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5Stress-StrainMaterial Laws

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Lecture 5: STRESS-STRAIN MATERIAL LAWS

TABLE OF CONTENTS

Page

5.1 Introduction . . . . . . . . . . . . . . . . . . . . . 535.2 Constitutive Equations . . . . . . . . . . . . . . . . . 53

5.2.1 Material Behavior Assumptions . . . . . . . . . . . 53

5.2.2 The Tension Test Revisited . . . . . . . . . . . . 53

5.3 Characterizing a Linearly Elastic Isotropic Material . . . . . . . 55

5.3.1 Determination Of Elastic Modulus and Poissons Ratio . . . 55

5.3.2 Determination Of Shear Modulus . . . . . . . . . . . 55

5.3.3 Determination Of Thermal Expansion Coefficient . . . . . 55

5.4 Hookes Law in 1D . . . . . . . . . . . . . . . . . . . 56

5.4.1 Elastic Modulus And Poissons Ratio In 1D Stress State . . . 56

5.4.2 Shear Modulus In 1D Stress State . . . . . . . . . . . 565.4.3 Thermal Strains In 1D Stress State . . . . . . . . . . 58

5.5 Generalized Hookes Law in 3D . . . . . . . . . . . . . . 59

5.5.1 Strain-To-Stress Relations . . . . . . . . . . . . . 59

5.5.2 Stress-To-Strain Relations . . . . . . . . . . . . . 510

5.5.3 Thermal Effects in 3D . . . . . . . . . . . . . . 511

5.6 Generalized Hookes Law in 2D . . . . . . . . . . . . . . 512

5.6.1 Plane Stress . . . . . . . . . . . . . . . . . 512

5.6.2 Plane Strain . . . . . . . . . . . . . . . . . . 513

5.7 Example: An Inflating Balloon . . . . . . . . . . . . . . 514

5.7.1 Strains and Stresses in Balloon Wall . . . . . . . . . . 514

5.7.2 When Will the Balloon Burst? . . . . . . . . . . . 515

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5.2 CONSTITUTIVE EQUATIONS

5.1. Introduction

Recall from the previous Lecture the following connections between various quantities that appear

in continuum structural mechanics:

internal forces stresses

MP

strains displacements size & shape changes (5.1)

displacements strains MP

stresses internal forces (5.2)

Of these, we have studied mechanical stresses in Lecture 1 and strains in Lecture 4. How are they

linked? Through the material properties of the structural body. This is pictured by the MP symbol

above the appropriate arrow connectors. Material behavior is mathematically characterized by the

so-calledconstitutive equations, also calledmaterial laws.

5.2. Constitutive Equations

In this Lecture we will restrict detailed examination of constitutive behavior to elastic isotropicmaterials. More complex behavior (for example: orthotropy, plasticity, viscoelasticity, and fracture)

are studied in senior and graduate level courses in Structural and Solid mechanics.

5.2.1. Material Behavior Assumptions

There is a very wide range of materials used for structures, with drastically different behavior. In

addition the same material can go through different response regimes: elastic, plastic, viscoelastic,

cracking and localization, fracture. As noted above, we will restrict our attention to a very specific

material class and response regime by making the following behavioral assumptions.

1. Macroscopic Model. The material is mathematically modeled as acontinuumbody. Features

at the meso, micro and nano levels: crystal grains, molecules, and atoms, are ignored.

2. Elasticity. This means the stress-strain response isreversibleand consequently the material

has a preferrednaturalstate. This state is assumed to be taken in theabsence of loadsat a

reference temperature. By convention we will say that the material is thenunstressedand

undeformed. On applying loads, and possibly temperature changes, the material develops

nonzero stresses and strains, and moves to occupy adeformedconfiguration.

3. Linearity. The relationship between strains and stresses is linear. Doubling stresses doubles

strains, and viceversa.

4. Isotropy. Thepropertiesof thematerial are independent of direction. This is a good assumption

formaterialssuch as metals, concrete, plastics, etc. It isnotadequate forheterogenousmixtures

such as composites or reinforced concrete, which are anisotropicby nature. The substantial

complications introduced by anisotropic behavior justifies its exclusion from an introductory

treatment.

5. Small Strains. Deformations are considered so small thatchanges of geometry are neglected

as the loads are applied. Violation of this assumption requires the introduction of nonlinear

relations between displacements and strains. This is necessary for highly deformable materials

such as rubber (more generally, polymers). Inclusion of nonlinear behavior significantly

complicates the constitutive equations and is therefore left for advanced courses.

