Dynamic analysis is a branch of the theory of mechanisms and machines, which studies the movement of the links of a mechanism under the action of a given system of forces. The main goal of dynamic analysis is to establish general relationships between the forces (moments of forces) acting on the links of the mechanism and the kinematic parameters of the mechanism, taking into account the masses (moments of inertia) of its links. These dependencies are determined from the equations of motion of the mechanism.

With all the variety of problems of dynamic analysis, they are divided into two main types: in problems of the first type, it is determined under the action of what forces a given movement of the mechanism occurs (the first problem of dynamics); in problems of the second type, according to a given system of forces acting on the links of the mechanism, their kinematic parameters are found (the second problem of dynamics).

The law of motion of the mechanism in analytical form is set in the form of the dependences of its generalized coordinates on time. The simplest problems of dynamics are solved for mechanisms with rigid links and one degree of freedom using the classical methods of the theory of mechanisms and machines. However, modern technical practice requires the solution of more complex problems in which the dynamics of high-speed machines and mechanisms is investigated taking into account the elastic properties of the materials of their links, the presence of gaps in their kinematic chains and other factors. In such cases, the problems of dynamics of mechanical systems with several degrees of freedom (or with an infinite number of degrees of freedom) are solved using the complex mathematical apparatus of multidimensional systems of ordinary differential equations, partial differential equations or integro-differential equations.

Forces acting on the links of the mechanism and their classification

The forces acting on the links of the mechanism can be divided into the following groups.

driving forcesF d (or a pair of forces with a moment M d ) – these are forces whose elementary work on possible displacements of the points of their application is positive The driving forces are applied to the drive links from the side of the motors. They are designed to set machines in motion, overcome resistance forces and implement a given technological process. Internal combustion engines, electric, hydraulic, pneumatic, etc. are used as drive motors.

Resistance forcesF c (or a pair of resistance forces with a moment M with ) – these are forces whose elementary work on possible displacements of the points of their application is negative. Resistance forces impede the movement of the mechanism. They are divided into useful resistance forces. (F ps, Mps), to overcome which this mechanism is intended, and the forces of harmful resistance (F BC, Мвс), causing unproductive energy consumption of driving forces.

The forces of useful resistances are due to technological processes, therefore they are called through technological or production resistance... Usually they are attached to the output links of the executive machines. The forces of harmful resistance are mainly the forces of friction in kinematic pairs and the forces of resistance of the medium. The concept of "harmful forces" is conditional, since in some cases they ensure the operability of the mechanism (for example, the movement of the roller ensures its adhesion to the road surface).

Link weight forcesF g, depending on their direction of action relative to the direction of the driving forces, can be beneficial or harmful when they respectively facilitate or impede the movement of the mechanism.

Forces of inertiaF and or moments of inertia M and, arising from a change in the speed of movement of the links, can be both driving forces and forces of resistance, depending on the direction of their action relative to the direction of movement of the links.

In the general case, the driving forces and resistance forces are functions of kinematic parameters (time, coordinates, velocity, acceleration of the point of application of the force). These functions for specific motors and working machines are called them mechanical characteristics, which are set analytically or graphically.

In fig. 1.20 shows the mechanical characteristics M d = = Мд (ω) electric motors of various types.

direct current with parallel excitation(the excitation winding of the motor is connected parallel to the armature winding) has the form of a linear monotonically decreasing dependence of the moment Md on the angular speed of rotation of the shaft ω (Fig. 1.20, a). An engine with such a mechanical characteristic operates stably over the entire range of angular velocities from.

Mechanical characteristic of the electric motor direct current with series excitation(the field winding is connected in series with the armature winding) is represented by a nonlinear dependence M d = Md (ω) shown in Fig. 1.20, b.

Mechanical characteristic asynchronous dc motor(fig. 1.20, v) is described by a more complex dependency. The characteristic has ascending and descending parts. The area of sustainable operation of electro-

Rice. 1.20

of the engine is the descending part of the characteristic. If the moment of resistance M s becomes greater than the maximum moment of driving forces M d, the engine stops. Such a moment M with called overturning moment M def. The angular speed ω = = ωnom, at which the engine develops maximum power, is called the nominal angular speed, and the corresponding moment M d = M nom - nominal torque... Angular velocity ω = ωs. with which M q = 0 is called synchronous angular velocity.

The mechanical characteristics of working machines are often ascending curves (Fig. 1.21). This is the type of characteristics of compressors, centrifugal pumps, etc.

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Dynamic analysis of mechanisms

1. Problems of kinetostatics

The design of new mechanisms is usually accompanied by the calculation of their elements for strength, and the dimensions of the links are set in accordance with the forces that act on them.

If in the kinematics of mechanisms, in which only the geometry of motion was considered, the outline of the links was neglected, fixing only the characteristic dimensions, such as, for example, the distance between the centers of the hinges and other dimensions that determine the relative motion of the links, then when calculating the strength, it is necessary to have an idea of the link in three-dimensional space. The forces acting on the elements of the kinematic pairs, which appear as a result of technological and mechanical resistance, determine the stresses in the links, if the dimensions of the latter are chosen, or determine the dimensions of the links, if the stresses of the material of the links are specified.

Thus, the calculation of mechanisms for strength should be preceded by the determination of forces, therefore one of the main tasks of kinetostatics is to determine the forces that act on the elements of kinematic pairs and cause deformations of the links during operation.

The methods for calculating the forces acting on the links of the mechanism without taking into account the forces of inertia are united under the name of the statics of mechanisms, and the methods for calculating the forces taking into account the forces of inertia of the links, determined approximately, are the kinetostatics of the mechanisms. In practice, the methods of static and kinetostatic calculations of mechanisms are no different if we consider the inertial forces as given by external forces.

Kinetostatics combines methods for calculating the forces acting on the links of the mechanism, taking into account the forces of inertia.

2. Forces acting on the mechanism

2.1 Classification of forces

During the operation of the machine, specified external forces are applied to its links, which include: driving force, force of technological resistance, gravity of links, mechanical or additional resistance and inertial forces resulting from the movement of the link. The unknown forces will be the reactions of constraints acting on the elements of the kinematic pairs.

The forces acting on the links are conventionally divided into 2 groups: driving forces P dv and resistance forces P C.

Driving forces are called forces that produce positive work, i.e. the directions of the driving force and the velocity of the point of its application either coincide or form an acute angle.

However, in some cases, the force applied to the driving link can turn into a force of resistance and, therefore, will produce negative work. As an example, we can point to heat engines in which the force acting on the piston produces negative work when the gas mixture is compressed.

In an internal combustion engine, for example, the driving force will be the resultant from the pressure forces when the combustible mixture is ignited.

Resistance forces are called forces that impede the movement of the links of the mechanism. The work of these forces is always negative, i.e. the direction of the force and velocity of the point of its application are either opposite or form an obtuse angle. Distinguish between the forces of useful resistance and harmful resistance. In working machines, the force of useful resistance is, for example, the resistance to metal cutting, the resistance to compression of gases. The forces of harmful resistance are the forces of friction, the forces of resistance of the medium.

In addition to these forces, it is necessary to take into account the gravity forces (weight forces) of the G links, which are applied in their centers of gravity, the inertial forces of the links and the forces of the coupling reactions.

The forces of inertia P u appear when the link moves unevenly. The forces of inertia, just like the forces of weight, can do both positive and negative work.

The forces of the reaction of the connection R, acting in kinematic pairs, are introduced when considering any link isolated from the mechanism. When considering the entire mechanism as a whole, bond reactions should be considered internal forces, i.e. balanced in pairs.

Mechanical or additional resistance F in machines occurs mainly in the form of resistance forces that appear during the relative movement of the elements of kinematic pairs, or, in other words, friction forces, in the form of resistance of the medium, for example, aerodynamic resistance, resistance force due to the stiffness of flexible links, for example, ropes, chains, belts, etc. Friction forces appear under the action of normal reactions acting in kinematic pairs, and are known forces. Friction forces, as a rule, produce negative work, because they are always directed in the direction opposite to the speed of the relative motion of the elements of the kinematic pairs. This type of additional resistance that accompanies the operation of machines is most important, because in many cases, almost all the energy spent on driving a machine is spent on overcoming frictional forces. In view of this, the friction forces will be considered separately.

