Tmm crank mechanism. Sliding crank mechanism of working machine

1. Structural Analysis mechanism

1.1 Determination of the degree of mobility of the mechanism

Where N= 3 - the number of moving links of the mechanism

- the number of kinematic pairs of the fifth class

- the number of kinematic pairs of the fourth class

In a given mechanism, four pairs of the fifth grade

Rotational pairs

3.0 translational pairs

No fourth grade couples

1.2 Determining the class of the mechanism

To do this, we divide the mechanism into Assur groups.

We define the Assur group of the second class formed by links 2 and 3. The leading link remains, which forms the mechanism of the first class.

Class I movement Class II movement

Order 2

Formula of the structure of the mechanism

I (0.1) II (2.3)

The class of the connecting group is the second, therefore the considered mechanism belongs to the second class.

2 Geometric synthesis of the mechanism

2.1 We draw the mechanism in extreme positions

2.2 Determine the linear dimensions of the crank and connecting rod

Crank speed n1 = 82 rpm

Slider stroke S = 0.575 m

The ratio of the length of the crank to the length of the connecting rod

Eccentricity to Crank Length Ratio

2.3 During one revolution of the crank s;

The slider will cover the distance S, at S = 2AB

Determine the length of the link;

Determine the position of point M on the AB link from the ratio

; VM= 0.18 x 1.15 = 0.207 m;

3 Building a plan of the crank-slider mechanism

To build a plan of the crank-slider mechanism, draw a circle with a radius of AB, then draw a horizontal AC. We divide the circles into 12 parts (for 12 positions of the mechanism). Next, we postpone the segments B0C0, B1C1 ... B11C11 on the AC horizontal. We connect the center of circle A with points B0, B1 ... B11. At each of the 12 crank positions, set aside the BMi segment (where i is the number of the crank position). Connecting points М0, М1 ... М11, we obtain the trajectory of movement of point M.

4 Determination of the speeds of points O, A, B, M for four positions.

Position 1:

Determine the speed of point B

Consider

Determine From the triangle ABC

Consider

We determine the PC through

We define AR

Determine BP

We define Ð J

Determine MR

Determine the speeds of points A, C and M from the formula

We define

We check:

Position 2:

Determine the speed of point B

Consider

By the sine theorem, we define:

Determine From the triangle OAB

By the sine theorem, we define the AC

Consider

We determine the PC through

We define AR

Determine BP

We define Ð J

We define the MR

We define Ð Y

We check:

Position 3:

Since the speeds VB, VC and VM are parallel and points B, C and M cannot lie on the same perpendicular to the direction of these speeds, at the moment the instantaneous center of speeds of the connecting rod BC lies at infinity, its angular velocity, and he makes an instant forward motion. Hence, at the moment:

Position 4:

Determine the speed of point B

Consider

By the sine theorem, we define:

We define Ð B from triangle ABC

By the sine theorem, we define the AC

Consider

We determine the PC through

We define AR

Consider

Determine BP

We define Ð J

We define the MR

Determine the speeds of points A, B and M from the formula

We define Ð Y

We check:

5. Construction of diagrams of displacements, speeds and accelerations.

Let it be required to construct a kinematic diagram of the distances, speeds and accelerations of the slider C of the crank-slider mechanism. Crank AB with a length of l = 0.29 m rotates with a constant angular velocity n1 = 82 rpm.

Crank- slide mechanism serves to convert rotary motion into translational motion and vice versa. It consists of bearings 1, crank 2, connecting rod 3 and slider 4.

The crank makes a rotational movement, the connecting rod is plane-parallel, and the slider is reciprocating.

Two bodies movably connected to each other form a kinematic pair. The bodies that make up a pair are called links. Usually, the law of motion of the driving link (crank) is set. The construction of kinematic diagrams is carried out within one period (cycle), the steady motion for several positions of the leading link.

We build on a scale in twelve positions, corresponding to successive turns of the crank every 300.

Where S = 2r - the actual value of the slide, equal to twice the value of the crank.

- the stroke of the slider on the mechanism diagram.

Where is the time scale

Segment 1 on the time axis is divided into 12 equal parts corresponding in the selected scale to the rotation of the crank at the angles: 300, 600, 900, 1200, 1500, 1800, 2100, 2400, 2700, 3000, 3300, 3600 (at points 1-12). Let us postpone the vertical segments from these points: 1-1S = B0B1, 2-2S = B0B2, etc. These distances increase until the extreme right position of the slider B, and decrease from position B. If points 0s, 1s, 2s ... 12s are connected in series with a curve, then we get a diagram of displacements of point B.

