The momentum conservation law for a moving small volume W of a liquid particle (with impenetrable walls) is

where on the right side is the sum of all the forces acting on the selected volume, and for simplicity we will first assume that there is no internal mass inflow (M "= 0). Limiting ourselves to the consideration of the body force F m (for example, centrifugal or gravity acting per unit mass, [n / kg]) and pressure forces P (acting per unit area, [n / m 2]), we write

.

Given that (the integral is taken over a liquid particle, that is, over a given amount of liquid), and by converting the surface integral of pressure into a volume one, one can rewrite the equation in the form

. (1.15)

This is the law of conservation of momentum in integral form.

Based on an arbitrary choice of the volume of a liquid particle, we can proceed to the differential form:

. (1.16)

This is the law of conservation of momentum in the form of Lagrange.

The derivative dV/dt included in the equation is a substantive derivative that describes the change in the velocity of a liquid particle.

Using the connection between the substantial (total) time derivative and the partial time derivative of speed (change in speed at a given point), obtained earlier, we arrive at another differential form of the momentum conservation equation (the Euler form):

. (1.17)

This is Euler's equation, it was obtained by him back in 1755. This equation expresses the law of conservation of momentum (momentum).

In projections onto the axes of the Cartesian system, this equation has the form

Let us write down the obtained equations of motion in another form - in the form of momentum transfer. To do this, we perform the following transformations using the continuity equation:

, But ,

Then and hence

. (1.18)

In the Cartesian coordinate system, these equations have the form

These equations, as in the case of the continuity equation, can be obtained in yet another way. Let us single out a fixed elementary parallelepiped with sides dx, dy, dz in the flow of the moving mass and calculate the mass of fluid flowing through it during the time dt.

Let us single out an elementary parallelepiped with edges dx, dy, dz in a gas or liquid flow. The selected volume is affected by mass forces (for example, inertial, gravitational), surface forces - pressure and friction. Let's find the projections of these forces on the x-axis (Fig. 1.5):

a) we apply body forces in the center of the element with volume dw.

Its projection on the x-axis is:

similarly for other axes;

b) pressure force. On the left side of the element along the x axis, the specific pressure is equal to Р, the force Pdydz acts on the area dydz. On the opposite side, the specific pressure is , and this side is affected by a force . The "-" sign indicates that the force acts against the direction of the x-axis. The resultant of these forces is equal to their algebraic sum:

. (1.19)

According to the second law of mechanics, the resultant is equal to the product of the mass of the element ρdW and its acceleration dV x /dt:

where is local, - convective change in the value of V x, d / dt - substantive derivative:

Equating equations (1.19) and (1.20), we obtain:

Similarly, we write down the equations for the projections of forces on the y and z axes:

This is the equation of motion. It is often written as

In the case of stationarity of the process, the first terms of the equation will be equal to zero.

Consider now this law for a real fluid, considering viscosity (internal friction). Let's start by considering the equations of motion for an isothermal fluid and once again recall that the continuity equation is also valid for a real fluid, since its derivation was based only on the law of conservation of matter. Let us use the equation written in the form of a law for the momentum transfer of an ideal fluid and add terms to it that are responsible for the momentum transfer as a result of the action of viscous forces.

The main momentum vector K of a system of material particles is equal to the integral of the products of their elementary masses dm and the particle velocity vectors V:

.

Let us apply to the volume W of mass m the theorem on the change of the principal momentum vector. Equating the total time derivative of the main momentum vector to the main vector of external mass F and surface P forces, we obtain

, (13)

where p n is the resulting component of the internal forces in the gas (pressure forces and viscous stresses) applied to the surface S of the volume W.

We calculate the total derivative of the main vector, and for simplicity we will first assume that there is no internal mass inflow (M "=0), then

To transform the surface integral on the right side of (13) into a volume integral, we rewrite it as:

where p x, p y , p z is the stress vector applied to the positive sides of the site, and we apply the vector analysis formulas:

(1.23)

Then we will have

. (1.24)

Substituting into (1.16) the values ​​of the quantities included in it and transferring all the terms to one line, we obtain

. (1.25)

Using the assumption that the volume W is arbitrary and equating the integrand to zero, we obtain

Projecting both parts of the equality onto the directions of the coordinate axes, we get:

(1.27)

These equations of continuum dynamics are "in stresses", or "momentum equations".

