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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:
2. Kinematic analysis.
3. Kinetostatic analysis.
Each section performs a specific set of calculations required for the 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 (that is, 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 the 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 calculation, 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;
Crank- slide 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, visual 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 and 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 using 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
Objectives of kinetostatic analysis:
· Determination of the force of useful resistance 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 axis, as in 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 plot 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 connections, 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 rotate 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 the same 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 permissible, 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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Introduction
1. Literary review
3. Kinematic analysis of the mechanism
4. Kinetostatic analysis of the mechanism
Conclusion
Design and research of the crank-slider mechanism of the screen
The length of the explanatory note was 37 sheets, 4 illustrations, 10 tables, 2 annexes, 3 sources used.
The object of the course design is the crank-slider mechanism. IN term paper the study of the crank-slider mechanism was carried out. Structural, kinematic, kinetostatic analyzes were carried out.
In the structural analysis, the composition of the crank-slider mechanism was determined. In the kinematic analysis, the speeds and accelerations of the points of the mechanism are determined by the methods of plans and kinematic diagrams. IN kinetostatic analysis a force calculation was carried out using the force plan method and the Zhukovsky method.
Introduction
The purpose of the course work is to consolidate and systematize, expand theoretical knowledge, and also develop the calculation and graphic skills of students.
The development of modern science and technology is inextricably linked with the creation of new machines. In this regard, the requirements for new developments are becoming more and more stringent. The main ones are: high performance, reliability, manufacturability, minimum dimensions and weight, ease of use and efficiency.
A rationally designed machine must meet social requirements - safety of service and creation best conditions for service personnel, as well as operational, economic, technological and production requirements. These requirements represent a complex set of tasks that must be solved in the process of designing a new machine.
The design object of this course work is a crank-slider mechanism.
The theory of mechanisms and machines is a science that studies the structure (structure), kinematics and dynamics of mechanisms in connection with their analysis and synthesis.
The aim of the theory of mechanisms and machines is the analysis and synthesis of typical mechanisms and their systems.
The problems of the theory of mechanisms and machines are diverse, the most important of them can be grouped into three sections: analysis of mechanisms, synthesis of mechanisms and the theory of machines-automatic machines.
The analysis of the mechanism consists in the study of the kinematic and dynamic properties of the mechanism according to its given scheme, and the synthesis of the mechanism in the design of the scheme of the mechanism according to its given properties.
From all of the above, it follows that the theory of mechanisms and machines, in conjunction with courses in theoretical mechanics, machine parts, engineering technology, strength of materials, is a discipline directly dealing with the problems outlined earlier. These disciplines are fundamental in the training of specialists working in the field of mechanical engineering.
When solving problems of designing kinematic schemes of mechanisms, it is necessary to take into account structural, metric, kinematic and dynamic conditions ensuring the reproduction of a given law of motion by the designed mechanism.
Modern methods of kinematic and kinetostatic analyzes are linked to their structure, i.e., the method of formation.
Structural and kinematic analyzes of mechanisms are aimed at studying the theory of the structure of mechanisms, studying the movement of bodies that form them, from a geometric point of view, regardless of the forces that cause the movement of these bodies.
The dynamic analysis of mechanisms aims to study the methods of determining the forces acting on the bodies that form the mechanism, during the movement of these bodies, the forces acting on them, and the masses that these bodies possess.
1. Literary review
In the study of the mechanism, methods of calculation and design of modern automated and high-performance machines are used. A rationally designed machine must meet the requirements for safe operation and the creation of the best conditions for the operating personnel, as well as the operational, economic, technological and production requirements. These requirements represent a complex set of tasks that must be solved in the process of designing a new machine.
The solution to these problems at the initial stage of design consists in the analysis and synthesis of the designed machine, as well as in the development of its kinematic scheme, which ensures, with sufficient approximation, the reproduction of the required law of motion.
To accomplish these tasks, it is necessary to first study the basic provisions of the theory of machines and general methods kinematic and dynamic analysis and synthesis of mechanisms, as well as acquire skills in applying these methods to the study and design of kinematic schemes of mechanisms and machines different types.
A machine is a device created by man for the study and use of the laws of nature in order to facilitate physical and mental labor, increase its productivity and facilitate it by partial or complete replacement a person in his labor and physiological functions.
