Introduction to Materials Science Prof. Michael Roth

Transcription

1 Introduction to Materials Science Prof. Michael Roth Chapter 3 Crystal Binding and Elasticity

2 Introduction What holds a crystal together? Atoms in a solid are bound together by Coulomb forces; Others, like gravitational or magnetic forces, are too weak. e e < Repulsive force > - short range Pauli Exclusion Principle (exchange type interaction) Short range : become repulsive when two atoms begin to overlap with each other. <Attractive force> - long range Van der Waals interaction Electrostatic Coulomb interaction Covalent bonding: sharing of electrons (locally) 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. <U 0, cohesive energy> needed to separate atoms in a solid infinitely far apart.

3 Cohesive energies of atoms (from Kittel)

4 Force and energy of inter-atomic binding Cohesive energy = energy of free atoms crystal energy E 0 : 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. -3Å), so that the forces under consideration are short range. E Fr ( ) = r

5 General binding principles The short range repulsive energy can be approximated by two different formulae which give similar results: B (i) U = (the Lennard - Jones potential) r 1 r (ii) U =λexp (the Born - Mayer potential, QM) ρ 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.

6 Types of binding 1. Covalent crystals, diamond (electronic overlap, chemical bonding) Diamond SiC Binding energy kcal/mole Ionic crystals, Na + Cl - (electrostatic charge) 3. Metals (sea of electrons) 4. Inert gases, van der Waals forces 5. Hydrogen bond (special case of closed-shell bonding) NaCl LiF Na Fe Ar CH 4 H O HF

7 Covalent bonding example of H 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.

8 Covalent bonding example of diamond/silicon A similar situation holds for carbon which has a configuration [He]s p with four empty positions in the p subshell. When other atoms are nearby, the s and p 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 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 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.

9 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 ev (ionization energy) Na + + e e + Cl Cl ev (electron affinity) Na + + Cl NaCl ev (electrostatic energy) The cohesive energy with respect to neutral atoms can be calculated as 7.9 ev ev ev, i.e. Na + Cl NaCl ev (cohesive energy).

10 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 U ij r ij ρ = λe ± q Here r ij 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 r 1 N ij a q ρ U = Uij = Uij = N λe ±, r i j j j ij 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. r ij Na Cl

11 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 R ρ q U = N zλe α R Here R is the distance between the nearest neighbors, z is the number of the nearest neighbors, and α is the Madelung constant: ( ± 1) α = p j i ij where p ij is defined by r ij p ij R. 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, α = ln = n n 1 x where we took into account the logarithm expansion into series. 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 (α NaCl 1.75). n= 1 ( ) ln(1 + x) = 1 n

12 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 R0 R0 zλ ρ αq ρ αρq e + = 0 or Re 0 = ρ R zλ 0 This relationship determines the equilibrium separation R 0 is terms of the parameters ρ and λ of the repulsive potential. The cohesive energy per atom of the ionic solid can be written as follows: αnρq αnq αnq ρ U0 = = 1 R0 R0 R0 R0 Let us estimate the magnitude of the cohesive energy in NaCl. The Madelung constant, α = The interatomic distance is R 0 = a/.81å. The charge q = e. The repulsive interaction has a very short range of the order of ρ = 0.1R 0. As follows from the eq. above, U0 αe 0.1R ev ( Note : in SI e e /4 πε 0; ε 0 = 10 /4 πc ) N R0 R0 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).

13 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 Attract e - Attract Na + Na + Repel Attract e - Attract 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.

14 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? 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.

15 Crystal Binding van der Waals Consider two inert gas atoms (1 and ) 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: <p 1 > = 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 p 1. According to electrostatics this dipole moment produces an electric field, which induces a dipole moment on atom. This dipole moment is proportional to the electric field which is in turn proportional to the p 1 /R 3, so that p e R p 1 e p ~ E ~ p R 1 3 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 pp 1 p1 = 3 6 R R The time averaged potential is determined by the average value of which does not vanish, even though <p 1 > is zero. Since the actual values of p 1 are not permanent (fluctuating), the potential is spherically symmetric. p 1

16 Crystal Binding van der Waals The van der Waals potential is thus p1 A U ~ = 6 6 R R It decreases as R 6 with the separation between the atoms. Van der Waals bonding is relatively weak; the respective cohesive energy is of the order of 0.1eV/atom. 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/R 1, 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 1 6 σ σ U 4ε = R R (repulsive interaction arises due to Pauli exclusion principle), where 4εσ 6 A and 4εσ 1 B. This potential is known as Lennard-Jones potential. atom1 atom Electron charge distribution overlaps electron charge distribution UR ( ) 4ε 1/4 6 R σ

17 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. 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 H H H C N sugar N C C N C C N N H H H O N C C C N C H C H sugar adenine thymine

18 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.