2.2 External forces and mechanical characteristics of machines

External forces can be constant, such as gravity, resistance to metal cutting at a constant chip section, etc., or depending only on the position of the link on which they act (the forces of pressure of gases acting on the piston of an internal combustion engine or compressor, resistance, encountered by the punch of the press when piercing holes, etc.), from the speed of the link (the moment of the electric motor, the frictional force of lubricated bodies, etc.), from time to time. In addition, a machine may be subject to forces that depend on a number of the above independent variables. Determination of a specific magnitude of the external force is possible only if its characteristic is set.

So for the main mechanism of a four-stroke internal combustion engine, the law of change in the gas pressure P in the cylinder is set by the indicator diagram - the dependence P = ѓ (H) (Fig. 1)

A full engine cycle ends within two crank turns. During the first half of the revolution, the combustible mixture FO is sucked in, during the second half of the revolution this mixture is compressed OD, along the curve DA - ignition of the mixture, along the curve AB - expansion of the ignited mixture (working stroke) along the curve BF - exhaust.

Plotting along the H axis the displacement x taken from the plan of the mechanism, it is easy to find the corresponding ordinate on the indicator diagram.

The excess pressure P from the piston is the difference between the gas pressure in the cylinder and the atmospheric pressure, proportional to the ordinate, measured from the atmospheric pressure line.

The force acting on the piston is determined from the formula:

where d is the piston diameter.

For a single-acting compressor, the law of change in the gas pressure in the cylinder is also given by the indicator diagram (Fig. 2).

kinetostatics gear machine sliding

FCD curve - gas compression,

DA - exhaust,

AB - expansion of the gas remaining in the dead volume,

BF - suction of a new portion of gas

Force scale factor

where is the ordinate corresponding to the variable x.

The diagram of the change in the power on the motor shaft or the average torque depending on the number of revolutions is called the mechanical characteristic of the motor (Fig. 3).

2.3 Determination of inertial forces

During the operation of the mechanism, inertial forces arise. They cause additional pressure in the kinematic pairs. These forces are especially large in high-speed machines.

The forces of inertia are determined by the given weight of the links and their accelerations. The method of determination depends on the type of movement of the link.

The first case: the link makes a plane-parallel movement (connecting rod). It is known that the elementary inertial forces in this case are reduced to the resultant force P u and to the moment of inertia forces M u.

The force of inertia P u is applied at the center of gravity of the link and is equal to:

where m is the mass of the link

a s - linear acceleration of the center of gravity of the link.

Moment of inertia:

where J s is the moment of inertia of the link relative to the center of gravity,

Angular acceleration of the link.

The minus sign indicates that the inertial force P u is directed in the direction opposite to the acceleration a s, and the moment M u is directed in the direction opposite to the angular acceleration.

The magnitude and direction of accelerations are determined from a kinematic calculation. And the value m, J s must be specified.

The force P u and the moment M u can be replaced by one resultant force P u applied at the swing point (Fig. 4).

For this, the force of inertia P u must be transferred to a distance equal to

The value of this shoulder is found in the following way: a triangle is transferred from the acceleration plan (Figure 3.3) to the AB link

the segment having found the point "K" (swing point), we apply the vector of inertia force in it, directed in the direction opposite to the acceleration vector of the center of gravity.

The second case: the link makes a rotational movement (Fig. 5)

a) With non-uniform rotation and with a mismatch between the center of gravity and the axis of rotation, there is an inertial force Pu and a moment of inertia forces. When reducing the force and moment, the shoulder SK is determined by the formula (3.4):

where SK is the distance from the center of gravity to the swing point.

b) With uniform movement, P and is placed in the center of gravity.

M and = 0 because = 0.

c) The center of gravity coincides with the axis of rotation = 0, then P and = 0; M and = 0.

The third case: the link makes a translational movement (slider) (Fig. 6).

Here, M and = 0. If the movement of the link is uneven, then there is an inertial force

If the moment of inertia of the link is not specified in the assignment for course design, it can be approximately determined by the formula:

where m is the mass of the link,

l - link length,

K - coefficient 810

One of the tasks of the dynamics of mechanisms is to determine the forces acting on the elements of kinematic pairs, and the so-called balancing forces. Knowledge of these forces is necessary for calculating mechanisms for strength, determining engine power, wear of rubbing surfaces, establishing the type of bearings and their lubrication, etc. power calculation of the mechanism is one of the essential stages of machine design.

By balancing forces, it is customary to understand forces that balance the given external forces and the inertial forces of the links of the mechanism, determined from the condition of uniform rotation of the crank. The number of balancing forces that must be applied to the mechanism is equal to the number of initial links or, in other words, to the number of degrees of freedom of the mechanism. So, for example, if a mechanism has two degrees of freedom, then two balancing forces must be applied in the mechanism.

3. Force analysis of mechanisms. Determination of reactions in kinematic pairs

Force analysis of mechanisms is based on solving a straight line, or the first, problem of dynamics - for a given movement to determine the acting forces. Therefore, the laws of motion of the initial links in the force analysis are assumed to be given. External forces applied to the links of the mechanism are usually also considered given and, therefore, are subject to determination only of the reaction in kinematic pairs. But sometimes the external forces applied to the initial links are considered unknown. Then the force analysis includes the determination of the forces at which the adopted laws of motion of the initial links are fulfilled. When solving both problems, Alambert's principle D is used, according to which a link of a mechanism can be considered as being in equilibrium if inertial forces are added to all external forces acting on it. Equilibrium equations in this case are called kinetostatic equations in order to distinguish them from ordinary equations statics, ie equations of equilibrium without taking into account the forces of inertia.Usually, the links of plane mechanisms have a plane of symmetry parallel to the plane of motion.Then the main vector of the inertia forces of the link P u and the main moment of the inertia forces of the link are determined by the formulas:

where m is the mass of the link;

Center of mass acceleration vector.

In the kinetostatic calculation of the mechanism, it is necessary to determine the reactions in the kinematic pairs and either the balancing force or the balancing moment of the pair of forces.

The force calculation of the mechanisms will be carried out on the assumption that there is no friction in the kinematic pairs, and all the forces acting on the mechanism are located in the same plane.

One of the well-known methods of force calculation is the method of considering each link of the mechanism in equilibrium. With this method, the mechanism is divided into separate links.

First, the equilibrium of the extreme link is considered, counting from the main (leading), then the equilibrium of the link connected to the extreme, etc. The balance of the main link is considered last.

Considering a separately taken link in equilibrium, it is necessary to apply to it all external forces (P DV, R PS, R I, G), including the reactions of bonds with which the disconnected links act on the taken link.

Let us describe the calculation method using the example of a four-link mechanism. First, consider link 3 (rocker) in equilibrium, applying all the acting forces to it, including the reactions of links. (Fig. 7)

The reaction in the rotational pair "C" is unknown neither in magnitude nor in direction.

To determine this reaction, we replace it with two components (Fig. 7b), one of which is directed along the connecting rod (2), the second component - along the rocker arm (3).

The value can be found from the equilibrium condition of the link under consideration.

Link (3) is in equilibrium under the influence of the following forces R PS; P from; G 3; R 03; ; ...

We compose the equation of the moments of all forces relative to point D

If after determining this value it turns out to be negative, then its direction will be opposite to the chosen one. The component can be found by considering in equilibrium a separately taken link (2) (Fig.8a).

From the equilibrium condition of the link (2), we can write

The remaining unknown reaction R12 can be found graphically by building a plan of the forces of this link (Fig. 3.8b).

The equilibrium equation for the link (2) is as follows:

From an arbitrarily chosen pole, we lay off the force in the form of a vector on a scale, to it we geometrically add a vector representing the force G on the same scale, etc.

The vector gives us the magnitude of the reaction R 12 to scale.

For this, we consider the crank AB in equilibrium. (fig. 9).

