DEFORMED STATE (“FLAT PROBLEM”)

Plane stressed and plane deformed states are characterized by the following features.

1. All stress components do not depend on one of the coordinates common to all components and remain constant when it changes.

2. In planes normal to the axis of this coordinate:

a) tangential stress components are equal to zero;

b) normal stress is either zero (plane stress state) or equal to half the sum of two other normal stresses (plane deformed state).

Let's take the y-axis as the axis mentioned earlier. From the previous it is clear that this axis will be the main one, i.e. it can also be denoted by the index 2. Moreover, , and do not depend on y; at the same time and , and therefore, and are equal to zero.

For a plane stressed state = 0. For a plane deformed state (this feature of a plane deformed state will be proven later).

The significant difference between plane stress and plane strain states should always be taken into account.

In the first, in the direction of the third axis, there is no normal stress, but there is deformation; in the second, there is normal stress, but there is no deformation.

A plane stress state can exist, for example, in a plate subject to forces applied to its contour parallel to the plane of the plate and distributed evenly across its thickness (Fig. 3.16). The change in the thickness of the plate in this case does not matter, and its thickness can be taken as unity. The stressed state of the flange when drawing a cylindrical blank from a sheet material can be considered flat with sufficient accuracy.

A plane deformed state can be assumed for sections of a cylindrical or prismatic body of great length, distant from its ends, if the body is loaded with forces that do not change along its length and are directed perpendicular to the generatrix. In a flat deformed state, for example, a beam can be considered to be subject to settlement in the direction of its thickness, when deformation along the length can be neglected.

All stress state equations for a plane problem are significantly simplified and the number of variables is reduced.

The equations for a plane problem can be easily obtained from those derived earlier for a volumetric stress state, taking into account that = 0 and taking = 0, since inclined areas should be considered only parallel to the y-axis, i.e., normal to areas free from stress in a plane stressed state or free from deformations in a plane deformed state (Fig. 3.17).

In this case

Denoting the angle (see Fig. 3.17) between the normal to the inclined platform and the axis (or axis if the stress state is given in the main axes 1 and 2) through , we obtain from where .

Taking into account the above, by directly substituting into the corresponding expressions (3.10) and (3.11) for the volumetric stress state, we obtain the normal and shear stresses in the inclined area (see Fig. 3.17).

Fig.3.15. Plane stress state (a), stress on an inclined platform (b)

Normal voltage

Shear stress

. (3.41)

From expression (3.41) it is easy to see that it has a maximum at sin 2 = 1, i.e. at = 45°:

. (3.42)

The magnitude of the principal stresses can be expressed in terms of components in arbitrary axes using equation (3.13), from which we obtain

. (3.43)

In this case, for a plane stress state = 0; for plane deformed state

Knowing the stress state in the main axes, it is easy to move to any arbitrary coordinate axes (Fig. 3.18). Let the new coordinate axis x make an angle with the axis, then, considering it as a normal to the inclined platform, we have for the latter according to equation (3.40)

but for the axis the voltage is the voltage, hence

This expression can be transformed like this:

(3.44)

The new axis will be inclined to axis 1 at an angle (+90°); therefore, replacing in the previous equation by ( + 90°), we get

We determine the voltage from expression (3.41):

. (3.46)

Denoting the average voltage by , i.e., taking

, (3.47)

and taking into account equation (3.42), we obtain the so-called transformation formulas, which express the stress components as a function of the angle:

(3.48)

When constructing the Mohr diagram, we take into account that since we are considering areas parallel to the y-axis (i.e., axis 2), the direction cosine is always equal to zero, i.e., angle = 90°. Therefore, all corresponding values ​​will be located on the circle defined by equation (3.36 b) when substituting = 0 into it, namely:

, (3.49)

or taking into account expressions (3.47) and (3.42)

. (3.49a)

This circle is shown in Fig. 3.19 is a Mohr diagram. The coordinates of some point P located on the circle determine the corresponding values ​​and Let us connect point P with point . It is easy to see that the segments 0 2 P = ;

Рр= , Ор= , and therefore sin = .

Comparing the obtained expressions with equations (3.48), we can establish that

Р0 2 А = 2 , Р0 2 А = .

Thus, knowing the position of the inclined platform, determined by the angle, one can find the values ​​of the stresses and acting in this area.

Fig.3.17. Mohr's diagram

,

then the segment OP expresses the total voltage S.

