Types of clamping devices and their calculation. Mechanical engineering textbooks Cam mechanisms. Types of cam mechanisms. Advantages and disadvantages. Main purpose
FEDERAL AGENCY FOR EDUCATION OF THE RUSSIAN FEDERATION
STATE EDUCATIONAL INSTITUTION OF HIGHER PROFESSIONAL EDUCATION
"TYUMEN STATE OIL AND GAS UNIVERSITY"
INSTITUTE OF TRANSPORT
department Machine parts
OVERVIEW OF THE MAIN TYPES OF MECHANISMS
METHODOLOGICAL INSTRUCTIONS
To practical training By Theories of mechanisms and machines for students of specialties NR-130503 PST-130501 NB-130504 01, MSO- 190207
full-time and part-time full and reduced forms of education
Tyumen 2007
Approved by the editorial and publishing council
Tyumen State Oil and Gas University
Compiled by: Associate Professor, Ph.D. Zabanov Mikhail Petrovich
professor, d.t.s. Babichev Dmitry Tikhonovich
assistant, Pankov Dmitry Nikolaevich
© State educational institution of higher professional education
"Tyumen State Oil and Gas University"
In the course of the lesson, it is necessary to familiarize yourself with the main groups and types of mechanisms, their graphic images. Learn to represent a real mechanism in the form of a diagram.
The report must depict and describe the classic types of mechanisms.
Mechanical engineering is the leading branch of modern technology. The progress of mechanical engineering is determined by the creation of new high-performance and reliable machines. The solution of this most important problem is based on the complex use of the results of many scientific disciplines and, first of all, the theory of mechanisms and machines.
With the development of machines, the content of the term "machine" has changed. For modern machines, we give the following definition: A machine is a device created by a person for converting energy, materials and information in order to facilitate physical and mental labor, increase its productivity and partially or completely replace a person in his labor and physiological functions.
According to the functions performed by machines, they are divided into the following classes:
1) Energy machines
2) Transport vehicles
3) Technological machines
4) Control and control machines
5) Logic machines
6) Cybernetic machines
The definition of the term "mechanism" has been repeatedly changed as new mechanisms have appeared.
A mechanism is a system of bodies designed to convert the movement of one or more rigid bodies into the required movements of other bodies. If, in addition to solid bodies, liquid or gaseous bodies are involved in the transformation of motion, then the mechanism is called hydraulic or pneumatic, respectively. In terms of functionality, mechanisms are divided into the following types:
1) Mechanisms of motors and converters
2) Gears
3) Actuators
4) Management, control and regulation mechanisms
5) Mechanisms for feeding, transporting and sorting processed products and objects
6) Mechanisms for automatic counting, weighing and packaging of finished products
The main feature of the mechanism is the transformation of mechanical movement. The mechanism is part of many machines, since the transformation of energy, materials and information usually requires the transformation of the movement received from the engine. It is impossible to equate the concepts "machine" and "mechanism". Firstly, in addition to the mechanisms in the machine, there are always additional devices associated with the control of mechanisms. Secondly, there are machines in which there are no mechanisms. For example, in recent years, technological machines have been created in which each executive body is driven by an individual electric or hydraulic motor.
When describing the mechanisms, they were divided into separate groups based on their design (lever, cam, friction, gear, etc.)
Mechanisms are formed by successive attachments of links to the initial mechanism.
LINK - one or more parts fixedly connected to each other, included in the mechanism and moving as one.
INPUT LINK - a link to which the movement is reported, which is converted by the mechanism into the required movements of other links. The input link is connected to the engine or to the output link of another mechanism.
OUTPUT LINK - a link that makes a movement for which the mechanism is intended. The output link is connected to the actuating device (working body, instrument pointer), or to the input link of another mechanism.
The links are connected to each other movably by means of kinematic pairs: rotational (hinge) and translational (slider).
TRAJECTORY points(link) - the line of movement of the point in the plane. It can be a straight line or a curve.
LINKAGES
Lever mechanisms are mechanisms that include rigid links interconnected by rotational and translational kinematic pairs. The simplest lever mechanism is two-link mechanism, consisting of a fixed link-rack 2 (Fig.1.1 ) and a movable lever 1 , which has the ability to rotate around a fixed axis (usually this is the initial mechanism).
Fig.1.1 Two link linkage
TO two-link lever mechanisms include the mechanisms of many rotary machines: electric motors, bladed turbines and fans. The mechanisms of all these machines consist of a rack and a link (rotor) rotating in fixed bearings.
More complex lever mechanisms are mechanisms consisting of four links, the so-called four-link mechanisms.
