3 when it = 0.1. Here R,- =1, Re = 4 and Warm = 1.50826 so
that we = 0.862\Ilmex. Since we > \Ilem, the two term perturbation solution involves
decrease of g(R) when R = R,-

4.5 Boundary Layer Type Solution for the Ex-
tended Model

In this section the case [we — w,] > \Ilmex will be examined by means of perturbation
theory for small 64. This has the added benefit of clarifying the proper treatment of
the case [we — w,] = (1 — e)\IImam with 0 < 6 << 1. As discussed at the end of previous
section if [we — w,] > ‘I’max then there is no go(r) that satisfies the given boundary
conditions so a regular perturbation treatment is not possible. Moreover if [we — w,] =
(1 — e)\IIe,ax with 0 < 6 << 1 then the standard perturbation procedure does not yield
a useful approximation if Er is not sufficiently small. To get a useful approximation
in such cases we will proceed as follows: From (4.69) and (4.88) it follows that there
is a region near 7' = R,- in which g(r) changes rapidly thus motivating a boundary

layer type analysis. Indeed from (4.69) one can see that if g’(r) is of order 1/5

100

then the first term in (4.69) cannot be neglected while computing the first term of
the perturbation expansion. i.e., there is a boundary layer at r = R,. A boundary
layer correction (inner solution) at r = R,- will be constructed by using the stretching

transformation

 

(4.98)

This inner solution will be denoted ((7). Note that away from 7‘ = R._-, the first
term in (4.69) is small, whereupon a standard perturbation can be used to get an
approximate form of the outer solution. This outer solution will be denoted C(r).
However there will be only one boundary condition, namely g(Re) = we, for the outer
solution to meet. The additional integration constant generated by the outer solution

will be determined by patching the outer solution with the inner solution so that
G000 = (00") (4-99)
and

go '(7‘) = (60"), (4-100)

for some 7‘ E [R4, Re] where 00(7) is the first term of the outer solution and (g(r)
is the first term of the inner solution. Indeed it will be shown that if 64 and d are
suitably restricted, then f can be chosen uniquely. The geometric meaning of (4.99)
and (4.100) is that the one—term outer and one-term inner solutions can be pieced up
together smoothly; i.e., they have a common tangent at the point of contact. The
terminology “patching” is used to indicate that the inner and outer solutions are
reconciled at a single point, as opposed to some form of matching over a region [7],
[8], [13]-

We remark that the phenomena of interest are not associated with a singular
perturbation in the usual sense. Namely, the small parameter 54 does not annihilate

higher order derivatives of the governing equations in the limit 64 —’ 0.

101

If more terms are used in the outer and inner solutions, then patching can proceed
as follows. For each new term that is added to the inner and outer solutions there
will be one new integration constant. Now 7" and the constants (including the new
one) can be chosen so that derivatives of the inner and outer solutions match up to
one more order at f . For example if a two—term inner and outer expansions are used

we require that

(Co + 01W) = (C0 + (1)0) (4-101)
and

(90' + 91’)(7‘) = (<6 + (0(7‘) (4-102)
and

(90 ” + 91")(5) = ( 6' + (1')”), (4-103)

where G1 and C1 are the second terms in the outer and inner solutions respectively.
Note that f and all constants will be computed independently from any previous
determination involving less terms in the expansions. In this sense this method is not
iterative for computing patching constants. Technically, this process can be imple-
mented for any order.

The procedure is illustrated with the two term form of the outer solution
C(r) = 00(7) + 6 G1(7‘) + C(62). (4.104)
As usual we impose that
000%) = 1.00, (4.105)
and

0,02.) = 0, j 2 1. (4.106)

102

The outer solution C(r) must satisfy (4.69). By using the expansions (4.80) (for d)
and (4.104) in (4.69), the 0(1) equation is given by

 

 

 

3 I
T 00 = d‘e. (4.107)

4/4 + (70;)2

Thus,
r4 — 1,2 —
Go(r) = arctan 82 + Co, (4.108)
0

where

_ R4 _ '2

Co = we — arctan 0&2 0. (4.109)

Hence it is found that the first term of the outer solution is given by (4108) up to
constant do which will be computed by patching.

Note for a given do that the maximal domain of the one—term outer solution, Go(r),
is [\/d?o,oo). Moreover the derivative of Go(r) at r = \/d?o is infinite. Therefore as
do varies by virtue of a change in boundary conditions, the domain of the first term
of the outer solution also changes. Indeed by using (4.108) and (4.109) the leftmost
endpoints of the one-term outer solutions has the following coordinates as do varies

between R? and R3,:

(7‘, g) = (\fdro we — arctan W ). (4.110)

The graphs of the one-term outer solutions are depicted in Figure 4.9 for trial values

do = 1, 2.25, 4, 6.25, 9, 12.25. Note that as the trial value do increases the graph of the

one-term outer solutions move to the right. The dashed line gives (4.110) at Wthh
point each one-term outer solution has a vertical tangent.