If an element of a stressed body, in the inclined face of which stress is considered, is drawn so that the main stress is directed parallel to the axis, then the normal N drawn to this inclined face, and therefore the direction of stress, will be parallel to the segment CP.

Continuing the line P0 2 to the intersection with the circle, at point P" we obtain a second pair of values ​​for another inclined area, for which " = + 90°, i.e. for an area perpendicular to the first, with the direction of the normal ". Directions of normals N and N" can be taken, respectively, as the directions of the new axes: and , and the stresses and " - respectively, as the coordinate stresses and . Thus, it is possible to determine the stress state in arbitrary axes without using formulas (3.44)-(3.46). Absolute values ​​of stresses git" are equal to each other according to the law of pairing.

It is not difficult to solve the inverse problem: given the stresses in two mutually perpendicular areas , and , t" (where t" = t) find the principal stresses.

We draw coordinate axes n and (Fig. 3.19). We plot points P and P" with coordinates corresponding to the given stresses , and , . The intersection of the segment PP" with the axis will determine the center of the Mohr circle 0 2 with diameter PP" = 2 31. Further, if we construct axes N, N" (or something same, , ) and rotate the figure so that the directions of these axes are parallel to the directions of stresses and at the considered point of a given body, then the directions of the axes and diagrams will be parallel to the direction of the main axes 1 and 2.

We obtain the differential equilibrium equation for the plane problem from equations (3.38), taking into account that all derivatives with respect to y are equal to zero, and also equal to zero and :

(3.50)

When solving some problems related to plane, it is sometimes convenient to use polar coordinates instead of rectangular coordinates, determining the position of a point by the radius vector and polar angle, i.e., the angle that the radius vector makes with the axis.

The equilibrium conditions in polar coordinates can be easily obtained from the same conditions in cylindrical coordinates by equating

And taking into account that the derivatives are equal

(3.51)

A special case of a plane problem is one where the stresses also do not depend on the coordinate (stress distribution symmetrical relative to the axis). In this case, the derivatives with respect to and stress and will vanish, and the equilibrium conditions will be determined by one differential equation

. (3.52)

It is clear that tensions are the main ones here too.

Such a stressed state can be assumed for the flange of a round billet when drawing a cylindrical cup without pressing it.

Type of stress state

The stressed state at any point of a deformable body is characterized by three main normal stresses and the directions of the main axes.

There are three main types of stress states: volumetric (triaxial), in which all three principal stresses are not equal to zero, plane (biaxial), in which one of the principal stresses is zero, and linear (uniaxial), in which only one principal stress is different from zero.

If all normal stresses have the same sign, then the stress state is called the same, and when stresses of different signs are called opposite.

Thus, there are nine types of stress states: four volumetric, three plane and two linear (Fig. 3.18).

The stress state is called homogeneous when at any point of the deformable body the directions of the main axes and the magnitude of the main normal stresses remain unchanged.

The type of stress state affects the ability of a metal to deform plastically without breaking and the amount of external force that must be applied to achieve a deformation of a given magnitude.

For example, deformation under conditions of the same volumetric stress state requires more effort than under conditions of different stress states, all other things being equal.

Control questions

1.What is voltage? What characterizes the tense state of a point, the body as a whole?

2. What do the indices express in the notation of the components of the stress tensor?

3. Give the sign rule for the components of the stress tensor.

4. Write down the Cauchy formulas for stresses on inclined platforms. What is the basis for their conclusion?

5.What is a stress tensor? What components are included in the stress tensor?

6.What are the eigenvectors and eigenvalues ​​of the stress tensor called?

7.What are principal stresses? How many are there?

8. Give the rule for assigning indices to the main normal stresses.

9.Give a physical interpretation of the principal normal stresses and the principal axes of the stress tensor.

10. Show diagrams of the main normal stresses for the main processes of metal milling - rolling, drawing, pressing.

11.What are stress tensor invariants? How many are there?

12.What is the mechanical meaning of the first invariant of the stress tensor?

13.What is called the intensity of tangential stresses?

14..What are principal shear stresses? Find their sites

15..How many areas of principal tangential stresses can be indicated at some point of a deformable body?

16.What is the maximum shear stress, the normal stress on the site along which it acts?

17.What is an axisymmetric stress state? Give examples.

18. Show diagrams of the main normal stresses for the main processes of metal machining - rolling, drawing, pressing.

19.What do plane stress and plane deformed states have in common and what is the difference between them? Which of these states is a simple shift?

