Introduction to Materials ScienceGraduate students (Applied Physics)

Prof. Michael Roth

Chapter 3Crystal Binding and Elasticity

IntroductionWhat holds a crystal together?vAtoms in a solid are bound together by Coulomb forces;Others, like gravitational or magnetic forces, are too weak.

< Repulsive force > - short rangev Pauli Exclusion Principle (exchange type interaction)v Short range : become repulsive when two atoms begin to overlap with each other.

<Attractive force> - long rangev Van der Waals interactionv Electrostatic Coulomb interactionv Covalent bonding: sharing of electrons (locally)v Metallic: sharing electrons (globally)

The overall force is the sum of these two. Each forcegives rise to a potential energy of the two atoms .The PE due to attraction is negative, whereasthat due to repulsion is positive. The two atoms willbond together if the potential energy is a minimum.

<U0, cohesive energy> – needed to separate atoms in a solid infinitely far apart.

e e

Cohesive energies of atoms (from Kittel)

Force and energy of inter-atomic binding

E0: 1÷10 eV/atom, except for inert gases (~ 0.1 eV/atom). The cohesive energy controlsthe melting temperature. A typical value of the equilibrium distance is of the order of afew angstroms (e.g. 2-3Å), so that the forces under consideration are short range.

( ) EF rr

∂= −

∂

Cohesive energy = energy of free atoms − crystal energy

General binding principlesThe short range repulsive energy can be approximated by two different formulae whichgive similar results:

where B, λ and ρ are constants and r is the distance between the atoms. This potentialincreases very rapidly at short distances. (The long range attractive potential varies muchmore slowly with r).The origin of the repulsive force is similar in all solids and is mainly due to the Pauli exclusion principle. The elementary statement of this principle is that two electrons cannot occupy the same orbital. As ionsapproach each other close enough, the orbits of the electrons begin to overlap, i.e. some electronsattempt to occupy orbits already occupied by others. This is, however, forbidden by the Pauli exclusionprinciple. As a result, electrons are excited to unoccupied higher energy states of the atoms, increasingthe total energy of the system and giving repulsive contribution to the interaction. The repulsive interactionis not easy to treat analytically from the first principles. In order to make some quantitative estimates it isoften assumed that this interaction can be described by a central field repulsive potential as given above.

The attractive interatomic forces reflect the presence of bonds between atoms in solids, thenature of the long range attractive energy is different in different solids. There are severaltypes of bonding, depending on the physical origin and nature of the bonding force involved.

12(i) (the Lennard-Jones potential)

(ii) exp (the Born-Mayer potential, QM)

BUr

rU

=

= λ − ρ

Types of binding

170283

DiamondSiC

1. Covalent crystals, diamond(electronic overlap, chemical bonding)

180240

NaClLiF

2. Ionic crystals, Na+Cl- (electrostaticcharge)

2696

NaFe

3. Metals (sea of electrons)

1.82.4

Ar2

CH4

4. Inert gases, van der Waals forces

127

H2OHF

5. Hydrogen bond (special case ofclosed-shell bonding)

Bindingenergy

kcal/mole

Covalent bonding – example of H2 moleculeTwo atoms can form a bond by sharing some or all of their valence electrons thus reducing theoverall potential energy of the combination. For example, when the 1s shells from two hydrogenatoms overlap, in the combination each has a closed K shell, which is stable (see right figure).The electron sharing results in greater concentration ofnegative charge in the region between the two nuclei,where the two electrons spend the majority of their time,and a net attraction between the electrons and the twonuclei, which is the origin of the covalent bond.

Covalent bonding – example of diamond/siliconA similar situation holds for carbon which has a configuration[He]2s22p2 with four empty positions in the 2p subshell. Whenother atoms are nearby, the 2s and 2p subshells becomeindistinguishable and we can consider that the C atom has 4electrons in its L shell out of a possible 8. Like H, the C atomcan share electrons with adjacent atoms. In the case of diamond,it shares with 4 adjacent C atoms and the whole structure canform an extended network. In realty, the atoms do not form aplanar array, but occupy corners of a tetrahedron with preciseAngle of 109.5° between the bonds (coordination number = 4).Elemental semiconductors, e.g. silicon and germanium, as wellas II, III, V, VI elements and their oxides, carbides, phosphidesand sulfides also form a covalently bonded diamond structure.

