Mashina detallari va loyihalash asoslri fanidan elektron o’quv uslubiy majmua (ingliz tilida)

5430100-agricultural Mechanization for students of educational fields

MACHINE PARTS AND DESIGN BASICS FANIDAN EDUCATIONAL AND METHODICAL COMPLEX

REPUBLIC OF UZBEKISTAN MINISTRY OF HIGHER AND SECONDARY SPECIAL EDUCATION
KARSHI INSTITUTE OF ENGINEERING AND ECONOMICS

5430100-agricultural Mechanization for students of educational fields MACHINE PARTS AND DESIGN BASICS FANIDAN EDUCATIONAL AND METHODICAL COMPLEX

QARSHI — 2020

ABSTRACT

Collection of lectures on the subject «Parts of machines» is compiled according to the standards Programme.
Information’s about combinations, mechanical transmissions, shafts, axis, muffs and bearings are offered.

REPUBLIC OF UZBEKISTAN MINISTRY OF HIGHER AND SECONDARY SPECIAL EDUCATION
KARSHI INSTITUTE OF ENGINEERING AND ECONOMICS

Registered: “CONFIRM”

____ Vice-rector for academic Affairs
_ O.N. Bozorov
2020 year “
____ 2020 year

IN THE DISCIPLINE MACHINE ELEMENTS AND DESIGN PRINCIPLES THE WORKING TRAINING PROGRAM

Field of expertise: 400000 – Agriculture and water management-technical sphere;

Field of education: 430000-Agricultural machinery;
450000-Irrigation and land reclamation.
Areas of education: 5430100-Agriculture mechanization

KARSHI-2020 y

Developed in accordance with the working curriculum, working curriculum and curriculum

Originator: A. A. Nazarov — Department of General technical Sciences associate Professor, p. F. F. D.

Repeaters:
Alikulov T. Head of the Department «higher mathematics» T-Karmii T-F. F. N.,
Donaev B.- Department of «General technical Sciences» t-F. F. N., associate Professor

Working curriculum of the Department «General technical Sciences» 2020 the year was discussed at the meeting on August 1, _ _ _ _ _ _ _ and recommended for discussion at the faculty Council” technology».

Head of the Department___________ X. Faizullaev
Department of General technical Sciences founder: editorial office of the newspaper «xabar»), Faculty of engineering technology» In the methodological Commission application №, 2020 I.) and the Institute of methodology In the Council statement №, 2020 me.) Discussion and training it is recommended to use in the process.

Introduction.
In the higher education system, it is important to provide the educational process with modern educational and methodological complexes that have high qualification and initiative abilities, are able to independently solve professional and life problems, and train personnel who can quickly adapt to new technologies and technologies.
Educational and methodological complex for the discipline «machine parts and design basics» — includes the formation of knowledge, skills, skills and competencies that must be mastered by students on the basis of the state educational standard and scientific program , obtaining guaranteed results based on the integrated design of the educational process, independent receipt and study and control, educational and methodological sources, didactic tools and materials, electronic educational resources, learning technologies, methods and criteria aimed at developing the student’s creative abilities.

Relevance of educational science and its role in professional education
The purpose of the subject «machine parts and design basics» is to train qualified specialists for their effective and correct use while constantly providing agriculture and water management with new equipment.
Therefore, when training highly qualified personnel for agriculture and water management, in accordance with the requirements, future specialists should be sufficiently equipped with modern achievements of science and technology in their activities
The study of the science” machine parts and design basics » of a future specialist in the field of agriculture and water management creates a fundamental basis for the independent solution of technical problems that arise in the process of mastering and scientific and technical development of the science of the future specialty.
Relationship between science and other Sciences
For deep mastery of the science of machine parts and design basics, complete solution of theoretical and practical problems based on a number of fundamental Sciences
including: higher mathematics, physics, computer science, engineering computer graphics, materials science, etc
Sections of higher mathematics on vector analysis, function eccentricity, differential equations, strings, indefinite and exact integrals, calculations using integrals of form surfaces
In computational mathematics and computer science technology: solving practical and laboratory problems, course and calculation work introduction of methods of computational mathematics and computer technology in the implementation of;
The project on engineering computer graphics is related to the preparation of design drawings in accordance with standard requirements, the introduction of Metrology and standardization

  • experimental determination of mechanical properties of materials based on materials science and selection of material for parts;
  • analysis of mechanical physical properties of parts based on mechanics, molecular physics, electromagnetism, and optics.
    The role of science in production
    State and development of agriculture and water management, efficiency of machines and installations used, high productivity, ease of operation, safety. As well as the availability of highly qualified specialists who operate and provide technical services
    Purpose and objectives of the discipline
    The main goal of discipline is teaching students the tasks, structures, working principles, application areas and disciplines of the failure of the machines used in agriculture and water sector, as well as the formation of a student’s knowledge, skills and skills corresponding to the profile direction calculation and design based on the criteria of performance used in the water and agricultural sectors
    The task of science is to provide General technical, fundamental knowledge necessary for the study of methods for calculating and designing efficiency according to the criteria of working capacity, based on the analysis of the structure of components and mechanical transmissions, their operating conditions, which are common to almost all machines, apparatuses and equipment
    The following requirements apply to students ‘ knowledge, skills and qualifications in the subject
    Student:
  • have an understanding of mechanical transmissions on machines, types, designs and applications of parts that perform General functions, as well as their main parameters
  • knowledge and use of theoretical and practical knowledge on calculation and design based on the features of loading machine parts, their performance and criteria
  • total number of machines used in water supply and agriculture tasks, types, designs and applications of parts that perform their functions, as well as their main parameters, must have the appropriate skills of the direction profile for calculation and design
    Modern information and pedagogical technologies in science teaching
    Innovative pedagogical technologies in science teaching include the following interactive methods, including discussion-discussion, collective discussion or making a list of problems, study, situation analysis, discussion or discussion, critical thinking, role-playing games, working in small groups, mental attack, cluster pen, snap), fish skeleton, boomerang, scarab, cascade, Viennese diogram, pinboard, «T-scheme”, blitz survey,» Why?»technologies, lecture methods, BBHB I know, I want to know, I learned), concept and insert tables are widely used
    When preparing texts of lectures on science in foreign countries, including in the Commonwealth countries, new ones were published. Electronic textbooks, manuals, and lecture texts distributed over the Internet are used. Also, during the lectures, the use of electronic lectures, multimedia slides and videos corresponding to the topics is provided
    Practical classes use sets of electronic exercises and questions, computer games on science using computers, test questions and answers, and laboratory classes-computer electronic models of devices and equipment, as well as technological process, virtual laboratories
    This education essentially involves the full development of all participants in the educational process. This implies an approach to the design of education, of course, based on the learning goals related primarily to future specialized activities, and not the personality of a particular recipient of education
    Educational technology should embody all the features of the system: the logic of the process, the interconnectedness, the integrity of all its links
    It is an education aimed at the formation of procedural qualities of the individual, activation and intensification of the student’s activity, disclosure of all his abilities and opportunities, initiative in the learning process
    The dialogical approach. This approach suggests the need to create learning relationships. It increases the creative activity of a person, as well as self-reflection and self-reflection
    Organization of joint education. The democratic, equal, educational and educational meaning of the activity means the need to pay attention to the implementation of joint work on the formation and evaluation of the achieved results
    Problem-based learning. One of the ways to activate educational activities through problematic provision of educational content. At the same time, independent creative activity is provided, aimed at the objective contradiction of scientific knowledge and ways to solve it, the formation and development of dialectical thinking, their creative application to practical activities
    Application of modern means and methods of providing information-application of new computer and information technologies in the educational process
    Methods and methods of training. Lecture Introduction, topic, visualization), problem-based learning, case stage, pinboard, paradox and design methods, practical work
    Forms of training organization: dialogue, polylogue, frontal, collective and group communication based on cooperation and mutual learning
    Means of training: along with traditional forms of learning tutorial, the text of the lecture — computer and information technology
    Method of communication: direct interaction with the audience based on operational feedback
    Feedback methods and tools: diagnostics of training based on the analysis of the results of observation, blitz survey, intermediate and current, final control
    Management methods and tools: planning of training sessions in the form of a technological map that defines the stages of training, joint actions of the teacher and the listener in achieving the goal, monitoring not only classroom sessions, but also independent work outside the classroom
    Monitoring and evaluation: planned monitoring of learning outcomes both during training and throughout the course. At the end of the course, the student’s knowledge is evaluated using test tasks or written task options
    Distribution of classes on the subject” machine parts and design
    basics » by subject and hours
    The total time of training 256 clock
    Including
    Total audience hours 126 clock
    lecture 54 clock
    Practical class 54 clock
    Laboratory 18 clock
    Self-study 130 clock

MAIN PART
Ma’ruza mashg‘ulotlari
Lectures 1-module. General questions concerning the calculation and design of machine parts
1-subject. General information about science
Achievements in mechanical engineering and the role of science «machine parts and design basics», new directions in the design of modern machines, the relationship of science with goals, objectives and other Sciences, General concepts and classification of machine parts
Applicable educational technologies: Dialogic approach, problem-based learning. Mental attack, blitz, concept analysis, discussion, self-control
2-the Theme. Stages of creating machines
The main requirements for them when designing machine parts. Structural materials used in mechanical engineering.
Applicable educational technologies: Dialogic approach, problem-based learning. Mental attack, glare, fish skeleton, debate, self-control.
3-subject. Standardization work in design and construction
General rules of standardization in mechanical engineering. Main goals of standardization. Standardization system, normative documents on standardization
Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control.
4-subject. Factors of the employee’s ability to work and its provision
Exclusivity, bikernet, resistance to eating, heat resistance and vibration resistance. Features of calculating machine parts for strength
Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control
5-subject. The load on the machine parts and the stresses arising from them
The types of loads that affect the details. Operating and permissible voltage in them
Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control
2-module. Mechanical transmission
6-the theme. Expansions. General concepts of mechanical transmissions.
Initial data on mechanical transmissions. Kinematics and classification of mechanical transmission power.
Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control
7-subject. Friction transmission and variable speed
Kinematics of friction gears and forces in them. The concept of variators Design types of variators
Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control
8-subject. Mechanisms with a curved link. Belt transmission.
General information about tape transfers. Geometry and kinematics of belt tension. Polishing tape
Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control.
9-subject. Tape slide and CP curves.
The forces and stresses in the tape distribution networks. Efficiency of belt drives and their calculation
Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control.
10-the theme. Chain transmission
General concepts of chain drives. Forces in chain transmission networks Efficiency of chain transmission and its calculation.
Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control.
3-module. Gear.
11-subject. General concepts of gears.
Types of threaded joints, applications, advantages and disadvantages Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control.
12-subject. Geometry and kinematics of gears
Geometry and kinematics of involute cylindrical gears. The materials used for the manufacture of gear wheels
Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control.
13-subject. The forces and stresses on the gears.
Types of tooth enamel. The degree of precision of gear manufacturing and its impact on the quality of operation
Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control.

14-subject. Calculation of the tooth tension of the wheel teeth for strength
Fixed bending stress. Accepted restrictions. Source data for the project invoice
Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control.
15-the theme. Calculation of the contact stress on the strength of the teeth teeth teeth gear teeth
Acceptable contact voltages. Accepted restrictions. Source data for the project invoice. Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control.
4-module. Worm extensions.
16-subject. Worm extensions. General information.
Types, scope, advantages and disadvantages. Geometric parameters of the hood and how to prepare them. Their kinematic parameters. Materials used in their manufacture
Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control.
17-subject. The calculation of the worm extension strength
The efficiency of the worm gear. Tensile forces. Evaluation of stretching and its use. Assessment of the strength of teeth. The calculation of the cooling and lubrication of the worm gear under heat
Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control.
18-subject. Reducers.
Types of gearboxes-scope of application. Selection of gearboxes.
Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control.
5-module. Shafts and wasps, as well as their supports bearings)
19-subject. The trees and wasps.
General information about shafts and arrows. Structural design of valves. And the materials used in their manufacture
Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control.
20-the subject. The basis of the calculation of the shafts.
Efficiency of shafts and arrows. Calculated and accurate calculation of shafts.
Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control.
21-subject. Bearings.
Rolling bearings and their calculation basics Bearings. General information. Roller bearings. General information and classification. Conditional calculation of rolling bearings.
Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control.
22-subject. Roller bearings.
Roller bearings. General information about rolling bearings and their classification. General information about rolling bearings and their classification. Operating conditions of rolling bearings and their impact on performance. Practical calculation of rolling bearings their choice). Selection of bearings for dynamic load-carrying capacity C. Checking and selecting rolling bearings according to static load Co. Features of calculating the load of radial bearings.
Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control.
6-module. General concepts of couplings and joints.
23-subject. Couplings.
Couplings with permanent mounting. Controlled couplings. Coupling with an elastic element. The automatic clutch. The selection of the coupling
Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control.
24-subject. Joints. Essential compounds.
General view of welded joints. Performance of welded joints and their calculation.
Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control.
25-subject. Brocade nail joints.
Parchin-a General concept about the joints of the nails. Parchin is able to process nail joints and calculate them.
Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control.
26-subject. Decomposing compounds.
General concepts of veneer and Spline combinations. Operability of veneer and spline joints and their calculation.
Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control.
27-subject. Threaded joint.
Types of threaded joints and their application. Forces in carved joints and calculate their strength.
Applicable educational technologies: Dialogic approach, problem-based learning. Psychic attack, blitz, concept analysis, fish skeleton, discussion, self-control.

Recommended topics for practical training 4th-5th semester  
  1. Learn about the practice of selecting material for machine parts.
    Applicable educational technologies: problem-based education. Blitz surveys, discussion, BBB, insert
  2. The implementation of the kinematic schemes of a mechanical gear. Applicable educational technologies: Dialogic approach, problem-based learning. Working in small groups, disputes, self-control
  3. Perform kinematic calculation of the mechanical drive.
    Applicable educational technologies: Dialogic approach, problem-based learning. Disputes, self-control.
  4. Calculation of kinematic and power indicators of friction gears and variators. Applicable educational technologies: Dialogic approach, problem-based learning. Disputes, self-control
  5. Determination of forces and stresses in belt transmission networks, as well as calculation of their geometric cross-sections.
    Applicable educational technologies: Dialogic approach, problem-based learning. Disputes, self-control.
  6. Calculation of V-belt transmission.
    Applicable educational technologies: Dialogic approach, problem-based learning. discussion, self-control.
  7. Determine the performance criteria of the chain transmission and calculate them. Applicable educational technologies: problem education, debate, self-control .
  8. Calculation of the design dimensions of a cylindrical straight gear, conical and conical evolutionary gear. Applicable educational technologies: problem education, debate, self-control .
  9. Selecting the material for gears and determining the number of allowable stresses. Applicable educational technologies: Dialogic approach, problem-based learning, working in small groups, discussion, role-playing games, self-control.
  10. Calculation of the contact stress strength of teeth in cylindrical straight and bevel gears. Applicable educational technologies: problem education, discussion, why, self-control.
  11. Calculation of the strength for normal stress when bending teeth in cylindrical straight and bevel gears. Applicable educational technologies: problem-based education, discussion, cluster, insert table, self-monitoring.
  12. Calculation of bending strength and contact stress of teeth in bevel gears. Applicable educational technologies: problem-based education, discussion, cluster, insert table, self-monitoring.
  13. Calculation of the geometric dimensions of the worm gear. Applicable educational technologies: problem-based education, discussion, cluster, Insert table, self-monitoring.
  14. The calculation of the worm gear for strength on contact and bending stress. Applicable educational technologies: problem-based education, discussion, cluster, Insert table, self-monitoring.
  15. Check that the worm gear is heated, cooled, and lubricated. Applicable educational technologies: problem-based education, discussion, cluster, insert table, self-monitoring.
  16. Practice to choose gear. Applicable educational technologies: problem-based education, discussion, cluster, Insert table, self-monitoring.
  17. Determining the design dimensions of the gearbox parts. Applicable educational technologies: problem-based education, discussion, cluster, insert table, self-monitoring.
  18. Design calculation of shafts and their construction. Applicable educational technologies: problem-based education, discussion, cluster, insert table, self-monitoring.
  19. The execution of line reducers at the initial stage. Applicable educational technologies: problem-based education, discussion, cluster, insert table, self-monitoring.
  20. Accurate calculation of shafts. Applicable educational technologies: problem-based education, discussion, cluster, insert table, self-monitoring.
  21. selection of rolling bearings and calculation of their service life. Applicable educational technologies: problem-based education, discussion, cluster, Insert table, self-monitoring.
  22. selection of rolling bearings and calculation of their service life. Applicable educational technologies: problem-based education, discussion, cluster, Insert table, self-monitoring.
  23. selection of the design of the shaft supports. Applicable educational technologies: problem-based education, discussion, cluster, Insert table, self-monitoring.
  24. The selection of the coupling for a mechanical drive. Applicable educational technologies: problem-based education, discussion, cluster, Insert table, self-monitoring.
  25. Determine the performance criteria of welded and rivet joints of nails and calculate them according to these criteria for strength.
  26. Determine the performance criteria for veneer and spline joints and calculate their strength based on these criteria. Applicable educational technologies: problem-based education, discussion, cluster, insert table, self-monitoring.
    Recommended topics for laboratory classes
  27. Study of threaded joints. Applied technical means and method of performing the work: threaded joints and a dinometric key using a computer, a virtual laboratory.
  28. The study spanih joints. Applied technical means method of performing the work: several types of dowels and measuring instruments using a virtual laboratory on a computer.
  29. Study of the design of chain screws. Applicable techniques, procedures, and method of performing the work. Dummy simulator, jgut, bandage. It is practiced on the simulator.
  30. Investigation of the design of longitudinal cross beams with ponasonnymi belts. Applied technical means and method of work execution. Dummy simulator, jgut, bandage. It is practiced on the simulator.
  31. Research of mechanical power conductors. Applied technical means and method of work execution. Dummy simulator, jgut, bandage. Practicing on the simulator
  32. Research on the design of a single-cylinder gearbox. Applied technical means and method of work execution. Dummy simulator, jgut, bandage. It is practiced on the simulator.
  33. Study design the worm gear.Applied technical means and method of work execution. Dummy simulator, jgut, bandage. It is practiced on the simulator.
  34. Research on the design of a cylindrical gearbox with two-stage direct transmission. Applied technical means and method of work execution. Dummy simulator, jgut, bandage. It is practiced on the simulator.
  35. Research on the construction of rounded nomads. Applied technical means and method of work execution. Dummy simulator, jgut, bandage. It is practiced on the simulator.
    The purpose of independent education is to consolidate the knowledge and skills acquired by students in the process of teaching under the guidance of a teacher, using textbooks, teaching AIDS, educational and methodological complexes, Internet information, educational and visual and multimedia materials.

