Method of possible movements. Application of the principle of possible movements. General displacement formula

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Establishing the general condition of equilibrium of a mechanical system. According to this principle, for the equilibrium of a mechanical system with ideal connections it is necessary and sufficient that the sum of virtual work $A\_i$ only active forces at any possible displacement of the system were equal to zero (if the system was brought to this position with zero velocities).

The number of linearly independent equilibrium equations that can be compiled for a mechanical system, based on the principle of possible displacements, is equal to the number of degrees of freedom of this mechanical system.

Possible movements of a non-free mechanical system are called imaginary infinitesimal movements allowed at a given moment by the constraints imposed on the system (in this case, the time explicitly included in the equations of non-stationary constraints is considered fixed). Projections of possible displacements onto Cartesian coordinate axes are called variations Cartesian coordinates.

Virtual movements are called infinitesimal movements allowed by connections during “frozen time”. Those. they differ from possible movements only when the connections are reonomic (explicitly dependent on time).

If, for example, the system is subject to $l$ holonomic rheonomic connections:

$f\_(\backslash alpha)(\backslash vec\; r,\; t)\; =\; 0,\; \backslash quad\; \backslash alpha\; =\; \backslash overline(1,l)$

These are possible movements $\backslash Delta\; \backslash vec\; r$ are those that satisfy

$\backslash sum\_(i=1)^(N)\; \backslash frac(\backslash partial\; f\_(\backslash alpha))(\backslash partial\; \backslash vec(r))\; \backslash cdot\; \backslash Delta\; \backslash vec(r)\; +\; \backslash frac(\backslash partial\; f\_(\backslash alpha\; ))(\backslash partial\; t)\; \backslash Delta\; t\; =\; 0,\; \backslash quad\; \backslash alpha\; =\; \backslash overline(1,l)$

And virtual $\backslash delta\; \backslash vec\; r$:

$\backslash sum\_(i=1)^(N)\; \backslash frac(\backslash partial\; f\_(\backslash alpha))(\backslash partial\; \backslash vec(r))\backslash delta\; \backslash vec(r)\; =\; 0,\; \backslash quad\; \backslash alpha\; =\; \backslash overline(1\; ,l)$

Virtual movements, generally speaking, have no relation to the process of movement of the system - they are introduced only in order to identify the force relationships existing in the system and obtain equilibrium conditions. A small amount of displacement is needed so that the reactions of ideal connections can be considered unchanged.

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Literature

• Buchgolts N. N. Main course theoretical mechanics. Part 1. 10th ed. - St. Petersburg: Lan, 2009. - 480 p. - ISBN 978-5-8114-0926-6.
• Targ S. M. Short course in theoretical mechanics: Textbook for universities. 18th ed. - M.: graduate School, 2010. - 416 p. - ISBN 978-5-06-006193-2.
• Markeev A.P. Theoretical mechanics: textbook for universities. - Izhevsk: Research Center "Regular and Chaotic Dynamics", 2001. - 592 p. - ISBN 5-93972-088-9.