The bond energy is highest for covalent materials. Due to the strong bonding forces, covalently bonded materials have high melting points and are very hard. Diamond is the hardest known material. The directional nature of the bonding means they are nonductile and undergo brittle fracture. The electrons are all locked into the bonds therefore they are not free to move underan electric field, thus these materials are insulators or very poor conductors.

filled L shell

hybrid bonds

109.5°

Ionic bonding – example of NaClThe ionic bond results from electrostatic interaction of oppositely charged ions. Let us takeNaCl as an example. In the crystalline state, each Na atom loses its single valence electronto a neighboring Cl atom (a), producing Na+ and Cl− ions which have filled electronic shells(b). As a result, an ionic crystal is formed containing positive and negative ions coupled bya strong electrostatic interaction (c).

Na + 5.1 eV (ionization energy) → Na+ + e−

e− + Cl → Cl− + 3.6 eV (electron affinity)

Na+ + Cl− → NaCl + 7.9 eV (electrostatic energy)

The cohesive energy with respectto neutral atoms can be calculatedas 7.9 eV - 5.1 eV + 3.6 eV, i.e.

Na + Cl → NaCl + 6.4 eV (cohesive energy).

Ionic binding - crystalThe structure of NaCl is two interpenetrating fcc lattices of Na+ and Cl−ions as shown in the Fig. Thus each Na+ ion is surrounded by 6 Cl−ions and vice versa. This structure suggests that there is a strongattractive Coulomb interaction between nearest-neighbors ions,which is responsible for the ionic bonding.

To calculate binding energy we need to include Coulomb interactionswith all atoms in the solid. Also we need to take into account therepulsive energy, which we assume to be exponential. Thus theinteraction between two atoms i and j in a lattice is given by

Cl

Here rij is the distance between the two atoms, q is the electric charge on the atom, the (+) sign is taken for the like charges and the (–) sign for the unlike charges.The total energy of the crystal is the sum over i and j so that

In this formula ½ is due to the fact that each pair of interactions should be counted only once. The second equality results from the fact in the NaCl structure the sum over j does not depend on whether the reference ion i is positive or negative, which gives the total number of atoms. The latter divided by two gives the number of molecules N, composed of a positive and a negative ion.

Na

2ijr

ijij

qU e rρλ −= ±

2

,

12 2

ijra

ij ijiji j j j

N qU U U N e rρλ −

= = = ±

∑ ∑ ∑

Ionic Binding – Madelung constantWe assume for simplicity that the repulsive interaction is non-zero only for the nearest neighbors(because it drops down very quickly with the distance between atoms). In this case we obtain

Here R is the distance between the nearest neighbors, z is the number of the nearest neighbors,and α is the Madelung constant:

where pij is defined by rij ≡ pijR . The value of the Madelung constant plays an important role in thetheory of ionic crystals. In general it is impossible to compute the Madelung constant analytically.A powerful method for calculating lattice sums was developed by Ewald – Ewald summation.

Example below - a one-dimensional lattice of ions of alternating sign:

In this case,

where we took into account the logarithm expansion into series .

In three dimensions calculating the series is much more difficult. The values of the Madelungconstants for various solids are calculated, tabulated and can be found in literature (αNaCl ≈ 1.75).

2R qU N z e Rρλ α− = − ⋅

( 1)j i ijp

α≠

±= ∑

1 1 1 12 1 ... 2ln22 3 4 5

α = − + − + − = ( ) 1

1ln(1 ) 1

nn

n

xxn

∞−

=

+ = −∑

Ionic binding – cohesive energyNow we calculate the equilibrium distance between the nearest neighbors for the NaCl - typelattice. At the equilibrium the derivative dU/dR = 0, so that

or

This relationship determines the equilibrium separation R0 is terms of the parameters ρ and λ ofthe repulsive potential. The cohesive energy per atom of the ionic solid can be written as follows:

Let us estimate the magnitude of the cohesive energy in NaCl. The Madelung constant, α = 1.75.The interatomic distance is R0 = a/2 ≈ 2.81Å. The charge q = e. The repulsive interaction has avery short range of the order of ρ = 0.1R0. As follows from the eq. above,

We see that the typical value of the binding energy per pair of atoms is about 8 eV. This impliesthat ionic bond is very strong. Experimentally, this strength is characterized by the relatively highmelting temperatures. For example, the melting temperature of NaCl is about 1100°C, while themelting temperatures for the Na metal is about 400°C (weaker metallic bond).