Information and methodological support of the program.
In the process of teaching this science,literature published abroad and in the Republic, electoral literature, virtual laboratories, technical equipment on laboratory topics, various slides, Wikipedia, articles in scientific journals, lecture texts, educational and methodological complexes on science and Internet materials are used. Current assessment-laboratory, practical classes and independent training tasks to determine and evaluate the level of knowledge and practical skills in the subjects of science. an oral survey, testing, interview, Test work, Colloquium, homework check, and other similar forms are conducted. Intermediate assessment-conducted in written, oral, test form to determine and evaluate the level of knowledge and practical skills of the student after completing the relevant section of the curriculum including several subjects of the subject) on the basis of a modular system during the semester.
MAIN LITERATURE

  1. Richard G., Budinas J., Keith Nisbett Tent edition Shigley mechanical Engineering design Megraw-hill new York, USA, 2015
  2. Sobitova.A. «machine Parts». Textbook. T.: «national encyclopedia of Uzbekistan», 2014 / -444 P.
  3. Tozhiboev R. Juraev A., Mansurov R. «machine Parts» textbook T.: «Science and technology» 2010. -216 P.
  4. Suleymanov I. S. «Details of the machine» T. «Teacher», 1981. — 358 P.
  5. Tozhiboev R. N. Dzhuraev A. «Details of the machine». T. «Teacher». 2002. — 268 P.
  6. Guzenkov P. G. «Detailed machine» M. «Three schools», 1982. -358 p.
  7. Botirmukhammedov J. K. «Details of the machine» T. «Teacher». 2001. -194 P.
    ADDITIONAL LITERATURE
  8. Mirziyoyev sh. M. we will jointly restore a free and prosperous Democratic Republic of Uzbekistan. Tashkent, Uzbekistan, 2016-56 years.
  9. Mirziyoyev Sh.Critical analysis, strict discipline,and personal responsibility should be the daily rule of every Manager. Tashkent, Uzbekistan, 2017 -104
  10. Mirziyoyev sh. M. the rule of law and ensuring human interests-the key to the development of the country and the welfare of the people Tashkent, Uzbekistan, 2017 -48 b.
  11. Mirziyoyev sh. M. strategy of actions on five priority directions of development of Uzbekistan. T. Uzbekistan, 2017. «Gazeta.uz I’m sorry.
  12. Batirmukhammedav J. K. «machine parts, lifting and transport machines», T. «teacher», 1994. — two Hundred and seventy-four P.
  13. Nazarov A. A. text of the lecture on «machine parts» — Vol.: 2018 — 104 b.
  14. Nasirov S. complete the course project on the subject «machine parts». — Against, 2008. -223 b.
    FOREIGN LITERATURE
  15. A Text-book on Applied Mechanics: Specially Arranged for the Use of Science and Art, City and Guilds of London Institute, and Other Engineering Students, Volume 2 Andrew Jamieson 2012 y.
  16. Applied Mechanics of Solids. Allan F. Bower 2009 y
  17. Technical mechanics. Edward R. Maurer 2011y
  18. Kinematics of planar manipulator of parallel structure with three degrees of freedom in Mathcad environment. F.A. Doronin 2015 y
  19. Sugimoto K., Duffy J., Hunt K.H., «Special Configurations of Spatial Mechanisms and Robot Arms», Mechanism and Machine Theory, Vol. 17, 1982, pp. 119-132.
  20. K.H. Hunt. Kinematic structures of parallel manipulators with actuation. ASME, Engineering, 1983, vol. 105, No. 4, pp. 201-210.
  21. Mohammed M., Duffy J. A Direct Determination of the Instantaneous Kinematics of Fully Parallel Robot Manipulators. / ASME J. Mech., Trans., Autom. Des., 1985, V. 1072): p. 226-229
  22. Koichi Sugimoto. Kinematic and dynamic analysis of parallel manipulators by means of motor algebra. ASME Journal of Mechanisms, Transmissions, and Automation in Design, 109:3-7, March 1987.
  23. Gosselin, C., and Angeles. J., 1988, “The Optimum Kinematic Design of a Planar Three-Degree-of-Freedom Parallel Manipulator”, Journal of Mechanisms, Transmissions, and Automation in Design, Vol.110, pp. 35–41.
  24. Liu, X.-J., Wang, J. and Gao, F., 2000, “Performance of the Workspace for Planar 3-DOF Parallel Manipulators”, Robotica, Vol.18, pp. 563–568.
  25. Gallant, M., and Boudreau, R., 2002, “The Synthesis of Planar Parallel Manipulators with Prismatic Joints for An Optimal, Singularity-Free Workspace”, Journal of Robotics Systems, Vol.19, No.1, pp. 13–24.
  26. R. D. Gregorio, “Kinematics of the 3-upu wrist,” Mechanism and Machine Theory, vol. 38, pp. 253–263, 2003.
  27. Hall, N.G., Potts, C.N. and Sriskandarajah, C., Parallel machine scheduling with a common server. Discr. Appl. Math., 2000, 102, 223–243.
  28. Han, B.T. and Cook, J.S., An efficient heuristic for robot acquisition and cell formation. Ann. Oper. Res., 1998, 77, 229–252.

Internet sites
http://www.libs.ru.
http://www.rsl.ru

http://www.sputnikplus.ruhttp://www.ziyonet.edu.uzhttp://www.study.uzhttp://specural.com

A.A.Nazarov

DETAILS OF MACHINES

(a collection of the texts of lectures)

KARSHI – 2018

REPUBLIC OF UZBEKISTAN MINISTRY OF HIGHER AND SECONDARY SPECIAL EDUCATION

KARSHI INSTITUTE OF ENGINEERING AND ECONOMICS

Department of General technical Sciences

MACHINE DETAILS
(text package of lectures)

Karshi-2018
Report 1. HISTORY OF THE SCIENCE OF MACHINE PARTS DETAILS, DESIGNS, AND THEORIES

Plan:

  1. Introduction. The main goals and objectives of the science of machine parts.
  2. Basic requirements for machine parts.
  3. the Main sections of science and the problems that are studied in them.
  4. Classification classification) by the subject of the machine part.
  5. The role of science in mechanical engineering.

Keywords: part, Assembly, mechanism, machine, gearbox, coupling and bearing

Introduction.
A tool designed to perform a particular job is a set of mechanisms made up of several parts called machines. A part that is made of the same material as the machine and is not split into separate parts is called a part. For example, a nut, bolt, stud, And similar parts. Currently, the power of engines and electric generators exceeds 1500,000 kW. The load capacity of the car is up to 250 vehicles, cranes — up to 680 vehicles, locomotive trains-4500 vehicles. The speed of land transport is 150-300 km/h, and air transport is 1000 km / h. The machine has so penetrated into human life that nothing can be imagined without machines and mechanisms. A part of a machine that is designed to perform a specific task and consists of several parts is called a nodal part. Examples of assemblies are a gearbox, coupling, bearing, and others. This means that the machine will consist of nodes, and nodes will consist of parts.
The first book called «details of the machine», Prof. V. L. Kirpichev began Printing in 1881 in St. Petersburg. Russian scientists L. Euler, N. P. Petrov, N. E. Dzhukovsky, S. A. Chaplygin, N. I. Gokhman and foreign scientists: O. Reynolds, A. Sommerfeld, A. Mitchell, H. Hugens, R. Willis, V. Lewis, E. Buckingham, H. merrita, K. Bach , R. Shtribek, A. the palmgrens made a Big contribution.

Basic requirements for the machine and its parts. Basically, the following requirements apply to machines:

  1. high labor Productivity.
  2. Efficiency
  3. Reliable operation
  4. The structure is simple and disappointing
  5. Sizes are small and light
  6. Right to transport vehicles
  7. The beauty of appearance
    The main requirement for machines and its manufacturability: Technologically advanced means that by spending less cocktails, it is said to make a high-quality machine
    Classification classification) of machine parts)
    1.1. Details of combinations or combinations. Compounds are mainly divided into two groups: compounds that can be separated, and compounds that cannot be separated.
    1.2. Mechanisms of transmission. These include coupling shafts.
    1.3. the parts that drive the gears are pulleys, sprockets.
    1.4. Shock and Shock mechanisms. This basically serves to cause one movement to flip another movement
    1.5. Spring and springs. They mainly serve to reduce vibrations and shocks.
    1.6. Flywheels, pendulums, Yucca, sabbaths. They are mostly used for uniform motion.
    1.7. Device prevent contamination and lubrication systems. They serve to ensure the long term operation of the machine.
  8. 8. The management arrangements.
    SECURITY QUESTIONS:
  9. What does the mechanism say?
  10. What is the difference between a car and an engine?
  11. What are the requirements for the car?
  12. What are the three different groups of materials used in the design of machine parts?
  13. What is meant by adaptability?
  14. what do you mean by standardization in mechanical engineering?
  15. What do you understand when it comes to designing parts?
  16. «Technical task», «technical recommendation», «technical sketch» and so on.

Lecture 2. THE FUNCTION OF THE COMPONENTS AND THEIR SOFTWARE
PLAN:
1.Strength.

  1. Elasticity.
  2. Resistance to vibration.
  3. The temperature resistance.
  4. Corrosion resistance.
  5. Reliability.
    Keywords: gear, shaft, shaft, motor, resonance. Strength, virginity, heat resistance, vibration and abrasion resistance are the main characteristics that determine the performance of parts. Force. The ability of a part to work smoothly and flawlessly under normal operating conditions is called its strength. The voltage that occurs in the parts can be from zero to maximum: shafts of one-sided, slowly rotating gears) Mechanisms with a voltage from minus to plus for example, shafts and shafts): The strength calculation basically begins with the determination of the allowable stress and the determination of the stress reserve coefficient. The total reserve ratio is as follows. n = n1n2n3 where: n1; n2; n3; — spare coefficients of each part.
    Rigidity. Force alone is not enough for some parts, especially those that are exposed to force. For example, under the influence of a certain force and torque, a rotating shaft may bend more than allowed, even if it is strong. Such a shaft should not be used because the distance between gears, such as gears, is limited. Excessive bending of the shaft will cause these parts to fail earlier than expected.
    Basic requirements for the rigidity of parts:
    a) the parts are uneven and the workmanship is accurate:
    b) the distance between the connected parts does not change:
    c) in dynamic forces no resonance phenomenon)
    g) provision of technological conditions:
    d) ensure the operation of the machine as a whole:
    Vibration resistance. Increasing the speed of the machine and reducing the weight of the parts allow for various vibrations. In this respect, the phenomenon of resonance is particularly dangerous. It is known that a resonance event occurs when the specific frequency of vibration generated by the part itself coincides with the frequency of vibration under the action of an external force.
    Thermostability. When designing machines, it is important to ensure that the heat generated in them does not exceed the norm. Here Q is the amount of heat generated by the machine: Q1 — the amount of heat dissipated outside the machine:
    Resistance to erosion. The service life of friction parts during operation is determined by the degree of wear. It is determined by the degree of rapid or slow wear of the part. Rapid or slow wear of the part depends on its operating conditions, the level of lubrication, the amount of contact voltage, and other factors.

where: — relative pressure
— friction speed.
Reliability. They say that the parts work for a certain period of time without losing their appearance. Reliability is as follows. Integrity, durability, maintainability and durability. A partial or complete loss of the ability to process parts is called a failure.
Determination of constant voltage.
Depending on the force acting on the part and the type of material used to determine the permissible stress, the tensile strength can be taken as the ultimate strength for plastic materials) or the ultimate endurance for materials with a variable load cycle). For plastic materials at the same time for aggressive materials: where is the acceptable safety factor.
plastic materials

endurance for materials: cycle.
Where: — or the ultimate endurance for materials with a variable load cycle.
This is the of reliability .
elastic limit.
the limit of readability age resistance
tensile strength.
CONTROL QUESTIONS

  1. What an opportunity to handle the details?
  2. what is reliability?
  3. How is the permissible voltage determined?
  4. What is the difference between the operating limit and the allowable stresses?
  5. What is contact voltage and in which parts it is present?
  6. What is meant by inspection of design and calculations?
    LECTURE 3. JOINTS. RIVET JOINT.
    PLAN:
  7. General concept of joints.
  8. Types of riveted nail joints.
  9. Calculation of riveted nail joints.
  10. The preparation of the riveted seams of the nail.
  11. Advantages, disadvantages and application of riveted nail joints.
  12. Materials from which riveted nails are made

Keywords: rivet nail, head, standard, steel.
Links: 1. 2. 3. 4.

 Joints are divided into one-piece and split. If it is necessary to break the connecting elements in order to disassemble the components or the machine, this joint is called an all-in-one joint, otherwise-a split joint. If the joints are riveted and welded, the joints are unbroken; dowel, grooved, and bolted joints are split.

The rivet joint..
The rivet is a round nail with a round head. The second head is installed on the head after mounting on the parts.
Figure 1. Riveted joints.
a) reinforced: C) reinforced-compacted.
Rivet joints have been used mainly since the 1930s.
Reinforced joints are mainly used in the construction of metal structures.
Advantages of riveted joints over welded ones:
a) Less damage when separating parts:
b) Resistant to vibrations.
c) High strength:
Disadvantages include the following:
a) Metals are expensive.
b) special inserts are Required.
c) it is more expensive to Attach thick parts.
d) 20-30% of the cost compared to expensive welded joints.
Types of rivet joints.
According to state standards, rivets are divided into the following types:
Semicircular head rivets are mainly used for reinforced and reinforced joints. Semi hidden rivets are mainly used to connect parts up to 4 mm thick. Hidden rivets are mainly used in steam boilers.
Flat head rivets are mainly used in areas with high corrosion:
Hollow rivets are used where the weight of the nails is small.
a) — round: b) — semi — conical head: C) — hidden head: d) — low round: d) — flat head: d) — hollow riveted nails.
Flat head rivets are mainly used in areas with high corrosion:
Hollow rivets are used where the weight of the nails is small.
a) — round: b) — semi — conical head: C) — hidden head: d) — low round: d) — flat head: d) — hollow riveted nails.

Calculation of rivet nail joints. Calculation of the rivet section of the nail

Where: — is the force acting on one riveted nail
— cutting surface with riveted nails
— permissible shear stress.

for rivets nails in the second cut.
number of riveted nails

  1. Calculation of crushing the rivet nail.
    S- the thickness of the part
    If, in addition to the bending moment, the riveted nail is also affected by the transverse force Q and the tension force N, the forces acting on the riveted nail are equal.

Where: — is the total number of rivet nails in the seam
— force N corresponds to one common nail
— Q force value per riveted nail:
For this case, the strength of the riveted nail is checked using.

SECURITY QUESTIONS

  1. What joints are called inseparable joints?
  2. What are rivets?
  3. What riveted the nails?
  4. what materials are used to make riveted nails?
  5. what are the classifications of rivets by seam and construction?
  6. What is the relationship between the diameter, pitch, and thickness of the rivet nail material?
  7. What is the strength of rivets?
  8. what are the advantages and disadvantages of a riveted nail joint?
    LECTURE 4. WELDING AND ADHESIVE JOINTS.
    PLAN:
  9. The General concept of a welded joint.
  10. The types of welding.
  11. Calculation of welded joints.
  12. Adhesive compositions.
  13. The calculation of the adhesives.
    Keywords: welding, welding, electrode, strength coefficient, glue.
    Links: 1. 2. 4.

Welding is mainly divided into the following types:
a) electric Arc welding:
b) electric Welding or fusion welding of parts)
c) Contact welding. Electric welding is divided into 3 groups: 1. Automatic welding: 2. semi-Automatic welding: 3. Manual welding
Automatic welding is inexpensive, high-performance and high-quality. In semi-automatic welding, some sections are welded manually, and the quality of welding depends largely on the skill of the welder. Electric arc welding was invented by 7 Russian inventors N. N. Benardos, 1882 and G. Slavyanov, 1888). the Electrodes are manufactured according to GOST 9467-60. Their brands are E 34, E 42, E42, A: E 50 A, E 55 and so on.
The numbers in the letter E indicate the strength of the weld when welding with this electrode. kg / mm2). Here, the letter A increases the viscosity of the weld when welding with an electronic electrode.
Contact welding is mainly used for welding materials up to 3 mm thick.
Basically, they are divided into roller and point.
Advantages, disadvantages and applications of welded joints compared to riveted joints,
Advantages:
I. the Price is not very high 20/30%) compared to a cheap riveted nail.

  1. The strength and density of the joints.
  2. The ability to automate the welding.
  3. The possibility of welding even thick pieces.
    Disadvantages:
  4. The quality of the seam mainly depends on the welder, this defect is eliminated by automatic welding.
  5. The curvature of the parts. In places that are not welded in the plane.
  6. Loss of weld strength due to envelopes and vibrations.
    Area of use.
    Welded joints, frame frames, gear housings, pulleys, gears and crankshafts are used. It is also widely used in construction.
    Types of welded joints.
    Triple welded seams. This is mainly done by electric welding
    Figure 5. The triple seam And the seam with no sharp edges.
    Corner seams: a) regular seam: b) concave seam: C) convex seam: d) special seam: «T» — cement welds. In this case, the parts are welded vertically on top of each other.
    F igure 6. » T » — cement welds.

