Marder M.P. Condensed Matter Physics

Подождите немного. Документ загружается.

Nonlinear Elasticity 323

Figure 12.1. Rubber consists of a tangle of polymers, joined occasionally at nodes, but

otherwise free to slide about. However, there are forces between all the homopolymers

which strongly prefer a certain overall

fixed

density. The typical distance between nodes

where polymers bind together is

CRi.

conservation of density provided

λ,λ,λ

=1.

(12.7)

Since ds

/dx = λ* describes stretches along x, with similar expressions for y and

the metric tensor in Eq. (12.3) and strain tensor in Eq. (12.5) become

8aß = X

Jaß (12.8)

^Ε

αβ

= \(\

-\)δ

αβ

(12.9)

The trace of any tensor such as Ε

β is invariant under rotations of coordinate axes.

This suggests taking the energy proportional to

ΊτΕ =

^{Χ

Χ]

-3)

(12.10)

and imposing incompressibility through Eq. (12.7) to obtain the free energy per

volume for rubber due to Mooney (1940),

9 1 \

-j-

-| 3 1 G is a constant with dimensions of

energy

per ( \ 2 11)

X y

XjXj

I volume.

G is called the elastic modulus of rubber.

Example: Stretching a Circular Sheet. Consider stretching a circular rubber

sheet, of initial radius

in the

x—y

plane and thickness t along z. When the radius

increases to R =

RoX

the thickness must decrease to ÎRQ/R

to preserve

volume. So Eq. (12.11) becomes

= -nRU

£)

+(t)

The

free energy per volume describes the en-

ergy

attributed to a volume element in the

ref-

erence

frame. So when summing up to get

the

free energy of the whole object, just

mul-

tiply

by the original volume, not the stretched

volume.

(12.12)

A comparison of

Eq.

(12.12) with experiment appears in Figure 12.2. Notice that G

for this sample is on the order of

5 ·

J m

-3

= 0.5 MPa. This is around 10

times

324

Chapter 12. Elasticity

smaller than elastic moduli for for materials such as glass or metal described in Ta-

ble 12.2. The reason rubber is so floppy can be understood in part by returning to

Eq. (5.73), which describes the free energy of polymers from a microscopic point

of view. The overall energy scale is set by kßT times the number of polymer seg-

ments per volume. The number of polymer segments per volume is the density of

molecular units per volume (around 10

/cm

) divided by the number of molecular

units per polymer chain between linking points (around 100). Assembling these

values gives a typical elastic modulus of 0.4 MPa.

■

—r

Y^^"

—B-

— g

□

12 3 4 5 6

Extension

R/RQ

Figure 12.2. Force per length applied to circular rubber sheet, originally

cm thick, as it is

stretched uniformly beyond original size. Squares are data from Treloar (1975), p. 89, and

line is a plot offeree per length (\/R)(d3

/dR), using Eq. (12.12), scaled to match data.

Deviation of theory from experiment at large values of R is due to the fact that polymer

chains stiffen faster than quadratically when stretched too much.

12.2.2 Larger Extensions of Rubber

Figure 12.2 indicates that for large enough extensions, the Mooney theory no

longer applies, and rubber stiffens at a higher rate. In searching for a higher-order

term to add, symmetry provides a guide. The term should be invariant under rota-

tions of coordinate axes; that is, the energy of rubber should not depend upon an

arbitrary decision that some direction is x. To form rotationally invariant quantities

from the Lagrangean strain tensor, observe that

J(e) = det

E-el

Here / is a 3 x 3 unit tensor and

is a constant.

(12.13)

is invariant under coordinate rotations because determinants have this property.

The invariance obtains for all values of e, so it must hold for the quantity that

multiplies each power of

as well. This observation leads to three

strain invariants

Α=Σ

(12.14)

Linear Elasticity 325

h = z2

{

aßE

- E

Eßß } (12.15)

aß

= det|£|. (12.16)

The first invariant J\ is just the trace of E, which formed the basis of the

Mooney free energy. The third invariant gives the volume of the rubber, which

is constant, and so cannot enter into an expression for the energy. That leaves Jj.