The crank is under the influence of the force of the weight G 1, the reaction of the connecting rod (2) to the crank R 21, the force of inertia P u 1.

Under the influence of these forces, the cranks will generally not be in equilibrium. For balance, it is necessary to apply a balancing force P y, or a balancing moment M y.

These balancing forces and moments are the reactive forces or torque from the engine.

Let the balancing force be directed along the normal to the crank and apply at point B. From the equilibrium condition of link AB, one can compose an equation for the sum of the moments of all forces relative to point A.

The balancing force can also be found by a method in which the whole mechanism is considered in equilibrium.

The equilibrium condition of the mechanism can be expressed by the following equation:

The sum of the powers of all forces applied to the mechanism, taking into account the forces of inertia and balancing forces, is equal to zero.

The instantaneous power of the force applied at the i-th point is proportional to the moment of this force relative to the end of the vector of the rotated velocity of the given point (Fig. 10).

The balancing force can be found from the equilibrium equation. It is often convenient to find Py with the help of Zhukovsky's auxiliary lever when a polar velocity plan is constructed for the mechanism, rotated by 90 °. In the latter case, acting external forces should be applied to the ends of the found velocity vectors.

After that, considering the rotated plan of velocities as a rigid lever rotating around the pole P, one can write the equilibrium equation of the lever in the form of the sum of the moments of forces relative to the pole:

The equilibrium equation of the plan of velocities, considered as a rigid lever, is identical to the equation of powers.

If, in addition to forces, moment M is also applied to the links of the mechanism (Fig. 11), then it can be considered as a pair of forces, the component of which is equal to:

The found forces P are applied at the corresponding depicting points of the velocity plan.

4. Friction in kinematic pairs

4.1 Sliding friction

Friction losses in a mechanism mean friction losses in its kinematic pairs. There are two main types of friction: sliding friction and rolling friction. In the lower kinematic pairs, sliding friction occurs, in the higher - only rolling friction or rolling friction together with sliding friction.

If the surfaces of moving bodies A and B (Fig. 12) are in contact, then the friction arising in this case is called dry. If the surfaces are not in contact (Fig. 13) and there is a layer of lubricant between them, then such friction is called liquid friction. There are also cases when there is semi-dry (dry prevails), or semi-liquid, friction.

4.2 Dry friction

Basic laws:

1. In a certain range of speeds and loads, the sliding friction coefficient can be considered constant, and the friction force - F proportional to the normal pressure:

where f is the coefficient of sliding friction,

N is normal pressure.

2. The coefficient of sliding friction depends on the material and the state of the rubbing surfaces.

3. The forces of friction are always directed in the direction opposite to the relative velocities.

4. The coefficient of friction at rest is slightly greater than the coefficient of friction during motion.

5. With an increase in the speed of movement, the friction force in most cases decreases, approaching a certain constant value; at low speeds, the friction coefficient is almost independent of speed.

6. With increasing specific pressure, the coefficient of friction in most cases increases. At low specific pressures, the friction coefficient is almost independent of the specific pressure and the contact area.

7. With an increase in the time of preliminary contact, the friction force increases.

4.3 Fluid friction

With dry friction, there is a large expenditure of work, which turns into heat, and wear of the rubbing surfaces. To eliminate these phenomena, a layer of lubricant is introduced between the rubbing surfaces. In this case, subject to certain conditions, the lubricant layer can completely separate the rubbing surfaces (Figure 3.13).

4.4 Sliding friction on the horizontal plane

A translational kinematic pair, consisting of a horizontal guide 2 and a slider 1, is shown in Figure 14. Let the following forces act on the slider 1: P D is the driving force, G is the weight of the load or the load acting on the slider, N is the normal reaction, F 0 - friction force (tangential reaction) at rest. With a moving slider, instead of the friction force F 0, the friction force F acts during movement, and, moreover, the total reaction.

The angle of deviation of the full reaction from the normal in the direction opposite to the movement of the slide is called the angle of friction.

Considering that

Consequently, the coefficient of friction is equal to the tangent of the angle of friction.

4.5 Friction in the kinematic pair of stud - bearing

In the presence of a gap, the trunnion, under the action of M D, rolls from its lowest position to a new position, which is characterized by the onset of equilibrium between the driving forces and the resistance forces. In fig. 15, the following designations are adopted: - radius of the stud, Q - external load, R - bearing reaction acting on the stud, - friction angle, - radius of the friction circle.

The forces Q and R form a pair of forces, the moment of which is the moment of resistance; at any given moment, it balances the moment of the driving forces, i.e. ...

Moment of forces of resistance

Frictional moment,

where; - radius of the thorn;

Due to the smallness of the angle, the value. Consequently, the radius of the friction circle is equal to the displacement of the total reaction R from the external load Q.

So, the moment of friction forces

5. Coefficient of efficiency of the mechanism

Mechanical efficiency machines call the ratio of the absolute value of the work of useful resistances A PS. to the work of the driving forces A D for the period of steady motion:

From the equation of motion of the machine at steady motion we find.

After substitution in expression (1), we obtain the following expression for the efficiency:

where is the loss factor.

The efficiency is the greater, the less the work of harmful resistances. Having determined, for example, the instantaneous efficiency in twelve positions of the lever mechanism for one revolution of the steady motion, it is possible to construct a graph of the function. In practice, they usually use the arithmetic mean of the efficiency over the period of steady motion:

The machine can have very low instantaneous efficiency in certain positions of the mechanism. The instantaneous efficiency of the linkage can be expressed as the power ratio:

where N P.S. - instantaneous power of useful resistance forces for each position of the mechanism;

N D - instantaneous power of driving forces for the corresponding position of the mechanism.

To. P. D. Group of series-connected mechanisms or machines. A number of machines or mechanisms included in the unit can be connected in series (Fig. 16 a), in parallel (Fig. 16 b)

The total efficiency of a machine with a series connection of mechanisms is equal to the product of their efficiency.

In general

To. P. D. Group of parallel connected mechanisms or machines. This connection is characterized by the branching of the total energy flow.

The total efficiency is equal to:

Figure 16

6. Determination of reactions in kinematic pairs taking into account friction

The calculation performed in the first part without taking into account friction gives the values of the reactions in the kinematic pairs of the mechanism in the first approximation. The determination of forces with allowance for friction is a further refinement and is usually (and in our case) carried out by the method of successive approximation. To perform the second approximation, the values of the sliding friction coefficients in all pairs and the diameters of the pins of the rotary pairs are specified. The methodology for calculating the mechanism with and without friction is the same. The only difference is that the reaction forces in the translational pairs deviate from their previous normals by the friction angle and are directed against the velocity vector of the translational pair. In rotational - the lines of their action will pass tangentially to the circles of friction, these reactions can be replaced by the reaction applied in the center of the hinge, while it is necessary to apply to this hinge a moment of friction determined by the formula:

where r is the friction radius determined by the formula:

where D y is the diameter of the pins,

Angle of friction.

R in formula (3.13) is the reaction in the given hinge obtained in the first part, disregarding the friction forces. The direction of the moment is opposite to the angular velocity of the link relative to the given joint.

6.1 Force analysis of gear mechanisms

For the overwhelming majority of gear drives, steady-state operation is essential. Therefore, in transmissions of this type, the moments from inertia forces will be equal to zero (excluding oscillations caused by variable stiffness and step errors).

The pressure between the involute profiles is transmitted along the line of engagement, which coincides with their common normal.

If a resistance moment M C is applied to the driven wheel, then the resistance force is:

Force P C is applied to the drive wheel 1; driving force is applied to the driven wheel 2. It follows from the formula that, if, then the force P C of pressure between the teeth is constant both in magnitude and in direction; it increases with an increase in the angle of engagement.

In the center of the driving wheel 1, we apply two equal and oppositely directed forces P C. Forces R * - pressure in the wheel bearings; two other forces R form a pair of forces, the moment of which is equal to the moment M D. Substituting the value of P C from the formula, we obtain

The pair applied to the wheel 2 overcomes the resistance moment M C applied to this wheel.