Calculation of welded joints

  1. For butt-joint. where — is the length taken to calculate the seam
    — thickness of the sheet at the welding point
    — permissible voltage for the welding material.
    — strength factor where

This means that when the sheets are joined in threes, the strength of the weld is almost equal to the strength of the sheet. If for some reason it is necessary to increase the strength of the seam, then the seam is expanded by turning it to one side.
A welded joint is a three-post joint.

  1. Calculation of the upper weld. When two parts to be joined, such as two sheets, are placed on top of each other and welded, a joint is formed with an overlay. a) face-to-face weld.
    Mostly and
    B) Side weld.
    The side seam is calculated using the following formula

Calculation of contact welding. If the sheets are welded with three contacts, the strength of the weld is equal to the strength of the sheet. Sheets can be welded in two ways, one of which is spot welding and the other is sheet welding. In this case the diameter of each point is selected depending on the thickness of the sheet S
bo’lsa if d=1,2 S+4mm
bo’lsa if d=1,5 S+5mm

The strength of such joints is calculated taking into account the point shift.

where z – is the number of joint points
i – the number of planes that can be cut at each poin.
Welding the tape by contact method.

Here is the width of the weld - seam length

Adhesive composition.
Adhesives are used for bonding various materials. Advantages: 1) bonding of various materials metal plastics, etc.) 2) density: 3) corrosion resistance: 4) the ability to glue very thin materials. Disadvantages: reduces heat resistance. Currently, the following adhesives are most commonly used BF: VK-I: VX- » MPF-I). The calculation of adhesive joints is mainly carried out on the displacement of parts, the permissible shear stress by the method of material resistance.

CONTROL QUESTIONS

  1. What are the advantages of welded joints riveted with nails before?
  2. What are the types of welded joints?
  3. The calculation of the welds?
  4. How is the contact-welded joint calculated?
  5. what are the disadvantages of a welded joint compared to a riveted one?
  6. How do I determine the allowable stress value for a weld?
  7. What are the advantages of automatic welding before welding ash?
  8. what are the types of contact welding?
  9. What brand of electrodes is used for welding?
  10. What are the types of welds for three or three welds?
    LECTURE 5. THREAD JOINTS.
    PLAN:
  11. General concept of threaded joints.
  12. Types of thread.
  13. Basic geometric dimensions of the thread.
  14. Basic methods of thread preparation and non-opening.
    Keywords: thread, inch, table top, washer, tire

Threaded joints are one of the most common split joints. Thread is common for bolts, screws, studs, nuts, and other parts. The main elements of threaded joints are threaded. It is cut mainly in the form of a helical line. Stretch the screw to make a triangle

Where: — the lift — angle thread
— pitch the inside
— diameter of the thread

Advantages:

  1. Sufficiently reliable under heavy loads:
  2. it is easy to Divide them:
  3. Threaded parts working under different conditions can produce many:
  4. Relatively cheap:
  5. All sizes have standardized on
    Disadvantages:
  6. Insufficient resistance to variable forces:
  7. Some technical difficulties in the production of special threaded parts
  8. Loosening of joints at work.
    Classification of threads.
    The thread can be cylindrical or conical depending on the shape. Depending on the profile there are 5 types of thread
  9. Triangular thread
  10. Buttress thread
  11. 3. Trapezoidal thread.
  12. Rectangular thread
  13. Halfway there

Depending on the rotation of the thread, there are right and left threads. Threads are divided into small and large. Depending on the income of the threads: one-way, two-way, three-way. Depending on the work performed by the carvers, they can connect parts and move from one shaft to another. Basic geometric dimensions of the thread.
d – external diameter of the thread
d1 – average thread diameter
d2 – internal thread diameter
S – refueling step
S1 – thread path distance along the axis of the screw
for single thread S=S1
for multithreaded threads S1=nS
Here n- is the amount of revenue
— working thread height
— angle of the thread profile
— angle of gradient
Types of thread..

  1. Metric thread. This is the most common thread. It can be large or small. Step threads are often used.
  2. Inch thread. These threads are also used in the CIS countries, =550 . Inch thread is mainly used in machines imported from abroad. 1 inch = 25.4 mm.
    (1 Dyuym=25,4mm)
  3. Tubular thread. This is a small inch thread. For example, if the thread surface diameter on the outside of the pipe is set to one inch, it will be 33.25 mm instead of the usual 25.4 mm, and if set to half an inch, it will be 16.66 mm.
  4. Trapezoidal thread. =300
    Used in jacks, screws-nuts. The efficiency is high.
  5. The carving on the pillars. =270
  6. Rectangular thread. =0
  7. A semicircular thread. =300
    The following methods can be used to prevent threads from opening:
  8. Counterweight and spring washer. This increases the total resistance of the thread for additional details.
  9. Use a splint or wire. In this case, the nut is attached to the rod Use a splint or wire. In this case, the nut is attached to the rod.
  10. Using the method of welding. The nut is welded to the part.

Calculation of thread strength.
The following ratio is used to calculate stream compression

Where: is the number of threads on the nut height N. This formula also applies to the screw thread.
Thread cutting is found in the following formulas
for the screw
for the nut
where is the coefficient that takes into account the thread type.
for triangular thread K=0,8
for rectangular thread K=0,5
for trapezoidal thread Mostly K=0,65

Steel screws, bolts, and studs have 12 strength levels defined by two numbers: 3,6, 4,6, 4,8, 5,6, 5,8, 6,6, 6,8, 6,9, 8,8, 10,9, 12,9, 14,9. Multiplying the first number by 100 represents the lowest strength limit, s m in MPa); multiplying the first and second numbers by 10 gives an excellent readability limit in MPa). When selecting threaded parts, the load value and the operating conditions that affect it are taken into account. Carved parts are mainly made of various grades of steel. These include CT 3, 10, 20, 35, 45 and other steels for General purpose threaded parts, as well as alloy steels 40G, 35X, 38XA, 45G, 40G2, 40X, 30XGSA for heavy threaded parts. 16xsn and similar steels are included. Plastic threaded parts are used in electrical machines and in many places.
Strength and mechanical properties of screws, bolts and studs at normal temperatures
Level strength Ultimate
strength
σ m MPa
Yield limit
σoq MPa

Steel grade
Bolt, Screw, Stad
Nut

3.6
4.6
5.6
6.6
8.8
10.9
4
5
6
8
10
12
300-490
400-550
500-700
600-800
800-1000
1000-1200
180
240
300
360
640
900
St3; 10
20
30; 35
35; 45; 40G
35Х; 38ХА; 45G
40G2; 40Х; 30ХGСА; 6 ХСN

SECURITY QUESTIONS

  1. What are the classifications and functions of threads?
  2. What threads are used for cutting?
  3. When is a small metric thread used?
  4. What is the relationship between the axial force and the rotational force in the thread?
  5. What materials are used in carving?
  6. What can be done to prevent the thread from loosening?
  7. As the bolts are tightened?
  8. Where are the studs?

LECTURE 6. SCREW JOINTS.
PLAN:
1.Advantages, disadvantages and areas of use of double helical transmission.

  1. Determine the tightening torque of the screw.
  2. Calculate the threaded joint.
  3. Determine the ratio of the nut to the surface area.
  4. screw Detection F. I. C ..
    Keywords: screw-nut, kinematic pair, torque and friction moment. A kinematic double screw nut is used in machines and mechanisms to convert rotational motion into translational motion. In a loaded pair, the torque is converted to the axial force on the screw or nut.
    Figure -10. Direction of the forces acting on the screw pair.
    Where: Q- is the force directed along the axis i
    P- rotational force Teeth friction
    Tish – the friction force
    • angle of friction — the average diameter of the thread.
      • normal reaction force
      • coefficient of friction
    • the angle of the thread.
      Replace the force N and the force of the tooth with the force R acting on the thread. The projection of forces on the x-x axis gives the following.

from here will be.

The resulting equation applies only to rectangular profile threads.
For triangular threads. Nish=fN
Where

Figure-11

When rectangular and triangular thread and force =0 , then
N1=Q .
In that case
Where coefficient of friction given .
— specified angle of friction.
Thus, for threads with triangular and trapezoidal profiles, the rotational force is equal to.

Torsion moment.
When the nut or screw is turned, the following force is applied to the wrench.

where R- is the force applied to the switch.
L- is the length of the key
M- rotational force P – by the average diameter of the thread.

V – is the tensile force of the bolt this force is replaced by the axial force.
Mish – the moment of friction of the nut surface on the part.

The cutting surface of the nut has a ring shape, and the relative pressure exerted on it by the expression.

Where: D1 — outer diameter of the ring, the value of which is equal to the key size.
d0 – diameter of the hole where the bolt is inserted.

Figure 12. The scheme for determining the torque
The diameter of the nut ring is found by differentiating the strength of the elementary friction generated in the ring as radiusi and thickness d of the elementary ring.

Figure 13. Scheme for determining the moment of friction that occurs in the nut ring.
Elementary moment of friction

Friction moment at the edge of the nut

;
In this case, the torque

Self-braking of a pair of propellers.
All hardened threads are automatically slowed down.
State of braking by a double-drive screw.
Where: — is the angle of the thread lift.
— reduced friction angle.
It is small on a fine metric thread, so braking on a fine metric thread is reliable.
Here is useful work from a pair of an-screw
Where: An — The EFFICIENCY. screw pair is defined as follows
A3 — is the work done by a pair of screws An=QS;
In this case, it ; will be

SECURITY QUESTIONS

  1. Where are screw pair extensions used?
  2. How to determine the torque on the screws?
  3. Where are twin screw extensions used?
  4. How are the tension bolt joints calculated?
  5. What material and voltage are acceptable?
  6. How to determine the efficiency of the nut and screw transmission?
  7. how do I determine the state of the nut and screw gear braking?
  8. how is the force acting on the wrench when turning the nut or screw determined? 9. How to determine the friction angle of the screw pair drive?
  9. How to determine the moment of friction of the nut surface on the part?

LECTURE 7. Key joint .
PLAN:

  1. Advantages, disadvantages and application of keyway joints.
  2. Types of dowels.
  3. Calculate the prismatic key.
  4. Calculate the segmented key.
  5. Materials used in the manufacture of dowels.
    Keywords: key, segment, prismatic, and potential dowels. Keyway joints are mainly used for connecting cylindrical parts. These joints are mainly used to attach gears, pulleys, couplings, and other similar parts to the shaft.
    Advantages.
  6. the Design is simple, the design is reliable.
  7. Easy to assemble and disassemble.
  8. Low cost.
    Disadvantages.
  9. Low load capacity of the connected parts.
  10. it is Difficult to place the parts accurately.
  11. Cutting out the shafts, that is, opening the space for the dowel, which, in turn, reduces the strength of the shaft.
    Keyway joints can be stretched or not strained. Stuffed dowels are used for tension joints, and prismatic dowels are used for tension-free joints.
  12. the Joints formed during the prismatic visit of the key require high-precision processing of both the key and the shaft, since in such cases the side collars of the key must evenly touch the side collars of the shaft.
  13. Panels are used in the form of prismatic rods. Two types of wedge dowels are used: a) key without head according to GOST 8792-58. b) key with heads according to GOST 8793-58.
  14. tangential dowels are installed on the shaft in 2-3 places 1: 100). This, in turn, puts the shaft in 2-3 places, rather than in one place. This, in turn, reduces the probability of failure of the shaft.
  15. Segment dowels, as well as prismatic ones, transmit the load by means of side clamps, the hole for the dowel is opened by a disc cutter. The advantage of these dowels is huge. They are widely used in machine tools.

Materials used in the manufacture of keys, and the permissible stresses for them
Standard keys are made of steel ST 5, St 6, 45, 50, 55, 60 according to GOST 8757-68, 8786-68.
Fixed anchors allowable load on the joints,
Here: — Yield strength of steel materials::

In the manufacture of keys according to GOST 8787-68 . steel keys made of steel grade 45 are used in gearboxes:
For keys, the permissible shear stress is in the range of .
Select a key from the tables and check for strength

  1. Calculate the crushing of the prismatic key.
    Rotation force that crushes pin transmission surface where f=0,05h chamfer key.
    Without it,
    Here T=M- is the torque
    (0,95h-f)- working slot in Spline
    — voltage in the allowable pressure.
    — working length of veneer
  2. Calculation of the segmental key for crushing

Here l – is the length of the key
(h-t)- working Spline in the loop
Calculating intersections
where — is the width of the key
— Working Spline in the corner
Calculating intersections
where — is the working length of the key.
— coefficient of friction

CONTROL QUESTIONS

  1. What is the key?
  2. What pins are used in engineering?
  3. What materials are used pins?
  4. How do I determine the key size?
  5. how are dowels determined for strength?
  6. How is the segment key calculated for strength?
  7. When are segment dowels used?
  8. what are the advantages of a key?
  9. which dowels are used in stretched joints?
  10. What are the advantages of tangential keys?

LECTURE 8. Spline joint .
PLAN:

  1. Advantages, disadvantages and areas of use.
  2. The types of Spline.
  3. The calculation of the Spline.
  4. Materials from which the Spline is made.
  5. Calculation of the Evolvent profile Spline.
    Keywords: Spline, involute, hub, heat treatmen If shallow grooves are laid on the outside of the shaft and on the surface of the hole in which the part is mounted, and one of the parts is installed so that the line passes into the bottom of the other, a Spline joint is formed.

The spline joint has the following advantages over key joints:

  1. The details are well centered on the shafts, the axial displacement is clearly aimed.
  2. a Spline Of the same size can cause a large torque, because the contact surfaces of the parts are large. 3. Works well under dynamic loads, since the forces are evenly distributed between the teeth.
  3. The shaft is less weakened.
  4. the loop Length is reduced.

The main drawback of the Spline layout is the complexity of its preparation and high cost. The compositions of the Spline are divided into: Depending on the nature of the Association, it can be mobile or stationary. Depending on the shape of the tooth, it has straight teeth, triangular teeth, and involute teeth. Depending on the center of the shaft relative to the shaft, the outer and inner diameters, and the side surfaces of the teeth.

Figure 14. a) involute profile tooth and b) triangular profile tooth
Make sure that the parts are aligned with the shaft and the shaft is centered on the Spline diameters D or d. However, when centering the sidewalls, the load is evenly distributed between the Splines. As a result, the joint can operate at higher loads. The dimensions of the Spline are selected from the tables according to GOST, depending on the diameter of the shaft, the type of Spline.
Calculation of the strength of splined joints . splice joints are mainly tested for crushing teeth.

Picture-15

Here — is the number of teeth
— average tooth diameter
— the surface of the crushing of the tooth.
For a rectangular profile Spline

where — is the working length of the tooth for evolevent
Spline profile
where: m – tooth module.

The permissible value of the crushing stress for the Spline is determined by the operating conditions of the joint and the heat treatment of its components. When the surface of the Spline is not subjected to heat treatment. When heat treatment of the surface of the Spline:

.::

SECURITY QUESTIONS

  1. How can I determine the spline?
  2. What are the advantages of the spline?
  3. As centered rectangular spline?
  4. How are the spline strength calculated?
  5. what are the disadvantages of spline joint?
  6. What are the different types of spline?
  7. What are the splines depending on the shape of the tooth?
  8. what kind of deformation is checked for rectangular splines?
  9. How do I determine the permissible value of the spline crushing stress?
  10. How do I determine the grinding surface of the tooth for the Evolvent profile splines?

LECTURE 9. The trees and wasps.
PLAN:

  1. The trees and wasps are places where they are used.
  2. Types of shafts.
  3. Calculation of shafts.
  4. Count the axes.
  5. Preparing shafts and expenditure.

Keywords: Shaft, arrow, compensation, step, ring. comb

Shafts and axles are the main parts used to install gears, pulleys, and similar rotating parts. The shaft is mainly used for transmitting torque. The shaft is torsion, the shaft is only bent. Depending on the geometry of the shaft and the shaft can be straight, an elbow or flexible. It is smooth, stepped in length. It is divided into solid and hollow shafts.

Figure 16. Main types of shafts and shafts: a-smooth transmission shaft b-step shaft
Shaft and shaft construction elements
The position of the shaft and axis at the base is called the groove
They are divided into the neck, heel, and inner part of the neck. Compensation mainly for axial forces. The base areas of compensation are called the lower part of compensation. Compensation is mainly divided into 3 groups: integral, ring and comb.

Figure 17. Compensation: a-whole: b-ring: v-scallop:
The shaft and transmission surfaces of the axis are cylindrical and conical. The shafts are mainly designed for fatigue strength and stiffness.

Calculation of shafts for strength
Depending on the degree of accuracy, the calculation of shaft strength can be divided into the following methods. 1. The approximate method. 2. The approximate method. 3. Specific method relatively accurate method)

  1. The approximate diameter of the shaft

where T- is the torque, Nmm
— Permissible torsional stress for steel shafts of grades 5,6,35,40,45,40 X
.

  1. The approximate method. The diameter of the shaft in the most dangerous section is determined by the following formula, taking into account the torque T and the torque M acting on the shaft.

where is the equivalent moment
value total bending moment value:
bending moment in the design section of the shaft in the horizontal and vertical planes, the value of which is taken from the shaft diagram:
T – torque,
3. Relatively accurate method:
a) When checking the fatigue strength of the shaft, the safety margin of the shaft is determined..

where =1,5…2,0 a valid value of the factor of safety:

  • the values of safety factors for normal and applied voltages: value of the bending stress resistance limit:
  • the value of the resistance limit for the applied voltage:
  • the strength limit of the selected shaft material, the value of which is taken from the table:
    • normal voltage on the test shaft.
    • rated voltage on the shaft segment to be checked
    • moment of resistance and polar resistance of the shaft section:
    • coefficients that take into account the effect of stress on the value of the safety margin
  • the safety factor of the material and the diameter of the shaft.
  • constant value of voltage cycles.
  • coefficient that takes into account the influence of the variable part of the voltage cycle on the value of the safety margin.. Calculation of shafts for static strength.