To plot the diagrams of speeds and accelerations, the method of graphical differentiation is used. The velocity diagram is constructed as follows.

Under the displacement diagram, we plot the coordinates v and t, and on the continuation of the v-axis to the left, the selected pole distance HV = 20mm is arbitrarily laid.

From the point Pv draw straight lines parallel to the tangent curve S, respectively, at the points of the point 0s, 1s, 2s… 12s. These straight lines cut off the segments on the V-axis: 0-0v, 0-1v, 0-2v ... proportional to the speeds at the corresponding points of the diagram. We demolish the points to the ordinates of the corresponding points. We connect a number of obtained points 0v, 1v, 2v ... with a smooth curve, which is a diagram of speeds. The time scale remains the same, the speed scale:

We build the acceleration diagram in the same way as the velocity diagram. Acceleration scale

Where Ha = 16mm is the selected pole distance for the acceleration diagram.

Since the speed and acceleration are the 1st and 2nd derivatives of time displacement, but relative to the upper diagram, the lower one is a differential curve, and relative to the lower one, the upper one is an integral curve. So the velocity diagram for the displacement diagram is differential. When constructing kinematic diagrams for verification, you should use the properties of the derivative:

- an increasing graph of displacements (speed) corresponds to positive values of the graph of speed (equations), and a decreasing one - negative;

- the point of maximum and minimum, i.e. the extreme values of the displacement (velocity) graph correspond to zero values of the velocity (acceleration) graph;

- the inflection point of the displacement (velocity) graph corresponds to the extreme values of the velocity (acceleration) graph;

- the inflection point on the displacement diagram corresponds to the point where the acceleration is zero;

- the ordinates of the beginning and the end of the period of any kinematic diagram are equal, and the tangents drawn at these points are parallel.

To plot the movement of the slider B, select the coordinate axes s, t. On the abscissa axis, we put off the segment l = 120mm, representing the time T of one full turnover crank

They made a geometric calculation of the links of the crank-slider mechanism, determined the lengths of the crank and slider, and also established their ratio. The crank mechanism was calculated in four positions and the speeds of the points were determined using the instantaneous center of speeds for the four positions. Diagrams of displacements, velocities and accelerations were built. Found that there is some error due to the construction and rounding in the calculations.

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Introduction

2.1.1 Marking the mechanism

2.1.2 Calculation of speeds

2.1.3 Calculation of accelerations

Conclusion

Introduction

The theory of mechanisms solves the problems of structure, kinematics and dynamics of machines in connection with their synthesis and analysis.

In this work, an analysis is carried out, since the existing mechanism is being investigated.

The course project in the discipline "Theory of Mechanisms and Machines" provides for the calculation of the mechanism in three main sections:

1. Structural analysis.

2. Kinematic Analysis.

3. Kinetostatic analysis.

In each section, a certain set of calculations is performed that are necessary to study this mechanism.

Structural analysis gives general idea about the structure of the investigated mechanism. This section does not provide for a large amount of calculations, but only gives initial information about the parts and about the entire mechanism as a whole. This information will be needed in the future when calculating the mechanism.

Kinematic analysis is based on the results of structural analysis and provides for the calculation of kinematic characteristics. In this section, the positions of the mechanism at different times are built, the speeds, accelerations, displacements of points and links of the mechanism are calculated. Calculations are carried out by various methods, in particular, the method of plans (i.e., the solution of equations by the vector method), the method of kinematic diagrams, in which diagrams of kinematic characteristics are plotted, and the mechanism is studied using them.

Kinetostatic analysis or force calculation allows you to calculate those forces and reactions that act on the mechanism, and not only external forces such as gravity, but also forces of an exclusively internal nature. These are forces - the reactions of bonds that are formed when any links are excluded. In the force analysis, the same methods are partially used as in the kinematic analysis, but in addition to them, the method of N.E. Zhukovsky, which allows you to check the correctness of the work.

All methods used in the work are simple and accurate enough, which is not unimportant in engineering calculations of this kind.