The second term of the outer solution is now computed as follows. By using the

expansions (4.80) and (4.104) in (4.69), the 0(64) equation is given by

40', _ 4?,

G" + [4 + r2(G;,)2) 3/‘2 ‘ 5’

 

(4.111)

103

 

 

 

 

2.2 """"" |" """" r """"" r --------------- r ....... ..1
00 2r -------------------------- -/ ................... ..
18 -------------------- 7 -:- ----------------- * ------- *'
/'/ '
1.6 ---------- ,—-< .................................. _
F/
1.4 ‘ 1 ‘ 1 .
1 15 2.5 3 35 4
I’

Figure 4.9: One-term outer solutions with R,- = 1, Re 2 4, w,- = 0 and we = 3 for
the following six trial values do, do = l,2.25,4,6.25,9,12.25. Here do = 1 is on the
far left and do = 12.25 is on the far right. The curves order themselves sequentially
in do. The envelope of vertical tangency is shown as a dashed line

which can be solved for G"1

, z (41 — Ga) [4 + ”(C-‘02P”
1 4 r3 '

 

(4.112)

Computing G], from (4.108) and then substituting into (4.112) gives after some sim-

 

plification
d-1 7'3 2 (IO 1'5
G' = 2 [_____ _ _____—]. 4113‘
1 [r4 — 4413/2 94 — 4812 ( ’
Therefore
2 - - ./T 275.4; 2 -
(11(7) :1 1044]” +80 _ d1 7' 0 0’ +01, (4.114)

2 r2—do r4—dg

where Cl is an integration constant and is determined on the basis of boundary

condition (4.106) yielding

(4.115)

Note that (4.115) is the same as (4.95). Hence (4.114) is also same as (4.92). Hence

two-term outer solution is given by

7‘4 al2

C(r) = arctan J? 0 + 00 (4.116)
0

 

1 2 d
+é[-logl:—_—0

JI,/—_'r4_dg_aor2
2 r2—do

r4—dg +Cll’

 

where 00 and Cl are given by (4.109) and (4.115) respectively.

The boundary layer equation (inner equation) can be found by substituting (4.98)
in (4.69)

 

 

 

 

((5) + \/4 + (R1. +2 8):? (<’<2>)2 ___ W (4.117)
Let

((2) = (0(2) + 2 (1(7) + 0(22) (4118)

be the form of the inner solution. Since d is a constant, the same value must apply

to both inner and outer expansion. Thus d is again given by expansion (4.80). For

the first term of the inner solution we set 51 = 0. The resulting equation is

, 1 J
(0(8) + E = 1%., (4.119)
Thus,
d— 1
(0(8) = (5‘;- — R7) 8 + 214-, (4.120)

where the boundary condition g(R,) = (0(0) = 1,1),- is used to determine the integration
constant.

Note that by using (4.98) first term of the inner solution is given by

Co“) = (1% “ i) (r 2310+ 1121.

 

(4.121)

105

The second term of the inner solution is computed as follows. By using the expansions

(4.80) and (4.118) in (4.117), the 0(a) equation is given by

, d1 {id-03 s
Q®=§—7?+E' (um
Therefore
d— d— 2 2 _
(1(8)— 18 3 03 5 +01. (4.123)

— 71: ‘ TR? + fi—
By using the boundary condition (1(0) = 0 it is found that CI = 0. Again by using

(4.98) the second term of the inner solution can be written as

C1(r) = (éiRg — 3%) (3:52—31)2 +21%, (T LR‘). (4.124)

 

Consequently the two-term inner solution is given by

((2) = w. + (1% - i) (7" La) (4.125)

 

 

_ 2 _
The patching procedure will be demonstrated for the one term case where ((r) is
given by (0(1") in (4.121) and G(r) is given by Go(r) in (4.108) with Co given by
(4.109). At issue then is the determination of the single constant do which appears
in both (4.108) and (4.121). Indeed it will be shown that if d is sufficiently (small

then do is uniquely determined by patching the one-term inner and outer solutions

so that (4.99) and (4.100) hold for some 1" E [Rh R0]. To begin for given R47, R0, 1,0,-

 

 

and 1,00, let
a — 3 1 — —-1— (4 126)
"m R? m mR1’ '
and
R3
d” = m 2 (4.127)

106

Where m = (100 - W/(Ro - R1)-

amaz given by (4.126) is a consequence of the requirement that G6(Re) = m which
in turn provides a natural upper bound for a values. Note for an arbitrary we — w,-
that amaa: given by (4.126) may not be positive. However if |we — 1M > \Ilmx or
|we — wil = (1 — e)\Ilmex with 0 < 6 << 1, then amex > O. The above restriction
on we — w.- once used in (4.127) also implies that R? g (1*. Finally it is easy to see
from (4.127) that d" 3 R3. Therefore R? g d“ 3 R2 if |we — 1(le > (1 — e)\IImax for
sufficiently small 6 > 0.

The following claim establishes that do can be chosen uniquely so that one-term
inner and outer solutions can be patched on the basis of (4.99), (4.100).

Claim 4.5.1 If 0 < a _<_ amaz) then there exists a unique do 6 [R2, d‘] and a
unique 1" E [R,-,Re] such that Go(7") = co(f) and go’('F) = (60‘) for this unique
1* e [R1, Re].

Proof of Claim 4.5.1: This claim will be proven in two steps.

Step 1: For each do 6 [R2,d‘] there is a unique 7" E [\/d—_o,Re] such that 06(1‘) =
Th, where in is the slope of the line passing through (Roi/21') and (7", Go(f)).