The Mooney-Rivlin theory of rubber posits that

1 G

—

—J, -\-BJo· Subject to the constraint that the volume J3 (12.17)

2 remains constant.

A large number of observations can be explained reasonably well by this addition

to the energy. A statistical theory that derives an energy of approximately this form

from statistical fluctuations of an incompressible solid has been obtained by Xing

et al. (2007).

12.3 Linear Elasticity

In the nineteenth century, the theory of linear elasticity was viewed as one of the

crowning achievements of physics. For a multitude of purposes, ranging from

study of waves motion to the design of buildings and structures, it remains the

most useful description of mechanical motion of solids.

To obtain the theory, return to Eq. (12.1) and recall that the displacement of ma-

terial points is given by u(r) = ?(r)

—

r. The theory of linear elasticity is legitimate

when the displacement

is small enough that it need be retained in the Lagrangian

strain tensor E of

Eq.

(12.5) only to linear order. To see how this works, write out

the metric tensor of

Eq.

(12.3) in terms of

ü:

it becomes

ÖM

ö/

3 = <W + —^ +-— + > ^-

· (12.18)

^ dr

The last term in Eq. (12.18) is what linear elasticity discards. After dispensing

with this nonlinear term, the Lagrangian strain tensor is simply called the strain

tensor, is conventionally denoted by lower case e, and equals

_ du

aß = ih l·-« ■ (12.19)

To obtain the theory of linear elasticity, simply assume that the energy of

solid

is given by a quadratic functional of the strain tensor. The most general functional

of this form is

?=Σ dr \e

ßC

aßl5

. (12.20)

αβηδ

The reason that the free energy is a quadratic functional of the strain tensor is that

one wants the undistorted solid with e

ß = 0 to correspond to a state of minimum

326

Chapter 12. Elasticity

energy.

any terms linear

in e

were added

the theory, this would not

be the

case.

Because the strain tensor is symmetric under interchange

its indices, without

loss

generality one can take the tensor

C to be

invariant under the interchanges

a*->ß,

<->

and also

aß ^ ηδ.

(12.21)

does not have these symmetries, then replace

with ^[C

g-ys +CßayS +6 more

terms],

which does have them. See Problem

Therefore,

has

most 21 components. Equation (12.20) can also

rewrit-

ten

= \ / (ff -e

(7„ii Generalization

"work equals force times

(\2 22)

i 2 ap ap,

distance

aß

with the stress tensor

given

= 7^

Cgß-fS

δ-

(12.23)

7(5

12.3.1 Solids

Cubic Symmetry

Although 21 independent components

the tensor

are allowed

the most gen-

eral case, solids with more than triclinic symmetry have simpler tensors. For exam-

ple,

solid with cubic symmetry can have only three independent elastic constants.

Because cubic symmetry implies that the solid must

symmetrical under reflec-

tion about the

x-y, x-z,

and

y-z

planes, no constant C

allowed

which some

index assumes

value

an odd

number

times.

For

example,

xyyy

must

zero

because

is the coefficient of

e^e^,

and

flips sign when

—>

—x, but the energy

of the crystal must be invariant under this change. Furthermore,

cubic crystal has

threefold axes that lead

it to be

symmetric under

—»

—>

Therefore,

any

coefficient

equal

to all

those that can be obtained from

by cyclic permu-

tation

its indices. Three independent parameters survive these considerations;

they may

taken

xxxx

, C

xxyy

and

C^.^.

The free energy becomes

^xxxx

[

= I ctr

—

-\-l.l^

xxy

[£xx€

-+-

+4C

,xy

[e^y

+ e

+ e^l

(12.24)

The reason that the last two terms have factors

2 and 4, respectively, reflects

the numbers

times they appear when one sums freely over

αβη

and

<5.