Equal and oppositely directed forces R * and Q * form a pair with the moment

This pair tends to rotate the transmission rack (frame) (in our case, clockwise). To prevent this from happening, the rack must be secured. The moment created by the pair in question is called the reactive moment.

Obviously, with a variable M C, the directions of the pressure forces between the teeth and in the shaft bearings will be constant. This is one of the advantages of involute engagement, as it ensures smooth transmission operation.

Since the profiles of the teeth in the process of their engagement have relative sliding, then friction forces arise between them, the resultant F of which is directed against the sliding speed

The magnitude of this force

where f is the coefficient of sliding friction of the profiles.

External frictional force

Consequently, the power of the friction forces in the engagement is variable and increases as the point M of contact of the profiles moves away from the pole of the engagement.

Friction forces are also generated in the shaft bearings, which are proportional to the pressures R and Q in these bearings. The magnitudes of these friction forces depend on a number of factors (on the lubrication conditions of the contacting surfaces, on their elastic properties, which determine the distribution law of specific pressures, on the sliding speed of the bearing surfaces, etc.). The resultant of these forces, where f n 1 is the coefficient of friction, taking into account the operating conditions of the shaft in the bearings. This force is applied at one of the points of the bearing surface of the shaft at a distance r B from its axis.

The power of the friction forces in the supports

It can be seen from the formulas that if, then the power of the friction forces in the supports is constant.

Using this formula, you can determine the moment M D and power N D of the engine, which must be connected to the transmission drive shaft, if M C and i 12 are specified

The values of the coefficients f and f n depend on a large number of different factors and can vary over a very wide range. For example, the coefficients of friction of profiles depend not only on the materials and the accuracy of their processing, but also on the lubricant; in addition to sliding friction, rolling friction occurs between the profiles; if the transmission works in an oil bath, then work is spent on mixing the oil, etc.

6.2 Determination of moments in the planetary mechanism without regard to friction

Consider the question of determining the moments in the planetary mechanism, the links of which rotate uniformly. In the planetary mechanism shown in (Fig. 18), the sun wheel 1, the carrier 2 and the crown wheel 4 rotate around the central axis C. The tangential component P 31 of the reaction to the satellite 3 from the side of the sun wheel 1, disregarding the friction force, is applied at the gearing pole A. B the opposite side is directed by the force P 13. At point B, the reaction components P 34 and P 43 act, and in the center of the satellite, P 23 and P 32.

We will consider such planetary mechanisms in which the satellite is not an output link, i.e. M 3 = 0. Then and because:

where k is the number of satellites of the mechanism.

From the equilibrium of link 2 we have:

Taking into account (3.15) and (3.16), we rewrite (3.17):

Let us write down the equilibrium condition for link 4:

Therefore, taking into account the condition: Р 43 = -Р 13 from (3.19) we have:

Therefore, if one of the moments acting in the planetary mechanism is known, then knowing the radii of the initial circles, using formulas (3.18) and (3.19), we can determine the unknown moments.

The problem of determining the moments can be solved using the general plan of angular velocities. Consider the method for determining the moments.

Let a general plan of angular velocities be built for a planetary gearbox with corrected gears (Fig. 19)

Power supplied to link 1.

The power taken from the carrier.

Since losses are not taken into account, then:

Since under the action of the moments, the planetary mechanism in the steady-state equilibrium mode is in equilibrium, then the equality takes place

where M 4, when should be understood as the moment that must be applied to the link 4 in order to keep it from rotating.

From (3.21) we get:

6.3 Determination of the efficiency of the planetary gear

Efficiency d. mechanical transmission depends on many factors, of which the loss of power in the meshing of pairs of gear wheels is of the greatest importance. Let us determine the efficiency of the planetary gearbox when transferring moments from link 1 to link 2 according to the formula:

where is called the power transmission ratio. Here and are the moments acting on links 2 and 1 taking into account the friction in the engagement - the kinematic gear ratio.

6.4 Force analysis of cam mechanisms

Since the driven link (rod-pusher) moves with variable speed, the schemes of action of the forces applied to the cam mechanism in different sections of the interval of its movement are different.

In the interval of the working movement, a useful resistance force R is applied to the driven link, directed against the speed of the link. The force R is usually always given; it can be constant or variable.

If a force closure of the higher pair is realized in the mechanism, then the elastic force P P of the spring acts on the driven link in the same direction, which is compressed at this time.

Due to the uneven movement of the bar, an inertial force occurs:

where is the weight of the bar, is its acceleration; the force of Ra is directed opposite to the acceleration of the bar. Since the mass of the bar is constant, the law (graph) of the force change coincides with the law (graph) of the change in the acceleration of the bar.

The resultant Q of all forces applied to the bar is equal to:

If we neglect the friction in the cam - rod pair, then the direction of the force P of the cam pressure on the rod coincides with the normal to the cam profile. If we do not take into account the friction in the guide C, then, in order for the rod to move according to a given law, it is necessary that in each position of the mechanism the force P of the cam pressure on the rod would be equal to

where - the angle between the force and the direction of movement of the bar - the angle of transmission of movement.

If you do not take into account the friction in the bearings of the cam shaft, then the driving moment on the cam shaft

where is the radius vector of the cam profile.

Self-braking. Taking into account the frictional forces in the force calculation of the mechanism, it is possible to reveal such relationships between the parameters of the mechanism, at which, due to friction, the movement of the link in the required direction cannot begin regardless of the magnitude of the driving force.

In most mechanisms, self-braking is unacceptable, but in some cases it is used to prevent spontaneous movement in the opposite direction (jack, some types of lifting mechanisms, etc.).

Pressure angle. The angle of pressure on the link from the link side is the angle between the direction of the pressure force (normal reaction) on the link from the link side and the speed of the point of application of this force. The angle of pressure on the link from the link side is indicated by. Often, however, only one pressure angle is considered. Then the indices are omitted in the notation.

4. Analysis of the movement of the mechanism under the action of forces

Dynamic pressures are additional forces that arise in kinematic pairs when the mechanism moves. These pressures cause vibrations of some parts of the mechanism, they are variable in magnitude and direction. The frame of this mechanism also experiences dynamic pressures that have a detrimental effect on its fastenings and thereby disrupt the connection between the frame and the foundation. Also, dynamic pressures increase the frictional forces at the bearing points of rotating shafts and increase bearing wear. Therefore, when designing mechanisms, they try to achieve full or partial damping of dynamic pressures (the problem of balancing the inertial forces of mechanisms).

The link of the mechanism will be considered balanced if its main vector and the main moment of inertia forces of material points are equal to zero. Each link of the mechanism can be unbalanced separately, but the mechanism as a whole can be balanced in whole or in part. The problem of balancing inertial forces in mechanisms can be divided into two tasks: 1) balancing the pressures in the kinematic pairs of the mechanism 2) balancing the pressures of the mechanism as a whole on the foundation.

Balancing the rotating links is essential. The slight imbalance of rapidly rotating rotors and electric motors causes high dynamic bearing pressures.

The problem of balancing rotating bodies consists in choosing their masses in such a way that the additional inertial pressures on the supports will be completely or partially canceled out.

Resulting centrifugal inertia force:

The resulting moment of all inertial forces of the body relative to the plane passing through the center of mass.

where m is the mass of the whole body,

Distance of the center of mass of the body from the axis of rotation;

Centrifugal moment of inertia about the axis of rotation and a plane perpendicular to the axis of rotation and passing through the center S of mass of the body.

When the body rotates, the angle between the vectors and keeps the same value all the time. If the resulting inertial force and the resulting moment of inertia forces are zero, then the body will be completely balanced, which means that the rotating body does not exert any dynamic pressure on the supports.

These conditions will be met only when the center of mass of the body lies on the axis of rotation, which will be one of its main axes of inertia. If equalities (4.1) and (4.2) are simultaneously satisfied, then the centrifugal moment of inertia will be equal to zero. If the condition (4.1) is satisfied, then the body is considered to be balanced statically, if the condition (4.2) is satisfied, then the body is considered to be dynamically balanced.