Friction force per unit surface according to Newton's law

(μ is the coefficient of dynamic viscosity, N × s / m 2).

.

.

Summing up the forces, we obtain the projection onto the x-axis of the resultant forces applied to the volume dW:

. (1.28)

Let's get the equations of motion, taking into account the viscosity, using the approach shown in Fig. 1.5. Let us add the friction force, having determined it from the consideration of a flat laminar flow, in which the velocity V x changes only in the direction of the y axis. In this case, the friction force s occurs only on the side faces of the element (Fig. 1.6).

Near the left side, the velocity of particles is less than in the element itself, so here in the section "y" the friction force is directed against the movement and is equal to - sdxdz. At the right side, the speed of movement is greater than in the element itself, so here in the section "y + dy" the friction force is directed towards the movement and is equal to

Here is the friction force per unit surface, according to Newton's law.

Substituting this expression into the previous equation and taking μ = const, we get .

In general, when V x changes in three directions, the projection of the friction force on the x-axis is given by

.

Summing up the forces, we obtain the projection onto the x-axis of the resultant forces applied to the volume dW:

. (1.29)

Using again the notion of a substantial derivative

according to the second law of mechanics we get:

Similarly, we write the equations for the projections of forces on the y and z axes (taking into account that ):

These equations of motion are called the Navier-Stokes equations. The differential equation of motion in the Navier-Stokes form describes the motion of a viscous compressible fluid or gas and is valid for both laminar and turbulent motion.

In the case of the “ideal gas” hypothesis, the Navier-Stokes equations of motion turn into the Euler equations:

(1.30)

In the case of stationarity of the process, the first terms of the equation will be equal to zero. For two- and one-dimensional motion, the Navier-Stokes and Euler equations are simplified accordingly.

Law of energy conservation. The law of conservation of energy not only establishes the invariance of all energy for any given mass of liquid or gas, but also reflects the mutual transformation of various forms of motion of matter, and primarily mechanical energy into thermal energy. To calculate these transformations, the energy balance equation is used, which is derived from the general thermodynamic law of conservation of energy, which for an individual (impenetrable) volume of a moving medium is formulated as follows:

- the change in the total energy of the allocated volume of liquid or gas per unit time is equal to the sum of the work of the mass and surface external forces applied to it on the surfaces limiting this volume, and the heat supplied from the outside for the same time.

This law is expressed by the integral equality

Where is the specific total energy; U = c v T is the specific internal energy; is the resultant of body forces, is the resultant component of internal forces in the gas (pressure forces and viscous stresses) applied to the surface S of the selected volume W; q is the specific amount of energy (usually heat) supplied per unit time to the working fluid in the allocated volume.

Taking into account the arbitrariness of the allocated volume W, we obtain the differential form of this law:

The need to introduce the energy equation follows from the fact that two equations - continuity (scalar) and motion (vector) - contain three unknown quantities: one vector (velocity) and two scalar (pressure p and density r), therefore for gas (W = var ) the number of required quantities is one more than the number of equations. If we add the energy equation, then another unknown quantity will be added - temperature T. The system of equations will turn out to be closed by adding the equation of state, and then the problem of aerogasdynamics (under given boundary and initial conditions) becomes defined.

If an ideal incompressible fluid is considered, then it is assumed that there is no heat transfer and friction in the fluid. In this case, the motion is adiabatic in each liquid particle. Therefore, the law of conservation of energy translates into the statement that the energy of each liquid element remains constant:

It follows that the energy equation is not used to describe the motion of an ideal incompressible fluid.

From the theorem on the change in the momentum of the system, the following important consequences can be obtained:

1) Let the sum of all external forces acting on the system be equal to zero:

if the sum of all external forces acting on the system is equal to zero, then the momentum vector of the system will be constant in magnitude and direction.

2) Let the external forces acting on the system be such that the sum of their projections on some axis (for example Ox) is equal to zero:

Then it follows from the equation that in this case . Thus, if the sum of the projections of all acting external forces on some axis is equal to zero, then the projection of the momentum of the system on this axis is a constant value.