For

example,

shows up as the coefficient of

XVA>

xyyx

, C

yxyx

and

yxxy

A conventional notation

for

elastic constants cuts

half the number

sum-

mations that must explicitly

mentioned

defining

Ζχχ €yy

&ZZ

^yz ^zx ^xy

ϊ Ϊ [ I I

(12.25)

β2 e-i <?4 e

Linear Elasticity

327

Table 12.1. Elastic constants for cubic crystals

Element Cu

(GPa)

C44

(GPa)

C12

(GPa)

Element Cu

(GPa)

C44

(GPa)

C12

(GPa)

Ar(80K)

(78

C (diamond)

Ge (undoped)

Ge («-doped,

Sb)

Ge(/>doped,10

Ga)

(0.4

(1.6

Kr (115K)

cmVmole)

/mole)

108

2.77

123

190

2.47

346

169

1040

230

129

128.8

118.0

0.0235

0.0311

600

3.71

2.85

48.8

28.3

0.98

45.3

42.3

2.06

100

75.3

550

117

67.1

65.5

65.3

0.01085

0.0217

270

1.88

1.35

14.8

1.37

161

1.48

122

170

135

47.7

39.0

0.0197

0.0281

260

3.15

1.60

41.4

Li (195K)

(6 K)

O (54.4

Si (undoped)

Si («-doped, 10

As)

Xe(156K)

13.4

459

7.59

1.62

247

245

2.60

224

347

2.96

165

162.2

14.7

262

517

230

2.98

9.6

111

4.30

0.93

122

28.4

0.275

71.6

76.5

1.60

79.2

78.7

5.74

82.6

157

43.2

1.48

11.3

168

6.33

0.85

153

132

2.06

173

251

2.44

65.4

9.9

156

203

120

1.90

Notice that the Cauchy relation C\2 = C44 is badly violated for almost all entries. Source:

Landolt and Bernstein (New Series), vol. 11.

*~-xxxx

^xxyy

*~-xxzz

^yzxx

*~zxxx

^xyxx ^C

C\\ C]2 C13 C4J C51 Cßi etc.

(12.26)

so that the free energy may be rewritten

3":

aß=\

ßeß. (12.27)

The three constants C\\,C\2, and C44 for cubic crystals appear in Table 12.1.

It was shown by Cauchy and Saint Venant that if all the atoms composing a

solid interact pairwise through central forces—that is, forces that are directed from

the center of one ion to another as in Eq.

(11.1

)—then there is an additional symme-

try requiring C44 = Cn—the Cauchy relation. Because many computer simulations

of solids assume forces of this form, it is important to observe that for all real cubic

solids the relation is badly violated, as shown in Table 12.1. This fact caused con-

siderable controversy in the nineteenth century, because the elastic anisotropy of

solids revealed the underlying lattice on a macroscopic scale, but seemed inconsis-

tent with the only imaginable force laws. The paradox was first resolved by Born

(1914),

who introduced the possibility of angular forces between atoms, such as

discussed in Section 11.8.

328

Chapter 12. Elasticity

Bulk modulus. The bulk modulus of any solid is defined as B =

/dV

Uni-

form dilation of a solid is obtained by setting e

= e

SV/3V,

while the

other components of the strain tensor vanish. From Eq. (12.24), for a cubic crystal,

the free energy has the form

•5

= IV [Cu +2Ci

] [6V/V]

, (12.28)

and therefore the bulk modulus is given by

B = \[C

+2Q

}. (12.29)

12.3.2 Isotropie Solids

Many solids are effectively isotropie. In some cases, as in glass, it is difficult

to find any length scale much above the atomic for which there is any preferred

orientation. In other cases, such as commercial cast metals or ceramics, a solid

may be composed of so many crystalline grains of varying orientation that for

scales much above the grain size, no trace of crystalline anisotropy remains in the

elastic response. In this case, the number of elastic constants reduces to two.