Static unbalance is measured by static torque.

G is the weight of the rotating body, n.

The dynamic imbalance of a rotating body is measured by the value

In practice, an unbalanced body is balanced with counterweights. Rotating bodies, for which the total length a is much less than their diameter, have insignificant centrifugal moments of inertia; therefore, it is sufficient to balance such bodies only statically.

Suppose body A is statically unbalanced. In the simplest case, the counterweight is placed on a line through the center of gravity S, on the other side of the axis of rotation at a distance from it. (fig. 21)

We find the mass of the counterweight from equation (4.1):

Instead of installing a counterweight, you can remove some of the mass. The amount of the removed mass is determined by the formula (4.5). Sometimes the plane of attachment of the counterweight cannot be chosen constructively in the plane of rotation in which the unbalanced masses are located. In this case, it is possible to install two counterweights in two planes perpendicular to the axis of rotation, usually called correction planes, but it is necessary to exclude the possibility of pressure on the supports not only from the resulting inertial force, but also from the moments of inertia forces. The masses and counterweights are determined in accordance with formulas (4.1) and (4.2) from the equations

Adding the masses of these counterweights, we get

Full balancing of the rotating body can also be achieved using two counterweights located in arbitrarily chosen planes 1 and 2 and at arbitrary distances from the axis of rotation.

Rotating bodies are usually performed so that they are balanced on their own. Most often, rotating bodies are made in the form of one or more cylinders having a common axis coinciding with the axis of rotation of the body. However, in many cases such a shape cannot be made and a rotating body without counterweights is unbalanced. To determine the size and position of the counterweights, it is necessary to select the balanced part of the body according to the drawing and determine for the remaining parts - knees, cams, etc. their centers of gravity, considering that the masses of these parts are concentrated in them.

Suppose that for any body all of its unbalanced masses are reduced to three unbalanced masses (Fig. 22). Using the method of bringing the vector to a given center, it is possible to balance any number of masses rotating in different planes with two counterweights. Let the centers of gravity of masses and are located in three planes perpendicular to the axis of rotation. The conditions for the absence of pressure on the bearings from the main vector and the main moment relative to the center of reference O 1 of centrifugal inertia forces are expressed by the equations:

We build polygons of vectors of forces and vectors of moments (Fig. 22 d, e). The balancing vector in the first case is the vector depicted in plane 2 by the vector (Fig. 22 c) and in the second - the vector (Fig. 22 e) depicting the rotated moment of a pair of vectors located in plane 1 and located in plane 2. Each of them is equal in size. Thus, the given masses and will be completely balanced by two masses located along in plane 1 and along the resultant in plane 2. It follows from the above that:

1.) any number of rotating masses located in the same plane of rotation is balanced by one counterweight located in the same plane, subject to the equilibrium condition

2.) any number of masses lying in different planes of rotation is balanced by two counterweights installed in two arbitrary planes perpendicular to the axis of rotation, subject to two equilibrium conditions:

To balance the flat mechanism on the foundation, it is necessary and sufficient to select the masses of the links of this mechanism so that the general center of mass of its moving links remains stationary:

and the centrifugal moments of inertia of the masses of the links relative to the x and z axes, y and z were constant:

If these conditions are met, the main vector of inertia forces and the main moments of inertia forces relative to the x and y axes will be balanced. The main moment of inertia forces about the z-axis, perpendicular to the plane of movement of the mechanism, is balanced by the moment of driving forces and resistance forces on the main shaft of the machine. In practice, when balancing the mechanisms, the specified conditions (4.9) and (4.10) are partially satisfied.

Suppose, for example, given the mechanism of a four-link articulated link ABCD (Fig. 23), it is required to balance only the main vector of inertial forces. Let us denote the masses of links AB, BC and CD, respectively, through and; the lengths of the links - through and and the distance of the centers of gravity and these links from points A, B and C - through and. To satisfy condition (4.9.), It is necessary that the general center of mass of the mechanism is on the straight line AD, either between points A and D, or behind them. In this case, the center S of mass of the mechanism during its movement will remain stationary and, therefore, the main vector of the inertial forces of the mechanism will be balanced.

The masses of the links and the positions of their centers of gravity must be selected so that

If the mechanism consists of n moving links, then when solving the problems of selecting the mechanism masses that satisfy the condition of the balance of the main vector of the mechanism's inertia forces, we have 2n unknown quantities; the equations connecting these quantities can be made (n-1). After an arbitrary choice of (n + 1) quantities, the remaining quantities receive certain values. In the investigated mechanism, the number of moving links is n = 3, the number of selected values is 2n = 6, and the number of independent equations is n-1 = 2. Thus, setting, for example, the values of m 3 and s 3, from equation (4.12) we obtain the value m 2 s 2, in which one of the unknowns can be specified and another is obtained. Substituting the obtained values into equation (4.11), we determine the value of m 1 s 2, in which you can also set one value. From equations (4.11) and (4.12) with different initial tasks, three variants of the schemes of a balanced four-link mechanism can be obtained. 23 (a, c, e). Therefore, if we assume that the location of the center of gravity of the link behind its hinges corresponds, as it were, to the installation of a counterweight, then we can say that the problem of balancing the main vector of inertia forces of the four-link hinge mechanism can be solved by installing counterweights on two of its links.

In a similar way, it is possible to solve the problem of selecting the masses of individual links to balance the articulated six-link link and any mechanism formed by layering two-thread groups. Having given equations (9.) can be replaced by one vector equation

Where r s is a vector defining the position of the common center of mass.

Condition (4.13) is satisfied in particular when r s = 0; this condition leads to a method of selecting mechanisms with symmetrically located links of equal masses.

Figure 24 shows the diagrams of symmetrical crank-slider and articulated four-link mechanisms. In cases where the placement of links in symmetrical mechanisms is very cumbersome or the selection of masses is constructively impractical, the method of installing counterweights is used.

Let, for example, it is required to balance only the main vector of the inertia forces of the crank-slider mechanism, the diagram of which is shown in Figure 25. Let us denote the masses of the crank 1, connecting rod 2 and slider 3 by m 1, m 2, m 3 and we will consider them concentrated, respectively, in the centers gravity S 1, S 2 and B links. We install a counterweight on the AB line at point D and determine its mass m pr from the condition that the center of gravity of masses m pr, m 2 and m 3 coincides with point A. From the equation of static moments relative to point A, we have

The mass of the counterweight installed at point C of the crank is determined from the condition that the center of gravity of the masses and coincides with point O. From the equation of static moments relative to point O we find

The radii s and from the counterweights are chosen arbitrarily. After installing the counterweights, the center of mass of the mechanism in all its positions will coincide with the point O and, therefore, will remain motionless during the entire operation. Thus, the two counterweights completely balance all the inertial forces of the mechanism under consideration. However, such a complete balancing of the inertial forces of crank-slider mechanisms is rarely used in practice, since at a small value of the radius c, the mass turns out to be very large, which leads to the appearance of additional loads in the kinematic pairs and links of the mechanism. With a large value of the radius c, the overall dimensions of the entire mechanism increase greatly. Therefore, they are often limited to only an approximate balancing of inertial forces. So, in crank-slider mechanisms, the method of installing a counterweight on the crank is the most common method of approximately balancing inertial forces. In these mechanisms, in practice, balancing of only the mass of the crank and part of the mass of the connecting rod is often used.

When solving some questions of the dynamics of a mechanism with one degree of freedom, it is possible to apply the law of change in kinetic energy, which is formulated as follows: the increment of the kinetic energy of the mechanism at its final displacement is equal to the algebraic sum of the work of all given forces.

where is the kinetic energy of the mechanism in an arbitrary position

Kinetic energy of the mechanism in the initial position

Algebraic sum of works of all forces and moments applied to the mechanism

For plane-parallel movement:

where is the moment of inertia of the link relative to the axis passing through the center of mass S

By the nature of the change in kinetic energy, the complete cycle of a machine unit generally consists of three parts: acceleration (start-up), steady-state and run-down (stop) (Fig. 4.6). The time t p is characterized by an increase in the speed of the leading link, and this is possible when>, and during the runout time<, т.е. кривая зависимости кинетической энергии в первом случае монотонно возрастает, во втором случае - монотонно убывает.