Let do 6 [R2, d*] and let S : [\/d?o, Re] -—2 R“ be the function defined by

5(2) = G60“) (7‘ - R1) - 00(2) + 2124-, (4-128)

where the dependence of Go(r) on do follows from (4. 108), (4.109). Equivalently

 

 

Jor—( R1) . .
S r): -——-——— we — ,- 4.129
< . T 2W —( w) ( >
— arctan r4 — .3 + arctan R3 _ .0
63 402 '

Clearly S (r) is a continuous function for r E [Vt-1:, Re]. Since Go(\/dTo) —- w. and

\/d-_—o — R,- are finite and lim er—o + go ’(7‘) = + 00, it follows that

lim S(=) +00. (4.130)

do

107

Also note that do S (1* gives

 

lim S(r) = 230 (R0 — R2) —
r—vRa R0‘/R3—dg

Hence by the intermediate value theorem there exists 7" 6 [V do, Re] such that S (7‘) =

(2120 — 212.) s 0. (4.131)

O. This proves that there is a line passing through (R), wi); that is, tangent to Go (7')
at (7‘, 00(7)) .
Note that

_—2do 3r4—d02
"' r2 (7.4_J02)3/2

5'0)

 

(r —- R.) < O. (4.132)

Hence S (r) is a strictly decreasing function. Thus 2? must be unique.

Step 2: Let 0 < d g (1*. Then there exists a do 6 [R2, d‘] such that the one—term
inner solution, (o('r) , coincides with the tangent line whose existence is guaranteed
by step one. Moreover such a do is unique.

We introduce two families of rays, each family of which is centered at (124,102)-
The first family is the one—term inner solutions (4.120) where S is given by (4.98).

The slope Bw/BR of any member of this family is

m1(do) = agfla — i) (4.133)

This family is parameterized by do.
The second family consists of the rays passing through (Rafi/'11) that osculate with
the one-term outer solution at r = 7“ E [V do, Re] (the existence of such a family is

shown in step 1). This family is also parameterized by do. The slope Bw/BR of any

member of this family is

 

 

 

m2(0lo) = if]: + (4.134)
1 -4 _ ‘2
(arctan _ 0 — arctan ° 0 ),
M2.- 4.2 .2

108

where r = r is point of contact.

Let U : [R2, d‘] ——> R“ be the function defined by

U(d0) = "MW—0) — m2(d_ol- (4-135)

Clearly U (do) is a continuous function on [R2, d*]. Since m1(R,2) = 0 and m2(R,2) =

00 it follows that

U(R,2) < 0. (4.136)
Conversely
. 1 d" 1
U(d ) _ E(F€§ —- E) — (4.137)

1 , F4 - (d*)2 [133— (d*)2
f _ Ri (we — wi + arctan W — arctan W).

Since 1" ——> Re as do ——> d*, (4.137) gives

 

 

It _ 1 d2 1 11/)O _ 1&1:
U(d)—E(-fi§-E)—Ro_&, (4.138)
or equivalently
* l d* 1

where the inequality is a consequence of the restriction 0 < 67 3 am. Hence again
by the intermediate value theorem there exists a do 6 [R2,d‘] for which U (do) = 0,
i.e. m1(do) = m2(do). This proves that the one-term inner solution (o(r) and the
one-term outer solution Go(r) are tangent to each other at (f, Go(f)) for this choice

of do. Moreover since

 

U’(do) = (4.140)

ah? _ (‘77:173) (fi‘fl) >0.

109

 

 

 

 

 

 

 

 

 

 

 

 

 

 

L we [ a | f do
1.5 0.1 1.220537 1.259226
1.5 0.01 1.055384 1.057055
1.5 0.001 1.012634 1.012155
3 0.1 1.289941 1.613179
3 0.01 1.089118 1.180059
3 0.001 1.027634 1.055379
15.0826 0.1 1.713057 2.929204
15.0826 0.01 1.246533 1.553191
15.0826 0.001 1.080849 1.168164

 

 

 

 

 

 

 

Table 4.1: Numerically computed values of 7" and do for one term patching for several
we and a values when R,- =1, Re=4 and 1111‘ =0

such a do must be unique. This completes the proof of Claim 4.5.1 demonstrating
that do can be chosen uniquely so that one-term inner and outer solutions can be
patched on the basis of (4.99), (4.100). D

The Table 4.1 shows the values of r' and do for one term patching for several we
and a values when R,- =1, Re = 4 and w, = 0.

Figure 4.10 shows the one-term patched solution (dotted line), the one—term regu-
lar perturbation solution (dashed line) and the numerical solution (solid line) corre-
sponding to the values R,- = 1, Re = 4, w.- = O, we = 1.5 and 6: = 0.1. Note that
\Ilmax z 1.50826; hence [we — 1M = 0.995\Ilmex. This suggests when |we -— w,| z \I'max
that this new construction gives a satisfactory result even with patching only one-term
inner and outer solutions.

The patching procedure will be demonstrated for the two term case where C (r) is
given by (4.125) and 0(7) is given by (4.116). At issue then is the determination of
the constants do and d1, which appear in both (4.116) and (4.125), and the patching
location 7" E [R,-, Re]. The constants do, d1 and 7" are computed numerically so that

(4.101), (4.102) and (4.103) hold.