To show that one constant disappears relative to the cubic case, it is sufficient

to recognize that isotropie solids possess all the symmetries of their cubic coun-

terparts, but with many additional symmetries that may simplify the free energy

beyond Eq. (12.24). One could impose, for example, the condition that the free

energy remain invariant under infinitesimal variations about the z axis. An alterna-

tive is to demand the free energy to remain invariant under 45° rotations about the

z axis. In this case, the strain tensor e

ß(r) transforms into e'

g{r'), where

ß(7) = Y

s(r')R

Sß

(12.30a)

7(5

with

/I -1 0 \

= Rr aadR=-= I \ 1 0 . (12.30b)

\0 0 y/l)

Substitute Eq. (12.30) into Eq. (12.24), subtract the free energy of the rotated

state from that of the unrotated state, and demand that the result vanish (Problem

3).

After carrying out the matrix multiplications and some algebra, the result is

0 =

{2C

ΧΧΧΧ

)(β

2β

~ β

ΧΧ

)

(β

2β

~ <?

) (12.31)

^ ^-xxxx — ^xxyy τ Ά^-xyxy

■

\ l L.5L)

Substituting Eq. (12.32) into Eq. (12.24) allows the free energy to be written

\ I dr A{ Σ e

) +2μ^

iß■

whereλ = Cmï andμ =

Crvrv

(12.33)

Linear Elasticity

329

is now rotationally invariant, so there is no point in trying to apply additional ro-

tations to simplify it further. The numbers λ and μ are called the Lamé constants;

although convenient for analytical work, they are not the conventional constants

with which to report experimental properties of isotropie elastic media, so the pre-

sentation of typical values is deferred.

Equations of Motion and Equilibrium. Knowing the free energy associated with

any possible deformation of the solid and observing that while it is in motion there

will also be a kinetic energy

T= dr±p\U(r)\

, P is the mass density.

(12.34)

one can find the equation of motion for ü, by computing

pü

(r) = --—p- = Σ -^-

αβ(~τ), (12.35)

àu

{r) y dr

using Eq. (12.23) for the stress tensor,

Cgß = y

„ Cgß-yS

δ· (12.36)

Because the acceleration of small sections of mass is given by the divergence

of the stress tensor, the stress tensor is physically interpreted as giving the forces

that each section of

the

body exerts upon its neighbor. To see why, integrate (12.35)

over any small volume V bounded by closed surface Σ, to get

dr pii

= / ί/Σ y~] Πβθβα Employ the divergence theorem, with η

a (12.37)

J ' component of the unit normal to surface Σ.

Taking the small volume to be a tiny cube oriented along the coordinate axes,

Eq.

( 12.37)

shows that the total force on the material inside the cube is provided

by the appropriate components of the stress tensor σ multiplied by the areas of the

cube faces.

Specifically, if one imagines taking a knife and using it to sever bonds in a

small two-dimensional region, say perpendicular to the x axis, then o

gives the

force per unit area required to pull the faces of the region together along x, and σ^,

and σ

χζ

give the forces per unit area required to stretch the faces in the directions

perpendicular to x so that each atom is directly across from the atom that was its

neighbor in equilibrium. The directions of the stresses on each face of a small

cube within a solid are depicted in Figure 12.3. The first symmetry of Eq. (12.21)

implies that σ

β = σβ

. A glance at Figure 12.3 shows that this requirement is

equivalent to demanding that all torques vanish on small volume elements of the

solid.

In the special case of an isotropie solid, one obtains

αβ

= \δ

β Y] β

ΊΊ

+ 2μβ

αβ

(12.38)

\ r ι

-αβ = , ,

α0

~ Ν Y" σ

ΊΊ

—σ

αβ

Just invert the matrix of Eq.(

12.38).

(12.39)

2μ(3λ + 2μ) ^

Ί1

2μ

αρ

330 Chapter 12. Elasticity

i *

Figure 12.3. Imagine singling out a small cube from within a solid. Every face is po-

tentially under stress in three directions, and the sign conventions for the directions of the

stresses are shown

here.

The total force on each face is given by the stress in each direction

times the area of

the

face.

and the equation of motion Eq. (12.35) becomes

P^ki = (λ + μ) V( V

■

ii) + /Λ7

Μ.