These results express law of conservation of momentum of the system. It follows from them that internal forces cannot change the total momentum of the system. Let's look at some examples:

a) The phenomenon of bestowal or rollback. If we consider a rifle and a bullet as one system, then the pressure of the powder gases when fired will be an internal force. This force cannot change the total momentum of the system. But since the propellant gases, acting on the bullet, give it a certain amount of movement directed forward, they must simultaneously give the rifle the same amount of movement in the opposite direction. This will cause the rifle to move backward, i.e. so-called return. A similar phenomenon occurs when firing from a gun (rollback).

b) The operation of the propeller (propeller). The propeller informs a certain mass of air (or water) of motion along the axis of the propeller, throwing this mass back. If we consider the discarded mass and the aircraft (or ship) as one system, then the forces of interaction between the propeller and the medium as internal cannot change the total momentum of this system. Therefore, when a mass of air (water) is thrown back, the aircraft (or ship) receives the corresponding forward speed, such that the total momentum of the system under consideration remains equal to zero, since it was zero before the start of movement.

A similar effect is achieved by the action of oars or paddle wheels.

c) Jet propulsion. In a rocket projectile (rocket), gaseous products of fuel combustion are ejected at high speed from a hole in the tail of the rocket (from the nozzle of a jet engine). The pressure forces acting in this case will be internal forces, and they cannot change the total momentum of the rocket system - fuel combustion products. But since the escaping gases have a certain amount of motion directed backwards, the rocket receives a corresponding forward speed.

d'Alembert principle.

All methods for solving problems of dynamics that we have considered so far are based on equations that follow either directly from Newton's laws, or from general theorems that are consequences of these laws. However, this path is not the only one. It turns out that the equations of motion or the equilibrium conditions of a mechanical system can be obtained by assuming other general propositions instead of Newton's laws, called the principles of mechanics. In a number of cases, the application of these principles makes it possible, as we shall see, to find more efficient methods for solving the corresponding problems. In this chapter, one of the general principles of mechanics, called d'Alembert's principle, will be considered.

Suppose we have a system consisting of n material points. Let's single out some of the points of the system with mass . Under the action of external and internal forces applied to it and (which include both active forces and coupling reactions), the point receives some acceleration with respect to the inertial reference frame.

Let us introduce into consideration the quantity

having the dimension of force. A vector quantity equal in absolute value to the product of the mass of a point and its acceleration and directed opposite to this acceleration is called the force of inertia of the point (sometimes the d'Alembert force of inertia).

Then it turns out that the movement of a point has the following general property: if at each moment of time we add the force of inertia to the forces actually acting on the point, then the resulting system of forces will be balanced, i.e. will

.

This expression expresses the d'Alembert principle for one material point. It is easy to see that it is equivalent to Newton's second law and vice versa. Indeed, Newton's second law for the point in question gives . Transferring the term here to the right side of the equality, we arrive at the last relation.

Repeating the above reasoning with respect to each of the points of the system, we arrive at the following result, which expresses the d'Alembert principle for the system: if at any moment of time to each of the points of the system, in addition to the external and internal forces actually acting on it, the corresponding inertia forces are applied, then the resulting system of forces will be in equilibrium and all the equations of statics can be applied to it.

The significance of the d'Alembert principle lies in the fact that when it is directly applied to problems of dynamics, the equations of motion of the system are compiled in the form of well-known equilibrium equations; which makes a uniform approach to solving problems and usually greatly simplifies the corresponding calculations. In addition, in conjunction with the principle of possible displacements, which will be discussed in the next chapter, the d'Alembert principle allows us to obtain a new general method for solving problems of dynamics.

Applying the d'Alembert principle, it should be borne in mind that only external and internal forces act on a point of a mechanical system, the movement of which is being studied, and , arising as a result of the interaction of the points of the system with each other and with bodies that are not included in the system; under the action of these forces, the points of the system and move with the corresponding accelerations. The forces of inertia, which are mentioned in the d'Alembert principle, do not act on moving points (otherwise, these points would be at rest or move without acceleration, and then there would be no inertial forces themselves). The introduction of inertial forces is just a technique that allows you to compose the equations of dynamics using simpler methods of statics.

Let us turn to the basic equation of the dynamics of rotational motion

and consider a special case when external forces either do not act on the body at all, or they are such that their resultant does not give a moment about the axis of rotation Then

But if the change in a quantity is zero, then, therefore, the quantity itself remains constant:

Rice. 66. Salto-mortale.