Rz qeR

0 2

20

0ρλ αρ

−− + =

0 220

R qR ez

ρ αρλ

−=

2 2 2

20 0 0 0

10N q Nq NqUR R R R

α ρ α α ρ = − = − −

22 2 7 20 0

0 00 0

0.11 8 ( : / 4 ; 10 / 4 )U Re eV Note in SI e e cN R R

α πε ε π

≈ − − ≈ − ⇒ =

Metallic bindingMetal atoms have only a few valence electrons which are looselybound to the nucleus. When many atoms are brought together,these valence electrons are lost from individual atoms and areshared by all the atoms, i.e. they are delocalized and form anelectron gas which fills the space between the atoms (see fig.).Attraction between the negative charge of this electron gas andthe positively charged metal ions is enough to hold the structuretogether. The bonding is non-directional, so the metal ions try toget as close as possible together leading to close-packedcrystal structures with high coordination numbers. The electrongas acts like a ‘glue’. The main source of the ‘glue’ is lowering ofthe energy of the valence electrons in a metal as compared tothe free atoms (explain based on the uncertainty principle).

Na+ Na+

e-

e-

Attract

Attract

Attract

AttractRepelRepel

Because there is no directionality in the bonds, the metal ions are able to move with respect to each other, so the metals tend to be ductile. The large number of ‘free’ electrons in the gas can easily move under the influence of an electric field; therefore, they are good conductors of electricity. The free s-electrons can easily transfer energy through the crystal → good thermal conductors. Transition metals like Fe, Ni, Ti, Co have also 3d electrons which are more localized and create covalent bonds.

Crystal Binding – van der Waals (molecular) bindingFirst, we consider crystals of inert gases, which are characterized by van der Waals (or molecular)bonding. The electron distribution in such crystals is very close to that in free atoms. The noblegases such as neon (Ne), argon (Ar), krypton (Kr) and xenon (Xe) are characterized by filledelectron shells and a spherical distribution of electronic clouds in the free atoms. In the crystal theinert gas atoms pack together within the cubic fcc structure.

What holds atoms in an inert gas crystal together?

n The electronic distribution cannot be significantly distorted from free atoms - the cohesive energy of an atom in a crystal is less than 1% or less of the ionization of an atomic electron.

n Therefore, not much energy is available to distort the free atom charge distributions.n But, we will find that small distortions of the electron clouds, which cause instantaneous and

temporary attractive forces (due to electrostatic charge), will be enough to hold these crystals together. Small, induced dipoles hold these solids together.

n These are called van der Waals forces.

Crystal Binding – van der WaalsConsider two inert gas atoms (1 and 2) separated by distance R. The average charge distributionin a single atom is spherically symmetric, which implies that the average dipole moment of atom 1is zero: <p1> = 0. Here the brackets denote the time average of the dipole moment. However, atany moment of time there may be a non-zero dipole moment caused by fluctuations of theelectronic charge distribution. We denote this dipole moment by p1. According to electrostatics thisdipole moment produces an electric field, which induces a dipole moment on atom 2. This dipolemoment is proportional to the electric field which is in turn proportional to the p1/R3, so that

ee

p1

Rp2

12 3~ ~ pp E

RThe dipole moments of the two atoms interact with each other. The energy is therefore reduced due to this interaction. The energy of the interaction is proportional to the product of the dipole moments and inversely proportional to the cube of the distance between the atoms, so that

p p pR R

21 2 1

3 6− = −

The time averaged potential is determined by the average value of which does not vanish, even though <p1> is zero. Since the actual values of p1 are not permanent (fluctuating), the potential is spherically symmetric.

p21

Crystal Binding – van der WaalsThe van der Waals potential is thus

It decreases as R6 with the separation between the atoms.n Van der Waals bonding is relatively weak; the respective cohesive energy is of the order of

0.1eV/atom.n This attractive interaction described above holds only for a relatively large separation

between atoms. At small separations very strong repulsive forces caused by the overlap of the inner electronic shells start to dominate. It appears that for inert gases this repulsive interaction can be fitted quite well by the potential of the form B/R12, where B is a positive constant. Combining this with the attractive term we obtain the total potential energy of two atoms at separation R which can be represented as

(repulsive interaction arises due to Pauli exclusion principle),

where 4εσ6 ≡ A and 4εσ12 ≡ B. This potential is known as Lennard-Jones potential.