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Nominal stress

= P/A 0

0

Yield

Linear elasticbehavior(Hooke's law isvalid over thisresponse region)

Strainhardening

Maxnominalstress

Nominalfailure

stress

Localization

Mild SteelTension Test

Nominal strain = L /L

Elasticlimit

Undeformed state

L0 PPgage length

Figure 5.1. Typical tension test behavior of mild steel, which displays a well defined yield

point and extensive yield region.

Brittle(glass, ceramics,concrete in tension)

Moderatelyductile(Al alloy)

Nonlinearfrom start(rubber,polymers)

Figure 5.2. Three material response flavorsas displayed in a tension test.

Nominal stress

= P/A0

0

Mild steel(highly ductile)

High strength steel

Nominal strain = L /L

Tool steel

Conspicuous yield

Note similarelastic modulus

Figure 5.3. Different steel grades have approximately the same elastic modulus, but very

different post-elastic behavior.

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5.3 CHARACTERIZING A LINEARLY ELASTIC ISOTROPIC MATERIAL

5.2.2. The Tension Test Revisited

The first acquaintance of an engineering student with lab-controlled material behavior is usually

throughtension testscarried out during the first Mechanics, Statics and Structures sophomore

course. Test results are usually displayed as axial nominal strain versus axial nominal strain, as

illustrated in Figure 5.1 for a mild steel specimen taken up to failure. Several response regionsare indicated there: linearly elastic, yield, strain hardening, localization and failure. These are

discussed in the aforementioned course, and studied further in courses on Aerospace Materials.

It is sufficient to note here that we shall be mostly concerned with the linearly elastic regionthat

occurs before yield. In that region the one-dimensional (1D) Hookes law is assumed to hold.

Material behavior may depart significantly from that shown in Figure 5.1. Three distinct flavors:

brittle, moderately ductile and nonlinear-from-start, are shown schematically in Figure 5.2 . Brittle

materials such as glass, rock, ceramics, concrete-under-tension, etc., exhibit primarily linear be-

havior up to near failure by fracture. Metallic alloys used in aerospace, such as Aluminum and

Titanium alloys, display moderately ductile behavior, without a well defined yield point and yield

region: the stress-strain curve gradually turns down finally dropping to failure. Some materials,

such as rubbers and polymers, exhibit strong nonlinear behavior from the start. Although such

materials may be elastic there is no easily identifiablelinearlyelastic region.

Even for a well known material such as steel, the tension test behavior can vary signi ficantly

depending on combination with other components. Figure 5.3 sketches the response of mild steel

with high-strength steel used in critical structural components, and with tool steel. Mild steel is

highly ductile and clearly exhibits an extensive yield region. Hi-strength steel is less ductile and

does not show a well defined yield point. Tool steel has little ductility, and its behavior displays

features associated with brittle materials. The trade off between ductility and strength is typical.

Note, however, that all three grades of steel haveapproximately the same elastic modulus, which is

the slope of the stress-strain line in the linear region of the tension test.

5.3. Characterizing a Linearly Elastic Isotropic Material

For anisotropicmaterial in the linearly elastic region of its response,fournumerical properties are

sufficient to establish constitutive equations. Those equations are associated with the well known

Hookes law, originally enunciated by Robert Hooke by 1660 for a spring, and later expressed

in terms of stresses and strains once those concepts appeared in the XIX Century. These four

properties: E, ,Gand , are tabulated in Figure 5.4.

5.3.1. Determination Of Elastic Modulus and Poissons Ratio

The experimental determination of the elastic modulus Eand Poissons ratio makes use of a

uniaxial tension test specimen such as the one pictured in Figure 5.5. See slides for operational

details.

5.3.2. Determination Of Shear Modulus

The experimental determination of the shear modulusGmakes use of a torsion test specimen such

as the one pictured in Figure 5.6. See slides for operational details.

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E Elastic modulus, a.k.a. Young's modulus Physical dimension: stress=force/area (e.g. ksi)

Poisson's ratioPhysical dimension: dimensionless (just a number)

G Shear modulus, a.k.a. modulus of rigidity Physical dimension: stress=force/area (e.g. MPa)

Coefficient of thermal expansion Physical dimension: 1/degree (e.g., 1/ C)

E, andG are not independent. They are linked by

E =2G (1+), G =E/(2(1+)), =E /(2G)1

Figure 5.4. Four properties that fully characterize the thermomechanical response

of an isotropic material in the linearly elastic range.

