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Introduction to Materials Science
Graduate students (Applied Physics)
Prof. Michael Roth
Chapter 3
Crystal Binding and Elasticity
Introduction
What holds a crystal together?
e
v Atoms in a solid are bound together by Coulomb forces;
Others, like gravitational or magnetic forces, are too weak.
< Repulsive force > - short range
v Pauli Exclusion Principle (exchange type interaction)
v Short range : become repulsive when two atoms begin to overlap with each other.
<Attractive force> - long range
v Van der Waals interaction
v Electrostatic Coulomb interaction
v Covalent bonding: sharing of electrons (locally)
v Metallic: sharing electrons (globally)
The overall force is the sum of these two. Each force
gives rise to a potential energy of the two atoms .
The PE due to attraction is negative, whereas
that due to repulsion is positive. The two atoms will
bond together if the potential energy is a minimum.
<U0, cohesive energy> – needed to separate atoms in a solid infinitely far apart.
e
Cohesive energies of atoms (from Kittel)
Force and energy of inter-atomic binding
Cohesive energy = energy of free atoms − crystal energy
E0: 1¸10 eV/atom, except for inert gases (~ 0.1 eV/atom). The cohesive energy controls
the melting temperature. A typical value of the equilibrium distance is of the order of a
few angstroms (e.g. 2-3Å), so that the forces under consideration are short range.
F (r ) = -
¶E
¶r
General binding principles
The short range repulsive energy can be approximated by two different formulae which
give similar results:
B
(i) U = 12 (the Lennard- Jones potential)
r
æ rö
(ii) U = l exp ç - ÷ (the Born-Mayer potential, QM)
è rø
where B, λ and ρ are constants and r is the distance between the atoms. This potential
increases very rapidly at short distances. (The long range attractive potential varies much
more 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 ions
approach each other close enough, the orbits of the electrons begin to overlap, i.e. some electrons
attempt to occupy orbits already occupied by others. This is, however, forbidden by the Pauli exclusion
principle. As a result, electrons are excited to unoccupied higher energy states of the atoms, increasing
the total energy of the system and giving repulsive contribution to the interaction. The repulsive interaction
is not easy to treat analytically from the first principles. In order to make some quantitative estimates it is
often 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, the
nature of the long range attractive energy is different in different solids. There are several
types of bonding, depending on the physical origin and nature of the bonding force involved.
Types of binding
Binding
energy
kcal/mole
1. Covalent crystals, diamond
(electronic overlap, chemical bonding)
Diamond
SiC
170
283
NaCl
LiF
180
240
3. Metals (sea of electrons)
Na
Fe
26
96
4. Inert gases, van der Waals forces
Ar2
CH4
1.8
2.4
5. Hydrogen bond (special case of
closed-shell bonding)
H2O
HF
12
7
2. Ionic crystals, Na+Cl- (electrostatic
charge)
Covalent bonding – example of H2 molecule
Two atoms can form a bond by sharing some or all of their valence electrons thus reducing the
overall potential energy of the combination. For example, when the 1s shells from two hydrogen
atoms overlap, in the combination each has a closed K shell, which is stable (see right figure).
The electron sharing results in greater concentration of
negative 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 two
nuclei, which is the origin of the covalent bond.
Covalent bonding – example of diamond/silicon
A similar situation holds for carbon which has a configuration
[He]2s22p2 with four empty positions in the 2p subshell. When
other atoms are nearby, the 2s and 2p subshells become
indistinguishable and we can consider that the C atom has 4
electrons in its L shell out of a possible 8. Like H, the C atom
can share electrons with adjacent atoms. In the case of diamond,
it shares with 4 adjacent C atoms and the whole structure can
form an extended network. In realty, the atoms do not form a
planar array, but occupy corners of a tetrahedron with precise
Angle of 109.5° between the bonds (coordination number = 4).
Elemental semiconductors, e.g. silicon and germanium, as well
as II, III, V, VI elements and their oxides, carbides, phosphides
and sulfides also form a covalently bonded diamond structure.
filled L shell
hybrid bonds
109.5°
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 under
an electric field, thus these materials are insulators or very poor conductors.
