The last question in the theory of elasticity that I will discuss is an attempt to calculate elastic material constants, based on some properties of the atoms that make up this material. We will consider a simple case ionic cubic crystal type sodium chloride. The size or shape of the deformed crystal changes. Such changes lead to an increase potential energy crystal. To calculate the change in strain energy, you need to know where each atom goes. To do full energy as little as possible, the atoms in the lattice of complex crystals rearrange very in a complex way. This makes it quite difficult to calculate the strain energy. But it is still possible to understand what happens in the case of a simple cubic crystal. The disturbances inside the crystal will be geometrically similar to the disturbances on its external faces.

The elastic constants of a cubic crystal can be calculated as follows. First of all, we will assume the existence of some law of interaction between each pair of atoms in the crystal. Then we calculate the change internal energy crystal when it deviates from its equilibrium shape. This will give us a relationship between energy and strain, which is quadratic in strain. By comparing the energy obtained in this way with equation (39.13), it is possible to identify the coefficients for each term with elastic constants C¡jkl .

In our example, we will assume the following simple law of interaction: between neighboring atoms there are central forces, meaning that they act along a line connecting two adjacent atoms. We expect that the forces in ionic crystals should be of precisely this type, since they are based on a simple Coulomb interaction. (At covalent bond the forces are usually more complex, because they also lead to lateral pressure on neighboring atoms; but we don’t need all these complications.) In addition, we are going to take into account only the force of interaction of each atom with nearest to him and next nearby neighbors. In other words, we will make an approximation in which we neglect the forces between distant atoms. In fig. 39.10, and the forces in the plane are shown hu, which we will take into account. It is also necessary to take into account the corresponding forces in the planes yz And zx.

Since we are only interested in elastic constants, which describe small deformations, and, therefore, in the expression for energy we need only terms that are quadratic in deformations, we can assume that the forces between each pair of atoms change linearly with displacement. Therefore, for clarity, we can imagine that each pair of atoms is connected by a “linear” spring (Fig. 39.10, b). All springs between the sodium and chlorine atoms must have the same elastic constant, say k 1 . The springs between two sodium atoms and two chlorine atoms may have different constants, but I want to simplify our reasoning, so I will assume that these constants are equal. Let us denote them by k 2 . (Later, when we see how the calculations go, you can go back and make them different.)

Let us now assume that the crystal is perturbed by a uniform deformation described by the tensor e¡j. IN general case it will have components containing x, y and z, but for greater clarity we will consider only deformations with three components: e xx, e xy And e yy. If one of the atoms is chosen as the origin of coordinates, then the displacement of any other atom is given by an equation like (39.9):

Let's call the atom with coordinates x=y=0"atom 1", and the numbers of its neighbors are shown in Fig. 39.11. Denoting the lattice constant by A, we get X- and y-components of displacement u x , u y written out in table. 39.1

Now we can calculate the energy stored in the springs, which is equal to the product k 2 /2 per square of tension of each spring. Thus, the energy of a horizontal spring between atoms 1 and 2 will be equal

Note that up to first order (1-atomic displacement 2 does not change the length of the spring between atoms 1 And 2. However, in order to obtain the deformation energy of the diagonal spring, the one that goes to the atom 3, we need to calculate the change in length due to both vertical and horizontal displacements.

For small deviations from the origin of the cube, the change in the distance to the atom 3 can be written as a sum of components their And u y V diagonal direction:

Using the quantities their and u y. we can obtain an expression for the energy

For the total energy of all springs in a plane xy we need the sum of eight terms of type (39.43) and (39.44). Referring to this energy as U 0, we get

To find the total energy of all the springs associated with an atom 1, we must make some addition to equation (39.45). Although we only need X- and y-components of deformation, some additional energy associated with diagonal neighbors outside the plane contributes to them hu. This additional energy is equal to

Elastic constants are related to energy density w equation (39.13). The energy that we have calculated is associated with one atom, or rather it is doubled energy per atom, because each of the two atoms connected by a spring should account for 1/2 of its energy. Since there are 1/a 3 atoms per unit volume, then w And Uo related by the relation

To find elastic constants WITH¡jkl , you just need to square the sums in brackets in equation (39.45), add (39.46) and compare the coefficients for e¡j e kl with the corresponding coefficients in equation (39.13). For example, collecting terms with e 2 xx And e 2 y y, we find that its multiplier is equal to

