MECHANICS OF ELASTOMERSkentaromaruyama.sakura.ne.jp/.../files/elastomers_eng.pdfcalled elastomers. Concerning an ideal elastomer (ideal rubber), the extension caused by one-dimensional

MECHANICS OF ELASTOMERS

――Mechanics of Rubber Balloons and Tubes,

and A Try for Mechanics of Blood Vessels――

MASAAKI MARUYAMA

Preface

This book is intended for people who want to explain the phenomenon

observed during the inflation of a rubber balloon. This is the English version

of the Japanese edition1) published in June 2012.

According to the experiment2), the phenomenon of the inflation of a rubber

balloon shows three stages. In the first stage, the bigger the balloon becomes,

the higher the internal pressure becomes toward a peak. In the second stage,

the internal pressure decreases after the peak, though the balloon becomes

bigger and bigger. In the third stage, the balloon continues inflating, and the

internal pressure changes into increasing again. And the balloon bursts.

The tensile-test of vulcanized rubber by JIS K-6251 (Japan Industrial

Standards) gives us a stress-extension curve. This curve shows a monotone

increasing function, so the decreasing of the stress is not shown in the curve.

This tensile-test is a one-dimensional (uniaxial) tensile-test. And the rubber

balloon inflation experiment is equivalent to a two-dimensional (biaxial)

tensile-test. There are two kinds of extensions.

In chapter 1 of this book, the two kinds of extension ratios of an ideal

elastomer (ideal rubber) have been studied and transformation formulas of

the extension ratios have been derived. In addition, C-type equation that is

available to express a one-dimensional stress-extension curve has been

offered.

In chapter 2, by the use of the transformation formulas and C-type equation,

the internal pressure-extension relations in the inflations of rubber balloons

have been calculated. In appropriate conditions, the calculated relation

curves show the three stages mentioned above.

In chapter 3 through chapter 7, the transformation formulas and C-type

equation have been also applied to the calculations of rubber ball inflations

and rubber tube inflations. And interesting results have been obtained.

１）丸山昌明,「エラストマーの力学：ゴム風船とゴムチューブの力学および血管の力学への試

み」，2012, Stocked in National Diet Library（Japan）

２）安達健、松田和久，「ゴム風船の力学実験」，物理教育:Physics Education Society of Japan

Vol.29 (No.1) p.26-27 (1981)

ⅰ

In chapter 8, the calculation method for rubber tubes have been applied to

blood vessels as a try for mechanics of blood vessels.

In the Japanese edition, though I am not a professional in medical science,

I tried to discuss about the inflation of blood vessels in the light of the results

of the calculations of rubber tube inflations. After that, I have been known

that there is no peak of the internal pressure in the inflation of blood vessels.

So, in this English version, instead of the previous discussion, I have tried to

show some C-type equations that may be available to calculate the inflation

of blood vessels which have no peak of the internal pressure.

I hope that C-type equation and the calculation methods in this book can

contribute to some of the studies of elastomers and blood vessels.

MASAAKI MARUYAMA

July, 2013

ⅱ

Contents

Preface ⅰ

1 Extension and Stress of Elastomers 1

1.1 One-Dimensional Pull and Two-Dimensional Pull 1

1.1.1 One-Dimensional Pull 1

1.1.2 Two-Dimensional Pull 3

1.2 One-Dimensional Stress-extension Relation 5

1.2.1 A-Type Equation 5

1.2.2 B-Type Equation 7

1.2.3 C-Type Equation 8

1.2.4 One-Dimensional Test and Two-Dimensional Test 11

2 One-Layer Calculation Method for Rubber Balloons 12

2.1 Derivation of Equations 12

2.2 Use of A-Type Equation 13

2.3 Use of B-Type Equation 15

2.4 Use of C-Type Equation 17

3 Two-Layer Calculation Method for Rubber Balls 21


3.2 Use of The C-Type Equation at n=2 24


4 Four-Layer Calculation Method for Rubber Balls 30




5 One-Layer Calculation Method for Rubber Tubes 39



ⅲ


6 Two-Layer Calculation Method for Rubber Tubes 49




7 Four-Layer Calculation Method for Rubber Tubes 65




8 A Try for Mechanics of Blood Vessels 77


8.2 Use of C-Type Equation 82

8.3 C-Type Equation for The Inflation of Arteries 84

ⅳ

1 Extension and Stress of Elastomers

1.1 One-Dimensional Pull and Two-Dimensional Pull

Vulcanized rubber and polymers that exhibit rubber-like elasticity are

called elastomers. Concerning an ideal elastomer (ideal rubber), the extension

caused by one-dimensional (uniaxial) pull have been compared with the other

extension caused by two-dimensional (biaxial) pull. And transformation

formulas of the extension ratios have been derived.

1.1.1 One-Dimensional Pull

An elastomer block is shown in Figure 1.1. Its initial (unstressed) lengths

are Ａ０ ,Ｂ０ and Ｃ０ in the x, y and z direction respectively. The one-

dimensional (uniaxial) pull force ＦＸ acting on the block in the x direction

stretches the length Ａ０ to ＡＸ. The extension ratio λＸ is the ratio of the

stressed length ＡＸ to the unstressed length Ａ０. Then, ＡＸ is represented

by the following equation.

（１-１）ＡＸ＝Ａ０λＸ

On the assumption ① that the volume of the elastomer does not change in

the stretched condition, the cross-sectional area of the block is reduced to one-

λＸths. Furthermore, on the assumption ② that the elastomer is isotropic,

Ｂ０ and Ｃ０ are reduced to one-λＸ１/２ths. Then, the lengths ＢＸ and ＣＸ

are represented by the following equations.

（１-２）ＢＸ＝Ｂ０λＸ－１/２

（１-３）ＣＸ＝Ｃ０λＸ－１/２

The other one-dimensional pull force ＦＹ acting on the block in the y

direction stretches the length Ｂ０ to ＢＹ , and the extension ratio λＹ is

expressed as the ratio of ＢＹ／Ｂ０.

1

Figure 1.1 One-dimensional (uniaxial) pull

Figure 1.2 Two-dimensional (biaxial) pull

2

And the other one-dimensional pull force ＦＺ acting on the block in the z

direction stretches the length Ｃ０ to ＣＺ, and the extension ratio λＺ is

expressed as the ratio of ＣＺ／Ｃ０ . Then the stressed lengths are expressed

as

（１-４）ＡＹ＝Ａ０λＹ－１/２

（１-５）ＢＹ＝Ｂ０λＹ

（１-６）ＣＹ＝Ｃ０λＹ－１/２

（１-７）ＡＺ＝Ａ０λＺ－１/２

（１-８）ＢＺ＝Ｂ０λＺ－１/２

（１-９）ＣＺ＝Ｃ０λＺ

where λ indicates the one-dimensional extension ratio caused by one-

dimensional pull force.

1.1.2 Two-Dimensional Pull

Figure 1.2 shows the two-dimensional (biaxial) pull. Keeping the pull force

ＦＸ acting on the block in the x direction, we can exert the other pull force

ＦＹ in the y direction. The force ＦＹ stretches the length ＢＸ to ＢＸＹ. The

extension ratio λＹ is the ratio of ＢＸＹ／ＢＸ. And ＢＸＹ is represented by

the following equation.

（１-１０）ＢＸＹ＝ＢＸλＹ＝Ｂ０λＸ－１/２λＹ

On the assumptions ① and ② which are mentioned in section 1.1.1 , the

lengths ＡＸＹ and ＣＸＹ are represented by the following equations.

（１-１１）ＡＸＹ＝ＡＸλＹ－１/２＝Ａ０λＸλＹ

－１/２

（１-１２）ＣＸＹ＝ＣＸλＹ－１/２＝Ｃ０λＸ

－１/２λＹ－１/２

In this case, λＸ and λＹ are the one-dimensional extension ratios, and these

are not true extension ratios caused by two-dimensional pull.

3

The true extension ratios based on the initial lengths are represented by the

following equations.

（１-１３） α＝ＡＸＹ／Ａ０＝λＸλＹ－１/２

（１-１４） β＝ＢＸＹ／Ｂ０＝λＸ－１/２λＹ

（１-１５） γ＝ＣＸＹ／Ｃ０＝λＸ－１/２λＹ

－１/２ or γ＝１／αβ

α,β and γ are the true extension ratios in the x , y and z directions

respectively. But γ is a dependent variable determined from α and β.

These extension ratios are the two-dimensional extension ratios caused by

two-dimensional pull.

From equations （１-１３） and （１-１４）, λＸ and λＹ are represented

by the following equations.

（１-１６） λＸ＝（α２β）２/３＝α４/３β２/３

（１-１７） λＹ＝（αβ２）２/３＝α２/３β４/３

Equations（１-１６）and （１-１７）are transformation formulas of the

extension ratios. These equations transform λＸ and λＹ into α and β.

We can define the elastomer which above equations are applied to as an

ideal elastomer.

The tensile-test by JIS K-6251 is one-dimensional (uniaxial) tensile-test

that is generally used. So we can get much information about one-dimensional

stress-extension relations. But two-dimensional (biaxial) tensile-test is not

generally used because of its difficulty. We can get little information about

two-dimensional stress-extension relations, but we can calculate about a two-

dimensional stress-extension relations (for example, about a rubber balloon

inflation) by the use of the corresponding one-dimensional stress-extension

relation and the above transformation formulas. In order to do, it is necessary

to obtain the proper expression that can represent the corresponding one-

dimensional stress-extension relation.

4

1.2 One-Dimensional Stress-extension Relation

To obtain the proper expression for the one-dimensional stress-extension

relation, three kinds of expressions have been investigated. Two of them, A-

type equation and B-type equation, are generally known. The remainder, C-

type equation, is newly offered in this book.

1.2.1 A-Type Equation

A-type equation is derived from Hooke's Law. Hooke's Law is expressed

as

（１-１８）ｆ＝Ｋ（λ－１）

where ｆ is the true stress, λ is the one-dimensional extension ratio, (λ－

１) is the strain (extension rate) and Ｋ is the proportional constant. The

true stress ｆ is defined as the total force ＦＴ divided by the true cross-

sectional area Ｓ. Then ＦＴ is represented by the following equation.

（１-１９）ＦＴ＝Ｓｆ＝ＳＫ（λ－１）

But Ｓ decreases in inverse proportion to λ because of the assumption ①

that the volume of the elastomer does not change. Then Ｓ is expressed as

（１-２０）Ｓ＝Ｓ０／λ

where Ｓ０ is the initial cross-sectional area.

The stress ｆ０ is defined as the total force ＦＴ divided by the initial cross-

sectional area Ｓ０. The stress ｆ０ is not the true stress, but it is generally

used. Then the total force ＦＴ is expressed as

（１-２１）ＦＴ＝Ｓ０ｆ０＝ＳＫ（λ－１）.

From equations（１-２０）and（１-２１）, we have the following equation.

5

（１-２２）ｆ０＝Ｋ（１－λ－１）

Equation（１-２２） is called A-type equation in this book. Both ｆ０ and

Ｋ are expressed in [ Pa ] or [N／m２]. λ is dimensionless. A typical curve

expressed by A-type equation is shown in Figure 1.3.

Figure 1.3 Typical curves expressed by A-type and B-type equations

A heavy solid curve：A-type equation

ｆ０＝Ｋ（１－１／λ） at Ｋ＝3

A fine solid curve：B-type equation

ｆ０＝Ｋ（λ－１／λ２） at Ｋ＝1

6

0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

10.0

0 1 2 3 4 5 6 7 8

ｆ０

λ

1.2.2 B-Type Equation

From the statistical mechanics of ideal rubber, the entropy change ⊿Ｓ

per unit volume on stretching is expressed as

（１-２３） ⊿Ｓ＝－Ｃ（λ２＋２／λ－３）

where λ is the one-dimensional extension ratio, and C is constant.

The derivative of the function of energy F=T⊿Ｓ with respect to λ gives us

（１-２４）ｆ０＝－２ＣＴ（λ－１／λ２）

whereｆ０ is the stress, T is the absolute temperature. The reader who wants

to know these details is referred to the book, 久保亮五著「ゴム弾性〔初版復

刻版〕」裳華房 p.69~85（１９９６）, published in Japan.

Equation（１-２４）can be rewritten as

（１-２５）ｆ０＝Ｋ（λ－λ－２）

where ｆ０ is the stress based on the initial cross-sectional area, λ is the

one-dimensional extension ratio and Ｋ is the proportional constant.

Equation（１-２５）is called B-type equation in this book. A typical curve

expressed by B-type equation is shown in Figure 1.3.

In Figure 1.3, when λ becomes larger, B-type equation shows larger stress

than A-type equation. That is, when λ becomes larger, the differential

coefficient of B-type equation becomes almost 1, but the differential coefficient

of A-type equation becomes smaller toward zero, though both equations show

the same differential coefficient at λ=1.

In the case of B-type equation, the following fact is considered. In the un-

stretched condition the polymer chains of an ideal elastomer (ideal rubber)

coiled and twisted. In the stretched condition the chains uncoil to a

considerable extent and tend to become oriented along the direction of

elongation (and occasionally crystallized partially).

7

In the case of A-type equation, the above fact is not considered. Thus B-

type equation shows larger stress than A-type equation in the stretched

condition. So B-type equation can express the one-dimensional stress-

extension relation better than A-type equation.

But both A-type and B-type equations are not sufficient to calculate rubber

balloon inflations. These details are described in chapter 2.

1.2.3 C-Type Equation

A new equation is offered in this section. It is expressed as

（１-２６）ｆ０＝Ｋ｛１－λ－１＋ｃ（λ－１）ｎ｝

where ｆ０ is the stress based on the initial cross-sectional area, λ is the

one-dimensional extension ratio, Ｋ and ｃ are constant and ｎ is 2 or 3.

(Another value of ｎ may be chosen, if it is needed.) Both ｆ０ and Ｋ are

expressed in [ Pa ] or [N／m２]. λ, ｃ and ｎ are dimensionless.

Equation（１-２６）is called C-type equation in this book, and term

ｃ（λ－１）ｎ is a correction term added to A-type equation. Then C-type

equation becomes A-type equation at ｃ＝０.

Considering a rubber balloon inflation, ｃ should be a far smaller positive

value than 1, (０＜ｃ≪１). Then term ｃ（λ－１）ｎ shows a very small

value at λ＜２. But it shows a suitable large value at λ＞２ because ｎ

is 2 or 3. The fact mentioned in the end of section 1.2.2 is considered also in

the case of C-type equation. And C-type equation can show larger stress than

A-type equation in the stretched condition.

Several curves expressed by C-type equation at ｎ＝２ are shown in

Figure 1.4. In this figure, λ is shown from 1 through 25 on the horizontal

axis. It is considered that putting α＝β＝５ gives us λ＝２５ in equation

（１-１６）. And several curves expressed by C-type equation at ｎ＝３ are

shown in Figure 1.5. λ is shown from 1 through 25. And the curves of A-

type and B-type equations are shown in Figures 1.4 and 1.5 for reference.

From observation of Figures 1.4 and 1.5, it is seen that the combination of

values of ｃ and ｎ determines the shape of the curve. And it is expected

8

that, by the choice of proper values of Ｋ, ｃ and ｎ, we can fit the curve

expressed by C-type equation into the stress-extension curve obtained by a

one-dimensional tensile-test.