5.3.3. Determination Of Thermal Expansion Coefficient

The experimental determination of the thermal expansion coefficient can be made byt using a

uniaxial tension test specimen such as the one pictured in Figure 5.7. See slides for operational

details.

5.4. Hookes Law in 1D

Once the valuesofE, , Gand areexperimentally determined (forwidelyusedstructural materials

they can be simply read off a manual), they can be used to construct thermoelastic constitutive

equations that link stresses and strains as described in the following subsections.

5.4.1. Elastic Modulus And Poissons Ratio In 1D Stress State

The one-dimensional Hookes law relates 1D normal stress to 1D extensional strain through two

materialparameters introducedpreviously: themodulus of elasticityE, also calledYoungs modulus

and Poissons ratio . The modulus of elasticity connects axial stress to axial strain :

=E, E =

, =

E. (5.3)

Poissons ratiois defined as ratio of lateral strain to axial strain:

=

lateral strainaxial strain = lateral strainaxial strain . (5.4)

The sign is introduced for convenience so that comes out positive. For structural materials

lies in the range 0.0

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5.4 HOOKES LAW IN 1D

PP

P

= P/A(uniform over cross section)xx

(b)

(a)

xxyy

zz

00

00

0 0

Strainstate

xx 00

0

0 0

0 0 0

Stressstate

at all points in the gaged region

y

zx Cartesian axes

Cross section of areaA

gaged length

Figure 5.5. Speciment for determination of elastic modulusEand Poissons ratio in the linearly elastic

response region of an isotropic material, using an uniaxial tension test.

TT

T

Circularcross section

(b)

(a)

0

0

0

0

0 0 0

Strainstate

xy

xy

xy

xy

yx

yx

yx

yx

0

0

0

0

0 0 0

Stressstate

at all points in the gaged region. Both the shear stress = as wellas the shear strain = vary linearly as perdistance from thecross section center(Lecture 7). They attain maximum values on thespecimen surface.

y

zx Cartesian axes

gaged length

For distribution of shear stresses andstrains over the cross section, cf. Lecture 7

Figure 5.6. Speciment for determination of shear modulusG in the linearly elastic response region of an

isotropic material, using a torsion test.

5.4.2. Shear Modulus In 1D Stress State

The shear modulusGconnects a shear strainto the corresponding shear stress :

= G , G =

, =

G. (5.5)

Correspondingmeans that if = xy , say, then = xy , and similarly for the other shear

components. The shear modulus is usually obtained from a torsion test. It turns out that the

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gaged length

x

Figure 5.7. Speciment for determination of coefficient of thermal expansionby heating a

tension test specimen and allowing it to expand freely.

3 material propertiesE, and G for an elastic isotropic material are not independent, but are

connected by the relations

G = E

2(1 + ), E= 2(1 + ) G, =

E

2G 1. (5.6)

which means that if two of them are known by measurement, the third one can be obtained from

the relations (5.6). In practice the three properties are often measured independently, and the

(approximate) verification of (5.6) gives an idea ofhow isotropicthe material is.

5.4.3. Thermal Strains In 1D Stress State

A temperature change ofTwith respect to abaseorreferencelevel produces a thermal strain

T = T, (5.7)

in which is the coefficient of thermal dilatation, measured in 1/ For 1/C. This is typically

positive: >0 and very small:

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5.5 GENERALIZED HOOKES LAW IN 3D

(a)

a b

c

xx

xx

yy

yy

zz

zz

Initial shape

y

x

z

(c)

ba

c

Final shapeafter test (2)



(b)

a b

c

(d)

a b

c

Figure5.9. Derivation of three-dimensional Generalized Hookes Law for normal stresses and strains.Three tension tests are assumed to be carried out along {x,y,z}, respectively, and strains superposed.

Solving forgives

= E T. (5.10)

SinceEandare positive, a rise in temperature, i.e., T >0, will produce a negative axial stress, and the bar

will be incompression. This is an example of the so-calledthermally induced stressor simplythermal stress.

It is the reason behind the use of expansion joints in pavements, rails and bridges. The effect is important in

orbiting vehicles such as satellites, which undergo extreme (and cyclical) temperature changes from full sun

to Earth shade.