110

 

 

 

 

 

 

 

 

 

— (solid line) numeric
— — (dashed line) one-term poMbalion ~
----- (dotted line) one-term patched

 

 

 

 

1.5

2 2.5

R

3

3.5 4

Figure 4.10: Oneterm patched, one-term regular perturbation and numeric solutions
when R,- =1, Re = 4, wi = 0, we =1.5 < \Ilmex % 1.50826 and (I = 0.1. The

patching point is at r = 1.220537

 

 

 

 

 

 

 

 

 

 

 

 

[ we 1 a 72 do | all ]
1.5 0.1 1.114935 0.769605 6.607128
1.5 0.01 1.007296 0.932538 16.428066
1.5 0.001 1.000533 0.984481 35.746076
3 0.1 1.233646 1.202777 8.082971
3 0.01 1.081715 1.094041 18.477852
3 0.001 1.029278 1.042663 39.633449

15.0826 0.1 1.586166 2.056238 26.629071
15.0826 0.01 1.245361 1.442835 50.379749
15.0826 0.001 1.093927 1.173599 87.871407

 

 

 

 

 

 

 

 

Table 4.2: Numerically computed values of 1", do and d1 for two term patching for
several we and E! values when R,- = 1, Re = 4 and w.- = 0

111

 

 

1.5

 

g(R)

 

 

 

0.5 - ' 1 J

 

 

— (solid line) numeric
----- (dotted line) one-term patched
— — (dashed line) two-term patched

1 1 .5 2 2.5 3 3.5 4
R

 

 

 

 

 

 

Figure 4.11: One-term patched, two-term patched and numeric solutions when R,- =
1, Re = 4, w,- = O, we = 1.5 < \Ilm z 1.50826 and ('1 = 0.1 The patching point
is at r = 1.220537 for the one-term patched solution and the patching point is at
r = 1.114934699 for the two-term patched solution

Table 4.2 shows the values of 1‘, do and d1 for two term patching for several we
and d values when R,- =1, Re = 4 and 1121:: 0.

Figure 4.11 shows the one-term patched solution (dotted line), the two-term
patched solution (dashed line), and the numeric solution (solid line) corresponding to
the values R.- =1, Re = 4, 1111: O, we =1.5 and ii = 0.1. It is clear from Figure
4.11 that two-term patched solution is better than the one-term patched solution.

Figure 4.12 shows the one-term patched solution (dotted line), the two-term
patched solution (dashed line), and the numeric solution (solid line) correspond-
ing to the values R,- = 1, Re = 4, w,- = 0, we = 3 and (i = 0.1. Note that
we = 3 > \IImax z 1.50826 and hence regular perturbation cannot be used. However
two-term patched solution (dotted line) gives an excellent approximation to numerical
solution (solid line). The graphs of three solutions are enlarged around the patching

point for the one term patched solution.

112

 

 

 

 

 

 

 

0 5 _ — (solid line) numeric
' ' ----- (dotted line) one-term patched l
— -— (dashed line) two-term patched

1 1 .5 2 2.5 3 3.5 4
R

 

 

 

 

 

 

Figure 4.12: Two—term patched and numeric solution when R.- = 1, Re = 4, w. = 0,
we = 3 > \Ilmee z 1.50826 and (1 = 0.1. The patching point is at r = 1.2336459

16 r . . . r

. V '1
12» l ._ 2
,/

1o- :

 

 

 

 

 

 

 

g(R)

ONAOQ

 

 

 

 

 

— (solid line) numeric
----- (dotted line) one-term patched
— — (dashed line) two-term patched

1 1 .5 2 2.5 3 3.5 4
R

 

 

 

 

 

 

 

Figure 4.13: One—term patched, two-term patched and numeric solution when R,- = 1,
Re 2 4, w,- = O, we = 10\I1max and a = 0.01. The patching point is at r = 1.246533
for the one-term patched solution and the patching point is at r = 1.245361 for the
two-term patched solution

113

f‘w‘r

We close this section by depicting the graph of the one-term patched solution
(dotted line), the two—term patched solution (dashed line), and the numeric solution
(solid line) corresponding to the values R,- = 1, Re = 4, w, = 0, we = 10111mam and
6: = 0.01. The two-term patched solution (dashed line) gives an excellent approxi-
mation to the numerical solution (solid line). The graphs of all three solutions are

enlarged around the patching points.

4.6 Twist, Energy and Torque in the Extended

Model

The relation between energy and twist in the extended model can be obtained by
using the numerical solution discussed in Section 4.3. Recall that the strain-energy
density in the extended material model is given by (4.1). Hence when (37 = 07* the

potential energy functional (3.4) becomes

R0 A
E = nh/ [0(11 — 3) + 61(1f + 11— 6)] r dr. (4.141)
R,-
By using 11 = 3 + (rg’)2 and If 2 i1 = 4+ (rg’)2 + 1 it follows that (4.141)
becomes
Re
E = 71h / [a (rg')2 + 3(2 4 + (rg')2 — 4)] r dr. (4.142)
R.-

Letting a = 01/61, it follows from (4.142) that
He ______
E = nhd/ [51(7‘9')2 + (2\/4 + (rg’)2 -— 4)] r dr. (4.143)
R,

Divide the interval [R4, Re] into N equal subintervals and label the end points as
R,- = ro,r1, . . . ,rN = Re. For given 1/14, we and 64 the numerical solution presented
in Section 4.3 provides the values of g, E g(rj). Once 93-, j = 0, . . . , N are known the

centered difference approximation (4.71) can be used to estimate 93- _=-: g’(rj). Then

114

 

0|

A

Normalized Energy
1» u

 

 

 

 

Amount of Twist

Figure 4.14: Normalized energy E/(27rhd) versus twist when R, = 1, Re = 4 for
the four values of 67, 5: = 0, 0.001, 0.01, 0.1 are depicted as the curves E1, E2, E3, E4
respectively

(4.143) can be integrated numerically by using the standard composite Simpson’s
rule. The normalized energy (energy /27rhd) is computed by the outlined method for
several values of we and (1 when R; = 1, Re = 4 and w,- = 0. The normalized
energy versus twist graph is depicted in the following figure for the values of 51 =
0.1, 0.01, 0.001. Note that since 51 = 0 corresponds to the material model discussed
earlier in Part I it is possible to sketch the normalized energy versus twist and then
compare with the extended material model. The energy versus twist graph is labeled
as E1, E2, E3, E4 for 51 = O, 0.001, 0.01, 0.1 respectively. It is clear from Figure 4.14
that the stOred energr is an increasing function of 52 for a fixed amount of twist.
This is to be expected since increasing 6: at fixed 61 represents an increase in energy
penalization which would, in general, stiffen the material response.