Section 1. Structural analysis of the mechanism

Structural analysis allows you to understand the structure of the mechanism. The main objectives to be achieved in this section are:

1) Determination of the structure of the mechanism;

2) Calculation of the mobility of the mechanism;

3) Determination of the class of the mechanism;

Sliding crank mechanism working machine shown in Fig. 1.1, it consists of: 0 - rack; 1 - crank; 2 - connecting rod; 3 - slider.

The total number of links in the mechanism is N = 4.

Let us determine the mobility of the mechanism according to the Chebyshev formula:

W = 3n - 2P 5 - P 4, (1.1)

where n is the number of moving links (n = 3), Р 5 is the number of pairs of the fifth class, Р 4 is the number of pairs of the fourth class.

Let's depict the block diagram of the mechanism:

Rice. 1.2 Block diagram

The number of pairs of the fifth class Р 5: (0; 1), (1; 2), (2; 3),

The number of pairs of the fourth class P 4 = 0.

Mechanism mobility (1.1):

Let's write down the formula for the structure of the mechanism:

Mechanism class - II.

Section 2. Kinematic analysis of the mechanism

crank slider kinematic lever

In this section, the tasks of the kinematic analysis of the crank-slider mechanism of the working machine are solved, namely: the marking of the mechanism is constructed for its twelve positions; the positions of the centers of mass of the links are determined; plans for speeds and accelerations are made; the values of speed, acceleration and displacement of the output link are determined; the extreme positions of the mechanism are determined; kinematic diagrams are built.

2.1 Kinematic analysis by the method of plans

Kinematic analysis by the method of plans (graphic-analytical method) is quite simple, clear and has sufficient accuracy for engineering calculations. Its essence is that the relationship between speeds and accelerations is described by vector equations, which are solved graphically.

2.2.1 Marking the mechanism

The movement marking is a movement in twelve positions at specific points in time. The layout of the mechanism is built on the basis of the initial data. When constructing a markup, the main task is to maintain the proportions of the dimensions of the links and the overall design of the mechanism.

To build the markup, it is necessary to calculate the scale factor, which allows you to maintain all proportions and relate the actual dimensions of the mechanism to the dimensions used in the graphic part. The scale factor is determined from the ratio of the actual size of the mechanism (expressed in meters) to the size on the sheet in the graphic part (expressed in millimeters). Find the value of the scale factor using the actual size of the crank, equal to 0.280 m, and the size of the crank on the sheet in the graphic part, which we take 70 mm

where is the real size of the crank.

Using the resulting scale factor, we calculate the remaining dimensions of the links of the mechanism.

The same for all other sizes. The size calculation results are shown in Table 1.

Table 1

Based on the dimensions obtained, we build twelve positions of the mechanism, strictly observing all proportions and the basic structure. The markup of the mechanism is built on the first sheet of the graphic part of the course project. In fig. 2.1.1 presents the mechanism in twelve positions.

Rice. 2.1.1 Mechanism in twelve positions

2.1.2 Calculation of speeds

The calculation of speeds is made for all twelve positions of the mechanism. Linear and angular velocities of all links are calculated, as well as the velocities of the centers of mass.

Calculation of speeds and construction of plans will be carried out for position No. 2 of the mechanism.

Angular crank speed:

Using the value of the angular speed of the crank, we determine the speed of point A:

where is the length of the OA link.

Let us write down the vector equation for the speed of point B:

In this equation, we know the directions of the velocity vectors V B, V A, V AB. The speed of point B is directed along the t-t guide, the speed of point A is directed perpendicular to the crank OA, and the speed of link AB is directed perpendicular to this link. Knowing the direction of the speeds and the value of the speed of point A, we solve equation (2.1) graphically (Figure 2.1.2). To do this, we will initially determine the value of the scale factor, which is necessary for the constructions. It is defined similarly to the scale factor found in clause 2.1.1:

where pa is a segment representing the speed of point A on the plan of speeds (pa is chosen arbitrarily).

After determining the scale factor, we solve the vector equation (2.1) (Fig. 2.1.2). To do this, mark the point p v - a pole, from it we draw a segment p v a equal to the value of the speed of point A and directed perpendicular to the crank OA. From the end of the constructed segment, we draw a line of action of the relative velocity, which is directed perpendicular to AB, at the point of intersection of this vector with the t-t direction, there will be point b. The vector p v b determines the speed of point B, it is directed from the pole p v.