Ionic bonding – example of NaCl
The ionic bond results from electrostatic interaction of oppositely charged ions. Let us take
NaCl as an example. In the crystalline state, each Na atom loses its single valence electron
to 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 by
a 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 respect
to neutral atoms can be calculated
as 7.9 eV - 5.1 eV + 3.6 eV, i.e.
Na + Cl → NaCl + 6.4 eV (cohesive
energy).
Ionic binding - crystal
The 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 strong
attractive Coulomb interaction between nearest-neighbors ions,
which is responsible for the ionic bonding.
To calculate binding energy we need to include Coulomb interactions
with all atoms in the solid. Also we need to take into account the
repulsive energy, which we assume to be exponential. Thus the
interaction between two atoms i and j in a lattice is given by
Uij = l e
-
rij
r
2
q
±
Na
Cl
rij
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
N
1
U = å U ij = a
2 i, j
2
æ - rij r q 2 ö
åj Uij = N åj ç le ± rij ÷
è
ø
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.
Ionic Binding – Madelung constant
We 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
2
R
æ
ö
- r
q
U = N ç zle
-a ×
R ÷ø
è
Here R is the distance between the nearest neighbors, z is the number of the nearest neighbors,
and a is the Madelung constant:
a =å
j ¹i
( ±1)
pij
where pij is defined by rij º pijR . The value of the Madelung constant plays an important role in the
theory 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,
é
1
1
1
1
ù
a = 2 ê1 - + - + - ...ú = 2ln 2
2 3 4 5
ë
û
where we took into account the logarithm expansion into series
¥
ln(1 + x ) = å ( -1)
n =1
n -1
xn
.
n
In three dimensions calculating the series is much more difficult. The values of the Madelung
constants for various solids are calculated, tabulated and can be found in literature (aNaCl » 1.75).
Ionic binding – cohesive energy
Now we calculate the equilibrium distance between the nearest neighbors for the NaCl - type
lattice. At the equilibrium the derivative dU/dR = 0, so that
-
zl
r
e
-
R0
r
+
aq2
=0
R e
2
0
or
-
R0
r
ar q 2
=
zl
R
This relationship determines the equilibrium separation R0 is terms of the parameters r and l of
the repulsive potential. The cohesive energy per atom of the ionic solid can be written as follows:
2
0
r ö
1
ç
÷
R02
R0
R0 è R0 ø
Let us estimate the magnitude of the cohesive energy in NaCl. The Madelung constant, a = 1.75.
The interatomic distance is R0 = a/2 » 2.81Å. The charge q = e. The repulsive interaction has a
very short range of the order of r = 0.1R0. As follows from the eq. above,
U0 =
a N rq 2
U0
a e 2 æ 0.1R0 ö
»ç1÷ » -8 eV
N
R0 è
R0 ø
-
a Nq 2
=-
a Nq 2 æ
(Note : in SI e 2 Þ e 2 / 4pe 0 ; e 0 = 107 / 4p c 2 )
We see that the typical value of the binding energy per pair of atoms is about 8 eV. This implies
that ionic bond is very strong. Experimentally, this strength is characterized by the relatively high
melting temperatures. For example, the melting temperature of NaCl is about 1100°C, while the
melting temperatures for the Na metal is about 400°C (weaker metallic bond).
Metallic binding
Metal atoms have only a few valence electrons which are loosely
bound to the nucleus. When many atoms are brought together,
these valence electrons are lost from individual atoms and are
shared by all the atoms, i.e. they are delocalized and form an
electron gas which fills the space between the atoms (see fig.).
Attraction between the negative charge of this electron gas and
the positively charged metal ions is enough to hold the structure
together. The bonding is non-directional, so the metal ions try to
get as close as possible together leading to close-packed
crystal structures with high coordination numbers. The electron
gas acts like a ‘glue’. The main source of the ‘glue’ is lowering of
the energy of the valence electrons in a metal as compared to
the free atoms (explain based on the uncertainty principle).
Repel
eAttract
Attract
Na+
Na+
Attract
Repel
Attract
eBecause 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 selectrons 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) binding
First, 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 noble
gases such as neon (Ne), argon (Ar), krypton (Kr) and xenon (Xe) are characterized by filled
electron shells and a spherical distribution of electronic clouds in the free atoms. In the crystal the
inert gas atoms pack together within the cubic fcc structure.