In terms of the functions performed by machines, machines can be divided into the following groups:
a) power machines (motors and generators);
b) working machines (transport and technological machines);
c) information machines (mathematical and control machines);
d) cybernetic machines.
With the development of modern science and technology, systems of automatic machines are being used more and more widely. A set of automatic machines, interconnected and designed to perform a certain technological process is called an automatic line. Modern advanced and perfect machines usually represent a collection of many devices, in the operation of which the principles of mechanics, thermal physics, electrical engineering and electronics are put.
A mechanism is an artificially created system of bodies designed to transform the movement of one or more bodies into the required movements of other bodies. According to their functional purpose, the mechanisms of the machine are usually divided into mechanisms of motors and converters; transmission mechanisms; executive mechanisms; mechanisms of management, control and regulation; mechanisms for feeding, transporting, feeding and sorting processed media and objects; mechanisms for automatic counting, weighing and packaging of finished products.
Despite the difference in the functional purpose of the mechanisms of certain types, there is much in common in their structure, kinematics and dynamics. Therefore, in the study of mechanisms with various functional purposes, it is possible to apply general methods based on the basic principles of modern mechanics.
The main types of mechanisms:
1) rod mechanisms are used to transform motion or transfer of force in machines;
2) in many cases there is a need to design mechanisms that include elastic links in the form of springs, springs, elastic beams, etc.;
3) gear mechanisms are used to transfer rotary motion between shafts with parallel or non-parallel axes;
4) cam mechanisms are used to communicate periodic or limited episodic movement to the slave link of the mechanism according to a given
specific or chosen law;
5) as flexible links transmitting motion from one rigid body in the mechanism to another, are practically used of various shapes cross-section belts, ropes, chains, threads, etc .;
6) frictional mechanisms - mechanisms in which the transfer of motion between contacting bodies is carried out due to friction;
7) movement mechanisms with stops;
8) wedge and screw mechanisms are used in of various kinds clamping devices or in devices in which it is required to create large forces on the output side with limited forces acting on the input side;
9) wider possibilities in the sense of reproducing the laws of motion of the driven links in comparison with purely lever, gear or other mechanisms are given by the so-called combined mechanisms, in which lever, gear, cam and other mechanisms are combined in various combinations;
10) mechanisms of variable structure are used, if necessary: to protect the links of mechanisms from accidental overloads; carry out the required movements of the driven links, depending on the presence or absence of payloads; change the speed or direction of movement of the driven link of the mechanism without stopping the engine and in many other cases;
11) mechanisms with a given relative movement of the links;
12) hydraulic mechanisms - a set of translational or rotary mechanisms, a source of injection working fluid, control and regulating equipment;
13) pneumatic mechanisms are piston or rotary mechanisms in which movement is carried out by energy compressed air, i.e. gas in these mechanisms is used as an energy carrier;
The most important stage in the design of machines is the development of the structural and kinematic schemes of the machine, which largely determine the design of individual units and parts, as well as performance cars .
In this course work, the crank-slider mechanism will be considered.
The crank mechanism is one of the most common. It is the main mechanism in all piston (engines internal combustion, compressors, pumps, gas expansion machines), agricultural (mowers, reapers, combines) and forging machines and presses.
In each functional purpose, the design must take into account the specific requirements for the mechanism. However, the mathematical relationships describing the structure, geometry, kinematics and dynamics of the mechanism will be practically the same for all different applications. The main or main difference between TMM and academic disciplines that study methods for designing special machines is that TMM focuses on the study of synthesis and analysis methods that are common for a given type of mechanism and do not depend on its specific functional purpose.
The rocker crank-slider mechanism is a crank-slider mechanism with an infinitely long connecting rod, which has been structurally transformed into a slider-stone. Its guide, the curtain, is one piece with the slider making a harmonious movement. Therefore, the movement of the slider is proportional to the cosine of the crank angle. This mechanism, also called a rocker sinus mechanism, is used in small piston pumps and compressors, devices for harmonious movement of the slider or for determining values proportional to the sine or cosine of the crank angle, etc.
Depending on the purpose and operating conditions, mechanisms with higher pairs can be divided into a number of types, of which the main ones are cam, gear, friction, Maltese and ratchet.
The cam mechanism is a mechanism, the upper pair of which is formed by links called - cam and pusher. They differ in the shape of their elements. The shape of the pusher element can be arbitrary, and the shape of the cam element is chosen such that, for a given pusher element, the required law of motion of the driven link is provided. The simplest cam mechanism - three-link, consisting of a cam, a pusher and a rack; its leading link is usually the cam.