Figure 1.4 Several curves expressed by C-type equation atｎ＝2

Fine solid curves：C-type equation

ｆ０＝Ｋ｛１－λ－１＋ｃ（λ－１）ｎ｝ at Ｋ＝3 , ｎ＝2 and

ｃ＝0.005, 0.01, 0.02, 0.05, 0.1 in order from the bottom


ｆ０＝Ｋ（１－１／λ） at Ｋ＝3

A dotted curve：B-type equation

ｆ０＝Ｋ（λ－１／λ２） at Ｋ＝1

9

0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

45.0

50.0

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

ｆ０

λ

Figure 1.5 Several curves expressed by C-type equation atｎ＝3

Fine solid curves：C-type equation

ｆ０＝Ｋ｛１－λ－１＋ｃ（λ－１）ｎ｝ at Ｋ＝3 , ｎ＝3 and

ｃ＝0.0002, 0.0005, 0.001, 0.002, 0.005 in order from the bottom


ｆ０＝Ｋ（１－１／λ） at Ｋ＝3

A dotted curve：B-type equation

ｆ０＝Ｋ（λ－１／λ２） at Ｋ＝1

10

0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

45.0

50.0

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

ｆ０

λ

1.2.4 One-Dimensional Test and Two-Dimensional Test

Even if we obtain the C-type equation that can express a one-dimensional

stress-extension curve, it is uncertain that this C-type equation is sufficient

to calculate about two-dimensional extensions, for example, rubber balloon

inflations.

The reason is as follows: It generally occurs that the coiled and twisted

polymer chains of an elastomer uncoil to a considerable extent in the

stretched condition. But it hardly occurs in two-dimensional extension that

uncoiled polymer chains tend to become oriented along one direction. Thus

the degree of stress increasing in two-dimensional extension may be lower

than the degree of that in one-dimensional extension.

We may handle this problem according to the following steps.

Step 1: We can obtain the values of Ｋ, ｃ and ｎ in the C-type equation,

when the curve expressed by C-type equation is nearly fitted into the stress-

extension curve obtained by a one-dimensional tensile-test. These values

are Ｋ１, ｃ１ and ｎ１.

Step 2: We do the inflation-test of a balloon or a ball or a tube which is made

of the same kind of material tested in Step 1, and we obtain the internal

pressure-extension curve. This test is equivalent to a two-dimensional tensile-

test.

Step 3: By the use of the C-type equation determined in Step 1, we can

calculate the internal pressure-extension curve according to the methods

described in chapter 2 through 7 in this book. When we cannot fit the

calculated curve into the curve obtained in Step 2, we can change the values

of Ｋ１, ｃ１ and ｎ１. And we calculate again and again until we can nearly

fit it. And we obtain the proper values of Ｋ, ｃ and ｎ. These values are Ｋ

２ , ｃ２ and ｎ２ . This C-type equation is sufficient to calculate two-

dimensional extensions.

Step 4: Repeating from Step 1 through Step 3, we can obtain many pairs of

｛Ｋ１, ｃ１, ｎ１｝ and ｛Ｋ２, ｃ２, ｎ２｝. Comparing ｛Ｋ２, ｃ２, ｎ２｝with

｛Ｋ１, ｃ１, ｎ１｝, if we can find the rules (or transformation formulas), we

may know ｛Ｋ２, ｃ２, ｎ２｝ from ｛Ｋ１, ｃ１, ｎ１｝ obtained in Step 1

without the test in Step 2.

11

2 One-Layer Calculation Method for Rubber Balloons

2.1 Derivation of Equations

The wall thickness of a rubber balloon is usually thin, so we can ignore the

difference between two extension ratios of the inside and the outside surfaces

during the inflation. So we can consider the wall to be one-layer.

Figure 2.1 shows a cross-section view of a rubber balloon. The wall-

thickness is exaggerated. The initial sizes of the balloon are expressed as

（２-１）｛Inside radius｝＝Ｒ０

（２-２）｛Outside radius｝＝Ｒ０ｔ０ , ｛Outside radius ratio｝＝ｔ０

（２-３）｛Wall thickness rate｝＝ｔ０－1

（２-４）｛Wall thickness｝＝Ｔ０＝Ｒ０（ｔ０－1）

where the subscript indicates the initial (unstressed) state, and the outside

radius ratio ｔ０ is the ratio of the outside radius to the inside radius.

Figure 2.1 A cross-section view of a rubber balloon

12

When the balloon inflates, the internal pressure is Ｐ, that is the increase

from the atmospheric pressure, and the inside radius increases to Ｒ０α. The

balloon inflation keeps spherical. Then the extension ratio α has the same

value in every direction on the inside surface. And α is the two-

dimensional extension ratio.

When the internal pressure is Ｐ , the force ＦＰ acting on the inside

surface is represented by the following equation.

（２-５）ＦＰ＝πＲ０２α２Ｐ

The force ＦＴ caused by the wall extension is the product of the initial

cross-sectional area Ｓ０ and the stress ｆ０ based on the initial cross-

sectional area. Then it is represented by the following equation.

（２-６）ＦＴ＝Ｓ０ｆ０＝πＲ０２（ｔ０

２－１）ｆ０

Under the condition of equilibrium of the forces,ＦＰ＝ＦＴ, we have the

following equation.

（２-７） α２Ｐ＝（ｔ０２－１）ｆ０

Equation（２-７）is the basic equation of the one-layer calculation method

for rubber balloons. Then it is necessary to know about ｆ０.

2.2 Use of A-Type Equation

Substituting the expression of A-type equation（１-２２） for ｆ０ in

equation（２-７）gives us the following equation.

（２-８） α２Ｐ＝（ｔ０２－１）Ｋ（１－λ－１）

Equation（２-８）contains two kinds of extension ratios , α and λ. α is

13

the two-dimensional extension ratio of the balloon and λ is the one-

dimensional extension ratio of the material. We need to transform λ into

α. The balloon inflation is isotropic, thus we can put λ＝λＸ＝λＹ and α

＝β in equation（１-１６）, and we have the following equation.

（２-９） λ＝λＸ＝（α２β）２/３＝α２

Substituting this expression for λ in equation（２-８）gives us the following

equation.

（２-１０）Ｐ＝Ｋ（ｔ０２－１）（α－２－α－４）

Equation（２-１０）expresses the relation between the internal pressure Ｐ

and the extension ratio α. Both Ｐ and Ｋ are expressed in [ Pa ] or [N／

m２]. And ｔ０ and α are dimensionless.

The derivative of equation（２-１０）with respect to α gives us

（２-１１）Ｐ′＝Ｋ（ｔ０２－１）（－２α－３＋４α－５）.

The differential coefficient Ｐ′becomes zero at α＝２１/２, therefore the

internal pressure Ｐ has a peak (maximal value) at α＝２１/２≒1.4 .

We can calculate equation（２-１０）and draw the relation curve in Excel.

A typical curve expressed by equation（２-１０）is shown in Figure 2.2. In

this figure, α is shown from 1 through 5.5 on the horizontal axis. Ｋ is put

at 100. The wall thickness of the balloon is five percent of the inside radius

(ｔ０＝1.05). This curve shows a peak at α＝２１/２. The position of this peak

agrees with the result of the experiment1) of rubber balloon inflations. But

this curve does not show the re-increasing of the internal pressure in the

range of α＞４. This is different from the result of the above experiment.

Therefore A-type equation（１-２２）is not sufficient to calculate rubber

balloon inflations.

１）安達健、松田和久，「ゴム風船の力学実験」，物理教育:Physics Education Society of Japan

Vol.29 (No.1) p.26-27 (1981)

14

Figure 2.2 Results of the one-layer calculation method (1)

A heavy solid curve is expressed by equation（２-１０）

using A-type equation at Ｋ＝100 and ｔ０＝1.05

A fine solid curve is expressed by equation（２-１３）

using B-type equation at Ｋ＝33 and ｔ０＝1.05

2.3 Use of B-Type Equation

Substituting the expression of B-type equation（１-２５） for ｆ０ in


（２-１２） α２Ｐ＝（ｔ０２－１）Ｋ（λ－λ－２）

15

0.0

1.0

2.0

3.0

4.0

0 1 2 3 4 5

Ｐ

α

Equation（２-１２）contains two kinds of extensions of α and λ. And we

need to transform λ into α.

Substituting the expression of equation（２-９）for λ in equation（２-１２）

gives us the following equation.

（２-１３）Ｐ＝Ｋ（ｔ０２－１）（１－α－６）

Equation（２-１３）expresses the relation between the internal pressure Ｐ

and the two-dimensional extension ratio α . Both Ｐ and Ｋ are

expressed in [ Pa ] or [N／m２]. And ｔ０ and α are dimensionless.

The derivative of equation（２-１３）with respect to α gives us

（２-１４）Ｐ′＝Ｋ（ｔ０２－１）６α－７ .

This differential coefficient Ｐ′does not become zero, that is, the internal

pressure Ｐ does not have a peak (maximal value). It is different from the

result of the experiment of rubber balloon inflations mentioned in section 2.2.

Therefore B-type equation（１-２５）is not sufficient to calculate rubber

balloon inflations.

We can calculate equation（２-１３）and draw the relation curve in Excel.

A typical curve expressed by equation（２-１３）is shown in Figure 2.2. Ｋ

is put at 33 so that the differential coefficient of equation（２-１３）at α=1

becomes almost the same as that of equation（２-１０）. The wall thickness

of the balloon is five percent of the inside radius (ｔ０＝1.05).

Comparing the two curves in Figure 2.2, it is seen that B-type equation

shows stronger stress than A-type equation in the range of α＞1.3. That is,

B-type equation shows too strong stress to calculate about two-dimensional

extensions. The reason is the same as the fact and the reason described in the

end of section 1.2.2 and the beginning of section 1.2.4.

16

2.4 Use of C-Type Equation

Substituting the expression of C-type equation（１-２６） for ｆ０ in


（２-１５） α２Ｐ＝（ｔ０２－１）Ｋ｛１－λ－１＋ｃ（λ－１）ｎ｝

Equation（２-１５） contains two kinds of extension ratios of α and λ.

And we need to transform λ into α.

Substituting the expression of equation（２-９）for λ in equation（２-１

５）and putting ｎ＝2 give us the following equation.

（２-１６）Ｐ＝Ｋ（ｔ０２－１）α－２｛１－α－２＋ｃ（α２－１）２｝

Equation（２-１６）expresses the relation between the internal pressure Ｐ

and the two-dimensional extension ratio α . Both Ｐ and Ｋ are

expressed in [ Pa ] or [N／m２]. And ｔ０, α and ｃ are dimensionless.

We can calculate equation（２-１６）and draw the relation curves in Excel.

Several curves expressed by equation（２-１６）are shown by fine solid curves

in Figure 2.3. In this figure, α is shown from 1 through 5.5 on the horizontal

axis. Ｋ is put at 100. The wall thickness of the balloon is five percent of the

inside radius (ｔ０＝1.05).

Substituting the expression of equation（２-９）for λ in equation（２-１

５）and putting ｎ＝3 give us the following equation.

（２-１７）Ｐ＝Ｋ（ｔ０２－１）α－２｛１－α－２＋ｃ（α２－１）３｝

Equation（２-１７）also expresses the relation between the internal pressure

Ｐ and the two-dimensional extension ratio α.

Several curves expressed by equation（２-１７）are shown by fine solid line

in Figure 2.4. In this figure, α is shown from 1 through 5.5 on the horizontal

axis. Ｋ is put at 100. The wall thickness of the balloon is five percent of the

inside radius (ｔ０＝1.05).

17

Figure 2.3 Results of one-layer calculation method (2)

Fine solid curves are expressed by equation（２-１６）

using C-type equation at ｎ＝2 .

Other conditions are Ｋ＝100, ｔ０＝1.05, and

ｃ＝0.005, 0.01, 0.02, 0.03, 0.05 in order from the bottom.


using A-type equation at Ｋ＝100 and ｔ０＝1.05.

18

0.0

1.0

2.0

3.0

4.0

5.0

6.0

0 1 2 3 4 5

Ｐ

α

Figure 2.4 Results of one-layer calculation method (3)

Fine solid curves are expressed by equation（２-１７）

using C-type equation at ｎ＝3 . Other conditions are Ｋ＝100,

ｔ０＝1.05, and ｃ＝0.0001, 0.0002, 0.0005, 0.001, 0.003, 0.01, 0.02

in order from the bottom.


using A-type equation at Ｋ＝100 and ｔ０＝1.05.

A dotted curve is the bottom fine solid curve in Figure 2.3.

19

0.0

1.0

2.0

3.0

4.0

5.0

6.0

0 1 2 3 4 5

Ｐ

α

From observation of Figures 2.3 and 2.4, it is seen that the two of the curves

are similar in shape to the curves obtained by the experiment1) of rubber

balloon inflations. These are the bottom fine solid curve in Figure 2.3 and the

second fine solid curve from the bottom in Figure 2.4.

Two C-type equations which are used for the above two curves are as follows.

（２-１８）ｆ０＝Ｋ｛１－λ－１＋0.005（λ－１）２｝

（２-１９）ｆ０＝Ｋ｛１－λ－１＋0.0002（λ－１）３｝

Ｋ is put at 100 in the above calculation, but actually we should determine

Ｋ from results of experiments.

In this book, equations（２-１８）and（２-１９）are considered as the two

kinds of standard equations which express the one-dimensional stress-

extension relations of ideal elastomers. And these equations will be used for

the calculations of two-dimensional extension in following chapters.

１）安達健、松田和久，「ゴム風船の力学実験」，物理教育:Physics Education Society of Japan

Vol.29 (No.1) p.26-27 (1981)

20

3 Two-Layer Calculation Method for Rubber Balls


When a rubber balloon has a thick wall (rather than a balloon, it is better

to call it a ball), we cannot ignore the difference between two extension ratios

of the inside and the outside surfaces during the inflation. But we can handle

this problem by considering the wall to be two-layer. And the two-layer

calculation method for rubber balls is described in this chapter.

Figure 3.1 A cross-section view of a rubber ball

Figure 3.1 shows a cross-section view of a rubber ball with a two-layer wall.

We call the inside layer the 1st layer and the outside layer the 2nd layer. In

each layer, the inside radius timesｔ０ is the outside radius. Thus the 1st layer

wall thickness timesｔ０ is the 2nd layer wall thickness. ｔ０ is the outside

radius ratio. The initial sizes of the ball are expressed as follows.

21

In total：

（３-１）｛Inside radius｝＝Ｒ０

（３-２）｛Outside radius｝＝Ｒ０ｔ０Ｔ＝Ｒ０ｔ０２

｛Total outside radius ratio｝＝ｔ０Ｔ＝ｔ０２

（３-３）｛Total wall thickness rate｝＝ｔ０Ｔ－1＝ｔ０２－1

（３-４）｛Total wall thickness｝＝Ｔ０Ｔ＝Ｒ０（ｔ０Ｔ－1）＝Ｒ０（ｔ０２－1）

In the 1st layer：

（３-５）｛1st layer inside radius｝＝Ｒ０

（３-６）｛1st layer outside radius｝＝Ｒ０ｔ０

｛1st layer outside radius ratio｝＝ｔ０

（３-７）｛1st layer thickness rate｝＝ｔ０－1＝ｔ０Ｔ1/2－1

（３-８）｛1st layer thickness｝＝Ｔ０１＝Ｒ０（ｔ０－1）

In the 2nd layer：

（３-９）｛2nd layer inside radius｝＝Ｒ０ｔ０

（３-１０）｛2nd layer outside radius｝＝Ｒ０ｔ０２

｛2nd layer outside radius ratio｝＝ｔ０

（３-１１）｛2nd layer thickness rate｝＝ｔ０－1＝ｔ０Ｔ1/2－1

（３-１２）｛2nd layer thickness｝＝Ｔ０２＝Ｒ０ｔ０（ｔ０－1）

When the ball inflates, the internal pressure is Ｐ, that is the increase from

the atmospheric pressure, the 1st layer inside radius Ｒ０ increases to Ｒ０α

１ and the 2nd layer outside radius Ｒ０ｔ０２ increases to Ｒ０ｔ０

２α２. α１ is

the extension ratio of the 1st layer inside surface and α２ is that of the 2nd

layer outside surface. We assume that α１ represents the extension ratio of

the whole 1st layer and α２ represents that of the whole 2nd layer.

On the assumption that the volume of the rubber material of the ball does

not change during the inflation, we can write the following equation

concerning the volumes of the all layers before and after the inflation.

（４／３）π（Ｒ０ｔ０２）３－（４／３）πＲ０

３

＝（４／３）π（Ｒ０ｔ０２α２）

３－（４／３）π（Ｒ０α１）３

22

From this equation, we have the following equations which express the

relations between α１ and α２. These are the same meaning equations.

（３-１３）（α１３－１）／（α２

３－１）＝ｔ０６

（３-１４） α１＝｛ｔ０６（α２

３－１）＋１｝１/３

（３-１５） α２＝｛ｔ０－６（α１

３－１）＋１｝１/３

From equation（３-１３）we can see α２＜α１ because of ｔ０＞１. By the

use of equation（３-１４）we can transform α１ into α２, and by the use

of equation（３-１５）we can transform α２ into α１. The ball inflation keeps

spherical. Then α１has the same value in every direction on the inside

surface and α２ has also the same value in every direction on the outside

surface.



（３-１６）ＦＰ＝πＲ０２α１

２Ｐ

The force ＦＴ１ caused by the 1st layer extension is the product of the

initial cross-sectional area Ｓ０１ of the 1st layer and the stress ｆ０１. Then it

is represented by the following equation.

（３-１７）ＦＴ１＝Ｓ０１ｆ０１＝πＲ０２（ｔ０

２－１）ｆ０１

The force ＦＴ２ caused by the 2nd layer extension is the product of the

initial cross-sectional area Ｓ０２ of the 2nd layer and the stressｆ０２. Then it


（３-１８）ＦＴ２＝Ｓ０２ｆ０２＝πＲ０２ｔ０

２（ｔ０２－１）ｆ０２

Under the condition of equilibrium of the forces, ＦＰ＝ＦＴ１＋ＦＴ２ , we

have the following equation.

（３-１９） α１２Ｐ＝（ｔ０

２－１）ｆ０１＋ｔ０２（ｔ０

２－１）ｆ０２

23

Equation（３-１９）is the basic equation of the two-layer calculation method

for rubber balls. Then it is necessary to know about ｆ０１ and ｆ０２.

3.2 Use of The C-Type Equation at n=2

First, we use equation（２-１８）which is the C-type equation at n=2 to

express ｆ０１ and ｆ０２. Those are the following equations.

（３-２０）ｆ０１＝Ｋ｛１－λ１－１＋ｃ（λ１－１）２｝

（３-２１）ｆ０２＝Ｋ｛１－λ２－１＋ｃ（λ２－１）２｝

In these equations, ｃ is put at 0.005, λ１ is the one-dimensional extension

ratio of the material of the 1st layer, and λ２ is that of the 2nd layer.

When we substitute these expressions for ｆ０１ and ｆ０２ in equation（３-

１９）and transform λ１ and λ２ into α１ and α２ by the use of the

equations λ１＝α１２ and λ２＝α２

２ , which are derived from equation（２

-９）, we can write the following equation.

（３-２２）Ｐ＝Ｋ（ｔ０２－１）α１

－２［１－α１－２＋ｃ（α１

２－１）２

＋ｔ０２｛１－α２

－２＋ｃ（α２２－１）２｝］

In this equation, α１ and α２ are considered as one variable because any one

of them is determined from the other one by equations（３-１４）or（３-１

５）. Equation（３-２２）expresses the relation between the internal pressure

Ｐ and the extension ratio α１ or α２. Both Ｐ and Ｋ are expressed in

[ Pa ] or [N／m２]. And ｔ０, α１ and α２ are dimensionless.

We can calculate equation（３-２２）and draw the relation curves in Excel.

Several curves expressed by equation（３-２２）are shown in Figure 3.2. In

this figure, α１ is shown from 1 through 7 on the horizontal axis. Ｋ is put

at 100. The wall thickness conditions are as follows： The outside radius

ratios are ｔ０Ｔ＝ｔ０２＝1.10, 1.20, 1.50 and 2.00, that is to say, the total wall

thicknesses are 10%, 20%, 50% and 100% of the inside radius.

24

Figure 3.2 Results of the two-layer calculation method

The relation curves expressed by Equation（３－２２）at Ｋ＝100,

ｃ＝0.005 and ｎ＝2

Fine solid curves：The relations of Ｐ to α１

Heavy solid curves：The relations of Ｐ to α２

The wall thickness conditions： In order from the bottom,

ｔ０Ｔ＝ｔ０２＝1.10, 1.20, 1.50, 2.00,

that is to say, ｔ０＝1.0488, 1.0954, 1.2247, 1.4142

25

0

5

10

15

20

25

30

35

40

45

0 1 2 3 4 5 6 7

Ｐ

α1 , α2

From observation of Figure 3.2, it is seen that the larger ｔ０Ｔ becomes,

the smaller α２ becomes. For example, when the total wall thickness is 10%

of the inside radius and α１(the extension ratio of the inside surface) is 1.40,

α２(that of the outside surface) is 1.322 (at near the peak of the internal

pressure). But when the total wall thicknesses is 100% of the inside radius

and α１ is 1.60, α２ is 1.115. Of cause, this result is also estimated from

equation（３-１３）.

Some comparisons between the one-layer calculation method (by equation

（２-１６）) and the two-layer calculation method (by equation（３-２２）)

are shown in Figure 3.3. In this figure, the curves are calculated at Ｋ＝100,

ｃ＝0.005 and ｎ＝2 in both methods. And the wall thicknesses are 5%, 10%,

20% and 50% of the inside radius in both methods.

From observation of Figure 3.3, we can see the following. When the wall

thickness is 5% of the inside radius, the bottom dotted curve (one-layer

calculation method) agrees to the bottom solid curve (two-layer calculation

method). But when the wall thickness becomes 10% of the inside radius or

more, the peaks of the dotted curves are higher than the corresponding peaks

of the solid curves. Especially when the wall thickness is 50% of the inside

radius, the peak of the dotted curve is much higher than the peak of the solid

curve by 40%.

The reason is as follows： In the one-layer calculation method, it is assumed

that the extension ratio of the outside surface is equal to that of the inside

surface in spite of the large wall thickness.

Therefore we had better to use the two-layer calculation method when the

wall thickness is larger than 5% of the inside radius.

26

Figure 3.3 Comparison between the one-layer and the two-layer

calculation methods

Dotted curves：By the one-layer calculation method

The relations of Ｐ to α expressed by equation（２-１６）

at Ｋ＝100, ｃ＝0.005 and ｎ＝2 .


ｔ０＝1.05, 1.10, 1.20, 1.50

Solid curves：By the two-layer calculation method

The relations of Ｐ to α２ expressed by equation（３－２２）

at Ｋ＝100, ｃ＝0.005 and ｎ＝2


ｔ０Ｔ＝ｔ０２＝1.05, 1.10, 1.20, 1.50

27

0

5

10

15

20

25

30

35

0 1 2 3 4 5 6 7

Ｐ

α , α2


Next, we use equation（２-１９）which is the C-type equation at n=3 to

express ｆ０１ and ｆ０２. Those are the following equations.

（３-２３）ｆ０１＝Ｋ｛１－λ１－１＋ｃ（λ１－１）３｝

（３-２４）ｆ０２＝Ｋ｛１－λ２－１＋ｃ（λ２－１）３｝


ratio of the material of the 1st layer, and λ２ is that of the 2nd layer.

When we substitute these expressions for ｆ０１ and ｆ０２ in equation（３-

１９）and transform λ１ and λ２ into α１ and α２ by the use of equations

λ１＝α１２ and λ２＝α２

２, we can write the following equation.

（３-２５）Ｐ＝Ｋ（ｔ０２－１）α１

－２［１－α１－２＋ｃ（α１

２－１）３

＋ｔ０２｛１－α２

－２＋ｃ（α２２－１）３｝］

In this equation,α１ and α２ are determined from the other one by

equations（３-１４）or（３-１５）. Equation（３-２５）expresses the relation

between the internal pressure Ｐ and the extension ratio α１ or α２.

We can calculate equation（３-２５）and draw the relation curves in Excel.

The calculations have been done in the range of 1≦α１≦7. These calculated

curves are shown by heavy solid curves in Figure 3.4. In equation（３-２５）,

Ｋ is put at 100. The wall thickness conditions are as follows： Total outside

radius ratios are ｔ０Ｔ＝ｔ０２＝1.05, 1.10, 1.20, 1.50 and 2.00, that is, the

total wall thicknesses are 5%, 10%, 20%, 50% and 100% of the inside radius.

To compare the effect of ｎ＝3 with ｎ＝2 in C-type equation, the curves

calculated by equation（３-２２）are also shown by fine solid curves in Figure

3.4. All these curves are expressed as the relations between the internal

pressure Ｐ and the extension ratio α２.

From observation of Figure 3.4, it is seen that the peaks of heavy solid

curves are much the same as the corresponding peaks of fine solid curves.

But, after the peak, the each heavy solid curve has a smaller minimal value

28

and has steeper slopes of decreasing and increasing of the internal pressure

than the corresponding fine solid curve. These differences depend on the

combination of the values of ｃ and ｎ. Which combination is the better?

It is depends on a result of an experiment.

Figure 3.4 Comparison between n=3 and n=2 in C-type equation

Heavy solid curves：The relations of Ｐ to α２ expressed

by equation（３－２５）at Ｋ＝100, ｃ＝0.0002 and ｎ＝3

Fine solid curves：The relations of Ｐ to α２ expressed

by equation（３－２２）at Ｋ＝100, ｃ＝0.005 and ｎ＝2

The wall thickness conditions： In order from the bottom

ｔ０Ｔ＝ｔ０２＝1.05, 1.10, 1.20, 1.50, 2.00

29

0

5

10

15

20

25

30

35

40

45

50

55

0 1 2 3 4 5 6 7

Ｐ

α２

4 Four-Layer Calculation Method for Rubber Balls


In this chapter a four-layer calculation method is studied to improve the

accuracy of the calculation for rubber balls. Figure 4.1 shows a cross-section

view of a rubber ball having a four-layer wall. We call the layers the 1st, the

2nd, the 3rd and the 4th layers in order from the inside. The initial sizes of

the rubber ball are shown in Table 4.1.

Figure 4.1 A cross-section view of a rubber ball

Table 4.1 The initial sizes of the rubber ball

Inside

radius

Outside radius Wall thickness

1st layer Ｒ０Ｒ０ｔ０Ｒ０（ｔ０－1）

2nd layer Ｒ０ｔ０Ｒ０ｔ０２Ｒ０ｔ０（ｔ０－1）

3rd layer Ｒ０ｔ０２Ｒ０ｔ０

３Ｒ０ｔ０２（ｔ０－1）

4th layer Ｒ０ｔ０３Ｒ０ｔ０

４Ｒ０ｔ０３（ｔ０－1）

Total Ｒ０Ｒ０ｔ０Ｔ,(ｔ０Ｔ＝ｔ０４) Ｒ０（ｔ０Ｔ－1）

30

In each layer, the inside radius times ｔ０ is the outside radius. Thus the 1st

layer wall thickness times ｔ０ is the 2nd layer wall thickness. And the 2nd

layer wall thickness times ｔ０ is the 3rd layer wall thickness. And so on.

ｔ０ is the outside radius ratio of each layer. ｔ０Ｔ is the total outside radius

ratio.

When the ball inflates, the internal pressure is Ｐ, that is the increase

from the atmospheric pressure, the1st layer inside radius increases to Ｒ０

α１ , the 2nd layer outside radius increases to Ｒ０ｔ０２α２ and the 4th layer

outside radius increases to Ｒ０ｔ０４α３. Then α１ is the extension ratio of

the 1st layer inside surface, α２ is that of the 2nd layer outside surface and

α３ is that of the 4th layer outside surface. We assume that α１ represents

the extension ratio of the whole 1st layer, α２ represents that of the whole

2nd and 3rd layers and α３ represents that of the whole 4th layer.

On the assumption that the volume of the rubber material of the ball does


concerning the volumes of the 1st and 2nd layers before and after the inflation.

（４／３）π（Ｒ０ｔ０２）３－（４／３）πＲ０

３

＝（４／３）π（Ｒ０ｔ０２α２）

３－（４／３）π（Ｒ０α１）３


relation between α１ and α２.

（４-１）（α１３－１）／（α２

３－１）＝ｔ０６

（４-２） α１＝｛ｔ０６（α２

３－１）＋１｝１/３

（４-３） α２＝｛ｔ０－６（α１

３－１）＋１｝１/３

And furthermore we can write the following equation concerning the volumes

of the all layers before and after the inflation.

（４／３）π（Ｒ０ｔ０４）３－（４／３）πＲ０

３

＝（４／３）π（Ｒ０ｔ０４α３）

３－（４／３）π（Ｒ０α１）３

31


relation between α１ and α３.

（４-４）（α１３－１）／（α３

３－１）＝ｔ０１２

（４-５） α１＝｛ｔ０１２（α３

３－１）＋１｝１/３

（４-６） α３＝｛ｔ０－１２（α１

３－１）＋１｝１/３

Equations（４-１）through（４-６）are the similar to equations（３-１３）

through（３-１５）. We can transform any one of α１, α２ and α３ into the

other one we want by the above equations.



（４-７）ＦＰ＝πＲ０２α１

２Ｐ

The forceＦＴ１ caused by the 1st layer extension is the product of the initial

cross-sectional area Ｓ０１ of the first layer and the stress ｆ０１. Then it is

represented by the following equation.

（４-８）ＦＴ１＝Ｓ０１ｆ０１＝πＲ０２（ｔ０

２－１）ｆ０１

The force ＦＴ２ caused by the 2nd and the 3rd layers extension is the

product of the initial cross-sectional area Ｓ０２ and the stress ｆ０２. Ｓ０２ is

the sum of the initial cross-sectional areas of the 2nd and the 3rd layers. Then

it is represented by the following equation.

（４-９）ＦＴ２＝Ｓ０２ｆ０２＝πＲ０２ｔ０

２（ｔ０４－１）ｆ０２

The force ＦＴ３ caused by the 4th layer extension is the product of the

initial cross-sectional area Ｓ０３ of the 4th layer and the stressｆ０３. Then it


（４-１０）ＦＴ３＝Ｓ０３ｆ０３＝πＲ０２ｔ０

６（ｔ０２－１）ｆ０３

32

Under the condition of equilibrium of the forces, ＦＰ＝ＦＴ１＋ＦＴ２＋ＦＴ３ ,

we have the following equation.

（４-１１） α１２Ｐ＝（ｔ０

２－１）ｆ０１＋ｔ０２（ｔ０

４－１）ｆ０２

＋ｔ０６（ｔ０

２－１）ｆ０３

Equation（４-１１）is the basic equation of the four-layer calculation method

for rubber balls. Then it is necessary to know about ｆ０１, ｆ０２ and ｆ０３.


We use the C-type equation at n=2 as the expressions of ｆ０１, ｆ０２ and

ｆ０３. Those are the following equations.

（４-１２）ｆ０１＝Ｋ｛１－λ１－１＋ｃ（λ１－１）２｝

（４-１３）ｆ０２＝Ｋ｛１－λ２－１＋ｃ（λ２－１）２｝

（４-１４）ｆ０３＝Ｋ｛１－λ３－１＋ｃ（λ３－１）２｝


ratio of the material of the 1st layer, λ２ is that of the 2nd and 3rd layers

and λ３ is that of the 4th layer.

When we substitute these expressions forｆ０１, ｆ０２ and ｆ０３ in equation

（４-１１）and transform λ１, λ２ and λ３ into α１, α２ and α３ by the

use of the equations λ１＝α１２, λ２＝α２

２ and λ３＝α３２ , which are

derived from equation（２-９）, we can write the following equation.

（４-１５）Ｐ＝Ｋ（ｔ０２－１）α１

－２［１－α１－２＋ｃ（α１

２－１）２

＋ｔ０２（ｔ０

２＋１）｛１－α２－２＋ｃ（α２

２－１）２｝

＋ｔ０６｛１－α３

－２＋ｃ（α３２－１）２｝］

In this equation, α１, α２ and α３ are considered as one variable, because

any one of them is determined from the other one by the equations（４-２）,

（４-３）,（４-５）and（４-６）. Equation（４-１５）expresses the relation

between the internal pressureＰ and the extension ratio α１ or α２ or α３.

33

We can calculate equation（４-１５）and draw the relation curves in Excel.

The calculations have been done in the range of 1≦α１≦7. Ｋ is put at 100

and ｃ is put at 0.005. The wall thickness conditions are as follows： The

total outer radius ratios are ｔ０Ｔ＝ｔ０４＝1.10, 1.20, 1.50 and 2.00, that is to

say, the total wall thicknesses are 10%, 20%, 50% and 100% of the inside

radius. And the calculated curves have been shown by solid curves in Figure

4.2. These curves are expressed as the relation between Ｐ and α３ (α３ is

the extension ratio of the outside surface), because we can measure α３ more

easily than α１.

To compare the four-layer calculation method with the two-layer calculation

method, the curves calculated by equation（３-２２）are shown by dotted

curves in Figure 4.2. These curves have been calculated under the same

conditions as the conditions used above, and have been shown in the range of

1≦α１≦7.5 to distinguish easily from the solid curves.

From observation of Figure 4.2, it is seen that the peaks of solid curves are

almost the same as the corresponding peaks of dotted curves.

But when the total outside radius ratio ｔ０Ｔ is 1.50 or more and the

extension ratio of the outside surface (α２ or α３) is 2 or more, there are some

differences between the two-layer and the four-layer calculation methods.

Whether we should choose the four-layer calculation method or not depends

on a need of the accuracy of the calculation.

In the four-layer calculation method, the each layer ’s share of the internal

pressure is shown in Figures 4.3. In this figure, these curves are shown as

the relation between Ｐ and α１ for comparison. These curves have been

calculated by equation（４-１５）at Ｋ＝100, ｃ＝0.005 and ｔ０Ｔ＝ｔ０４＝

2.00.

From observation of Figures 4.3, the each layer ’s share of the internal

pressure at the peak is estimated about one-quarter, so the each layer ’s share

at the peak is approximately equal. From equation（４-４）, the extension

ratio of the 4th layer α３ is less than α１ . But, from Table 4.1, the initial

wall thickness of the 4th layer is ｔ０３ times as large as that of the 1st layer.

34

Figure 4.2 Comparison between the two-layer and the four-layer

calculation methods (at n=2)

Solid curves：The four-layer calculation method

The relation of Ｐ to α３ expressed by equation（４-１５）

at Ｋ＝100, ｃ＝0.005 .

The wall thickness conditions：In order from the bottom,

ｔ０Ｔ＝ｔ０４＝1.10, 1.20, 1.50, 2.00

Dotted curves：The two-layer calculation method

The relation of Ｐ to α２ expressed by equation（３－２２）

at Ｋ＝100, ｃ＝0.005 .


ｔ０Ｔ＝ｔ０２＝1.10, 1.20, 1.50, 2.00

35

0

5

10

15

20

25

30

35

40

45

0 1 2 3 4 5 6 7

Ｐ

α2 , α3

Figure 4.3 The each layer ’s share of the internal pressure in the four-

layer calculation Method

A heavy solid curve：Total internal pressure by equation（４－１５）

at Ｋ＝100, ｃ＝0.005, ｎ＝2 and ｔ０Ｔ＝ｔ０４＝2.00

A fine solid curve：The 1st layer’s share

A chain curve：The sum of the 2nd layer’s and 3rd layer’s shares

A dotted curve：The 4th layer’s share

36

0

5

10

15

20

25

30

35

40

45

0 1 2 3 4 5 6 7

Ｐ

α1


Next, we use the C-type equation at n=3 as the expressions of ｆ０１, ｆ０２

and ｆ０３. Those are the following equations.

（４-１６）ｆ０１＝Ｋ｛１－λ１－１＋ｃ（λ１－１）３｝

（４-１７）ｆ０２＝Ｋ｛１－λ２－１＋ｃ（λ２－１）３｝

（４-１８）ｆ０３＝Ｋ｛１－λ３－１＋ｃ（λ３－１）３｝


ratio of the material of the 1st layer, λ２ is that of the 2nd and 3rd layers

and λ３ is that of the 4th layer.

When we substitute these expressions forｆ０１, ｆ０２ and ｆ０３ in equation

（４-１１）and transform λ１, λ２ and λ３ into α１, α２ and α３ by the

use of the equations λ１＝α１２, λ２＝α２

２ and λ３＝α３２ ,which are

derived from equation（２-９）, we can write the following equation.

（４-１９）Ｐ＝Ｋ（ｔ０２－１）α１

－２［１－α１－２＋ｃ（α１

２－１）３

＋ｔ０２（ｔ０

２＋１）｛１－α２－２＋ｃ（α２

２－１）３｝

＋ｔ０６｛１－α３

－２＋ｃ（α３２－１）３｝］

In this equation, α１, α２ and α３ are considered as one variable, because

we can transform any one of the extension ratios α１, α２ and α３ into the

other one we want by the use of equations（４-２）,（４-３）,（４-５）and

（４-６）. Equation（４-１９）expresses the relation between the internal

pressure Ｐ and the extension ratio α１ or α２ or α３.

We can calculate equation（４-１９）and draw the relation curves in Excel.

In equation（４-１９）, Ｋ=100 and ｃ=0.0002. The wall thickness conditions

are as follows： The total outside radius ratios are ｔ０Ｔ＝ｔ０４＝1.10, 1.20,

1.50 and 2.00, that is to say, the total wall thicknesses are 10%, 20%, 50% and

100% of the inside radius. The calculations have been done in the range of 1

≦α１≦7. And these are shown by heavy solid curves in Figure 4.4.


calculated by equation（４－１５）are also shown by fine solid curves in

37

Figure 4.4. All these curves are expressed as the relation between Ｐ and

α３. Figure 4.4 is quite similar to Figure 3.4. Then we can have the same

discussion as the discussion held at the end of section 3.3.

Figure 4.4 Comparison between n=3 and n=2 in C-type Equation

Heavy solid curves：The relation of Ｐ to α３ expressed

by equation（４－１９）at Ｋ＝100, ｃ＝0.0002 and ｎ＝3

Fine solid curves：The relation of Ｐ to α３ expressed

by equation（４－１５）at Ｋ＝100, ｃ＝0.005 and ｎ＝2


ｔ０Ｔ＝ｔ０４＝1.10, 1.20, 1.50, 2.00

38

0

5

10

15

20

25

30

35

40

45

0 1 2 3 4 5 6 7

Ｐ

α３

5 One-Layer Calculation Method for Rubber Tubes


When the wall thickness of a rubber tube is thin, we can ignore the


during the inflation. So we can consider the wall to be one-layer. This tube

may be considered as a tube-shaped balloon.

Figure 5.1 shows a rubber tube that is cut to one-quarter. The length

direction is x direction, and the circumferential direction is y direction.

Figure 5.1 A cross-section view of a rubber tube

39

The initial sizes of the tube are expressed as

（５-１）｛Length｝＝Ｌ０

（５-２）｛Inside radius｝＝Ｒ０

（５-３）｛Outside radius｝＝Ｒ０ｔ０ , ｛Outside radius ratio｝＝ｔ０

（５-４）｛Wall thickness rate｝＝ｔ０－1

（５-５）｛Wall thickness｝＝Ｔ０＝Ｒ０（ｔ０－1）

where the subscript indicates the initial (unstressed) state, and the outside

radius ratio ｔ０ is the ratio of the outside radius to the inside radius.

When the tube inflates, the internal pressure is Ｐ, that is the increase

from the atmospheric pressure, the initial length Ｌ０ becomes Ｌ０α, and

the inside radius Ｒ０ becomes Ｒ０β. α is the extension ratio in the x

direction, and β is the extension ratio in the y direction on the inside

surfaces. The shape of the tube is not spherical. The force ＦＰＸ acting in the

x direction is different from the force ＦＰＹ acting in the y-direction. Then α

is different from β.

The force ＦＰＸ acting in the x direction is caused by the internal pressure

Ｐ. Then ＦＰＸ is represented by the following equation.

（５-６）ＦＰＸ＝πＲ０２β２Ｐ

The force ＦＰＹ acting in the y direction is caused by the internal pressure

Ｐ. Then ＦＰＹ is represented by the following equation.

（５-７）ＦＰＹ＝２Ｒ０βＬ０αＰ

The force ＦＴＸ caused by the extension of the tube in the x direction is the

product of the initial cross-sectional area πＲ０２（ｔ０

２－１） and the stress

ｆ０Ｘ of the material of the tube. And it is represented by the following

equation.

（５-８）ＦＴＸ＝πＲ０２（ｔ０

２－１）ｆ０Ｘ

40

The force ＦＴＹ caused by the extension in the y direction is the product of

the initial cross-sectional area ２Ｒ０（ｔ０－1）Ｌ０ and the stress ｆ０Ｙ of the

material of the tube. And it is represented by the following equation.

（５-９）ＦＴＹ＝２Ｒ０（ｔ０－1）Ｌ０ｆ０Ｙ

Under the condition of equilibrium of the forces in the x direction, ＦＰＸ＝

ＦＴＸ, we have the following equation.

（５-１０） β２Ｐ＝（ｔ０２－１）ｆ０Ｘ

Under the condition of equilibrium of the forces in the y direction, ＦＰＹ＝

ＦＴＹ, we have the following equation.

（５-１１） αβＰ＝（ｔ０－1）ｆ０Ｙ

Equations（５-１０）and（５-１１） are the basic equations of the one-

layer calculation method for rubber tubes. And it is necessary to know about

ｆ０Ｘ and ｆ０Ｙ .


We use the C-type equation at n=2 as the expressions of ｆ０Ｘ and ｆ０Ｙ.

Those are the following equations.

（５-１２）ｆ０Ｘ＝Ｋ｛１－λＸ－１＋ｃ（λＸ－１）２｝

（５-１３）ｆ０Ｙ＝Ｋ｛１－λＹ－１＋ｃ（λＹ－１）２｝

In these equations, ｃ is put at 0.005, λＸ is the one-dimensional extension

ratio of the material of the tube in the x direction, and λＹ is that in the y

direction.

When we substitute the expression of equation（５-１２） for ｆ０Ｘ in

41

equation（５-１０）and transform λＸ into α and β by equation（１-１

６）,λＸ＝α４/３β２/３,we can write the following equation.

（５-１４）

Ｐ＝Ｋ（ｔ０２－１）β－２｛１－α－４/３β－２/３＋ｃ（α４/３β２/３－１）２｝

This is the equation derived from the condition of equilibrium of the forces in

the x direction.

Furthermore when we substitute the expression of equation（５-１３） for

ｆ０Ｙ in equation（５-１１）and transform λＹ into α and β by equation

（１-１７）, λＹ＝α２/３β４/３ ,we can write the following equation.

（５-１５）

Ｐ＝Ｋ（ｔ０-1）α－１β－１｛１－α－２/３β－４/３＋ｃ（α２/３β４/３－１）２｝

This is the equation derived from the condition of equilibrium of the forces in

the y direction.

We have to obtain the solutions that satisfy the equations（５-１４）and

（５-１５）. We can calculate, and draw the relation curves in Excel.

For example, from equations（５-１４）and（５-１５）, we can obtain the

following equation.

（５-１６）

０＝（ｔ０＋１）β－１｛１－α－４/３β－２/３＋ｃ（α４/３β２/３－１）２｝

－α－１｛１－α－２/３β－４/３＋ｃ（α２/３β４/３－１）２｝

This equation expresses the relation between α and β. Then we can

obtain the many pears of values of α and β that satisfy the equation（５-

１６）. And substituting these pears of values for α and β in equation

（５-１４）or（５-１５） give us the values of the internal pressureＰ.

The calculation has been done in the range of 0.995≦α≦5.0. Ｋ is put at

100 and ｃ is put at 0.005. The outside radius ratio ｔ０ is 1.05, then the

wall thickness is 5% of the inside radius. The relations of Ｐ to α and β

are shown by fine and heavy solid curves in Figure 5.2.

42

Figure 5.2 Results of the one-layer calculation method for a rubber tube

The relation curves expressed by equations（５-１４）and（５-１５）

at Ｋ＝100, ｃ＝0.005, ｎ＝2 andｔ０＝1.05 are shown as follows：

A fine solid curve：The relation of Ｐ to α

A heavy solid curve：The relation of Ｐ to β


at Ｋ＝100, ｃ＝0 andｔ０＝1.05 are shown for reference as follows：

A fine dotted curve：The relation of Ｐ to α

A heavy dotted curve：The relation of Ｐ to β

(These are equivalent to the use of A-type equation.)

43

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

0 1 2 3 4 5 6 7 8

Ｐ

α , β

Figure 5.3 An enlarged view of the vicinity of α＝1 in Figure 5.2

From observation of Figures 5.2, the each of two kinds of solid curves has

its own peak (maximal value) of the internal pressure. And the each curve

shows the re-increasing of the internal pressure after the peak. These are

similar to the balloon inflation. (For reference, two kinds of dotted curves ( at

ｃ＝0 ) do not show the re-increasing of the internal pressure after the peak,

because these are equivalent to using A-type equation.)

But, it is different from the balloon inflation that the extension ratio β in

y direction is much larger than the extension ratio α in x direction. For

example, at the peak of the internal pressure, α is 1.13 and β is 1.70.

The reason is based on the shape of the tube which is not spherical.

When Ｘ is defined as the force acting on an initial unit area in x direction,

Ｘ is expressed as Ｘ＝πＲ０２β２Ｐ／πＲ０

２（ｔ０２－１）. And when Ｙ is

44

0.0

0.2

0.4

0.6

0.8

0.995 1.000 1.005

Ｐ

α , β

defined as the force acting on an initial unit area in y direction, Ｙ is

expressed as Ｙ＝２Ｒ０βＬ０αＰ／２Ｒ０（ｔ０－1）Ｌ０ . Therefore the ratio

Ｙ／Ｘ is expressed as the following equation.

（５-１７）

Ｙ／Ｘ＝βα（ｔ０２－１）／｛β２（ｔ０－1）｝＝α（ｔ０＋１）／β

When we letｔ０＝1.05, α≒1 and β≒1, the ratio Ｙ／Ｘ is about 2.05,

that is, the force acting in the y direction is larger than two times of the force

acting in the x direction. So α＜β. Furthermore, from equation（５-１７）,

the larger ｔ０ becomes, the larger the ratio Ｙ／Ｘ becomes.

Figure 5.3 shows an enlarged view of the vicinity of α＝1 in Figure 5.2.

From observation of Figures 5.3, at the early stage in the tube inflation, a

little shrink in the x direction (length direction) occurs. This is the result of

the above calculation.

As you see in chapter 1, the extension of an elastomer block in the y direction

causes the shrink of the length in the x direction. For example, in equation

（１-１１）, ＡＸＹ＝ＡＸλＹ－１/２＝Ａ０λＸλＹ

－１/２, let λＸλＹ－１/２＜ 1

(that is, λＸ＜λＹ１/２ ), then α＝ＡＸＹ／Ａ０＜ 1 . This expression means

the shrink of the length in x direction.

In chapter 6, the shrinks of tubes with more thick walls have been

calculated. The results show that the thicker the wall becomes, the larger the

degree of the shrink becomes. In a tube inflation experiment, whether the

shrink of the tube is observed or not is very interesting.


Next, we use the C-type equation at n=3 as the expressions of ｆ０Ｘ and

ｆ０Ｙ. Those are the following equations.

（５-１８）ｆ０Ｘ＝Ｋ｛１－λＸ－１＋ｃ（λＸ－１）３｝

（５-１９）ｆ０Ｙ＝Ｋ｛１－λＹ－１＋ｃ（λＹ－１）３｝

45

In these equations, ｃ is put at 0.0002. λＸ is the one-dimensional extension

ratio of the material of the tube in the x direction, and λＹ is that in the y

direction.

When we substitute the expression of equation（５-１８）for ｆ０Ｘ in

equation（５-１０）and transform λＸ into α and β by the use of equation

（１-１６）,λＸ＝α４/３β２/３ , we can write the following equation.

（５-２０）

Ｐ＝Ｋ（ｔ０２－１）β－２｛１－α－４/３β－２/３＋ｃ（α４/３β２/３－１）３｝

This is the basic equation derived from the condition of equilibrium of the

forces in the x direction.

Furthermore when we substitute the expression of equation（５-１９） for

ｆ０Ｙ in equation（５-１１）and transform λＹ into α and β by the use of

equation（１-１７）, λＹ＝α２/３β４/３ ,we can write the following equation.

（５-２１）

Ｐ＝Ｋ（ｔ０-1）α－１β－１｛１－α－２/３β－４/３＋ｃ（α２/３β４/３－１）３｝

This is the basic equation derived from the condition of equilibrium of the

forces in the y direction.

We can calculate and obtain the solutions that satisfy the equations（５-２

０）and（５-２１）. For example, from equations（５-２０）and（５-２１）,

we can obtain the following equation.

（５-２２）

０＝（ｔ０＋１）β－１｛１－α－４/３β－２/３＋ｃ（α４/３β２/３－１）３｝

－α－１｛１－α－２/３β－４/３＋ｃ（α２/３β４/３－１）３｝

This equation expresses the relation between α and β. Then we can

obtain the many pears of values of α and β that satisfy the equation（５-

２２）. And substituting these pears of values for α and β in equation

（５-２０）or（５-２１） give us the values of the internal pressureＰ.

46


100 and ｃ is put at 0.0002. The outside radius ratio ｔ０ is 1.05, that is, the

wall thickness is 5% of the inside radius. The relations of Ｐ to α and β

are shown by fine and heavy solid curves in Figures 5.4.


calculated by equation（５-１４）and（５-１５）are also shown by fine and

heavy dotted curves in Figure 5.4.


almost the same as the corresponding peaks of dotted curves.

But, after the peak, the each solid curve has a smaller minimal value and has

steeper slopes of decreasing and increasing of the internal pressure than the

corresponding dotted curve. This is quite similar to Figure 3.4. Then we can

have the same discussion as the discussion held at the end of section 3.3.

An enlarged view of the vicinity of α＝1 in Figure 5.4 is, of course, the

same as Figure 5.3

47

Figure 5.4 Comparisons between n=3 and n=2 in C-type equation

The relation curves expressed by equations（５-２０）and（５-２１）

at Ｋ＝100, ｃ＝0.0002, ｔ０＝1.05 and ｎ＝3 are shown as follows：

A fine solid line：The relation of Ｐ to α

A heavy solid line：The relation of Ｐ to β


at Ｋ＝100, ｃ＝0.005, ｔ０＝1.05 and ｎ＝2 are shown as follows：

A fine dotted line：The relation of Ｐ to α

A heavy dotted line：The relation of Ｐ to β

48

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

0 1 2 3 4 5 6 7

Ｐ

α , β

6 Two-Layer Calculation Method for Rubber Tubes


When the wall thickness of a rubber tube is thick and we cannot ignore the


during the inflation, we can consider the wall to be two-layer. And the two-

layer calculation method for rubber tubes is described in this chapter.

Figure 6.1 shows a rubber tube that is cut to one-quarter. This figure shows

a two-layer wall. The length direction is x direction, and the circumferential

direction is y direction. We call the inside layer the 1st layer and the outside

layer the 2nd layer. In each layer, the inside radius timesｔ０ is the outside

radius. The outside radius ratioｔ０ is the ratio of the outside radius to the

inside radius. Thus the first layer wall thickness timesｔ０ is the second layer

wall thickness.

Figure6.1 A cross-section view of a rubber tube

49

The initial sizes of the tube are expressed as follows.

In total：

（６-１）｛Length｝＝Ｌ０

（６-２）｛Inside radius｝＝Ｒ０

（６-３）｛Outside radius｝＝Ｒ０ｔ０Ｔ＝Ｒ０ｔ０２

｛Total outside radius ratio｝＝ｔ０Ｔ＝ｔ０２

（６-４）｛Total wall thickness rate｝＝ｔ０Ｔ－1＝ｔ０２－1

（６-５）｛Total wall thickness｝＝Ｔ０Ｔ＝Ｒ０（ｔ０Ｔ－1）＝Ｒ０（ｔ０２－1）

In the 1st layer：

（６-６）｛1st layer inside radius｝＝Ｒ０

（６-７）｛1st layer outside radius｝＝Ｒ０ｔ０

（６-８）｛1st layer thickness rate｝＝ｔ０－1＝ｔ０Ｔ1/2－1

（６-９）｛1st layer thickness｝＝Ｔ０１＝Ｒ０（ｔ０－1）

In the 2nd layer：

（６-１０）｛2nd layer inside radius｝＝Ｒ０ｔ０

（６-１１）｛2nd layer outside radius｝＝Ｒ０ｔ０２

（６-１２）｛2nd layer thickness rate｝＝ｔ０－1＝ｔ０Ｔ1/2－1

（６-１３）｛2nd layer thickness｝＝Ｔ０２＝Ｒ０ｔ０（ｔ０－1）


from the atmospheric pressure, the length Ｌ０ increases to Ｌ０α, and the

1st layer inside radius Ｒ０ increases to Ｒ０β１ and the 2nd layer outside

radius Ｒ０ｔ０２ increases to Ｒ０ｔ０

２β２. α is the extension ratio in the x

direction. β１ is the extension ratio in the y direction on the 1st layer inside

surface and β２ is that on the 2nd layer outside surface. We assume that

β１ and β２ represent the extension ratios of the whole 1st and the whole

2nd layers respectively.

On the assumption that the volume of the rubber material of the tube does


concerning the volumes of the all layers before and after the inflation.

50

πＲ０２（ｔ０

４－１）Ｌ０＝πＲ０２（ｔ０

４β２２－β１

２）Ｌ０α


relation between β１ and β２.

（６-１４） β１＝（ｔ０４β２

２－ｔ０４α－１＋α－１）１/２

（６-１５） β２＝（ｔ０－４β１

２－ｔ０－４α－１＋α－１）１/２

We can transform β１ into β２ by equation（６-１４）, and we can transform

β２ into β１ by equation（６-１５）

The forces caused by the internal pressure

The force ＦＰＸ acting in the x direction caused by the internal pressure Ｐ


（６-１６）ＦＰＸ＝πＲ０２β１

２Ｐ

The force ＦＰＹ acting in the y direction caused by the internal pressure Ｐ


（６-１７）ＦＰＹ＝２Ｒ０β１Ｌ０αＰ

The forces caused by the tube extension

In the 1st layer：

The force ＦＴ１Ｘ caused by the 1st layer extension in the x direction is the



ｆ０１Ｘ. And it is represented by the following equation.

（６-１８）ＦＴ１Ｘ＝πＲ０２（ｔ０

２－１）ｆ０１Ｘ

51

The force ＦＴ１Ｙ caused by the 1st layer extension in the y direction is the

product of the initial cross-sectional area ２Ｒ０（ｔ０－1）Ｌ０ and the stress

ｆ０1Ｙ. It is represented by the following equation.

（６-１９）ＦＴ１Ｙ＝２Ｒ０（ｔ０－1）Ｌ０ｆ０１Ｙ

In the 2nd layer：

The force ＦＴ２Ｘ caused by the 2nd layer extension in the x direction is the

product of the initial cross-sectional area πＲ０２ｔ０

２（ｔ０２－１） and the

stressｆ０２Ｘ. It is represented by the following equation.

（６-２０）ＦＴ２Ｘ＝πＲ０２ｔ０

２（ｔ０２－１）ｆ０２Ｘ

The force ＦＴ２Ｙ caused by the 2nd layer extension in the y direction is the

product of the initial cross-sectional area ２Ｒ０ｔ０（ｔ０－1）Ｌ０ and the

stress ｆ０２Ｙ. It is represented by the following equation.

（６-２１）ＦＴ２Ｙ＝２Ｒ０ｔ０（ｔ０－1）Ｌ０ｆ０２Ｙ

Equilibrium of the forces：

The condition of equilibrium of the forces in the x direction is ＦＰＸ＝ＦＴ１Ｘ

＋ＦＴ２Ｘ. From equations（６-１６）,（６-１８）and（６-２０）, we have


（６-２２） β１２Ｐ＝（ｔ０

２－１）ｆ０１Ｘ＋ｔ０２（ｔ０

２－１）ｆ０２Ｘ

The condition of equilibrium of the forces in the y direction is ＦＰＹ＝ＦＴ１Ｙ

＋ＦＴ２Ｙ. From equations（６-１７）,（６-１９）and（６-２１）, we have


（６-２３） β１αＰ＝（ｔ０－1）ｆ０１Ｙ＋ｔ０（ｔ０－1）ｆ０２Ｙ

52

Equations（６-２２）and（６-２３） are the basic equations of the two-layer

calculation method for rubber tubes. And it is necessary to know about ｆ

０１Ｘ , ｆ０２Ｘ , ｆ０１Ｙ and ｆ０２Ｙ .


We can write the following equations as the expressions of ｆ０１Ｘ , ｆ０２Ｘ ,

ｆ０１Ｙ and ｆ０２Ｙ by the use of the C-type equation at n=2.

（６-２４）ｆ０１Ｘ＝Ｋ｛１－λ１Ｘ－１＋ｃ（λ１Ｘ－１）２｝

（６-２５）ｆ０２Ｘ＝Ｋ｛１－λ２Ｘ－１＋ｃ（λ２Ｘ－１）２｝

（６-２６）ｆ０１Ｙ＝Ｋ｛１－λ１Ｙ－１＋ｃ（λ１Ｙ－１）２｝

（６-２７）ｆ０２Ｙ＝Ｋ｛１－λ２Ｙ－１＋ｃ（λ２Ｙ－１）２｝

In these equations, ｃ is put at 0.005. λ１Ｘ and λ２Ｘ are the one-

dimensional extension ratios of the material of the tube in the x direction, and

λ１Ｙ and λ２Ｙ are those in the y direction.

We can substitute the above expressions for ｆ０１Ｘ , ｆ０２Ｘ , ｆ０１Ｙ and

ｆ０２Ｙ in equations（６-２２）and（６-２３）, and transform λ１Ｘ , λ２Ｘ ,

λ１Ｙ and λ２Ｙ into α, β１ and β２ by the following equations ：

λ１Ｘ＝α４/３β１２/３, λ２Ｘ＝α４/３β２

２/３ , λ１Ｙ＝α２/３β１４/３ and λ２Ｙ＝

α２/３β２４/３ which are derived from equations （１-１６）and（１-１７）.

Then we can write the following equations.

（６-２８） In the x direction：

Ｐ＝Ｋ（ｔ０２－１）β１

－２［１－α－４/３β１－２/３＋ｃ（α４/３β１

２/３－１）２

＋ｔ０２｛１－α－４/３β２

－２/３＋ｃ（α４/３β２２/３－１）２｝］

（６-２９） In the y direction：

Ｐ＝Ｋ（ｔ０－1）α－１β１－１［１－α－２/３β１

－４/３＋ｃ（α２/３β１４/３－１）２

＋ｔ０｛１－α－２/３β２－４/３＋ｃ（α２/３β２

４/３－１）２｝］

53

Equations（６-２８）and（６-２９）express the relation of Ｐ to α, β１ and

β２ . In these equations, β１ and β２ are considered as one variable

because each of them is determined from the other one by the equation （６

-１４）or（６-１５）.

We have to obtain the solutions that satisfy equations（６-２８）and（６-

２９）. We can calculate, and draw the relation curves in Excel.

For example, from equations（６-２８）and（６-２９）, we can obtain the

following equation.

（６-３０）

０＝（ｔ０＋１）β１－１［１－α－４/３β１

－２/３＋ｃ（α４/３β１２/３－１）２

＋ｔ０２｛１－α－４/３β２

－２/３＋ｃ（α４/３β２２/３－１）

２｝］

－α－１［１－α－２/３β１－４/３＋ｃ（α２/３β１

４/３－１）２

＋ｔ０｛１－α－２/３β２－４/３＋ｃ（α２/３β２

４/３－１）２｝］

Transforming β２ in this equation into β１ by equation（６-１５） give us


（６-３１）

０＝（ｔ０＋１）β１－１［１－α－４/３β１

－２/３＋ｃ（α４/３β１２/３－１）２

＋ｔ０２｛１－α－４/３（ｔ０

－４β１２－ｔ０

－４α－１＋α－１）－１/３

＋ｃ（α４/３（ｔ０－４β１

２－ｔ０－４α－１＋α－１）１/３－１）２｝］

－α－１［１－α－２/３β１－４/３＋ｃ（α２/３β１

４/３－１）２

＋ｔ０｛１－α－２/３（ｔ０－４β１

２－ｔ０－４α－１＋α－１）－２/３

＋ｃ（α２/３（ｔ０－４β１

２－ｔ０－４α－１＋α－１）２/３－１）２｝］

This equation expresses the relation between α and β１. Then we can

obtain the many pears of values of α and β１ that satisfy equation（６-３

１）. And we can obtain the values of β２ from the pears of values of α and

β１ by equation（６-１５）.

Substituting these values for α, β１ and β２ in equation（６-２８）or（６

-２９）give us the values of Ｐ.

54


100 and ｃ is put at 0.005. The total outside radius ratio ｔ０Ｔ is 1.05, so the

total wall thickness is 5% of the inside radius. The relation curves of Ｐ to

α, β１ and β２ are shown in Figure 6.2. When the total outside radius

ratio (ｔ０Ｔ ) is more than 1.05, the relation curves are shown in Figures 6.3

and 6.4.

Figure 6.2 Results ( 1 ) of the two-layer calculation method

The relation curves expressed by equations（６-２８）and（６-２９）

at Ｋ＝100 and ｃ＝0.005, ｔ０Ｔ＝ｔ０２＝1.05

A fine solid curve：The relation of Ｐ to α

A heavy solid curve：The relation of Ｐ to β１

A dotted curve：The relation of Ｐ to β２

55

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

0 1 2 3 4 5 6 7 8

Ｐ

α , β１ , β2

Figure 6.3 Results ( 2 ) of the two-layer calculation method

The relation curves expressed by equations（６-２８）and（６-２９）

at Ｋ＝100 and ｃ＝0.005

Fine solid curves：The relations of Ｐ to α

Heavy solid curves：The relations of Ｐ to β１

Dotted curves：The relations of Ｐ to β２


ｔ０Ｔ＝ｔ０２＝1.20, 1.50, 2.00,

that is to say, ｔ０＝1.0954, 1.2247, 1.4142

56

0.0

5.0

10.0

15.0

20.0

25.0

0 1 2 3 4 5 6 7 8 9

Ｐ

α , β１ , β２

Figure 6.4 An enlarged view of the vicinity of α＝1 in Figure 6.3

From observation of Figure 6.2, it is seen that β２ (the extension ratio in

the y direction on the outside surface) is smaller than β１ (the extension

ratio in the y direction on the inside surface) at the same internal pressure.

But the difference between β１ and β２ is a little because the wall thickness

is thin (ｔ０Ｔ＝ｔ０２＝1.05).

From observation of Figure 6.3, it is seen that the larger ｔ０Ｔ becomes,

the smaller α and β２ become at the peak of the internal pressure, but the

larger β１ becomes.

From observation of Figure 6.4, it is seen that, at the early stage in the tube

inflation, the largerｔ０Ｔ becomes, the larger the degree of shrink in the x

direction becomes. The reason is as follows: From equation（５-１７）,

57

0.0

5.0

10.0

15.0

20.0

25.0

0.98 0.99 1.00 1.01 1.02 1.03 1.04 1.05

Ｐ

α , β１ , β２

the larger ｔ０ becomes, the larger the ratio Ｙ／Ｘ becomes. For example,

at the early stage in a tube inflation, when we let ｔ０Ｔ＝1.50 ( the total wall

thickness is 50% of the inside radius ), α≒1 and β２≒1, the ratio Ｙ

／Ｘ is about 2.50, that is to say, the force acting in the y direction is about

2.5 times as large as the force acting in the x direction. Then, the larger

extension occurs in the y direction, and this larger extension causes the larger

shrink in the x direction as you see in chapter 1.

In an experiment of a tube inflation, whether the shrink in the x direction

at the early stage in the inflation is observed or not is very interesting.

Comparisons between the one-layer and the two-layer calculation methods

are shown in Figure 6.5. These curves are calculated at Ｋ＝100 and ｃ＝

0.005. And the wall thicknesses are 5%, 10%, 20% and 50% of the inside

radius in both methods.

From observation of Figure 6.5, we can see the following. When the wall

thickness is 5% of the inside radius, the bottom dotted curve (one-layer

calculation method) almost agrees to the bottom solid curve (two-layer

calculation method). But when the wall thickness becomes 10% or more, the

peaks of the dotted curves are higher than the corresponding peaks of the

solid curves. Especially when the wall thickness is 50% of the inside radius,

the peak of the dotted curve is much higher than the corresponding peak of

the solid curve by 20%. The reason is as follows： In the one-layer calculation

method, it is assumed that the extension ratio of the outside surface is equal

to that of the inside surface in spite of large wall thickness.

Therefore we had better to use the two-layer calculation method when the

wall thickness is larger than 5% of the inside radius.

58

Figure 6.5 Comparison between the one-layer and the two-layer

calculation methods

Dotted curves：By the one-layer calculation method

The relations of Ｐ to β expressed by equations（５-１４）and

（５-１５）at Ｋ＝100 and ｃ＝0.005


ｔ０＝1.05, 1.10, 1.20, 1.50

Solid curves：By the two-layer calculation method

The relations of Ｐ to β２ expressed by equations（６-２８）and

（６-２９）at Ｋ＝100 and ｃ＝0.005


ｔ０Ｔ＝ｔ０２＝1.05, 1.10, 1.20, 1.50

59

0.0

2.0

4.0

6.0

8.0

10.0

12.0

14.0

16.0

0 1 2 3 4 5 6 7 8

Ｐ

β , β２


Next, we can write the following equations as the expressions of ｆ０１Ｘ , ｆ

０２Ｘ , ｆ０１Ｙ and ｆ０２Ｙ by the use of the C-type equation at n=3.

（６-３２）ｆ０１Ｘ＝Ｋ｛１－λ１Ｘ－１＋ｃ（λ１Ｘ－１）３｝

（６-３３）ｆ０２Ｘ＝Ｋ｛１－λ２Ｘ－１＋ｃ（λ２Ｘ－１）３｝

（６-３４）ｆ０１Ｙ＝Ｋ｛１－λ１Ｙ－１＋ｃ（λ１Ｙ－１）３｝

（６-３５）ｆ０２Ｙ＝Ｋ｛１－λ２Ｙ－１＋ｃ（λ２Ｙ－１）３｝

In these equations, ｃ is put at 0.0002. λ１Ｘ and λ２Ｘ are the one-


λ１Ｙ and λ２Ｙ are those in the y direction.

We can substitute the above expressions for ｆ０１Ｘ , ｆ０２Ｘ , ｆ０１Ｙ and

ｆ０２Ｙ in equations（６-２２）and（６-２３）, and transform λ１Ｘ , λ２Ｘ ,

λ１Ｙ and λ２Ｙ into α, β１ and β２ by the following equations ：

λ１Ｘ＝α４/３β１２/３, λ２Ｘ＝α４/３β２

２/３ , λ１Ｙ＝α２/３β１４/３ and λ２Ｙ＝

α２/３β２４/３ which are derived from equations （１-１６）and（１-１７）.

Then we can write the following equations

（６-３６） In the x direction：

Ｐ＝Ｋ（ｔ０２－１）β１

－２［１－α－４/３β１－２/３＋ｃ（α４/３β１

２/３－１）３

＋ｔ０２｛１－α－４/３β２

－２/３＋ｃ（α４/３β２２/３－１）３｝］

（６-３７） In the y direction：

Ｐ＝Ｋ（ｔ０－1）α－１β１－１［１－α－２/３β１

－４/３＋ｃ（α２/３β１４/３－１）３

＋ｔ０｛１－α－２/３β２－４/３＋ｃ（α２/３β２

４/３－１）３｝］

Equations（６-３６）and（６-３７）express the relation of Ｐ to α, β１ and

β２ , but β１ and β２ are not two independent valuables because each of

them is determined from the other one by the equation （６-１４）or（６-１

５）.

60

We have to obtain the solutions that satisfy equations（６-３６）and（６-

３７）. We can calculate, and draw the relation curves in Excel.

For example, from equations（６-３６）and（６-３７）, we can obtain the

following equation.

（６-３８）

０＝（ｔ０＋１）β１－１［１－α－４/３β１

－２/３＋ｃ（α４/３β１２/３－１）３

＋ｔ０２｛１－α－４/３β２

－２/３＋ｃ（α４/３β２２/３－１）

３｝］

－α－１［１－α－２/３β１－４/３＋ｃ（α２/３β１

４/３－１）３

＋ｔ０｛１－α－２/３β２－４/３＋ｃ（α２/３β２

４/３－１）３｝］

Transforming β２ in this equation into β１ by equation（６-１５） give us


（６-３９）

０＝（ｔ０＋１）β１－１［１－α－４/３β１

－２/３＋ｃ（α４/３β１２/３－１）３

＋ｔ０２｛１－α－４/３（ｔ０

－４β１２－ｔ０

－４α－１＋α－１）－１/３

＋ｃ（α４/３（ｔ０－４β１

２－ｔ０－４α－１＋α－１）１/３－１）３｝］

－α－１［１－α－２/３β１－４/３＋ｃ（α２/３β１

４/３－１）３

＋ｔ０｛１－α－２/３（ｔ０－４β１

２－ｔ０－４α－１＋α－１）－２/３

＋ｃ（α２/３（ｔ０－４β１

２－ｔ０－４α－１＋α－１）２/３－１）３｝］

This equation expresses the relation between α and β１. Then we can

obtain the many pears of values of α and β１ that satisfy equation（６-３

９）. And we can obtain the values of β２ from the pears of values of α and

β１ by equation（６-１５）.

Substituting these values for α, β１ and β２ in equation（６-３６）or（６

-３７）give us the values of Ｐ.


100 and ｃ is put at 0.0002. And the total outside radius ratios are 1.20 and

2.00, that is to say, the total wall thicknesses are 20% and 100% of the inside

radius. The relation curves of Ｐ to α, β１ and β２ are shown in Figure

6.6.

61

Figure 6.6 Results of the two-layer calculation method

The relation curves expressed by equations（６-３６）and（６-３７）

at Ｋ＝100 and ｃ＝0.0002



Dotted curves：The relations of Ｐ to β２


ｔ０Ｔ＝ｔ０２＝1.20, 2.00, that is to say, ｔ０＝1.0954, 1.4142

62

0.0

5.0

10.0

15.0

20.0

25.0

0 1 2 3 4 5 6 7 8

Ｐ

α , β１ , β２


Heavy solid curves：The relations of Ｐ to α２

expressed by equations（６-３６）and（６-３７）

at Ｋ＝100, ｃ＝0.0002 and ｎ＝3

Fine solid curves：The relations of Ｐ to α２

expressed by equations（６-２８）and（６-２９）

at Ｋ＝100, ｃ＝0.005 and ｎ＝2


ｔ０Ｔ＝ｔ０２＝1.20, 1.50, 2.00

that is to say, ｔ０＝1.0954, 1.2247, 1.4142

63

0.0

5.0

10.0

15.0

20.0

25.0

0 1 2 3 4 5 6 7

Ｐ

β２

Figure 6.6 is similar to Figure 6.3. And the larger ｔ０Ｔ becomes, the

smaller α and β２ become at the peak of the internal pressure, but the

larger β１ becomes. Of course, the shrink in the x direction (length direction)

occurs at the early stage of the tube inflation in Figure 6.6 as it is shown in

Figure 6.4.

To compare the effect of ｎ＝3 with ｎ＝2 in C-type equation, three pairs

of curves are shown in Figure 6.7. Heavy solid curves are expressed by

equations（６-３６）and（６-３７）at ｎ＝3. Fine solid curves are expressed

by equations（６-２８）and（６-２９）at ｎ＝2.

From observation of Figure 6.7, it is seen that these shapes of curves are

quite similar to the curves in Figure 3.4. Then we can have the same

discussion held at the end of section 3.3.

64

7 Four-Layer Calculation Method for Rubber Tubes


In this chapter, a four-layer calculation method is investigated to improve

the accuracy of the calculation of rubber tubes. Figure 7.1 shows a rubber

tube that is cut to one-quarter. This tube has a four-layer wall. The length

direction is x direction, and the circumferential direction is y direction. We

call the layers the 1st, the 2nd, the 3rd and the 4th layers in order from the

inside. The initial sizes of the rubber tube are shown in Table 7.1.

Figure 4.1 A cross-section view of a rubber tube

65

Table 7.1 The initial sizes of the rubber tube

Inside

radius

Outside radius Wall thickness

1st layer Ｒ０Ｒ０ｔ０Ｒ０（ｔ０－1）

2nd layer Ｒ０ｔ０Ｒ０ｔ０２Ｒ０ｔ０（ｔ０－1）

3rd layer Ｒ０ｔ０２Ｒ０ｔ０

３Ｒ０ｔ０２（ｔ０－1）

4th layer Ｒ０ｔ０３Ｒ０ｔ０

４Ｒ０ｔ０３（ｔ０－1）

Total Ｒ０Ｒ０ｔ０Ｔ,(ｔ０Ｔ＝ｔ０４) Ｒ０（ｔ０Ｔ－1）

The initial length of this tube is Ｌ０. In each layer, the inside radius times

ｔ０ is the outside radius. Thus the 1st layer wall thickness timesｔ０ is the

2nd layer wall thickness. And the 2nd layer wall thickness timesｔ０ is the

3rd layer wall thickness. And so on. ｔ０ is the outside radius ratio of each

layer. ｔ０Ｔ is the total outside radius ratio. Then ｔ０Ｔ＝ｔ０４.


from the atmospheric pressure, the length Ｌ０ increases to Ｌ０α, the 1st

layer inside radius increases to Ｒ０β １ , the 2nd layer outside radius

increases to Ｒ０ｔ０２β２ and the 4th layer outside radius increases to Ｒ０

ｔ０４β３. Then α is the extension ratio in the x direction. β１ is the

extension ratio in the y direction on the 1st layer inside surface, β２ is that

on the 2nd layer outside surface and β３ is that on the 4th layer outside

surface. We assume that β １ represents the extension ratio (in the y

direction) of the whole 1st layer , β２ represents that of the whole 2nd and

3rd layers and β３ represents that of the whole 4th layer.

On the assumption that the volume of the rubber material of the tube does


concerning the volumes of the 1st and the 2nd layers before and after the

inflation.


４－１）Ｌ０＝πＲ０２（ｔ０

４β２２－β１

２）Ｌ０α


66


（７-１） β１＝（ｔ０４β２

２－ｔ０４α－１＋α－１）１/２

（７-２） β２＝（ｔ０－４β１

２－ｔ０－４α－１＋α－１）１/２

We can transform β１ into β２ by equation（７-１）, and we can transform

β２ into β１ by equation（７-２）.

Furthermore, we can write the following equation concerning the volumes of

the all layers before and after the inflation.


８－１）Ｌ０＝πＲ０２（ｔ０

８β３２－β１

２）Ｌ０α


relation between β１ and β３.

（７-３） β１＝（ｔ０８β３

２－ｔ０８α－１＋α－１）１/２

（７-４） β３＝（ｔ０－８β１

２－ｔ０－８α－１＋α－１）１/２

We can transform β１ into β３ by equation（７-３）, and we can transform

β３ into β１ by equation（７-４）.

The forces caused by the internal pressure Ｐ



（７-５）ＦＰＸ＝πＲ０２β１

２Ｐ

The force ＦＰＹ acting in the y direction caused by the internal pressure

Ｐ is represented by the following equation.

（７-６）ＦＰＹ＝２Ｒ０β１Ｌ０αＰ

67

The forces caused by the tube extension

In the 1st layer




ｆ０１Ｘ . It is represented by the following equation.

（７-７）ＦＴ１Ｘ＝πＲ０２（ｔ０

２－１）ｆ０１Ｘ


product of the initial cross-sectional area ２Ｒ０（ｔ０－1）Ｌ０ and the stress

ｆ０1Ｙ . It is represented by the following equation.

（７-８）ＦＴ１Ｙ＝２Ｒ０（ｔ０－1）Ｌ０ｆ０１Ｙ

In the 2nd and the 3rd layers

The force ＦＴ２Ｘ caused by the 2nd and 3rd layers extension in the x

direction is the product of the initial cross-sectional area πＲ０２ｔ０

２（ｔ０４

－１） and the stress ｆ０２Ｘ . It is represented by the following equation.

（７-９）ＦＴ２Ｘ＝πＲ０２ｔ０

２（ｔ０４－１）ｆ０２Ｘ

The force ＦＴ２Ｙ caused by the 2nd and 3rd layers extension in the y

direction is the product of the initial cross-sectional area ２Ｒ０ｔ０（ｔ０２－

1）Ｌ０ and the stress ｆ０２Ｙ . It is represented by the following equation.

（７-１０）ＦＴ２Ｙ＝２Ｒ０ｔ０（ｔ０２－1）Ｌ０ｆ０２Ｙ

In the 4th layer

The force ＦＴ３Ｘ caused by the 4th layer extension in the x direction is the

product of the initial cross-sectional area πＲ０２ｔ０

６（ｔ０２－１） and the

stress ｆ０３Ｘ . It is represented by the following equation.

（７-１１）ＦＴ３Ｘ＝πＲ０２ｔ０

６（ｔ０２－１）ｆ０３Ｘ

68

The force ＦＴ３Ｙ caused by the 4th layer extension in the y direction is the

product of the initial cross-sectional area ２Ｒ０ｔ０３（ｔ０－1）Ｌ０ and the

stress ｆ０３Ｙ . It is represented by the following equation.

（７-１２）ＦＴ３Ｙ＝２Ｒ０ｔ０３（ｔ０－1）Ｌ０ｆ０３Ｙ

Equilibrium of the forces

The condition of equilibrium of the forces in the x direction is ＦＰＸ＝ＦＴ１Ｘ

＋ＦＴ２Ｘ＋ＦＴ３Ｘ. From equations（７-５）, （７-７）,（７-９）and（７-

１１）, we have the following equation.

（７-１３） β１２Ｐ＝（ｔ０

２－１）ｆ０１Ｘ＋ｔ０２（ｔ０

４－１）ｆ０２Ｘ

＋ｔ０６（ｔ０

２－１）ｆ０３Ｘ

The condition of equilibrium of the forces in the y direction isＦＰＹ＝ＦＴ１Ｙ

＋ＦＴ２Ｙ＋ＦＴ３Ｙ. From equations（７-６）, （７-８）,（７-１０）and（７

-１２）, we have the following equation.

（７-１４） β１αＰ＝（ｔ０－1）ｆ０１Ｙ＋ｔ０（ｔ０２－1）ｆ０２Ｙ

＋ｔ０３（ｔ０－1）ｆ０３Ｙ

Equations（７-１３）and（７-１４） are the basic equations of the four-layer

calculation method for rubber tubes. Then it is necessary to know about ｆ

０１Ｘ , ｆ０２Ｘ , ｆ０３Ｘ , ｆ０１Ｙ , ｆ０２Ｙ and ｆ０３Ｙ.



ｆ０３Ｘ , ｆ０１Ｙ , ｆ０２Ｙ and ｆ０３Ｙ by the use of the C-type equation at n=2.

69

（７-１５）ｆ０１Ｘ＝Ｋ｛１－λ１Ｘ－１＋ｃ（λ１Ｘ－１）２｝

（７-１６）ｆ０２Ｘ＝Ｋ｛１－λ２Ｘ－１＋ｃ（λ２Ｘ－１）２｝

（７-１７）ｆ０３Ｘ＝Ｋ｛１－λ３Ｘ－１＋ｃ（λ３Ｘ－１）２｝

（７-１８）ｆ０１Ｙ＝Ｋ｛１－λ１Ｙ－１＋ｃ（λ１Ｙ－１）２｝

（７-１９）ｆ０２Ｙ＝Ｋ｛１－λ２Ｙ－１＋ｃ（λ２Ｙ－１）２｝

（７-２０）ｆ０３Ｙ＝Ｋ｛１－λ３Ｙ－１＋ｃ（λ３Ｙ－１）２｝

In these equations, ｃ is put at 0.005. λ１Ｘ, λ２Ｘ and λ３Ｘ are the one-


λ１Ｙ, λ２Ｙ and λ３Ｙ are those in the y direction.

When we substitute the above expressions for ｆ０１Ｘ , ｆ０２Ｘ , ｆ０３Ｘ ,

ｆ０１Ｙ , ｆ０２Ｙ and ｆ０３Ｙ in equations（７-１３）and（７-１４）, and

transform λｉＸ and λｉＹ into α and βｉ (where ｉ=1, 2, 3 ) by the

equations of λｉＸ＝α４/３βｉ２/３ and λｉＹ＝α２/３βｉ

４/３ (where ｉ=1, 2, 3 )

which are derived from equations （１-１６）and（１-１７）, we can write

the following equations.

（７-２１） In the x direction：

Ｐ＝Ｋ（ｔ０２－１）β１

－２｛１－α－４/３β１－２/３＋ｃ（α４/３β１

２/３－１）２｝

＋Ｋｔ０２（ｔ０

４－１）β１－２｛１－α－４/３β２

－２/３＋ｃ（α４/３β２２/３－１）２｝

＋Ｋｔ０６（ｔ０

２－１）β１－２｛１－α－４/３β３

－２/３＋ｃ（α４/３β３２/３－１）２｝

（７-２２） In the y direction：

Ｐ＝Ｋ（ｔ０－1）β１－１α－１｛１－α－２/３β１

－４/３＋ｃ（α２/３β１４/３－１）２｝

＋ｔ０Ｋ(ｔ０２－1）β１

－１α－１｛１－α－２/３β２－４/３＋ｃ（α２/３β２

４/３－１）２｝

＋ｔ０３Ｋ(ｔ０－1）β１

－１α－１｛１－α－２/３β３－４/３＋ｃ（α２/３β３

４/３－１）２｝

Equations（７-２１）and（７-２２） express the relation of Ｐ to α, β１ ,

β２ and β３ . In these equations, β１ , β２ and β３ are considered as

one variable because any one of them is determined from the other one by the

equations （７-１）through（７-４）.

70

We have to obtain the solutions that satisfy equations（７-２１）and（７-

２２）. We can calculate, and draw the relation curves in Excel.

For example, when we eliminate Ｐ from the equations（７-２１）and（７-

２２） and transform β２ and β３ into α and β１ by equations（７-２）

and（７-４）, we have the equation which expresses the relation between α

and β１. And we can obtain the many pears of values of α and β１ that

satisfy this equation. From these values of α and β１, we can obtain the

values of β２ and β３ by equations（７-２）and（７-４）. And from the

values of α, β１, β２ and β３, we can obtain the values of Ｐ by equation

（７-２１）or（７-２２）.


100 and ｃ is put at 0.005. The total outside radius ratios are 1.3 and 2.0,

that is to say, the total wall thicknesses are 30% and 100% of the inside radius.

The relation curves of Ｐ to α, β１ , β２ and β３ are shown in Figure 7.2.

From this figure, it is seen that the extension ratios are β１＞β２＞β３＞α

at the peak of the internal pressure. This figure is similar to Figure 6.3. Of

course, in Figure 7.2, the shrink in the x direction occurs at the early stage of

the tube inflation as it is shown in Figure 6.4.

To compare the four-layer calculation method with the two-layer calculation

method, the relations of Ｐ to α３ expressed by equations（７-２１）and（７

-２２）are shown by solid curves in Figure 7.3. And the relations of Ｐ to

α２ expressed by equations（６-２８）and（６-２９）are also shown by dotted

curves. These corresponding curves have been calculated under the same

wall thickness conditions in the range of 0.98≦α≦5.0. The bottom one of

dotted curves agrees to the bottom one of solid curves exactly in this figure.


much the same as the corresponding peaks of dotted curves. But when ｔ０Ｔ

is 2.0, there are some differences between the two-layer and the four-layer

calculation methods. Whether we should choose the four-layer calculation

method or not depends upon a need of the accuracy of the calculation.

71

Figure 7.2 Results of the four-layer calculation method (at n=2)

The relation curves expressed by equations（７-２１）and（７-２２）

at Ｋ＝100 and ｃ＝0.005



Broken curves：The relations of Ｐ to β２

Dotted curves：The relations of Ｐ to β３


ｔ０Ｔ＝ｔ０４＝1.30, 2.00, that is to say, ｔ０＝1.0678, 1.1892

72

0.0

5.0

10.0

15.0

20.0

25.0

0 1 2 3 4 5 6 7 8 9 10

Ｐ

α , β１ , β２ , β３

Figure 7.3 Comparison between two-layer and four-layer calculation

methods (at n=2)

Solid curves：Four-Layer Calculation Method

The relations of Ｐ to α３ expressed by

equations（７-２１）and（７-２２）at Ｋ＝100 and ｃ＝0.005


ｔ０Ｔ＝ｔ０４＝1.20, 1.30, 1.50, 2.00

Dotted curves：Two-Layer Calculation Method

The relations of Ｐ to α２ expressed by

equations（６-２８）and（６-２９）at Ｋ＝100 and ｃ＝0.005


ｔ０Ｔ＝ｔ０２＝1.20, 1.30, 1.50, 2.00

73

0.0

5.0

10.0

15.0

20.0

25.0

0 1 2 3 4 5 6 7

Ｐ

β２ , β３


Next, we can write the following equations as the expressions of ｆ０１Ｘ , ｆ

０２Ｘ , ｆ０３Ｘ , ｆ０１Ｙ , ｆ０２Ｙ and ｆ０３Ｙ by the use of the C-type equation

at n=3.

（７-２３）ｆ０１Ｘ＝Ｋ｛１－λ１Ｘ－１＋ｃ（λ１Ｘ－１）３｝

（７-２４）ｆ０２Ｘ＝Ｋ｛１－λ２Ｘ－１＋ｃ（λ２Ｘ－１）３｝

（７-２５）ｆ０３Ｘ＝Ｋ｛１－λ３Ｘ－１＋ｃ（λ３Ｘ－１）３｝

（７-２６）ｆ０１Ｙ＝Ｋ｛１－λ１Ｙ－１＋ｃ（λ１Ｙ－１）３｝

（７-２７）ｆ０２Ｙ＝Ｋ｛１－λ２Ｙ－１＋ｃ（λ２Ｙ－１）３｝

（７-２８）ｆ０３Ｙ＝Ｋ｛１－λ３Ｙ－１＋ｃ（λ３Ｙ－１）３｝

In these equations, ｃ is put at 0.0002.

The calculation can be done according to the process described in the above

section, therefore the details of the derivation of equations are omitted. The

results of the four-layer calculation method (at n=3) are shown in Figure 7.4.

This figure is similar to Figure 6.6.

And comparison between n=3 and n=2 in C-type equation are shown in

Figure 7.5. This figure is similar to Figures 6.7, 4.4 and 3.4. Then we can have

the same discussion as the discussion held at the end of section 3.3.

74

Figure 7.4 Results of the four-layer calculation method (at n=3)

The curves calculated by the use of the C-type equation

at Ｋ＝100 , ｃ＝0.0002 and ｎ＝3



Broken curves：The relations of Ｐ to β２

Dotted curves：The relations of Ｐ to β３


ｔ０Ｔ＝ｔ０４＝1.30, 2.00, that is to say, ｔ０＝1.0678, 1.1892

75

0.0

5.0

10.0

15.0

20.0

25.0

0 1 2 3 4 5 6 7 8 9

Ｐ

α , β１ , β２ , β３


Heavy solid curves：The relations of Ｐ to β３

expressed by the equations at Ｋ＝100, ｃ＝0.0002 and ｎ＝3

Fine solid curves：The relations of Ｐ to β３

expressed by equations（７-２１）and（７-２２）

at Ｋ＝100, ｃ＝0.005 and ｎ＝2


ｔ０Ｔ＝ｔ０２＝1.20, 1.50, 2.00

76

0.0

5.0

10.0

15.0

20.0

25.0

0 1 2 3 4 5 6 7

Ｐ

β3

8 A Try for Mechanics of Blood Vessels


In this chapter, the four-layer calculation method for rubber tubes is

applied to a blood vessel. I am not a professional in medical science. So this

may be a mere try. When a blood vessel wall is considered to be a four-layer,

the cross-section view is shown in Figure 8.1. This figure shows a blood vessel

cut to one-quarter. Of cause you can change a four-layer wall into a two-layer

wall or a one-layer wall.

Figure 8.1 A cross-section view of a blood vessel

77

The length direction is x direction, and the circumferential direction is y

direction. We call the layers the 1st, the 2nd, the 3rd and the 4th layers in

order from the inside. The initial sizes of the blood vessel are shown in Table

8.1. The initial sizes mean the unstressed sizes under the atmospheric

pressure.

Table 8.1 The initial sizes of the blood vessel

Inside radius Outside radius Wall thickness

1st layer Ｒ０Ｒ０ｔ１Ｒ０（ｔ１－1）

2nd layer Ｒ０ｔ１Ｒ０ｔ１ｔ２Ｒ０ｔ１（ｔ２－1）

3rd layer Ｒ０ｔ１ｔ２Ｒ０ｔ１ｔ２ｔ３Ｒ０ｔ１ｔ２（ｔ３－1）

4th layer Ｒ０ｔ１ｔ２ｔ３Ｒ０ｔ１ｔ２ｔ３ｔ４Ｒ０ｔ１ｔ２ｔ３（ｔ４－1）

Total Ｒ０Ｒ０ｔ１ｔ２ｔ３ｔ４Ｒ０（ｔ１ｔ２ｔ３ｔ４－1）

The initial length of this blood vessel is Ｌ０. The 1st layer inside radius times

ｔ１ is the 1st layer outside radius, that is the 2nd layer inside radius. The

2nd layer inside radius times ｔ２ is the 2nd layer outside radius, that is the

3rd layer inside radius. And so on. Then ｔ１,ｔ２,ｔ３ andｔ４ are the

outside radius ratios of the 1st , the 2nd , the 3rd and the 4th layers

respectively. When these outside radius ratios are put at 1.05 or less, we can

ignore the difference between two extension ratios of the inside and the

outside surfaces of each layer during the inflation as it is shown in Figure 6.5

in chapter 6. Then we can apply this four-layer calculation method to the

blood vessel whose total wall thickness is about 20% or less of the inside

radius, because the total outside radius ratio of this blood vessel is ｔ０Ｔ＝

ｔ１ｔ２ｔ３ｔ４≦1.05４＝1.216.

The relation of extension ratios

When the blood vessel inflates, the internal pressure is Ｐ, that is the

increase from the atmospheric pressure, the length Ｌ０ increases to Ｌ０α,

the 1st layer inside radius increases to Ｒ０β１, the 2nd layer inside radius

increases to Ｒ０ｔ１β２, the 3rd layer outside radius increases to Ｒ０ｔ１ｔ２

ｔ３β３ and the 4th layer outside radius increases to Ｒ０ｔ１ｔ２ｔ３ｔ４β４ .

78

Then α is the extension ratio in the x direction in the all layers. β１ , β２ ,

β３ and β４ are the extension ratios in the y direction on the 1st layer inside

surface, the 2nd layer inside surface, the 3rd layer outside surface and the

4th layer outside surface respectively. We assume that β１ , β２ , β３ and

β４ represent the extension ratios (in the y direction) of the whole 1st , the

whole 2nd , the whole 3rd and the whole 4th layers respectively.

On the assumption that the volume of the material of the blood vessel wall

does not change during the inflation, we can write the following equation

concerning the volumes of the 1st layer before and after the inflation.

πＲ０２（ｔ１

２－１）Ｌ０＝πＲ０２（ｔ１

２β２２－β１

２）Ｌ０α

From this equation, we have the following equation which expresses the


（８-１） β２＝（ｔ１－２β１

２－ｔ１－２α－１＋α－１）１/２

We can transform β２ into β１ by equation（８-１）.

Next, we can write the following equation concerning the volumes of the

1st, the 2nd and the 3rd layers before and after the inflation.


２ｔ２２ｔ３

２－１）Ｌ０＝


２ｔ２２ｔ３

２β３２－β１

２）Ｌ０α


relation between β１ and β３.

（８-２）

β３＝（ｔ１－２ｔ２

－２ｔ３－２β１

２－ｔ１－２ｔ２

－２ｔ３－２α－１＋α－１）１/２

By the use of equation（８-２）we can transform β３ into β１.

Furthermore, we can write the following equation concerning the volumes

of the all layers before and after the inflation.

79


２ｔ２２ｔ３

２ｔ４２－１）Ｌ０＝


２ｔ２２ｔ３

２ｔ４２β４

２－β１２）Ｌ０α


relation between β１ and β４.

（８-３） β４＝（ｔ１－２ｔ２

－２ｔ３－２ｔ４

－２β１２

－ｔ１－２ｔ２

－２ｔ３－２ｔ４

－２α－１＋α－１）１/２

By the use of equation（８-３）we can transform β４ into β１.

The forces caused by the internal pressure Ｐ



（８-４）ＦＰＸ＝πＲ０２β１

２Ｐ

The force ＦＰＹ acting in the y direction caused by the internal pressure Ｐ


（８-５）ＦＰＹ＝２Ｒ０β１Ｌ０αＰ

The forces caused by the blood vessel extension

In the 1st layer


product of the initial cross-sectional area πＲ０２（ｔ１


ｆ０１Ｘ in the x direction of the material of the 1st layer. And it is represented

by the following equation.

（８-６）ＦＴ１Ｘ＝πＲ０２（ｔ１

２－１）ｆ０１Ｘ


80

product of the initial cross-sectional area ２Ｒ０（ｔ１－1）Ｌ０ and the stress

ｆ０1Ｙ in the y direction of the material of the 1st layer. It is represented by


（８-７）ＦＴ１Ｙ＝２Ｒ０（ｔ１－1）Ｌ０ｆ０１Ｙ

In the 2nd layer

The force ＦＴ２Ｘ caused by the 2nd layer extension in the x direction is the

product of the initial cross-sectional area πＲ０２ｔ１

２（ｔ２２－１） and the

stressｆ０２Ｘ. And it is represented by the following equation.

（８-８）ＦＴ２Ｘ＝πＲ０２ｔ１

２（ｔ２２－１）ｆ０２Ｘ

The force ＦＴ２Ｙ caused by the 2nd layer extension in the y direction is the

product of the initial cross-sectional area ２Ｒ０ｔ１（ｔ２－1）Ｌ０ and the

stress ｆ０２Ｙ. It is represented by the following equation.

（８-９）ＦＴ２Ｙ＝２Ｒ０ｔ１（ｔ２－1）Ｌ０ｆ０２Ｙ

By the same way as the above, the following equations are written.

In the 3rd layer

（８-１０）ＦＴ３Ｘ＝πＲ０２ｔ１

２ｔ２２（ｔ３

２－１）ｆ０３Ｘ

（８-１１）ＦＴ３Ｙ＝２Ｒ０ｔ１ｔ２（ｔ３－1）Ｌ０ｆ０３Ｙ

In the 4th layer

（８-１２）ＦＴ４Ｘ＝πＲ０２ｔ１

２ｔ２２ｔ３

２（ｔ４２－１）ｆ０４Ｘ

（８-１３）ＦＴ４Ｙ＝２Ｒ０ｔ１ｔ２ｔ３（ｔ４－1）Ｌ０ｆ０４Ｙ

81

Equilibrium of the forces

The condition of equilibrium of the forces in the x direction is ＦＰＸ＝ＦＴ１

Ｘ＋ＦＴ２Ｘ＋ＦＴ３Ｘ＋ＦＴ４Ｘ. From equations（８-４）,（８-６）,（８-８）,

（８-１０）and（８-１２）, we have the following equation.

（８-１４）

β１２Ｐ＝（ｔ１

２－１）ｆ０１Ｘ＋ｔ１２（ｔ２

２－１）ｆ０２Ｘ

＋ｔ１２ｔ２

２（ｔ３２－１）ｆ０３Ｘ＋ｔ１

２ｔ２２ｔ３

２（ｔ４２－１）ｆ０４Ｘ

The condition of equilibrium of the forces in the y direction isＦＰＹ＝ＦＴ１Ｙ

＋ＦＴ２Ｙ＋ＦＴ３Ｙ＋ＦＴ４Ｙ. From equations（８-５）,（８-７）,（８-９）,（８

-１１）and（８-１３）, we have the following equation.

（８-１５）

β１αＰ＝（ｔ１－1）ｆ０１Ｙ＋ｔ１（ｔ２－1）ｆ０２Ｙ

＋ｔ１ｔ２（ｔ３－1）ｆ０３Ｙ＋ｔ１ｔ２ｔ３（ｔ４－1）ｆ０４Ｙ

Equations（８-１４）and（８-１５） are the basic equations of the four-

layer calculation method for blood vessels. Then it is necessary to know

about ｆ０１Ｘ , ｆ０２Ｘ , ｆ０３Ｘ , ｆ０４Ｘ ,ｆ０１Ｙ , ｆ０２Ｙ , ｆ０３Ｙ and ｆ０４Ｙ.

8.2 Use of C-Type Equation


ｆ０３Ｘ , ｆ０４Ｘ ,ｆ０１Ｙ , ｆ０２Ｙ , ｆ０３Ｙ and ｆ０４Ｙ by the use of C-type

equation（１-２６）at n=2. (But, if you want, you can chose another value

of ｎ in C-type equation.)

（８-１６）ｆ０１Ｘ＝Ｋ１Ｘ｛１－λ１Ｘ－１＋ｃ１Ｘ（λ１Ｘ－１）２｝

（８-１７）ｆ０２Ｘ＝Ｋ２Ｘ｛１－λ２Ｘ－１＋ｃ２Ｘ（λ２Ｘ－１）２｝

（８-１８）ｆ０３Ｘ＝Ｋ３Ｘ｛１－λ３Ｘ－１＋ｃ３Ｘ（λ３Ｘ－１）２｝

（８-１９）ｆ０４Ｘ＝Ｋ４Ｘ｛１－λ４Ｘ－１＋ｃ４Ｘ（λ４Ｘ－１）２｝

82

（８-２０）ｆ０１Ｙ＝Ｋ１Ｙ｛１－λ１Ｙ－１＋ｃ１Ｙ（λ１Ｙ－１）２｝

（８-２１）ｆ０２Ｙ＝Ｋ２Ｙ｛１－λ２Ｙ－１＋ｃ２Ｙ（λ２Ｙ－１）２｝

（８-２２）ｆ０３Ｙ＝Ｋ３Ｙ｛１－λ３Ｙ－１＋ｃ３Ｙ（λ３Ｙ－１）２｝

（８-２３）ｆ０４Ｙ＝Ｋ４Ｙ｛１－λ４Ｙ－１＋ｃ４Ｙ（λ４Ｙ－１）２｝

We can substitute the above expressions for ｆ０１Ｘ , ｆ０２Ｘ , ｆ０３Ｘ ,

ｆ０４Ｘ , ｆ０１Ｙ , ｆ０２Ｙ , ｆ０３Ｙ and ｆ０４Ｙ in equations（８-１４）and（８

-１５）, but λｉＸ and λｉＹ (where ｉ=1, 2, 3, 4 ) are the one-dimensional

extension ratios of the materials of the blood vessel. So we have to transform

λｉＸ and λｉＹ into α and βｉ by the use of equations λｉＸ＝α４/３βｉ２/３

and λｉＹ＝α２ /３βｉ４ /３ (where ｉ=1, 2, 3, 4 ) which are derived from

equations （１-１６）and（１-１７）. Then we have the two equations which

express the relations of Ｐ to α, β１, β２, β３ and β４ in the x

direction and the y direction.

Furthermore, when we eliminate Ｐ from the above two equations and

transform β２ , β３ and β４ into α and β１ by equations（８-１）,（８

-２）and（８-３）, we have the equation which expresses the relation between

α and β１. And we can obtain the many pears of values of α and β１ that

satisfy this equation in the manner described in section 7.2.

From these values of α and β１, we can obtain the values of β２ , β３ ,

and β４ by equations（８-１）,（８-２）and（８-３）. And from the values

of α, β１, β２ , β３ and β４, we can obtain the values of Ｐ by the above

two equations which express the relations of Ｐ to α, β１, β２, β３ and

β４ in the x direction and the y direction.

However, I do not know the values of constants ＫｉＸ , ＫｉＹ , ｃｉＸ and

ｃｉＹ (where ｉ=1, 2, 3, 4 ) in equations（８-１６）through（８-２３）.

These are possibly known by tensile-tests of the materials of blood vessels or

by experiments of blood vessel inflations. But there may be some difference

of the physical property of blood vessels between in a living state and in a

dead state.

I hope someone continues this calculation.

83

8.3 C-Type Equation for The Inflation of Arteries

In the Japanese edition, I discussed about blood vessel inflations in the

light of the results of the calculations of rubber tubes. In the case of the

inflation of rubber tubes, there is a peak (maximal value) of the internal

pressure. But, in the case of the inflation of blood vessels, it is said that there

is no peak.

In this section, instead of the discussion in the Japanese edition, I will show

some C-type equations that may be available to calculate the inflation of blood

vessels which have no peak of the internal pressure.

Shadwick introduced and explained the physical property of arteries which

are composite structures that derive their non-linear properties from the

combination of both rubbery and stiff fibrous constituents. (The reader who

wants to know these details is referred to the article, R. E. Shadwick,

“MECHANICAL DESIGN IN ARTERIES,” The Journal of Experimental

Biology 202, 3305-3313 (1999))

According to the above article, in the inflation of an artery, the initial

stiffness of the artery wall represents the elasticity of the elastin (rubber-like

material), while the much higher stiffness at high strain (extension)

represents the contribution of fully tensed collagen fibers.

I think, the above physical property of the artery wall may be expressed by

C-type equation（１-２６）, which is written as follows.

（８-２４）ｆ０＝Ｋ｛１－λ－１＋ｃ（λ－１）ｎ｝

＝Ｋ（１－λ－１）＋Ｋｃ（λ－１）ｎ

This equation represents a one-dimensional stress-extension relation. The

value of Ｋ determines the value of the differential coefficient (dfo/dλ) of this

equation at λ＝１. This differential coefficient is Young’s modulus, and it

represents the initial stiffness of the artery wall. The values of ｃ and ｎ

determine the stiffness at high strain. The larger the values of ｃ and ｎ

become, the higher the stiffness becomes at high strain.

84

Figure 8.2 One-dimensional stress-extension curves expressed by C-type

equation（８-２４）at ｎ= 3

Heavy solid curves：The relations of ｆ０ to λ at Ｋ＝2 and

ｃ＝0.5, 5, 25 (from the right), then Ｋｃ＝1, 10 , 50

Fine solid curves：The relations of ｆ０ to λ at Ｋ＝20 and

ｃ＝0.05, 0.5, 2.5 (from the right), then Ｋｃ＝1, 10 , 50

Chain curves：The relation of ｆ０ to λ at Ｋ＝50 and

ｃ＝0.02, 0.2, 1 (from the right), then Ｋｃ＝1, 10 , 50

85

0

10

20

30

40

50

60

70

80

90

100

110

120

130

140

150

0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5

ｆｏ

λ

Figure 8.3 Results of the calculations of one-layer tubes which have no

peak of internal pressure and ｔ０＝1.05

These curves are calculated by equations（５-２０）and（５-２１）using

the same values of Ｋ, ｃ and ｎ as the values shown in Figure 8.2.

Heavy solid curves：The relations of Ｐ to β at Ｋ＝2, ｎ= 3 and

ｃ＝0.5, 5, 25 (from the right), then Ｋｃ＝1, 10 , 50

Fine solid curves：The relations of Ｐ to β at Ｋ＝20, ｎ= 3 and

ｃ＝0.05, 0.5, 2.5 (from the right), then Ｋｃ＝1, 10 , 50

Chain curves：The relations of Ｐ to β at Ｋ＝50, ｎ= 3 and

ｃ＝0.02, 0.2, 1 (from the right), then Ｋｃ＝1, 10 , 50

86

0

0.5

1

1.5

2

2.5

3

3.5

4

0.5 1 1.5 2 2.5 3 3.5

Ｐ

β

Figure 8.2 shows several curves calculated by equation（８-２４）at ｎ=3.

(Of course, you can choose another value of ｎ in this equation, if you need.)

In this figure, you can see the shapes of the curves which depend on the values

of Ｋ and ｃ. And when ｎ is larger than 3, the curves have the steeper

slopes of increasing of the stress.(These are not shown in this figure.) Then

you may choose the proper values of Ｋ, ｃ and ｎ to express the average

physical property of an artery wall.

Figure 8.3 shows several curves calculated by equations（５-２０）and（５

-２１） which are derived in chapter 5 for one-layer rubber tubes. The

calculations have been done by the use of the same values of Ｋ, ｃ and ｎ

as the values shown in Figure 8.2. And these several curves in Figure 8.3

show the relations of the internal pressure Ｐ to the extension ratio β in

the circumferential direction of one-layer tubes. But these tubes are not tubes

such as the rubber tubes. These tubes have no peak of the internal pressure.

The left one of three heavy solid curves in Figure 8.3 maybe express the

inflation of an artery. It has low initial stiffness and the much higher stiffness

at high strain (extension). And if the deterioration of the collagen fibers in the

artery wall occurs, the center one of three heavy solid curves maybe express

the inflation of the artery.

Futhermore, a point to be paid attention is as follows. It is likely that the

physical property of an artery wall is not isotropic. So we may need two kinds

of the expressions of ｆ０ for the circumferential direction and the length

direction of the artery. That is to say, we may need two groups of the values

of Ｋ, ｃ and ｎ in equation（８-２４）.

I hope that the proper values of Ｋ, ｃ and ｎ in C-type equation will be

estimated to express the physical property of an artery by reference to the

above Figures 8.2 and 8.3.

87

Documents

MECHANICS OF ELASTOMERSkentaromaruyama.sakura.ne.jp/.../files/elastomers_eng.pdfcalled elastomers. Concerning an ideal elastomer (ideal rubber), the extension caused by one-dimensional