What holds atoms in an inert gas crystal together?
n
n
n
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.
Therefore, not much energy is available to distort the free atom charge distributions.
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.
These are called van der Waals forces.
Crystal Binding – van der Waals
Consider two inert gas atoms (1 and 2) separated by distance R. The average charge distribution
in a single atom is spherically symmetric, which implies that the average dipole moment of atom 1
is zero: <p1> = 0. Here the brackets denote the time average of the dipole moment. However, at
any moment of time there may be a non-zero dipole moment caused by fluctuations of the
electronic charge distribution. We denote this dipole moment by p1. According to electrostatics this
dipole moment produces an electric field, which induces a dipole moment on atom 2. This dipole
moment is proportional to the electric field which is in turn proportional to the p1/R3, so that
p2 ~ E ~
e
e
p1
p2
R
p1
R3
The 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
p1p2
p12
- 3 =- 6
R
R
The time averaged potential is determined by the average value of p1 which does not vanish,
even though <p1> is zero. Since the actual values of p1 are not permanent (fluctuating), the
potential is spherically symmetric.
2
Crystal Binding – van der Waals
The van der Waals potential is thus
p12
A
R6
R6
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
éæ s ö12 æ s ö6 ù
U = 4e êç ÷ - ç ÷ ú (repulsive interaction arises due to Pauli exclusion principle),
è R ø ûú
ëêè R ø
U~-
=-
where 4es6 º A and 4es12 º B. This potential is known as Lennard-Jones potential.
atom1
atom2
electron charge distribution
U (R )
4e
6
Electron charge distribution overlaps
1/4
2
æRö
ç ÷
ès ø
Crystal Binding – hydrogen bond
Hydrogen has one electron and it should form a covalent bond with only one atom. Under certain
conditions atom of hydrogen is attracted to two other atoms. The binding energy is small, ~ 0.1 eV.
n
n
n
n
n
Additional form of molecular bond
H atom forms links between two most electronegative atoms (F, O, N …).
Partially covalent and partially ionic
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.
• e.g. ice, glue, DNA
Bonding in DNA molecule
C
N
sugar
N
N
C
C
O
C
H
C
H
C
H
C
N
C
C
H
adenine
H
C
N
N
H
H
H
N
sugar
thymine
Crystal binding - electronegativity
Electronegativity denotes the relative electron
attracting power of an atom. It is not the same as
the electron affinity; the latter measures the
amount of energy released when an electron
from an external source "falls into" a vacancy
within the outermost orbital of the atom to yield an
isolated negative ion.
The products of bond formation, in contrast, are
not ions and they are not isolated; the two nuclei
are now drawn closely together by attraction to
the region of high electron density between them.
Any shift of electron density toward one atom
takes place at the energetic expense of stealing it
from the other atom.
Electronegativity is also not a measurable
quantity like the ionization potential.
By convention, electronegativities are measured
on a scale on which the highest value, 4.0, is
arbitrarily assigned to fluorine ( proposed by
Linus Pauling). It is based on a study of bond
energies in a variety of compounds.
The greater the electronegativity difference
between two elements A and B, the more
polar will be their molecule A-B. 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 Crystals
The behavior of materials when subjected to forces can be understood by a consideration of
atomic bonding. The Young’s Modulus or Elastic Modulus, Y, of a solid indicates its ability to
deform elastically. When a solid is subjected to tensile forces, F, acting on opposite faces as in
figure (a), it experiences a stress σ defined as the force per unit area F/A where A is the area
on 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 distance
from their equilibrium positions, the deformation is elastic and when the force is removed they
will return to them. The stress and strain are related by Y = s/e.
Displacement δr results in a net attractive
force δFN (fig. (b)). FN is the interatomic
force. The ‘area’ of atomic cell is ~ r02,
therefore s ~ dFN/r02 and ε = δr/r0 and
1 é dFN ù
dFN d 2U
Y= ê
But,
,
=
2
r0 ë dr úû r = r0
dr
dr
where U is the atom potential energy. Now
U
1 é d 2U ù
Y = ê 2 ú
@ f bond
,
r0 ë dr û r = r
r03
0
where Ubond is the minimum of U(r) and f is
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 Crystals
Material
Melting point, (Tm,°C)
Young modulus, Y (psi)
1 psi ~ 102 N/m2 ~ 104 Pa
W
SiC
MgO
Al2O3
Fe
Cu
NaCl
Pb
Polystyrene
Nylon
Rubber
3180
3410
2800
2045
1539
1083
801
327
<300
<300
<300
50
50
35
50
30
16
5
2.2
0.4
0.4
0.4
Elasticity_stress_1
n
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’, s. We denote by sij 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 faces
must be equal and opposite to those shown in the figure.
s11, s22, s33 are the normal components of stress, and s12, s21, s23, etc. are the shear components.