The acceptable value of the yield stress of the shaft:

Calculation of shafts for stiffness:
a) the maximum bending

where: l – is the distance between the shaft and the base
Rotational and radial forces on R-and T-shaped shafts.
E – the modulus of elasticity
J- moment of inertia of the shaft section:
b) the maximum torsion

where — . is the torque

  • shift module
  • inertia of the polar moment.
  • the value depends on the type of work performed by the shaft
  • taken in the interval

Calculation of the strength of wasps

The arrow-like shafts. Only Mbur = 0 is accepted, since It does not increase the torque, it is calculated for bending.
      it is obtained in the range.
Forces acting on the shaft.

The shafts are mainly affected by the rotational forces Ft, perpendicular to the axis Fz and longitudinal Fa.

  1. Forces generated on cylindrical gears with bevel gear
  2. Forces arising in a conical transmission
  3. Forces generated by the worm gear.

CONTROL QUESTIONS

  1. What is the difference between shaft and axle?
  2. What parts of shafts are called sapphires, ceilings, necks and heels?
  3. what materials are the shafts and shafts made of ?
  4. How are shafts calculated?
  5. how are shafts calculated for stiffness and fatigue?
  6. what forces act on the shafts?
  7. As arrows are calculated on the force?
  8. How to determine the permissible value of the bending stress of the axes?
  9. How are shafts calculated for stiffness?
  10. Which shafts are called special shafts?

LECTURE 10. MECHANICAL transmission. Friction transmissions.
PLAN:

  1. General concept of mechanical transmissions.
  2. Classification of extensions.
  3. Kinematic calculation of transmissions.
  4. Forces acting on mechanical transmissions and their determination.
  5. EFFICIENCY. Definition of mechanical transmissions.
    Keywords: machine, friction, roller, contact, EFFICIENCY.

Rotary motion is the most common motion in machines and mechanisms and has the following advantages: Provides smooth movement and requires a little effort on the friction. Creates a simple and compact extension. A transmitter is a device that transmits energy from a source to a consuming organ. Mechanical, pneumatic, hydraulic and electric transmissions are used in mechanical engineering.
Classification of transmission
The extension is divided into 2 groups depending on the operating mode:

  1. Friction gears friction and belt gears. 2. Gears gears, worm gears, chain gears. It is divided into 2 groups depending on the combination of leading and leading links. 1. Transfers by contact — friction, gear and worm: 2. Flexible coupling-belt, chain. Flexible couplings are used when there is a large distance between the drive shaft and the drive shaft.

The kinematic transmission calculation
All transmissions provide power transmission to the drive and driven shafts and angular speeds to the drive shafts. These two parameters are necessary to complete the design calculations in any extension.

Figure 18. Diagram of the direction of moments in the transmission: Wheels a in working position: wheels b in uncoupled position.
Additional transmission characteristics
A) mechanic C. P. D.

in multi-speed transmissions

where: — each extension is EFFICIENCY. gears
worms, belts and other extensions, bearings, couplings.

B) rotation speed

where D- is the diameter of the wheel, pulley, pulley, etc., mm.
V) rotational force

where: N – power, kVt
G) torque.

where i- is the number of revolutions

For multi-stage extensions
where: — the number of revolutions of each gear.
CONTROL QUESTIONS

  1. What is mechanical transmission?
  2. How are design and inspection calculations performed?
  3. What are the classification of a manual transmission?
  4. Classified as mechanical transmission?
  5. How to determine the torque in the mechanical transmission?
  6. how do I determine effectiveness?
  7. How is the speed determined?
  8. what materials are the shafts made of?
  9. What is the function of the reducer?
  10. What is the function of the multiplier?

LECTURE 11. Friction transmission .
PLAN:

  1. Advantages, disadvantages and use of friction transmission.
  2. Classification of friction transmission.
  3. The eating of cats.
  4. Geometric parameters of the cylindrical friction transmission.
  5. Calculation of the friction transmission.
    Keywords: cathode, variator, modulus of elasticity, radius of curvature. If the movement of the drive shaft is transmitted to the shaft by friction, this elongation is called friction transfer.
    The state of operation of the transmission.

where: R- is the rotational force
Tish- is the friction force at the point of contact:

Classification of friction transmission

Depending on the design, there are two types of extensions:
  1. the number of revolutions is constant.
  2. The number of turns in the process, changes equally such transmission is called a (CVT).

Fig. 20. diagram of the test variator
The shaft is divided into 3 types depending on the location of the shafts. 1. The axis — parallel cylindrical friction transmission. 2. Axes intersect by less than 90% — conical friction gears. 3. Axis cross each other is the variator face.
The extension can be open or closed depending on the operating conditions.
The extension can be open or closed depending on the operating conditions.
Advantages:

  1. The construction is simple, easy to clean.
  2. Works smoothly and noiselessly when working.
  3. the number of revolutions can be changed without stopping the movement. variators.
    Disadvantage:
  4. Work surfaces are uneven and wear out quickly.
  5. Creates high loads on shafts and bearings.
  6. The number of revolutions is changed due to slipping of the rollers.
    Application.
    Friction transmissions are mainly used in presses, hammers, tools. They are mainly used in areas with a power of 50 kW and a rotation speed of 25 m / s.
    Friction roller wear.
  7. Wear pitting corrosion) due to fatigue, which occurs mainly during closed extensions. If sand or metal shavings get in the way of the oil, the roller will get worse. 2. Badass. This propagation occurs due to the oil film breaking on the roller surfaces when the rollers rotate rapidly. 3. Mechanical wear. During the operation of the rollers, their surfaces wear out, changing their geometric dimensions.
    Fig.21. Diagram of the contact stresses the sump.
    Calculation of the cylindrical friction transmission.
    Number of revolutions.

where is the slip coefficient.
The distance between arrows

Diameter of the guide roller.

The diameter of the guide roller.

                             Transmission voltage
The extension must meet the following condition

where: friction force
f- coefficient of friction.
T – force of the spring compressing the roller.-
Rotation power.

in this case
hence the force of the spring
where: K- is the voltage coefficient
in strong gears K=1,25…1,5 in devices K=3…5
Calculation of cylindrical friction transmission for strength.

  1. Calculate the contact voltage
    The contact voltage is calculated using the Gers formula

where: q — is the normal pressure on the contact line:
Eg – is a given modulus of elasticity.
— the specified radius of curvature.

where: v2-width of a large skating rink.

Calculate the pressure in the contact line.
If the materials of friction gears do not comply with Hooke’s law fiber, rubber), then the materials are considered resistant to corrosion.

where a — is the width of a large skating rink

  • calculation of the permissible value of the permissible load.

SECURITY QUESTIONS

  1. Where are friction gears used and what is their classification?
  2. what are the advantages and disadvantages of friction transmission?
  3. How to calculate cylindrical transmission by friction?
  4. Describe the principle of operation of variators?
  5. what condition is necessary for the operation of the friction transmission?
  6. How to determine the Gers formula for contact voltage?
  7. How to determine the spring shear force?
  8. Explain how the laboratory variable works?
  9. If the materials of the friction transfer rollers do not comply with Hooke’s law, how is the wear of the rollers calculated?
    LECTURE 12. Toothed gearing .
    PLAN:
  10. Advantages, disadvantages and application of gears.
  11. Classification of gears.
  12. Erosion of the surface of the gear.
  13. Materials used in the manufacture of teeth.
  14. methods of preparation of teeth and the degree of accuracy.

Keywords: gear, involute, module, diameter of the divider. Links: 1. 2. 3. 4.

The mechanism by which motion is transmitted from one shaft to another by means of gears is called a gear train. While gears with a diameter of less than 1 mm are used in precision measuring instruments, it can be seen that in heavy industry they can reach several tens of meters in diameter. These are teeth with an involute profile according to Euventa, proposed by Eyler in 1760.

Advantages.

  1. It can transmit large several thousand kW power at a speed of up to 150 m per second, and the number of transmissions reaches several hundred:
  2. External dimensions are relatively small:
  3. the Force on the supports is small:
  4. High efficiency 0.97… 0.98):
  5. there is No slip event that negatively affects the number of gears:
  6. Allows you to use different materials:
    Disadvantages.
  7. relative complexity of cooking:
  8. Noise during operation:
  9. the Damage caused by the forces acting on impact is more noticeable.
    Features of the transmission skew geometry
    When the transmission speed is V> 6 m / s, it is recommended to use bevel or Chevron gears, because in order for straight gears to work satisfactorily at this speed, the accuracy of their preparation must be very high. ; ; where: — normal module
  • side surface module
  • normal step
  • side surface step
  • angle of deviation of the tooth:

It is said that the dividing diameter of the bevel wheel

where: z — is the number of teeth.

Classification of gear. Depending on the location of the gear shafts, there are the following: 1. Cylindrical-the axes are parallel: 2. Conic-axes intersect at 900: 3. Chervyakli-arrows pass over each other.

Depending on the profile of your teeth you can: Molitalia, cycloidal, gear transmission Novikov. There are two types of bites: Internal and external bite. Depending on the design, it can be closed or open.
Change the shape of the tooth by moving the toothpick
Zmin =17÷18 The shape of the teeth in small-toothed gears is shown in figure 23 to prevent them from loosening. Attempts are made to reduce the number of teeth in order to reduce the geometric dimensions of the gear. Reducing the number of teeth leads to a decrease in the coating coefficient, which in turn leads to a decrease in the strength of the tooth. Usually this value is: Zmin = 17 ÷ 18.
Materials used in the manufacture of gear wheels.
Currently, gears are mainly made of steel, cast iron and plastic. Due to the need to reduce the size of gears when used in high-power machines, most of them are made of various steels, such as 40, 45, 50, 40, G2, 50G, 40X, 30 XGS and other steel grades. It is desirable that wheels made of steel 18-HGT, 20 X2N4A, 15X, 20X, 12XZA are operated under the influence of impact and changes in direction or speed.
The efficiency of the gears and their wear. Two main forces act on the attached teeth.
One of them is the force along the contact line A1, A2, perpendicular to the involute surfaces of the teeth. The second is the friction force created by slipping between the teeth:
. Of these, the main stresses that determine the performance of teeth are the contact stress on the tooth surface and the bending stress in the lower part of the tooth.
Under erosion of the surface of the teeth should be understood as follows.
a) Crushing due to fatigue. b) erosion of abrasive particles and under normal friction conditions. c) in the case of a heavy duty transmission where one wheel tooth is detached from the surface and adheres to the surface of the other wheel: g) displacement due to plastic deformation: d) cases of displacement of the outer hard layer of heat-treated teeth:

CONTROL QUESTIONS

  1. What is the classification of gears?
  2. what is the main advantage of a gear train over other gears?
  3. Why is the gear extension with Evolvent profile widely used?
  4. What is the bite module?
  5. Write down the geometric dimensions of the tooth?
  6. How do I find the minimum cost of a tooth?
  7. what is the purpose of dental correction according to the Evolvent profile?
  8. how are the forces acting on the tooth determined? 9. What gear parameters are standardized?

LECTURE 13. THE THEORY OF GEARING.
PLAN:

  1. The theory of gearing.
  2. Calculate the contact force of the tooth.
  3. Geometric parameters of the gear train.
  4. Determine the forces acting on the tooth.
  5. Advantages and disadvantages of the Novikov tooth.

In a gearbox, one wheel tooth is formed by attaching it to the other wheel tooth. Keeping the number of revolutions constant gives the bite theory.
In the theory of engagement.

where: — angular velocity of the gear
— angular speed of the wheel
— is the radius of rotation of the gear
— the radius of rotation of the wheel.
The dividing circle is used to determine the geometric dimensions of the tooth elements. The following equation can be constructed for the length of this circle on each wheel

where: z – is the number of teeth on the wheel:
Pz — tooth pitch:
d — diameter of the dividing circle.

The rotation speed of point C relative to the centers O1 and O2.

Let’s use the similarity of triangles AEC and BCO1
that is from here
we use similarities.

from here
in this
case number
of circles.

Using .,
Acceptable values of contact and bending stresses.

where: ; — Values are taken from the table depending on the material of the gears, their heat treatment and the size of the tooth surface.
; — coefficients that take into account the service life of the transmission.

The basic dimensions of transmission

  1. The distance between the axial.

where: — the ratio of the distance between the shafts for bevel gears =430; for straight gears

  • torque on the drive wheel shaft, ;
  • coefficient that takes into account the unevenness of the load distribution on the tooth surface.
  • coefficient that takes into account the position of the transmission wheel relative to the base points in the symmetrical position = 0,4….0,5, in the unbalanced position = 0,25…0,4, in the cantilever position =0,2…0,25
  1. The transmission module.
    m=0,01…0,02)aw, mm
  2. The Total number of teeth and the angle of inclination of the driving wheels.
    The minimum value of the tilt angle for inclined gears.
  3. Total number of extension teeth. must be. bo’lishi kerak
  4. The number of teeth of the driving and driven gears.

The calculated z1 value is rounded for straight and inclined gears.

Number of gears driven:

The diameters of rotation of the extension wheels.
a) the Diameter of the dividing circle.
;
b) Outer diameter.

v) Inner diameter.

  1. The forces generated by the gears.
    Circular power h
    A force that tends to the center
    Longitudinal force
  2. Bending stress of the wheel teeth.
    A) for steering wheel

b) for steering wheel
.

SECURITY QUESTIONS

  1. What do you mean by gear theory?
  2. What methods of teeth whitening are currently available?
  3. How is the module calculated in a bevel gear?
  4. What is the advantage of gear transmission Novikov with its use?
  5. what are the advantages of the Novikov gear transmission?
  6. Why is the tooth tilted?
  7. How large is the thermal rotation of the tooth in elongation?
  8. what is the minimum cost of a tooth?
  9. what materials are teeth made of?
  10. How does tooth surface erosion occur?

LECTURE 14. Worm gearing .
PLAN:

  1. Advantages and disadvantages of worm gears and applications.
  2. The classification of the worm gear.
  3. The sliding speed of the transmission.
  4. Calculate the power of the transmission.
  5. Materials used in worm gear.
    Keywords: worm, worm wheel, sliding speed.
    Links: 1. 2. 3 Worm gear is usually used when the geometric axes of the shafts intersect at right angles in the X-X and Y-y spaces.

1 – cylindrical worm 2-globular worm.
The leading link-worm can be cylindrical and globoid.
Advantages:

  1. Smooth and silent operation.
  2. the Ability to get a large number of gears with a compact transmission.
  3. Worm gear has braking properties.
  4. The possibility of obtaining a large number of revolutions. in some cases, the number of revolutions is up to 1000)
    Disadvantages:
  5. EFFICIENCY. in relation to other extensions. low 0.8… 0.82)
  6. Large heat release in the bite area during operation.
  7. The need for the manufacture of the teeth of the worm gear from the stable to the spreading of solid materials.
  8. Excessive wear of the cord.
    Area of use.
    Worm gears are mainly used where the power is low. In places where the power does not exceed 50 kW. They are mainly used in lifting machines and trolleybuses.
    The number of revolutions is as follows.

where: — z1 and z2 are the number of worm gears and the number of worm gear teeth. n1 and n2 — the number of revolutions of the worm and the number of revolutions of the worm wheel.
Classification of worm gears.
The outer surface of the worms can be cylindrical or spherical. Globoid worms are difficult to make, so they are expensive. Worms move left and right during rotation. There are 3 types of worms, depending on their location relative to the worm wheel.

Figure 25. Position of the worm wheel . a) down, b) sideways, C) up: depending on the type of worm screws, they can be Archimedean, tortuous, and involute.

Reasons for transmission failure.
Materials.
One of the main characteristics of the worm gear is a high sliding speed, which, in turn, negatively affects the hydrodynamic lubrication. The main cause of damage is the destruction of the surface, pinning and erosion of teeth.
Worms made of cemented steels 15X, 15XA, 20X, 12 XNZ, 20 XF). Their size is HRC-56-62 In some cases, it is made of 40,45,40 X, 40XN steels. This extension is used in manual transmissions.
Calculation of transmission strength.
The gear wheel is designed for bending.

where — gear wheel module

  • strength coefficient of the worm gear teeth.
  • angle of the contact line.
    The contact force is calculated using the Gers formula.
  • normal load acting on the contact line
  • modulus of elasticity.
  • the specified radius of curvature.
    Forces acting on the worm gear.

Picture. 26
The force of rotation on the wheel is equal to the axial force on the worm.

the rotational force on the worm is equal to the axial force on the wheel.

radial force-separating force.

The sliding speed in the extension

Picture 27
The speed of rotation of the worm.

Wheel speed.

where: and — are the diameters separating the worm and the wheel, mm
and — angular speeds of the worm and wheel, mm.

That’s all you need to do

The higher the sliding speed, the faster the transmission wears out
The coefficient of performance

where: — the number of double bearings on the banner.
— bearings efficiency =0,999
— efficiency taking into account hydraulic losses. =0,97…0,98
— taking into account the bite -=0,97…0,98
— efficiency. in a screw pair
.

CONTROL QUESTIONS

  1. What are the advantages of worm gear before gear?
  2. How is the sliding speed determined in the worm gear?
  3. at what time is the self-braking condition met?
  4. what materials are worms and worm wheels made of?
  5. What parameters are used to calculate the contact force of the worm wheel?
  6. What is the procedure for calculation of the worm gear?
  7. How to determine the efficiency of the worm gear?
  8. When used mnogodetnaya worms?
  9. What are the parameters of the worm gear is standardized?
  10. How to reduce heating of the oil?