19 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 L o, the stress σ stretches it by an amount δl. The strain is the fractional increase in length δl/l o. 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 = σ/ε. Displacement δr results in a net attractive force δf N (fig. (b)). F N is the interatomic force. The area of atomic cell is ~ r 0, therefore σ ~ δf N /r 0 and ε = δr/r 0 and 1 dfn df Y = N du But,, r dr = 0 r= r dr dr 0 where U is the atom potential energy. Now 1 du Ubond Y = f, 3 r0 dr r r= r 0 0 where U bond 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.

20 Elastic Properties of Crystals Material W SiC MgO Al O 3 Fe Cu NaCl Pb Polystyrene Nylon Rubber Melting point, (T m, C) <300 <300 <300 Young modulus, Y (psi) 1 psi ~ 10 N/m ~ 10 4 Pa

21 Elasticity_stress_1 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. Consider a unit cube within the body with edges parallel to the axes Ox 1, Ox, Ox 3. The force transmitted across each face may be resolved into three components. Force per unit area is called the stress, σ. We denote by σ ij the component of force in the +Ox i direction across the face perpendicular to Ox j. 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. σ 11, σ, σ 33 are the normal components of stress, and σ 1, σ 1, σ 3, etc. are the shear components. The σ ij thus defined form a nd rank tensor (see proof in the next slide). An assumption that the unit cube should be in static equilibrium imposes conditions on the σ ij. Let us take moments about an axis parallel to Ox 1 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 Ox 1 face give no moment, and the equilibrium condition is σ 3 = σ 3. In a similar way, σ 31 = σ 13 and σ 1 = σ 1, and so we can write: σ ij = σ ji.

22 Elasticity_stress_ If a set of quantuties T ij relate the components of two vectors p i, q i by an equation of the form p i = T ij q j, the T ij obey the tensor transformation law, i.e. form a tensor. Does σ ij relate two vectors by similar equation? We select any small surface element of area δs containing a point P within the stressed body. l is a unit vector perpendicular to it. pδs is the force transmitted across the area. How does pδs 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 σ ij, and resolving forces parallel to Ox 1 we have p 1 ABC = σ 11 BOC + σ 1 AOC +σ 13 AOB, or p 1 = σ 11 l 1 + σ 1 l + σ 13 l 3. Similarly, p = σ 1 l 1 + σ l + σ 3 l 3, and p 3 = σ 31 l 1 + σ 3 l + σ 33 l 3. Hence, we may write p i = σ ij l j Since σ ij relates to the two vectors p i and l j in a linear way it is a tensor.

23 Elasticity_stress_3 σ ij is a symmetrical tensor, as most physical properties, and consequently it can be referred to the principal axes: σ σ σ σ σ1 σ σ 3 0 σ 0 σ σ σ 0 0 σ Uniaxial stress σ 0 0 Biaxial stress σ Hydrostatic pressure σ p p p Pure shear stress 0 σ 0 σ σ σ 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.

24 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 = x. Then P Q = x + u. In studying strain we are concerned only with relative displacements. The strain of PQ is defined as increase in length P ' Q' PQ u = = original length PQ x The strain at the point P is defined as u du e = lim = x 0 x dx. Two-dimensional strain u1 u1 u u ui e11 =, e1 =, e1 =, e = or, collectively, eij = ( i, j = 1, ). x1 x x1 x xj Geometrical meaning: PQ = [ x i ]. After deformation P Q = [ x i ] + [ u i ]. [ u i ] is the difference in displacement. Then, since the components of [u i ] are functions of position, we may write u1 u1 u1 = x1+ x x1 x ui, or, briefly, ui = x j = eij x j u u xj u = x1+ x x1 x As [ u i ] and [ x i ] are both vectors it follows that [e ij ] is a tensor.

25 Elasticity_strain_ Further insight is given by considering the distortion of a rectangular element at P. For PQ 1 we put x = 0, and the equations become u1 u1 = x1 = e11 x1 x 1. u u = x1 = e1 x 1 x1 The meanings of u 1, u are indicated in the Fig. e 11 measures the extension per unit length resolved along Ox 1, for u1 u1 = = e11 x1 x1 e 1 measures the anticlockwise rotation of PQ 1 by u u tan ϑ = ϑ = e1 ( u1, u x1). x1+ u1 x1 In a similar way, e extension of PQ and e 1 clocwise rotation of PQ to P Q. The bottom fig. shows anticlockwise rotation of a rigid body by a small angle φ which, from the geometrical meaning of e ij can be described by an 0 φ antisymmetric tensor e ij = - no distortion, but [e ij ] does not vanish. φ 0