The sij thus defined form a 2nd rank tensor (see proof in the next slide).
An assumption that the unit cube should be in static equilibrium imposes conditions on the sij. Let
us take moments about an axis parallel to Ox1 passing through the center of the cube. Since the
stress is homogeneous, all three components pass through the mid-point of the face. The normal
and shear components on the Ox1 face give no moment, and the equilibrium condition is s23 = s32.
In a similar way, s31 = s13 and s12 = s21, and so we can write: sij = sji.
n
n
Elasticity_stress_2
If a set of quantuties Tij relate the components of two vectors pi, qi by an
equation of the form pi = Tijqj, the Tij obey the tensor transformation law,
i.e. form a tensor. Does sij relate two vectors by similar equation?
We select any small surface element of area dS containing a point P
within the stressed body. l is a unit vector perpendicular to it. pdS is the
force transmitted across the area. How does pdS change when l is altered
in direction? To answer it, we consider the equilibrium of a tetrahedron-shaped
element of the body OABC shown in the figure below.
ABC represents a variable surface element ^ l, and
the force across it is p ´ (area ABC). The forces on the
three faces at right angles may be specified by sij, and
resolving forces parallel to Ox1 we have
p1×ABC = s11×BOC + s12×AOC +s13×AOB,
or
p1 = s11 l1+ s12 l2+ s13 l3.
Similarly,
p2 = s21 l1+ s22 l2+ s23 l3,
and
p3 = s31 l1+ s32 l2+ s33 l3.
Hence, we may write
pi = sij lj
Since sij relates to the two vectors pi and lj
in a linear way it is a tensor.
Elasticity_stress_3
sij is a symmetrical tensor, as most physical properties,
and consequently it can be referred to the principal axes:
és 11
ê
ês 21
ê
ês 31
ë
s 12
s 22
s 32
s 13 ù
ú
s 23 ú
ú
s 33 ú
û
é
Uniaxial stress ê s
ê
ê0
ê
ê0
ë
Pure shear stress
0
0
0
é
ê0
ê
ês
ê
ê0
êë
Þ
ù
0ú
ú
0ú
ú
0ú
û
s
0
0
és 1 0 0 ù
ê
ú
ê0 s 2 0 ú
ê
ú
ê0 0 s 3 ú
ë
û
é
Biaxial stress ê s 1
ê
ê0
ê
ê0
êë
ù
0ú
ú
0ú
ú
0ú
úû
é
ê -s
ê
ê 0
ê
ê 0
êë
ù
0 ú
ú
0ú
ú
0 ú
úû
0
s
2
0
0
s
0
Hydrostatic pressure éê - p
ê
ê 0
ê
ê 0
ë
0
-p
0
ù
0ú
ú
0ú
ú
0ú
úû
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.
0ù
ú
ú
0ú
ú
- pú
û
Elasticity_strain_1
1. One-dimensional strain
We let OP = x grow to OP’ = x + u on stretching, where u Is the displacement. Let a point
Q, close to P, stretch to Q’ and let PQ = Dx. Then P’Q’ = Dx + Du. In studying strain we are
concerned only with relative displacements. The strain of PQ is defined as
increase in length P ' Q '- PQ Du
=
=
original length
PQ
Dx
The strain at the point P is defined as
Du du
e = lim
=
Dx ®0 Dx
dx
2. Two-dimensional strain
¶u
¶u
¶u
¶u
e11 = 1 , e12 = 1 , e21 = 2 , e22 = 2
¶x1
¶x2
¶x1
¶x2
¶ui
(i , j = 1,2).
¶x j
Geometrical meaning: PQ = [Dxi]. After deformation P’Q’ = [Dxi] + [Dui]. [Dui] is the difference
in displacement. Then, since the components of [ui] are functions of position, we may write
¶u
¶u
ü
Du1 = 1 Dx1 + 1 Dx2 ï
¶x1
¶x2
¶ui
ï
Dx j = eij Dx j
ý , or , briefly , Dui =
¶u2
¶u2
¶
x
j
Du2 =
Dx1 +
Dx2 ï
ïþ
¶x1
¶x2
or , collectively ,
As [Dui] and [Dxi] are both vectors it follows that [eij] is a
tensor.
eij =
Elasticity_strain_2
Further insight is given by considering the
distortion of a rectangular element at P. For
PQ1 we put Dx2 = 0, and the equations become
¶u1
ü
Dx1 = e11Dx1 ï
¶x1
ï
ý.
¶u2
Du2 =
Dx1 = e21Dx1 ï
ïþ
¶x1
Du1 =
The meanings of Du1, Du2 are indicated in the Fig.
e11 measures the extension per unit length resolved
along Ox1, for
Du1 ¶u1
=
= e11
Dx1 ¶x1
e21 measures the anticlockwise rotation of PQ1 by
Du2
Du2
tanJ =
Þ J»
= e21 ( Du1, Du2 = Dx1 ).
Dx1 + Du1
Dx1
In a similar way, e22 – extension of PQ2 and e12 –
clocwise rotation of PQ2 to P’Q’2.
The bottom fig. shows anticlockwise rotation of a rigid body by a small angle
f which, from the geometrical meaning of eij can be described by an
é0 - f ù
ú - no distortion, but [eij] does not vanish.
antisymmetric tensor éëeij ùû = ê
êf
0 úû
ë
Elasticity_strain_3
Any second rank tensor can be expressed by a sum of a symmetric and antisymmetrical tensor:
eij = eij + wij,
where
eij = ½ (eij + eji) and wij = ½ (eij – eji).
[eij] so defined is a symmetrical tensor, for
eij = ½ (eji + eij) = eji, and
wij = – ½ (eji – eij) = – wji.
We, therefore, define the symmetrical part of [eij], that is [eij], as the strain. Thus, in full,
ée 11 e 12 ù é
e11
ê
ú=ê
êe 12 e 22 ú ê 21 (e12 + e21 )
ë
û ë
1
2
(e12 + e21 )ù
ú
e22 úû
The diagonal components of [eij] are
the extensions per unit length parallel
to Ox1 and Ox2. e12 measures the
tensor shear strain.
ée 11 e 12 e 13 ù é
e11
ê
ú ê
êe 12 e 22 e 23 ú = ê 12 (e12 + e21 )
ê
ú ê
êe 13 e 23 e 33 ú ê 12 (e12 + e21 )
ë
û ë
(e12 + e21 )ù
ú
1
e22
(e12 + e21 )ú
2
ú
1
ú
(e12 + e21 )
e11
2
û
1
2
(e12 + e21 )
1
2
- in three dimensions
Elasticity –
Symmetry
Elasticity – Thermal Expansion
The strain of a crystal, eij, is not a property in the same sense like the dielectric constant. The strain
is a response of the crystal to an influence. The influence may be a stress (elasticity) or an electric
field (piezoelectricity). In both cases, the magnitudes and directions of the principal strains are
determined by the influence, as well as by the symmetry of the crystal. The strain tensor, like the
stress 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 must
conform to the crystal symmetry.
eij = aijDT,
where aij are the coefficients of thermal expansion. Since [eij] is a symmetric
tensor, [aij] is a symmetric tensor as well and may be referred to its principal
axes: e1 = a1DT, e2 = a2DT, e3 = a3DT, where a1, a2, a3 are the principal
expansion coefficients.