In the remaining terms we will encounter a slight complication. Since we cannot distinguish the works e xx e yy from e yy e xx, then the coefficient for it in the expression for energy equal to the sum two terms in equation (39.13). Coefficient at e xx e yy in equation (39.45) is equal to 2k 2, so we get

However, due to the symmetry of the expression for energy, when rearranging the first two values ​​with the last two, we can assume that With xxyy - With y uhx, That's why

In the same way you can get

Note, finally, that any member containing once the icon X or y,equal to zero, as was found earlier from symmetry considerations. Let's summarize our results:

So, it turns out that we are able to relate macroscopic elastic constants to atomic properties, which manifest themselves in constant k 1 And k2. In our particular case From xyxy = From xxyy. These terms for a cubic crystal, as you probably noticed from the calculations, turn out to be Always equal, whatever forces we take into account, but only given that, that the forces act along the line connecting each pair of atoms, that is, as long as the forces between the atoms are like springs and do not have a lateral component (which undoubtedly exists with a covalent bond).

Our calculations can be compared with experimental measurements elastic constants. In table Figure 39.2 shows the observed values ​​of the three elastic coefficients for some cubic crystals. You probably noticed that With xxyy, generally speaking, it's not the same With xyxy . The reason is that in metals like sodium and potassium, the interatomic forces are not directed along the line connecting the atoms, as assumed in our model. Diamond also does not obey this law, because the forces in diamond are covalent forces that have special property directionality: “springs” prefer to bind atoms located at the vertices of the tetrahedron. Such ionic crystals, like lithium fluoride or sodium chloride, etc., have almost all physical properties, assumed in our model; according to the data in table. 39.2, constant With xxyy And With xyxy they are almost equal. Only silver chloride for some reason has the disadvantage of obeying the condition With khhuu - With huhu.

The last issue in elasticity theory that I will discuss is the attempt to calculate the elastic constants of a material based on some properties of the atoms that make up that material. We will consider the simple case of an ionic cubic crystal such as sodium chloride. The size or shape of the deformed crystal changes. Such changes lead to an increase in the potential energy of the crystal. To calculate the change in strain energy, you need to know where each atom goes. To make the total energy as low as possible, the atoms in the lattice of complex crystals rearrange themselves in very complex ways. This makes it quite difficult to calculate the strain energy. But it is still possible to understand what happens in the case of a simple cubic crystal. The disturbances inside the crystal will be geometrically similar to the disturbances on its external faces.

The elastic constants of a cubic crystal can be calculated as follows. First of all, we will assume the existence of some law of interaction between each pair of atoms in the crystal. Then we calculate the change in the internal energy of the crystal when it deviates from the equilibrium shape. This will give us a relationship between energy and strain, which is quadratic in strain. By comparing the energy obtained in this way with equation (39.13), it is possible to identify the coefficients for each term with elastic constants.

In our example, we will assume the following simple law of interaction: between neighboring atoms there are central forces, meaning that they act along a line connecting two adjacent atoms. We expect that the forces in ionic crystals should be of precisely this type, since they are based on a simple Coulomb interaction. (With a covalent bond, the forces are usually more complex, since they also lead to lateral pressure on neighboring atoms; but we do not need all these complications.) In addition, we are going to take into account only the force of interaction of each atom with its nearest and next nearby neighbors. In other words, we will make an approximation in which we neglect the forces between distant atoms. In fig. 39.10, and shows the forces in the plane that we will take into account. It is also necessary to take into account the corresponding forces in the and planes.

Fig. 39.10. The interatomic forces we take into account (a) and the model in which atoms are connected by springs (b).

Since we are only interested in elastic constants, which describe small deformations, and, therefore, in the expression for energy we need only terms that are quadratic in deformations, we can assume that the forces between each pair of atoms change linearly with displacement. Therefore, for clarity, we can imagine that each pair of atoms is connected by a “linear” spring (Fig. 39.10b). All springs between the sodium and chlorine atoms must have the same elastic constant, say . The springs between two sodium atoms and two chlorine atoms may have different constants, but I want to simplify our reasoning, so I will assume that these constants are equal. Let's denote them by . (Later, when we see how the calculations go, you can go back and make them different.)