20. Give the formulas of stress theory known to you in the main coordinate system

21.What is a stress ellipsoid? Write down its equation and indicate the order of construction. What is the shape of the stress ellipsoid for hydrostatic pressure, plane and linear stress states?

22. Write down an equation to find the principal normal stresses and three systems of equations to find the principal axes T a.

23..What is a spherical tensor and stress deviator? To calculate what quantities are used the second and third invariants of the stress deviator?

24.Show that the main coordinate systems of the stress tensor and deviator coincide.

25. Why are stress intensity and tangential stress intensity introduced into consideration? Explain their physical meaning and give geometric interpretations.

26.What is a Mohr diagram? What are the radii of the main circles?

27.How will the Mohr diagram change when the average voltage changes?

28. What are octahedral stresses?

29. How many characteristic areas can be drawn through a point of a body in a stressed state?

30. Equilibrium conditions for a volumetric stress state in rectangular, cylindrical and spherical coordinates.

31. Equilibrium equations for a plane problem.

BIBLIOGRAPHY

1. Ilyushin A. A. Plasticity. Part I. M.-L., GTI, 1948. 346 p. (33)

2. Pavlov I. M. On the physical nature of tensor representations in the theory of plasticity. – “Izvestia of universities. Ferrous metallurgy", 1965, No. 6, p. 100–104.

3. Sokolovsky V.V. Theory of plasticity. M., “Higher School”, 1969. 608 p. (91)

4. Storozhev M.V. and Popov E.A. Theory of metal pressure processing. M., “Mechanical Engineering”, 1971. 323 p. (99)

5. Timoshenko S.P. Theory of Elasticity. Gostekhizdat, 1934. 451 p. (104)

6. Shofman L.A. Fundamentals of calculation of the stamping and pressing process. Mashgiz, 1961. (68)

Plane stress state

In the case of a plane stress state, one of the three stresses is zero.

Volumetric stress state of strength of material

Relationship between stress and strain

IN resistance of materials, when studying deformations in the case of a volumetric stress state, we assume that the material

obeys Hooke's law and that the deformations are small. Consider an element whose edge sizes are equal

a x b x c, and

For simplicity of reasoning, we consider all stresses to be positive. Due to rib deformation

elements change their length and become equal to a+^a; in+^in; s+^s.

The ratio of the increments in the lengths of the edges of the elements to their original length will give

main relative elongations in the main directions:

The total relative deformation of the element ax in xc in the direction of edge a will be expressed as the sum

Similarly, one can find the total relative deformations in the direction of edges b and c.

These three formulas are called the generalized Hookey law. Volumetric deformation can be expressed as follows:

The change in volume depends only on the sum of the principal stresses, and not on their ratio. Therefore the same

a change in volume will result in an elementary cube, on the faces of which the same stresses will act

Elastic deformation energy

Potential energy of elastic deformation in evidence is the energy accumulated in the body during

its elastic deformation caused by external forces.

Specific energy (elastic deformation energy per unit volume) is equal to:

This energy consists of 2 parts: 1) energy spent on changing volume, and 2) energy,

spent on changing the shape.

Energy of volume change:

Theories of strength strength

Theories of strength, in strength of materials, seek to establish a strength criterion for a material in a complex stress state (volume or plane). In this case, the studied stress state of the calculated part is compared with a linear stress state - tension or compression.

The limiting state of plastic materials is taken to be the state in which noticeable residual (plastic) deformations begin to appear.

For brittle materials, or those in a brittle state, the limiting state is considered to be one in which the material is on the border of the appearance of the first cracks, i.e. at the border of violation of the integrity of the material.

The strength condition for the volumetric stress state is as follows:

The safety factor (n) for a given stress state is a number indicating how many times all components of the stress state should be simultaneously increased so that it becomes the limiting state.

The equivalent stress Oeq is a tensile stress under a linear (uniaxial) stress state that is equally dangerous with a given volumetric or plane stress state.

Formulas for equivalent stress, expressing it through principal stresses, are established by strength theories depending on the strength hypothesis adopted by each theory.

There are several theories of strength or hypotheses of limiting stress states:

The first theory, or the theory of the greatest normal stresses, and the second theory, or the theory of the greatest linear deformations, are currently not used in practical calculations. The third theory, or the theory of maximum tangential stresses. The theory is based on the hypothesis that two stress states - complex and linear - are equivalent in strength if the highest shear stresses are the same.