We determine the numerical value of the speeds by measuring the obtained segments and multiplying them by the scale factor:

We calculate the angular velocities by the formulas:

where is the length of the connecting rod (m).

The position of the centers of mass on the plane of velocities will be determined by the similarity property:

The speed of the center of mass of the connecting rod is:

In this work, the calculation of speeds for all twelve positions is carried out. The calculation is carried out similarly to the considered provision. The vectors of all velocities come out of the same pole. The calculation results (full speed plan) are presented on the first sheet of the graphic part of the project. The values of all speeds of the links of the mechanism and points of the links are presented in table 2.

table 2

2.1.3 Calculation of accelerations

The calculation of accelerations is carried out for two positions of the working stroke of the mechanism, in which the force of the useful resistance is not equal to zero. Accelerations are determined similarly to the velocities, the calculation of which was done above (section 2.1.2).

Initially, we will determine the acceleration of point A of the crank. It is constant and equal to the product of the square of the angular velocity of the crank by its length:

We will find accelerations by the method of plans, for this we write down the vector equation for the acceleration of point B:

where and are the normal and tangential components of the acceleration of the AB link, respectively.

Let's solve equation (2.10) graphically. To do this, we take the scale factor of the acceleration plan, equal to:

We build a plan of accelerations according to the direction of the vectors:

Directed from point A to point O 1;

Directed from point B to point A;

Directed perpendicular to the AB link;

The direction is given by the t - t rail.

Let us determine the normal component of the acceleration of the AB link:

To build an acceleration plan:

· Choose the pole pa;

· Build the vector of acceleration of point A;

· From the end of the vector we build a ray parallel to the link AB, and on this ray we lay off the segment an equal to:;

· Draw a straight line perpendicular to AB through point n, mark the point of its intersection with the t-t guide - point b;

· Segment p and b - acceleration of point B on the plane of accelerations.

The acceleration of the centers of mass is determined by the principle of similarity:

The acceleration plan for position # 2 is shown in fig. 2.1.4

Rice. 2.1.4 Acceleration plan for position # 2

We calculate the numerical values of the accelerations by the formulas:

The obtained values of all accelerations for the positions of the mechanism No. 8 and No. 10 are shown in Table 3.

Table 3

2.2 Kinematic analysis by the method of diagrams

The method of kinematic diagrams allows you to visually see how the displacement, speed and acceleration change during the cycle of the mechanism.

Let's take the scale factor equal.

To build the charts, we need a time scale factor and a rotation angle scale factor. We calculate these coefficients using the formulas:

where t c - cycle time,; L = 180 mm.

The displacement diagram is shown in Fig. 2.2.1

Fig 2.2.1. Displacement diagram

We transfer the speeds of the output link to the speed diagram taking into account the obtained scale factors. We connect the obtained values of speeds with a line, and as a result we have a diagram for the speed of the output link in twelve positions of the mechanism (Fig. 2.2.2).

The speed diagram is built on the first sheet of the graphic part.

Rice. 2.2.2. Speed chart

The acceleration diagram is constructed using the graphical differentiation method. For this:

· The diagram of speeds is approximated by a broken line;

· Transfer the abscissa axis from the velocity diagram to the acceleration diagram and continue it beyond the origin (to the left);

· Set aside the segment H = 20 mm;

On the velocity diagram, define the point 1 /, then connect it to the point O with a straight line:

· From point P we draw a ray parallel to the chord O1 /. Get point 1 //;

Segment O1 // depicts the average acceleration in the time interval (0; 1);

· To find the point of the acceleration diagram, it is necessary to restore the perpendicular from the middle of the time interval (0; 1) and project point 1 // onto this perpendicular;

· These constructions are repeated for the entire time interval.

Let's define the scale factor of the acceleration diagram:

Rice. 2.2.3. Acceleration diagram

Section 3. Kinetostatic analysis of the mechanism

Goals kinetostatic analysis:

· Determination of the useful resistance force in the considered positions of the mechanism;

· Determination of reactions in kinematic pairs;

· Determination of the balancing moment by the method of plans;

· Determination of the balancing moment by the “rigid lever” method N.Ye. Zhukovsky

3.1 Force calculation by the method of plans

Force calculation by the method of plans allows you to determine the reactions in kinematic pairs and the balancing moment. This method is simple, intuitive and accurate enough for engineering calculations.