So, if external forces do not act on the body (or their resulting moment relative to the axis of rotation is zero), then the angular momentum of the body relative to the axis of rotation remains unchanged. This law is called the law of conservation of angular momentum about the axis of rotation

Let us give several examples illustrating the law of conservation of angular momentum.

The gymnast, while jumping over his head (Fig. 66), presses his arms and legs to the body. This reduces its moment of inertia,

and since the product must remain unchanged, the angular velocity of rotation increases, and in the short period of time while the gymnast is in the air, he manages to make a complete revolution.

The ball is tied to a thread wound around a stick; as the length of the thread decreases, the moment of inertia of the ball decreases and, consequently, the angular velocity increases.

Rice. 67 Rotation of a man standing on Zhukovsky's bench. speed up if he lowers his arms and slow down if he raises them.

Rice. 68. If we raise the bicycle wheel above our head and bring it into rotation, then we ourselves, together with the platform, will begin to rotate in the opposite direction.

A number of interesting experiments can be done by standing on a platform rotating on a ball bearing (Zhukovsky's bench). On fig. 67 and 68 show some of these experiences.

Comparing the equations derived in the last paragraphs with the laws of rectilinear translational motion, it is easy to see that the formulas that determine the rotational motion about a fixed axis are similar to the formulas for rectilinear translational motion.

The following table compares the main quantities and equations that determine these movements:

(see scan)

Gyroscopes. Reactive gyroscopic effect. A rigid body rotating at high angular velocity around an axis of complete symmetry (free axis) is called a gyroscope. According to the law of conservation of the angular momentum vector, the gyroscope seeks to keep the direction of its axis of rotation unchanged in space and exhibits the greater stability (i.e., the greater the resistance to rotation of the axis of rotation), the greater its moment of inertia and the greater the angular velocity of rotation.

When we hold some massive motionless body on outstretched arms and tell it to move, for example, from left to right, then the force of inertia developed by the body moves us in the opposite direction. The manifestation of the forces of inertia of a rotating gyroscope, when we turn its axis of rotation, turns out to be more complex and unexpected at first glance. So, if we, holding the horizontally directed axis of rotation of the gyroscope in our hands, begin to raise one end of the axis and lower the other, i.e. turn the axis in the vertical plane, then we will feel that the axis puts pressure on the hands not in the vertical, but in the horizontal plane, pressing one of our hands and pulling the other. If, when viewed from the right, the rotation of the gyroscope is seen to occur in a clockwise direction (i.e., the moment of momentum of the gyroscope is directed horizontally to the left), then an attempt to raise the left end of the axis, lowering the right end down, causes the left end of the axis to move in the horizontal plane away from us, and the right - on us.

Such a reaction of the gyroscope (the so-called gyroscopic effect) is explained by the desire of the gyroscope to keep its angular momentum unchanged and, moreover, to keep it unchanged not only in magnitude, but also in direction. Indeed, in order for the angular momentum of the gyroscope to remain geometrically unchanged during the rotation of the axis of rotation of the gyroscope in the vertical plane by the angle a (Fig. 69), the gyroscope must acquire additional rotation around the vertical axis with the moment of momentum such that geometrically

For this reason, a rotating gyroscope, balanced on a movable axis by a weight (Fig. 70), acquires additional

rotation around the vertical axis, if the weight that balanced the gyroscope is slightly moved away from the fulcrum of the axis (when outweighing, the weight imparts a certain inclination to the axis, which causes the gyroscope axis to rotate around the fulcrum in the direction that corresponds to the direction of the vector in Fig. 69).

For the same reason, the axis of the top acquires, as a result of the overturning action of gravity, a circular motion, which is called precession (Fig. 71).

So, if a couple of forces are applied to a rotating gyroscope, tending to rotate it about an axis perpendicular to the axis of rotation, then the gyroscope will actually turn, but only around the third axis, perpendicular to the first two. To turn a rotating gyroscope (for example, in the direction as shown in Fig. 72), it is necessary to apply a torque to the axis of the gyroscope in a plane perpendicular to the direction of rotation.

Rice. 71. Scheme of the movement of the top.

A more detailed analysis of the phenomena similar to those described above shows that the gyroscope tends to position its axis of rotation in such a way that it forms the smallest possible angle with the axis of forced rotation and that both rotations take place in the same direction.