The steady motion is longer. During this phase, useful work is performed for which the mechanism is designed. Therefore, the total steady-state motion time can consist of any number of motion cycles corresponding to one or more crank revolutions.

We have two options for steady motion.

The first option: the kinetic energy T of the mechanism is constant during the entire mode of motion. Example: A gear system rotating at constant angular velocities has constant kinetic energy.

The second option: characterized by the frequency of movement of the drive shaft of the mechanism with small fluctuations in T within the period. The frequency can include one or two crank turns, for example, for an engine, the frequency of changing T is two crank turns.

The entire flow of energy supplied to the machine, as well as the kinetic energy of the machine itself during its operation, can be balanced as follows:

where is the work of forces driving

The work of the forces of useful resistance

Work of friction forces

The work of gravity

The work of the forces of inertia

For the time of steady motion, when at the end of the cycle and at the beginning of the next cycle, the speed is the same, i.e. work and are equal to zero, i.e.

Neglecting the friction force, we have

This equation is the basic energy equation for the steady periodic motion of the mechanism.

In the general case, the angular velocity of the driving link within the cycle of steady motion is a variable value.

Changes in the angular velocity of the drive link cause additional (dynamic) pressures in the kinematic pairs, which reduce the overall efficiency of the machine, the reliability of its operation and durability. In addition, fluctuations in speed will impair the work process of the machine.

The fluctuation of speed is a consequence of two factors - a periodic change in the reduced moment of inertia of the mechanism and the periodic nature of the action of forces and moments.

In addition to periodic fluctuations of speeds, fluctuations in the mechanism can also occur non-periodic, i.e. non-repetitive, caused by various reasons, such as a sudden change in load.

The first type of vibration is regulated within the permissible non-uniformity of movement by placing an additional mass (flywheel) on the shaft.

In the second case, the regulation problem is solved by installing a special mechanism called a regulator.

The limits of the permissible change in the angular velocity are established empirically. The unevenness of the movement of the machine is characterized by the ratio of absolute unevenness to its average speed

Usually set and, where

Having the following relationships:

We solve together two equations (4.14) and find:

Or, neglecting the value due to its smallness, we obtain:

Periodic unevenness of the machine, as a rule, is a harmful effect and can be tolerated for most machines only within certain limits. These harmful phenomena in machines are expressed, for example, in the following: jerks during the movement of transport vehicles, breakage of threads in textile machines, overheating of the windings of electric motors, flickering of light due to uneven rotation of the armature of an electric current generator, insufficient cleanliness and accuracy of surface treatment of parts on metal-cutting machines , heterogeneity and unequal thickness of welded seams when welding with automatic welding machines, sheet breakage during extrusion of products on presses, etc.

The permissible unevenness of the machine stroke is set by the coefficient d and depends on the purpose of the machine. These values are established by many years of experience in operating the machine.

Thus, and differ from the given average angular velocity by, which at q = 1/25 is only 2%, and at q = 1/50 the greatest deviation will be only 1% of. From this it can be seen that even with relatively large q, the movement of the driving link of the machine is sufficiently uniform.

The movement of the driving link is the closer to uniform, the greater the reduced moment of inertia or the reduced mass of the mechanism. An increase in the reduced masses and moment of inertia is made practically by fitting a flywheel with a certain mass and moment of inertia onto the machine shaft.

When analyzing the operation of a machine and determining the law of motion of the initial link of a mechanism with one degree of freedom, it is convenient to operate not with real masses that move with variable speeds, but with masses, or equivalent, conventionally transferred to any link of the mechanism.

In the same way, the forces or moments applied to individual links can be conditionally replaced by a force or moment applied to any link of the mechanism.

Reduced force is a force whose power is equal to the sum of the powers of all forces applied to the links.

The link to which the reduced force is applied is called the reduction link.

The power of any force applied at the "" point, based on the previous section, can be defined as the moment of this force relative to the end of the velocity vector

Power can be written in terms of the reduced moment of forces

The reduced mass is such a fictitious mass concentrated at the point of the reduction link, the kinetic energy of which is equal to the kinetic energy of the entire mechanism

where is the reduced moment of inertia of the link,

Angular speed of the link,

The speed of point B of the reference link.

Reduced moment of inertia

The moment of inertia reduced to the main shaft (drive link) is called such a conditional moment of inertia, possessing which the main shaft has in a given position of the machine kinetic energy equal to the kinetic energy of the entire mechanism.

Most machines work, as a rule, in a steady state, which is characterized by the fact that the machine receives from the engine in 1 cycle as much energy as it consumes during the same time to produce the work for which it is intended.

A cycle is a period of time after which all parameters characterizing the operation of the machine are repeated (periodic repetition of speeds, accelerations, loads, etc.). The movement of the links of the machine is thus periodic. The concept of a steady motion does not at all mean that the driving link of the machine moves uniformly.

Consider the equation of motion of the reduction link:

It follows from this equation that for uniform motion (i.e., when e = 0) at any moment of the cycle, the following conditions must be met:

those. changing the moment must follow the law of changing the work, which in practice cannot be accessed by simple means.

Thus, even with

So, for example, the crank of a planer, which includes a rocker mechanism, or a crank press, which includes a crank-slider mechanism, will not move evenly even without load.

Equality of moments is extremely rare in practice. Due to these reasons, the steady motion of machines occurs with a periodic change in speed, which changes within the cycle in the aisles:

Most machines work, as a rule, in a steady state, which is characterized by the fact that the machine in one cycle spends the same work that it receives from the engine in a cycle, that is, a prerequisite for steady motion is.

The physical role of a flywheel in a car can be imagined as follows. If, within a certain angle of rotation of the initial link of the mechanism, the work of the driving forces is greater than the work of the resistance forces, then the initial link rotates at an accelerated rate and the kinetic energy of the mechanism increases.

In the absence of a flywheel, the entire increase in kinetic energy is distributed between the masses of the links of the mechanism. The flywheel increases the total mass of the mechanism and therefore, with the same increase in kinetic energy, the increase in angular velocity without a flywheel will be greater than with a flywheel.

...

Lecture N6

Dynamics of mechanisms.

Dynamics tasks:

The direct problem of dynamics (power analysis of the mechanism) is, according to a given law of motion, to determine the forces acting on its links, as well as the reactions in the kinematic pairs of the mechanism.
The inverse problem of dynamics is to determine the true law of motion of the mechanism by the given forces applied to the mechanism.

The dynamic analysis of mechanisms can also include the tasks of balancing and vibration protection.

Let us first tackle the solution of the inverse problem of dynamics, considering all the links of the mechanisms to be rigid.

Various forces are applied to the mechanism of the machine unit during its movement. These are driving forces, resistance forces (sometimes they are called useful resistance forces), gravity forces, friction forces, and many other forces. The nature of their action may be different:

A) some depend on the position of the links of the mechanism;

B) some from changes in their speed;

C) some are permanent.

By their action, the applied forces impart a particular law of motion to the mechanism.

Forces acting in machines and their characteristics

Forces and pairs of forces (moments) applied to the mechanism of the machine can be divided into the following groups.

1. Driving forces and momentsmaking positivework during its action or in one cycle, if they change periodically. These forces and moments are applied to the links of the mechanism, which are called leading.

2. Forces and moments of resistanceperforming negativework during its operation or in one cycle. These forces and moments are divided, firstly, into forces and moments of useful resistance, which perform the work required from the machine and are applied to the links, called driven, and, secondly, into forces and moments of resistance of the medium (gas, liquid), in which the links of the mechanism move. The forces of resistance of the medium are usually small in comparison with other forces, therefore, in the future they will not be taken into account, and the forces and moments of useful resistance will be called simply forces and moments of resistance.

3. Forces of gravity moving links and spring forces. In some areas of the movement of the mechanism, these forces can perform both positive and negative work. However, for a complete kinematic cycle, the work of these forces is zero, since the points of their application move cyclically.