Electron charge distribution overlaps

p AUR R

216 6~ − = −

UR R

12 6

4 σ σε = −

electron charge distributionatom1 atom2

1/4

σR

6 2

U R( )4ε

Crystal Binding – hydrogen bondHydrogen has one electron and it should form a covalent bond with only one atom. Under certainconditions atom of hydrogen is attracted to two other atoms. The binding energy is small, ~ 0.1 eV.

n Additional form of molecular bondn H atom forms links between two most electronegative atoms (F, O, N …). n Partially covalent and partially ionicn Hydrogen bond - a type of bond formed when the partially positive hydrogen atom of a polar

covalent bond in one molecule is attracted to the partially negative atom of a polar covalent bond in another.

n • e.g. ice, glue, DNA

CC

CC

H

C

N

Nsugar

N

NN

HH O

H NC C

C

C

NC

H H

H

H

sugar

Bonding in DNA molecule

adenine thymine

Crystal binding - electronegativityElectronegativity denotes the relative electronattracting power of an atom. It is not the same asthe electron affinity; the latter measures theamount of energy released when an electron from an external source "falls into" a vacancywithin the outermost orbital of the atom to yield anisolated negative ion.

The products of bond formation, in contrast, arenot ions and they are not isolated; the two nucleiare now drawn closely together by attraction tothe region of high electron density between them. Any shift of electron density toward one atomtakes place at the energetic expense of stealing itfrom the other atom.

Electronegativity is also not a measurablequantity like the ionization potential.

By convention, electronegativities are measuredon a scale on which the highest value, 4.0, isarbitrarily assigned to fluorine ( proposed byLinus Pauling). It is based on a study of bondenergies in a variety of compounds.

The greater the electronegativity difference between two elements A and B, the more polar will be their molecule A­B. Even such ionic solids as alkali halides possess a certain amount of covalent character, so there is no such thing as a "purely ionic" bond. It has become more common to place binary compounds on a scale from 0 to 100, in which the degree of shading is a rough indication of the number of compounds at any point on the covalent-ionic scale.

Elastic Properties of CrystalsThe behavior of materials when subjected to forces can be understood by a consideration ofatomic bonding. The Young’s Modulus or Elastic Modulus, Y, of a solid indicates its ability todeform elastically. When a solid is subjected to tensile forces, F, acting on opposite faces as infigure (a), it experiences a stress σ defined as the force per unit area F/A where A is the areaon which F acts. If the original length of the solid is Lo, the stress σ stretches it by an amount δL.The strain is the fractional increase in length δL/Lo. If the atoms are moved only a small distancefrom their equilibrium positions, the deformation is elastic and when the force is removed theywill return to them. The stress and strain are related by Y = σ/ε.

Displacement δr results in a net attractive force δFN (fig. (b)). FN is the interatomic force. The ‘area’ of atomic cell is ~ r0

2, therefore σ ~ δFN/r0

2 and ε = δr/r0 and

But, ,

where U is the atom potential energy. Now

where Ubond is the minimum of U(r) and f is

N

r r

dFYr dr

00

1

=

= NdF d U

dr dr

2

2=

bond

r r

Ud UY fr dr r

0

2

2 30 0

1 ,=

= ≅

a constant depending on the crystal structure. Materials with higher bond energies tend to have higher elastic moduli. The same expression occurs in compression as well as tension.