Let us now assume that the crystal is perturbed by a uniform deformation described by the tensor . In general, it will have components containing , and , but for greater clarity, we will consider only deformations with three components: , and . If one of the atoms is chosen as the origin of coordinates, then the displacement of any other atom is given by an equation like (39.9):

(39.42)

Let's call the atom with coordinates “atom 1”, and the numbers of its neighbors are shown in Fig. 39.11. Denoting the lattice constant by , we obtain the - and -components of displacement , , written out in Table. 39.1.

Table 39.1 MOVEMENT COMPONENTS

Fig. 39.11. Displacement of the nearest and next nearby neighbors of atom 1. (The scale is highly distorted.)

Now we can calculate the energy stored in the springs, which is equal to the square of the stretch of each spring. Thus, the energy of a horizontal spring between atoms 1 and 2 will be equal to

Note that, up to first order, moving atom 2 does not change the length of the spring between atoms 1 and 2. However, to obtain the strain energy of the diagonal spring, the one going to atom 3, we need to calculate the change in length due to both the vertical, and due to horizontal movements. For small deviations from the origin of the cube, the change in the distance to atom 3 can be written as a sum of components and in the diagonal direction:

Using the quantities and we can obtain an expression for the energy

. (39.44)

For the total energy of all springs in the plane, we need the sum of eight terms of type (39.43) and (39.44). Denoting this energy by , we get

(39.45)

To find the total energy of all the springs associated with atom 1, we must make some addition to equation (39.45). Although we only need the - and - components of the deformation, some additional energy associated with the diagonal neighbors outside the plane also contributes to them. This additional energy is equal to

. (39.46)

The elastic constants are related to the energy density by equation (39.13). The energy that we calculated is associated with one atom, or more precisely, it is double the energy per atom, because each of the two atoms connected by a spring should account for 1/2 of its energy. Since there are atoms in a unit volume, they are related by the relation

To find the elastic constants, you only need to square the sums in brackets in equation (39.45), add (39.46) and compare the coefficients with the corresponding coefficients in equation (39.13). For example, collecting the terms with and , we find that its multiplier is equal to

.

In the remaining terms we will encounter a slight complication. Since we cannot distinguish the product from , its coefficient in the expression for energy is equal to the sum of the two terms in equation (39.13). The coefficient for in equation (39.45) is equal to , so we get

.

However, due to the symmetry of the expression for energy, when rearranging the first two values ​​with the last two, we can assume that , therefore

.

In the same way you can get

.

Note finally that any term containing once the symbol or is equal to zero, as was found earlier for reasons of symmetry. Let's summarize our results:

(39.47)

So, it turned out that we are able to connect macroscopic elastic constants with atomic properties, which manifest themselves in constants and . In our particular case. These terms for a cubic crystal, as you probably noticed from the calculations, always turn out to be equal, no matter what forces we take into account, but only under the condition that the forces act along the line connecting each pair of atoms, i.e. as long as the forces between atoms are similar to springs and do not have a lateral component (which undoubtedly exists with a covalent bond).

Our calculations can be compared with experimental measurements of elastic constants. In table Figure 39.2 shows the observed values ​​of the three elastic coefficients for some cubic crystals. You probably noticed that, generally speaking, is not equal to . The reason is that in metals like sodium and potassium, the interatomic forces are not directed along the line connecting the atoms, as assumed in our model. Diamond also does not obey this law, because the forces in diamond are covalent forces that have a special directionality: “springs” prefer to bind atoms located at the vertices of the tetrahedron. Ionic crystals such as lithium fluoride or sodium chloride, etc., have almost all the physical properties assumed in our model; according to the data in table. 39.2, constant and almost equal. Only silver chloride for some reason does not want to obey the condition.

Table 39.2 ELASTIC CONSTANTS OF CUBIC CRYSTALS. V (V )

Elasticity constants

Elasticity is quantitatively characterized by constants characteristic of each material. It is necessary to take into account that most properties, except density and heat capacity, are associated with the anisotropy of the structure. Elasticity is a pronounced anisotropic property. Therefore it is necessary to distinguish elasticity of crystals and anisopropic materials and elasticity of isotropic bodies.