This property of the gyroscope is used in the gyroscopic compass, which is widely used, especially in the navy. The gyrocompass is a rapidly rotating top (three-phase current motor, making up to 25,000 rpm), which floats on a special float in a vessel with mercury and whose axis is set in the plane of the meridian. In this case, the source of external torque is the daily rotation of the Earth around its axis. Under its action, the axis of rotation of the gyroscope tends to coincide in direction with the axis of rotation of the Earth, and since the rotation of the Earth acts on the gyroscope continuously, the axis of the gyroscope finally takes this position, that is, it is established along the meridian, and continues to remain in it completely just like a regular magnetic needle.

Gyroscopes are often used as stabilizers. They are installed to reduce pitching on ocean-going ships.

Stabilizers for single-rail railways were also designed; a massive, rapidly rotating gyroscope placed inside a single-rail car prevents the car from tipping over. Rotors for gyroscopic stabilizers are made in weight from 1 to 100 tons or more.

In torpedoes, gyroscopic devices, automatically acting on the steering control, ensure the straightness of the torpedo's movement in the direction of the shot.

Rice. 73. Precession of the earth's axis.

The daily rotation of the Earth makes it similar to a gyroscope. Since the Earth is not a ball, but a figure close to an ellipsoid, the attraction of the Sun creates a resultant that does not pass through the center of mass of the Earth (as it would be in the case of a ball). As a result, a torque arises, which tends to rotate the Earth's axis of rotation perpendicular to the plane of its orbit (Fig. 73). In this regard, the earth's axis is experiencing precessional motion (with a complete revolution in about 25,800 years).

His movements, i.e. value .

Pulse is a vector quantity coinciding in direction with the velocity vector.

The unit of momentum in the SI system: kg m/s .

The impulse of a system of bodies is equal to the vector sum of the impulses of all bodies included in the system:

Law of conservation of momentum

If additional external forces act on the system of interacting bodies, for example, then in this case the relation is valid, which is sometimes called the law of momentum change:

For a closed system (in the absence of external forces), the law of conservation of momentum is valid:

The action of the law of conservation of momentum can explain the phenomenon of recoil when shooting from a rifle or during artillery shooting. Also, the operation of the law of conservation of momentum underlies the principle of operation of all jet engines.

When solving physical problems, the law of conservation of momentum is used when knowledge of all the details of motion is not required, but the result of the interaction of bodies is important. Such problems, for example, are the problems of impact or collision of bodies. The law of conservation of momentum is used when considering the motion of bodies of variable mass, such as launch vehicles. Most of the mass of such a rocket is fuel. In the active phase of the flight, this fuel burns out, and the mass of the rocket rapidly decreases in this part of the trajectory. Also, the law of conservation of momentum is necessary in cases where the concept is inapplicable. It is difficult to imagine a situation where a motionless body acquires some speed instantly. In normal practice, bodies always accelerate and pick up speed gradually. However, during the movement of electrons and other subatomic particles, the change in their state occurs abruptly without staying in intermediate states. In such cases, the classical concept of "acceleration" cannot be applied.

Examples of problem solving

EXAMPLE 1

Exercise	A projectile with a mass of 100 kg, flying horizontally along a railway track at a speed of 500 m/s, hits a wagon with sand of mass 10 tons and gets stuck in it. What speed will the car get if it moves at a speed of 36 km/h in the direction opposite to the projectile?
Solution	The wagon+projectile system is closed, so in this case the momentum conservation law can be applied. Let's make a drawing, indicating the state of the bodies before and after the interaction. When the projectile and the car interact, an inelastic impact occurs. The law of conservation of momentum in this case will be written as: Choosing the direction of the axis coinciding with the direction of movement of the car, we write the projection of this equation onto the coordinate axis: where is the speed of the car after a projectile hits it: We convert units to the SI system: t kg. Let's calculate:
Answer	After hitting the projectile, the car will move at a speed of 5 m/s.