On Fig.1.2 shows the mechanism of a hinged four-link, consisting of three movable links 1, 2, 3 and one fixed link 4. Link 1 connected to the rack, can make a full turn and is called a crank. Such an articulated four-link, which has one crank and one rocker in its composition, is called crank-rocker mechanism, where the rotational motion of the crank by means of the connecting rod is converted into the rocking motion of the rocker. If the crank and the connecting rod are extended in one line, then the rocker arm will take the extreme right position, and when superimposed on each other, it will take the left position.
Rice. 1.2 Hinged four-link mechanism
An example of such a mechanism is the mechanism shown in Fig.1.3 , where is the link 1 – crank (input link), link 2 - connecting rod, link 3 - rocker. Point M S moving along a curve describes a trajectory . Some trajectories can be reproduced by lever mechanisms theoretically exactly, others - approximately, with a degree of accuracy sufficient for practice.
The mechanism under consideration, called the symmetrical Chebyshev mechanism, is often used as a circular guide mechanism, in which AB = BC = BM = 1. With the indicated relationships
Rice. 1.3 Crank-rocker mechanism
dot M connecting rod AB describes a trajectory symmetrical about the axis n - p . The angle of inclination of the axis of symmetry to the line of centers CO is determined by: РМСО = π - Ω / 2. Part of the trajectory of the point M is an arc of a circle with a radius of O 1 M, which can be used in mechanisms with stopping the output link.
Another example of a four-link is a widespread in technology slider-crank mechanism (Rice. 1.4 ).
Rice. 1.4 Crank-slider mechanism
In this mechanism, instead of a rocker arm, a slider is installed, moving in a fixed guide. This crank mechanism is used in reciprocating engines, pumps, compressors, etc. If the eccentricity e is equal to zero, then we get a central crank-slider mechanism or an axial one. At e non-zero crank-slider mechanism is called non-central or deaxial. Here, the rotation of the crank OA through the connecting rod AB is converted into a reciprocating motion of the slider. Naturally extreme positions of the slider , will be when the crank and connecting rod are in one line.
If in the considered mechanism we replace the fixed guide with a movable one, which is called the backstage, then we get four link rocker mechanism with rocker stone. An example of such a mechanism is the rocker mechanism of a planer ( Fig.1.5 ). Crank 1 , rotating around the axis, through the rocker 2 makes backstage 3 make a swinging motion. In this case, the rocker stone moves back and forth relative to the rocker.
Rice. 1.5 Four link rocker mechanism
The extreme positions of the backstage will be with the crank perpendicular to it. It is easy to build such positions: a circle is drawn with a radius equal to the length of the crank (the trajectory of the point A), and tangents are drawn from the axis of rotation of the backstage.
Thus, the links can make progressive , rotational or complex movement.
1. Crank-slider mechanism.
a) central (Fig. 1);
b) off-axis (deoxy) (Fig. 2);
e - eccentricity
Rice. 2
1-crank, because the link makes a complete revolution around its axis;
2-rod, not connected to the rack, makes a flat movement;
3-slider (piston), performs translational motion;
1 - crank;
2 - the backstage stone (sleeve), together with star 1, makes a complete revolution around A (w1 and w2 are the same), and also moves along star 3, causing it to rotate;
3 - rocker (scene).
4.Hydraulic cylinder
(kinematically similar to the rocker mechanism).
During the design process, the designer solves two problems:
· analysis(explores ready mechanism);
· synthesis(a new mechanism is being designed according to the required parameters);
Structural analysis of the mechanism.
Concepts about kinematic pairs and their classification.
Two links fixedly interconnected form a kinematic pair. All kinematic pairs are subject to two independent classifications:
1. Pairs are higher or lower:
a. Higher pairs are pairs in which contact is made along the line.
b. Lower pairs are pairs in which contact is made along the surface.
2. All pairs are divided into five classes, depending on the number of bonds imposed on the mobility of each of the links. The number of degrees of freedom is indicated by . The number of imposed connections is denoted by . In this case, the number of degrees of mobility can be determined by the formula: .
a. First class pair: ; .
b. Second class pair: ; .
c. Third class pair: ; .
d. Couple of the fourth class: ; .
e. Fifth grade couple: ; .
Examples of pair classification:
Consider the kinematic pair "screw-nut". The number of degrees of mobility of this pair is 1, and the number of imposed bonds is 5. This pair will be a pair of the fifth class, you can choose only one type of movement for a screw or nut, and the second movement will be accompanying.
Kinematic chain– links interconnected by kinematic pairs of different classes.
Kinematic chains are spatial and flat.
Spatial kinematic chains- chains, the links of which move in different planes.
Flat kinematic chains- chains, the links of which move in one or parallel planes.
Concepts about the degree of mobility of kinematic chains and mechanisms.