An excerpt characterizing the Principle of Possible Movements

While such conversations took place in the reception room and in the princess's rooms, the carriage with Pierre (who was sent for) and with Anna Mikhailovna (who found it necessary to go with him) drove into the courtyard of Count Bezukhy. When the wheels of the carriage sounded softly on the straw spread under the windows, Anna Mikhailovna, turning to her companion with comforting words, was convinced that he was sleeping in the corner of the carriage, and woke him up. Having woken up, Pierre followed Anna Mikhailovna out of the carriage and then only thought about the meeting with his dying father that awaited him. He noticed that they drove up not to the front entrance, but to the back entrance. While he was getting off the step, two people in bourgeois clothes hurriedly ran away from the entrance into the shadow of the wall. Pausing, Pierre saw several more similar people in the shadows of the house on both sides. But neither Anna Mikhailovna, nor the footman, nor the coachman, who could not help but see these people, paid no attention to them. Therefore, this is so necessary, Pierre decided to himself and followed Anna Mikhailovna. Anna Mikhailovna walked with hasty steps up the dimly lit narrow stone staircase, calling to Pierre, who was lagging behind her, who, although he did not understand why he had to go to the count at all, and even less why he had to go up the back stairs, but , judging by the confidence and haste of Anna Mikhailovna, he decided to himself that this was necessary. Halfway up the stairs, they were almost knocked down by some people with buckets, who, clattering with their boots, ran towards them. These people pressed against the wall to let Pierre and Anna Mikhailovna through, and did not show the slightest surprise at the sight of them.
– Are there half princesses here? – Anna Mikhailovna asked one of them...
“Here,” the footman answered in a bold, loud voice, as if now everything was possible, “the door is on the left, mother.”
“Maybe the count didn’t call me,” Pierre said as he walked out onto the platform, “I would have gone to my place.”
Anna Mikhailovna stopped to catch up with Pierre.
- Ah, mon ami! - she said with the same gesture as in the morning with her son, touching his hand: - croyez, que je souffre autant, que vous, mais soyez homme. [Believe me, I suffer no less than you, but be a man.]
- Right, I'll go? - asked Pierre, looking affectionately through his glasses at Anna Mikhailovna.

As is known from the course of theoretical mechanics, the equilibrium condition of an object can have a force or energy formulation. The first option represents the condition that the main vector and the main moment of all forces and reactions acting on the body are equal to zero. The second approach (variational), called the principle of possible displacements, turned out to be very useful for solving a number of problems in structural mechanics.

For a system of absolutely rigid bodies, the principle of possible displacements is formulated as follows: if a system of absolutely rigid bodies is in equilibrium, then the sum of the work of all external forces on any possible infinitesimal displacement is zero. Possible (or virtual) is a movement that does not violate the kinematic connections and continuity of bodies. For the system in Fig. 3.1, only rotation of the rod relative to the support is possible. When turning through an arbitrary small angle, the forces and do work According to the principle of possible displacements, if the system is in equilibrium, then there must be . Substituting here the geometric relations we obtain the equilibrium condition in the force formulation

The principle of possible displacements for elastic bodies is formulated as follows: if a system of elastic bodies is in equilibrium, then the sum of the work of all external and internal forces on any possible infinitesimal displacement is zero. This principle is based on the concept of the total energy of an elastic deformed system P. If the loading of a structure occurs statically, then this energy is equal to the work done by external U and internal W forces when transferring the system from a deformed state to its original state:

With the specified translation, external forces do not change their value and perform negative work U= -F. In this case, the internal forces are reduced to zero and do positive work, since these are the forces of adhesion of material particles and are directed in the direction opposite to the external load:

Where - specific potential energy of elastic deformation; V is the volume of the body. For a linear system, where . According to the Lagrange-Dirichlet theorem, the state of stable equilibrium corresponds to the minimum of the total potential energy of the elastic system, i.e.

The last equality fully corresponds to the formulation of the principle of possible movements. Energy increments dU and dW can be calculated for any possible displacements (deviations) of the elastic system from the equilibrium state. To calculate structures that satisfy the requirements of linearity, the infinitesimal possible displacement d can be replaced by a very small final displacement, which can be any deformed state of the structure created by an arbitrarily chosen system of forces. Taking this into account, the resulting equilibrium condition should be written as

Work of external forces

Let us consider the methodology for calculating the work of external forces on actual and possible displacement. The rod system is loaded with forces and (Fig. 3.2, a), which act simultaneously, and at any point in time the ratio remains constant. If we consider it a generalized force, then from the value at any time we can calculate all other loads (in this case). The dashed line shows the actual elastic displacement arising from these forces. We denote this state by index 1. We denote the movement of the points of application of forces and in the direction of these forces in state 1 by and .

In the process of loading a linear system with forces, the forces increase and the displacements and increase in proportion to them (Fig. 3.2, c). The actual work of forces and on the displacements they create is equal to the sum of the areas of the graphs, i.e. . Writing this expression as , we obtain the product of the generalized force and the generalized displacement. In this form you can submit

Next, we will consider the work of external forces on a possible displacement. As a possible displacement, let us take, for example, the deformed state of the system resulting from the application of a force at some point (Fig. 3.2, b). This state, corresponding to the additional movement of the points of application of forces and at a distance and , will be denoted by 2. The forces and , without changing their value, perform virtual work on the displacements and (Fig. 3.2, c):

As you can see, in the designation of movement, the first index shows the state in which the points and directions of these movements are specified. The second index shows the state in which the forces that cause this movement act.