EXAMPLE 2

Exercise	A projectile with mass m=10 kg had a speed v=200 m/s at the top point . At this point, it broke into two pieces. A smaller part with a mass m 1 =3 kg received a speed v 1 =400 m/s in the same direction at an angle to the horizon. With what speed and in what direction will most of the projectile fly?
Solution	The trajectory of the projectile is a parabola. The speed of the body is always directed tangentially to the trajectory. At the top of the trajectory, the velocity of the projectile is parallel to the axis. Let's write the momentum conservation law: Let's pass from vectors to scalars. To do this, we square both parts of the vector equality and use the formulas for: Given that and also that , we find the speed of the second fragment: Substituting the numerical values ​​of physical quantities into the resulting formula, we calculate: The direction of flight of most of the projectile is determined using: Substituting numerical values ​​into the formula, we get:
Answer	Most of the projectile will fly at a speed of 249 m / s down at an angle to the horizontal direction.

EXAMPLE 3

Exercise

The mass of the train is 3000 tons. The coefficient of friction is 0.02. What should be the size of the steam locomotive for the train to pick up a speed of 60 km / h 2 minutes after the start of movement.

Solution

Since an (external force) acts on the train, the system cannot be considered closed, and the law of conservation of momentum does not hold in this case. Let's use the law of momentum change: Since the friction force is always directed in the direction opposite to the movement of the body, in the projection of the equation on the coordinate axis (the direction of the axis coincides with the direction of the train movement), the friction force impulse will enter with a minus sign:

1. If the main vector of all external forces of the system is zero (), then the amount of motion of the system is constant in magnitude and direction.

2. If the projection of the main vector of all external forces of the system on any axis is equal to zero (
), then the projection of the momentum of the system onto this axis is a constant value.

The theorem on the motion of the center of mass.

Theorem The center of mass of the system moves in the same way as a material point, the mass of which is equal to the mass of the entire system, if all external forces applied to the considered mechanical system act on the point.

, hence

The moment of momentum of the system.

moment of momentum systems of material points about some center is called the vector sum of the moments of momentum of individual points of this system relative to the same center

moment of momentum systems of material points
about some axis
passing through the center , is called the projection of the momentum vector
on this axle
.

The moment of momentum of a rigid body relative to the axis of rotation during rotational motion of a rigid body.

Let us calculate the angular momentum of a rigid body relative to the axis of rotation.

The angular momentum of a rigid body about the axis of rotation during rotational motion is equal to the product of the angular velocity of the body and its moment of inertia about the axis of rotation.

Theorem on the change in the angular momentum of the system.

Theorem. The time derivative of the angular momentum of the system, taken relative to some center, is equal to the vector sum of the moments of external forces acting on the system relative to the same center.

(6.3)

Proof: The theorem on the change in the angular momentum for
points looks like:

,

Let's put it all together equations and get:

or
,

Q.E.D.

Theorem. The time derivative of the moment of momentum of the system, taken with respect to any axis, is equal to the vector sum of the moments of external forces acting on the system with respect to the same axis.

To prove it, it suffices to project the vector equation (6.3) onto this axis. For axis
it will look like this:

(6.4)

Theorem on the change in the angular momentum of the system relative to the center of mass. (no proof)

For axes moving forward along with the center of mass of the system, the theorem on the change in the angular momentum of the system relative to the center of mass retains the same form as with respect to a fixed center.

Module 2. Strength of materials.

Topic 1 tension-compression, torsion, bending.

Deformations of the considered body (structural elements) arise from the application of an external force. In this case, the distances between the particles of the body change, which in turn leads to a change in the forces of mutual attraction between them. Hence, as a consequence, there are internal efforts. In this case, the internal forces are determined by the universal section method (or the cut method).

It is known that there are external forces and internal forces. External forces (loads) are a quantitative measure of the interaction of two different bodies. These include reactions in bonds. Internal forces are a quantitative measure of the interaction of two parts of the same body located on opposite sides of the section and caused by the action of external forces. Internal forces arise directly in the deformable body.

Figure 1 shows the calculation scheme of a bar with an arbitrary combination of external load forming an equilibrium system of forces:

From top to bottom: elastic body, left cutoff, right cutoff Fig.1. Section method.

In this case, the reactions of bonds are determined from the known equations of equilibrium of the statics of a solid body:

where x 0 , y 0 , z 0 is the base coordinate system of the axes.

Mental cutting of a beam into two parts by an arbitrary section A (Fig. 1 a) leads to the equilibrium conditions for each of the two cut-off parts (Fig. 1 b, c). Here ( S') And ( S"} - internal forces arising, respectively, in the left and right cut-off parts due to the action of external forces.