The number of links freely floating in space is denoted as . For links, the degree of mobility can be determined by the formula: . We form a kinematic chain from these links, connecting the links with each other in pairs of different classes. The number of pairs of different classes is denoted by , where is the class, that is: is the number of pairs of the first class, which has , and ; - the number of pairs of the second class, which has , and ; - the number of pairs of the third class, which has , and ; - the number of pairs of the fourth class, which has , and ; - the number of pairs of the fifth class, for which , and . The degree of mobility of the formed kinematic chain can be determined by the formula: .
We form a mechanism from the kinematic chain. One of the main features of the mechanism is the presence of a rack (case, base), around which the remaining links move under the action of the leading link (links).
The degree of mobility of the mechanism is usually denoted as. We turn one of the links of the kinematic chain into a rack, that is, we take away all six degrees of mobility from it, then: - the Somov-Malyshev formula.
In a flat system, the maximum number of degrees of freedom is two. Therefore, the degree of mobility of a flat kinetic chain can be determined by the following formula: . The degree of mobility of a flat mechanism is determined by the Chebyshev formula: , where is the number of moving links. Using the definition of higher and lower kinematic pairs, the Chebyshev formula can be written as follows: .
An example of determining the degree of mobility:
Classification of mechanisms
The number of types and types of mechanisms is in the thousands, so their classification is necessary to select one or another mechanism from a large number of existing ones, as well as to synthesize the mechanism.
There is no universal classification, but 3 types of classification are most common:
1) functional. According to the principle of the technological process, the mechanisms are divided into mechanisms: setting the cutting tool in motion; power supply, loading, removal of parts; transportation, etc.;
2) structural and constructive. It provides for the separation of mechanisms both by design features and by structural principles. This type includes mechanisms: crank-slider; rocker; lever-toothed; cam-lever, etc.;
3) structural. Simple, rational, closely related to the formation of a mechanism, its structure, methods of kinematic and force analysis, was proposed by L.V. Assur in 1916 and is based on the principle of constructing a mechanism by layering (attaching) kinematic chains (in the form of structural groups) to the initial mechanism. According to this classification, any mechanism can be obtained from a simpler one by attaching kinematic chains to the latter with the number of degrees of freedom W= 0, which are called structural groups, or Assur groups.
Ministry of Transport of the Russian Federation
Federal Agency of Sea and River Transport
Crimean branch
FGBOU VPO
"State Maritime University named after Admiral F.F. Ushakov"
Department "Fundamental disciplines"
Theory of mechanisms and machines
course project
Flat linkage
Explanatory note
The project was developed by: Art. gr. _
_____________________________
Project leader: prof. Burov V.S.
Sevastopol 2012
1. Kinematic analysis of a flat lever mechanism .............................................................. ........ 3
1.1. Construction of the movement in 12 positions.................................................... ................................... 3
1.2. Construction of plans for instantaneous velocities .............................................................. ............................... 4
1.3. Construction of plans for instantaneous accelerations .......................................................... .............................. 5
1.4. Construction of displacement diagram .................................................................. ........................................ 8
1.5. Constructing a velocity diagram .............................................................. .............................................. 9
1.6. Building an acceleration diagram .................................................................. ............................................. 9
2. Force analysis of a flat lever mechanism .............................................. ..................... 10
2.1. Determination of the loads acting on the links of the mechanism .............................................. ..... 10
2.2. Force calculation of a group of links 7, 6 .............................................. ............................................. 12
2.3. Force calculation of a group of links 4, 5 .............................................. ............................................. 13
2.4. Force calculation of the group of links 2, 3....................................... ............................................. 14
2.5. Force calculation of the leading link ....................................................... ................................................. 15
2.6. Force calculation of the leading link by the Zhukovsky method.................................................. .............. 15
3. Synthesis of the gear mechanism ............................................... ................................................. ...... 16
3.1. Determination of the geometrical parameters of the gear mechanism............................................... 16
3.2. Building a plan of linear velocities .............................................. .................................... 19
3.3. Building a plan of angular velocities .............................................. ....................................... 20
4. Synthesis of the cam mechanism ............................................... ................................................. .21
4.1. Plotting analogues of accelerations .................................................................. ................................. 21
4.2. Plotting speed analogues .................................................................. ................................. 22
4.3. Plotting analogues of displacements .......................................................... ........................... 22
4.4. Finding the minimum initial radius of the cam .............................................................. ........ 22
4.5. Building a cam profile ............................................................... ................................................. .. 23
Bibliography................................................ ................................................. ...................... 24
1. Kinematic analysis of a flat lever mechanism.
Given:
Scheme of a flat lever mechanism.
Geometrical parameters of the mechanism:
l OA \u003d 125 mm;
l AB \u003d 325 mm;
l AC \u003d 150 mm;
It is necessary to build a mechanism in 12 positions, plans for instantaneous speeds for each of these positions, plans for instantaneous accelerations for any 2 positions, as well as diagrams of displacements, speeds and accelerations.
