Given (Figure 2.10): j 1, w 1 \u003d const, l BD, l DC, l AB, l BC, m l [m / mm ] .
Speed V B\u003d w 1 l A Bpoint 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 \u003d 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 \u003d V B / Рв[m / s mm ].
Point B's absolute acceleration is equal to normal acceleration a p VA(since w 1 \u003d const, e 1 \u003d 0 and a t B \u003d 0) a B \u003d a p VA \u003d 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 \u003d a B /p in[m / s mm], where p in - a segment of arbitrary length, representing acceleration on the plan a B.
Point C acceleration:
(1 way),
where a p CB \u003d V 2 CB / l CB[m / s 2]
A segment representing this acceleration in the acceleration plan:
n sv \u003d a n sv /m a[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 indepicting 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 SVparallel to the SV link and postpone the segment n svdepicting on scale m athis 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 - direction of 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 \u003d (p from)m a[m / s 2]
The angular acceleration e 2 is determined as:
e 2 = a t CB / l CB= (t CB)m a / l SV[1 / s 2]
Direction e 2 shown in the diagram of the mechanism.
To find the speed of point D, you must 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 on the mechanism diagram. These figures are similarly located, i.e. when reading letter designations in one direction on the mechanism diagram, letters on the plan of speeds (accelerations) 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 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 a In and a C. It is required to construct a triangle D on the acceleration plan incd, similar to 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 indс, 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 indc ~ DBDC.
The obtained sides of the triangle r 1 and r 2 are equal in size to the sides of the desired
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a 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. Consequently, 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 …………………………
Articulated four-link
Given (fig.2.6): j1, w1 \u003d const, l1, l2, l3, lo \u003d lAD, ml [m / mm].
Crank mechanism
Given (fig.2.13): j1, w1 \u003d const, l1, l0 \u003d lAC, ml [m / mm]. Point B belonging to the first s
Kinematic synthesis of flat linkages
Kinematic synthesis - This is the design of the scheme of the mechanism according to its specified kinematic properties. When designing mechanisms, primarily on the basis of experience, in relation to
The condition for the existence of a crank in four-link mechanisms
The conditions for the existence of a crank in four-link mechanisms are determined by the Grashof theorem: if in a closed hinged four-link kinematic chain, the sum of the lengths
Application of Grashof's theorem to a kinematic chain with a translational pair
By increasing the size of the rotational pairs, it is possible to obtain translational pairs by expanding the pins. The size of the pivot pin D (Figure 2.19, b) can be taken large
Consider a crank mechanism, in which the line of movement
the slider is offset from the center of rotation of the crank. The "e" value is called offset or deaxial. Determine at what ratio of sizes
Crank mechanism
Consider two options for the rocker mechanism: with a swinging and a rotating rocker. To obtain a swing arm mechanism, it is necessary that the length of the rack is greater than the length of the crank,
Articulated four-link
Consider a hinged four-link link (Fig. 2.27), which is in equilibrium under the action of the given moments: driving МДв on the driving link 1 and the moment of resistance
Synthesis of four-link linkage mechanisms by link positions
Four-link mechanisms are often used to transfer various items from position to position. In this case, the carried object can be associated with both a connecting rod, that
Dynamic analysis and synthesis of mechanisms
The purpose of the dynamic research is to obtain the law of motion of the mechanism (its links) depending on the forces acting on it. When solving this problem, we will consider
I II III
I - the first link makes a rotational movement; II - link 2 makes a complex movement; III - link 3 moves progressively. To determine
Rack and pinion
If the center of one of the wheels is removed from infinity, then its circles will be transformed into parallel lines; point N1 of tangency of the generating line (it is also the common normal and
1. Structural analysis of the mechanism
1.1 Determination of the degree of movement 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 Definition of the mechanism class
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 \u003d 82 rpm
Slider stroke S \u003d 0.575 m
The ratio of crank length to connecting rod length
Eccentricity to Crank Length Ratio
2.3 During one revolution of the crank s;
The slider will cover the distance S, at S \u003d 2AB
Determine the length of the link;
Determine the length of the link;
Determine the position of point M on link AB from the ratio
; INM\u003d 0.18 x 1.15 \u003d 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 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 speed, 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. Building 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 \u003d 0.29 m rotates at a constant angular velocity n1 \u003d 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 \u003d 2r - the actual value of the slide, equal to twice the value of the crank.