Elastic Properties of CrystalsYoung modulus, Y (psi)1 psi ~ 102 N/m2 ~ 104 Pa

Melting point, (Tm,°C)Material

5050355030165

2.20.40.40.4

318034102800204515391083801327

<300<300<300

WSiCMgOAl2O3

FeCu

NaClPb

PolystyreneNylon

Rubber

Elasticity_stress_1n A body in which one part exerts a force on

neighboring parts is said to be in a state of stress. A stress is said to be homogeneous if the forces acting on the surface of an element of a fixed shape and orientation are independent of the position of the element in the body.

n Consider a unit cube within the body with edges parallel to the axes Ox1, Ox2, Ox3. The force transmitted across each face may be resolved into three components. Force per unit area is called the ‘stress’, σ. We denote by σij the component of force in the +Oxi direction across the face perpendicular to Oxj.

n Since the stress is homogeneous, the forces exerted on the cube across the opposite three facesmust be equal and opposite to those shown in the figure.

n σ11, σ22, σ33 are the normal components of stress, and σ12, σ21, σ23, etc. are the shear components.The σij thus defined form a 2nd rank tensor (see proof in the next slide).

n An assumption that the unit cube should be in static equilibrium imposes conditions on the σij. Letus take moments about an axis parallel to Ox1 passing through the center of the cube. Since thestress is homogeneous, all three components pass through the mid-point of the face. The normaland shear components on the Ox1 face give no moment, and the equilibrium condition is σ23 = σ32.In a similar way, σ31 = σ13 and σ12 = σ21, and so we can write: σij = σji.

Elasticity_stress_2If a set of quantuties Tij relate the components of two vectors pi, qi by anequation of the form pi = Tijqj, the Tij obey the tensor transformation law,i.e. form a tensor. Does σij relate two vectors by similar equation?

We select any small surface element of area δS containing a point Pwithin the stressed body. l is a unit vector perpendicular to it. pδS is theforce transmitted across the area. How does pδS change when l is alteredin direction? To answer it, we consider the equilibrium of a tetrahedron-shapedelement of the body OABC shown in the figure below.

ABC represents a variable surface element ⊥ l, andthe force across it is p × (area ABC). The forces on thethree faces at right angles may be specified by σij, andresolving forces parallel to Ox1 we have

p1⋅ABC = σ11⋅BOC + σ12⋅AOC +σ13⋅AOB,or p1 = σ11 l1+ σ12 l2+ σ13 l3.Similarly, p2 = σ21 l1+ σ22 l2+ σ23 l3,and p3 = σ31 l1+ σ32 l2+ σ33 l3.Hence, we may write pi = σij ljSince σij relates to the two vectors pi and ljin a linear way it is a tensor.

Elasticity_stress_3σij is a symmetrical tensor, as most physical properties,and consequently it can be referred to the principal axes:

Uniaxial stress Biaxial stress Hydrostatic pressure

Pure shear stress

The stress tensor is a field tensor – it does not represent a crystal property, but is akin to a ‘force’ impressed on the crystal, and like the electric field can have an arbitrary orientation in a crystal.

1

2

0 0

0 0

0 0 0

σ

σ

11 12 13 1

21 22 23 2

31 32 33 3

0 0

0 0

0 0

σ σ σ σ

σ σ σ σ

σ σ σ σ

⇒

0 0

0 0

0 0 0

σ

σ

0 0

0 0 0

0 0 0

σ

0 0

0 0

0 0

p

p

p

− −

−

0 0

0 0

0 0 0

σ

σ

−

Elasticity_strain_11. One-dimensional strainWe let OP = x grow to OP’ = x + u on stretching, where u Is the displacement. Let a pointQ, close to P, stretch to Q’ and let PQ = ∆x. Then P’Q’ = ∆x + ∆u. In studying strain we areconcerned only with relative displacements. The strain of PQ is defined as

The strain at the point P is defined as

2. Two-dimensional strain

Geometrical meaning: PQ = [∆xi]. After deformation P’Q’ = [∆xi] + [∆ui]. [∆ui] is the differencein displacement. Then, since the components of [ui] are functions of position, we may write

As [∆ui] and [∆xi] are both vectors it follows that [eij] is atensor.