1.1 Construction of 12 positions of a flat lever mechanism.
Draw a circle with radius OA. Then the scale factor will be:
We select the initial position of the mechanism and from this point we divide the circle into 12 equal parts. We connect the center of the circle (point O) with the obtained points. These will be the 12 positions of the first link.
Through t. O we draw a horizontal straight line X-X. Then we build circles of radius AB with centers at the previously obtained points. We connect the points B 0, B 1, B 2, ..., B 12 (the intersection of the circles with the line X-X) with the points 0, 1, 2, ..., 12. We get 12 positions of the second link.
From t. O we postpone up the segment b. We get the point O 1 . From it with a radius of O 1 D we draw a circle.
On the segments AB 0, AB 1, AB 2, ..., AB 12 from point A, we set aside a distance equal to AC. We get points С 0 , С 1 , С 2 , …, С 12 . Through them we draw arcs with a radius DC until they intersect with a circle centered at the point O 1 . We connect the points C 0, C 1, C 2, ..., C 12 with the received ones. These will be 12 positions of the third link.
Points D 0, D 1, D 2, ..., D 12 are connected to t. O 1. We get 12 positions of the fourth link.
From the highest point of the circle with the center at point O1, we lay off a horizontal segment equal to a. Through its end we draw a vertical line Y-Y. Further, from the points D 0 , D 1 , D 2 , ..., D 12 we build arcs with a radius DE to the intersection with the resulting straight line. We connect these points with the newly obtained ones. These will be 12 positions of the fifth link.
Considering the scale factor , the dimensions of the links will be:
AB \u003d l AB * \u003d 325 * 0.005 \u003d 1.625 m;
AC \u003d l AC * \u003d 150 * 0.005 \u003d 0.75 m;
CD= l CD * =220*0.005=1.1 m;
About 1 D \u003d l O1 D * \u003d 150 * 0.005 \u003d 0.75 m;
DE \u003d l DE * \u003d 200 * 0.005 \u003d 1 m;
a 1 \u003d a * \u003d 200 * 0.005 \u003d 1 m;
b 1 \u003d b * \u003d 200 * 0.005 \u003d 1 m.
1.2 Construction of plans for instantaneous velocities.
There are various methods for constructing a speed plan for a mechanism, the most common of which is the method of vector equations.
The velocities of the points O and O 1 are equal to zero, therefore, on the velocity plan, they coincide with the pole of the velocity plan p.
Position 0:
But the speed of t.B coincided with the pole p, therefore V B = 0, which means that the speeds of all other points will also coincide with the pole and will be equal to zero.
Plans of instantaneous speeds are constructed similarly for positions 3, 6, 9, 12.
Position 1:
The speed t.A is obtained from the equation:
The line of action of the velocity vector m.A is perpendicular to the link OA, and itself is directed in the direction of the link rotation.
On the plan of instantaneous speeds, we build a segment (pa) ┴ OA, its length (pa) = 45mm. Then the scale factor is:
The speed t.V is obtained from the equations:
, where V BA ┴ VA, and V BB0 ║X-X
From t.a on the speed plan we build a straight line ┴ to the link BC, and from t.r we draw a horizontal straight line. At the intersection we get point b. We connect t.a and t.b. This will be the velocity vector t.B (V B).
VB = pb* = 0.04*15.3 = 0.612
The speed of t.C is determined using the similarity theorem and the rule for reading letters. The rule for reading letters is that the order of writing letters on the plan of velocities or accelerations of a rigid link must exactly match the order of writing letters on the link itself.
From proportion:
You can determine the length of the segment ac:
We set aside a segment equal to 19.2 mm from t.a, we get t.s, we connect it with the pole, we get the velocity vector t.C (V C).
The speed t.D is determined by solving a system of geometric equations:
, where V DC ┴ DC, and V DO 1 ┴ DO 1
From t.c on the speed plan we build a straight line ┴ to the DC link, and from t.r we draw a straight line ┴ DO 1. At the intersection we get point d. We connect t.d with the pole, we get the velocity vector t.D (V D).
V D \u003d pd * \u003d 0.04 * 37.4 \u003d 1.496
We also find the speed i.e. from the solution of the system of equations:
, where V ED ┴ ED, and V EE 0 ║Y-Y
From t.d on the speed plan we build a straight line ┴ to the DE link, and from t.r we draw a vertical line. At the intersection we get i.e. We connect t.a and t.b. This will be the velocity vector t.B (V B).
V E \u003d pe * \u003d 0.04 * 34.7 \u003d 1.388
Similarly, plans for instantaneous speeds are constructed for 2, 3, 4, 5, 7, 8, 10, 11 positions of the mechanism.
1.3 Construction of instantaneous acceleration plans.
The accelerations of the points O and O 1 are equal to zero, therefore, on the acceleration plan, they will coincide with the pole of the acceleration plan π.