Gear mechanism, i.e. a mechanism, the upper pair of which is formed by toothed links, can be considered a special case of a cam mechanism, since a toothed link is, as it were, a multiple cam. Gear mechanisms are mainly used to transfer rotary motion between any two axes with a change in the angular velocity of the driven shaft.
A frictional mechanism is a mechanism in which the transfer of rotational motion between the links that form the upper pair is carried out due to friction between them. A simple friction mechanism consists of three links - two rotating circular cylinders and a rack.
Friction mechanisms are often used in continuously variable transmissions. With a constant angular speed of the disc, by moving the wheel-roller along its axis of rotation, it is possible to smoothly change not only its angular speed, but even the direction of rotation.
The Maltese mechanism converts the continuous rotation of the driving link - a crank with a pin into the intermittent rotation of the slave - "cross".
The ratchet mechanism with a leading pawl is used to convert the reciprocating movement into an intermittent rotary one in one direction. The leading rocker with a pawl gradually turns the ratchet wheel. The dog does not allow the wheel to rotate in reverse side... The top pair here is formed by a dog and a ratchet wheel.
Maltese and ratchet mechanisms are widely used in machine tools and appliances.
2. Structural analysis of the mechanism
The screen mechanism (Figure 1) consists of five links: 1 - OA crank, making a rotational movement; 2 - slider A, performing a reciprocating motion along the wings; 3 - rocker arm ABC, making a rocking motion around the hinge B; 4 - connecting rod CD; 5 - slider D, performing a reciprocating motion; as well as seven kinematic pairs.
Figure 1 - Diagram of the linkage mechanism
Determination of the degree of mobility of the mechanism
The degree of mobility of the mechanism is determined by the Chebyshev formula:
W = 3n - 2P 5 - P 4, (2.1)
Where n is the number of moving links for the mechanism, n = 5;
Р 5 - the number of kinematic pairs of the V class, Р 5 = 7;
Р 4 - the number of kinematic pairs of class IV, Р 4 = 0.
Substituting numerical values, we get:
W = 3 5 - 2 7 - 0 = 1.
Consequently, the degree of mobility of the mechanism, which shows the number of leading links in the investigated mechanism, is equal to 1. This means that one leading link is sufficient for the operation of the mechanism.
Breakdown of the mechanism into structural groups
According to the classification of I.I.Artobolevsky, we divide the mechanism under study into structural groups. The screening mechanism (Figure 1) consists of a driving link of 1 and two structural groups of II class of order 2.
Both structural groups belong to the third type: the first - (links 2 and 3), and the second - (links 4 and 5). Structural groups consist of 2 links and 3 kinematic pairs. The formula for the structure of the mechanism is as follows:
3. Kinematic analysis of the gear train
The drive of the linkage mechanism of the screen, consisting of a planetary gearbox and a gear train, is shown in Figure 2. The planetary gearbox, consisting of a carrier and four wheels with external gear, has gear ratio i H3 = 10. The gear wheels installed after the planetary gear have the following numbers of teeth: z 4 = 12, z 5 = 28.
Figure 2 - Lever mechanism drive
The gear ratio of gears 4 and 5 is determined by the formula
The total gear ratio of the entire drive is determined by the formula
Here are some parameters of the gear and planetary gear: m I = 3.5 mm; m II = 2.5 mm; the center distance of the gears - a w = 72 mm; angular speed of the drive shaft (motor shaft) - ω d = 150.00 rad / s. Let us determine the angular velocity of the driving link of the screening mechanism - ω 1 according to the formula:
ω 1 = ω d / i 15, (3.3)
ω 1 = 150 / 23.33 = 6.43 rad / s.
4. Kinematic analysis of the linkage
The purpose of the kinematic analysis is to determine the speeds and accelerations of the characteristic points of the lever-slider mechanism of the screen.
Building plans for the provisions of the mechanism
The parameters of the investigated mechanism (Figure 1) are shown in Table 1.
Table 1 - Mechanism parameters
| ω 1, rad / s |
||||||
The scale of the mechanism plan is determined by the formula
where l OA is the true length of the OA crank, m;
OA - the length of the OA crank in the drawing, mm.