' 'increase in length P Q PQ uoriginal length PQ x

− ∆= =

∆

0limx

u duex dx∆ →

∆= =

∆

1 1 2 211 12 21 22

1 2 1 2

, , , , , ( , 1,2).iij

j

u u u u ue e e e or collectively e i jx x x x x

∂ ∂ ∂ ∂ ∂= = = = = =

∂ ∂ ∂ ∂ ∂

1 11 1 2

1 2

2 22 1 2

1 2

, , , ii j ij j

j

u uu x xx x uor briefly u x e xu u xu x xx x

∂ ∂ ∆ = ∆ + ∆ ∂ ∂ ∂ ∆ = ∆ = ∆∂ ∂ ∂∆ = ∆ + ∆

∂ ∂

Elasticity_strain_2Further insight is given by considering thedistortion of a rectangular element at P. ForPQ1 we put ∆x2 = 0, and the equations become

The meanings of ∆u1, ∆u2 are indicated in the Fig.e11 measures the extension per unit length resolvedalong Ox1, for

e21 measures the anticlockwise rotation of PQ1 by

In a similar way, e22 – extension of PQ2 and e12 –clocwise rotation of PQ2 to P’Q’2.

11 1 11 1

1

22 1 21 1

1

.

uu x e xxuu x e xx

∂ ∆ = ∆ = ∆ ∂ ∂ ∆ = ∆ = ∆∂

1 111

1 1

u u ex x

∆ ∂= =

∆ ∂

2 221 1 2 1

1 1 1

tan ( , ).u u e u u xx u x

ϑ ϑ∆ ∆= ⇒ ≈ = ∆ ∆ ∆

∆ + ∆ ∆=

The bottom fig. shows anticlockwise rotation of a rigid body by a small angle φ which, from the geometrical meaning of eij can be described by an

antisymmetric tensor - no distortion, but [eij] does not vanish.0

0ije

φ

φ

− =

Elasticity_strain_3Any second rank tensor can be expressed by a sum of a symmetric and antisymmetrical tensor:

eij = εij + wij,where εij = ½ (eij + eji) and wij = ½ (eij – eji).[eij] so defined is a symmetrical tensor, for

εij = ½ (eji + eij) = εji, andwij = – ½ (eji – eij) = – wji.

We, therefore, define the symmetrical part of [eij], that is [εij], as the strain. Thus, in full,

111 12 11 12 212

112 22 12 21 222

( )

( )

e e e

e e e

ε ε

ε ε

+ = +

1 111 12 13 11 12 21 12 212 2

1 112 22 23 12 21 22 12 212 2

1 113 23 33 12 21 12 21 112 2

( ) ( )

( ) ( )

( ) ( )

e e e e e

e e e e e

e e e e e

ε ε ε

ε ε ε

ε ε ε

+ + = + + + +

The diagonal components of [eij] are the extensions per unit length parallel to Ox1 and Ox2. ε12 measures the tensor shear strain.

- in three dimensions

Elasticity –Symmetry

Elasticity – Thermal ExpansionThe strain of a crystal, εij, is not a property in the same sense like the dielectric constant. The strainis a response of the crystal to an influence. The influence may be a stress (elasticity) or an electricfield (piezoelectricity). In both cases, the magnitudes and directions of the principal strains aredetermined by the influence, as well as by the symmetry of the crystal. The strain tensor, like thestress tensor, does not have to conform to the crystal symmetry unless the influence itself conforms.

However, a strain may also be caused by a temperature change (thermal expansion). In this case,the influence has no orientation (it is represented by a scalar, T), and so the resulting strain mustconform to the crystal symmetry.

εij = αij∆T,

where αij are the coefficients of thermal expansion. Since [εij] is a symmetrictensor, [αij] is a symmetric tensor as well and may be referred to its principalaxes: ε1 = α1∆T, ε2 = α2∆T, ε3 = α3∆T, where α1, α2, α3 are the principalexpansion coefficients.

Calcite

CaCO3

α1=α2= -5.6

α3 = 25

Elasticity – Hooke’s Law_1The tensorial representation of the Hooke’s Law is now as follows:

σij = cijklεkl or εij = sijklσkl,where cijkl and sijkl are the stiffness constants and compliances of the crystal respectively.