Work of unit force F 2 on actual displacement

If we consider state 1 as a possible displacement for the force F 2, then its virtual work on displacement

Work of internal forces

Let us find the work of the internal forces of state 1, i.e., from the forces and , on the virtual displacements of state 2, i.e., resulting from the application of load F 2 . To do this, select a rod element with length dx (Fig. 3.2 and 3.3, a). Since the system under consideration is flat, only two forces S and Q z and a bending moment Mu act in the sections of the element. These forces for the cut element are external. Internal forces are the adhesive forces that provide the strength of the material. They are equal to the external ones in value, but are directed in the direction opposite to the deformation, therefore their work under loading is negative (Fig. 3.3, b-d, shown in gray). Let us sequentially calculate the work done by each force factor.

The work of longitudinal forces on displacement, which is created by forces S 2 resulting from the application of load F 2 (Fig. 3.2, b, 3.3, b),

We find the elongation of a rod with length dx using the well-known formula

where A is the cross-sectional area of ​​the rod. Substituting this expression into the previous formula, we find

In a similar way, we determine the work that the bending moment does on the angular displacement created by the moment (Fig. 3.3, c):

We find the angle of rotation as

where J is the moment of inertia of the cross section of the rod relative to the y-axis. After substitution we get

Let's find the work done by the transverse force on displacement (Fig. 3.3, d). Tangential stresses and shears from the shearing force Q z are not distributed linearly over the cross section of the rod (unlike normal stresses and elongations in previous loading cases). Therefore, to determine the shear work, it is necessary to consider the work done by the tangential stresses in the layers of the rod.

Tangential stresses from the force Q z, which act in a layer lying at a distance z from the neutral axis (Fig. 3.3, e), are calculated using the Zhuravsky formula

where Su is the static moment of the part of the cross-sectional area lying above this layer, taken relative to the y-axis; b is the width of the section at the level of the layer under consideration. These stresses create a shift of the layer by an angle, which, according to Hooke’s law, is defined as - shear modulus. As a result, the end of the layer is shifted by

The total work done by the tangential stresses of the first state acting at the end of this layer on the displacements of the second state is calculated by integrating the product of the cross-sectional area

After substituting here the expressions for and we get

Let us subtract from the integral quantities that do not depend on z, multiply and divide this expression by A, we obtain

Here a dimensionless coefficient is introduced,

depending only on the configuration and ratio of section sizes. For a rectangle = 1.2, for I-beam and box sections (A c is the cross-sectional area of ​​the wall or in a box section - two walls).

Since the work of each of the considered loading components (S, Q, M) on displacements caused by other components is equal to zero, then the total work of all internal forces for the considered rod element of length dx

In the cross section of an element of a spatial rod system there are six internal forces (S, Q, Q z, M x, Mu, M 2), therefore for it the expression for the total work of internal forces will have the form,

Here M x is the torque in the rod; J T is the moment of inertia of the rod during free torsion (geometric torsional rigidity). In the integrand, the subscripts “and” are omitted.

In formulas (3.3) and (3.4) S v Q yV Q zl , M x1 , M y1 , M g1 denote analytical expressions for diagrams of internal forces from the action of forces F(and F(,aS 2 , Q y 2 , Q z 2 , M x2, M y2, M g2 - descriptions of diagrams of internal forces from force F 2.

Theorems about elastic systems

The structure of formulas (3.3) and (3.4) shows that they are “symmetrical” with respect to states 1 and 2, i.e. the work of the internal forces of state 1 on the displacements of state 2 is equal to the work of the internal forces of state 2 on the displacements of state 1 But according to (3.2)

Consequently, if the work of internal forces is equal, then the work of external forces is equal. This statement is called the theorem on the reciprocity of work (Betti’s theorem, 1872).