Substituting the data, we get
m l = 
The procedure for constructing a plan for the provisions of this mechanism:
- we mark on the drawing the position of the centers of rotation of the crank t. O and the rocker mechanism t. C;
- we outline the trajectories of the points A and O of these parts;
- we will divide the trajectory of movement of the OA crank into 12 equal parts;
- from the obtained points A 0, A 1, A 2, ..., A 12 draw lines to point B;
- from point B we draw perpendiculars, taking the angle ABC equal to 90◦;
- determine the position of point C at certain positions of the OA crank;
- we postpone the segment of the SD on a scale in such a way that the point D lies on the ATS straight line;
- using the serif method, we determine the position of point D at certain positions of the OA crank;
- in a clockwise direction, put the OA crank in a new position and repeat the construction;
- we designate in the drawing the trajectories of the extreme points of the links and the position of the centers of mass of the links.
Building a diagram of the displacement of the working link
To construct kinematic diagrams by the method of graphic differentiation, 12 positions of the movement of the mechanism (along the crank OA) are considered.
Consider the movement of the output link. We will take the zero position as the starting point (it is also the last one). The abscissa axis is divided into 12 equal parts. Along the ordinate, we plot the distance traveled by point D in a straight line (on link 5) from the extreme left position to the extreme right position corresponding to a given moment in time. Using the obtained points, we construct a displacement diagram φ = φ (t) of the output link.
Determine the scale of displacement from the angle of rotation and in time:
where l is the distance in the drawing full turnover crank ОА, mm;
n is the number of revolutions per minute of rotation of the OA crank, rpm, determined by the formula
Taking the length of a full turn in the drawing 180 mm, we determine the scale
Let's take a smaller scale of displacements.
m s = 
Graphical differentiation of the diagrams of speeds and accelerations of the output link. Choosing an arbitrary pole distance H v = (40 ... 60 mm) = 50 mm, we calculate the scale of the velocity diagram m V
(4.5)
We replace the displacement curve with a set of chords, select the pole distance and build a coordinate system. To do this, on the graph of velocities parallel to the chords, we build straight lines passing through the pole. From the point of intersection of the straight line with the S axis, draw a straight line parallel to the t axis to the desired position. We connect the resulting points in series, resulting in a graph of the speeds of the output link. Similarly to the velocity diagram, having arbitrarily chosen the value of the pole distance H A equal to 40 mm, we calculate the scale of the acceleration diagram m A
(4.6)
Plotting an acceleration diagram is similar to plotting a velocity diagram.
Building speed plans for three positions
To plot, you need to know the speed of point A in the rotating motion of the link OA. Let's define it from the formula:
V A 1 = 
To build plans of speeds, we will select the positions of the mechanism: the first, seventh and tenth. For all positions, the construction is similar, therefore, we describe the construction algorithm. Let's define the characteristic points for construction: pivot points - A1, B6, D6, C3; and basic - A3, D4. Let's compose the vector equations of the velocities of these points:
(4.8)
(4.9)
We build a speed plan. OA crank moves at a constant speed. From the pole - P of the speed plan in the direction of rotation of the crank perpendicular to OA, we postpone the speed vector (Pa 1), conventionally taking its length equal to 80 mm. Then we determine the scale of the speed plan:
m V = 
In accordance with the system of equations (4.8), we make the corresponding constructions. To do this, draw a straight line parallel to BA through point a 1, and draw a straight line perpendicular to AB from the pole P, since the speed B6 is zero. Thus, we get point a 3. Since point C belongs to the ABC link, then on the plan of velocities it can be found using the similarity theorem. We determine its location by the ratio of the lengths of the ABC lever and the ratio of the lengths of the speeds a 3 to 6 c 3. Then, we use the system of vector equations (4.9). Having found a point with 3, we set aside from it the perpendicular to the SD connecting rod. Draw a straight line parallel to the VD line from the pole; since the speed of the point b 6 is equal to zero, then we thereby obtain the point d 4. The positions of the velocity vectors of the centers of mass are determined from the similarity theorem. Since the center of mass of the link OA is at point O, then on the plan of velocities it will be at point P. The position of the center S 4 on the plan of velocities is determined on the line with 3 d 4, in the middle of the segment. On the segment b 6 a 3, we find from the proportion (4.11) the position of the point S 3:
For all three positions, we will calculate the speeds from the graphical construction, taking into account their recalculation to the actual size, by measuring the length of the vectors corresponding to the speeds and multiplying them by the scale of the speed plan:
Table 2 - Actual values of the speeds of the characteristic points of the linkage mechanism in three positions
| Mechanism position |
Point speed |
Vector length from the plan (pn), mm |
||
Construction of acceleration plans for three positions
Let us compose a system of vector equations for the acceleration of the linkage mechanism by analogy with the vector equations of velocities:
(4.13)
(4.14)
Let's define the normal acceleration of point A of the link OA. Since the link rotates at a constant speed, there is no tangential acceleration. Then we have:
Let us give an algorithm for constructing a plan of acceleration analogs using the example of the first position. We carry out the rest of the constructions in the same way.