These 81 coefficients (in each case) are 4th rank tensors. To attach a physical meaning tothe cijkl we imagine a set of stress components applied to the crystal and chosen in such away that all the components of strain, except for one normal or a pair of shear components,vanish. If only ε12 and ε21 components exist, σij = cij12ε12 + cij21ε21 = (cij12 + cij12)ε12. Then,cijkl = cijlk in general and, by considering other special cases we find that cijkl = cjikl. The sameis valid for sijkl , which reduces the number of independent coefficients from 81 to 36, andthey represented by two arrays of matrices, (sij) and (cij):

Elasticity - Hooke’s Law_2Consider a crystal which in the unstrained state has the form of a unit cube and suppose it issubjected to a small homogeneous strain with components εi. Now let the strain components allbe changed to εi + dεi. The work done by the stress components σi acting on the cube faces is

dW = σi dεi (i = 1, 2, …, 6).If the deformation process is isothermal and reversible, the work done is equal to theincrease in the Gibbs free energy dG and we may write, per unit volume,

dG = dW = σi dεi .If Hooke’s Law is obeyed this becomes dG = cij εj dεi . Hence,

Differentiating both sides of this equation with respect to εj we have

But since G is a function of only of the state of the body, specified by the strain components, theorder of differentiation is immaterial, and the left-hand side of this equation is symmetrical withrespect to i and j. Hence,

cij = cjisij = sji

The symmetry of (cij) & (sij) matrices reduces the number of independent coefficients from 36 to 21.

Further reduction of the number of independent coefficients is due to presence of crystal symmetry.

ij ji

G c εε

∂=

∂

.ijj i

G cε ε

∂ ∂= ∂ ∂

Elacticity_8

Elasticity - Hooke’s Law_4For isotropic crystals, we write outthe equations of strain componentsin terms of the stress componentsand vice versa in matrix form and,for comparison, in the form used inelasticity books:

Y – Young’s ModulusG – Rigidity Modulusν – Poisson’s Ratio

s11 = 1/Y; s12 = −ν /Y; 2(s11−s12) = 1/G

G = Y/{2(1+ν)}

c11 = 2µ + λ

c12 = λ

( )( )( )

( ){ }

( ){ }

( ){ }

1 1 2 3

1 11 1 12 2 12 3 2 2 3 1

2 12 1 11 2 12 3

3 3 1 23 12 1 12 2 11 3

4 11 12 44 4

5 11 12 5

6 11 12 6 5 5

6 6

1

1

1

2 12

12

1

Y

s s sYs s s

s s s Ys s

Gs s

s sG

G

ε σ ν σ σ

ε σ σ σ ε σ ν σ σε σ σ σ

ε σ ν σ σε σ σ σ

ε σε σ

ε σ

ε σ ε σ

ε σ

= − += + + = − + = + + = − + = + +

= − = = − = − =

=

( )( )( )

( )( )

( )

1 1 2 31 11 1 12 2 12 3

2 12 1 11 2 12 3 2 1 2 3

3 12 1 12 2 11 33 1 2 31

4 11 12 424 4

15 11 12 52

5 51

6 11 12 62 6 6

2

2

2

c c cc c cc c c

c c

c c

c c

σ µ λ ε λε λεσ ε ε εσ ε ε ε σ λε µ λ ε λεσ ε ε ε

σ λε λε µ λ εσ ε

σ µεσ ε σ µεσ ε σ µε

= + + += + + = + + = + + + = + + = + + + = − =

= − = = − =

Elasticity - Hooke’s Law_5Volume compressibility – proportional decreasein volume of a crystal when subjected to unithydrostatic pressure If we put σkl = −pδkl, then

εij = − psijklδkl = − psijkk.For dilatation ∆ we have

∆ = εii = − psiikk,and so the volume compressibility, − ∆/p, is siikk.

In matrix notation, the volume compressibility iss11 + s22 + s33 + 2(s12 + s23 + s31),

and is thus the sum of nine coefficients. For acubic or isotropic crystal it is evidently

3(s11 + 2s12).

For isotropic materials it is customary to definethe reciprocal of the volume compressibility asthe Bulk Modulus,

B = 1/{3(s11 + 2s12))} = Y/{3(1 – 2ν)}.K = 1/B – Bulk Compressibility

The change of volume of a unit cube is called dilatation:

∆ = (1 + ε1)(1 + ε2)(1 + ε3) = ε1 + ε2 + ε3

Referred to general axes the dilatation is given by ∆ = εii