5.5. Generalized Hookes Law in 3D

5.5.1. Strain-To-Stress Relations

We now generalize the foregoing equations to the three-dimensional case, still assuming that the

material is elastic and isotropic. Condider a cube of material aligned with the axes {x,y,z}, as

shown in Figure 5.9. Imagine that threetension tests, labeled (1), (2) and (3) respectively, are

conducted alongx ,yandz, respectively. Pulling the material by applyingxxalongxwill produce

normal strains

(1)xx = xx

E, (1)yy =

xx

E, (1)zz =

xx

E. (5.11)

Next, pull the material by yy alongyto get the strains

(2)yy = yy

E, (2)xx =

yy

E, (2)zz =

yy

E. (5.12)

Finally pull the material byzz alongzto get

(3)zz = zz

E, (3)xx =

zz

E, (3)yy =

zz

E. (5.13)

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In the general case the cube is subjected tocombinednormal stresses xx , yy and zz . Since we

assumed that the material is linearly elastic, the combined strains can be obtained bysuperposition

of the foregoing results:

xx = (1)

xx + (2)

xx + (3)

xx = xx

E

yy

E

zz

E=

1

E xx yy zz .yy =

(1)yy +

(2)yy +

(3)yy =

xx

E+

yy

E

zz

E=

1

E

xx + yy zz

.

zz = (1)

zz + (2)

zz + (3)

zz = xx

E

yy

E+

zz

E=

1

E

xx yy + zz

.

(5.14)

The shear strains and stresses are connected by the shear modulus as

xy = yx = xy

G=

yx

G, yz = zy =

yz

G=

zy

G, zx = xz =

zx

G=

xz

G. (5.15)

The three equations in (5.14), plus the three in (5.15), may be collectively expressed in matrix form

as

xxyyzzxyyzzx

=

1E

E

E

0 0 0

E 1E E 0 0 0

E

E

1E

0 0 0

0 0 0 1G

0 0

0 0 0 0 1G

0

0 0 0 0 0 1G

xxyyzzxyyzzx

. (5.16)

5.5.2. Stress-To-Strain Relations

To get stresses if the strains are given, the most expedient method is to invert the matrix equation

(5.16). This gives

xxyyzzxyyzzx

=

E(1 ) E E 0 0 0E E(1 ) E 0 0 0E E E(1 ) 0 0 0

0 0 0 G 0 0

0 0 0 0 G 0

0 0 0 0 0 G

xxyyzzxyyzzx

. (5.17)

Here Eis aneffectivemodulus modified by Poissons ratio:

E = E

(1 2)(1 + )(5.18)

The six relations in (5.17) written out in long form are

xx = E

(1 2)(1 + )

(1 ) xx + yy + zz

,

yy = E

(1 2)(1 + )

xx + (1 ) yy + zz

,

zz = E

(1 2)(1 + )

xx + yy + (1 ) zz

,

xy = Gxy, yz = Gyz , zx = Gzx .

(5.19)

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The combination

av = 13

(xx + yy + zz ) (5.20)

is called themean stress, oraverage stress. The negative ofavis thepressure: p= av .

The combination v = xx + yy + zz is called thevolumetric strain, ordilatation. The negativeofv is known as thecondensation. Both pressure and volumetric strain areinvariants, that is,

their value does not change if axes {x,y,z}are rotated. An important relation between pressure

and volumetric strain can be obtained by adding the first three equations in (5.19), which upon

simplification and accounting for (5.20) andp = avrelates pressure and volumetric strain as

p= E

3(1 2)v = Kv . (5.21)

This coefficientKis called thebulk modulus. If Poissons ratio approaches 12, which happens for

near incompressible materials,K .

Remark 5.1. In the solid mechanics literature p is also defined (depending on authors preferences) as

p = av = 1

3(xx + yy + zz ), which is the negative of the above one. If so, p = +Kv . The definition

p= avis the most common one influid mechanics.

5.5.3. Thermal Effects in 3D

To incorporate theeffect of a temperature change Twith respect toa baseor reference temperature,

add Tto the three normal strains in (5.14)

xx =

1

E

xx yy zz

+ T,

yy = 1

E

xx + yy zz

+ T,

zz = 1

E

xx yy + zz

+ T.