Calcite
CaCO3
a1=a2= -5.6
a3 = 25
Elasticity – Hooke’s Law_1
The tensorial representation of the Hooke’s Law is now as follows:
sij = cijklekl
or
eij = sijklskl,
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 to
the cijkl we imagine a set of stress components applied to the crystal and chosen in such a
way that all the components of strain, except for one normal or a pair of shear components,
vanish. If only e12 and e21 components exist, sij = cij12e12 + cij21e21 = (cij12 + cij12)e12. Then,
cijkl = cijlk in general and, by considering other special cases we find that cijkl = cjikl. The same
is valid for sijkl , which reduces the number of independent coefficients from 81 to 36, and
they represented by two arrays of matrices, (sij) and (cij):
Elasticity - Hooke’s Law_2
Consider a crystal which in the unstrained state has the form of a unit cube and suppose it is
subjected to a small homogeneous strain with components ei. Now let the strain components all
be changed to ei + dei. The work done by the stress components si acting on the cube faces is
dW = si dei
(i = 1, 2, …, 6).
If the deformation process is isothermal and reversible, the work done is equal to the
increase in the Gibbs free energy dG and we may write, per unit volume,
dG = dW = si dei .
If Hooke’s Law is obeyed this becomes dG = cij ej dei . Hence,
¶G
= cij e j
¶e i
Differentiating both sides of this equation with respect to ej we have
¶ æ ¶G ö
ç
÷ = cij .
¶e j è ¶e i ø
But since G is a function of only of the state of the body, specified by the strain components, the
order of differentiation is immaterial, and the left-hand side of this equation is symmetrical with
respect to i and j. Hence,
cij = cji
sij = 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.
Elacticity_8
Elasticity - Hooke’s Law_4
For isotropic crystals, we write out
the equations of strain components
in terms of the stress components
and vice versa in matrix form and,
for comparison, in the form used in
elasticity books:
Y – Young’s Modulus
G – Rigidity Modulus
n – Poisson’s Ratio
s11 = 1/Y; s12 = −n /Y;
ìe1 = s11s 1 + s12s 2 + s12s 3
ï
ïe 2 = s12s 1 + s11s 2 + s12s 3
ïe 3 = s12s 1 + s12s 2 + s11s 3
ï
íe = 2 s - s s
( 11 12 ) 4
ï 4
ïe = 2 ( s - s ) s
11
12
5
ï 5
ïîe 6 = 2 ( s11 - s12 ) s 6
2(s11−s12) = 1/G
1
ì
e
=
{s1 -n (s 2 + s 3 )}
1
ï
Y
ï
ïe = 1 s - n (s + s )
3
1 }
ï 2 Y{ 2
ï
ïe 3 = 1 {s 3 - n (s 1 + s 2 )}
ï
Y
í
ïe = 1 s
ï 4 G 4
ï
1
ïe 5 = s 5
G
ï
ï
1
ïe 6 = s 6
G
î
G = Y/{2(1+n )}
c11 = 2m + l
c12 = l
ìs 1 = c11e 1 + c12e 2 + c12e 3
ï
ïs 2 = c12e 1 + c11e 2 + c12e 3
ïs 3 = c12e 1 + c12e 2 + c11e 3
ï
ís = 1 c - c e
ï 4 2 ( 11 12 ) 4
ïs = 1 ( c - c ) e
ï 5 2 11 12 5
ïîs 6 = 21 ( c11 - c12 ) e 6
ìs 1 = ( 2m + l ) e1 +
le 2 +
le 3
ï
ïs 2 =
le 1 + ( 2m + l ) e 2 +
le 3
ï
ïs =
le 1 +
le 2 + ( 2m + l ) e 3
í 3
ïs 4 = me 4
ï
ïs 5 = me 5
ïs = me
6
î 6
Elasticity - Hooke’s Law_5
Volume compressibility – proportional decrease
in volume of a crystal when subjected to unit
hydrostatic pressure If we put skl = −pdkl, then
eij = − psijkldkl = − psijkk.
The change of volume of a unit cube is
called dilatation:
D = (1 + e1)(1 + e2)(1 + e3) = e1 + e2 + e3
For dilatation D we have
D = eii = − psiikk,
and so the volume compressibility, − D/p, is siikk.
In matrix notation, the volume compressibility is
s11 + s22 + s33 + 2(s12 + s23 + s31),
and is thus the sum of nine coefficients. For a
cubic or isotropic crystal it is evidently
3(s11 + 2s12).
For isotropic materials it is customary to define
the reciprocal of the volume compressibility as
the Bulk Modulus,
B = 1/{3(s11 + 2s12))} = Y/{3(1 – 2n)}.
K = 1/B – Bulk Compressibility
Referred to general axes the dilatation
is given by D = eii