3.1.1 Determination of the useful resistance force

The order of construction of markings for the force calculation of the mechanism does not differ from its construction in the section of kinematic analysis, therefore, no additional explanations are required here. After building the markup, we turn to the power diagram, which must be transferred from the source data to the sheet. In doing so, it is important to define. the magnitude of the resistance forces in each position of the marking and establish their compliance with these provisions. On the marking of the mechanism there are marks of the position of point B of the slider. Let us direct the ordinate axis of the desired graph parallel to the trajectory of point B from its zero position towards the other extreme position. Let us direct the abscissa axis perpendicular to this axis. In this case, along the ordinate axis, in essence, the movement of point B is plotted, along the abscissa, the same as on the original graph, the resistance force P is plotted.

In the selected coordinate system, it is necessary to apply scales along both axes and then a coordinate grid in the same way as it was done on the original graph in the assignment for the course project. Having read the coordinates of a number of characteristic points of the original graph, we build these points in the coordinate system prepared for this, and then we connect the plotted points sequentially with each other, which gives the desired graph.

Lowering the perpendiculars from the trajectory marks to the ordinate axis of the graph, we obtain the abscissas P in the desired positions of the marking of the working stroke of the mechanism. Note that the scale along the ordinate axis of the graph is equal to the marking scale (Figure 3.1.1 a)

Let's find the resistance forces:

for the 2nd position:

P s_ 2 = 1809 N,

For the 4th position:

P s_ 4 = 1298 N.

Fig 3.1.1а Determination of the useful resistance force

3.1.2 Force analysis of the structural group

We transfer link AB from the marking of the mechanism and at point A we free it from links, discarding link 1 and replacing the action of this link with a reaction, which, in turn, has normal and tangential components.

We apply the forces of gravity, inertia, useful resistance, and bond reactions to the links of the group. In the loading diagram (Fig. 3.1.1), the forces are depicted by segments of arbitrary magnitude, but strictly maintaining the directions of these forces. We direct the forces of inertia in the direction opposite to the acceleration of the corresponding points. The force of the useful resistance is directed in the direction opposite to the direction of the speed of the slider in the selected position.

Rice. 3.1.1. Loading scheme of the structural group for position No. 2

Determine the force of inertia of the slider in position No. 7:

The forces of inertia of the AB link:

Let us write down the sum of the moments relative to the slider B:

From equation (3.3) we express:

Let's write down the sum of all forces acting on the group:

Let's solve equation (3.5) graphically (Fig. 3.1.4). Let's choose a scale factor. We select a pole through which we draw a straight line parallel to the loading scheme and lay on it the segment depicting. We sequentially construct the vectors of all forces in accordance with equation (3.5) so that the unknown reactions are constructed in the last turn. The intersection of the lines of action of these two vectors will give the solution to this equation. In fig. 3.1.2 shows the plan of forces for the trailed group in position No. 2 of the mechanism.

Rice. 3.1.2. Force plan for the trailed group

To determine the numerical values of the unknown reactions and it is necessary to measure the segments that indicate the reaction data on the plan of forces and multiply them by the scale factor.

The obtained values of calculations and constructions are entered into the table.

3.1.3 Force calculation of the initial mechanism

The power calculation of the crank allows you to determine the balancing moment.

For the calculation, we transfer the initial link from the marking, discard the stand and replace it with an unknown reaction R 01. Let's load the crank by gravity and link reactions (Fig. 3.1.3).

The balancing moment M ur is determined from the equilibrium equation of the crank in the form of moments relative to the point O 1.

From equation (3.6) we express the moment M ur and find its numerical value:

To find the unknown reaction R 01, we compose the equation of all forces acting on the link, and solve it by the method of plans:

Rice. 3.1.4. Initial Mechanism Force Plan

Reaction R 01:

3.2 Force calculation by the "rigid lever" method N.Ye. Zhukovsky

The main task of the force calculation by the Zhukovsky "rigid lever" method is to check the correctness of the construction of plans of forces and the determination of reactions in kinematic pairs.

From an arbitrary point, taken as a pole P, we build a plan of accelerations for position No. 8 and turn it 90 0 clockwise relative to its normal position. The speed plan for position # 8 was built in section 2.1.2. We transfer these forces to the ends of the velocity vectors of the points at which the forces applied to the mechanism act, keeping their exact directions.