(5.22)

No change in the shear strain-stress relation is needed because if the material is linearly elastic and

isotropic, a temperature change only produces normal strains. The stress-to-strain matrix relation

(5.16) expands to

xxyyzzxyyzzx

=

1E E E 0 0 0

E

1E

E

0 0 0

E

E

1E

0 0 0

0 0 0 1G

0 0

0 0 0 0 1G

0

0 0 0 0 0 1G

xxyyzzxyyzzx

+ T

1

1

1

0

0

0

. (5.23)

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Inverting this relation provides the stress-strain relations that account for a temperature change:

xxyyzzxyyzzx

=

E(1 ) E E 0 0 0E E(1 ) E 0 0 0E E E(1 ) 0 0 0

0 0 0 G 0 0

0 0 0 0 G 0

0 0 0 0 0 G

xxyyzzxyyzzx

E T

1 2

1

1

1

0

0

0

,

(5.24)

in which Eis defined in (5.18). Note that if all mechanical normal strainsxx ,yy , andzz vanish,

but T = 0, the normal stresses given by (5.24) are nonzero. Those are calledinitial thermal

stresses, and are important in engineering systems exposed to large temperature variations, such as

rails, turbine engines, satellites or reentry vehicles.

5.6. Generalized Hookes Law in 2D

Two specializations of the foregoing 3D equations to two dimensions are of interest in the appli-

cations: plane stressand plane strain. Plane stress is more important in Aerospace structures,which tend to be thin, so in this course more attention is given to that case. Both specializations

are reviewed next.

5.6.1. Plane Stress

In this case all stress components with azcomponent are assumed to vanish. For a linearly elastic

isotropic material, the strain and stress matrices take on the formxx xy 0

yx yy 0

0 0 zz

,

xx xy 0

yx yy 0

0 0 0

(5.25)

Note that the zz strain, often called thetransverse strainor thickness strainin applications, in

general will be nonzero because of Poissons ratio effect. The strain-stress equations are easily

obtained by going to (5.14) and (5.15) and settingzz = yz = zx = 0. This gives

xx = 1

E

xx yy

, yy =

1

E

xx + yy

, zz =

E

xx + yy

,

xy = xy

G, yz = zx = 0.

(5.26)

The matrix form, omitting known zero components, is

xxyyzzxy

=

1E

E

0

E

1

E

0

E

E

0

0 0 1G

xx

yyxy

. (5.27)

Inverting the matrix composed by the first, second and fourth rows of the above relation gives the

stress-strain equations xxyyxy

=

E E 0E E 0

0 0 G

xxyyxy

. (5.28)

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in which E=E/(1 2). Written in long hand,

xx = E

1 2(xx + yy ), yy =

E

1 2(yy + xx ), xy = Gxy. (5.29)

5.6.2. Plane Strain

In this case all strain components with azcomponent are assumed to vanish. For a linearly elastic

isotropic material, the strain and stress matrices take on the formxx xy 0

yx yy 0

0 0 0

,

xx xy 0

yx yy 0

0 0 zz

(5.30)

Note that the zz stress, which is called thetransverse stressin applications, in general will not

vanish. The strain-to-stress relations can be easily obtained by settingzz = yz = zx = 0 in

(5.19). This gives

xx = E(1 2)(1 + )

(1 ) xx + yy

,

yy = E

(1 2)(1 + )

xx + (1 ) yy

,

zz = E

(1 2)(1 + )

xx + yy

,

xy = Gxy, yz = 0, zx = 0.

(5.31)

which in matrix form, with the zero components removed, is

xx

yyzzxy

= E(1 ) E 0

E E(1 ) 0E E 0

0 0 G

xxyyxy

. (5.32)

Inverting the system provided by extracting the first, second and fourth rows of (5.32) gives the

stress-to-strain relations, which are omitted for simplicity.

The effect of temperature changes can be incorporated in both plane stress and plane strain relations

without any difficulty.

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Backup Material Only Not Covered In Lecture

Initial (reference)diameterD ,underinflation pressurep

0

0

0

0

Final (deformed)diameterD = D

inflationpressure

p = p + p

For simplicity assumeaspherical ballon shape

for all calculations

f

f

D = D0f

0D

Figure 5.10. Inflating balloon example problem.

5.7. Example: An Inflating Balloon

This is a generalization of Problem 3 of Recitation #2. The main change is that all data is expressed

and kept in variable form until the problem is solved. Specific numbers are plugged in at the end.

This will be the only problem in the course where some features of nonlinear mechanicsappear.

These come in by writing the governing equations in both the initial and final geometries, withoutlinearization.