LECTURE 15. CHAIN TRANSMISSION.
PLAN:

  1. Advantages and disadvantages of chain drives and applications.
  2. The classification of the transmission.
  3. Geometric dimensions of the transmission.
  4. Calculate transmission wear.
  5. Scanners transfer.
    Keywords: chain, drum, block-crankcase.
    Links: 1. 2. 3
    A chain drive consists of two gears sprockets) mounted on two parallel shafts, a guide and a drive that are connected to each other by an infinite chain attached to these sprockets.
    Picture 28
    Advantages.
  6. the distance between the arrows can be much larger Amax = 5 m.
  7. EFFICIENCY. high 0.97… 0.98
  8. there is less load on the shafts than on the belt drive.
  9. it is Possible to switch from one shaft to several shafts.
    Disadvantages.
  10. Relatively high cost.
  11. During the movement of rotation is not smooth.
  12. Makes noise during operation.
  13. Precision is Required when assembling the transmission.
    Chain drives are mainly used in lathes, conveyors, and agricultural machinery.
    Classification of transmission
    Depending on the type of chain roller, bushing, and gear. It is divided into chains single-row and multi-row depending on the load. Depending on the number of stars normal — one star, special — several stars. The extension can be open or closed.

Chain chain details
Chains come with bushings, rollers, and gears. Both the roller chain and the pinion are distributed equally. However, roller chains are more sensitive to stretching increased pitch) when the joint is spread out, making more noise than toothed chains, but they are lighter in weight. In General, it is better to use a toothed chain at high speeds.
GOST are made of chain: single strand standard S, single enhanced PDP, single-row stub PV, single-row reinforced PRT, and additionally produced 2-3-4-5-6 row chain. Sprockets are mainly made of cast iron. SCH 18-36, SCH-21-40, SCH 24-44, SCH 28-48 and steel 15,15 X and refined steel 40.40 X. The crankcase is used to protect the transmission from dust and grease.
Chains must be equipped with special tensioning devices:

Tensioner.
Picture -29

The number of revolutions of the chain drive.
The path that the star passes through the chain as it rotates, and in this case the speed of the chain
,
where: — the chain pitch
— the number of teeth of the leading and leading sprockets.
— angular velocity of the leading and guiding stars.
Number of revolutions

Main geometric dimensions of the chain drive:
Between the axial distance A=30…50)t
The length of the chain

  1. After selecting the chain step, the distance between the axes is determined as follows.

Forces acting on the chain.
Rotation force
where: — the diameter dividing the star.
The tension force of the chain.

where: q- is the weight of the chain at a distance of 1 meter.
Kf — coefficient of the suspension chain.
Centrifugal tension.

where: — g = 9.81: the speed of the v-chain. Voltage in the main working network circuit.

shaft voltage.
— shaft voltage coefficient
Calculation of wear of a chain drive. The wear resistance of the chain is checked by the specific pressure «R» between the roller and the bushing.

where: R- is the average pressure in the chain link.
R- circular force.

  • cutting surface of the chain loop.
  • chain utilization rate.
    .
    Where:
    Kdin – dynamic voltage coefficient
    KA – arrow coefficient
    KS- coefficient that takes into account the lubrication system
    KO- coefficient that takes into account the angle of the chain
    KR – coefficient that takes into account the operation mode of the circuit
    Kt – coefficient that takes into account the type of chain tension.

SECURITY QUESTIONS

  1. what are the advantages and disadvantages of a chain drive?
  2. What are the different types of chain tension?
  3. How do I determine the geometric dimensions of a chain drive?
  4. How are chains calculated so that they can be eaten?
  5. how is the number of revolutions of the chain drive determined?
  6. how to determine the forces acting on the chain neck?
  7. How is the chain length determined?
  8. What materials are made inverted tooth chains?
  9. How do I determine the diameter of an asterisk?
  10. How is the distance between the arrows determined?

LECTURE 16. BELT TRANSMISSION.
PLAN:

  1. Advantages and disadvantages of belt drive and application areas.
  2. Types of flat belt drives.
  3. Kinematic calculation of belts.
  4. Calculation of flat-belt transmission.
  5. The device of belt tension.
    Keywords: belt, pulley, speed, slip coefficient.
    Links: 1. 2. 3 The most simple are considered to be transmission tension belts, consisting of leading, driven pulleys and a tape stretched on them
    Picture 31. Scheme of a belt transmission.
    a) — flat tape: b) — tape: C) — round tape: d) — multi-tape
    The belt drive is also used for several revolutions of 100 kW and 25..50 m / s. the number of revolutions i = 10… 15. They are widely used mainly in electric motors, automobiles, lathes, conveyors.
    = 0,94…0,98.
    Advantages.
  6. Move over medium distances.
  7. Smooth and silent operation.
  8. Ability to work even at high speeds.
  9. Low cost compared to chain transmission.
    Disadvantages:
  10. Dimensions.
  11. Sliding tape.
  12. Tension in shafts and bearings.
  13. The use of additional tensioning devices.
  14. The need to protect the tape from oil.
  15. Short service life.

Flat belt transmission.
Flat belt drives are mainly used when the axes are parallel and overlap.
Types of flat belt drive.

  1. Cross extension. In this case, if the drive pulley rotates clockwise, the driven pulley rotates counterclockwise.
  2. The floor cross-transmission.
  3. Angular transmission.
  4. Idling rotating pulley gear.

The continuously rotating drive shaft O1 has a wide pulley 1, and the drive shaft O2 has two pulleys: one of which is a pulley 2 that is attached to the shaft by a pin, and the other is a salt rotating pulley that is easily installed on the O2 shaft. either the working pulley 2 or the only rotating pulley 3 can be connected while the pulley belt is rotating, so the drive shaft can rotate or stop at any time, regardless of whether the guide shaft is rotating continuously.
5. Step pulley transmission.
Tape structures.
Leather strap. They are made from the skins of cattle. They work for a long time and withstand high loads well. Rotation speed 40… 45 m/s.
Rubber belt. Widely used in mechanical engineering, produced 3 types:
Type A belt V>20m/s;
Type B tape V>15m/s gacha;
V-type tape up to V> 20 m / s;
Cotton fiber belts. They are flexible and work well on light pulleys of small diameter. Does not apply to areas with tattoos.
The belt is made of wool. They are flexible and work well on light pulleys of small diameter. Does not apply to areas with tattoos.
Special belts. They are made of synthetic fibers. In the future, these feeds will replace the ones listed above. They are strong, light, elastic, and can reach speeds of up to 100 m / s.

Kinematic calculations of the tapes.
Belt strength and tension.
; ; ; ;
where — is the belt tension force

  • power in the working network leading to the belt.
  • power in the network with belt drive idling.
  • torque on the belt.

Belt tension

where: — the cut surface
of the belt.
Centrifugal force.

wher: — strap weight
g= 9,81
v – speed, m/s
centrifugal tension.

Where: — is the relative pressure of the belt.
The bending of the belt tension

where: E- the Modulus of elasticity defined by the band.
S- the thickness of the belt.
D- diameter of the small pulley.
Number of revolutions.
; ;
where: — the coefficient of relative slip of the belt

But keep in mind that this formula can only be used when there is no slippage, when in fact, when the elongation goes, big or small, the tape will always slide. When investigating slippage, it should be divided into:
a) Elastic sliding. This slippage is due to the fact that the tension of different tape wires is not the same.
b) Tula slip skidding) resulting from belt overload:

Calculation of flat belt.
Gravity coefficient.

Practical useful voltage.

where: — useful belt tension.
K1 – the coefficient of adhesion of the belt to the pulley.

where α0 – is the angle of attachment of the belt with a small pulley.
In open transfers

 login allowed. 

For cross extension:

K2 – speed coefficient.
K2=1,04-0,0004 v2
where: v – is the feed speed: m/s.
K3 – coefficient that takes into account the position of the transmission relative to the horizon determined by the table
K4 – coefficient that takes into account the transmission load mode taken from the table
The cross section of the tape is defined as follows:

The width of the tape is determined by the thickness of the tape. found by default
The number of passes of the tape in a second

where: L – is the length of the strip.
To find the small diameter of the pulley, M. A. Severin Formula:

where: nmax — the number of revolutions of the shaft speed, rev / min.
N – N-power, HP due to horsepower.
The resulting diameter is selected according to OST 1655 or GOST 3133-56.
Centre distance.

Length of belt in open belt drive.

Cross on the belt drive.

V-belt drive.
The V- belt drive has the following advantages: quiet and smooth operation, good grip on the pulley, transmits a large torque. The voltage in the shafts is low, the transition from one shaft to several shafts without a tensioner.
Disadvantage-the price of the pulley is high, EFFICIENCY
Low =0,92…0,95 and short operating time.
Belt length

Calculation of V-belt gears.
Number of belts
Where: rotational force, kg:

  • allowable voltage, kg/sm2
  • belt cutting surface, sm2

Allowable voltage.

The thermal coefficient K2 is found by the empirical formula.

It is appropriate to assume that under automatic tension K2 = 1.

.
CONTROL QUESTIONS

  1. What are the different types of strips of the cross section?
  2. What materials are made of tape?
  3. How do I determine the geometric dimensions of the tape?
  4. What material is the pulley made?
  5. Why is the groove on the ponasimon pulley deeper than on the belt?
  6. Determine the order of the ponasimon ribbon?
  7. When toothed belts are used, what are their advantages?
  8. How do I determine the distance between the axes on the tape?
  9. what are the advantages and disadvantages of gears?
  10. What materials are made of tape?
  11. How to determine the angle of attachment of the belt to the pulley?

LECTURE. 17. COUPLINGS.
PLAN:

  1. Designation and classification of clutches.
  2. The vertical and the compensation coupling.
  3. Couplings-disk couplings.
  4. Clutch hinge.
  5. Gear couplings and their calculation.
    Keywords: Coupling, loop, punch, flange.
    Links: 1. 2. 3

Home Information. Classification of couplings.
The main function of the couplings is to turn them into a single shaft by connecting individual parts of the shaft along the length to transmit torque. One of the simplest couplings is the built-in coupling, which is designed to combine two separate shafts into one shaft to perform a single function.
Depending on the effect. Fixed clutches retain the shafts in one piece. The couplings are separated by a special control system.
The self-regulating couplings are automatically switched off when the operating mode is changed. Depending on the nature of the work, they are divided into: Rigid couplings that move from one shaft to another without changing the shock and vibration of the first shaft. Elastic couplings, such as these, reduce the impact, vibration, and vibration of the first shaft to the second shaft during the opening of the mechanism.
The main characteristic of the clutch is the torque that leads to this.
Mxis=KM
Here: K – is the coefficient that takes into account the mode of operation of the coupling.
When moving the electric motor:
If the load is flat K=1,15 … 1,4
If the load is variable K=1,5 … 2,0
if the load is pulsed K=2,5 … 4,0
Couplings are selected according to GOST depending on the shaft diameter and torque.
Rigid and compensating couplings.
Fixed couplings are divided into simple and compensating groups. They are firmly attached to the shaft by a single shaft. It is mainly mounted on slowly rotating shafts. The most commonly used single couplings are bushings and flanges. The buffer coupling. Holds the shafts with a dowel or a slot with a bushing.
The bushing is tested for strength.

where D and d are the diameters of the clutch.

  • permissible torsion stress 45 steel.
    The coupling flange. These couplings consist of two semi-couplings. They are connected using a bolt.

Compensating coupling. These couplings compensate for radial, axial, and angular misalignments that occur during Assembly.
a – radial displacement.
b – axial movement.
v– compensates for angle shifts

Gear couplings. They consist of two half-couplings with teeth of an involute profile on the inner side. The clutch is triggered when the tooth is inserted. The gear coupling is cast according to GOST 5006-55.

where: Mgost — maximum torque selected by GOST.
K1-safety coefficient. If the half-coupling causes the car to stop, then K1 = 1.0, in case of an accident, K1 = 1.2. K2-coefficient that takes into account the operating conditions of the machine K2 = 1.0 when working on a flat surface, K2 = 1.2 at variable load; K2 = 1.5 when working with impact in severe conditions.
Couplings-disk coupling. They consist of two semi-couplings with floating disks efficiency η = 0,97 … 0,98.
The wheels are mainly made of 45 l steel.
Power transmitted by a single tap,

Here R- is the maximum pressure.
h – is the height of the fist.

The coupling is hinged. The shaft is installed when the axes intersect below d = 450. They consist of two forks and a crosspiece. When the angular velocity of the first shaft does not change, the angular velocity of the second shaft changes in one revolution. They are widely used in universal joints of automobiles. Shaft diameters are selected at d = 10 … 40 mm GOST — 5147 – 69.

Where: MGOST – is the permissible torque of the coupling at δ=0 according to GOST.
Snake spring couplings.
It consists of two identical half-couplings, tightly inserted into the ends of the shafts and fastened with dowels. On the outer surfaces of the half-couplings there are teeth, between which 5-8 sections of the serpentine belt spring are placed.
To prevent the springs from falling, the coupling is inserted into a casing consisting of two phases bolted together. A sealant is placed between the cornea and the half-coupling of the gear. The teeth of serpentine spring couplings have a shape that provides an elastic joint with a constant or variable element. This reduces the impact caused by uneven operation variable load on the drive shaft.

SECURITY QUESTIONS

  1. What are the groups of couplings according to the principle of operation?
  2. How does the fixed coupler?
  3. As a special clutch?
  4. How are the couplings selected?
  5. What is the principle of cumulative couplings?
  6. How special clutches and how they work?
  7. Where the use of a flexible coupling?
  8. how are the couplings arranged?
  9. How do I determine the size of the coupling?
  10. When are couplings used?
    LECTURE 18. Elastic sleeve finger coupling MUVP.
    PLAN:
  11. Elastic coupling MUVP).
  12. Couplings.
  13. Clutches, friction.
  14. Self-propelled clutch.
  15. Calculation and selection of storage couplings.
    Keywords: Clutch, disc, sleeve, finger, rubber.
    Literature: 1. 2. 3.4.

To align the ends of the connecting shafts with each other and soften the impact caused by sudden movement of the machine and the load, elastic and flexible elements are added to the drive couplings. Couplings are made of cast iron Sh21-40 or steel 35 or 35A. The fingers are made of 45 steel. These couplings are usually installed after electric motors.To align the ends of the connecting shafts with each other and soften the impact caused by sudden movement of the machine and the load, elastic and flexible elements are added to the drive couplings. Couplings are made of cast iron Sh21-40 or steel 35 or 35A. The fingers are made of 45 steel. These couplings are usually installed after electric motors.
The tie fingers are bent.

where:: lb; db and D1 – are the finger and half-coupling sizes.
z – the number of fingers.
[σeg] = 80…90 N/mm2 – the permissible value of the permissible stress when bending the finger.
The rubber is bent.
where: lb – is the length of the rubber sleeve.
[σez] =2 N/mm2 the permissible value of the allowable compression stress.

Suspension couplings.
The clutch is used to quickly switch the shafts on and off while the machine is running. Depending on the principle of operation, it is divided into clutches and clutches. The clutch Cam. They consist of two semi-couplings with fists at the edges. Made of 20 X and 20 XN steel.
Cam Profile:
a) rectangular:
b) trapezoidal:
v) triangular:
g) asymmetric.
The main drawback of clutch clutches is the inability to add fast rotating shafts. If there is a difference in the speed of rotation, it must be added without loading.
The meal is calculated as follows.
where: 0,75 – is a coefficient that takes into account the fact that the loads falling on the teeth of the fist are distributed unevenly.
D1; b and h – are the sizes of the coupling.
[P] = 25…35 N/mm2 – permissible pressure in the fist.
Teeth are bent.

where: the moment of resistance to the flexion of the fist.
a – the average width of the fist. [σeg] ≈ [P].
Friction clutch.
This coupling serves to separate the shafts into a single smooth joint. Depending on the type of friction surface, it can be disk, conical or cylindrical. dry and oily depending on the lubrication conditions. Today, multi-disc couplings are widely used.
Friction coupling diagram
Self-propelled couplings.
Self-propelled couplings are used for automatic switching on and off of shafts. They are divided into the following groups. At the moment the clutch is cumulative. Angular velocity-centrifugal clutches. In the direction of rotation — the coupling.
Clutch is cumulative. These couplings are automatically switched off if they exceed the required power. Centrifugal clutch. They will split when you change the rotation mode. Mixed grip. This coupling performs the function of the three couplings mentioned.

CONTROL QUESTIONS

  1. Where the use of a flexible coupling?
  2. How are the drive and exhaust couplings arranged?
  3. How do I determine the size of the coupling?
  4. When are magnetic couplings used?
  5. Where are the floating clutch?
  6. Where are safety couplings used?
  7. Where the use of complex clutch?

LECTURE 19. BEARINGS. Roller bearings.
PLAN:

  1. Advantages, disadvantages and applications of rolling bearings.
  2. Classification and marking of bearings.
  3. Fundamentals of the theory.
  4. Bearing calculation.
  5. Determination of load-bearing capacity.
    Keywords: bearing, ball, coil, cone, needle.
    Links: 1. 2. 3 Rolling bearings are made of rollers or castors. Bearings, rings and rollers are made of shxb, SHX9, SHX15, SHX 15 GS chrome steels. Their hardness is 62-65, and they are polished after grinding. It currently has an internal diameter of 1.5 mm to 2.6 m and weighs 0.5 g. 3.6 tons of bearings.
    Advantages.
  6. Requires less force on the friction.
  7. Resistant to corrosion and fire.
  8. Low consumption of lubricants.
  9. Relatively cheap, because they are manufactured according to standards.
  10. The service is cheap.
    Disadvantages.
  11. it Has a high hardness due to the small working contact surface.
  12. Less operating time under heavy loads.
  13. Changing the life of bearings operating in the same mode.
  14. The complexity of the installation some parts like crankshafts).

Classification and labe.
Classification and labeling. Rolling bearings are divided into classes based on the following characteristics. rolled and rolled as rolling parts. Your own coils.
Depending on the perception of the load, it is divided into radial, radial, support and support. Depending on the size of the yard, only one or two rows will fit. It is not self-leveling and self-leveling spherical.
By external dimensions-divided into series. As for the outer diameter, ultralight series, light series, medium series and heavy series. By width-divided into thin, medium, wide and very wide series. All bearings are marked, for example, bearing «1121», the last 2 digits multiplied by five give the inner diameter of the bearing, i.e. 21×5 = 105 mm. At the end, 3 is the zero row, and 4 is the zero type.
Asosiy nazariyasi.
Distributed unevenly. If a load R acts on the bearing, then the loads acting on the bearings are R0; R1; Assuming R2 n Rn, the total load on the bearing is determined as follows.

where: ;2 ;…n — kuchlar orasidagi burchaklar. Basic theory. Loads on the support shafts are. If a load The load distribution between elements of the vibration in the radial bearing

where: R- is the radial load acting on the bearing:
D and d are the outer diameter of the outer ring and the diameter of the ceiling in sm:
— Grip angle: for radial bearing
friction losses that occur mainly between rings, ash and separators.

where: Q – is the force acting on the bearing, kg.
f – applied coefficient of friction f= 0,002…0,02.
d – shaft diameter, mm.
n – the number of revolutions of the shaft, rev/min.
Causes of bearing failure. Elements of rolling bearings fail mainly due to damage. Properly coated, it will withstand many adverse conditions. If the bearing is running under heavy load, its outer rings quickly fail. The failure of the separators is mainly due to the high pressure created by the solvents, which at high speeds is high due to the centrifugal force. The abrasive failure is mainly caused by poor closing of the bearing.
Calculation of bearings.
The rolling bearing catalogues show the durability and load capacity characteristics for each bearing size, which is called the performance coefficient and is indicated by the letter S table. The required efficiency factor S is found by the following empirical formula.

where: Q – is the number of conditional loads.
n – number of revolutions of the rotating ring per minute:
h- endurance, in hours.
Q – is determined by the following formula for radial bearings.

where: R – is the actual radial load,
A – actual load on the axis,
m – A – coefficient that converts the load to R load.
KB – margin of safety.
Kx — is a coefficient that takes into account the dependence of the resistance on the rotation of the inner or outer ring.
KT – CT is a factor that takes into account the reduced bearing life when operating at high temperatures.
The following formula is used when selecting a support bearing..

For example, let the axial load A on the shaft affect the first bearing A, in this case the applied load.

for the second bearing

The value of the axial component of the radial load is given by the following formula.

where: β – is the wetting angle.
The oils used in rolling bearings can be divided into two groups:
A – liquid oils-mineral oils:
B – oil consistency-thick oils:
When the number of revolutions of the rotating ring per minute is less than 50 and the shaft is working intermittently for example, on trucks the bearings must withstand static loads..
Depending on the static load capacity, the bearings are selected as follows:
Q ≤ Qst
Where: Q – loading that affects the bearing.
Qst – static load capacity of the bearing permissible static load, the value of which is indicated in the table.

CONTROL QUESTIONS

  1. Marked as rolling bearings?
  2. Describe the classification of rolling bearings?
  3. how are bearings calculated?
  4. What is the basic theory of rolling bearings?
  5. When are radial bearings installed?
  6. What materials made the bearings?
  7. What series of rolling bearings?

LECTURE. 20. The sliding bearing.
PLAN:

  1. Advantages and disadvantages of plain bearings and applications.
  2. Fundamentals of the theory.
  3. Lubricants.
  4. Bearing calculation.
  5. Materials used in the manufacture of plain bearings. Keywords: plain bearing, liner, bronze, babbit, cast iron, organic and mineral oils. Links: 1. 2. 3

Rotating shafts and shafts touch fixed bearings with their ceilings or necks, which are a kinematic pair of bearings that rotate with the ceiling. The force transmitted to the base by the boom or shaft is perceived by the bearings. Sometimes a short shaft load is placed on one bearing rather than two bearings: in this case, the bearing is made much longer l ≈ 12d, the shaft diameter is d, and the two edges of the bearing are processed in the middle. l1: the length remains unprocessed.
The most important part of the bearing is the liners. These inserts, which directly absorb juice pressure, must be well protected against corrosion and be flexible enough to withstand shock loads without breaking.

Advantages.

  1. Provides a clear direction of the shafts, as the number of parts is small compared to rolling bearings.
  2. an oily cushion is formed Between the insert and the shaft, which in turn absorbs shock and vibration.
  3. works Well at high speeds, rolling bearings shorten the service life.
  4. The bearing consists of two parts, so the bearings are installed when they are not available.
    Disadvantages:
  5. High friction when starting and stopping the engine.
  6. Wear of shafts and shaft bearings.
  7. Consumes a lot of oil.
  8. the design of the bearing hub at high loads and speeds is also much more difficult to use.

Application areas:

  1. Forging, vibrating work places hammers, piston engines.
  2. On the crank shafts:
  3. Where the diameter of the shafts is large.
  4. When the shafts are working with a large rotation centrifuge.
  5. On the machine spindles.
  6. In slowly rotating engines.
    Basic theory.
    Friction in the supports. Depending on the operating mode, there are 3 types of lubrication systems. 1. Semi-dry: 2. Dry: 3. Borderline.

Fig. 33. the Relationship between the shear pressure P and the shear rate v ,
With the wetting speed, the speed of the solid surface obeys the law of a straight line, the speed from O to V: Fig. 33, a). The speed and thickness of oil change according to the parabola law Fig. 33, b). In cylindrical bearings, the oil thickness is formed in the form of a pallet Fig. 33, b). Theory of friction. The friction force in plain bearings is so great that if the coefficient of friction is high Fig. 34.
Graph of the coefficient of friction depending on the characteristic mode. μ – oil viscosity:
R – specific pressure,
ω – angular rotation speed, 1 / s, 1/sek
f – f — coefficient of frictio.
The moment of friction and the force on the bearing are as follows.

Where: R – is the force acting on the bearing, kg:
f – coefficient of friction.
d – diameter of shaft, mm.
n – is the number of revolutions, Rev / min.

Calculation of bearings..
Contact voltage. .. In slowly rotating shafts, i.e. when the rotation speed does not exceed V = 0.1 m / s, the heat release is negligible, and in this case the surface is calculated at a limited voltage.

Here: — relative range:
∆ — diameter interval:
[σ] = 3…5 NV – permissible voltage, input is accepted depending on the material in kg / cm2.
— conditional relative pressure acting on the sole.
l – length of the support.
d – diameter of the shaft.
E – modulus of elasticity.
Prevent wear and tear. If V> 0.1 m / s, this is checked by two criteria.

Reasons for failure

  1. Abrasive wear.
  2. Load when the bearing is hot.
  3. Fatigue spot corrosion occurs mainly in piston engines.
    Materials Bronze.
    Bronze bearings are mainly used at medium speeds and heavy loads. BROF 10-1 BR. OTSS-6-3 and Manager) Babbit. They are used as anti-friction materials B89, B 83). Cast iron. Bearings with a low content of cast iron are used on shafts with low spin. Cast iron anti-slip ASCH-1 and others. Cermet. Cermet materials are produced by pressing. Copper and iron are supplemented with copper or lead. Non-metallic materials. These are plastics, wood plastics, wood, rubber.
    Lubricant Two types of oil are used for lubrication:
    a) — organic and b) mineral oils.
  4. Lubricating fluid must moisten and compact the surface of the rubbing parts: the phenomenon that occurs at the border between liquid and solid media, depending on material bodies sliding and fluid properties. These properties are possessed by fats, especially animal fats.
  5. The oil should be viscous, i.e., triturating the place:
  6. the friction Surfaces should form a thin crack in the form of a tail.
    SECURITY QUESTIONS
  7. when and where are plain bearings used?
  8. What is the basic theory of plain bearings?
  9. what materials are plain bearings made of?
  10. what oils are used in plain bearings?
  11. Advantages, disadvantages and application of plain bearings.
  12. Basic theories.
  13. Lubricants.
  14. Bearing calculation.
    Textbooks.
  15. Guzenkov P. G. Details of the machine M: Higher school, 1989 361 p.
  16. Ivanov M. I. Details of the machine M: Mashinostroenie 1989 336 p.
  17. Reshetov D. N. Details of the machine M: mechanical engineering 1989 496 s
  18. Iosilevich G. B. Details of the machine M: mechanical engineering 1988 367 p.
    Textbooks.
  19. Machine parts. Calculate the design catalog). M: Mashinostroenie 1968 vol. 1-3
  20. Reference book of the machine Builder. M: mechanical engineering 1968 vol. 4.
  21. Noshiro S. N. The text of the lectures on machine parts. Karshi 1999
    Additional literature.
  22. Machine parts. Atlas of construction M: mechanical engineering, 1968
  23. Chernavsky S. A. I. D. R. design of mechanical transmissions. M: mechanical engineering, 1980
  24. Silvie G. M., etc. Collection of tasks and calculation methods for the course of machine parts. M: mechanical engineering, 1975.
  25. Dunaev P. F., Lelikov O. P. Detail of machines. Course design. M .: High school. 1984

TOPICS OF COURSE PROJECTS
The course is conducted in the form of individual work with students to complete project tasks. The purpose of course design is to apply machine learning lessons. the project is a form of independent work of the student under the guidance of a teacher.
Based on this, students create assignments and manuals for course projects. The course should consist of the order of projects, not all structures that can be copied, but parts or elements of the structure.
Typical tasks include conveyor drives drum or sprocket shafts with supports, mechanical tracks, machine drives, individual units with gearboxes or radiators, truck drives, testing machines, robots, and other objects. The exact types, scope, and content of tasks are determined by the Department.
Subject of the course project:

  1. Calculate the single-stage cylindrical gearbox installed in the extension cable.
  2. transmission Design consisting of a single-stage cylindrical gearbox with a curved gear and a chain drive;
  3. transmission Design consisting of a single-stage cylindrical gearbox with a curved gear and a belt drive;
  4. The construction of the transmission consisting of spur single-stage conical reducer and chain transmission;
  5. Calculation of the worm gear driving the vehicle.
  6. Transmission and its design.
  7. Calculation and design of a two-stage cylindrical gearbox that drives the sliding mechanism of the bridge crane. 8. Calculation and design of the gearbox that drives the electrical circuit.
  8. Calculation and design of a cone-cylindrical reducer that drives a screw conveyor.
  9. Design of worm gears.

ON THE SUBJECT OF MACHINE PARTS AND DESIGN BASICS
CONTROL QUESTION

  1. Engine, extension cable with ponason tape, 2-speed 3-axis closed cylinder transmission and General f of this engine.I. K. to determine.
  2. Engine, flat belt extension, 2-stage shaft consisting of 3-axis cylindrical longitudinal beams. General Full Name determine the value of s. Select the missing values.
  3. Engine, intake, closed cone-cylindrical drive drawing and General EFFICIENCY. determine. Select the missing values.
  4. Engine, clutch, closed 2-speed cone-cylinder transmission, as well as a common f-shaped transmission. I. K. determine. Select the missing values.
  5. Draw angles consisting of an engine, intermediate, closed worm gears, and open cylindrical gears. General F. I. C. define. Select the missing values.
  6. The motor consists of a clutch closed and open conical cylindrical gears, as well as F. I. to determine the overall value of S. Select the missing values.
  7. Draw the angles consisting of the engine, intermediate, closed cylindrical and open bevel gears, as well as the F. I. determine the total value of C. Select the missing values.
  8. A drawing of the drive consists of motor, coupling, a closed cylinder and open the angle gear. F. viewing angle.I. determine the total value of C. Select the missing values.
  9. Draw the angles consisting of the engine, clutch, 2-speed 3-axis closed cylindrical gears, as well as F. I. S. General values are selected.
  10. Drawing and drawing of the drive, consisting of the engine, wear belt, and closed worm gears, as well as the General Name of the steering. I. K. determine. Select the missing values. Google Tarjimon
  11. the Engine, the drive drawing consisting of a ponasny belt, as well as closed worm gears, and the General EFFICIENCY. determine. Select the missing values.
  12. Engine, drive drawing, consisting of a ponasnoy belt transmission, closed cylindrical transmission, as well as a General EFFICIENCY. determine. Select the missing values.
  13. Draw angles consisting of an engine, a flat belt, and closed cylindrical gears. General Full Name determine the value of s. Select the missing values.
  14. Drive drawing consisting of a motor, flat belt, worm gear, and a common f-shaped transmission. I. K. define. Select the missing values.
  15. the Engine consists of open bevel gears on a cylindrical extension, a step-down belt, a closed cylindrical gear reducer, and an open bevel drive. General Full Name determine the value of s. Select the missing values.
  16. the Engine, the drive belt, and the drive belt consisting of closed gears, and the General EFFICIENCY. determine. Select the missing values.
  17. In the first case, U1 = 2, Up = 5, Ush = 4, T1 = 50 Nm. Determine the values of T2, T3, T4 in the drive shafts. Select the missing values.
  18. In the first case, U1 = 2.5, Up = 4, n2 = 400 min-1. Let the house values be determined, n1. Select the missing values.
  19. In the first case, U1 = 4, Up = 2.5, n2 = 250 min-1. Determine the total number of cases transferred and the value of n3.
  20. In the first case, U1 = 2, U2 = 20, n2 = 500 min-1. Define UI, n1, n3 values. Select the missing values.
  21. In the first case, T2 = 200 Nm, U1 = 2.5, Up = 4. Determine the values T1, T3. Select the missing values.
  22. in the first question, T2 = 500 Nm, H2 = 1000 min-1. Determine the values of P1, P2, and P3. Select the missing values.
  23. In the first case, T2 = 200 Nm, U1 = 20, U2
  24. In the first case, the house = 40, n2 = 100 min-1, z3 = 20, z4 = 80. Define the values U2, n1, n3.
  25. in the first case, R2 = 7 kW. Determine the values of P1, P3. Select the missing values.
  26. in the first question, determine the values R2 = 5, R3, R4. Select the missing values.
  27. In the first case, R2 = 4 kW. Determine the values of P1, P3. Select the missing values.
  28. In the first case, T3 = 400 Nm n3 = 200 min-1. Determine the values of P1, P2, and P3. Select the missing values.
  29. In the first case, z2 = 80, m = 4, d1 = 100 mm, n2 = 250 min-1, n3 = 50 min- Determine the total number of gears and the values of N1.
  30. In the first case, z1 = 20, m = 5, d2 = 400 mm, n2 = 250 min-1, n3 = 50 min- Determine the total gear ratio of the house and the values of n1.
  31. Advantages and disadvantages of belt drives.
  32. Advantages and disadvantages of tension belts. Kinds.
  33. Angle of coverage in a tape extension is the essence 1.
  34. the figure shows the distribution of forces G ‘1 and G’ 2 in tape transmission networks..
  35. What voltage is used in the belt transmission. Draw a sketch diagram. This is the purpose of using the equipment.
  36. How to distribute the bending stress in the tape circuits. Draw a sketch of the drawing.
  37. what forces affect the stretching of the tape in the networks.
  38. What characteristics should the working surfaces of rolling bearings have? 41. Advantages and disadvantages of slide bearings. In which cases rolling bearings are used.
  39. what kind of friction can be in rolling bearings. These tensions differ from each other.
  40. Write a conditional calculation of rolling bearings when working on dry and liquid friction, what are conditional rolling bearings when working on dry and liquid friction.
  41. The force on the stretching increases or decreases at transition from shaft to shaft. Explain the reason.
  42. What are the cycles of stress change. Draw a sketch of the drawing.
  43. Draw a sketch diagram of voltage changes using symmetric and pulse cycles.
  44. How to determine the number of tapes in pansemna elongation.
  45. Types of loose tape. The feeds differ from each other. Structure ponzini tape.
  46. condition for the formation of friction in the liquid.
  47. What parts and components will be studied in the course of machine parts.
  48. what forces arise in the process of coupling straight cylindrical gears. This is the name of the forces. The recipe is shown in the diagram.
  49. Advantages and disadvantages of rolling bearings.
  50. Types of circulation elements in rolling bearings.
  51. in which cases choose rolling bearings for dynamic load.
  52. Which series are divided into rolling bearings. These series are different from each other.
  53. Advantages and disadvantages of rolling bearings. Draw a sketch drawing of rounded bearings with an indication of the drawing elements.
  54. in which cases choose rolling bearings for dynamic load.
  55. The choice of rolling bearings for static and dynamic load.
  56. Advantages and disadvantages of rolling bearings. Sketch diagram.
  57. the Dimensions of the gears must be the same as they have for mutual coupling.The advantages and disadvantages of gear drive.
  58. how to glue gears so that their dimensions are the same a necessary.
  59. what dimensions should be the same, so that the gears are linked together.
  60. What are the dimensions must be the same on gears.
  61. What are the dimensions must be identical to gears mated together. Explain the reason.
  62. What are the dimensions must be the same on gears.
  63. Variators. Using. Level of management. Draw a sketch diagram.
  64. what forces are formed in the process of gluing conical stretch marks. Draw a sketch diagram, specify the name and direction of the same forces.
  65. Advantages and disadvantages of bevel gears.
  66. in the cone transmission, the values R e, 2 are given.we Draw a sketch and define the values de1, de2 and draw on the drawing.
  67. in the cone transmission, the values 1, de1 are given. Specify on the sketch. Determine the values of re, dE2.
  68. With the help of some modules expect a quick ejaculation, and the geometric dimensions of the conical stretch marks.
  69. The geometric dimensions of the conical elongation.
  70. Pros and cons of worm gears.
  71. Geometric dimensions of the worm in the worm extension.
  72. On the schematic drawing indicate the direction of the forces generated during bonding of worm gears.
  73. Draw a sketch of the worm wheel, specify the dimensions dam2, df1, and write the formulas.
  74. when drawing a worm sketch, draw the diameters df,D.
  75. Worm elongation. What materials are used for making worm and worm wheels? Explain the reason.
  76. Draw a worm wheel and show the drawing diameters d2, da2, dat2.
  77. In multi-stage elongation draw a sketch) of the total F. I. is defined As the total number of signals?
  78. Friction extensions. Using. The pros and cons. Draw a sketch diagram.
  79. when drawing a sketch drawing of a cylindrical gear wheel, the diameters DF, d are indicated in the drawing, and formulas are also written.
  80. Geometric dimensions of the cylindrical wheel.
  81. Structure and principle of operation of simple variators. Explain the reason.
  82. Simple structure of variator and how it works.
  83. Movement due to coupling with the transmission. Circuitry.
  84. Advantages and disadvantages of chain extensions. Scheme. Kinds.
  85. Types of chains in chain drives. In which cases chains are made.
  86. Movement due to coupling with the transmission. Scheme.
  87. Movement due to coupling with the transmission.
  88. What extensions are used in mechanical engineering. Types of mechanical transmissions. Draw a sketch diagram.

TEST QUESTIONS ON THE DISCIPLINE MACHINE PARTS AND DESIGN BASICS

  1. What mechanisms are used to build machines?
    a) mobile, transmitting, Executive
    b) mobile, Executive, computing
    C) drive, transmission, counter
    d) performer, counter, transmitter
  2. Find the part description.
    a) made of the same material
    b) designed to perform a single task
    C) General function
    d) from simple pieces
  3. specify the details Correctly.
    a) worm, nut, pulley
    B) gearbox, pinion, key
    C) coupling, water repellent, chain
    d) gear, sprocket, bearing
  4. What details does the science of «car parts» study?
    a) common for all types of machines
    B) on the main machines
    C) performs the same functions
    d) universal type
  5. what stretch marks are studied in the discipline «car parts»?
    a) mechanical transmissions
    b) hydraulic extensions
    c) electrical extensions
    D) pneumatic actuators
  6. What is the car part fan?
    a) technical
    b) humanitarian
    c) social
    d) economic
  7. Omitted word: «after several operations to assemble parts) on the prepared structural element…… called.»
    a) node
    b) fluidity
    C) details
    d) exclusivity
  8. The omitted word will mean: «the sum of the parts attached to the collection operations is called…………………»
    a) node
    b) mechanism
    C) exclusivity
    d) car
  9. What do machines-engines say?
    a) converts energy into useful mechanical work
    b) changes its shape and condition
    C) converts energy into useful heat work
    d) converts mechanical energy into thermal work
  10. the Word omitted: «from the same material without Assembly operations) on the prepared construction element………… called.»
    a) details
    b) fluidity
    c) node
    d) exclusivity
    11.Specify the main requirements for machine parts.
    a) operability, low maintenance, safe, easy to prepare
    b) precise working, durable, increasing power
    c) light, cheap, durable, rust-resistant
    d) accurate, safe, flawlessly working, poorly repaired, durable
  11. Specify the main signs of working parts.
    a) strength, bicarness, heat resistance, vibration resistance, food resistance
    b) strength, strength, hardness, abrasion resistance, purity
    c) strength, rust resistance, excellence, dedication, hardness
    d) heat resistance, strength, malleability, bicarness, coolness
    13.In what case is the contact voltage generated?
    a) when two balls are compressed with each other
    B) when two prisms are compressed with each other
    C) when two planes are compressed with each other
    d) when the cone and plane are compressed with each other
  12. Determine the unit of measurement of the contact voltage.
    a) MPa
    b) mm
    c) kg
    d) N•m
    15.Specify the factors that determine the ability of the part to work.
    a) detail
    b) load affecting parts
    C) manufacturing technology of parts
    d) part size
  13. what is the advantage of the design element?
    a) resistance to deformation
    b) non-return to the initial form
    c) absolute hardness
    d) item detail
  14. what is bikrlik description.
    a) modulus of elasticity
    b) volume
    C) heat
    d) measurement
  15. Unit of measurement of normal voltage.
    a) MPa
    b) ATM.
    C) Time
    D) x
  16. the safety Factor depends on several factors.
    a) 3
    b) 2
    c) 1
    d) 4
    20.What is the unit of measurement P of the consistency elongation coefficient.?
    a) without measurement
    b) ATM.
    c) m
    d) Pascal
    21.What stresses are formed in the joints of brocade nails?
    a) intersection, splitting
    b) twisting, bending
    C) bending, stretching
    d) crushing, compression
    22.What alloys are included in ferrous metals?
    a) steel, cast iron
    b) bronze, aluminum
    c) brass, copper
    d) babbit, lead
  17. from what material do brocade nails?
    a) mild steel, aluminum, brass
    b) steel 40, 45, 50
    c) cast iron
    d) made of bronze
  18. since when have brocade nails been used?
    a) 1930
    b) 1820
    c) 1950
    d) 1991
    25.Where are hidden parchment nails mostly used?
    a) used in steam boilers
    b) used in places where there is a lot of rust
    c) used for connecting parts up to 4mm thick
    d) used in fortified and reinforced joints
  19. Which of the following compounds are irreplaceable?
    a) welding
    b) Bolt
    c) dowels
    d) Shlisselburg
  20. what visible combinations are formed with brocade nails?
    a) top-top, years, three, three three, three three, three three, three three, three three, three three
    b) only on top
    C) three-three
    D) only three-three with one vertex
    a) 28. What form of rod used waste paper when exposed to a variable force on the overlap?
    b) boat
    b) kavarik
    c) text
    d) normal
  21. Where and from what voltage are the corner balls fed?
    a) from cutting of the voltage passing through the bisector of the triangular
    B) from the tension of the incisor passing through the hypotenuse of the triangle
    C) from the normal voltage passing through the triangle catheter
    d) from the cutting stress passing through the base of the triangle
  22. Contact welding of the tape affects the joint of the rod F=25 kN. Sheet width l=100 mm, weld width b=5 mm. Calculate the voltage in the rod.
    a) 50 m
    b) 40 m
    c) 80 m
    d) 100 m
  23. the Thickness of the three parts to be welded δ=5 mm. The tensile strength of the weld F=10 kN, the length of the weld b=20 mm. Determine the voltage in the welding rod?
    A) 100 m
    b) 80 m
    c) 110 m
    d) 120 m
    32.What voltage is created at the point of welding the seam?
    a) intersection
    b) twisting
    c) crushing
    d) stretch
  24. What is the voltage generated in triple welding when it affects the lick force?
    a) stretching
    b) twisting
    c) bending
    d) Eilisia
  25. Specify the main advantages of the carved joint.
    a) reliable, easy to disassemble and assemble, dimensions are standardized
    b) strong, durable, reliable, easy to disassemble and assemble
    c) easy to assemble, durability is high, dimensions are standardized
    d) dimensions are standardized, easy to disassemble and assemble, high strength
  26. the Flange coupling transmits the torque T=10 nm. Semi-finished products are attached with 10 bolts. The diameter of the bolt arrangement D0=200 mm. Find the force that falls on each bolt.
    a) 10 N
    b) 25 N.
    c) 15 N
    d) 20 N
    36.Weight of the load suspended on the hook F=6.28 kN; diameter of the threaded part of the hook d1=10 mm. Determine the voltage generated on the threaded part of the hook.
    a) 80 m
    C) 40 m
    C) 75 MPa
    D) 100 m
  27. Pressure in a sealed container with a hinged lid with bolts P=2.0 MPa, the diameter of the lid D=100 mm, the number of bolts 8 PCs. Find the force that will fall on each bolt from the pressure in the vessel.
    a) 1962 N.
    b) 2574 N.
    C) 1250 N.
    d) 1712 N.
  28. What thread profiles are used in fasteners?
    a) triangular
    b) trapezoidal
    C) rectangle
    d) volume
  29. What kind of stress is created in the center of the trolley thread?
    a) stretching
    b) cutting
    C) flexor
    d) shredder
  30. Specify the criteria for calculating the mounting rail.
    a) cutting threaded rods
    b) crushing of threaded work surfaces
    C) feeding the working surfaces of the screw thread
    d) impregnation of working surfaces with a nut
  31. How much does the number of resume sheets in the mounting joints cost?
    a) 1
    b) 2
    c) 3
    d) 4
  32. the angle of the metric profile is zero?
    a) 600
    b) 550
    d) 500
    e) 650
  33. One inch equals several mm?
    a) 25.4
    b) 15
    c) 22
    d) 34
  34. Specify the unit of measurement of the modulus of elasticity.
    a) kN/km2
    b) kN.
    c) mm
    d) kN.
  35. What is the unit of measurement of contact voltage?
    a) MPa
    b) time
    c) N.
    d) ATM
  36. Causes of normal urinary tension.
    a) torque
    b) normal bending force
    C) bending moment
    d) compression force
  37. Joints. What compounds can be distinguished?
    a) threaded joints, key joints, spline joints
    b) wedge joints, brocade nail joints, threaded joints, welded joints, awl joints
    C) welded joints, threaded joints, key joints, spline joints
    d) carved joints, key joints, wedge joints
  38. Specify the main advantage of helical and hypoid gears.
    A) transmits motion to several shafts
    B) consists of gears
    C) transmits the movement to the cross shafts
    d) the number of databases can be large
  39. Specify the main drawback of the veneer joint.
    a) the presence of a socket on the shaft
    b) cannot be selected
    C) vertical installation on the shaft
    d) violated the Flammability of the shaft and cupcake
  40. Why is it enough to calculate the specific stress when checking the strength of the prismatic veneer joint?
    a) the dimensions are selected in this way
    b) other stresses are not generated
    C) due to low speed
    d) due to an imbalance in the shaft

GEARBOX, DRIVE, BELT DRIVE, SHAFT-GEAR, WHEEL, GREASE, BEARING, SHAFT, GEAR, GEAR, ELECTRIC MOTOR, PULLEY.

This design and explanatory note provides a kinematic calculation of a multi-stage drive, which includes an electric motor, a belt, a cylindrical gearbox and an open gear. The strength calculation of the cylindrical and gear transmission was also performed, and the bearings of the input and output shafts of the closed transmission were selected. The selection of gearbox lubrication was made. The stresses arising in the supports of the output shaft, as well as the torsional and bending moments on this shaft, are calculated, and their diagrams are given.
The note contains:
• 14 tables;
• 11 drawings;
• 2 apps;
• 55 sheets.

INTRODUCTION
In the chemical technology of organic materials, multi-stage drives are widely used, which can include belt, worm, chain, gear, etc.transmission.
Belt drives have the following advantages:
• simple design;
• smooth and quiet operation;
• low requirements for the accuracy of the transmission parts location;
• overload protection due to the possibility of belt slipping on the pulley.
Along with the advantages of belt drives have some significant disadvantages. These are:
• large dimensions;
• variable gear ratio due to belt slipping on the pulley;
• heavy load on shafts and supports;
• low durability of the belts.
The transmitted power is usually no more than 50 kW, and the gear ratio is up to 6.
Gear drives are widely used in all branches of mechanical engineering and instrument making due to a number of their advantages:
• constant gear ratio;
• no slippage;
• high load-bearing capacity with relatively small dimensions and weight;
• long life; work in a wide range of loading modes;
• relatively low loads on shafts and supports;
• high efficiency, easy maintenance and maintenance;
• The disadvantages of gears include:
• high requirements for precision wheel manufacturing and gear Assembly; the need for increased rigidity of housings, shafts, supports;
• noise, especially at high speeds and low accuracy; vibrations;
• has a low damping capacity.

  1. BRIEF DESCRIPTION OF THE DRIVE OPERATION
    The source of mechanical energy in this drive is an asynchronous motor 4A160M8. A V-belt drive pulley is installed on the motor shaft, by which the rotation is transmitted to a driven pulley mounted on the input shaft (worm) of the worm gear. The belt drive has a gear ratio of IRP = 2.6. Belt drives have the following advantages: simplicity of design; smooth and silent operation; low requirements for the accuracy of the location of transmission parts; overload protection due to the possibility of slipping the belt on the pulley. Along with the advantages of belt drives have some significant disadvantages: large dimensions; variable gear ratio due to slipping of the belt on the pulley; high load on the shafts and supports; low durability of the belts.
    The cylindrical gearbox serves to increase the torque by reducing the angular velocity of rotation and has a gear ratio of PPI = 2.5. the advantages of oblique cylindrical gears include: smooth engagement and quiet operation; increased kinematic accuracy, which is especially important for dividing devices, high efficiency, low loads on supports and shafts .
    A significant disadvantage due to the geometry of the teeth is the occurrence of axial forces, as well as the high cost and complexity of manufacturing wheels
    Further torque is transmitted to the spur gear transmission having a gear ratio RPMs = 2.5. In this section of the drive also increases the torque and the shaft gear wheels get power 9 kW at an angular speed of 4.7 s-1.
  2. SELECTION OF THE ELECTRIC MOTOR AND KINEMATIC CALCULATION OF THE DRIVE
    Kinematic scheme of the drive and shaft numbers Fig. 1 2.1. Motor choice
    The efficiency of the drive is determined by the formula

Where: Efficiency of individual kinematic pairs (belt, cylindrical, gear, bearings). The efficiency values are selected as average values from the recommended range [1].

The required power of the electric motor is found taking into account the losses, arising in the drive:

Approximate value of the total gear ratio of the drive

where are the approximate values of the drive gear ratios (selected as the average values from the recommended range for the respective gears) [1].

Approximate value of the angular velocity of the motor shaft
where is the angular velocity on the driven (slow-moving) shaft, с-1.

Approximate value of the motor shaft speed

Select an electric motor with a power and actual speed of the MPV close to the value of the MPV.OR [1]. The selected engine is 4A160M8..
In the future, the calculation is based on the selected one

2.2. Kinematic calculation of the drive
Angular velocity of the motor shaft

Total gear ratio of the drive:

We divide the UO into separate drive stages
where are the gear ratios of individual stages.

We determine the angular velocities of the drive shafts (Fig. 1):

Determine the speed of the drive shafts:

We determine the power on the drive shafts::

We determine the torques on the drive shafts:

The calculation results are summarized in table 1.
Table 1
Summary table of results of kinematic calculation of the drive.
№ shafts Power P, kW The angular velocity ω, s-1 Rotation speed p, min-1 Torque T, Nm
Engine 10,59 76,4 2900 139
1 10,59 76,4 730 139
2 10,1 29,4 280 342
3 9.65 11,7 112.3 821
4 9.0 4,7 44.9 1900

  1. CALCULATION OF OPEN GEARS
    3.1. Calculation of V-belt transmission The main criteria for the performance and calculation of belt drives are: the traction capacity, determined by the amount of transmitted circumferential force, and the durability of the belt, which in normal operation is limited to the destruction of the belt from fatigue. The main calculation of belt drives is the calculation of traction capacity. The calculation of the belt durability is performed as a test. Industry mass-produced V-belts and V-belts: woven with polyamide coating and rubberized with cord-cord bearing layer. Due to its strength, elasticity, low sensitivity to moisture and temperature fluctuations, and low cost, rubberized belts are widely used. Therefore, the design calculation for rubberized belts is given below. The calculation is performed in the following order:
    Select the belt section
    (Fig. 2). Belt drive diagram

Pic. 2
The belt section is selected using the nomogram [3] depending on the power transmitted by the drive pulley, P1=PH=10.6 kW and its rotation speed n1=pH=730 rpm. Thus, we select the section of the UA .
We determine the minimum allowable diameter of the drive pulley d1min, mm, depending on the torque on the motor shaft TDV, N•m, and the selected belt section
Тen = 139 N•m,
d1min = 63 mm.
We accept the estimated diameter of the drive pulley d1 = 140 mm. Determine the diameter of the driven pulley d2:
d2 = d1*u(1 – ε),
where u = 2,6 – is the V-belt gear ratio;;
ε = 0,015– the coefficient of friction
d2 = 140•2,6(1 – 0,015) =358 мм.
We round the d2 value to the standard value and take it to 355 mm..
2. Determine the actual gear ratio if and check its deviation Δu from the specified u:

 .                     
   3. Determine the approximate center distance а, mm:
                  а ≥ 0,55(d1 + d2) + h(H),                               

where h(H) = 8 – is the cross-section height of the V-belt.
а = 280 mm.
4. Determine the estimated length of the belt l, mm:

We round the l value to the standard value and take it to 1400 mm.