Similarly the relation between torque and twist in the extended material model

can be obtained as follows. Torque M acting on the outer surface is given by virtue

115

 

 

 

 

 

 

2.5» ............. ............ .5 ............
5 5 5 M4
. 2 ------------ 4
3 I I 1
E = - =
1315' "M --------- a
Q E = = 1.112
E 11 ------------ :-- —r—#—T— fl
2° ; : M1
0.5 L' """ é ---------- A ------------------------- ..l
0 1 1 1
0 1 2 3 4
AmountofTwist

Figure 4.15: Normalized torque M / (27rhd) versus twist when R,- = 1, Re = 4 for the
four values of 64, d = 0,0.001,0.01, 0.1 are depicted as the curves M1,M2,M3,M4
respectively

of (2.42), (4.2), (4.4) as

 

 

 

 

271'
M = h / 1523094120) (19 = (4.144)
0
2n 2‘ I
h R3 [5 Re g’(Re) + 2R” (12") ]
O \/4 + (R09’(Ro))2
Equivalently
3 I
M = 2mm R3 g’(Re) + R29 (R2) 1. (4.145)
\/ 4 + (Rog’(Ro))"’

Since the torque acting on the outer surface R = Re depends on 9’ (Re), the value
of g’ (Re) must be estimated numerically. By using the numerical solution given in
Section 4.3 and one sided centered difference the value of 9’ (Re) is estimated for fixed
R=R,-, R=Re, Er, w,=0, and we.

The normalized torque (torque / 27rhci) versus twist graph is depicted in Figure

4.15 for the values of ii = 0.1, 0.01, 0.001. Note that since 51 = 0 corresponds to the

116

earlier material model it is possible to sketch normalized torque versus twist and then
compare with the extended material model. The torque versus twist graph is labeled
as M1, M2, M3, M4 for 54 = 0, 0.001, 0.01, 0.1 respectively. It is clear from the Figure

4.15 that torque is also an increasing function of 64 for a fixed amount of twist.

117

CHAPTER 5

Conclusions and Discussions

In this concluding chapter the main work of this thesis is discussed and summarized.

A preliminary discussion from the well known conventional theory of hyperelastic-
ity is presented at the outset. Then the solution of the pure azimuthal shear problem
in this context is presented in order to provide context for this thesis.

In Chapter 3 the pure azimuthal shear problem is formulated in the new framework
that is proposed by Pence and Tsai. This theory contains a material parameter
k = 64/04“ where d can be viewed as a macroscopic stiffness and 04"“ can be viewed
as a substructural stiffness. It is confirmed that if k —+ 0, then the pure azimuthal
shear problem in the new framework reduces to the standard pure azimuthal shear
problem for a neo-Hookean material. Since the governing field equations are too
complicated for a general k, the pure azimuthal shear problem is studied for small I:
by perturbation method and a numerical solution is presented for any k. The exact
solution is also presented when k = 1 and a necessary restriction [\III _<_ \llme,x for
a smooth solution is determined. A variational formulation of the problem is also
presented and direct methods of the calculus of variations are used to investigate
minimum energy configurations for arbitrarily large twist \II when k = 1.

In Chapter 4 a more generalized material model is considered. It is characterized

by an additional stiffness parameter oz. The first order nonlinear ODE governing the

118

pure azimuthal shear problem is derived by using equilibrium and internal balance
equations. An existence and uniqueness result for the pure azimuthal shear problem
for any 01* = d: > 0, a > O and total twist \II > 0 is obtained by using the integral
equation and initial value problems corresponding to the governing nonlinear ODE.

A numerical solution is presented for any 04/43 and regular perturbation is used to
investigate solution possibilities when 04/61 is small for three qualitatively different
ranges of \II: either \I' < \Ilmex or \I' > \Ilmex or \I1 = (1 — e)\IImex with 0 < 6 << 1.
A boundary layer type analysis is used to get a useful approximation for the latter
two ranges of \II. In particular using the properties of inner and outer solutions a
patching method is proposed. This patching method gives satisfactory agreement
with the numerical solution. The relation between torque, twist and energy in the
extended material model is also studied.

The pure azimuthal shear problem is complex in this new framework in either ma-
terial model. The internal balance equation is a second order tensor equation. Even
though it has been proved that the internal balance equation subject to requirement
det E = 1 has a unique solution the proof is not fully constructive. The nonlin—
ear governing equations offer many conceptual, analytical and numerical challenges.
In this thesis we have advanced in all of these areas and examined in depth many
interesting phenomena such as occurrence of singular surfaces, boundary layer type

behavior and uniqueness and regularity of the solution.