- slider stroke 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 \u003d B0B1, 2-2S \u003d B0B2, etc. These distances increase to the extreme right position of the slider B, and from position B they decrease. If points 0s, 1s, 2s ... 12s are connected in series with a curve, then we get a diagram of the displacement of point B.
For plotting speed and acceleration diagrams, the graphical differentiation method 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 \u003d 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 on the V axis the segments: 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:
The acceleration diagram is built in the same way as the velocity diagram. Acceleration scale
Where Ha \u003d 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, use the properties of the derivative:
- an increasing graph of displacements (speed) corresponds to positive values \u200b\u200bof the graph of speed (equations), and a decreasing one - negative;
- the point of maximum and minimum, i.e. the extreme values \u200b\u200bof the displacement (velocity) graph correspond to zero values \u200b\u200bof the velocity (acceleration) graph;
- the inflection point of the displacement (velocity) graph corresponds to the extreme values \u200b\u200bof 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 end of the period of any kinematic diagram are equal, and the tangents drawn at these points are parallel.
To plot the movement of the slider B, select the coordinate axes s, t. On the abscissa axis, we put the segment l \u003d 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 the 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.
Perm State Technical University
DEPARTMENT "Mechanics of Composite Materials and Structures".
COURSE PROJECT
ON THEORYMECHANISMS AND MACHINES
Theme:
The task:
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
Design gear transmission………………………………………...20
Instructions for performing calculations for the course project on TMM …… .23
References ………………………………………………………… ... 24
Structural analysis 3 crank-slide mechanism
1. Let's depict 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 us write down the structural formula of the mechanism I \u003d\u003e 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 \u003d m, 
AB \u003d mm
Kinematic analysis solves three problems:
the task of the provisions;
speed problem;
acceleration problem.
Provisions problem
Designing a crank-slider mechanism, Let's find the extreme positions of the mechanism: the beginning and end of the working stroke. We find the beginning of the working stroke by the formula:
l - length of the OA crank
g - connecting rod length AB
We find the end of the working stroke by the formula:
Working stroke
S \u003d S "- S" \u003d 2r [m];
Let's build a mechanism to scale
1
=
AB / OA \u003d [m / mm]
Find the length AB:
AB \u003d 
AB / 1 \u003d [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 construct the connecting rod curve. To do this, we find the center of gravity of each link and connect it with a smooth line.
Machine position plans are used to determine the speeds and accelerations at given positions.
Speed \u200b\u200bproblem
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 \u003d V A + V AB; V B \u003d V X + V BX
where V X \u003d 0; V A OA; V AB AB; V BX BX
The values \u200b\u200bof the vectors V BA, V B, V S 2 will be determined by construction. Let's choose the scale of the speed plan
[ms -1 / mm].
Ge pa - a segment characterizing the value of the speed in the drawing \u003d mm. From an arbitrary point p - the pole of the speed plan, 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 point on the plan of speeds is determined from the proportion:
By 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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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 course design is a crank-slider mechanism. In the course work, a study of the crank-slider mechanism was made. Structural, kinematic, kinetostatic analyzes were carried out.
In the structural analysis, the composition of the crank-slide mechanism was determined. IN kinematic analysis the speeds and accelerations of 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 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 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 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 consists in designing 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 set forth 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 conditionsensuring 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 causing 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 diagrams, providing with sufficient approximation the reproduction of the required law of motion.
To perform 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 increasingly used. A set of automatic machines, interconnected and designed to perform a certain technological processis 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 the 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, it is possible to apply general methods based on the basic principles of modern mechanics in the study of mechanisms with different functional purposes.
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, which 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 at a given
specific or chosen law;
5) as flexible links transmitting motion from one rigid body in the mechanism to another, are practically used 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 provided 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: \u200b\u200bto 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 crucial 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 basic mechanism in all piston (engines internal combustion, compressors, pumps, gas expansion machines), agricultural (mowers, cutters, combines) and forging machines and presses.
For 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 for 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 structurally turned 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 angle of rotation of the crank. This mechanism, also called a yoke sinus mechanism, is used in small piston pumps and compressors, devices for the harmonious movement of the slider or for determining the values \u200b\u200bproportional 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, frictional, 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 is 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 back 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 roar 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 \u003d 3n - 2P 5 - P 4, (2.1)
Where n is the number of moving links for the mechanism, n \u003d 5;
Р 5 - the number of kinematic pairs of the V class, Р 5 \u003d 7;
Р 4 - the number of kinematic pairs of class IV, Р 4 \u003d 0.