We begin the construction of the plan by constructing the acceleration of point A. Let us postpone it in scale from the pole P, and the direction of the vector from A to O. Determine the scale of the accelerations, taking arbitrarily in the drawing the length of the acceleration a 1 = 80 mm:
m a = 
Let us determine the angular velocities of the ABC and SD links. We find their values by formula (4.17), and are directed parallel to the corresponding links from the base point.
(4.17)
We find the angular velocity for each link from the velocity plan. Let's summarize the obtained values in Table 3.
Table 3 - Angular speeds of links and normal accelerations
| Position |
Speed |
Value, m / s |
Normal acceleration |
Meaning, |
Scale value, mm |
The construction is carried out using a system of vector equations. The tangential accelerations are directed perpendicular to the links. Considering all this, we will construct a plan of accelerations for the positions of the mechanism: 1, 7, 10. Point with 3 is found by analogy with the plan of speeds. We find the Coriolis acceleration by the formula:
(4.18)
(4.19)
The obtained values are summarized in table 4. It is deposited in the direction of rotation 90 o from the velocity vector. The relative speed has a direction parallel to the motion, putting the vectors in order. Find point a 3 and d 4.
Table 4 - Calculation of Coriolis acceleration
The results of all calculations by the graphical method and differentiation are summarized in Table 5.
Table 5 - Convergence table
We find the discrepancies in the values of speeds and accelerations by the formulas:
(4.20)
(4.21)
where is the value of the acceleration from the plan, m / s 2;
- value of acceleration from the diagram, m / s 2;
V D4 - the value of the speed from the plan, m / s;
V pp D4 - the value of the speed from the diagram, m / s.
5. Kinetostatic analysis of the mechanism
The purpose of the kinetostatic analysis is to find the forces of inertia and to determine the reactions in kinematic pairs.
From the first sheet of drawings, we transfer the plan of the mechanism in the first position, and also transfer the plan of accelerations of this position and the plan of speeds turned 90 0 counterclockwise.
Determination of the weight of the links of the mechanism
The weight of the links is determined by the formula
G i = m i ∙ g, (5.1)
where g is the acceleration due to gravity, g = 9.81 m / s 2.
The obtained values are summarized in table 6.
Table 6 - Weight and mass of links
| Parameter |
|||||
| Weight, kg |
|||||
Determination of moments of inertia forces and inertia forces of links
Let's find the force of inertia of each link separately.
Force Ф И is directed opposite to the full acceleration of point S and can be determined by the formula
where m is the mass of the link, kg;
and S is the acceleration of the center of mass of the link, m / s 2.
Substituting numerical values, we get Ф 1 = Ф 2 = 0,
The moment of inertia M And a pair of inertial forces is directed opposite to the angular acceleration e of the link and can be determined by the formula
where is the moment of inertia of the link relative to the axis passing through the center of mass S and perpendicular to the plane of the link movement, kg ∙ m 2,
Determine the angular acceleration by the formula
Substituting the numerical values in the formulas (5.3-5.4), we get the values that we will enter in table 6.
Table 6 - Moments of inertia forces and inertia forces of links
| The quantities |
|||||
Determination of points of application of forces
Consider the groups of asura separately each to find reactions. The calculation will be carried out with the latter. For rotational pairs, the reactions are divided into two - parallel and perpendicular. Let us direct the force of useful resistance against the forces of inertia.
Determination of reactions in a kinematic pair
We start the calculation with the last structural group. We draw a group of links 4 and 5, transfer all external loads and reactions to this group. We consider this group to be in equilibrium and compose the equilibrium equation
The value is decomposed into two components: normal and tangential.