For a rod system loaded with force F 1 (Fig. 3.4, a), we take as a possible displacement the deformed state that arose when it was loaded with force F 2 (Fig. 3.4, b). For this system, according to Betti’s theorem 1- If we put , we get

This formula expresses Maxwell's theorem (1864) on the reciprocity of displacements: the displacement of the point of application of the first unit force in its direction, caused by the action of the second unit force, is equal to the displacement of the point of application of the second unit force in its direction, caused by the action of the first unit force. This theorem can also be applied to the system in Fig. 3.2. If we set = 1 N (section 3.1.2), we obtain the equality of generalized displacements .

Let's consider a statically indeterminate system with supports that can be used to set the required movement, which is accepted as possible (Fig. 3.4, c, d). In the first state, we will shift the support 1 by and in the second - we will set the rotation of the embedment by an angle - In this case, reactions will arise in the first state and , and in the second - i . According to the work reciprocity theorem, we write If we set (here dimension = m, and quantity is dimensionless), then we get

This equality is numerical, since the dimension of the reaction = N, a = N-m. Thus, the reaction R 12 in fixed bond 1, which occurs when bond 2 moves by one, is numerically equal to the reaction that occurs in bond 2 with a unit displacement of bond 1. This statement is called the reaction reciprocity theorem.

The theorems presented in this section are used for the analytical calculation of statically indeterminate systems.

Definition of movements

General displacement formula

To calculate the displacements that occur in the rod system under the action of a given load (state 1), an auxiliary state of the system should be created in which one unit force acts, doing work on the desired displacement (state 2). This means that when determining linear displacement, it is necessary to specify a unit force F 2 = 1 N, applied at the same point and in the same direction in which the displacement must be determined. If it is necessary to determine the angle of rotation of any section, then a single moment F 2 = 1 N m is applied to this section. After this, the energy equation (3.2) is drawn up, in which state 2 is taken as the main one, and the deformed state

state 1 is considered as virtual movement. From this equation the required displacement is calculated.

Let us find the horizontal displacement of point B for the system in Fig. 3.5, a. In order for the required displacement D 21 to be included in the equation of work (3.2), we take as the ground state the displacement of the system under the action of a unit force F 2 - 1 N (state 2, Fig. 3.5, b). We will consider the possible displacement to be the actual deformed state of the structure (Fig. 3.5, a).

We find the work of external forces of state 2 on displacements of state 1 as follows: According to (3.2),

therefore, the required displacement

Since (section 3.1.4), the work of the internal forces of state 2 on the displacements of state 1 is calculated using formula (3.3) or (3.4). Substituting expression (3.3) into (3.7) for the work of internal forces of a flat rod system, we find

For further use of this expression, it is advisable to introduce the concept of single diagrams of internal force factors, i.e. of which the first two are dimensionless, and the dimension . The result will be

Expressions for distribution diagrams of the corresponding internal forces from the acting load should be substituted into these integrals And and from force F 2 = 1. The resulting expression is called Mohr's formula (1881).

When calculating spatial rod systems, formula (3.4) should be used to calculate the total work of internal forces, then it will be

It is quite obvious that expressions for diagrams of internal forces S, Q y, Q z, M x, M y, M g and the values ​​of the geometric characteristics of sections A, J t, Jу, J, for the corresponding n-th section are substituted into the integrals. To shorten the notation in the notation of these quantities, the index “and” is omitted.

3.2.2. Special cases of determining displacements

Formula (3.8) is used in the general case of a flat rod system, but in a number of cases it can be significantly simplified. Let's consider special cases of its implementation.

1. If deformations from longitudinal forces can be neglected, which is typical for beam systems, then formula (3.8) will be written as

2. If a flat system consists only of bent thin-walled beams with a ratio l/h> 5 for consoles or l/h> 10 for spans (I and h are the length of the beam and the height of the section), then, as a rule, the bending deformation energy significantly exceeds energy of deformations from longitudinal and transverse forces, so they can not be taken into account in the calculation of displacements. Then formula (3.8) will take the form

3. For trusses, the rods of which, under nodal loading, experience mainly longitudinal forces, we can assume M = 0 and Q = 0. Then the displacement of the node is calculated by the formula

Integration is performed over the length of each rod, and summation is performed over all rods. Bearing in mind that the force S u in i-th rod and the cross-sectional area does not change along its length, we can simplify this expression:

Despite the apparent simplicity of this formula, the analytical calculation of displacements in trusses is very labor-intensive, since it requires determining the forces in all the rods of the truss from the effective load () and from the unit force () applied at the point whose displacement needs to be found.