.119

 

Appendix A

g(r) given by (3.115) is a strong

local minimum of (3.168) if

Consider the basic problem of calculus of variations: Find a piecewise—smooth function

y(z) on the interval [a, b] that minimizes the definite integral

b
an = / F(4,y(4>.y'<x)> 44. (4.1)

subject to
W) = 4. y(t)) = 13. (A2)

The following are well-known from the calculus of variations [28]:

Theorem A.0.1 For a general Lagrangian F (11:,y, y’), the Legendre condition

Flo/yr _>_ 0 for all :1: in [a, b] is a necessary condition for 3) to minimize J [y]

Proof: See [28], page 88. Cl

Let
[[h] E / [Ryy(:r)h2(a:)+2Ryy1(:r)h(:r)h'(:r)+Rylyr(x)h'(x)2] dx. (A3)

120

Then it follows by direct calculation the Euler-Lagrange Equation associated with
I [h] is
[P(r) h'(a:>1' — 0(4) h(m) = o. (4.4)

where

A A A

P(x) : Fan/(3’3): QC”) = F.yy($) " [FM/(17W: (A-5)

The differential equation (A4) is called Jacobi ’s accessory equation.

Definition A.0.2 A point e > a is said to be conjugate to point a relative to g(r)

if there is a nontrivial extremal h(m) of I [h] which vanishes at c as well as at a, i.e.,

h(a) = h(c) = 0.

Because P(r) and Q(:r) are defined for a particular g(r), conjugacy is relative to a

given extremal g.

Theorem A.0.3 Let J [y] attain a weak local minimum with the admissible extremal
37(2) and let the strengthened Legendre condition Rory: > 0 hold for all :1: in [a, b].

Then there are no points in (a, b) conjugate to point a relative to 3).

Proof: See [28], page 91. II]
Let

E($7y7vvw) = F($,y,’lU) _ F(r,g,v) _ (”LU — v)F,yI(:I:,y,v). (A6)

The quantity E (2:, y,v,w) is called the Weierstrass excess function.

Theorem A.0.4 If, for the admissible extremal 1), J [g] is a strong local minimum

for J [y] , it is necessary that
E(x. 19(4). 17(1). w) _>_ 0. (A.7)

for all x in [a, b] and for all [w] < 00. At a corner point, this equality is to hold for

both 37+ and 3):.

121

Proof: See [28], page 116. D
The strengthened form of the above necessary conditions are useful in developing

sufficiency conditions for minimizing J [y]:

(51) g(x) is a piecewise—smooth (admissible) extremal. i.e., it is a solution of Euler-
Lagrange Equation where it is smooth and satisfies the Erdmann’s corner con-

ditions at a corner point.
(32) Strengthened Legendre Condition: F yryr > 0 on [a, b].
(33) Strengthened Jacobi Condition: y(x) has no conjugate points to a in (a, b].

(54) Strengthened Weierstrass Condition: E(x,u,v,w) = F(x,u,w) — F(x,u,v) —

(w — v)Fyr(x,u,v) Z 0 for all (u,v) near (9,37) and all [w] < 00.

Theorem A.0.5 If F is C4 in its arguments and the admissible extremal g(x) sat—

isfies (51), (s2), (s3), and (34), then J[y] is a strong local minimum.

Proof: See [28], page 128. C]

Using the above theorem it will be proven that g(r) given by (3.115) is a strong
local minimum of (3.168) if |\II| g \IJmex.

For (3.168) the Lagrangian F is

F(r,g,g’) = r\/4 + (rg’)2 — 2r. (A.8)

It is clear that F is C4 in its arguments. By using the Erdmann’s Comer Conditions
(Fe: and F — 32’ F e: must be continuous even at the corner points of 3)(x)), it will
be shown that any extremal of (3.168) must be smooth; i.e., there is no corner point.
From (A.8)

8F _ r3g’

5? — \/4+ (rg’)2'

(A.9)

122

Let p = g; and q = g’_ be the values of g’ at the left and right side of a corner point

respectively. Then by the first Erdmann’s corner condition

3 3
_23_ = __2'9__, (4.10)
(/4+ (rp)2 \/4-l—(rq)2
Since r 71$ 0,
P q
. ___ = —, A.11
\/4+r2p2 «4+qu2 ( )
By taking the squares of both sides in (All)
2 2
P _ q
4+r2p2 — 4+r2q2’ (A'IZ)
which leads to
p2 = q2. (A.13)
If (A.13) is used in (All), then (A.13) simplifies to
p = q. (A.14)

Therefore any extremal of (3.168) must be smooth. Note that in general the second
Erdmann’s corner condition is necessary to find the values of p and q. But in this
case it is not necessary to check the second Erdmann’s comer condition.
In Section 3.4 it was shown that g(r) given by (3.115) is the only solution of the
Euler-Lagrange Equation if |\Il| S ‘Ilmax when k = 1. Therefore (31) is satisfied.
From (A.9) it follows that

 

62F 4r3

_ = _ .1
Since r > 0, on [R1, Re] (A.15) gives that

r2F 4 3

d—— = r > 0. (A.16)

 

89’? (4 + 7.29/2)3/2
Hence (s2) is satisfied. Note from (A.16) that an extremal g(r) must be at least C2

by Hilbert ’5 Theorem.

123

Jacobi’s accessory equation (A.4) for the current problem may be written as

4r3 ,
(4 + 7.2942)3/2 h (T)

 

= 0,1, (A.17)

where a1 is a constant and h(r) is a nontrivial function that vanishes at both c and
R,; i.e., ME) = h(c) = 0. From (A.17) it follows

(11 (4 + r2g’2)3/2

 

h’(T) = 4r, (A.18)
Consequently (3.114) in (A.18) yields
h'(r) = ___204173' , (A.19)
(.4 — 402/2
where d1 is the constant given by (3.118). Hence
h(r) = 7:43: + 42, (A20)

where a2 is another constant. By using h(R,) = 0 from (A.20) it follows that

01

a Z . A21
2 44742 ‘ ’
Whereupon (A.20) can be written as
1 1
h r = a —— — —— . A.22
(> .( .___R:,_d, ,__T,_d,) < )
Now if h(r) = 0, then either a1 = 0 giving h(r) E 0 or
1 1
. _— = ___, A.23
4—42—42 472—41 ( )

giving r = Re. Therefore there is no conjugate points to R,- in (Re, Re]. Consequently
(33) is satisfied.