We determine the direction and value of the moments of inertia acting on the mechanism. Since ab and on the plan of speeds coincide with AB on the marking of the mechanism, then

Rice. 3.2.1. "Rigid lever"

We compose the equilibrium equations of the plan of velocities as a conditional rigid lever in the form of moments of forces relative to the pole of the plan of velocities. The shoulders of the forces are taken directly from the lever without any transformation:

We define:

Balancing moment:

Let's define the error:

The error, therefore, it can be concluded that the calculation was made correctly.

Force calculation for position No. 4 is carried out in a similar way.

Power calculation of the trailed structural group in position No. 4

The force calculation of the mechanism in the 10th position is carried out in the same way. As a result of calculations, we get:

Conclusion

In this course project, the tasks of the kinematic and kinetostatic analysis of the mechanism were solved. In the course of the project, the following goals were achieved:

· Complete kinematic calculation of the mechanism;

· The values of speeds, accelerations and displacements of links and points of the mechanism are determined;

· Positions of the working stroke of the mechanism are found;

· Determined the forces and reactions acting on the mechanism;

The values obtained in calculations and calculations were verified by the Zhukovsky method. According to this, the methods determined the error in position No. 2 () and in position No. 4 (), which turned out to be less than the allowable one, which indicates the correct constructions and calculations.

Bibliography

1. N.N. Fedorov. Design and kinematics of flat mechanisms. Tutorial... Omsk, publishing house OmSTU, 2010.

2. N.N. Fedorov. Kinetostatics of flat mechanisms and dynamics of machines. Tutorial. Omsk, publishing house OmSTU, 2009.

3. Artobolevsky I. I. Theory of mechanisms and machines. Textbook for universities - Moscow: Nauka, 1988.

4. Kozhevnikov S.N. The theory of mechanisms and machines. -M .: Nauka, 2012.

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All topics in this section:

Graphic method of kinematic research
2.1.1 Basic equations for determining speeds and accelerations …………………………………………… ..25 2.1.2 Kinematics of four-link mechanisms …………………………

Articulated four-link
Given (fig.2.6): j1, w1 = const, l1, l2, l3, lo = lAD, ml [m / mm].

Crank mechanism
Given (Figure 2.13): j1, w1 = const, l1, l0 = lAC, ml [m / mm]. Point B, belonging to the first s

Kinematic synthesis of flat linkages
Kinematic synthesis- This is the design of the scheme of the mechanism according to its specified kinematic properties. When designing mechanisms, first of all, on the basis of experience, in relation to

The condition for the existence of a crank in four-link mechanisms
The conditions for the existence of a crank in four-link mechanisms are determined by the Grashof theorem: if in a closed hinged four-link kinematic chain, the sum of the lengths

Application of Grashof's theorem to a kinematic chain with a translational pair
By increasing the size of the rotational pairs, it is possible to obtain translational pairs by expanding the pins. The size of the pivot pin D (Figure 2.19, b) can be taken large

Consider a crank mechanism, in which the line of movement
the slider is displaced relative to the center of rotation of the crank. The "e" value is called offset or deaxial. Let us determine at what ratio of sizes

Crank mechanism
Consider two options for the rocker mechanism: with a swinging and a rotating rocker. To obtain a mechanism with a swinging arm, it is necessary that the length of the rack is greater than the length of the crank,

Articulated four-link
Consider a hinged four-link link (Fig. 2.27), which is in equilibrium under the action of the given moments: the driving МДв on the driving link 1 and the moment of resistance

Synthesis of four-link linkages by link positions
Four-link mechanisms are often used to transfer various items from position to position. In this case, the carried object can be associated with both a connecting rod, that

Dynamic analysis and synthesis of mechanisms
The purpose of the dynamic research is to obtain the law of motion of the mechanism (its links), depending on the forces acting on it. When solving this problem, we will consider

I II III
I - the first link makes a rotational movement; II - link 2 makes a complex movement; III - link 3 moves progressively. To determine

Rack and pinion
If the center of one of the wheels is removed from infinity, then its circles will be transformed into parallel lines; point N1 of tangency of the generating line (it is also the common normal and

slider mechanism

2.1. Structural diagram of the mechanism

Fig 2.1 Block diagram of the crank-slider mechanism

2.2. Identifying complex and spaced kinematic pairs

There are no spaced kinematic pairs in the crank-slider mechanism. Pair V complex, so we will consider it as two kinematic pairs.