Position 0:
The acceleration of point A is found:
On the plan of instantaneous accelerations, we build a segment πа ║ ОА, its length (πа)=70 mm. Then the scale factor is:
The direction of acceleration t.B and t.A ║ straight X-X, ┴ BA, therefore, the acceleration t.B will coincide with the end of the instantaneous acceleration vector t.A, which means that the accelerations of all other points of the mechanism will coincide with it.
Statement 7:
The acceleration of point A is found:
On the plan of instantaneous accelerations, we build a segment πа ║ ОА, its length (πа)=70 mm.
The acceleration of point B can be found by solving the vector equation:
From t.a we set aside a segment equal to 21 mm ║ AB, then from the end of the resulting vector we build a segment ┴ AB, and draw a horizontal line through the pole. Connecting the anguish of the intersection with the pole, we get the acceleration vector t.V.
We find the acceleration t.C using the similarity theorem and the rule for reading letters:
Hence
The acceleration of point D can be found by solving a system of vector equations:
From t.s we set aside a segment equal to 14.5 mm ║ DC, then from the end of the resulting vector we build a segment ┴ DC.
From t. π we build a segment equal to 1.75 mm ║ O 1 D, then draw a straight line ┴ O 1 D through the end of the resulting vector.
The acceleration of point E can be found by solving a system of vector equations:
The direction of acceleration of the point E ║ ED, so we draw a horizontal straight line through the pole, and from the so-end of the acceleration vector t.D we build a segment equal to 1.4 mm ║ ED, then we draw a straight line ┴ ED from the end of the resulting vector. Connecting the point of intersection of the line ║ ED and the line ┴ ED with the pole, we obtain the acceleration vector of the point E.
1.4 Construction of the displacement diagram of the output link.
The displacement diagram of the output link is obtained as a result of constructing segments that are taken from the drawing of a flat lever mechanism in 12 positions, taking into account the scale factor
1.5 Construction of the speed diagram of the output link.
The speed diagram of the output link is obtained as a result of graphical differentiation by the incremental method of the displacement diagram of the output link. This method is essentially the chord method. If the constant pole distance H is taken equal to the value of the interval Δt, then there is no need to conduct rays through the pole П, since in this case the segments h i are increments of the function S(t) on the interval Δt.
That is, a vertical segment is built on the displacement diagram from the first division to the intersection with the graph. Then, a horizontal segment is laid off from the intersection point until it intersects with the next division. Then, from the obtained point, a vertical segment is again laid off until it intersects with the graph. This is repeated until the end of the schedule. The resulting segments are built on the velocity diagram, taking into account the scale factor, but not from the first division, but half a division earlier:
1.6 Construction of the acceleration diagram of the output link.
It is constructed similarly to the speed diagram of the output link of the mechanism
2. Force analysis of a flat lever mechanism.
Given:
l OA = 125 mm;
l AB = 325 mm;
l AC = 150 mm;
l CD = 220 mm;
l O1 D = 150 mm;
l DE = 200 mm;
Fmax = 6.3 kN;
m K = 25 kg/m;
Diagram of useful resistance forces.
It is necessary to determine the reactions in kinematic pairs and the balancing moment on the input shaft of the mechanism.
2.1 Determination of loads acting on the links of the mechanism.
Let's calculate the force of gravity. The resultants of these forces are located at the centers of mass of the links, and the magnitudes are equal:
G 1 \u003d m 1 * g \u003d m K * l OA * g \u003d 25 * 0.125 * 10 \u003d 31.25 H
G 2 \u003d m 2 * g \u003d m K * l B A * g \u003d 25 * 0.325 * 10 \u003d 81.25 H
G 3 \u003d m V * g \u003d 20 * 10 \u003d 200 N
G 4 \u003d m 4 * g \u003d m K * l CD * g \u003d 25 * 0.22 * 10 \u003d 55 H
G 5 \u003d m 5 * g \u003d m K * l O 1D * g \u003d 25 * 0.15 * 10 \u003d 37.5 H
G 6 \u003d m 6 * g \u003d m K * l DE * g \u003d 25 * 0.2 * 10 \u003d 50 H
G 7 \u003d m 7 * g \u003d 15 * 10 \u003d 150 H
Let's find the force of useful resistance according to the diagram of forces of useful resistance. For the considered position of the mechanism, this force is equal to zero.
There are no data for calculating the forces of harmful resistance, so we do not take them into account.
To determine inertial loads, accelerations of links and some points are required, so we will use the acceleration plan for the considered position of the mechanism.
Let us determine the forces of inertia of the links. The leading link, as a rule, is balanced, that is, its center of mass lies on the axis of rotation, and the resultant of the inertia forces is zero. To determine the inertia forces of other links of the mechanism, we first determine the accelerations of their centers of mass:
and S2 \u003d * πS 2 \u003d 0.4 * 58.5 \u003d 23.4 m / s 2
and B \u003d * πb \u003d 0.4 * 64.9 \u003d 25.96 m / s 2
and S4 \u003d * πS 4 \u003d 0.4 * 65.7 \u003d 26.28 m / s 2
and D \u003d * πd \u003d 0.4 * 78.8 \u003d 31.52 m / s 2
and S6 \u003d * πS 6 \u003d 0.4 * 76.1 \u003d 30.44 m / s 2
and E \u003d * πe \u003d 0.4 * 74.5 \u003d 29.8 m / s 2
Now let's define the forces of inertia:
F I2 \u003d m 2 * a S2 \u003d 8.125 * 23.4 \u003d 190 H
F I3 \u003d m 3 * a B \u003d 20 * 25.96 \u003d 519 H
F I4 \u003d m 4 * a S4 \u003d 5.5 * 26.28 \u003d 145 H
F I6 \u003d m 6 * a S6 \u003d 5 * 30.44 \u003d 152 H
F I7 \u003d m 7 * a E \u003d 15 * 29.8 \u003d 447 H
To determine the moments of forces of inertia, it is necessary to find the moments of inertia of the masses of the links and their angular accelerations. For links 3 and 7, the masses are concentrated at points, for link 1, and the angular acceleration is zero, so the moments of inertia forces of this link is zero.
Let us assume that the mass distribution of links 2, 4 and 6 is uniform along their lengths. Then the inertia of the links relative to the points S i is equal to:
J S 2 \u003d m 2 * l 2 2 / 12 \u003d 8.125 * 0.325 2 / 12 \u003d 0.0715 kg * m 2
J S 4 \u003d m 4 * l 4 2 / 12 \u003d 5.5 * 0.22 2 / 12 \u003d 0.0222 kg * m 2
J S 6 \u003d m 6 * l 6 2 / 12 \u003d 5 * 0.2 2 / 12 \u003d 0.0167 kg * m 2
Angular accelerations of links 2, 4, 5 and 6 are determined by relative tangential accelerations, therefore:
Let's find the moments of inertia forces of 2, 4, 6 links:
M I2 \u003d J S 2 * \u003d 0.0715 * 82.22 \u003d 5.88 Nm
M I4 \u003d J S 4 * \u003d 0.0222 * 42.73 \u003d 0.95 Nm
M I6 \u003d J S 4 * \u003d 0.0167 * 35.6 \u003d 0.59 Nm
2.2 Force calculation of a group of links 6, 7.
Let's select a group of links 6, 7 from the mechanism, arrange all real loads and forces and moments of inertia forces.
Let us replace the action on the considered group of dropped links by forces. In ie, the slider 7 is acted upon by a force from the side of the rack - the guide of the slider. In the absence of friction, the interaction force is directed perpendicular to the contacting surfaces, i.e., perpendicular to the direction of movement of the slider, and it is not yet known to the left or right, therefore, we will first direct this force to the right. If after calculations it turns out that it is negative, then it is necessary to change the direction to the opposite.
Two numbers are put in the designation index: the first shows from which link the force acts, and the second shows which link this force acts on.
At point D from link 5, link 6 is affected by the force R 56 . Neither the magnitude nor the direction of this force is known, therefore we determine it by two components: we direct one along the link and call it the normal component, and the second perpendicular to the link and call it the tangential component. the preliminary direction of these components is chosen arbitrarily, and the actual direction is determined by the sign of the force after calculations.
The force of useful resistance also acts on the slider E, but it is equal to zero.
Let's place all the listed forces on the selected group of links and determine the unknown reactions in the kinematic pairs E, D - R E and R 56 .
First, we determine the tangential component of the force R 56 from the equilibrium condition of link 6. Equating to zero the sum of the moments of forces relative to point E, we obtain:
The moment of inertia forces must be divided by because the links are shown on a scale, and their values taken from the drawing are used in the calculations.
The normal component of the force R 56 and the force R E are found by a graphical method from a vector polygon constructed for a group of links 6, 7. It is known that in a force equilibrium, a polygon composed of force vectors must be closed:
Since the directions of the lines of action of the normal component of the force R 56 and R E are known, then by constructing a previously open polygon from known force vectors, it can be closed if we draw straight lines through the beginning of the first and the end of the last vector, parallel to the directions of the desired forces. The point of intersection of these lines will determine the magnitude of the desired vectors and their actual directions.
It can be seen from the constructions that the direction of the force R 76 is from n to m, and the force R 67 is from m to n.
R 56 \u003d * \u003d 1/4 * 209.7 \u003d 52.43 N
R E \u003d * \u003d 1/4 * 69.3 \u003d 17.33 N
2.3 Force calculation of a group of links 5.4.
Let's select a group of links 4, 5 from the mechanism, arrange all the real loads and forces and moments of inertia forces, reactions of the discarded links. At point D, the force R 65 acts, which is equal to R 56 and is directed opposite to it.
The unknowns are: the force of interaction of 4 and 2 links, the force of interaction of 5 links and racks.
At point C from link 2, link 4 is affected by the force R 24 . Neither the magnitude nor the direction of this force is known, therefore we determine it by two components: we direct one along the link and call it the normal component, and the second perpendicular to the link and call it the tangential component. the preliminary direction of these components is chosen arbitrarily, and the actual direction is determined by the sign of the force after calculations.
First, we determine the tangential component of the force R 24 from the equilibrium condition of link 4. Equating to zero the sum of the moments of forces about the point D, we get:
The normal component of the force R 24 and the force R O 1 are found by a graphical method from a vector polygon constructed for a group of links 5, 4. It is known that in a force equilibrium, a polygon composed of force vectors must be closed:
Let us determine the magnitude of reactions in kinematic pairs:
R 24 \u003d * \u003d 1 * 26.6 \u003d 26.6 N
R O 1 \u003d * \u003d 1 * 276.6 \u003d 276.6 N
2.4 Force calculation of a group of links 2, 3.
Let's select a group of links 2, 3 from the mechanism, arrange all real loads and forces and moments of inertia forces, reactions of the dropped links. At point C, the force R 24 acts, which is equal to R 24 and is directed opposite to it.
The unknowns are: the force of interaction of 1 and 2 links, the force of interaction of 2 links and the slider.
At point C from link 1, link 2 is affected by the force R 12 . Neither the magnitude nor the direction of this force is known, therefore we determine it by two components: we direct one along the link and call it the normal component, and the second perpendicular to the link and call it the tangential component. the preliminary direction of these components is chosen arbitrarily, and the actual direction is determined by the sign of the force after calculations.
First, we determine the tangential component of the force R 12 from the equilibrium condition of link 2. Equating the sum of the moments of forces with respect to point A to zero, we obtain:
The normal component of the force R 12 and the force R B are found by a graphical method from a vector polygon constructed for a group of links 2, 3. It is known that with a force balance, a polygon composed of force vectors must be closed:
Since the directions of the lines of action of the normal component of the force R 24 and R O 1 are known, then by constructing a previously open polygon from known force vectors, it can be closed if we draw straight lines through the beginning of the first and the end of the last vector, parallel to the directions of the desired forces. The point of intersection of these lines will determine the magnitude of the desired vectors and their actual directions.
Let us determine the magnitude of reactions in kinematic pairs:
R 12 \u003d * \u003d 1/2 * 377.8 \u003d 188.9 N
R B \u003d * \u003d 1/2 * 55.4 \u003d 27.7 N
2.5 Power calculation of the leading link.
The leading link is usually balanced, that is, its center of mass is on the axis of rotation. This requires that the inertia force of the counterweight mounted on the continuation of the crank OA is equal to the inertia force of the OA link:
m \u003d M 1 / l OA \u003d 3.125 / 0.125 \u003d 25 kg - mass per unit length.
From here it is possible to determine the mass of the counterweight m 1 , given its distance r 1 from the axis of rotation. At r 1 = 0.5 * l m 1 = M 1 (mass of the OA link).
At point A, a force R 21 acts on link 1 from link 2, the moment of which relative to point O is equal to the balancing moment.
In this case, at the point O, a reaction R O occurs, which is equal and opposite to the force R 21 . If the gravity force of the link is commensurate with the force R 21, then it must be taken into account when determining the reaction of the support O, which can be obtained from the vector equation:
2.6 Power calculation of the leading link by the Zhukovsky method.
To the plan of instantaneous speeds of the mechanism, rotated by 90 0 in the direction of rotation, we apply all the forces acting on the mechanism, and draw up an equation for the moments of the acting forces relative to the pole.
Answers to exam questions on TMM
Moscow State University
Engineering Ecology
Theory of machines and mechanisms (TMM)
Exam questions
for study groups of the day department.
1. Structure of mechanisms
1.1. Machine and mechanism. Classification of mechanisms according to functional and structural-constructive features.
ANSWER: According to the definition of Academician Artobolevsky:
Car- there are devices created by man to study and use the laws of nature in order to facilitate physical and mental labor, increase its productivity by partially or completely replacing it in labor and physiological functions.
Mechanism- a system of bodies designed to convert the movement of one or more bodies into the required movement of other rigid bodies. If liquid or gaseous bodies are involved in the transformation of motion, then the mechanism is called hydraulic or pneumatic. Usually, the mechanism has one input link that receives movement from the engine, and one output link connected to the working body or the instrument indicator. Mechanisms are flat and spatial.