(5.6)
The value is found from the equilibrium condition relative to the point D for the fourth link.
where, h 1, are the shoulders of forces to point D, determined from the drawing m.
(5.8)
We build a plan of forces, from where we determine the values,. We obtain the following values, taking into account the scale of forces m F = 10 N / mm:
Taking into account that the slider can also be considered separately, we get that the force is applied, etc., since the distance b = 0. We define directions.
Similarly, we compose the equilibrium equation for the second Asura group.
We are not looking for the reaction of the slider 2 to the rocker arm, because it is not that important.
We build a power polygon, from where we determine the unknown reactions. We get the following values, taking into account the scale of forces:
Determination of the balancing force
We draw the leading link and apply the acting loads. To keep the system in equilibrium, we introduce a balancing force, which is applied at point A perpendicular to the link AO. The diagram shows that the balancing force is equal to the reaction
Determination of the balancing force by the Zhukovsky method
We rotate the speed plan of the mechanism by 90 ° and apply acting forces and forces of inertia to it. Then we compose the equilibrium equation, considering the plan of velocities as a rigid body, relative to the pole.
Substituting the numerical values, we get
Determine the error in calculating the balancing force using the plan of forces method and Zhukovsky's method according to the formula
(5.11)
Substituting numerical values, we get
Conclusion
In this course work, the analysis of the crank-slider mechanism was carried out.
In the literature review, we got acquainted with the principles of operation of various mechanisms. As a result of the analysis, the following types of research were carried out: structural, kinematic, kinetostatic and gearing synthesis.
In the course of the structural analysis, the structure and degree of mobility of the mechanism were determined.
In kinematic analysis, velocities and accelerations were determined using two methods: the method of plans and the method of graphical differentiation. The velocities and accelerations of point D for the first position turned out to be equal to 0.28 m / s, 0.27 m / s and 5.89 m / s 2.5.9 m / s 2, respectively, errors - 2.1% and 1, 2%. For the seventh position, the speeds and accelerations are equal to 0.5 m / s, 0.5 m / s and 8.6 m / s 2, 8.5 m / s 2, the errors were 0% and 2.3%. For the tenth position, the speed and acceleration turned out to be 2.05 m / s, 1.98 m / s and 3.6 m / s 2, 3.7 m / s 2, the errors are 2.3% and 2.6%. It can be argued that the calculations were performed correctly, since the error for speeds does not exceed 5%, and for accelerations less than 10%.
In kinetostatic analysis, a force calculation was carried out by two methods. We used the method of plans of forces and the method of Zhukovsky. According to the method of plans of forces, F UR turned out to be equal to 910 N, and according to the method of Zhukovsky - 906 N, the error was 2.3%, which does not exceed the permissible norms. It can be concluded that the force plans method is more laborious than the Zhukovsky method.
List of sources used
1 Artobolevsky I.I. Theory of Mechanisms and Machines: Textbook. - 4th ed., Add. pererab.-M.: Nauka, 1988.-640 p.
2 Korenyako A.S. Course design on the theory of mechanisms and machines: -5th ed., revised .- Kiev: Vishcha school, 1970.- 332 p.
3 Kozhevnikov S.N. Theory of mechanisms and machines: Textbook.- 4th ed., Revised.-M.: Mechanical engineering, 1973.-592 p.
4 Marchenko S.I. Theory of mechanisms and machines: Lecture notes. - Rostov n \ D: Phoenix, 2003 .-- 256 p.
5 Kulbachny OI .. Theory of mechanisms and machines design: Textbook.-M .: Higher school, 1970.-228
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 grade.
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.
Consequently, 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. Consequently, p 5 = 4, but 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 lengths, we translate all natural dimensions of the links into segments, with the help of which we will build a 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:
. (2)
Velocity vector action line
is perpendicular to the axis of the crank 1, and the direction of action of this vector coincides with the direction of rotation of the crank.Speed module point A:
, (3)
- the angular velocity of the link OA; is the length of the OS. Angular Velocity
Given (Figure 2.10): j 1, w 1 = const, l BD, l DC, l AB, l BC, m l [ MMM ] .
Speed V B= w 1 l A B point B is directed perpendicular to link AB in the direction of its rotation.
To determine the speed of point C, we compose a vector equality:
C = B + SV
The direction of the absolute speed of point C is known - parallel to the line xx. The speed of point B is known, and the relative speed V C B is directed perpendicular to the link BC.