Equivalent stresses for volumetric stress state:

The third theory of strength gives satisfactory results for plastic materials that are equally resistant to tension and compression, provided that the principal stresses have different signs.

The main disadvantage of this theory is that it does not take into account o"s, which, as experiments show, has some influence on the strength of the material.

The fourth theory of strength is energy. It is based on the premise that the amount of potential energy of shape change accumulated at the time of the onset of a dangerous state (material fluidity) is the same both in a complex stress state and in a simple tensile state. Equivalent stress at volumetric stress state

The fourth theory of strength is well confirmed by experiments with plastic materials that have the same yield strength in tension and compression.

The theory of limit states (Mohr's theory) proceeds from the assumption that the strength in the general case of a stressed state depends mainly on the value of the sign of the largest O1 and the smallest Oz of the principal stresses. The average principal stress O2 only slightly affects the strength. Experiments have shown that the error caused by neglecting O2 in the worst case does not exceed 12-15%, and is usually less.

For volumetric stress state:

Let us select a parallelepiped with edges of infinitesimal length around a certain point K of the body. In the general case, normal and tangential stresses can act on the faces of this elementary parallelepiped. The set of stresses on all possible areas passing through a point is called stressed state of the material at a point. It has been proven that it is possible to arrange a parallelepiped in space in such a way that only normal stresses remain on its faces. Such edges are called main venues, and the voltages on them are principal stresses. The greatest principal stress is denoted σ 1, the least - σ 3, and the intermediate - σ 2, therefore.

There are three types of stress states: linear, plane and volumetric (Fig. 3.1).

Fig.1. Types of stress state at a point: A– linear; b– flat; V– volumetric

2. Plane stress state

Let us consider the plane stress state in more detail. Let us select from a thin plate of thickness t an infinitesimal element, along the lateral faces of which normal and shear stresses act (Fig. 2, A). We assume that stresses are distributed evenly across the thickness of the plate, so the specific size t does not affect further analysis. We will look at the element from the tip of the axis z, and the stresses on the side faces of the element are considered positive (Fig. 2, b).

Rice. 2. Plane stress state

According to law of pairing of tangential stresses, i.e., shear stresses on mutually perpendicular areas are equal in magnitude and directed in such a way that they tend to rotate the element in opposite directions.

The main areas (Fig. 3) make an angle a 0 with the original areas, the value of which is determined from the expression

Rice. 3. Main areas and main stresses

The main stresses, denoted as and, are calculated using the formula

Extreme shear stresses are equal to the half-difference of the principal stresses and act on sites inclined to the main sites at an angle of 45°

Deformations of an infinitesimal element in a plane stress state consist in changing the linear dimensions of the element and changing the shape of the element. If, in the general case, normal and tangential stresses act on the faces of the element, then relative linear deformations occur at the point of the body

and angular deformation ( relative shift) in the form of a shear angle (Fig. 4, b).

Fig.4. Plane stress state: A– voltage; b– deformations

There are dependencies between relative linear deformations and stresses at a point of an elastic body in the form of Hooke's law:

Here, is the modulus of longitudinal elasticity (modulus of elasticity of the first kind); is Poisson's ratio.

A special case of a plane stress state is one in which only tangential stresses act on mutually perpendicular areas (Fig. 5).

This case is called pure shear, and the original sites are called pure shear sites. The main areas turn out to be inclined to the pure shear areas at an angle of 45°, and the main stresses are numerically equal to the shear stresses, with one of the main stresses being tensile and the other being compressive. According to the accepted rule for designating principal stresses;

Deformations of an infinitesimal element in pure shear consist in the distortion of right angles by a value called shear angle(Fig. 4 and 5).

There is a proportional relationship between the shear angle and shear stresses, called Hooke's law for pure shear

where is the proportionality coefficient G – shear modulus(modulus of elasticity of the second kind), measured in the same units as stress, MPa, kN/cm 2.

Three characteristics of the elastic properties of an isotropic material turn out to be interconnected by a dependence, which is most often written in the following form:

Fundamentals of the theory of elasticity

Lecture 4

Plane problem of elasticity theory

Slide 2

In the theory of elasticity there is a large class of problems that are important in the sense of practical applications and at the same time allow for significant simplifications of the mathematical side of the solution. The simplification lies in the fact that in these problems one of the coordinate axes of the body, for example the z axis, can be discarded and all phenomena can be considered as occurring in one coordinate plane x0y of the loaded body. In this case, stresses, strains and displacements will be functions of two coordinates - x and y.