Classification of machines by functional purpose:
Energy (motors, generators).
Workers (transport, technological).
Information (control and management, mathematical).
Cybernetic.
Machines are made up of mechanisms.
According to the functional classification, there are:
Mechanisms of engines and converters;
Executive mechanisms;
transmission mechanisms;
Mechanisms of control, regulation, adjustment;
Feeding, feeding, sorting mechanisms;
Mechanisms for counting, weighing, packaging.
There is much in common in terms of the structure and methodology for calculating their mechanical parameters.
Structural-constructive classification:
Lever mechanisms;
Cam mechanisms;
Gear mechanisms (consist of gear wheels);
Combined.
1.2. Lever mechanisms. Advantages and disadvantages. Application in technical devices.
ANSWER: Lever mechanisms consist of bodies made in the form of levers, rods. These rods or levers interact with each other along the surface. Therefore, lever mechanisms are able to perceive and transmit significant forces.
They are used as the main technological devices. However, the reproduction of the required law of motion by such mechanisms is very limited.
1.3. Cam mechanisms. Types of cam mechanisms. Advantages and disadvantages. Main purpose.
ANSWER: The cam mechanism consists of a curvilinear body, the nature of the movement of which determines the movement of the entire mechanism. The main advantage is that, without changing the number of links, any law of motion can be reproduced by changing the profile of the cam. But in the cam mechanism there are links that touch at a point or along a line, which significantly limits the amount of force transmitted due to the appearance of very high specific pressures. Therefore, cam mechanisms are mainly used as a means of automating the technological process, where the cam plays the role of a hard program carrier.
1.4. gear mechanisms. Types of gear mechanisms. Main purpose.
ANSWER: gear mechanism a mechanism is called, which includes gear wheels (a body that has a closed system of protrusions or teeth).
Gear mechanisms are mainly used to transmit rotational motion with a change, if necessary, in the magnitude and direction of the angular velocity.
The transmission of motion in these mechanisms is carried out due to the lateral pressure of specially profiled teeth. To reproduce a given ratio of angular velocities, the tooth profiles must be mutually bending, that is, the tooth profile of one wheel must correspond to a well-defined tooth profile of the other wheel. Tooth profiles can be delineated by various curves, but the most common are mechanisms with an involute tooth profile, that is, with a tooth outlined along an involute.
To reproduce a constant ratio of angular velocities, mechanisms with round gears are used.
There are flat and spatial mechanisms. In a flat mechanism, the axes are parallel, while in a spatial one they intersect or cross. In a flat mechanism, the wheels have a cylindrical shape, in a spatial one they are conical (if the axes intersect).
Very varied. Some of them are a combination of only solid bodies, others are composed of hydraulic, pneumatic bodies or electrical, magnetic and other devices. Accordingly, such mechanisms are called hydraulic, pneumatic, electric, etc.
From the point of view of their functional purpose, mechanisms are usually divided into the following types:
Engine mechanisms convert various types of energy into mechanical work (for example, the mechanisms of internal combustion engines, steam engines, electric motors, turbines, etc.).
The mechanisms of converters (generators) convert mechanical work into other types of energy (for example, the mechanisms of pumps, compressors, hydraulic drives, etc.).
The transmission mechanism (drive) has as its task the transfer of movement from the engine to the technological machine or actuator, converting this movement into the necessary for the operation of this technological machine or actuator.
An actuator is a mechanism that directly affects the processed environment or object. Its task is to change the shape, state, position and properties of the processed medium or object (for example, the mechanisms of metalworking machines, presses, conveyors, rolling mills, excavators, hoisting machines, etc.).
Control, monitoring and regulation mechanisms are various mechanisms and devices for ensuring and controlling the dimensions of processed objects (for example, measuring mechanisms for controlling sizes, pressure, liquid levels; regulators that respond to the deviation of the angular velocity of the main shaft of the machine and set the specified speed of this shaft; a mechanism that regulating the constancy of the distance between the rolls of the rolling mill, etc.).
Mechanisms for supplying transportation, feeding and sorting of processed media and objects include mechanisms for screw augers, scraper and bucket elevators for transporting and supplying bulk materials, mechanisms for loading hoppers for piece blanks, mechanisms for sorting finished products by size, weight, configuration, etc.
Mechanisms for automatic counting, weighing and packaging of finished products are used in many machines, mainly producing mass piece products. It must be borne in mind that these mechanisms can also be actuators if they are included in special machines designed for these purposes.
This classification shows only the variety of functional applications of mechanisms, which can be significantly expanded. However, mechanisms that have the same structure, kinematics and dynamics are often used to perform various functions. Therefore, for studying in the theory of mechanisms and machines, mechanisms are singled out that have common methods for their synthesis and analysis of work, regardless of their functional purpose. From this point of view, the following types of mechanisms are distinguished.