2.3. Classification of kinematic pairs of the mechanism

Table 2.1

P / p No.	The numbers of the links forming a pair	Symbol	Name	Mobility	Higher / Inferior	Closure (Geometric / Power)	Open / Closed
			Rotational
			Rotational
			Rotational
			Rotational
			Rotational
			Rotational
			Translational

The investigated mechanism consists only of one-way kinematic pairs ( R 1 = 7, R= 7), where R 1 - the number of single-moving kinematic pairs in the mechanism, R- the total number of kinematic pairs in the mechanism.

2. 4. Classification of links of the mechanism

Table 2.2

P / p No.	Link numbers	Symbol	Name	Traffic	Number of vertices
				Absent
			Crank	Rotational

				Rotational

				Translational

The mechanism has: four () two-vertex () linear links 1,2,4,5; one (n 3 = 1) three vertex link, which is the base link; five () moving links.

Find the number of connections to the rack. The conveyor mechanism has three () attachments to the rack.

In the investigated complex mechanism, one elementary mechanism can be distinguished

Rice. 2.4 Crank-slider mechanism.

There are no mechanisms with open kinematic chains in the investigated crank-slider mechanism.

The mechanism contains only simple stationary mechanisms.

In the investigated mechanism, there are no fastening links. Link 3 is simultaneously included in two simple mechanisms - an articulated four-link mechanism and a crank-slider. Hence, for this link

We classify the mechanism. The investigated mechanism has a constant structure, is complex and of the same type. It consists of one elementary mechanism and two stationary simple ones, which contain only closed kinematic chains.

The mechanism exists in a three-movable space.

Formulas for determining the mobility of these mechanisms will take the form, respectively:

Let us determine the mobility of the hinged four-link link. This mechanism has: three () movable links 1,2,3; four () one-way kinematic pairs O, A, B, C.

Let's find the mobility of the crank-sliding mechanism. It has: () movable links 3,4,5 and four () kinematic pairs C, B, D, K. Its mobility is determined in the same way:

We determine the mobility of a complex mechanism by the formula:

We analyze the structural model of the machine tool. We check whether the investigated mechanism corresponds to the structure of the mathematical model. The mechanism has: seven () single-moving kinematic pairs; five () movable two-vertex () links, the base is; three attachments to the rack () and no anchoring links ().

Mathematical model:

;

Since the equations of the model have turned into identities, the investigated device has the correct structure and is a mechanism.

Let's select and carry out the classification of structural groups. The elementary mechanism is conventionally classified as a class I mechanism.

The class of the structural group is determined by the number of kinematic pairs included in a closed loop formed by internal kinematic pairs. The order of the group is determined by the number of external kinematic pairs. The type of a group is determined depending on the location of rotational and translational kinematic pairs on it.

2-order

It can be seen that the identified structural groups are completely similar in the species and quantitative composition of links and kinematic pairs. Each of the structural groups has: two movable links (), and the links are two-vertex () and, therefore, the base link also has two vertices (); three () one-way kinematic pairs, of which two are external ().

We check whether the selected structural groups correspond to mathematical models. Since the groups are similar, we check only one group, for example, OAB. Mathematical models of structural groups are as follows:

The crank-slider mechanism belongs to the II class.

3. Kinematic analysis of the mechanism

The kinematic analysis of any mechanism consists in determining: the extreme (dead) positions of the machine, including the determination of the trajectories of individual points; the speeds and accelerations of the characteristic points of the links according to the well-known law of motion of the initial link (generalized coordinate).

3.1 Determination of extreme (dead) positions of the mechanism

The extreme (dead) positions of the mechanism can be determined analytically or graphically. Since analytics gives higher accuracy, it is preferred when determining extreme positions.

For the crank-slider and articulated crank-rocker four-link, the extreme positions will be when the crank and the connecting rod are extended (), then folded () into one line.

Rice. 3.1 Determination of the extreme positions of the mechanism.

3.2 Determination of the positions of the links of the mechanism graphically.

Rice. 3.3 Construction of closed vector contours.

We place the structural diagram of the mechanism in a rectangular coordinate system, the origin of which is placed at point O. We connect the vectors with the links of the mechanism so that their sequence is two closed contours: OABCO and CBDC.