Substituting numerical values, we get:
W \u003d 3 5 - 2 7 - 0 \u003d 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 I.I.Artobolevsky's classification, we divide the mechanism under study into structural groups. The screening mechanism (Figure 1) consists of a leading link 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 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 \u003d 10. The gear wheels installed after the planetary gear have the following numbers of teeth: z 4 \u003d 12, z 5 \u003d 28.
Figure 2 - Lever mechanism drive
Gear 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 \u003d 3.5 mm; m II \u003d 2.5 mm; the center distance of the gears - a w \u003d 72 mm; angular speed of the drive shaft (motor shaft) - ω d \u003d 150.00 rad / s. Let us determine the angular velocity of the driving link of the screening mechanism - ω 1 by the formula:
ω 1 \u003d ω d / i 15, (3.3)
ω 1 \u003d 150 / 23.33 \u003d 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-slide 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 \u003d 
The procedure for constructing a plan of provisions this mechanism:
- mark in the drawing the position of the centers of rotation of the crank t. O and the rocker mechanism t. C;
- outline the trajectories of 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 will 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 point D lies on the ATS straight line;
- using the serif method, we determine the position of point D at certain positions of the crank OA;
- clockwise put the crank OA 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 leftmost position to the rightmost position corresponding to a given moment in time. Using the obtained points, we construct a displacement diagram φ \u003d 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 full turn 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 \u003d 
Graphical differentiation of the diagrams of speeds and accelerations of the output link. Choosing an arbitrary pole distance H v \u003d (40 ... 60 mm) \u003d 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 obtained 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 movement of the link OA. Let's define it from the formula:
V A 1 \u003d 
To build plans of speeds, we will choose the positions of the mechanism: the first, seventh and tenth. For all positions, the construction is similar, so we describe the construction algorithm. Let's define 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 plan of speeds in the direction of rotation of the crank perpendicular to OA, we postpone the velocity vector (Pa 1), conventionally taking its length equal to 80 mm. Then we determine the scale of the speed plan:
m V \u003d 
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, it can be found on the plan of velocities 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, 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 proportion (4.11) the position of 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, measuring the length of the vectors corresponding to the speeds and multiplying them by the scale of the speed plan:
Table 2 - Actual values \u200b\u200bof the speeds of the characteristic points of the linkage mechanism in three positions
| Mechanism position |
Point speed |
Vector length from the plan (pn), mm |
||
Building acceleration plans for three positions
Let us compose a system of vector equations for the acceleration of the lever 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 construction in the same way.
We begin the construction of the plan by constructing the acceleration of point A. Let us postpone it on a scale from the pole P, and the direction of the vector from A to O. Determine the scale of accelerations, taking arbitrarily in the drawing the length of the acceleration a 1 \u003d 80 mm:
m a \u003d 
Let us determine the angular velocities of the ABC and SD links. We find their values \u200b\u200bby the 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 \u200b\u200bin table 3.
Table 3 - Angular Velocities links and normal acceleration
| Position |
Speed |
Value, m / s |
Normal acceleration |
Value, |
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 \u200b\u200bare summarized in Table 4. It is deposited in the direction of rotation 90 o from the velocity vector. The relative velocity 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 \u200b\u200bof speeds and accelerations by the formulas:
(4.20)
(4.21)
where is the value of 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 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 \u003d m i ∙ g, (5.1)
where g is the acceleration due to gravity, g \u003d 9.81 m / s 2.
The obtained values \u200b\u200bare 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 \u003d Ф 2 \u003d 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 \u200b\u200bin the formulas (5.3-5.4), we get the values \u200b\u200bthat 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 decomposed into two - parallel and perpendicular. We direct the force of useful resistance against the forces of inertia.
Determination of reactions in kinematic pair
We start the calculation with the last structural group. We draw a group of links 4 and 5, we 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)
We find the value 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 \u003d 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 \u003d 0. We determine the 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. For the system to be 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 inertial forces to it. Then we compose the equilibrium equation, considering the velocity plan as a rigid body, relative to the pole.
Substituting numerical values, we get
Determine the error in calculating the balancing force using the plan of forces method and Zhukovsky's method using 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 the synthesis of gearing.
In progress structural analysis determine the structure and degree of mobility of the mechanism.
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 .. The theory of mechanisms and machines design: Textbook.-M .: Higher school, 1970.-228
