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
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
; VM= 0.18 x 1.15 = 0.207 m;
3 Building a plan of the crank-slider mechanism
To build a plan for 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 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 VВ, VС 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 ВС 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. Crank AB 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), 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 set aside vertical segments from these points: 1-1S = B0B1, 2-2S = B0B2, etc. These distances increase to the extreme right position of the slider B, and from position B decreases. If points 0s, 1s, 2s ... 12s are connected in series with a curve, then we get a diagram of the displacement of point B.
To plot speed and acceleration diagrams, use the method of graphical differentiation. 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 to the 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 the 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:
- the increasing graph of displacements (speed) corresponds to positive values of the graph of speed (equations), and decreasing - 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 equal to 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 graph of 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, speeds and accelerations were built. Found that there is some error due to the construction and rounding in the calculations.
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 volume 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. V 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. V kinematic analysis the speeds and accelerations of points of the mechanism are determined by the methods of plans and kinematic diagrams. V 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, minimal dimensions and weight, ease of use and cost-effectiveness.
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 automatic machines.
The analysis of the mechanism consists in the study of the kinematic and dynamic properties of the mechanism according to a 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
When studying 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 maintenance and the creation of the best conditions for the operating 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 solution to these problems at the initial design stage consists in the analysis and synthesis of the designed machine, as well as in the development of its kinematic diagram s, providing 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 purposes, 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 diagrams 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-slider 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, harvesters) 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 design methods special machines, in that TMM focuses on the study of methods of synthesis and analysis, common to a given type of mechanism, independent of 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 yoke 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 speed of the driven shaft.
A frictional mechanism is a mechanism in which the transfer of rotational motion between the links forming 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 swinging movement 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;
P 5 - the number of kinematic pairs of the V class, P 5 = 7;
P 4 - the number of kinematic pairs of class IV, P 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:
3. Kinematic analysis gear transmission
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 ratio i Н3 = 10. The gear wheels installed after the planetary gearbox have the following numbers of teeth: z 4 = 12, z 5 = 28.
Figure 2 - Drive linkage
Ratio gear wheels 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;
ОА - the length of the ОА crank in the drawing, mm.
Substituting the data, we get
m l = 
The procedure for constructing a plan of provisions 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◦;
- we determine the position of point C at certain positions of the OA crank;
- we postpone the segment of the CD on a scale in such a way that the point D lies on the line of the ATS;
- 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 distances traversed 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 of the complete revolution of the crank OA, 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 pole P, since the speed of 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 located 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 velocities from the graphical construction, taking into account their conversion to their actual size, by measuring the length of the vectors corresponding to the velocities and multiplying them by the scale of the velocity 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 plotting 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 Velocities links and normal acceleration
| 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;
- the value of the 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 the 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 the 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:
Considering that the slider can also be considered separately, we get that the force is applied in t.D, 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 so 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 according to 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, an 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 in comparison with 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
Perm State Technical University
DEPARTMENT "Mechanics of Composite Materials and Structures".
COURSE PROJECT
ON THEORYMECHANISMS AND MACHINES
Topic:
Exercise:
Option:
Completed: group student
Checked: Professor
Poezzhaeva E.V.
Perm 2005
Structural analysis of the mechanism ……………………………………………… 3
Kinematic analysis of the mechanism ………………………………………… ..4
Kinetostatic analysis of the mechanism …………………………………….… 9
Flywheel calculation …………………………………………………… ............ 12
Cam profiling ………………………………………………… 17
Gear design ……………………………………… ... 20
Instructions for performing calculations for the course project on TMM …… .23
References ………………………………………………………… ... 24
Structural analysis 3 crank-slide mechanism
1. Let's represent block diagram mechanism
OA - crank - rotates;
AB - connecting rod - makes a plane-parallel movement;
B - slider - makes a forward motion.
2. Let us find the degree of mobility of the mechanism according to the Chebyshev formula:
3. Let us decompose into Assur structure groups
4. Let's write structural formula mechanism I => II 2 2
5. Let's define the class, the order of the whole mechanism.
The investigated mechanism consists of a mechanism of the first class and a structural group of the second class of the second order (connecting rod and slider), therefore, the OAV hydraulic pump is a mechanism second class second order.
Kinematic analysis of the mechanism
Initial data: 
OA = m, 
AB = mm
Kinematic analysis solves three problems:
the task of the provisions;
speed problem;
acceleration problem.
Provisions problem
Designing a crank-slider mechanism, Let us find the extreme positions of the mechanism: the beginning and end of the working stroke. We will find the beginning of the working stroke by the formula:
l is the length of the OA crank
g - the length of the connecting rod AB
We find the end of the working stroke by the formula:
Working stroke
S = S "- S" = 2r [m];
Let's build a mechanism to scale
1
=
AB / OA = [m / mm]
Let's find the length AB:
AB = 
AB / 1 = [mm]
Let's show the movement of points in twelve positions of the mechanism. To do this, divide the circle into 12 equal parts (using the serif method).
Let's build the connecting rod curve. To do this, find the center of gravity of each link and connect it with a smooth line.
Machine position plans are used to determine speeds and accelerations at given positions.
Velocity problem
The kinematic analysis is performed by the graphic-analytical method, which reflects the visibility of the speed changes and provides sufficient accuracy. Leading link speed:
[ms -1]
Let's write the vector equations:
V B = V A + V AB; V B = V X + V BX
where V X = 0; V A OA; V AB AB; V BX BX
The values of the vectors V BA, V B, V S 2 will be determined by construction. Let's choose the scale of the plan of velocities
[ms -1 / mm].
Ge pa - a segment characterizing the value of the speed in the drawing = mm. From an arbitrary point p - the pole of the plan of velocities, we postpone the vector pa,
perpendicular OA. Draw a straight line through m. A perpendicular to AB. The point of intersection of the x-axis (selected in the directions of m. C) with this straight line will give m. C, connecting m. C with the pole, we obtain the velocity vector m. C. Determine the magnitude of the speed t. In:
[ms -1]
The position of the t. On the plan of speeds is determined from the proportion:
Connecting point S 2 with the pole p, we obtain the magnitude and direction of the speed of point S2:
[ms -1]
[ms -1]
Let's define:
[ms -1]
[ms -1]
[ms -1]
Let's define:
[s -1]
The direction 2 is determined by the transfer of the vector vba to point B relative to point A.
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Parameter |
Mechanism position |
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