5.7.1. Strains and Stresses in Balloon Wall

The problem is depicted in Figure 5.10. A spherical rubber balloon has initial diameter D0under

inflation pressure p0. This is called the initial configuration. The pressure is increased by pso

that thefinal pressure is pf = p0 + p. The balloon assumes a spherical shape withfinal diameter

Df = D0, inwhich >1. This will be called thefinal configuration. The initial wall thickness is

t0

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5.7 EXAMPLE: AN INFLATING BALLOON

Since the balloon is assumed to remain spherical and its thickness is very small compared to its

diameter, the above strains hold at all points of the balloon wall, and are the samein any direction

tangent to the sphere. If we choose the sphere normal as local zaxis, the wall is in aplane stress

state.

Next we introduce material laws. We will assume that rubber obeys the two-dimensional, plane

stress generalized Hookes law (5.29) with respect to the Eulerian strain measure, with effectivemodulus of elasticity Eand Poissons ratio .1 Settingx x = yy =

Ef and x y =0 therein and

accounting for the initial stress 0, we obtain the inplane normal stress in the final configuration:

xx = yy = f = 0 +E

1 2(Ef +

Ef) = 0 +

E

1 Ef = 0 +

E

1

1

. (5.34)

The normal inplane wall stress is the same in all directions, so it is called simply 0and f, for

initial andfinal configurations, respectively. The inplane shear stressvanishes in all directions.

AssumeD0, t0,Eand are given as data. An interesting question: what is the relation between p

(the excess or gage pressure) and the diameterDf = D0? And, is there a maximum pressure that

will cause the balloon to burst?

5.7.2. When Will the Balloon Burst?

To relate pand it is necessary to express the wall stresses 0and fin terms of geometry and

internal pressure. This is provided by equation (3.10) in Lecture 3, derived for a thin-wall spherical

vessel. In that equation replace p, Rand tby quantities in the initial andfinal configurations:

0 =p0R0

2t0=

p0D0

4t0, f =

pfRf

2 tf=

(p0 + p)Df

4 tf=

(p0 + p) D0

4 tf. (5.35)

All quantities in the above expressions are known in terms of the data, except tf. A kinematic

analysis beyond the scope of this course shows that

tf =

1 + 2

1

2 1

t0. (5.36)

We can check (5.36) by inserting two limit values of Poissons ratio:

= 0: tf = t0. This is correct since the thickness does not change.

= 1/2: tf = t0/2. Is this correct? If =1/2 the material is incompressibleand does not

change volume. The initial and final volume of the thin-wall spherical balloon are

V0 = D20 t0and Vf = D

2f tf =

2D20 tf, respectively. On setting V0 = Vf and

solving for tfwe get tf = t0/2.

To obtain pin terms of, replace (5.36) into (5.35), equate this to (5.34) and solve for p. The

result provided byMathematicais

p=4Et0(1 )(2 +

2(1 2)) +D0p0(1 )(4 + 2(2 4))

D04(1 )(5.37)

1 Thisis a very rough approximation since constitutive equations for rubber (and polymers in general) are highly nonlinear.

But getting closer to reality would take us into the realm of nonlinear elasticity, which is a graduate-level topic.

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1.5 21 2.5 3 3.5 4

1

0

2

3

4

5

6

7

=D /D

p (MPa)

=1/2

=0

f 0

Figure 5.11. Inflating pressure (in MPa) versus diameter expansion ratio

= Df/D0for a balloon withE= 1900 MPa,D0 = 50 mm,p0 = 0 MPa,

t0 = 0.18 mm, 1 4 and Poissons ratios = 0 and = 12

.

This expression simplifies considerably in the two Poissons ratio limits:

p|=0 =4E t0( 1) +D0 p0(2 )

D02 (5.38)

p|=1/2 =8E t0( 1) +D0 p0(2

3)

D04 (5.39)

Pressure versus diameter ratio curves given by (5.38) and (5.39) are plotted in Figure 5.11 for

the numerical values indicated there. Those values correspond to the data used in Problem 3 of

Recitation 2, in which = 1/2 was specified from the start.

Rubber (and, in general, polymer materials) are nearly incompressible; for example 0.4995 for

rubber. Consequently, the response depicted in Figure 5.11 for = 1/2 is more physically relevant

than the other one.

Do the response plots in Figure 5.11 tell you when an in flating balloon is about to collapse? Yes.

This is the matter of a (optional) Homework Exercise.

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Iast.lect05

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