We build a plan of speeds (Fig. 2.11) in accordance with the above equation. Moreover, m n = V B / Рв[m / s mm ].
Point B's absolute acceleration is equal to normal acceleration a p VA(since w 1 = const, e 1 = 0 and but t B = 0) a B = a p VA = w 2× l VA[m / s 2]
and is directed along the AB link from point B to point A.
Scale factor of the acceleration plan m a = a B / p in[m / s mm], where p in- a segment of arbitrary length, representing acceleration on the plan a B.
Acceleration of point C:
(1 way),
where a p CB = V 2 CB / l CB[m / s 2]
A segment depicting this acceleration in the acceleration plan:
n sv = a n sv / m but[mm ]
We choose the pole p of the acceleration plan. Draw a line from the pole along which the acceleration is directed a B(// AB) and postpone the selected segment p in depicting this acceleration on the plan (Fig. 2.12). From the end of the resulting vector, draw a line of direction of the normal component a n SV parallel to the SV link and postpone the segment n sv depicting on scale m but this is normal acceleration. From the end of the normal acceleration vector, draw a line of direction of the tangential component a t CB, and from the pole p - the direction of the absolute acceleration of point C ( ïï xx). At the intersection of these two directions, we get point C; in this case, the vector pС represents the required acceleration.
The module of this acceleration is:
a C = ( p with) m but[m / s 2]
The angular acceleration e 2 is determined as:
e 2 = a t CB / l CB= (t CB) m a / l CB[1 / s 2]
Direction e 2 shown in the diagram of the mechanism.
To find the speed of point D, you need to use similarity theorem, which is used to determine the speeds and accelerations of the points of one link, when the speeds (accelerations) of the other two points of this link are known: the relative speeds (accelerations) of the points of one link form on the plans of speeds (accelerations) figures similar to the figure of the same name in the mechanism diagram. These figures are similarly located, i.e. when reading letter designations in one direction on the mechanism diagram, letters on the speed (acceleration) plan follow in the same direction.
To find the speed of point D, you need to build a triangle, similar to the triangle on the mechanism diagram.
Triangles D cvd(on the plan of speeds) and DCBD (on the plan of the mechanism) are triangles with mutually perpendicular sides. Therefore, to construct a triangle D cvd draw perpendiculars to CD and BD from points with and in respectively. At their intersection, we get a point d, which we connect to the pole.
The acceleration of point D is also determined by the similarity theorem, since the accelerations of the other two points of link 2 are known, namely but In and but C. It is required to construct a triangle D in cd, similar to the triangle DBCD in the mechanism diagram.
To do this, we first build it on the mechanism diagram, and then transfer it to the acceleration plan.
Line segment " sun»The plan of accelerations is transferred to the segment of the same name CB on the mechanism diagram, putting it on the CB link from any point (C or B) (Fig. 2.10). Then along the segment “ sun»Triangle D is constructed on the mechanism in dс, similar to the triangle DBDC, for which a straight line "dс" is drawn from the point "C", parallel to the straight line DC, until it intersects with the straight line BD. We get D in dc ~ DBDC.
The obtained sides of the triangle r 1 and r 2 are equal in size to the sides of the desired
|
|
|
| |
triangle on the acceleration plan, which can be built using serifs (Figure 2.12). Next, you need to check the similarity of the location of the figures. So, when reading the letter designations of the vertices of the DBDC triangle on the clockwise mechanism diagram, we get the order letters B-D-C; on the plane of accelerations in the same direction, i.e. clockwise, we should get the same letter order in-d-c. Therefore, the solution is satisfied by the left intersection point of the circles r 1 and r 2.
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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 …………………………
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The condition for the existence of a crank in four-link mechanisms
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Application of Grashof's theorem to a kinematic chain with a translational pair
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Consider a crank mechanism, in which the line of movement
the slider is offset 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
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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
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I II III
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Rack and pinion
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1. Structural analysis of the 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
Mechanism structure formula
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 length of the link;
Determine the position of point M on the AB link from the ratio
; INM= 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 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 it makes instant translational 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. AB crank with a length of l = 0.29 m rotates with a constant angular velocity n1 = 82 rpm
The crank-slider 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 is the actual value of the slide travel, 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 you get a diagram of the displacement 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 chosen 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 postpone the segment l = 120mm, depicting the time T of one complete revolution of the 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.