4. Forces and moments applied to the body of the machine(i.e. to the rack) from the outside. In addition to the force of gravity of the body, these include the reaction of the base (foundation) of the machine to its body and many other forces. All these forces and moments, since they are applied to the stationary body (rack), do not perform work.

5. Forces of interaction between the links of the mechanism, that is, the forces acting in its kinematic pairs. These forces, according to Newton's third law, are always reciprocal. Their normal work is not commit and the tangential components, that is, the frictional forces, do the work, and the work of the friction force on the relative displacement of the links of the kinematic pair negative.

The forces and moments of the first three groups are classified as active. Usually they are known or can be estimated. All these forces and moments are applied to the mechanism from the outside, and therefore are external. All forces and moments of the 4th group are also external. However, not all of them are active.

The forces of the 5th group, if we consider the mechanism as a whole, without highlighting its individual parts, are internal. These forces are reactions to the action of active forces. The reaction will also be the force (or moment) with which the base (foundation) of the machine acts on its body (that is, on the rack of the mechanism). The forward reactions are unknown. They depend on active forces and moments and on the accelerations of the links of the mechanism.

The greatest influence on the law of motion of the mechanism is exerted by driving forces and moments, as well as forces and moments of resistance. Their physical nature, magnitude and nature of the action are determined by the working process of the machine or device in which the considered mechanism is used. In most cases, these forces and moments do not remain constant, but change their magnitude when the position of the links of the mechanism or their speed changes. These functional dependencies, represented graphically, or by an array of numbers, or analytically, are calledmechanical characteristicsand are considered known when solving problems.

When depicting mechanical characteristics, we will adhere to the following rule of signs: the force and moment will be considered positive if on the considered section of the path (linear or angular) they produce positive work.

Characteristics of the forces depending on the speed.In fig. 6.1 shows the mechanical characteristic of an asynchronous electric motor - the dependence of the driving moment on the angular speed of the rotor of the machine. The working part of the characteristic is the area ab, at which the driving moment sharply decreases even with a slight increase in the rotational speed.

The forces and moments that also act in such rotary machines as electric generators, fans, blowers, centrifugal pumps (Fig. 6.2) and many others depend on the speed.

Fig 6.3

With an increase in speed, the torque of the motors usually decreases, and the torque of machines that consume mechanical energy usually increases. This property is very useful, since it automatically contributes to the stable maintenance of the machine's motion mode, and the more pronounced it is, the greater the stability. Let's call this property of machines self-regulation.

Characteristics of displacement-dependent forces. Figure 6.3 shows a kinematic diagram of the mechanism of a two-stroke internal combustion engine (ICE) and its mechanical characteristics. Force, applied to the piston 3, always acts to the left. Therefore, when the piston moves to the left (the process of gas expansion), it does positive work and is shown with a plus sign (branch czd). When the piston moves to the right (gas compression process), the forcegets a minus sign (branch dac). If the fuel supply to the internal combustion engine does not change, then at the next revolution of the initial link (link 1 ) the mechanical characteristic will repeat its shape. This means that the strengthwill change periodically.

Work of force will be graphically represented by the area bounded by the curve(s c). In Figure 6.3, this area has two parts: positive and negative, with the first being larger than the second. Therefore, the work of force for the full period will be positive. Consequently, the force is the driving force, although it is alternating in sign. Let us note in passing that if a force, being alternating in sign, performs negative work in one period, then it is a force of resistance.

Forces that depend only on displacement act in many other machines and devices (in reciprocating compressors, forging machines, planing and slotting machines, various devices both with pneumatic drive and with spring motors, etc.), and the action of forces 6 can be both periodic and non-periodic.

At the same time, it should be noted that the moment of rotary-type machines does not depend on movement, that is, it does not depend on the angle of rotation of the rotor; the characteristics of such machines are shown in Figure 6.4, a, b ... In this case, for machines-motors, and for machines-consumers of mechanical energy (i.e. working machines).

If you change the fuel supply to the internal combustion engine, then its mechanical characteristics will take the form of a family of curves (Figure 6.5, a ): the greater the fuel supply (parameter h families), the higher the characteristic is. The family of curves also depicts the mechanical characteristic of the shunt motor (Fig. 6.5, b ): the greater the resistance of the motor field winding circuit (parameter h ), the more to the right the curve is placed. The characteristic of the hydrodynamic coupling also has the form of a family of curves (Figure 6.5, c): the greater the filling of the coupling with liquid (parameter h ), the more to the right and higher are the characteristics.

Thus, acting on the parameter h , it is possible to control the operating mode of the drive - thermal, electric or hydraulic, increasing its driving force or speed. At the same time, the control parameter h is associated with the magnitude of the flow of energy flowing through the machine, that is, it determines its load and performance.

The mechanism of a machine unit is usually a multi-link system loaded with forces and moments applied to its various links. To better imagine it, consider a power pumping unit driven by an asynchronous electric motor.

A fluid resistance force is applied to the piston 3, and a driving moment is applied to the rotor 4 of the electric motor. If the pump is a multi-cylinder, then a resistance force will act on each piston, so that the loading pattern will become more complex.

To determine the law of motion of the mechanism under the action of specified external (active) forces, it is necessary to solve the equation of its motion. The basis for the formulation of the equation of motion is the theorem on the change in the kinetic energy of the mechanism with W = 1, which is formulated as follows:

The change in the kinetic energy of the mechanism occurs due to the work of all forces and moments applied to the mechanism

=
(6.1)

In a flat mechanism, the links perform rotational, translational and plane-parallel movements, then the kinematic energy of the mechanism

(6.2)

for all moving links of the mechanism

=
(6.3)

The total work of all external forces and moments

(6.4)

After substitution, we get

(
+
) - =(
)

The transition from many unknowns to one is carried out using the methods of bringing forces and masses. To do this, we pass from the real mechanism to the model, i.e. we replace the entire complex mechanism with one conventional link.

In this example, the mechanism has one degree of freedom ( W = 1). This means that it is necessary to determine the law of motion of just one of its links, which will thus be the initial one.

Dynamic model

Position of the mechanism with W = 1 is completely determined by one coordinate, which is called the generalized coordinate. As a generalized coordinate, the angular coordinate of the link performing rotational movement is most often taken. In this case, the dynamic model will be presented in the form:

Generalized angular coordinate of the model

Model angular velocity

The total reduced moment (the generalized force is the equivalent of the entire given load applied to the mechanism)

The total reduced moment of inertia, which is the equivalent of the inertia of the mechanism.

In the case of reduction, the actually acting forces and moments are replaced by the total reduced moment applied to the dynamic model.

It should be emphasized that the replacement made should not violate the law of motion of the mechanism determined by the action of actually applied forces and moments.

The reduction of forces and moments should be based on the condition of equality of elementary works, i.e. the elementary work of each force on the possible displacement of the point of its application or moment on the possible angular displacement of the link on which it acts should be equal toelementary work of the reduced moment on the possible angular displacement of the dynamic model.

Let us consider, as an example, the reduction of forces and moments applied to the links of a machine unit (Fig. 6.6), assigning an angular coordinate as a generalized coordinate.

We define, replacing the applied force
... By the condition of equality of elementary works

deciding with respect to the desired value and dividing the possible displacements by time, we obtain

=

cos (
,
), where cos (

)= 1

=

=

=, where

for computer solutions,

Using speeds.

Similarly, we will reduce the forces to a dynamic model (link 1)
,
, and
.

=
cos (
,
) = 0, 0 m. To . cos (
,
) = 0.

=
=

Center of Mass Velocity Projection
on the y-axis.

We will find in the same way.

If we add all the reduced moments applied to the initial link algebraically, we gettotal reduced moment, which replaces all forces and moments acting on the mechanism.

(6.5)

Bringing the masses.

Bringing masses is done with the same purpose as bringing forces:

modify and simplify the dynamic scheme of the mechanism, i.e. to come to the corresponding dynamic model, and, consequently, to simplify the solution of the equation of motion.