3.2.3. Methodology and examples for determining displacements

Let us consider the calculation of the Mohr integral using the method of A. N. Vereshchagin (1925). The Mohr integral has the form (3.8), where diagrams of bending moments, longitudinal or transverse forces can appear as D 1, D 2. At least one of the diagrams () in the integrand expression is linear or piecewise linear, since it is built from a unit load. Therefore for

The first isintegral is numerically equal to the area of ​​the subgraph (shaded in Fig. 3.6), and the second is equal to the static moment of this area relative to the axis. The static moment can be written as , where is the position coordinate of the center of gravity of the area (point A). Taking into account what has been said, we get

(3.13)

Vereshchagin's rule is formulated as follows: if at least one of the diagrams is linear on a section, then the Mohr integral is calculated as the product of the area arbitrarily

When calculating structures in the Mathcad environment, there is no need to use Vereshchagin's rule, since the integral can be calculated by numerical integration.

Example 3.1(Fig. 3.7, a). The beam is loaded with two symmetrically located forces. Find the displacement of the points of application of forces.

1. Let’s construct a diagram of bending moments M 1 from forces F 1 . Support reactions Maximum bending moment under force

2. Since the system is symmetrical, the deflections under the forces will be the same. As an auxiliary state, we take the loading of the beam with two unit forces F 2 = 1 N, applied at the same points as the forces F 1

(Fig. 3.7, b). The diagram of bending moments for this loading is similar to the previous one, and the maximum bending moment M 2max = 0.5 (L-b).

3. Loading of the system by two forces of the second state is characterized by a generalized force F 2 and a generalized displacement, which create the work of external forces on the displacement of state 1, equal to . Let's calculate the displacement using formula (3.11). Multiplying the diagrams by sections according to Vereshchagin’s rule, we find

After substituting values we get

Example 3.2. Find the horizontal displacement of the movable support of the U-shaped frame loaded with force F x (Fig. 3.8, a).

1. Let's construct a diagram of bending moments from force F 1 Support reactions . Maximum bending moment under force F 1

2. As an auxiliary state, let us take the loading of the beam with a unit horizontal force F 2 applied at point B (Fig. 3.8, b). We construct a diagram of bending moments for this loading case. Support reactions A 2y = B 2y = 0, A 2x = 1. Maximum bending moment.

3. We calculate the displacement using formula (3.11). In vertical sections the product is zero. On the horizontal section, the M 1 diagram is not linear, but the diagram is linear. Multiplying the diagrams using Vereshchagin’s method, we get

The product is negative, since the diagrams lie on opposite sides. The resulting negative displacement value indicates that its actual direction is opposite to the direction of the unit force.

Example 3.3(Fig. 3.9). Find the angle of rotation of the cross-section of a two-support beam under the force and find the position of the force at which this angle will be maximum.

1. Let's construct a diagram of the bending moments M 1 from the force F 1. To do this, we will find the support reaction A 1. From the equilibrium equation for the system as a whole let's find the maximum bending moment under force Fj

2. As an auxiliary state, we take the loading of the beam with a unit moment F 2 = 1 Nm in the section whose rotation must be determined (Fig. 3.9, b). We construct a diagram of bending moments for this loading case. Support reactions A 2 = -B 2 = 1/L, bending moments

Both moments are negative, as they are directed clockwise. Diagrams are built on stretched fiber.

3. We calculate the angle of rotation using formula (3.11), multiplying over two sections,

By denoting , we can obtain this expression in a more convenient form:

The dependence of the rotation angle on the position of the force F 1 is shown in Fig. 3.9, c. Having differentiated this expression, from the condition we find the position of the force at which the angle of inclination of the beam under it will be greatest in absolute value. This will happen at values ​​equal to 0.21 and 0.79.