26 Elasticity_strain_3 Any second rank tensor can be expressed by a sum of a symmetric and antisymmetrical tensor: e ij = ε ij + w ij, where ε ij = ½ (e ij + e ji ) and w ij = ½ (e ij e ji ). [e ij ] so defined is a symmetrical tensor, for ε ij = ½ (e ji + e ij ) = ε ji, and w ij = ½ (e ji e ij ) = w ji. We, therefore, define the symmetrical part of [e ij ], that is [ε ij ], as the strain. Thus, in full, 1 ε11 ε 1 e11 ( e 1 + e1) = 1 ε1 ε ( e 1 e1) e + The diagonal components of [e ij ] are the extensions per unit length parallel to Ox 1 and Ox. ε 1 measures the tensor shear strain. 1 1 ε11 ε1 ε 13 e11 ( e 1 + e1) ( e 1 + e1) 1 1 ε1 ε ε 3 = ( e 1 + e1) e ( e 1 + e1) 1 1 ε13 ε3 ε 33 ( e 1 + e1) ( e 1 + e1) e 11 - in three dimensions

27 Elasticity Symmetry

28 Elasticity Thermal Expansion The strain of a crystal, ε ij, 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. ε ij = α ij T, where α ij are the coefficients of thermal expansion. Since [ε ij ] is a symmetric tensor, [α ij ] is a symmetric tensor as well and may be referred to its principal axes: ε 1 = α 1 T, ε = α T, ε 3 = α 3 T, where α 1, α, α 3 are the principal expansion coefficients. Calcite CaCO 3 α 1 =α = -5.6 α 3 = 5

29 Elasticity Hooke s Law_1 The tensorial representation of the Hooke s Law is now as follows: σ ij = c ijkl ε kl or ε ij = s ijkl σ kl, where c ijkl and s ijkl are the stiffness constants and compliances of the crystal respectively. These 81 coefficients (in each case) are 4 th rank tensors. To attach a physical meaning to the c ijkl 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 ε 1 and ε 1 components exist, σ ij = c ij1 ε 1 + c ij1 ε 1 = (c ij1 + c ij1 )ε 1. Then, c ijkl = c ijlk in general and, by considering other special cases we find that c ijkl = c jikl. The same is valid for s ijkl, which reduces the number of independent coefficients from 81 to 36, and they represented by two arrays of matrices, (s ij ) and (c ij ):

30 Elasticity - Hooke s Law_ 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 ε i. Now let the strain components all be changed to ε i + dε i. The work done by the stress components σ i acting on the cube faces is dw = σ i dε i (i = 1,,, 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 = σ i dε i. If Hooke s Law is obeyed this becomes dg = c ij ε j dε i. Hence, G = c ij ε j ε i Differentiating both sides of this equation with respect to ε j we have G = cij. ε j εi 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, c ij = c ji s ij = s ji The symmetry of (c ij ) & (s ij ) matrices reduces the number of independent coefficients from 36 to 1. Further reduction of the number of independent coefficients is due to presence of crystal symmetry.

31 Elacticity_8

32 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: 1 ε1 = σ1 ν σ + σ3 Y 1 ε1 = s11σ1 + s1σ + s1σ 3 ε = σ ν σ3 + σ1 Y ε = s1σ1 + s11σ + s1σ3 1 ε3 s1σ1 s1σ s ε 11σ 3 3 = σ3 ν σ1+ σ = + + Y ε = σ 1 ( s s ) ( s s ) ( s s ) ε4 = σ4 ε5 = G 11 1 σ 5 { ( )} { ( )} { ( )} Y Young s Modulus 1 G Rigidity Modulus ε6 = 11 1 σ 6 ε5 = σ5 ν Poisson s Ratio G 1 s 11 = 1/Y; s 1 = ν /Y; (s 11 s 1 ) = 1/G ε6 = σ6 G G = Y/{(1+ν)} σ1 = c11ε1 + c1ε + c1ε 3 σ 1 = ( µ + λ) ε1+ λε + λε3 c 11 = µ + λ σ = c1ε1 + c11ε + c1ε3 σ = λε 1+ ( µ + λ) ε + λε3 σ3 c1ε1 c1ε c11ε 3 c 1 = λ = + + σ 3 = λε1+ λε + ( µ + λ) ε 3 1 σ4 = ( c11 c1 ) ε 4 σ 4 = µε 4 1 σ5 = ( c11 c1 ) ε 5 σ 5 = µε5 1 σ6 = ( c11 c1 ) ε 6 σ 6 = µε 6

33 Elasticity - Hooke s Law_5 Volume compressibility proportional decrease in volume of a crystal when subjected to unit hydrostatic pressure If we put σ kl = pδ kl, then The change of volume of a unit cube is called dilatation: = (1 + ε 1 )(1 + ε )(1 + ε 3 ) = ε 1 + ε + ε 3 ε ij = ps ijkl δ kl = ps ijkk. For dilatation we have = ε ii = ps iikk, and so the volume compressibility, /p, is s iikk. In matrix notation, the volume compressibility is s 11 + s + s 33 + (s 1 + s 3 + s 31 ), and is thus the sum of nine coefficients. For a cubic or isotropic crystal it is evidently 3(s 11 + s 1 ). Referred to general axes the dilatation is given by = ε ii For isotropic materials it is customary to define the reciprocal of the volume compressibility as the Bulk Modulus, B = 1/{3(s 11 + s 1 ))} = Y/{3(1 ν)}. K = 1/B Bulk Compressibility