If an initial link with a generalized coordinate is taken as a dynamic model, then the kinetic energy of the model should be equal to the sum of the kinetic energies of all links of the mechanism, i.e. in the basis bringing the masses the initial link is subject to the condition of equality of kinetic energies.

The reduced moment of inertia is the parameter of the dynamic model, the kinetic energy of which is equal to the sum of the kinetic energies of the really moving links.

Let us write down the condition for the equality of the kinetic energy of an individual link, the entire mechanism and model for a separate link:

(6.6)

where for the model, for the real links of the mechanism

(6.7)

The transfer functions in brackets do not depend on, therefore, they can be determined further in the event that the law of motion of the model (initial link) is unknown. At
=

Where,

Let's define the given moments of inertia

All these moments of inertia do not depend on the angular position of the initial link. This group of links connected with the dynamic model by linear gear ratios are called links of the first group, and their moments of inertia are called moments of inertia of the first group.

Determine the moments of inertia of the 2nd and 3rd links

The moments of inertia of the first and second group of links and the total reduced moment of inertia of the considered installation are shown in Fig. 6.7

Control questions for the lecture N 6

Formulate the definition of direct and inverse problems of dynamics.
What is meant by the dynamic model of the mechanism?
What is the purpose of bringing forces and moments in the mechanism? What condition is the basis for bringing forces and moments?
What condition is the basis for the replacement of masses and moments of inertia during reduction?
Write the formula for the kinetic energy for the crank mechanism.

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Slide 2

Lecture plan

2 Force analysis of mechanisms. Forces acting on the links of the mechanism. The driving forces and the forces of production resistance. Mechanical characteristics of machines. Friction in mechanisms. Types of friction. Sliding friction. Friction on an inclined plane. Friction in the helical kinematic pair. Friction in a rotary kinematic pair. Rolling friction. Friction in ball and roller bearings. Inertial forces of links of flat mechanisms.

Slide 3

3 The dynamics of machines is a section of the general theory of mechanisms and machines, in which the movement of mechanisms and machines is studied taking into account the acting forces and properties of the materials from which the links are made - elasticity, external and internal friction, etc. machines, taking into account the forces and pairs of inertial forces of the links, the elasticity of their materials, the resistance of the medium to the movement of the links, balancing the forces of inertia, ensuring the stability of movement, regulating the course of machines.

Slide 4

4 POWER ANALYSIS OF MECHANISMS The movement of real mechanisms of machines occurs under the influence of various forces and is variable in time in accordance with the change in modes and the purpose of the machines. The purpose of the study of the movement of machines is to determine the modes of their movement in accordance with the requirements of production technology, operation and reliability. To do this, it is necessary to establish the permissible values of the forces acting on various links in the process of movement, the efficiency, displacement, speed and acceleration: the movement of the links and their individual points.

Slide 5

Forces and moments acting in the links of the mechanism

5 Driving forces Fd and Md. Forces and moments of resistance (Fс, Mс). The work of forces and moments of resistance per cycle is negative: Аc

Slide 6

Mechanical characteristics

6 The mechanical characteristics are stated in the data sheet. 1 is the speed with which the motor shaft rotates; 2 is the speed at which the main shaft of the driven machine will rotate. 1 and 2 must be matched to each other. For example, the number of revolutions n1 = 7000 rpm, and n2 = 70 rpm. To match the mechanical characteristics of the engine and the working machine, a transmission mechanism is installed between them, which has its own mechanical characteristics. up2 = 1 / 2 = 700/70 = 10

Slide 7

Mechanical characteristics of the machine as an example of a piston machine

7 Mechanical characteristics of a 3-phase induction motor (fig. 1). ICE indicator diagram (Fig. 2). H - stroke of the piston in the piston machine (distance between the extreme positions of the piston) Fig. 3. Pump indicator diagram (fig. 4) fig. 1 fig. 2 fig. 3 fig. 4

Slide 8

Friction in mechanisms

8 Friction is a complex physicochemical process, accompanied by the release of heat. This is because moving bodies resist relative motion. A measure of the intensity of resistance to relative displacement is the force (moment) of friction. Distinguish between rolling friction, sliding friction, as well as dry, boundary and liquid friction. If the total height of microroughnesses of the interacting surfaces is greater than the height of the lubricant layer, then dry friction. is equal to the height of the lubricant layer, then is the boundary friction. less than the height of the lubricant layer, then - liquid

Slide 9

Types of friction

9 By the object of interaction, external and internal frictions are distinguished. External friction is the opposition to the relative movement of the contacting bodies in the direction lying in the plane of their contact. Internal friction is the opposition to the relative movement of individual parts of the same body. On the basis of the presence or absence of relative motion, static friction and motion friction are distinguished. Friction at rest (static friction) - external friction, with relative rest of the contacting bodies. Friction of motion (kinetic friction) - external friction, with the relative movement of contacting bodies. By the type of relative motion of bodies, they are distinguished: slip friction - external friction with relative sliding of contacting bodies, rolling friction - external friction with relative rolling of contacting bodies.

Slide 10

10 According to the physical characteristics of the state of interacting bodies are distinguished: pure friction - external friction in the complete absence of any foreign impurities on the rubbing surfaces; dry friction - external friction, in which the rubbing surfaces are covered with films of oxides and adsorbed molecules of gases and liquids, and there is no lubrication; boundary friction - external friction, in which there is semi-fluid friction between the rubbing surfaces - friction in which there is a thin (about 0.1 micron or less) layer of lubricant between the rubbing surfaces; surfaces have a layer of grease with normal properties; fluid friction - friction in which the surfaces of rubbing solids are completely separated from each other by a layer of fluid.

Slide 11

Inclined friction

11 Sliding friction Scheme of action of forces when sliding on an inclined plane

Slide 12

Accounting for friction in a rotational kinematic pair.

Slide 13

13 1 - pivot rц - pivot radius Δ - clearance  - friction circle radius;  = О1С From ΔО1SK  = sin  О1С = О1К sin  Mc = Q12.О1С = Q12. rц.sin  At small angles sin ≈tg  = f. Then: Mc = Q12. rц.f Taking into account the friction in the rotary gearbox, the resulting reaction deviates from the common normal by the friction angle  and passes tangentially to the friction circle of radius .

Slide 14

Rolling friction

14 Rolling friction is the moment of forces that occurs during the rolling of one of two contacting and interacting bodies relative to the other, which opposes the rotation of a moving body.

Slide 15

Rolling friction coefficient

15 The rolling friction coefficient is the shoulder of a rolling friction pair, i.e. the distance the normal response is shifted. The rolling friction coefficient is f = Mmax / N. It is measured in linear units and is determined empirically.

Slide 16

Angle and cone of friction

Slide 17

Friction in ball and roller bearings

17 Rolling friction is the friction of motion of two rigid bodies in which their velocities at the points of contact are the same in value and direction. Such interaction and, accordingly, the type of friction is observed in ball and roller rolling bearings, in roller-guide mates.

Slide 18

Inertia forces of flat mechanisms

18 Forces and moments of inertia forces of the links arising when the speed of movement of the links and acting on the ties that hold the links. The forces of inertia impede movement during acceleration and facilitate movement during deceleration. The forces of inertia are determined by the product of mass by the acceleration vector of the center of inertia of the link.

Slide 19

Forces of inertia

19 Forces of inertia - proposed by D'Alembert for the force calculation of moving mechanical systems. When these forces are added to the external forces acting on the system, a quasi-static equilibrium of the system is established and it can be calculated using the equations of statics (kinetostatic method). Calculated expressions for the determination of inertial forces are familiar to you from the course of Theoretical Mechanics.

Slide 20

Self-test questions

20 1. The main features of the force analysis of mechanisms? 2. What forces and moments can arise in the links of the mechanism during movement? 3. What are the main characteristics of the machines. 4. What types of friction do you know, give their characteristics? 5. What is the difference between sliding friction and rolling friction? 6. How is the coefficient of friction determined?

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Dynamic analysis of the tmm mechanism. Presentation on the topic "dynamic analysis of mechanisms". The total work of all external forces and moments