When compiling mentally cut off parts, the equilibrium condition of the body is provided by the ratio:

Since the initial system of external forces (1) is equivalent to zero, we obtain:

{S ’ } = – {S ” } (3)

This condition corresponds to the fourth axiom of statics about the equality of forces of action and reaction.

Using the general methodology of the theorem poinsot about bringing an arbitrary system of forces to a given center and choosing the center of mass as the pole of reduction, sections A " , point WITH " , system of internal forces for the left side ( S ’ ) is reduced to the main vector and the main moment of internal efforts. Similarly, it is done for the right cut-off part, where the position of the center of mass of the section A"; is determined, respectively, by the point WITH"(Fig. 1 b, c).

Thus, the main vector and the main moment of the system of internal forces arising in the left conditionally cut-off part of the beam are equal in magnitude and opposite in direction to the main vector and the main moment of the system of internal forces arising in the right conditionally cut-off part.

The graph (epure) of the distribution of the numerical values ​​of the main vector and the main moment along the longitudinal axis of the beam and predetermine, first of all, specific issues of strength, rigidity and reliability of structures.

Let us determine the mechanism of formation of the components of internal forces that characterize simple types of resistance: tension-compression, shear, torsion and bending.

At the centers of mass of the studied sections WITH" or WITH"Let's set ourselves accordingly to the left (c", x", y", z") or right (c", x", y", z”) systems of coordinate axes (Fig. 1 b, c), which, in contrast to the base coordinate system x, y, z we will call "followers". The term is due to their functional purpose. Namely: tracking the change in the position of section A (Fig. 1 a) with its conditional displacement along the longitudinal axis of the beam, for example, when: 0 x' 1 a, ax' 2 b etc., where A And b- linear dimensions of the boundaries of the studied sections of the beam.

Let's set the positive directions of the projections of the main vector or and the main moment or onto the coordinate axes of the servo system (Fig. 1 b, c):

(N ’ , Q ’ y , Q ’ z ) (M ’ x , M ’ y , M ’ z )
(N ” , Q ” y , Q ” z ) (M ” x , M ” y , M ” z )

In this case, the positive directions of the projections of the main vector and the main moment of internal forces on the axis of the servo coordinate system correspond to the rules of statics in theoretical mechanics: for force - along the positive direction of the axis, for moment - counterclockwise rotation when viewed from the end of the axis. They are classified as follows:

N x- normal force, a sign of central tension or compression;

M x - internal torque, occurs during torsion;

Q z , Q at- transverse or shear forces - a sign of shear deformations,

M at , M z- internal bending moments, correspond to bending.

The connection of the left and right mentally cut off parts of the beam leads to the well-known (3) principle of equality in absolute value and the opposite direction of all the components of internal forces of the same name, and the equilibrium condition of the beam is defined as:

As a natural consequence of relations 3,4,5, the obtained condition is necessary for the like components of internal forces to form subsystems of forces equivalent to zero in pairs:

1. {N ’ , N ” } ~ 0 > N ’ = – N ”
2. {Q ’ y , Q ” y } ~ 0 > Q ’ y = – Q ” y
3. {Q ’ z , Q ” z } ~ 0 > Q ’ z = – Q ” z
4. {M ’ x , M ” x } ~ 0 > M ’ x = – M ” x
5. {M ’ y , M ” y } ~ 0 > M ’ y = – M ” y
6. {M ’ z , M ” z } ~ 0 > M ’ z = – M ” z

The total number of internal forces (six) in statically determinable problems coincides with the number of equilibrium equations for a spatial system of forces and is related to the number of possible mutual displacements of one conditionally cut off body part relative to another.

The desired forces are determined from the corresponding equations for any of the cut-off parts in the servo system of coordinate axes. So, for any cut-off part, the corresponding equilibrium equations take the form;

1. ix = N + P 1x + P 2x + … + P kx = 0 > N
2. iy = Q y + P 1y + P 2y + … + P ky = 0 > Q y
3. iz = Q + P 1z + P 2z + … + P kz = 0 > Q z
4. x (P i) = M x + M x (P i) + … + M x (P k) = 0 > M x
5. y (P i) = M y + M y (P i) + … + M y (P k) = 0 > M y
6. z (P i) = M z + M z (P i) + … + M z (P k) = 0 > M z

Here, for simplicity of designation of the coordinate system c"x"y"z" And s"x"y"t" replaced by a single oxuz.