A problem considered in two coordinates is called plane problem of elasticity theory.

Under the term " plane problem of elasticity theory“combine two physically different problems, leading to very similar mathematical dependencies:

1) the problem of a plane deformed state (plane deformation);

2) the problem of a plane stress state.

These problems are most often characterized by a significant difference between one geometric size and two other sizes of the bodies under consideration: large length in the first case and small thickness in the second case.

Plane strain

The deformation is called flat if the displacements of all points of the body can occur only in two directions in one plane and do not depend on the coordinate normal to this plane, i.e.

u=u(x,y); v=v(x,y); w=0 (4.1)

Plane deformation occurs in long prismatic or cylindrical bodies with an axis parallel to the z axis, along which a load acts on the lateral surface, perpendicular to this axis and not changing in magnitude along it.

An example of plane deformation is the stress-strain state that occurs in a long straight dam and a long arch of an underground tunnel (Fig. 4.1).

Figure – 4.1. Plane deformation occurs in the body of the dam and the roof of the underground tunnel

Slide 3

Substituting the components of the displacement vector (4.1) into the Cauchy formulas (2.14), (2.15), we obtain:

(4.2)

The absence of linear deformations in the direction of the z axis leads to the appearance of normal stresses σ z . From the formula of Hooke’s law (3.2) for the deformation ε z it follows that

from which we obtain the expression for stress σ z:

(4.3)

Substituting this relationship into the first two formulas of Hooke’s law, we find:

(4.4)

Slide 4

From the analysis of formulas (4.2) − (4.4) and (3.2) it also follows that

Thus, the basic equations of the three-dimensional theory of elasticity in the case of plane deformation are significantly simplified.

Of the three differential equations of Navier equilibrium (2.2), only two equations remain:

(4.5)

and the third turns into identity.

Since the direction cosine is everywhere on the lateral surface n=cos(v,z)=cos90 0 =0, Z v =0, then of the three conditions on the surface (2.4) only two equations remain:

(4.6)

where l, m are the direction cosines of the external normal v to the contour surface;

X,Y,X v, Y v– components of volumetric forces and intensity of external surface loads on the x and y axes, respectively.

Slide 5

The six Cauchy equations (2.14), (2.15) are reduced to three:

(4.7)

Of the six continuity equations for Saint-Venant deformations (2.17), (2.18), one equation remains:

(4.8)

and the rest turn into identities.

Of the six formulas of Hooke’s law (3.2), taking into account (4.2), (4.4), three formulas remain:

In these relations, new elastic constants have been introduced for the traditional form of elasticity theory:

Slide 6

Plane stress state

A plane stress state occurs when the length of the same prismatic body is small compared to the other two dimensions. In this case it is called thickness. Stresses in the body act only in two directions in the xOy coordinate plane and do not depend on the z coordinate. An example of such a body is a thin plate of thickness h, loaded along the side surface (rib) with forces parallel to the plane of the plate and uniformly distributed throughout its thickness (Fig. 4.2).

Figure 4.2 – Thin plate and loads applied to it

In this case, simplifications similar to those in the plane deformation problem are also possible. The stress tensor components σ z, τ xz, τ yz on both planes of the plate are equal to zero. Since the plate is thin, we can assume that they are equal to zero inside the plate. Then the stressed state will be determined only by the components σ x, σ y, τ xy, which do not depend on the z coordinate, i.e., they do not change along the thickness of the plate, but are functions of only x and y.

Thus, the following stress state arises in a thin plate:

Slide 7

In relation to stresses, the plane stress state differs from plane strain by the condition

In addition, from the formula of Hooke’s law (3.2), taking into account (4.10), for the linear deformation ε z we obtain that it is not equal to zero:

Consequently, the bases of the plate will be curved, as displacements will appear along the z axis.

Under these assumptions, the basic equations of plane deformation: differential equilibrium equations (4.5), conditions on the surface (4.6), Cauchy equations (4.7) and deformation continuity equations (4.8) retain the same form in the problem of a plane stress state.

The formulas of Hooke's law will take the following form:

Formulas (4.11) differ from formulas (4.9) of Hooke’s law for plane deformation only in the values ​​of the elastic constants: E and E 1 , v And v 1 .

Slide 8

In reverse form, Hooke's law will be written as follows:

(4.12)

Thus, when solving these two problems (plane deformation and plane stress state), you can use the same equations and combine the problems into one plane problem of the theory of elasticity.

There are eight unknowns in the plane problem of elasticity theory:

– two components of the displacement vector u and v;

– three components of the stress tensor σ x, σ y, τ xy;

– three components of the deformation tensor ε x, ε y, γ xy.

Eight equations are used to solve the problem:

– two differential equilibrium equations (4.5);

– three Cauchy equations (4.7);

– three formulas of Hooke’s law (4.9), or (4.11).

In addition, the resulting deformations must obey the equation of continuity of deformations (4.8), and on the surface of the body the equilibrium conditions (4.6) between internal stresses and intensities of external surface load X must be satisfied v, Y v.

Biaxial or flat is a stressed state of a body in which at all its points one of the principal stresses is equal to zero. It can be shown* that a plane stress state arises in a prismatic or cylindrical body (Fig. 17.1) with loose and unloaded ends, if a system of external forces normal to the axis is applied to the lateral surface of the body Oz and changing depending on z according to the quadratic law, it is symmetrical about the middle section. It turns out that in all cross sections of the body

and voltage a x, a y, x vary depending on z also, according to the quadratic law, symmetrical about the middle section. The introduction of these assumptions makes it possible to obtain a solution to the problem that satisfies conditions (17.13) and all equations of the theory of elasticity.

Of interest is the special case when the voltages do not depend on the variable z‘-

Such a stressed state is possible only under the action of a load uniformly distributed along the length. From the formulas of Hooke’s law (16.3) it follows that the deformations e x, e y, e z, y also do not depend on z, and deformations y and y zx taking into account (17.13) are equal to zero. In this case, the fourth and fifth of the deformation continuity equations (16.4), (16.5) are identically satisfied, and the second, third and sixth equations take the form

Integrating these equations and taking into account the third of the formulas of Hooke’s law (16.3) with a z = 0, we get

Cm.: Timoshenko S. P., Goodyear J. Theory of elasticity. Moscow: Nauka, 1975.

Thus, a plane stress state in a prismatic or cylindrical body with free ends, loaded with a constant surface load along the length of the body, is possible only in the special case when the sum of the stresses a x + a y varies depending on the variables x and at linear or constant.

If the distance between the end planes of the body (Fig. 7.1) is small compared to the dimensions of the sections, then we have the case of a thin plate (Fig. 17.5) loaded along the outer contour with forces symmetrically distributed relative to the middle plane of the plate according to a quadratic law. Since the plate thickness h is small, then with a slight error we can assume that for any symmetrical load on the stress plate relative to the median plane a x, a v, txv are evenly distributed throughout its thickness.

In this case, stresses should be understood as their average values ​​over the thickness, for example

It should also be noted that when introducing assumption (17.14), conditions (17.13) for stress to be zero

The considered case of the stressed state of a thin plate with assumptions (17.13) and (17.14) is often called generalized plane stress state.

Let us consider the basic equations of the theory of elasticity for this case.

Taking into account (17.13), the formulas of Hooke’s law (16.3) will be written in the form

The corresponding inverse relations have the form

Formulas (17.17) and (17.18) differ from formulas (17.7) and (17.9) of Hooke’s law for plane strain only in that in the latter, instead of the elastic modulus E and Poisson's ratio v include the given values E ( and v r

The equilibrium equations, Cauchy relations, deformation continuity equation and static boundary conditions do not differ from the corresponding equations (17.10), (17.3), (17.11), (17.12) for plane deformation.

Plane strain and the generalized plane stress state are essentially described by the same equations. The only difference is in the values ​​of the elastic constants in the formulas of Hooke's law. Therefore, both tasks are united by a common name: plane problem of elasticity theory.

The complete system of equations for a plane problem consists of two equilibrium equations (17.10), three geometric Cauchy relations (17.3) and three formulas of Hooke’s law (17.7) or (17.17). They contain eight unknown functions: three voltages a x, a y, % xy, three deformations e x, e y, y xy and two movements And And And.

If solving a problem does not require determining displacements, then the number of unknowns is reduced to six. To determine them, there are six equations: two equilibrium equations, three formulas of Hooke’s law and the deformation continuity equation (17.11).

The main difference between the two types of plane problem considered is the following. For plane strain ? z = 0,o z * 0, and the value c z can be found using formula (17.6) after the stresses about xio, are determined. Under a generalized plane stress state a z = 0, ? z Ф 0, and deformation ? z can be expressed in terms of stress o x and OU according to formula (17.16). moving w can be found by integrating the Cauchy equation