One important remark is that the notion of being conjugate is defined relative to
an extremal. Since there is no admissible extremal when [‘11] > \I/max the outlined

procedure cannot be used in that case.

124

It remains to show that Strengthened Weierstrass Condition (54) is also satisfied.

For the current problem the Weierstrass excess function E (A.6) becomes

3

E(r,u, v, w) = r\/4 + (rw)2 — “/4 + (rv)2 — (w - v) r v (A24)

4 + (rv)2,

 

 

or equivalently

E(r,u,v,w) = r(\/m — %). (A.25)
Note that

(v — w)2 2 0, (A26)
for all |w| < 00. Then (A.26) can be written as

v2 + w2 2 2vw. (A27)
Multiplying both sides of (A27) with 4r2(> 0) yields

(4r2)v2 + (4r2)w2 _>_ (4r2)2vw. (A.28)
Adding 16 + r4v2w2 to both sides of (A28) leads to

16 + 4r2v2 + 4r2w2 + r4v2w2 Z 16 + 8r2vw + r4v2w2. (A.29)

Note that (A.29) can be written as

(4 + r2142) (4 + r2102) 2 (4 + r2vw)2, (A.30)

or equivalently

(4 + r2102) > (4 + r2vw)2

_ m7- (A.31)

By taking square root of both sides of (A31)

125

(4 + r2vw)

4 .1. y 2 > ___—___, A.32
(ru) '— (/4+ (rv)2 ( )
or equivalently
(4 + r2vw)
2 _ __
4 + (rw) 4 + (rv)2 _ 0 (A33)

giving that E(r,u,v,w) Z 0 for all [w] < 00.
This completes the proof of (34). Consequently g(r) given by (3.115) is a strong

local minimum of (3.168) if [\IJI S \Ilmax. Cl

126

Appendix B

Alternate Proof of Theorem 4.2.3

In Section 4.2 we presented a proof of the Theorem 4.2.3 based on maximal solution
intervals for first order initial value problems. Here we present an alternative proof
based on best existence and uniqueness result for first boundary value problem for

second order nonlinear ODEs. By eliminating d, equation (4.4) can be written as

9"(7‘) + K239201045!) = 0, (31)

where

 

" 2 l2
2‘" 9 ] (B2)

+ 4d+a[4+r2g’2]3/2

Lemma B.0.6 f(r,g’, 62,61) is a Lipschitzian in g’ on U E [R,-, Re] x R X R+ x R+.

i.e., there exists a nonnegative constant L such that

|f(r.gi.a.c‘r) -f(r.9§.a.d)| .<_ L lgi -9£|, (33)
for every (r,g[,a,oz) and (r,g§,a,(3z) in U.

Proof: Note that if f (r, g’ ,a,d) has bounded partial derivative 8 f /8g’ in U then
it is a Lipschitzian with lipschitz constant is given by

, .
L = sup 8f(r, 9 ,la, a) . (B4)
(r,g’,a,d) E U 89

 

127

Now (8.2) gives

af(7',gl,01,51) _ 3 +24arg’2[a+a ‘/4+,.2 9:2]

 

 

89’ _ ;+ [4d +a(4+r29’2)3/2]2 ' (B.5)
Note from (B.5) that 8 f /dg’ is an even function of g’ and
8 f 3
9' 'Blzlleoo— 89’- _. (B6)

By using the limit in (8.6) and the fact that 8 f / 69’ is continuous on U it can be
proven that 8 f / 89’ is bounded on U.

For a given a and 61, it is enough to find the supremum of (9 f / 89’ over the infinite
strip 9 = {(r,sf)lr e [Ri,Rola g’ e (-oo,oo)}-

For each fixed r E [Ra R0] and e > 0, there exists N > 0 such that

3f 3
a—g’—;| <6 (13.7)

whenever |g’l > N, or equivalently

§_ (if
r €<6_g’<— 3+6 (B.8)

whenever |g’| > N. Since 7" E [R4,Ro],

3 8f 3
E—é <W<E+€ (B9)

whenever |g’| > N. Let m1 = |(3/Ro )——e| and m2: (3/R.) +6. Then (B. 9) implies

that

   

 

whenever lg’ I > N and r 6 [R13 R0], where M1 = max{m1,m2}. Consider the region
= {(7‘.g’)|7‘ E [R;,Ro], g' E [—N,N]}. Since 90 is a compact set and df/Bg’ is a

continuous function, there exists a positive number M2 such that

   

 

128

whenever (r, g') E wo. Let L = max{M1,M2}. Then

affix 9’, a, d)
09’

 

< L (8.12)

for every (r, g’ ,a, d) in U. Thus f(r,g’,oz,d) is a Lipschitzian in g’ on U. This

completes the proof of Lemma B.0.6 E1

Lemma B.0.7 Consider the following differential equation

w”(r) + §w'(r) = 0, (B.13)
subject to
W(Ra') = 211,- > 0, w'(R,-) 2 mi > 0, (B.14)

where L is the lipschitz constant given by Lemma 3.1 and /\ is a positive real number
less than 1. Then the solution w(r) of (B.13), (3.14) and its derivative w’ (r) are

strictly positive functions on [Rh 00).

Proof: Clearly the solution of (B.13) satisfying (B.14) is given by

 

i A i ._
w(r) = 112,: + A: ~— Tm eL(R‘ r)”. (B.15)
Note that
w’(r) = m,- eL(R‘—T)/A > 0 (BIG)

for every r E [a,oo). Since w’(r) > 0 on [IL-,oo) and w(Rl-) = wi > 0, it follows
that w(r) > 0 on [R,~,oo). Hence w(r) and w’ (r) are strictly positive functions on

[R4, 00). This completes the proof of Lemma B.0.7. [:1

Proof of Theorem 4.2.3:
Note that the second boundary value problem (B.1), (4.15) is equivalent to follow-

ing integral equation
Ra
g(r) 2 l(r) +/ H(r,s) f(s,g'(s)) ds for R.- _<_ r S R0, (B.17)
R4.

129

and

R0
g’(r) = mo + R.- H,(r, s) f(s,g'(s)) ds, (B.18)
where
l(r) = 1%“ + mo (r — R4), (B.19)

and H (r, s) is the Green’s function given by

_ . -< < <
H(r,s)= (3 R“) E’s—LR) (B20)
(T_R1-) ILST'SSSROa
and
H( )—3H ) (B21)
TT38 —87' (T’S' .

Note that both H and H, are nonnegative functions.
Let S be the space of continuously differentiable functions on [R,-, R0] with the

norm

 

llg(r)|l=max{ max 'gm' max @933}, (3.22)

3.51530 w(r) ’ 3.95120 w’(r)
where w(r) is the positive weight function given by Lemma B.0.7. The space S
is complete with this norm, and convergence in the norm of gn(r) to g(r) implies
uniform convergence of both gn(r) to g(r) and g;,(r) to g’ (r) Let T be the operator
defined by the right side of (B.17). In order to see that T maps S into S, we need

to verify that Tg(r) is continuously differentiable whenever g(r) is. Observe that

 

r — r R"
iii?) Tg( +3 Tg( ) =mo+/& Hr(r,s) f(s,g’(s))ds, (B23)

and that this limit function is continuous in r. To verify that T is a contraction

mapping, consider

|T91(7') " T920“

wb")

 

)I—i—Rorss’s—s’ss
Swm/R. Ha )|f( ,gl< >> f( ,92( ))Id . (13.24)

130

By using the Lipschitz condition (B3) equation (B24) becomes

lT91(7‘)‘T92(7‘)| < 1

MT) _ w(r)

 

 

R0
[34 H(r,s) L was) —g;<s>l ds (B25)

or equivalently

 

 

 

 

 

|T91(r) - T92(r)| 1 R" , lgi(8) - gé(8)|
w(r) g w(r) [Rt H(r,s) Lw (s) w’(s) ds, (B26)
and hence by (B22)
ITQIU") — T92(T)| < (B 27)
“1(7") — '
1 3°
H(r, 8) L w'(8) 613 “910") - 920‘)”-
w(r) R.

Note that since w(r) is a solution of (B.13) with w(R,-) > O and w’(R,) > 0, it is

also a solution of the following integral equation
Ra L
w(r) = w(R,) + w'(R,-) (r — R.) + H(r, s) X w'(s) ds. (B28)
R.-

Since w(R,-) + w'(R,-) (r — 12,-) 2 O, for every r E [R1330] equation (B28) implies
Ra L
w(r) 2 / H(r,s) — w’(s) ds, (B29)
R.- A
and since w(r) and /\ are positive, inequality (B29) may be written as

1r) Ana H(r, s) L w'(s) ds 3 A. (B30)

 

w

Consequently substitution from (B30) into (B27) gives

|T91(7") - T92(7‘)|
w(7‘)

 

S A ”910") - 92(7")||- (B31)

Similarly from (B.18) we obtain

|%(T91(7‘) - Tg2(r))|
w’(r)

 

g (13.32)

R,
$5 [R- mas) magic» — f(s,gé(8))| ds,

131

which by virtue of (B3) and (B22) yields

Isagm - T920)» <

 

 

1 R0 ,
f Hras) Lw (s) ds “gm — 92m“.
Now (B28) supplies
R0 L
w’(r) = w’(R,-) +/ Hr(r, s) — w'(s) ds.
R,- A
Since w’ (1%,) > 0 for every r E [R4, R0], equation (B34) implies
30 L
w’(r) > / H,(r,s) — w’(s) ds.
R; A
Since 211’ (r) and A are positive, inequality (B35) may be written as
1 fRoH (r s)Lw’(s)ds <A
wI(T) R‘ 7‘ 1 '

Consequently substitution from (B36) into (B33) gives

l%(Tgi(r) - T92(7‘))|

w’(7‘)

 

< A ”91(7‘) - 920")”-

Now (B31) and (B37) together imply that

||T91(7') - T920)” S A “910") — 920“)”-

(3.33)

(13.34)

(13.35)

(13.36)

(8.37)

(8.38)

Since 0 < A < 1, T is a contraction mapping on S. Then by the contraction

mapping theorem, T has a unique fixed point in S. Of course this means that the

second boundary value problem, (B.1), (4.15) has a unique solution on [Rh R0] for

every R0 > R,- > O. This completes the proof of Theorem 4.2.3. [I]

132

BIBLIOGRAPHY

133

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