(12.40)

Uniform Stresses. If a body is under uniform stress S in the z direction, as shown

in Figure 12.4, then from Eq. (12.39) one has immediately that

with

μ(3λ +

2μ)_

λ + μ '

(12.41)

(12.42)

Y is called the modulus of elasticity or Young's modulus. Simultaneously, the body

contracts in directions perpendicular to the applied stress by an amount

-λ

6yy

2μ(3λ + 2μ)

(12.43)

and the negative of this perpendicular contraction e

divided by the extension e

Poisson 's

ratio

Another conventional constant is defined when only o

= S is nonzero. Again

using Eq. (12.39), one finds that

S = 2Ge

= G——

(12.45)

and defines G = μ to be the shear modulus. Typical values of Y and v appear in

Table 12.2.

Linear Elasticity

331

§ = —y =

(A)

~]SL

S = —G = 2e„G

(JL

(B)

Figure 12.4. Geometries for the definitions of (A) Young's modulus Y and (B) the shear

modulus G.

Table 12.2. Elastic moduli for isotropie materials

Material Young's Modulus Y (GPa) Poisson Ratio v

Lead (cast) 5

Tin (cast) 27

Glass 55

Aluminum (cast) 68

Copper (cast) 76

Zinc (cast) 76

Copper (soft, wrought) 100

Iron (cast) 110

Copper (hard drawn) 120

Iron (wrought) 200

Carbon steel 200

Tungsten 400

0.5

0.3

0.16

0.3

0.4

0.3

0.4

0.3

0.4

0.3

The value of the elastic modulus depends very strongly upon the processing

of the material, which means that a hierarchy of structural details starting

at the atomic level and proceeding up to the scale of grains all are impor-

tant. Data are indicative of orders of magnitude. Source: Brady and Clauser

(1991) and Grigoriev and Meilkhov (1997).

Traveling Waves. An important feature of the equation of motion (12.40) is that

it supports traveling waves of two types, longitudinal and transverse. To study the

longitudinal waves, define

A(r, t) = V

u(r, t) and w(r, t) = V x u(r, t).

First take the divergence of Eq. (12.40) to obtain

(λ + 2μ)ν

Δ,

(12.46)

(12.47)

332

Chapter 12. Elasticity

and then take the curl to obtain

f) lì)

Q — n\7^yj Because the curl of a gradient vanishes. (ΥΣ 48)

Supposing

to be of the form

Üoe'

luJt

shows that Eq.

( 12.47)

describes longitu-

dinal waves, where k and

are parallel, moving with speed

α = ^ψ, (12-49)

while Eq. (12.48) describes transverse waves, where k and

ÜQ

are perpendicular,

moving with speed

(12.50)

12.4 Other Constitutive Laws

12.4.1 Liquid Crystals

Understanding the mechanical forces needed to deform liquid crystals is important

for technical applications, because, for example, in liquid crystal displays, electri-

cal fields twist the molecules of the liquid crystal and thereby alter their optical

properties. The question naturally arises as to how hard one needs to twist this liq-

uid in order to make it turn. The question will be answered by asking how elasticity

is modified while accommodating the symmetries of liquid crystals.

The calculation will be carried out for a nematic liquid crystal and will be

restricted to the case where deformations of the material occur over length scales

much larger than the molecules of which it is composed. At every point, the liquid

crystal is described by a unit vector n(r), whose direction indicates the local axis

of the nematic.

The energy needed to deform a nematic is therefore an integral over space of

some function involving gradients of the unit vector h. Because the gradients are

supposed to be small, the theory will stop with the simplest collection of terms

that gives a nonzero result. There are two basic symmetries allowing one to reduce

the number of terms that must be considered. First, the head and tail of molecules

making up a nematic are indistinguishable, so no physical quantity should be able

to distinguish between h and

—h

or to tell if

the

system has been reflected about any

plane. Second, the free energy of the nematic must be independent of the reference

frame in which it is described. In particular, if one picks up a jar containing a

nematic and rotates it slightly, the calculation of the free energy cannot change just

because the point of view has altered.

Matters would be particularly simple if terms linear in gradients of

composed

the free energy. Unfortunately, one easily shows that all of them vanish. The only