For the OABCO circuit: (3.1)

We represent the equation in projections on the coordinate axes.

1. Structural Analysis mechanism

The crank-slider mechanism is presented.

The number of degrees of the mechanism under study is determined by the Chebyshev formula:

(1)

where n - the number of moving links in the studied kinematic chain; p 4 and p 5- respectively, the number of pairs of the fourth and fifth class.

To determine the value of the coefficient n let's analyze the block diagram of the mechanism (Figure 1):

Figure 1 - Block diagram of the mechanism

The structural diagram of the mechanism consists of four links:

1 - crank,

2 - connecting rod AB,

3 - slider B,

0 - rack,

wherein links 1 - 3 are movable links, and rack 0 is a fixed link. It is represented in the composition structural diagram two articulated fixed supports and a slide guide 3.

Hence, n = 3.

To determine the values of the coefficients p 4 and p 5 find all the kinematic pairs that are part of the considered kinematic chain. We enter the results of the study in table 1.

Table 1 - Kinematic pairs

№	Kinematic pair (KP)	Kinema diagram tic pair	Kinema class tic pair	The degree of mobility
1	0 – 1		rotational
2	1 – 2		rotational	1
3	2 – 3		rotational	1
4	3 – 0	rotational		1

From the analysis of the data in Table 1, it follows that the investigated ICE mechanism with increased piston stroke consists of seven pairs of the fifth class and forms a closed kinematic chain. Hence, p 5 = 4, a p 4 = 0.

Substituting the found values of the coefficients n, p 5 and p 4 into expression (1), we get:

To identify the structural composition of the mechanism, we divide the scheme under consideration into Assur structural groups.

The first group of links 0-3-2 (Figure 2).

Figure 2 - Structural group of Assur

This group consists of two moving links:

connecting rod 2 and slider 3;

two leashes:

and three kinematic pairs:

1-2 - rotational pair of the fifth grade;

2-3 - rotational pair of the fifth class;

3-0 - forward pair of the fifth grade;

then n = 2; p 5 = 3, a p 4 = 0.

Substituting the revealed values of the coefficients into expression (1),

Consequently, the group of links 4-5 is an Assur structural group 2 of class 2 of order 2 of the type.

The second group of links 0-1 (Figure 3).

Figure 3 - Primary mechanism

This group of links consists of a movable link - crank 1, rack 0 and one kinematic pair:

0 - 1 - rotational pair of the fifth class;

then n = 1; p 5 = 1, a p 4 = 0.

Substituting the found values into expression (1), we get:

Consequently, the group of links 1 - 2 is indeed the primary mechanism with mobility 1.

Structural formula of the mechanism

MECHANISM = PM (W = 1) + SGA (class 2, order 2, type 2)

2. Synthesis of the kinematic scheme

For the synthesis of the kinematic scheme, it is first necessary to set the scale factor of the lengths μ ℓ. To find μ ℓ, it is necessary to take the natural size of the crank OС and divide it by the size of a segment of arbitrary length │OС│:

After that, using the scale factor of the lengths, we translate all natural dimensions of the links into segments, with the help of which we will build the kinematic diagram:

After calculating the dimensions, we proceed to building one position of the mechanism (Figure 4) using the serif method.

To do this, first draw a rack 0 on which the crank is fixed. Then we draw a horizontal line XX through the center of the circle that was drawn to build the rack. It is necessary for the subsequent finding of the center of the slider 3. Further, from the center of the same circle we draw two other radii

and . Then from there we draw a segment with a length at an angle to the horizontal line XX. The intersection points of this segment with the constructed circles will be points A and C, respectively. Then from point A we build a circle with a radius.

The point of intersection of this circle with line XX will be point B. Draw a guide for the slider, which will coincide with line XX. We build the slider and all the other details of the drawing are needed. We designate all points. The synthesis of the kinematic scheme has been completed.

3. Kinematic analysis of a planar mechanism

We proceed to building a plan of speeds for the position of the mechanism. To simplify the calculations, you should calculate the speeds and directions for all points of the mechanism position, and then build a speed plan.

Figure 4 - One of the positions of the mechanism

Let us analyze the scheme of the crank-slider mechanism: point O and O 1 are fixed points, therefore, the moduli of the speeds of these points are equal to zero (

The velocity vector of point A is the geometric sum of the velocity vector of point O and the speed of the relative rotational motion of point A around point O: