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dissertation entitled

THE DEFORMATIONS OF A CURVED ROD

presented by

JUNGHO LEE

has been accepted towards fulﬁllment
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_ Ph.D degree in Mathematics

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2105 aﬁWoDuohdd—pts

The deformations of a curved rod
By

J ungho Lee

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

The deformations of a curved rod
By

J ungho Lee

This thesis investigates the deformations of an elastic clamped-free curved rod
which has a general natural curvature under a vertical load on its free end. This

result can be applied to the modeling of spring elements and biological vessels.

The balance of forces of an elemental length is considered to establish the govern-
ing nonlinear differential equation. An energy method is taken to derive variational
formulation which will be used to prove existence of solution. Approximation solu-
tions are obtained by a perturbation method. Numerical solutions are also obtained
by applying a shooting method based on the Runge-Kutta. The behavior of a curved

rod exhibits non-linear phenomena such as maximal load, non-uniqueness and stabil-

ity.

To my parents and loving Wife.

iii

ACKNOWLEDGMENTS

I would like to express my gratitude to Professor C. Y. Wang, my thesis advisor,
for his patience, guidance and encouragement during the course of this work. I also
wish to extend my thanks to my thesis committee members, Professors Z. Zhou,
C. Chiu, J. Kurtz and C. E. Wei]. Special thanks to Professor Clifford E. Weil for
delicate editing.

I should also acknowledge my gratitude to Barbara Miler, the secretary of math-
ematic graduate office, for her kind help. Finally, I am very grateful to my wife,

Eunjeong Kim, for everything.

TABLE OF CONTENTS

LIST OF TABLES
LIST OF FIGURES
Introduction

1 Mathematical Formulation

vi

vii

1.1 Introduction ..................................
1.2 Physical assumptions and notation .....................
1.3 Derivation of equation ............................
1.4 Variational formulation ............................

2 Existence of Solution

2.1 Introduction ..................................
2.2 Abstract functional preliminaries ......................
2.3 Existence of solution for deformed clamped-free elastic curved rod . . . .

2.4 Upper bound of the Maximal deﬂection .................

3 The analytical solution

3.1 Introduction .................................
3.2 The Stability and Perturbation Analysis .................
3.2.1 The First Case .............................
3.2.2 The Second Case ............................
3.2.3 The Third Case .............................

4 The numerical solution

4.1 Introduction ................................
4.2 Numerical integration ..........................
4.3 Numerical results .............................
4.3.1 Simple initial curve cases .......................
4.3.2 General case of initial angle ......................
4.3.3 Deformation of initial curved rod under tensile loads (pulling) . . .

5 Conclusions and Discussions

BIBLIOGRAPHY

LIST OF TABLES

4.1 Maximum loads,PM, for the unique solution ................ 46
4.2 Load displacements when 310(3): 60 sin— —3, 60- — 0.1 ............ 62
4.3 Load displacements when 310(3): 60(1 — 2cos— “23), 60 = 0.1 ......... 63
4.4 Load displacements when y0(3)= 60(1 — "6”), 60- — 0.1, a = 1 ....... 64
4.5 Load displacements when 60(s)— - 60 sin— 27’s, 60— — 1, n— — 21 ........ 65
4.6 Load displacements when 60(3) — 60 cos— 2”,.9 60: 1, n- — 20. ...... 66
4.7 Load displacements when 60(s)— — 60 sin— :3, 60- — 2, n- — 21 ........ 67
4.8 Load displacements when 00(3 )— — 60 cos— 2",3 6011— — 2, n— - 20. ...... 68

4.9 Tensile Load displacements when 00(3 )- — 260 sin— 277,3 60— — 2, n— — 5, 21. . 69
4.10 Tensile Load displacements when 00(3 )— — 60 cos— 2",.9 60— — 2, n- —— 4, 20. . 69

vi

1.1
1.2

3.1

4.1

4.2

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
4.16
4.17
4.18
4.19
4.20
4.21
4.22
4.23
4.24
4.25
4.26
4.27
4.28
4.29
4.30

LIST OF FIGURES

The elastic curved rod; ............................
An elemental length of rod ..........................

The elastic curved rod ............................

Column conﬁgurations for simple sine curve ................
Column conﬁgurations for simple cosine curve ...............
Column conﬁgurations for exponential curve ................
Analytic solutions for a sine function case and load-displacement curves .
Analytic solutions for a cosine function case and load-displacement curves
Analytic solutions for an exponential function case and load-displacement
Maximum load vs initial amplitude for a sine curve ............
Maximum load vs initial amplitude for a cosine curve ...........
Maximum load vs initial amplitude for an exponential curve .......
Load-vertical displacement curve for a sine function ............
Load-vertical displacement curve for a cosine function ...........
Load-vertical displacement curve for all exponential function .......
Column conﬁgurations for sine curve (‘59 = l ................
Column conﬁgurations for sine curve 60 f 2 ................
Column conﬁgurations for cosine curve {59 = 1 ...............
Column conﬁgurations for cosine curve 60 = 2 . . .~ ............
Load-lateral displacement curve for a sine function (’59 = 1 ........
Load-lateral displacement curve for a sine function 60 f 2 ........
Load-lateral displacement curve for a cosine function (29 = 1 .......
Load-lateral displacement curve for a cosine function~60 = 2 .......
Load-vertical displacement curve for a sine function {59: 1 ........
Load-vertical displacement curve for a sine function 603 2 ........
Load-vertical displacement curve for a cosine function Q): 1 .......
Load-vertical displacement curve for a cosine function 60: 2 .......
Column conﬁgurations for sine curve under tensile load n = 5 ......
Column conﬁgurations for sine curve under tensile load n = 21 ......
Column conﬁgurations for cosine curve under tensile load, n = 4 .....
Column conﬁgurations for cosine curve under tensile load, 12 = 20 . . . .
Load-vertical displacement curve for a sine function under tensile load . .
Load-vertical displacement curve for a cosine function under tensile load

vii

39
39
39
41
42
43
44
45
45
47
48
49
51
51
52
52

59
60
60
61
61

Introduction

This thesis is concerned with the nonlinear equations governing the deformation of

an elastic, inextensible and thin rod under a vertical load.

The elastic rod theory has always been an essential subject in structural mechan—
ics. Moreover, the theory of an originally straight rod has long been recognized by

many authors. The straight rod could buckle if the axial compressive load exceeded

a certain value. Euler ﬁrst found this critical load in his famous article, “De Curvis
Elasticis”, 1744. This critical load can be predicted by a variety of linear meth-
ods, such as, beam theory, minimum energy method, imperfection method, dynamic

method, etc. (C. L. Dym 1974; G. J. Simitses 1976; H. Ziegler 1968)

Nowadays, resulting from advances in materials science, ﬂexible rods are being
used as springs, linkages, robotic arms and space antennas. Because large deflections
are usually encountered, the linear beam-column theory used in traditional buckling

analyzes is inadequate to describe the mechanics.

If the rod is slender enough, an excellent model is the elastica. The elastic rod
deforms mainly by bending, such that the local curvature is proportional to the 10-
cal moment (Euler-Bernoulli law). In nonlinear elasticity theory, nonlinearity, which
enters through the geometry, plays a very important role because it leads to singu-
lar phenomena, such as multiplicity of solutions accompanied by bifurcation, loss of
stability and sometimes chaos (Chow and Hale 1982; Thomson and Stewary 1994;

Tvergaad 1999).

The deformation of an originally straight elastic rod, one end ﬁxed and one end
free, can be represented implicitly by elliptic integrals using the end angle at the
top (S. P. Timoshenko and J. M. Gere 1961; G. Wempsner 1981). However, there
are cases where the rod may not be originally straight but naturally curved (i.e.,
curved even when no load is present) whether unintentional, such as imperfection
or intentional, such as a curved leaf spring. Due to the difﬁculty of the problem,

literature for naturally curved rods are relatively scarce.

The deformation of a curved rod with constant curvature (an arc of a circle) was
considered by several authors. Frisch-Fay (1962) studied the initially curved rod with
constant curvature (circular arc rod) by the direct application of elliptic integrals. Yin
and Wang (2003) also studied multiplicity of large deformations of the same initially

circular arc rod by using phase plane.

In this thesis, we consider deformations of an elastic curved rod which has a general
natural curvature.

In chapter 1, section 1.2, the method of equilibrium of elemental segment is used to
formulate the governing equation. In section 1.3, the variational formulation is set up
using the energy method. This principle leads to a minimization of total energy which
is used to prove the existence of the solution, that will be shown in chapter 2. The
work on the variational method for the nonlinear elliptic problem by F. E. Browder
(1965) serves as the proof of the existence of solution. Chapter 2 reveals the existence
of the solution of the boundary value problem which we established in chapter 1 and
also an upper bound of the maximal deﬂection is obtained. An analytical solution is
introduced in chapter 3 using the perturbation method. Chapter 4 shows a numerical
solution (by the shooting method). Chapter 5 reviews the outcome from chapter 1

through chapter 4.

CHAPTER 1

Mathematical Formulation

1 .1 Introduction

In section 1.2, the physical assumptions needed to derive a mathematical formulation
are provided. A non-linear ordinary differential equation is found using the bending
theory under the assumptions given in section 1.3. This formulation will be applied
in chapter 3 to determine the perturbation and numerical solutions. In section 1.4,
we derive the formulation using potential energy and this will be used in chapter 2

to prove the existence of equilibrium states for a given load.

1.2 Physical assumptions and notation

We consider an elastic curved rod of arc length L shown in Figure 1.1, ﬁxed at the
base and free on the top. The curved rod “0.” is the original natural shape and “b” is
the deformed shape after a force P’ presses down at the free end. The rod is assumed
to be thin compared with the other length parameters of the problem. By virtue
of the assumption of thickness, the rod can be treated as an elastica (A. E. H Love
(1944) [8]), deﬁned as a material where the local bending moment is proportional to

the local curvature changes.

Let a Cartesian coordinate system (2:, y) be located at the free end where 17 is parallel
to P’. Let s’ be the arc length from the free end to the cross-section of rod “b”,
s’ E [0, L]. Let 00(3’) be the local angle between the tangent to the free shape “a”
and the :r-axis and 9(3’) be the local angle between the tangent line of the deformed
90

d2
curved rod “b” and the x-axis. We assume that 00(3’) is twice differentiable and F
s

is continuous; i.e., 60(3’) E C2[0, L].

The following is the outline of notation that will be used.
3’ = Arc-length

s 2 Normalized arc-length

m = Bending moment

P’ = Force (load)

P = Normalized force (load)

E1 = Flexural rigidity

E = Strain energy

V 2 Potential energy

W 2 Work done

 

 

60 s $

it ’7 “b”

a

 

 

 

////"/////

Figure 1.1: The elastic curved rod;

“a” shape when the load is absent
“b” shape when load P’ is applied

 

Figure 1.2: An elemental length of rod

1.3 Derivation of equation
A local moment balance for an elemental segment (Figure 1.2) shows
m+dm+P’ds’ sin0=m (1.1)

where m is the local moment and m = —P’y. The elastica assumption, the local

moment is preportional to the local curvature changes, gives

_ (16 (£60
m — EI [ds’ ds’l (1.2)
Hence,
d0 ddo
EI —— — —— = —P’ 1.3
[ds’ ds’] 9 ( )

where E1 is the ﬂexural rigidity.

Under the assumption that the rod is inextensible and undergoes only pure
bending, it can be parameterized by its arc length 3’ E [0, L] and repreSented by
s’ +——> (a:(s’), y(s’)) where :1:(s’) and y(s’) denote the coordinate of the point 3’ on the

cross section. We constructed the governing equation for (22(3’), y(s’)) geometrically:

d1: , dy _ , ,
g — cos 0(3 ), 8—37 — s1n6(s). (1.4)

For simplicity, we normalize all lengths by L, the force by % and drop primes. Equa-

tions (1.3) and (1.4) become

E — E — Py, ds — cos 0, ds —- smd (1.5)

and the boundary conditions corresponding to equation (1.5) (Figure 1.1) are

a
d6’0 _d0

:r(0)=0, y(0)=0, 0(1)=00(1), BEN—7.1?

(0). (1.6)

We will use equation (1.5) to ﬁnd an analytic solution by using the perturbation

method in chapter 3.

Dlﬂ'erentlatlng the ﬁrst equation 1n (1.5) by s and usmg d_y = 81119 gives
3

d2
d—.S2(6 — 60) Z —P sini9 (1.7)

with boundary conditions:
9’(0) = 06(0)

(1.8)

Here ’ represents differentiation with respect to s.

In the integrating equation (1.7), we proceed as is usually done in the case of a
d6
pendulum. We multiply both sides by d; and integrate. Then, for some C
1 3
5(6’)2 : P c050 +/ 03(t) 0'(t) dt + C
o

and so

ds - i d0 (1.9)

ﬁ\/P c080 + f; 63(t) an) dt + C.

 

 

If the rod is straight(60 E 0), we can solve equation (1.9) with the aid of elliptic
integrals (see Timoshenko and Gere (1961) [12]). But otherwise, it is impossible to
solve for the integrand analytically. Thus numerical tools will be needed to obtain

the solution of the equation (1.5) in chapter 4.

1.4 Variational formulation

For the deformation of a clamped-free elastic curved rod shown in Figure 1.1, the

strain energy E is deﬁned by
1 1 .
E :/ —7’71.2 (18 (110)
o 2

from Timoshenko and Gere (1961) [12]-

Replacing m from equation (1.2) gives

1 1d6 MO?
E = — — — —— d 1.11
2 [0 (d3 d3 ) 8 ( )
and the corresponding vertical deﬂection of the load P in Figure 1.1 is given by

A = h — :r(1) (1.12)

1
zit—f cos6ds (1.13)
o

where h is the height of the undeformed curved rod in the direction of P.

The dimensionless potential energy V is

VzE—P-A
1 1 1
_—_-—/ (6'—-6[))2ds+P/ cos6ds—Ph (1.14)
2 0 o
1
_—./ [%(6'—63)2+Pcos6] ds—Ph (1.15)
0

Let 6 be the variation symbol. Since 60 is ﬁxed, 660 E 66fJ E 0 and 6(6’ — 66) = 66’

6V: /0[(6' —6') 66’— Psin666]ds

=[ (19'( )- 6"(1069(1) - (9(0) -96(0))50(0)l

—/l [(6" — 66’) + Psin 6] 66ds.
0

Using the boundary condition (1.8) and since (56 is arbitrary, we can get the same
governing equation as (1.7) from (IV 2 0.
This variational formulation will be used in chapter 2 to show the existence of a

solution.

CHAPTER 2

Existence of Solution

2. 1 Introduction

Chapter 2 is concerned with the proof of the existence of equilibrium states for the
clamped-free curved rod formulated in chapter 1. The variational formulation is used
to prove the existence of energy minimization for a given load. The proof is based
on a theorem published by F. E. Browder (1965). In section 2.2, some preliminary
functional tools are stated. In section 2.3, the existence of a weak solution to the
variational formulation is showed using the Browder theory outlined in section 2.2.

In section 2.4, an upper bound of the maximal deﬂection is obtained.

2.2 Abstract functional preliminaries

Deﬁnition 2.2.1 A complete normed space is called a Banach space B.

 

Deﬁnition 2.2.2 By a Hilbert space H we mean a Banach space B in which there

 

is deﬁned a function (2:,y) on B x B to R with the following properties:
z'}(0l11171+ 02152, y) = C31(51311.1/)+ 012(532, y)
2'2') (III. a) = (v.27)

iii) (56.33) = ||$||2-

Deﬁnition 2.2.3 A sequence {an} in a Banach space B is said to converge weakly

 

to u E B if the sequence {f*(u,,)} converge to f“(u) for every f” E B*, the dual space

of B. We denote this weak convergence by un —> u.

Deﬁnition 2.2.4 Let 81 and 82 be Banach spaces. Then 31 is said to be imbedded
in B2 if

i) every element of B1 is in B2

ii) every convergent sequence in 81 is convergent in Bg.

The imbedding of Bl in 32 is compact if the imbedding operator I : Bl —+ 32 deﬁned

by [(33) = :1: is compact.

10

Deﬁnition 2.2.5 (F. E. Browder, 1965 [3]) For a real Banach space B, a function

\I':B x B —> R is said to be semi-convex if

i) For each v E B and each c 6 1R, the subset Sm, deﬁned by

SC,” 2 {u E B|\I’(u, v) g c}

is convex.
ii) For every bounded subset S ofB and each weakly convergent sequence {vn}
to v in B, \Il(u,v,,) —+ \Il(u, v) uniformly for u in S.

iii) For each ﬁred v E B, 111(., v) : B —+ IR is continuous.

Theorem 2.2.6 (Eberhard Zeidler, 1995 [26]) If 13,, —-\ a: in a Hilbert space H which

is compactly imbedded in a Banach space B, then at" —> :1: in B.

Theorem 2.2.7 (F. E. Browder, 1965 [3]) Let B be a real reﬂexive Banach space,
\I1 : B x B ——> IR be semi-convex and E(u) = ‘IJ(u, u) If E(u) ~+ +00 as [lull ——> +00,

then E assumes a minimum on B.

11

2.3 Existence of solution for deformed clamped-

free elastic curved rod
The expression of potential energy derived in equation (1.14) in section 1.4 is
2

var) = [01 [l (9' —6;,)2 +Pcos9] d3 — Ph. (2.1)

1
Set 7(3 )2 6(s) — 60(s ). Then V(*y)= / [g 7'2 + Pcos(7 + 60)] ds+ Constant.
o

1
Let 1(7 )2/ [——— + P cos (7 + 60)] ds. To apply Theorem 2.2.7, we let
0
B = {u | u e H1(0,1), u(l) = 0} (2.2)

where H1(0, 1) is the Sobolev space of function in L2(0, 1) where the distributional
derivatives are in L2(0, 1), or equivalently, the space of absolutely continuous functions

with square integrable ﬁrst derivatives. Deﬁne j, <I> and f by

j(v) = [0 cos(v + 60) ds (2.3)
<I>(u, v) = /0 gum ds + Pj(v) (2.4)
f(u) = <I>(u, u) for u, v E B (2.5)

and note [(u) = f(u).

12

Lemma 2.3.1 The function <I>(-, ) is a semi-convex function on B x B.

Proof. i) We will show that for each v E B and each c E IR, the subset Sm, deﬁned
by Sc,” 2 {u E B | <I>(u,v) g c} is convex. Pick v E B and c E R and let 111, 212 be in
S”. For t E [0,1], deﬁne g(t) = <I>(tu1 + (1 — t)u2, v) — t<I>(u1,v) — (1—t)<I>(u2,v).
Then g(t) is a parabola with 9(0) 2 g(l) = 0. Since the second and the third terms

in the expression of g(t) are linear with respect to t,

(129 d2 1 r r 2
Et—f -—— ﬁ¢(tu1+(1—t)u2, ’U) = 0 (111— “2) d3 2 0

So g(t) g 0 for 0 S t g 1 and because ul and ug E SW,

<I>(tu1 + (1 — L)U2, ’U) St<I>(u1, U)+(1—L)CI)(U2,”U)

gt-c+(1—t)-c=c.

Thus tul + (1 — t)u2 E SW. Since ul, U2 are arbitrary, Sm, is convex.

ii) Now we will show that for each weakly convergent sequence vn to v in B,
(I>(u, vn) —> <I>(u, v) uniformly for u E B. Then the imbedding implies vn converges to

v in L2-norm. Assume vn —\ v in V. For every u E B,

|<I>(u,v,,) _ <I>(u,v)| : lpﬂvn) _ P](v)l

1
S P / [cos(vn + 60) - cos(v + 60)] ds.
0

We note that by the Mean Value Theorem, for any :17, y E IR,

|COS(I) — cos(y)|
Ix - vi

 

= | - sin c] for some c E [.L',y] or [31,113] E 1-

13

Therefore,

1
|<I)(u, vn) — <I>(u,v)| S P / |cos(vn + 60) — cos(v + 60)| ds
0

1
SP/ |vn+6O—(v+60)|ds
o

1
=P/ [vn—vlds
o

S P ll’vn - Ulla -> 0-
By imbedding vn —> v in L2, <I>(u, v,,) —> <I>(u, v) uniformly.

iii) We will Show that for a ﬁxed v E B, <I>(-, v) : B —+ IR continuous. Let an be a

sequence in B converging to v E B. Then

|<I>(u,,,v)— <I>(u, v )|= [2 /1(11:,221'2)ds
1 l
_<_ —/ In; — u'] In; + u'l ds
2 o

1
< 5 Hui. — u'urz liu’. + 11'th

1
5 I’ll;l — U]1,2 '11:, + U’llg —) 0.

Hence <I>(-, v) is continuous. From i), ii) and iii), (I) is semi-convex on B x B. E]

Lemma232 f() <I>,(u u)—)+oo as |u|12—>+oo onB.

1
Proof. Since Pj(u) = P / cos(u + 60) ds is bounded, f(u) —> +00 if and only if
0

1
f u"2 ds —> +00.
0

1
Since u(s) 2 11(1) —/ u’ds for any 8 E [0,1] and u(1) = 0,

1 2 1
[u(s)]2 g (/ lu'lds) S 1 -/ |u'|2ds.
o o

14

The integration yields ”UNI? g ||u’]|iz. Hence, |u|1_2 g x/2||u’||L2. Therefore,

1
/ u'2 ds —> +00 as |u|L2 -—+ +00. El
0

We may apply Theorem 2.2.7 to conclude the following.

Theorem 2.3.3 The functional f(u) (= <l>(u,u)) deﬁned by (2.5) attains its mini-

mum on V.

By the Theorem 2.3.3, [(u) = [01 [—;-u'2 + P cos(u + 60)] ds = f(u) attains its
minimum on B. Let w be a minimizer of I . We now show that 6 = w + 60 is in fact
in C2[0, 1] and satisﬁes equation (1.7) in section 1.3.

If if is any test function and ift E IR, since the maps t —> (1,!) + té)’2 and t —+
P cos(w + t6 + 60) are differentiable, the map t —> I (1,0 + tg) is diﬂ'erentiable and its

derivative at t = 0 must be zero; i.e.,
d 1
are) + t€)lt=o = j W — P sine/2 + 00):] ds = 0. (2.6)
o
If we deﬁne 6(3) 2 61(3) + 60(3) for s E [0,1], then we have
1
f [(6’ — 605' — P sin6§] ds = 0. (2.7)
0

So we see that 6 is a weak solution of equation (1.7) in section 1.3. But it is standard

that weak solutions of (1.7) are, in fact, classical solutions.

Since 1!) E B: {ulu E H1(0,1), u(1)= 0},

9(1) = #41) + 90(1) = 90(1) (2-8)

15

To see 6’(0) 2 66(0), for each n E N, choose {u(s) = (1 — s)" for s E [0,1]. Then
||£n||L2 —> 0 as n —> 00. Integrating by parts in equation (2.6) yields
1
—1/J'(0) —-/ (10" - P Sin(w + 60))En ds = 0
0
so lw’(0)| g ||i/1”— P sin(w+ 60)|[L2||§,,]]L2. Since the right hand side of this inequality
is independent of n, by letting n —-> 00, w’(0) = 0; that is,
(7(0) = W0) + 96(0) = 96(0) (2-9)

Finally by (2.8) and (2.9), 6 is a solution of equation (1.7) with the given boundary

conditions (1.8).

2.4 Upper bound of the Maximal deﬂection

Theorem 2.4.1 (D. S. Mitrinovic, (1970) [9]) Let h be a positive function with
piecewise smooth derivative on I = [0,1] such that h is convex on I and h’(0) g 0.

Let f be continuous and piecewise smooth on I with f(0) = 0. Then

if unaware: > .._2.
f01h(:r)d:c folf(:r:)2 d2: _ 4

 

Let us start with equation (1.5) found in section 1.3.

d6 d60
_ _ _ ___. _p
ds ds y

y(0) = 0 (210)
0(1) 2 60(1).

16

Multiple both sides of the ﬁrst equation by y and integrate from 0 to 1 to get

1 1 1
/ 6'y ds -/ 66yds = —-P/ y2 ds. (2.11)
0 0 0

By integration by parts,

9(1)”!!(1) — 6(0)y<0) — / ey'ds

— 60(1)y(l) + 60(0)y(0) +/0 60y'ds = —P/0 y2 ds. (2.12)

From the boundary conditions 6(1) 2 60(1) and y(0) = 0, equation (2.12) becomes

1 1 1
/ 6y'ds = P f y2 ds +/ 60y'ds. (2.13)
0 0 0

o i ' 1
Slnce 6 = arcsm y’ and arcsm y’ 2 y’ + —y’3,

6
1 1
f6y'dsz/ arcsiny’y’ds (2.14)
o 0
1 1
2 / (90+69'4ld3 (215)
o
l 1 l 2
2/ y'2 d8+6(/ y'2ds) by Holder’s inequality (2.16)
0 o
, 1
= My H? +601)“ (2.17)

where H - [I = L2(0,1) norm.
1

1
The right hand side of (2.13) is P/ y2 ds +/ 6oy'd8 S P llyl|2 + ”90“ “y,”
o 0

Hence we have,
I 1 I
Hi! II2 + 6 “31 II4 S P llyll2 + ”90” lly'll- (2-18)
By Theorem 2.4.1, Hy’I] 2 gllyll and hence
1 2 2
lly’ll2 + 5 III/ll“ s P (-) lly’ll2 + noon lly’ll- (2.19)

17

Move all terms to the left side and combine with ||y’|| to get

1 2 2 ,
“21’“ [g Ily’ll3 — (P (g) —1)uyn— 116.11] 3 o. (2.20)
Since ||y’|| 75 0, multiply by ”5+” to get
2 2
III/“3 — 6 P (;) — 1] Ily’ll — 61160)) s o. (2.21)

 

The equation in (2.21) is cubic for [ly’ll so more than one real solution can be found.
The following theorem will lead the estimate for [[y’l].

Theorem 2.4.2 (J. V. Uspensky, (1948) [13]) Suppose a, B E IR. If (13 +132 2 0,

then the equation 2:3 + 3012: + 26 = 0 has only one real root which is

$=(—B+\/m)i+(—B-W)i

and if a3 + (32 < 0, then the equation 2:3 + 3011: + 26 = 0 has three real roots and

they are

$1 = 2\/—a C0813)-

 

 

:132 = —2\/—ozcos(§ + g)
273 = —2\/—acos(§ — g)
where cosp: _’3 .
—a3
Set
P
a = —2 [(92 — 1] and s = —3||90||. (2.22)

18

Then the inequality (2.21) becomes
lly’ll3 + 3ally'll + 2B S 0- (2-23)
By Theorem 2.4.2, there are two cases.

Caselza3+ﬂ220

The equations in (2.22) yield

In this case, the cubic equation Ily’ ||3+3a||y’ || +213 2: 0 has only one real root; namely,
lly'll = (-i3+ va3+52)%+(—13— \/a3+ﬂ2)i, (2.24)

and this root is positive because (3 is negative. Since the leading coefﬁcient of the

inequality (2.23) is positive, the region to satisfy the inequality (2.23) is

o s lly’ll s (-B + WE + (a? — «(1311132)? 1225)

31(8) S lly'll- (2.26)

A bound on y(l) is obtained by substituting (2.24) into (2.26) as following.

11(1): (-/3+ x/a3+i32)% + (—/3— «await (2.27)

19

Case2: a3+ﬁ2<0.

The equations in (2.22) yield

0<—[( neon )1 +1] [’2‘]2 < P.

By Theorem 2.4.2, the equation [ly'll3 + Bally’ll + 213 = 0 has three real roots:

51:1 = 2\/—a cos 8

— ——2\/—aco:(f+ 1)
3 3
- ——2\/—a cos(g— g)

 

—, . . 99
. Slnce oz and are ne ative, 0 < < 1. So cos —

90 7F . . . . . .
cos(§ -— E) are posmve. Hence 21 1s a posmve real root and 2:2 and 223 are negatlve.

where cos (p = + g) and

Inequality (2.23) yields

 

0 < ||y’ I] < 2\/—a cos — — —%2\/—-acos( arccos ). (2.28)

_03

Therefore, a bound on y(1) is obtained by (2.26) and (2.28) as follows.

 

 

3

1 r c . ,3 i
— 1] 2 [COME—0133)] . (2.29)

(12 i
arccos r—1
y(1) S [ZV-a COS( ’0‘ )[

 

 

= m l<P)

When the rod is initially straight, 60 E 0 and B = 0. By (2.29),

3

rel—-
n
O
to
00 lwla

y(1)<2\/§((—;— 1)

=f<P)(§— 1)

10h—

(2.30)

which agrees with T. M. Atanackovic (1981) [1] result for the bound of an originally

Straight rod.

20

CHAPTER 3

The analytical solution

3. 1 Introduction

This chapter considers the approximate solution of the boundary value problem which
is found in section 1.3. In section 3.2, the perturbation method is applied to ﬁnd an
analytic solution when the curvature of the initial curve is small. We will compare

this solution to the numerical solution in chapter 4.

3.2 The Stability and Perturbation Analysis

In this section, we investigate the behavior of an initially curved rod by perturbation

analysis. Consider a small perturbation from the original curve described by

310(8) = 5of(8) C“)

where f (s) : 0(1) and (50 is the original amplitude and s is the normalized arc length

from the top end. The order symbol “0” is explained well in Neyfeh (1973) [10].

21

 

 

Figure 3.1: The elastic curved rod

............ initial curved rod
rod with load

Figure 3.1 is the picture of an initially curved rod and a curved rod after a load P is
applied. Here, 6 is the lateral distance at the free end. We now recall the differential

equation which was found in section 1.3.

$(0*00)+Psin6=0 (3.2)
0'(0) = 03(0), 6(1) = 00(1). (33)

We can express 1% as a function of f(s). If there is no load (P = 0), then 6 = 00,

y = yo and 6 = 60. From equation (1.5), (17:9 = sin 00. Hence we get
00 = sin-1(60f’(s)). (3.4)
Now we introduce a new variable 7(3) 2 0(3) — 90(3). Note that 7 is the difference

between the angle of the rod under load and the angle without load. The equation

22

(3.2) with our new variable 7 and the equation (3.4) become
d2 . _
21:2“) + PSIII(’)’ + 00) — 0 (35)
00 : sin—1(60f'(s)). (3.6)
The boundary condition (3.3) changes into

7'(O) = 0 and 7(1) = 0. (3.7)

For the initial deﬂection, we assume that 7 is small. The variable can be expanded

as follows.
7=€VO+€371+~~

where 6 << 1 is some small parameter and ”y,- is of order 1. Aposteriori analysis shows
the 62 term is absent. If 60 or 60 : 0(1); i.e., the initial imperfection is large, then
equation (3.5) is impossible to solve analytically in general, in terms of the parameter
6. Therefore we shall concentrate on the case when the initial imperfection is as small
as the deformation. Without loss of generality, we can set 60 = e in equation (3.1).

Equation (3.4) gives
ens) = sin—Hem» = . ms) + 0(3).

Since sin('y + 90) = sin 7 cos 60 + cos ’y sin 00

63 62 fl)2 62’“?
EVSXI— ( ,0

 

= (670 + 6371 -
we can expand equation (3.5) as follows.
I: 3 n 3 3P 3
5’70 +6 71+5P’70‘l'5 1371—5 E70
P P ,
— c3573f' — 633’)"0(f’)2 + ePf + 0(65) 2 0. (3.8)

23

The comparison of like powers in equation (3.8) yields

’76, + P70 2 —Pf, (3.9)
II P _ 1P 3 1P 2 I £13 I I 2
71 + 71 — 6 70 + 2 70 f + 2 Wolf) (3-10)

with the boundary conditions:

76(0) = 0, 10(1) = 0 (3.11)
71(0) = 0, 71(1) = 0- (3.12)

In order to ﬁnd solutions to equations (3.9) and (3.10) with the boundary conditions

(3.11) and (3.12), we need the following theorem.

Theorem 3.2.1 Let k # (n + %)7r for any n = 0,1,2,... . Suppose F(:1:) is a

sectionally continuous function on [0,1]. Then the boundary value problem

y”($) + ’12 y(i‘) = F (:13)

y'(0) = 0, y(1) = 0

(3.13)

has the unique solution

cos k2:

k/lsink(1—t)F(t)dt.
0

 

y(1:) 2 %/Or Sink(:I: — t)F(t)dt — k cos

Proof. By Churchill (1972) [5],

sin k2:

y(sc) = £f; sin k(:1: — t)F(t)dt — k cosh/0 cos k(1 — t)F(t) dt

 

is a solution of the equation (3.13). To prove the uniqueness, let yl and y2 be solutions

of equation (3.17). Then yl — yg is a solution of the homogeneous equation that arises

24

when F(:1:) = 0 in equation (3.13). Hence yl — y2 = Cl sin k1: + C2 cos kit for some C1
and Cg but (yl — y2)(0) = 0 implies C2 = 0. Moreover, since (yl — y-z)’(1) = 0 and

A¢n+%foranyn=0,1,2...,C1=0.Thusy1—y2_=_0. [3

Note that suppose k 2 (11+ %)7r for some n = 0, 1,2,. . . . Then we have two cases.
1
1) If/ cos(n + %)7rtF(t) dt = 0, then the boundary value problem (3.13) has
0

the solution
x , I 1
y(x) = —/ Sin[(n + §)7r(;1: — t)] F(t) dt + C cos(n + §)7rt
o
for any C E R.

1
(2) If/ cos(n + %)7rtF(t) dt 316 0, then the boundary value problem (3.13) has
0

no solution.

However, we will see our perturbation solutions of the particular choice of the
initial curve become singular when the load P = n + % for some n = 0, 1, 2, . . . . This

means our perturbation solutions will not be applicable near these loads.

With the aid of Theorem 3.2.1, the solution of the zeroth order equation (3.9)

with the boundary condition (3.11) is

— —\/F/8 Sin x/I—3(s - t)f’(f)alt

3:3?) m.-.

and the solution of the ﬁrst order equation(3. 10) with the boundary condition (3. 12)

(3.14)

 

= ﬁ/ssin VIE—Ks — t)F(t)dt
o

_ ”swig—)5 /lsin x/ﬂl — t)F(t)dt
COS 0

25

(3.15)

 

1 1 1

where F(t) = ——('y0)3 + 5(70)2f' + 5(70)(f')2. The maximum curvature, or normal-

6

ized maximum moment is

d0 _d_00

(”0(1) 3 6171(3)
33(1) ds

(1)+c——+e ——+...

ds ds
__ €f’(1) (170(1) + 63d"/1(3)
_ 1— (e f’(1))2 ds (ls

 

 

 

+6 +....

A moment balance gives the lateral displacement

1d)

y(1) = —F21_3(1)'

The vertical height can be obtain by the integration.

1
33(1) 2/ cosﬂds.
0
The vertical displacement A is

1 1
A 2 350(1) — 33(1) =/ c0560 ds —/ cosﬂds
o o

l 62 62 4 4
=/, [’(i‘§)+(%'%)l ‘13

1 72
= 62/ (f’n + J) d8
0 2

l fl ’73 1

1

I

4 I A __122_ _
+6 /0(f’71+/071 (f)70 6 0 24

4

' —'13)ds

(3.16)

(3.17)

(3.13)

3
where x0(s) : / cos 60 ds is the vertical height of the initial curve without load. For
0

an arbitrary function f (3), equations (3.14) — (3.18) are tedious to handle. However if

the initial imperfection curve f (s) is in the form of a sine, a cosine or an exponential

function, then a direct calculation is possible.

We will discuss the following three cases.
Case 1) f(3) = sin(n + %)7rs for n = 0,1,2,. .. .

26

case 2) f(s)=1—cos(n+%)7rsforn=0,1,2,....
case 3) f(s) = e‘” — 1 for a E R.
We note that all of these forms give f (0) = 0 so that our initial curve without load

satisﬁes geometric boundary condition at s = 0, 310(0) 2 60 f (0) = 0.

3.2.1 The First Case

case 1) Consider that the initial curve is in the form of a sine curve described by
, 1
f(s) = s1n(n + §)7rs for n = 0,1,2,... .

Then the zeroth order equation (3.9) becomes

76’ + p70 2 —P(n + %)7r cos(n + %)7rs

 

(3.19)
76(0) = 0, 70(1) 2 0
and the zeroth order solution is
1 1
70(3) 2 c (n + §)7r cos(n + §)7rs (3.20)
_p . .
where c = . As we d1scussed at the end of the Theorem 3.2.1, Since

P — ((n + %)77)2
1
1 1
/ cos(n + §)7rt cos(n + §)7rsdt # 0, the boundary value problem (3.20) has no
0
solution when P = (n+ %)2. Moreover our zeroth order solution 70(3) is singular near

P = (n + %n)2. Hence for P close to (n + %7r)2, our analytic approximation can not

be applied. The only way integrating equations (3.2)-—(3.3) is numerically.

The ﬁrst order equation (3.10) with the zeroth order solution (3.20) becomes

7" + P71: P(n + 1)3713 (£3 + C: + E)(cos(n +1)7rs)3
l 2 6 2 2 2
1 3 12 1 1 1
= P(n + §)37T3(% + :2- + 5;) 21-(3cos(n + §)7rs + cos3(n + 5)“)

27

and boundary solutions are 7((0) = 0, 71(1) = 0. The ﬁrst order solution is

1 c3 c2 c1 3 1
:P -—-3 3— -— ——|: ‘ —
71 (n+2)7r(6+2+ )4 P_(n+%)27r2cos(n+2)7rs
1 (3.21)

+
P — 9(n + %)27r2

 

 

cos 3(n + ins].

Using equation (3.16), the lateral displacement is

y(1) — 77(1) — F cc<n + para—1)"
1 c3 c2 c 3
3 _ 4 4 _ _ _ __ _1 n
+e(n+2)7r(6+2+2)4( ) (3.22)
I 1

 

 

(P — (n + $2712 _ P — 9(n + %)2712)'

We can obtain the vertical displacement from equation (3.18).

We could extend our result for more general case of sine function. Suppose f (s) =

1
an sin(n + -)7rs for some M E N. Then the zeroth order equation (3.9) with

[V]:

n 0
boundary condition (3.11) becomes

M

1 1

”+P/=—P§:n +_ +_
70 70 nzoa (n 2)7rs cos(n 2)7rs

76(0) = 0, 70(1) = 0
and the zeroth order solution is

M
I 1

70(8) 2 anm + §)7r cos(n + 2)“ (3.23)
n20

—P

, =0,1,2,....
lD—(n-l-%)27r2 n

 

where b7] 2

28

The ﬁrst order equation (3.10) with the zeroth order solution (3 23) becomes

M 1
70 + P71: % )7r cos(n + 2)7rs)3
n=0
M
+P 1 2
+3 ( gbnm )7r cos(n + -2-)7rs)
M 1
(Zan(:+ )7r cos(n + 2)7rs) (3.24)
1120
M
P I 1
+ 2(gbnm + -2—)7r cos(n + -2-)7rs)

[V]:

(

a (n+l)7rcos(n+1) )2
n 2 2 713 .
The right hand side of equation (3.24) is a linear combination of

71

II
o

1 l 1
cos(i + -2-)7rs - cos(j + §)7rs - cos(k + §)7rs, i,j, k = 0, 1, 2, - - . - M.

By the harmonic properties of the trigonometric functions, equation (3 24) w1th the

boundary condition (3.12) becomes

M M M 1 1
71I+P71:PZZZCUkE[COS(i+j_k+

0 2)“

J k:
I I
< +cos(i—j+k+ §)7r 7rs+cos(—i+j+k+— —7rs)

2
3
+cos(i i+j+k+2)7rs s]

 

( 71(0) : 0,710) = 0 for some Cijk E R.

and the ﬁrst order solution is

 

 

M M M 1 1
ZZCMLL 12 2cos(i+j—k+-—)7rs
i:0j:0k-0 (i+j— k+§)7r 2
+ 1 cos(i '+k+1)7r
,, . — — s
p — (1 - j + k + 52712 J 2 (3.25)
1
+

 

1
(,cs —i+ '+k+-7rs
P-(—i+j+k+%)27r2 0( J 2)

+ 1 cos(i+ '+k+§)7rs]
P—(z‘+j+k+:§)‘2vr2 J 2 "

29

The lateral and the vertical displacements are obtained from equation (3.16) and

(3.13).

This general sine case is very useful when the angle at the ﬁxed end is zero; i.e.,
00(1) = 0. This means that the angle of the rod at bottom end does not change during
deformation. From equation (1.5), 316(1) 2 sin 60(1) 2 0. Since 310(3) = 60 f(s) and
y0(0) = O, f (0) = 0. Hence f (9) satisﬁes these two boundary conditions: f (0) = O
and f’(1) = 0. Thus we can express f(s) = :0: an sin(n+ %)7rs. Therefore we can use

n=0

our general sine case to approximate the general imperfection curve with zero angle

at the ﬁxed end.

3.2.2 The Second Case

case 2) Consider that the initial curve is in the form of a cosine curve described by
1
f(s)=1—cos(n+ §)7rs for n = 0,1,2,3,....

The zeroth order equation (3.9) with boundary condition (3.11) becomes

76' + P70 = —P(n + %)7r sin(n + %)7rs
76(0) = 0, 70(1) = 0

and the zeroth order solution is

I I
70(3) = (n + §)7rc (sin(n + §)7rs + a sin VPs + b cos VPs) (3.26)
l 1
Where a = ——._ 71+ — 71,
1 P( I 2)
()2 ——P(n+§)7rtan\/P—seC\/P,

30

—P
P — (n + @2712.
The ﬁrst order equation (3.10) with the zeroth order solution (3.26) becomes

 

C:

1 3 1 3
71' + P71: P (n + §)37r3 [CE (sin(n + §)7rs + asin \/Ps + bcos \/P3)

2

1 2 1
+ 9— sin(n + —)7rs + asin \/Ps + bcos \fP—s sin(n + —)7rs
2 2 2 (3 27)

, 1
+ (sin(n + §)7rs + asin \/Ps + bcos \/Ps)

(sin(n 71%)“)?

The right hand side of equation (3.27) is composed by products of trigonometric

[\DI’B

functions. By the harmonic properties of trigonometric functions, we are able to
rewrite equation (3.24) as following:
II 1 3 3 - 1 - 1
71+ P71: P(n + 5) 71 (C1 sm(n + §)7TS + Cg sm3(n + §)7rs
+ Cg cos \/Ps + 0.; sin \/Ps + C5 cos 3\/Ps + C6 sin 3\/Ps
+ c7( sin(\/P — (2n +1)7r)s + sin(\/P + (2n +1)7rs))
(3.28)
+ Cg(cos(\/P — (2n +1)7r)s + cos(\/P + (2n +1)7r)s)
1 1
+ Cg(sin(2\/P — (n + §)7r)s + sin(2\/P + (n + §)7r)s)
1 1
+ C10( cos(2\/P — (n + §)7r)s + cos(2\/P + (n + §)7r)s) )

31

where

 

 

 

 

C __ 3c + 3c2 a2c2 + b2c2 + c3 + a2c3 b2c3
1 " 8 8 4 4 8 4 4
02 _ _2 _ 22. _ £3
8 8 24
C _ bc be2 + be3 a2bc3 + b3c3
3 — 4 2 4 8 8
C _a_c 322 85:3 ﬁ+ab2c3
4 — 4 2 4 8 8
C5 : _a2bc3 b_3_cj
8 24
a3c3 ab2c3
CS _ _ 24 8
3
ac ac ac
C __ _ _ _ _
7 8 4 8
c. _8 _ 2c: _ 95
8 4 8
a2c2 b202 a2c3 b2c3
c9 .. —— — — — — —
8 8 8
abc2 abc
CID — —4 _—

32

Hence the ﬁrst order solution is

13 3 Cl
71(3) =P(n+§) 7r [P_ (WWW

 

sin(n + l)7r8 + C2
2 2 P — 9(n + é)???

 

1
sin 3(n. + -2-)7rs

+ —C§—s sin P3 + 31—3 cos P8 — 30—15) cos 3\/Ps — 80—1: sin 3\/Ps

 

 

 

 

 

 

 

 

2W3 N13

+ 1 - (x/P‘ 3727; + 1))21r2 Sin“? _ (2” + ”723

+1-(\/P —C(82n +1))27r2 sin(\/P _ (2n + 1)”)3

+1_(\/p _Cén +1))27r2 COS(\/'I3 + (2n +1)7r)s

+1w—‘i‘énm2... cosW+<2n+1>w>s 429)
+ 14min + 4M2 SW“? ‘ (n + é—ms

+ 1 — (NF fin + 8%? mm"? + (n + i)”

+ 1 — (NF €12). + hm “(N-’3‘“ (" + 3W

+ 1 — Wit + W <n + w

+ A sin \/Ps + B cos \/Ps.

A and B are determined from boundary conditions: 71(0) = 0, 71(1) = 0. To obtain

the lateral and the vertical displacements, we can use equations (3.16) and (3.18).

3.2.3 The Third Case

Case 3) Consider that the initial curve is in the form of an exponential function

described by

f(s) = e‘” — 1 for a 6 IR.

33

Then the zeroth order equation (3.9) with boundary condition (3.11) becomes

76' + P70 : —Pae"‘s
76(0) = 0, ”70(1): 0

and the zeroth order solution is
70(3) 2 ca (603 + a sin \/Ps + b cos \/P8) (3.30)

a a —P
wherea=——, bz—tanVP—aeaseCVP, c: .
x/P x/P P—02

The ﬁrst order equation (3.10) with the zeroth order solution (3.30) is

 

3
71' + 13,1: p83 [688203 + a sin mes + b W083

+ g(e‘” + a sin \/Ps + b cos \/Ps)2e°s (3-31)

+ g(e‘” + a sin x/Ps + b cos \/Ps)e2as].

The right hand side of equation (3.31) is composed by products of exponential func-
tions and trigonometric functions. With the aid of the harmonic properties of trigono-
metric functions and the fact that cos \/P8 and sin x/Ps are the real and the imaginary

parts of 6"”35 respectively, we can rewrite equation (3.31) as follows
71' + P 71 = P 013(0163” + Cg Im[e(2“+‘/Fi)s] + 03 Re[e(2“+‘/—Pi)s]
+ C4 8‘” + C5 Im[e(a+2‘/Pi)5] + C6 Re[e(a+2‘/ﬁ)s]
(3.32)
+ C7 cos ﬁs + C8 cos 3\/Ps

+ C9 sin \/P8 + Clo sin 3\/Ps)

34

where

 

 

 

 

c +c2 +c3
c =— — —
1 2+ 2+ 6
3
be be3
c =— bc2 _—
2 + + 2
C_a22c2c2+b +a2c3 (22c3
4‘ 4 4 4 4
C __a2_c_2__*_abc2 a2c3+abc3
5- 4 2 4 2
We2 bzc3
C =—— ——
6 4 + 4
a21)c3 b3c3
C :
7 8 + 8
a21)c3 b303
C __ __
8 8 + 24
a3c3 abzc3
C =—
9 8 + 8
a363 ab2c3
C _—____
1° 24 + 8

Note : Re[ f] 2 real part of f and I m[ f] = Imaginary part of f.
Hence the ﬁrst order solution with boundary condition (3.12) is

C1 1 -
= P 03 (_e3as + C [m I: e(2a+\/Pz)s
P + 90:2 2 p + (20 + @02

1 . C4
+ C Re 8(2a+\/;)8:| + ——e‘”
3 lP+ (2a+ \sz')? P+a2

1 .
+ C [m 8(a+2\/Pz)s]
5 [P + (a + 2\/-Pz')2

1 .
+ C R8 e(a+2\/Pz)s]
6 [P + (a + 2\/Pi)2

 

 

(3.33)

 

+

 

 

C7 . C8 C9 C__1_0
3 Sin P8 — cos 3v P3 + 9 cos P8 — sin 3VP3
2\/P 8P 2\/P 8P )

+Asinx/Ps+Bc08\/Ps

where A and B are determined by 71(0) 2 0 and 71(1) 2 0. Once again, the lateral

and the vertical displacements are obtained from equation (3.16) and (3.18).

35

CHAPTER 4

The numerical solution

4. 1 Introduction

In section 4.2, the numerical integration method by “MATHEMATICA”, which is
based on Runge-Kutta’s method, is described. Numerical solutions of the boundary
value problem which was set up in section 1.3 are presented in the choice of a sine, a
cosine and an exponential initial imperfect curve with various amplitudes in section
4.3. Also, the comparison between the numerical solutions and the analytic solutions,
which we found in section 3.2, is in section 4.3. The results for the multiply peri-
odically curved rod are particularly important since it models springs, textiles and

biological vessels such as varicose veins.

4.2 Numerical integration

In order to ascertain the range of validity of our expansions obtained in chapter 3, we

integrate equation (1.5) for the exact solution numerically. For simplicity, equation

36

(1.5) with our new variable 7 = 6 — 00 becomes

d7

a — ‘P y

d .

a? = s1n(7+90) (4-1)
d3:

3 = COS(’)’ 'l' 90)

where 60 = sin'l(60 f (3)) The boundary conditions are

Now given P and y0(s) (or 60(3)), we guess 7(0) and integrate equations (4.1) with
(4.2) as the system of initial value problems numerically by the Runge-Kutta method
(MATHEMATICA). Then we check whether 7(1) 2 0 or not. If so, a solution is
obtained. If not, 7(0) is adjusted and the procedure is repeated. The accuracy is con-
trolled by built-in function “Accuracy Goal” in MATHEMATICA (See Mathematica
Book, Stephen Wolfram, 1996). It is controlled by adjusting the step size. In this
paper, we used the default accuracy 10—8. After obtaining the correct 7(0) 2 B, the
equations (4.1) and (4.2) can be integrated in terms of s for the shape of the deformed
rod. We note that since y is bounded (in fact, Iyl < 2), if P is large, 8—: becomes

singular and this stiffness makes 7(0) difﬁcult to shoot.

37

4.3 Numerical results

In section 4.3.1, the deﬂections of the initially curved rod under compressive loads
are studied numerically for various initial amplitude 60 when the initial curve is in
the form of a quarter sine curve, a quarter cosine curve and an exponential curve
studied in Chapter 3. The multiply periodically curved rods under compressive loads
and tensile loads are studied in section 4.3.2 and 4.3.3. The boundary value problem
is solved numerically by the shooting method (see section 4.2) using “MATHEMAT-
ICA”, except when the initial amplitude 60 is small. In that case, the problem can

also be determined from the analytic perturbation solution found in Chapter 3.

4.3.1 Simple initial curve cases

Figure 4.1-3 show the equilibrium shapes of the initially curved rod under various
compressive loads P when the initial imperfect curve is in the form of a quarter
sine (yo = 60 sin gs), a quarter cosine (yo 2 60(1 — cos§s)) and an exponential
(yo = (50(80'3 — 1)) curve respectively. We notice the rod bends immediately after a
small compressive load is applied (Figure 4.1-3(a)) in all cases. We can expect this
because y E 0 is not a solution of our boundary valued problem; thus not an Euler
buckling problem which is an eigenvalue problem with discrete, ﬁnite bucking loads.

For large compressive loads (Figure 4.1-3 (b),(c)), our numerical results show multiple

equilibrium solutions.

38

(a) 60 = 0.1, P = 2 (b) 60 = 0.1, P = 3.21

Figure 4 1 Column conﬁgurations when the initial imperfection curve 18

yo = 60 sin 22:8.
The dashed curve is the initial curve (when P — 0)

v .1

(3)60201, P26

-
------

(a)60=0.1,P=2 b)60:0.,1P.—_3.04

Figure 4 2 Column conﬁgurations when the initial imperfection curve is

yo 2 60(1— cos gs).
The dashed curve is the initial curve (when P —— 0)

31

c)60=0.1, P26

~“_--

 

(a)60=0.1,P=2 b)60=0.,1P=3.37
Figure 4 3 Column conﬁgurations when the initial imperfection curve 18

yo = 60 (808 —1),a = 1.
The dashed curve is the initial curve (when P — 0)

39

The best way to display these solutions is a load—lateral displacement curve. Figures
4.4-6 show our numerical results of compressive load P versus the lateral displacement
6 at the free end. In Figures 4.4-6, the curve with zero amplitude (60 = 0), shown
by the dotted line, is obtained from the Euler beam data in S. P. Timoshenko and
J. H. Gere (1961) [12] or from more accurate data in C. Y. Wang and L. T. Watson
(1980) [38]. The shape is a pitchfork and the bifurcation occurs at P = ”72 which
is the critical buckling load. However for 60 75 0, the load-displacement curve is no

longer pitchfork shape and no such bifurcation exists. We see, with initial curvature,

the pitchfork degenerates into upper and lower branches.

Figures 4.4-6 also show the perturbation solutions obtained in section 3.2 for a
quarter sine, a quarter cosine and an exponential case respectively, using the am-
plitude 60 as a parameter. Our approximate solutions, in their respective regions of

validity, are compared quite well with the exact numerical solutions.

Although there is no bifurcation when 60 ¢ 0, in Figures 4.4-6 we can notice that
there is only one solution when the load P is less then PM. PM is deﬁned as the
location where the slope of the curve is vertical occurring on the lower curve. For
P > PM, multiple equilibrium states exist. Let us denote PM as the maximum load

for the unique solution.(When 60 = 0, PM is the Euler critical bucking load.)

40

 

 

 

 

 

 

 

 

 

Figure 4.4: Analytic solutions for a sine function case and load-displacement curves
when 60 = 0.01, 0.05, 0.1 from the horizontal axis.

— — — — : Euler buckling curve
-------- : Perturbation solution

Initial curve : yo = 60 sin gs.

41

 

 

 

0.75}

 

 

 

 

 

-0.75L

 

 

 

Figure 4.5: Analytic solutions for a cosine function case and load-displacement curves
when 60 = 0.01, 0.05, 0.1 from the horizontal axis.

—— — — — : Euler buckling curve
-------- : Perturbation solution

Initial curve : yo = 60(1 — cos gs).

42

 

 

 

 

 

 

 

 

-—O.75} ‘xs

 

 

 

Figure 4.6: Analytic solutions for an exponential function case and load-displacement
curves when 60 = 0.01, 0.05, 0.1 from the horizontal axis.

— — —-— —— : Euler buckling curve
-------- : Perturbation solution

Initial curve : yo 2 60(6"8 — 1), a = 1.

43

In Figures 4.7-9, we show the relation between the maximum load for the unique

solution PM and the initial amplitude 6. For the sine function case, the maximum

load for the unique solution PM with 60 = 0.1 is 3.21. Notice that this value is larger
2

than the Euler critical buckling load, Z— x 2.467 (when 60 = 0). In general, the

larger the initial imperfection, the larger the maximum load PM for uniqueness.

 

PM
.15

U)

 

maximum load,
N

H

 

 

 

0L1 0i2 3 03 0.4 0.5
initial amplitude, 60

Figure 4.7: Maximum load PM for unique solution vs Initial amplitude 60
when 310(3) 2 60 sin gs.

Figures 4.7-9 look similar because the initial curves are similar in shape. However,
there are differences as illustrated in the maximum load PM in Table 4.1.

When the load P is larger than PM (the maximum load of the unique solution),
there are several solutions. However, not all solutions are stable or can be realized in
practice. Since the work done is deﬁned by the product of the load and the vertical

displacement in section 1.5, the load-vertical displacement curve is needed to study

44

PM
.5

maximum load,

PM
Mb

maximum load,

 

N (A)

H

 

 

 

0:

Figure 4.8: Maximum load PM for unique solution vs Initial amplitude 60

1‘

0:2 ii 03
initial amplitude,

when 110(3) 2 60(1— cos 12‘s).

50

0

.4

 

l\) W

[_3

 

4m

 

0.

l

03 03
initial amplitude,

60

O

.4

 

 

Figure 4.9: Maximum load PM for unique solution vs Initial amplitude 60
when y0(s) 2 60(6‘” — 1), a = 0.5,1,1.5 from bottom to top.

45'

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

60 0.01 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
PM,sine 2.61 2.91 3.2 3.45 3.69 3.92 4.14 4.36 4.6 4.83 5.07
PM,cosine 2.58 2.8 3.03 3.25 3.46 3.68 3.91 4.16 4.43 4.73 5.07
PM,exp(a = 0.5) 2.55 2.74 2.92 3.07 3.22 3.36 3.49 3.63 3.76 3.9 4.03
PM,exp(a = 1) 2.63 2.99 3.36 3.71 4.07 4.46 4.88 4.89
PM,exp(a = 1.5) 2.73 3.33 4.04

Table 4.1: Maximum loads,PM, for the unique solution

the stability of solutions.

Figures 4.10-12 show the relation between the load and the vertical displacement
A for a quarter sine, a quarter cosine and an exponential initial curve case respectively
with the initial amplitude 60 = 0.01, 0.05 and 0.1. For example, in Figure 4.10 with
6 = 0.1, the state C denotes the location where the slope is vertical and the load
is PM, separating the unique solution for P < PM from three possible equilibrium
solutions for P > PM. Since the area under the P — A curve represents the work done
on the elastica, a negative slope signiﬁes negative work for a positive displacement
increment. Thus the equilibrium states on the segment CE in Figures 4.10-12 are
statically unstable for a given constant load. For P < PM, there is only one solution

and it is stable. For PM < P < 6, there are three solutions: two stable (curves AB

and CD) and one unstable (curve CE).

46

 

0.8-

0.6-

0.4-

0.2-

 

 

_O.2_

 

Figure 4.10: Load-vertical displacement curve for a sine function
when 60 = 0.01, 0.05, 0.1 from the horizontal axis

47

0.8}
0.6}
0.4}

0.2}

 

 

 

-0.2;

 

Figure 4.11: Load-vertical displacement curve for a cosine function
when 60 = 0.01, 0.05, 0.1 from the horizontal axis

48

0-8} 0

0.61 »

1 j v v r

0.4 /

0.2 II

 

-0.2:

Figure 4.12: Load-vertical displacement curve for an exponential function
when 60 = 0.01, 0.05, 0.1 from the horizontal axis

49

4.3.2 General case of initial angle

In the previous section, we discussed the deformations of the initial curved rod when

the initial curve is in the form of a quarter sine curve described by
, 7r
310(5) = <50 513(53)-
In this section, we consider the initial curve of multiple waves such as
, 1
310(3) 2 60 Sin(n + g)“ for n 75 0.

From equation (3.4), the initial angle is 90 = arcsin(60 (n + %)7r cos(n + %)7rs). Hence,

when n is large, our choice of 60 is limited because of the domain of an arcsine function.

In fact, 0 < 60 < (7731')? To overcome this limitation, in this section we will consider
2

the initial angle (instead of lateral displacement y) between the tangent to the initial

curve and the positive x-axis described by

where 9(3) 2 0(1), 60 is the amplitude and s is the normalized arc length from the

top end. The initial curve can be obtained by integrating equation (1.5)

310(8) 2 [08 sin 60(3) ds.

Equations (4.1)-(4.3) are valid with the initial angle, 00(3) = 6:) 9(3), so the numerical
procedure will be the same in section 4.2. We consider this general initial angle cases

when g(s) is in the form of a sine curve and a cosine curve described by

00(3) 2 (i) sin "—2”— 3 and (90(8) 2 6:) cos % s.

(a)60:1,P-—-2 (b)£5},=1,P=3.73 (as; 1,P—6

   

Figure 4.13: Column conﬁgurations when the initial angle is
~ mr
00 = 60 sin —8, n = 21.

The dashed curve is the initial conﬁguration when P = 0.
The continuous curves are solutions when P > 0.

     

(b) 6}, = 2, P = 12.13

Figure 4.14: Column conﬁgurations when the initial angle is
~ 72.
00 = 60 sin —7rs, n = 21.

The dashed curve is the initial conﬁguration when P = 0.
The continuous curves are solutions when P > 0.

51

   

=1,P=3.2 (b)5},=i,P=3.3i (0)60=I,P=6

Figure 4.15: Column conﬁgurations when the initial angle is
~ mr
00 = 60 COS —S, n = 20.
The dashed curve is the initial conﬁguration when P = 0.
The continuous curves are solutions when P > 0.

     

(b) 5}, = 2, P = 15.15

Figure 4.16: Column conﬁgurations when the initial angle is
~ mr
00 = (50 COS —8, Tl 1' 20.
The dashed curve is the initial conﬁguration when P = 0.
The continuous curves are solutions when P > 0.

Figures 4.13-6 show the equilibrium shapes of the deformed rod with various

compressive loads P. Notice that the large curvature cases, Figures 4.14 and 4.16
can not be described by the lateral displacement y which would give multiple values.
While in the sine function case, the column bends immediately as in section 4.3.1

after a small compressive load P is applied (Figure 4.13-4(a)), in the cosine function

52

case, the column hardly bends (Figure 4.15-6(a)). Notice P is larger in Figure 4.15(a)
than in Figure 4.13(a). Even when P = 2, there is hardly any deformation for the

cosine function case.

For large compressive load P, our numerical results show multiple equilibrium

solutions in both sine and cosine initial angle cases.

As we mentioned in previous section, the best way to display these solutions
is a load-lateral displacement curve. Figures 4.17-20 show our numerical results of
compressive load P versus the lateral displacement 6 at the free end. We see that the

increase in the initial amplitude 67, suppresses the lateral displacement.

 

0.751

 

 

 

-O.75:

 

 

min mm m__k mm L A m

 

312 3 4 5J6
P

Figure 4.17: Load-lateral displacement curve for a sine function

when 00:6?)sinn3’5,6;)=1andn=21.

 

0.75*

0.25’

 

 

"0.25L 1
-O.5*
"0.75L

 

 

L L n x J A 1 L L m

2.5 5 7.5 10 12.5 15 17.5 20
P

 

Figure 4.18: Load-lateral displacement curve for a sine function

when 00 =6:)sin12’—’,6B=2andn=21.

 

0.75*
0.5“
0.254

 

 

Am

-O.25*

LJ.

-0.5:
—O.75*

 

 

i 2 43‘ 4 5 6
P

 

Figure 4.19: Load-lateral displacement curve for a cosine function

when 0026;)cos’l2—I,(5~()=landn=20.

54

 

0.75*
0.5*
0.25:

PMCM

 

-0.25~
-0.5~
—0.75L

 

 

2 4‘ ‘8 88160 142 14
P

 

Figure 4.20: Load-lateral displacement curve for a cosine function

when 60:6?)cos%,63=2andn=20.

Notice that maximum load for the unique solution PM (deﬁned in section 4.3.1) is
also an increasing function of amplitude 6:) of the initial angle. Since 6;) = 2 is a quite
large amplitude of the initial angle, corresponding PM = 15 is also large compared to

the Euler critical bucking load 142— when 60 = 0.

Figures 4.21-24 show relation between the load and the vertical displacement for
a sine and a cosine initial angle case with various amplitudes 6;). As discussed in
section 4.3.1, load-vertical displacement curves with negative slope are unstable to
given load, but are stable to given displacement. Hence for P < PM, there is only
one stable solution and for PM < P, there are two stable solutions and one unstable

solution.

55

O.75~

0.5

0.25’

 

—0.25-
—0.5~
—0.75»

0.75*
0.5
0.25L

-0.25*
-O.5:
-O.75*

 

 

Figure 4.21: Load-vertical displacement curve for a sine function

when 60 = 6;) sin "—2”, 6:): 1 and n = 21.

Figure 4.22: Load-vertical displacement curve for a sine function

when 90 =6()sin"—2”,6:) = 2andn= 21.

56

0.75:
0.5:
0.25L

 

-O.25:
—O.5
-O.75

0.75:
0.5»
0.25E

—0.25L
—0.5-
—0.75~

 

 

Figure 4.23: Load-vertical displacement curve for a cosine function

when 60:6?)cosn7",60=1andn=20.

Figure 4.24: Load-vertical displacement curve for a cosine function

when 00:6:Jcosn—‘2”—,6B=2andn:20.

4.3.3 Deformation of initial curved rod under tensile loads

(pulling)

We’ve studied the deformations of initial curved rods under compression loads. How-
ever, the deformation of sinusoidal shaped elastic rods or ﬁlaments under tensile end
load is also of interest because wavy ﬁlaments such as polymer tapes or thin ﬁlms
often act as reinforcing components. In this section, we investigate the deformations
of wavy elastic rods under tensile loads when the initial curvature is in the form of a
sine and a cosine curve described by 00(3) = 60 sin "—253 and 00(8) 2 6:) cos $3 . Since
we did not restrict the sign of P in the previous chapters, the governing equations, the
existence of solution and the numerical procedure are valid with negative loads which

indicate tensile loads. As we mentioned in section 4.2, the numerical results for large

P is difﬁcult to achieve. Therefore, we restrict our investigation to —150 g P S 0.

Figures 4.25-8 show the equilibrium shapes of the deformed rod with various
tensile loads P. We see that there is only one solution in all cases. This is expected
in practice.

Figures 4.29 and 4.30 show the load-vertical displacements of a sine and a cosine
case respectively with different periods. Since the slope of load-vertical displacement
curves are positive, these unique solutions are stable. We see that an increase in

period 72 makes the curved rod more difﬁcult to stretch.

58

   

(c) 60 = 2, P =-150

Figure 4.25: Column conﬁgurations when the initial curvature is
~ , mr
60 = 60 Sin ——3, n = 5.
The dashed curve is the initial conﬁguration when P = 0.

     

(b) 60:2,Pz-50 (c){sI,=2,P=-150
Figure 4.26: Column conﬁgurations when the initial curvature is

00 = 6:) sin PIS, n = 21.

The dashed curve is the initial conﬁguration when P = 0.

59

 

Figure 4.27: Column conﬁgurations when the initial curvature is
~ mr
90 = 60 cos—s, n = 4.

The dashed curve is the initial conﬁguration when P = 0.

Figure 4.28: Column conﬁgurations when the initial curvature is

 

(a) £5}, = 2, 10 (b) 8;, = 2, 50 (c) 5",, = 2, P = -150

00 = 6:) cos 17:3, n = 20.

The dashed curve is the initial conﬁguration when P = 0.

60

0.75:

 

 

 

 

 

 

£140 —120 —100 —80 L80‘ —4oi —20 0

I?

Figure 4.29: Load-vertical displacement curve for a sine function

0.75:

when 00:6;)sinE2-’5,6;)=2andn=5, 21.

dotted line is for n = 5, continuous line is for n = 21.

 

 

 

 

 

 

—140 —120 —1004 —80 —80 ~40‘ L20 0

P

Figure 4.30: Load-vertical displacement curve for a cosine function

when 00:6?)cosn7",6:,=2andn=4, 20.

dotted line is for n = 4, continuous line is for n = 20.

61

The following tables are the numerical results.

These tables are useful in the design of curved rods.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

[P 14(1) 1 14(1) 1
0 0.099997 0
0.25 0.111137 0.00596476
0.5 0.125045 0.00801437
0.75 0.142869 0.110027
1 0.166339 0.155673
1.25 0.198492 0.023002
1.5 0.244352 0.0360444
1.75 0.311642 0.0606765
2 0.407192 0.108193
2.3 0.54364 0.208331
2.5 0.624229 0.293291
2.7 0.686479 0.380751
3 0.748182 0.503721
3.1 0.762062 0.541328
3.21 -0.518811 -0.433703 0.774366 0.186288 0.124149 0.580689
3.3 -0.619304 -0.336942 0.782396 0.286569 0.0716537 0.611238
3.4 -0.671901 -0.286876 0.78953 0.356697 0.0507584 0.643607
3.5 -0.707015 -0.25289 0.795012 0.414373 0.03837523 0.674309
3.7 -0.751954 -0.20685 0.801961 0.510099 0.02513 0.731121
4 -0.787236 -0. 163956 0.804966 0.62472 0.0150489 0.806181
4.5 -0.806782 -0.122593 0.797845 0.769868 0.00762038 0.90927
5 -0.804339 -0.098113 0.782796 0.879179 0.00425761 0.991434
6 -0.775858 -0.0702033 0.744498 1.034703 0.00134464 1.113803

 

 

 

 

 

 

 

 

Table 4.2: Load displacements when 310(3) 2 60 sin 13, 60 = 0.1.

62

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

[P 1 41(1) l A(1)
0 0.1 0
0.5 0.116576 0.00154697
1 0.144021 0.00479434
1.25 0.165744 0.00800278
1.5 0.197826 0.0138167
1.75 0.249032 0.02586
2 0.336356 0.0547535
2.5 0.615264 0.238776
2.75 0.717279 0.364962
3 0.780539 0.481402
3.04 —0.394594 -0.345954 0.787969 0.125666 0.0968099 0.498758
3.1 -0.495216 -0.255141 0.797921 0.202278 0.0543911 0.523992
3.3 -0.613484 -0.161569 0.822877 0.335128 0.024172 0.602309
3.5 -0.672473 -0.117486 0.838075 0.43317 0.0142216 0.672201
3.75 -0.714758 -0.084393 0.847549 0.533213 0.00848242 0.749025
4 -0.738364 -0.0628129 0.849885 0.616581 0.00554114 0.816022
5 -0.755587 -0.0194278 0.825016 0.851704 0.00166677 1.01386
6 -0.731399 -0.000157625 0.782616 0.999285 0.0010355 1.1421

 

 

 

 

 

 

 

 

63

Table 4.3: Load displacements when y0(3) 2 60(1 — cos g8), 60 = 0.1.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(P MI) I I 4(1) l
0 0.171828 0
0.5 0.20244 0.000572008
1 0.252288 0.00402889
1.5 0.344325 0.0191823
1.75 0.42046 0.0411918
2 0.521468 0.0860133
2.25 0.631435 0.16125
2.5 0.723825 0.25734
2.75 0.788245 0.36759
3 0.828707 0.452285
3.25 0.851942 0.538037
3.37 -0.456431 -0.391681 0.858662 0.270604 0.212295 0.575893
3.4 -0.505185 -0.343627 0.859994 0.322238 0.175222 0.585067
3.5 -0.576808 -0.273182 0.863442 0.414475 0.128883 0.614527
3.75 -0.65853 -0.189956 0.867016 0.559764 0.0847637 0.682426
4 -0.698438 -0. 1438 0.865262 0.665073 0.0647537 0.742774
4.5 -0.731492 -0.0901078 0.85216 0.820543 0.0452444 0.844612
5 -0.737058 -0.058681 0.832475 0.933568 0.035691 0.926849
6 -0.718657 -0.0222741 0.787301 1.09006 0.0264628 1.05102

 

 

 

 

 

 

 

 

Table 4.4: Load displacements when y0(s) 2 60(1 — e”), 60 = 0.1, a = 1.

64

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

IP 1.1/(1) [4(1) ]
0 0.0270785 0.00000221297
0.5 0.0328974 0.000606045
1 0.0413892 0.00147322
1.5 0.0548153 0.00293975

2 0.0789926 0.00612161
2.5 0.132075 0.0162327
2.75 0.185735 0.0313352

3 0.268554 0.0654737
3.25 0.362912 0.123778

3.5 0.441086 0.193654
3.73 0334192 0.279615 0.493698 0.0936539 0.0636385 0.257702
3.75 -0366966 0.249876 0.497511 0.115357 0.0501395 0.263108

3.8 0410757 0.213253 0.506572 0.149441 0.0360215 0.276475

4 -0.498052 0.149203 0.536751 0.240862 0.0174799 0.32757

4.5 -0.58619 00920447 0.581996 0.396619 0.0071822 0.438827

5 0618006 0.0685928 0.60143 0.508969 000470395 0.529007

5.5 -0627822 -00557563 0.607149 0.596098 0.00386266 0.602742

6 -0626949 00476778 0.605261 0.666059 000358386 0.663816

 

65

n17
2

Table 4.5: Load displacements when 00(3) 2 60 sin ——3, 60 = 1, n = 21.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

[P l ly(1) 1 14(1)

0 0.0000000856317 0.000002607

0.5 -0.000390103 0.0001991

1 -0.000931541 0.0003967

1.5 -0.00171249 0.0005962

2 -0.0031412 0.0008014

2.5 -0.00642223 0.0010306

3 -0.0235675 0.001679

3.1 -0.0420512 0.0027374

3.2 -0.112005 0.0116957

3.31 -0.239978 0.0882 0.153655 0.0501359 0.00741319 0.0203006
3.35 -0.2754 0.0511843 0.226116 0.0664318 0.00332276 0.0435184
3.5 -0.372278 0.0226601 0.351446 0.126474 0.00174278 0.109911
3.75 -0.468444 0.0123533 0.45793 0.216787 0.0015731 0.203005
4 -0.525738 0.00871026 0.518799 0.295198 0.00161789 0.282327
4.5 -0.585713 0.00569711 0.58172 0.423472 0.00178714 0.411117
5 -0.610172 0.00433319 0.60746 0.523599 0.00197772 0.511191
5.5 -0.61765 0.00350914 0.615651 0.603832 0.00217275 0.591175
6 -0.616177 0.00299318 0.614628 0.669387 0.00237011 0.656398

 

 

Table 4.6: Load displacements when 60(8) = 60 cos $8, 60 = 1, n = 20.

66

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

P 19(1) 1 14(1) 1
0 0.0376598 0000000102097
2 0.0472483 000465041
4 0.0612113 0.011732
6 0.0814979 0.0237667
7 0.0942056 0.0328486
8 0.107963 0.0444768
9 0.121661 0.0584913
10 0.134124 0.0741878
11 0.144558 0.0905783
12 0.152757 0.106835
13 0.158878 0.122406
14 0.163252 0.137027
15 0.166206 0.150571
15.15 -0.158357 -0.150112 0.166547 0.0751958 0.0658936 0.152507
15.2 -0164617 0.143379 0.166654 0.0832148 0.0591782 0.153151
15.5 -0.178154 -0.127128 0.167247 0.104496 0.0456053 0.156937
16 -0.188176 -0.112711 0.16805 0.126018 0.0361411 0.163051
17 0197243 00953085 0.169039 0.156237 0.0274797 0.174527
18 -0200743 00840418 0.169373 0.17905 0.0234182 0.18508
20 -0201107 -00696276 0.168663 0.213417 0.0201788 0.203777

 

 

 

 

 

 

 

 

7177

Table 4.7: Load displacements when 00(5) 2 60 sin —2—5, 60 = 2, n = 21.

67

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

P I [90) 1 [4(1)

0 0000000045 000000018

0.5 000013167 0000338969

1 000026679 0000677929

1.5 000042378 000101725

2 000059509 0.001357

2.5 000077952 000169733

3 000099584 000203808

3.5 000122239 000237906

4 000147728 000272083

5 000210021 000340865

6 0.00293151 000410302

7 000412537 000481265

8 000598397 000555691

9 000936732 000640776

10 00174414 000775724

11 0.0488542 0.0145749

12 0105038 0.0425052

12.13 0110456 0.0498348 0.06324 0.0466361 0.0146401 0.0188897
12.15 0111254 0.0459384 0.0679607 0.047272 0.0136305 0.0206801
12.5 012371 0.0269527 0.0993749 0.0582093 0.0101732 0.0369864
13 0137571 0.0185487 0.12161 0.0730901 000953526 0.0542628
14 0156188 0.0119373 0.146773 0.0999112 000971974 0.0829002
15 0.167408 0.009026 0.16085 0.123109 0.0102547 0.106742

 

 

 

 

 

 

 

Table 4.8: Load displacements when 60(8) 2 60 cos g3, 60 = 2, n = 20.

68

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

[P [7125, y(1) [ A(1) [71:21, y(1) [ A(1) [
0 0.128172 0.000000145393 0.0376599 0.000000213969
—5 0.0964071 -0.10609 0.0238376 -0.00742142
-10 0.0693823 -0.172058 0.0168315 -0.0126775
-20 0.0566073 -0.271057 0.0102534 -0.02l3119
-30 0.0589025 -0.34274 0.0074728 -0.0290314
-40 0.0637537 -0.395796 0.00616602 -0.0363371
-50 0.0682371 -0.436672 0.00555063 -0.0433939
-60 0.0719733 -0.469098 0.00529626 -0.0502804
-70 0.0750453 -0.495296 0.00524272 ~0.0570417
-80 0.0771536 -0.517592 0.00530883 ~0.0637063
-90 0.0790197 -0.535877 0.00544222 -0.0702941
-100 0.0798021 -0.552871 0.00562379 -0.0768172
-150 0.0810182 -0.609828 0.00673994 -0.108718

 

 

 

Table 4.9: Tensile Load displacements when 00(3) 2 60 sin '12—”3, 60 = 2, n = 5, 21.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

[P |n=4, y(1) | A(1) [ n=20, y(1) 1 A(1) 1
0 0.0000000184059 0.0000000437453 -0.000000000873903 0.000000220194
-5 0.0280973 -0.0799036 0.000955954 -0.00338388
-10 0.0471055 -0.151637 0.0015669 -0.00676472
-20 0.0695157 -0.270606 0.00240398 -0.0135271
-30 0.0825142 -0.35777 0.00302517 -0.020293
-40 0.0906626 -0.421112 0.00354258 -0.0270623
-50 0.0956144 -0.468635 0.00400848 -0.0338329
-60 0.0987545 -0.504841 0.00442885 -0.0406042
-70 0.100585 -0.533291 0.00481973 -0.047374
--80 0.101577 -0.556022 0.00520272 -0.054138
-90 0.101252 —0.576191 0.00554701 -0.0606899
-100 0.101617 -0.590688 0.0058877 -0.0676502
-150 0.0927302 -0.651513 0.00739026 -0.101215

Table 4.10: Tensile Load displacements when 60(3) 2 60 cos "f3, 60 = 2, n = 4, 20.

69

CHAPTER 5

Conclusions and Discussions

In this chapter, previous works are discussed and summarized.

In chapter 1, the governing ordinary differential equation was formulated using
the balance of normal and tangential stresses. This provided the non linear boundary
value problem for the deformation of a clamped-free elastic curved rod. Hamilton’s
principle yielded a variational formulation which was used to prove the existence of

equilibrium states in chapter 2.

In chapter 2, the existence of equilibria were proved using the variational formula—
tion obtained in chapter 1. In section 2.4, the upper bound of the maximal deflection
was obtained and our result agreed with the work of T. M. Atanackovie (1981) [1]

when the initial out-of—straightness 60 is zero (originally straight rod).

In chapter 3, an analytic solution for the deformation was obtained using pertur-
bation and when the initial imperfect curve is in the form of a sine, a cosine and an
exponential function, the explicit expression of the solution was derived. The method

Can be applied when the initial curvature is given. Also, the procedure can be easily

70

modiﬁed for different boundary conditions; for example, pined-pined case, ﬁxed-ﬁxed

case, etc.

In chapter 4, the boundary valued problem formulated in chapter 1 was solved
using the numerical tool “MAT HEMATICA”. The shooting method was used to
ﬁnd exact numerical solutions. For the variety of choices of functions for the initial
curve and curvature along with various initial amplitude 60 and period 17., stable
and unstable equilibrium status were calculated. Perturbation in chapter 3 and the
numerical results in chapter 4 showed the theoretical development for existence of

equilibria in chapter 2 was appropriate.

We studied the deﬂection of clamped-free elastic curved rod which has a general
natural curvature. The behavior of an initially curved rod is quite different from the
Euler buckling rod which is an initially straight rod. The Euler buckling rod has
a conventional pitchfork bifurcation. The curved rod has no such bifurcation but
two branches which does not intersect the trivial axis since zero is not a solution.
Previously, Timoshenko and Gere (1961) [12] and George J. Simitses (1976) [11]
studied the initial curved rod when the initial curve is in the form of a sine function
with approximate curvature. Our analytic solution in chapter 3 is more accurate than
their work because we used the true curvature. We also investigated various choice of
functions for the initial curve and initial curvature analytically and numerically. The
asymptotic solutions were combined with exact numerical solutions in their respective

regions of validity.

71

Timoshenko and Gere (1961) [12] and George J. Simitses (1976) [11] obtained one
solution because their governing equations were linear and accurate only under the
condition that the deﬂection was small. However, our numerical result in chapter
4 conﬁrmed multiple solutions for large compressive load. We also investigated the
behavior of the multiply periodically curved rod in chapter 4 that models springs,
textiles and biological vessels. The tables of solutions for various choices of functions

for the initial curve and curvature along with various amplitude and period were

established.

The deﬂection of a thin curved rod is complex. The nonhomogeneous nonlinear
governing equations offer many conceptual, analytical, numerical challenges. In this
thesis, we have made advances in all of theses areas and examined in depth many
interesting phenomena such as non-uniqueness, maximum load for the unique solution
and instability. In spite of these results, some extensions of these are still possible.
These include studies in

i ) Different boundary conditions.
ii ) Extending the perturbation analysis to higher order.
iii) Estimating the maximum load for the unique solution.

iv) Three dimensional curved rods.

72

BIBLIOGRAPHY

73

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