Polycrystalline bodies and materials are generally isotropic; the anisotropy of their properties appears only as a result of molding or processing, for example, pressing, stamping, rolling, compaction, etc. Thus, anisotropy in the properties of ceramic tiles, tiles, steel sheets, etc. is formed. In what follows, only the elasticity of isotropic properties is considered, for which the concepts of oriented crystallographic axes, etc., are not applicable.

Taking into account the above, for most natural and artificial materials (rocks, ceramics, concrete, metals, etc.) at small deformations, the relationships between stresses “σ” and deformations “ε” can be considered linear (Fig. 5.2) and describe by generalized Hooke's law:

where E is the elastic modulus (Young's modulus).

Similarly, the shear stress "τ" is directly proportional to the relative shear strain or shear angle y (Fig. 5.3):

where G is the shear modulus.

Rice. 5.2. Classic stress-strain relationship:

A - ceramics; B - metals; C - polymers

Rice. 5.3. Elastic deformation of a solid body under shear

The elongation of the sample during tension is accompanied by a decrease in its thickness (Fig. 5.4). Relative change in thickness Δl/l to relative change in length Δd/d called Poisson's ratio "μ" or lateral compression ratio:

μ = (Δl/l) / (Δd/d).

Rice. 5.4. Elastic deformation of a solid body under tension

If, when a body is deformed, its volume does not change, and this can only occur during plastic or viscous flow, then μ = 0.5. However, in practice, this value is significantly lower than the theoretical indicator for different materials it is different. Elastic materials (concrete, ceramics, etc.) have low values ​​of Poisson’s ratio (0.15-0.25), plastic ( polymer materials) - higher (0.3-0.4). This is explained by the relationship between the forces of attraction and repulsion and the change in interatomic distance during deformation.

Young's modulus

Young's modulus, or longitudinal deformation modulus E, shows the critical stress that a material structure can have at its maximum deformation before failure; has a stress dimension (MPa).

Where: σ р – critical stress.

Polycrystalline materials usually exhibit deviations from linearity. σ = ƒ(ε,), unrelated to energy crystal lattice, but depending on the structure of the material. To assess the elastic properties of such materials, two elastic moduli are used: tangent E = tanα and secant V = tanβ, which is called the deformation modulus (Fig. 5.5).

Rice. 5.5. Schematic representation of refractory deformation:

a - deformation curve; b - point of destruction;

σ; - ultimate stress at failure; ε - deformation

The value of the elastic modulus of a two-phase system is the average between the values ​​of the elastic moduli of each of the phases, and analytical expressions to find it are similar to those used when different meanings linear KTE.

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ELASTIC MODULES (elastic constants), quantities characterizing elastic properties solids(see Elasticity). Elastic modulus is a coefficient depending on deformation on applied mechanical stress (and vice versa). In the simplest case of small deformations, this dependence is linear, and the modulus of elasticity is a coefficient of proportionality (see Hooke's law).

Number of elastic moduli for anisotropic crystals reaches 21 and depends on the symmetry of the crystal. Elastic properties an isotropic substance can be described by 2 constants (see Lamé constants) associated with Young's modulus E = ?/? (? - tensile stress, ? - relative elongation), Poisson's ratio? = ??y?/?х (?y - relative transverse compression, ?х - relative longitudinal elongation), shear modulus G = ?/? (? - shear angle, ? - tangential stress) and with bulk modulus K = ?/? (? - decrease in volume).

The elastic moduli of a given material depend on its chemical composition, pre-treatment, temperature, etc.

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ELASTICITY MODULES (elastic constants), quantities characterizing the elastic properties of solids (see Elasticity). Elastic modulus is a coefficient depending on deformation on applied mechanical stress (and vice versa). In the simplest case of small deformations, this dependence is linear, and the modulus of elasticity is a coefficient of proportionality (see Hooke's law).

The number of elastic moduli for anisotropic crystals reaches 21 and depends on the symmetry of the crystal. The elastic properties of an isotropic substance can be described by 2 constants (see Lamé constants) associated with Young's modulus E = ?/? (? - tensile stress, ? - relative elongation), Poisson's ratio? = ??y?/?х (?y - relative transverse compression, ?х - relative longitudinal elongation), shear modulus G = ?/? (? - shear angle, ? - tangential stress) and with bulk modulus K = ?/? (? - decrease in volume).

The elastic modulus of a given material depends on its chemical composition, pre-treatment, temperature, etc.