  1. Specify the value of the center distance for the standard length:
  2. Determine the belt girth angle of the drive pulley α1, deg: Angle α1 ≥ 120º.
    1. Determine the belt speed v, m / s:

where d1 – is the diameter of the drive pulley, mm;
n1 – speed of the drive pulley, rev/min;
[v] = 40 м/с – permissible speed.
v = 5,35 м/с.
8. Determine the frequency of belt runs U, с-1:
U = v/l ≤ [U],
where [U] = 30 с-1 – is the allowed frequency of runs.
U = 3.8 с-1 ≤ [U], which guarantees a service life of – 1000…5000 ч.
9. Determine the permissible power transmitted by a single V-belt [RP]:
[Pп] = [P0]Ср Сα Сl CZ= 1.849 кВт,
where [P0] = 2.7 кВт – the permissible reduced power transmitted by one belt, kW, which is selected depending on the type of belt, its cross– section, speed and diameter of the drive pulley; Ср = 0,9, Сα = 0,89, Сl = 0,95, CZ=0,90– correction factors.
10. Determine the number of wedges of the V-belt z:
z = Pном/[Pп] = 6
where Pном = 10.59 кВт – is the rated power of the engine;
[Pп] = 1,849 кВт – permissible power transmitted by belts.
11. Determine the pre-tension force F0, N:

 12. Determine the circumferential force transmitted by the V-belt Ft, N: 

 13.  Determine the tension forces of the leading F1 and driven F2 branches, N::  

F1 = F0 + Ft/2Z = 406,0 Н F2 = F0 – Ft/2Z = 168,0 Н.
14. Determine the force of the belt pressure on the shaft FOP, N:

3.2. Calculation of gear
Calculation of cylindrical spur gears is made in accordance with GOST 21354-75.. The main types of calculations are calculations for the contact endurance of active surfaces of teeth and calculation of teeth for bending endurance. Since the main cause of failure of the teeth of closed gears operating with good lubrication is fatigue contact painting, the design calculation of closed gears is performed for contact endurance, followed by checking the teeth for contact endurance and bending endurance. Open gears are designed to withstand bending stresses.
In this project, the calculation of teeth for contact endurance and bending endurance is performed at zero offs ( ).The calculations use a constant load mode, for which the equivalent number of cycles of stress change is greater than the base number of cycle ( ).in this case, the durability coefficient, which takes into account the influence of the service life and load mode, is assumed to be equal to

  1. Choose the material for making gears.
    When choosing steel grades, take into account the purpose and type of transmission, requirements for dimensions and weight, manufacturing technology, and economic feasibility.
    Table 2
    Properties of steel St45.
    Steel grade Mechanical property Heat treatment
    Hardness Ultimate strength GB, MPa GT yield strength, MPa
    HB HRC
    St45 235–262 – 780 540 Improvement
  2. Estimated value of the modulus m is calculated by the formula:

where – is the auxiliary coefficient, which for cylindrical spur gears is equal to
– torque on the gear shaft, Nm, which is taken from table 1:

– a coefficient that takes into account the load distribution over the width of the crown, which is found from the corresponding graph depending on the value [2].

 – number of gear teeth

where z2 – is the number of wheel teeth;
UIII – gear ratio.

– a coefficient that takes into account the shape of the tooth, which is determined according to the graph depending on the equivalent number of teeth:

– the ratio of the width of the ring gear [1]

– permissible bending stresses of the teeth, MPa, which is determined by the formula:

where – is the limit of the teeth ‘ bending endurance corresponding to the equivalent number of stress change cycles, MPa, which is calculated according to the formula

– the limit of tooth endurance during bending, corresponding to the base number of cycles of stress change, MPa, which is determined depending on the method of chemical and heat treatment [1].

– coefficient that takes into account the effect of grinding the transition surface of the tooth.. For teeth with an unpolished transition surface of the tooth, and for other cases, it is determined depending on the thermal or chemical-thermal treatment = 0,9; when quenching normalization, improvement = 1,1; cementation and nitrocementation = 0,7.
= 1,1;
– coefficient that takes into account the effect of deformation hardening or electrochemical treatment of the transition surface. For wheel teeth without deformation hardening or electrochemical treatment of the transition surface of the teeth, take = 1;
– coefficient that takes into account the impact of two-way load application. For one-way load application
= 1;
– durability coefficient. For long-running transmissions, it is accepted-
= 1;

Given all the coefficients found, we define
:

– a safety factor that is equal to

Table 3

A coefficient that takes into account the instability of the material properties of the gear wheel and the responsibility of the gear train. Coefficient that takes into account the method of obtaining a gear billet
.
It is determined depending on the method of thermal and chemical-thermal treatment and the specified probability of failure. When the probability of failure is 0.99 and volume quenching, normalization and improvement = 1,75; when cementation and nitrocementation = 1,55. = For forgings and stampings = 1;
rental = 1,15;
For cast workpieces = 1,3.

= 1,75
= 1

–a coefficient that takes into account the stress gradient and the sensitivity of the material to the stress concentration. When calculating the design of open gears, we accept
– coefficient that takes into account the roughness of the transition surface. For grinding and gear milling with a roughness not lower than RZ40, take = 1. when polishing, depending on the method of thermal hardening, take: for cementation, nitrocementation, nitriding = 1,05; for normalization and improvement = 1,2.
= 1,2;
– coefficient that takes into account the size of the gear wheel. It is determined depending on the diameter of the toothed wheel vertices according to a special graph [1]..
= 1.
Having determined all the values and coefficients included in the formula, we find :

  1. Defining the approximate value of the module m:

The resulting value is rounded to the standard value in accordance with GOST 9563-60 [1]:

  1. Determine the diameters of the initial (external) dividing circles of the gear and whee.
  2. The initial diameter of the pitch circle of the gear:

Diameter of the initial dividing circle of the wheel:

  1. Determine the center distance.
  2. Determining the circumferential speed
    where ω1 – is the angular velocity on the gear shaft с-1,
  3. Determine the degree of transmission accuracy. The degree of accuracy is chosen depending on the purpose of the transmission, its working conditions and production capabilities. Open cylindrical gears are usually performed on the 9th degree of accuracy
  4. Determine the working width of the gear and wheel crown
    we Will perform a test calculation of the teeth for bending endurance
  5. The calculated bending stress of the teeth is determined by the formula

where – is the specific calculated circumferential force.
For cylindrical spur gears

wherе – is the torque on the gear shaft, which is taken from table 1:

– coefficient that takes into account the load distribution between the teeth. The calculation of gears is initially made, assuming that there is one pair of teeth in the engagement. Then
= 1;
– coefficient that takes into account the load distribution over the width of the crown. See point 2.

– coefficient that takes into account the dynamic load that occurred in the engagement: = 1;
– coefficient that takes into account the shape of the tooth. See point 2.
= 4,05;
– coefficient that takes into account the overlap of teeth: = 1;
– coefficient that takes into account the slope of the tooth = 1;
Having determined all the values and coefficients included in the formula, we find the bending stress of the teeth:

The found value of the bending stress of the teeth corresponds to the calculation conditions.

  1. CALCULATION OF CLOSED TRANSMISSION (CYLINDRICAL GEAR)
    4.1. The choice of material gear
    In the designed gearboxes, it is recommended to use heat-treated medium-carbon non-alloyed steels 45, 40X. Steel is currently the main material for the manufacture of gears. In conditions of individual and small-scale production, gears with a material hardness not exceeding 350 NV are used. At the same time, the teeth are cut cleanly after heat treatment, high manufacturing accuracy and good working capacity of the teeth. We determine the steel grade: for the gear-40X, hardness ≥ 45hrce1; for the wheel 40X, hardness ≤350 NV2 [1, p. 49]. The difference in average hardness of Nv1sr – Nv2sr ≥ 70.
    We determine the mechanical characteristics of steel 40X: for the gear Tver-dost 269…302 NV1, heat treatment-improvement and hardening of HDPE. Determine the average hardness of the gear and wheel teeth

НВ1ср = 285,5 НВ2ср = (235 + 262)/2 = 248,5.
4.2. Determination of permissible contact stresses [σ]Н
The permissible contact stresses for strength calculations are determined separately for the gear teeth [σ]Н1 and wheel [σ]Н2.
We calculate the durability coefficient КHL. Lifetime operating time:
for wheel: N2 = 573ω2Lh,
N1=48,26∙107 cycles;
for gear: N1 = N2∙uзп,
N2=10, 72∙107 cycles.
The number of cycles of stress change NH0 corresponding to the endurance limit is found in table 3.3 [3] by interpolation:
NН01 = 25∙106 cycles;
NН02 = 25∙106 cycles.
Since N1>NN01 and N2>NN02, these are the koéfficients of longevity KNL1 = 1 and KHL2 = 1.
Since N1>NН01 and N2>NН02, these are the longevity coefficients КНL1 = 1 и КHL2 = 1.
We determine the permissible contact voltage [σ]Н0, , corresponding to the number of cycles of voltage change NН0 [3]
for gear:
[σ]Н01=1,8HВ1ср+67
[σ]Н01= 1,8∙285,5+67=580,9 Н/мм2;

for the wheel:
[σ]Н02=1,8HВ2ср+67
[σ]Н02= 1,8∙248,5+67=514,3 N/mm2.

We determine the permissible contact voltage: for gear
[σ]Н1=КHL1∙[σ]Н01
[σ]Н1= 1∙580,9=580,9 Н/мм2;

for the wheel:
[σ]Н2=КHL2∙[σ]Н02
[σ]Н2= 1∙514,3=514,3 N/mm2.
S Since Nv1sr-Nv2sr =285.5 – 248.5 = 20 … 50 NV, the bevel gear is calculated for strength at a lower permissible contact voltage.
4.3 Determination of permissible bending stresses [σ]F
We calculate the coefficient of durability КFL. . Lifetime operating time:
for wheel N2 = 10,72∙107 cycles;
for gear N1 =48,26∙107 cycles.
The number of stress change cycles corresponding to the endurance limit, NF0 = 4∙106 for both wheels.
Since N1>NF01 and N2>NF02, the durability coefficients are
КFL1 = 1 and КFL2 = 1.
We determine the allowable bending stress [3] corresponding to the number of cycles of stress change NF00:
for gear: [σ]F01 = 294,07 N/mm2 assuming that m<8 m;

for the wheel: [σ]F02 = 1,03HВ2ср = 1,03∙248,5 =255,96 N/mm2.
We determine the allowable bending stress:
for gear: [σ]F1 =294,07 N/mm2;

for the wheel: [σ]F2 =255,96 N/mm2.
Table 4
Creating a tabular response to the problem:
The transfer element Steel grade Heat treatment НВ1ср [σ]Н [σ]F
НВ2ср N/mm2
Gear 40Х У 285,5 580,9 294,07
Wheel 40Х У 248,5 514,3 255,96

4.4 design calculation of a closed gear train
1. Define the main parameter-the center distance аW, mm:

where Ка — is the auxiliary coefficient. For skew gears Ка = 43;
— coefficient of wheel rim width equal to 0.28…0.36-for a gear located symmetrically relative to the supports in the designed non-standard single-stage cylindrical gearboxes. Let’s take it equal to 0.30; u-gear ratio;
u — torque on the low;
Т2 — speed gear shaft, N/m;
[]Н — permissible contact voltage of a wheel with a less strong tooth or the average permissible contact voltage, N/mm2;
КН — coefficient of uneven load along the length of the tooth. For working teeth КН = 1.
(mm)
aw=230 mm
2. Defining the engagement module m, mm:

wher Кm — is the auxiliary coefficient. For skew gears Кm = 5,8;
— wheel dividing diameter, mm, d2=271,5 mm;
b2 = aaW — wheel rim width, mm, b2= 48 mm;
[]F — is the average allowable contact voltage, N/mm2. Thus, m = 2.16, округляя до rounding to the standard value, we take, m=2,5(mm).
3. Determining the angle of the teeth min for bevel gear:
,

   4. Determine the total number of teeth of the pinion and wheel helical gears: 
  1. We specify the actual value of the angle of inclination of the teeth for bevel gears: 6. Determine the number of gear teeth: . 7. Determine the number of teeth of the wheel: z2 = zΣ – z1 =90 - 26=64 . 8. Determining the actual gear ratio:

and check its deviation from the set value:

  9. Determining the actual center distance for skew gears: 

The geometric parameters of the transmission are shown in table 5.
Table 5
The geometric parameters of the transmission
Parameter Bevel gear The helical wheel
Diametr dividing

the tops of the teeth    


tooth depressions    

Width of the crown

4.5. Checking calculation

  1. Checking the center distance:
    .
  2. Check the contact stresses Н:
    .
    where К — is the auxiliary coefficient. For skew gears К = 376; — circumferential force in engagement, N;
    КН — coefficient that takes into account the load distribution between the teeth.
    Depends on the circumferential speed of the wheels , and the degree of transmission accurac, we assume equal to 8; КН=1,119 [1, с.62-63];
    КН — dynamic load factor, depending on the circumferential speed of the wheels and the degree of transmission accuracy, КН=1,01 [1, с.62].
    Substituting numeric data we get
  3. the wheelCheck the bending stresses of the gear teeth F1 and F2, N/mm2:

wher m — is the engagement module, mm; e m
b2 — width of the gear ring of the wheel, mm;
Ft — circumferential force in engagement, N;
KF — depends on the degree of transmission accuracy. coefficient that takes into account the load distribution between the teeth. КF = 1,0.
КF — — For skewed wheels,. coefficient of unevenness of the load along the length of the tooth. For working teeth of wheels КF = 1;
КF — is a dynamic load coefficient that depends on the circumferential speed of the wheels and the degree of transmission accuracy equal to 1,04, [3];
YF1 и YF2 — are the coefficients of the gear and wheel tooth shape. For oblique teeth are determined depending on the equivalent number of teeth
gears
.
and wheels
YF1 = 3,88 AND YF2 = 3,62;
— coefficient that takes into account the slope of the tooth;
[]F1 и []F2 — permissible bending stresses of the gear and wheel, N/mm2.

Making a table resp onse *, mm:
Table Checking calculation
Checking calculation
Parameter Acceptable values Calculated value Note (deviations)
Contact voltages Н, N/mm2 514,3 474,99 underload
Bending stress, N/mm2 F1 294,07 84,03 underload
F2 255,96 112,56 underload

4.6. Determination of forces in engagement
Table 7
Force values
Forces in engagement Force values, N
on the gear on the wheel
Circumferential Ft1 = Ft2 = 4787

Radial Fr1 = Fr2 = 1220

Axial Fa1 = Fa2 = 1742,7
Fa2 = Ft2tg =1742,7
4.7. Definition of cantilever forces
8 In the designed drives, open-toothed cylindrical and conical gears with straight teeth are designed, as well as belt and chain gears that determine the cantilever load on the output ends of the shafts. In addition, the cantilever load is caused by couplings connecting the motor to the gearbox or the gearbox to the working machine. The values of cantilever forces are shown in.
8 In the designed drives, open-toothed cylindrical and conical gears with straight teeth are designed, as well as belt and chain gears that determine the cantilever load on the output ends of the shafts. In addition, the cantilever load is caused by couplings connecting the motor to the gearbox or the gearbox to the working machine. The values of cantilever forces are shown in table 8
Table 8
Console force values
Type of open transmission Character forces Power value,N
On the gear On the wheel
Cylindrical straight-toothed Circumferential Ft1 = Ft2 = 105556

Radial  Fr1 = Fr2 =  

V-belt Radial

                                                                                                    Table 9

Partial calculation for the task
Design calculation
Parameter value Parameter Value
Center distance, aW 230 The angle of inclination of the teeth  13,717
Engagement module m 5
Width of the toothed crown:
gears b1
wheels b2
77
74 Diameter of the dividing circle:
gears и d
wheels а d2

    143.8
     329.4

Number of teeth:
gears and z1
wheels а z2
26
64 Diameter of the vertex circle:
gears da1
wheels da2

     143.8
     339.4

The appearance of the teeth inclined Diameter tooth depressions:
gears df1
wheels df2

     121.8
     317.4
  1. PRELIMINARY CALCULATION OF SHAFTS AND SELECTION OF STANDARD PRODUCTS (BEARINGS, CAPS, SEALS).
    The shafts are designed for mounting rotating parts on them and transmitting torque. Shaft designs are mainly determined by the parts that are placed on them, the location and design of bearing units, the type of seals and technical requirements. Shafts perceive stresses that change cyclically from the combined action of torsion and bending. At the initial stage of shaft design, only the torque is known, and the bending moment cannot be determined, because the distance between the supports and the acting forces is unknown. Therefore, in the design calculation of the shaft, its diameter is determined by the torsional stress, and the effect of bending is taken into account by lowering the permissible torsional stress.
    5.1. Determination of geometric parameters of shaft stages
    The gear shaft is a stepped cylindrical body, the number and size of the steps of which depend on the number and size of the parts installed on the shaft. Pic. 3 Input shaft We determine the estimated approximate geometric dimensions of each stage of the shaft, mm.
    Section I-output end of the shaft for installing the belt drive pulley. The diameter of the output end of the shaft is determined by the formula:

where – is the torque on the shaft in question, Nm;
– reduced permissible torsion stresses, MPa, for the output ends of the shaft are assumed to be equal to MPa;,

Section II – the section for installing the seal; the diameter is selected based on the standard values for parts using the empirical formula:

Section III – section for mounting bearings; the diameter is selected based on the standard values for parts using the empirical formula:

Taking into account the resulting diameter, we choose bearings according to GOST 333-79 (single-row tapered roller bearings [3].
Both shaft supports are performed on bearings 7212 GOST 333-79.
Table 10
Bearings mounted on the input shaft
Designation Main dimensions Load capacity, kN The load factor
d, mm D, mm Y b, mm с, mm α° Cr C0r Y
7212 60 110 1.547 23 19 2.5 72,9 58,4 1.710

Section IV – section for installing the wheel. The diameter is determined by the formula:

On the side of the output end of the shaft, an end cap with a hole for the lip seal is placed, selected depending on the diameter of the outer ring of the bearing [4], [3].
On the other hand, a blind end cap is placed, selected depending on the diameter of the outer ring of the bearing [4], [3].
5.2. Wheel shaft (output shaft)
Output shaft

Pic. 4
Section I-output end of the shaft for mounting the gear train. The diameter of the output end of the shaft is determined by the formula

where – is the torque on the shaft in question, Nm;
– reduced permissible torsion stresses, MPa, for the output ends of the shaft are assumed to be equal to MPa;

Section II – the section for installing the seal; the diameter is selected based on the standard values for parts using the empirical formula
To protect the bearings from the external environment and keep the grease in the support units, a lip seal is placed, selected depending on the shaft diameter according to GOST 8752-79 [5].
Section III – section for mounting bearings; the diameter is selected based on the standard values for parts using the empirical formula:

Taking into account the resulting diameter, we choose bearings according to GOST 333-79 (single-row tapered roller bearings) [3].
Both shaft supports are performed on bearings 7315 GOST 333-79.
Table 11
Bearings mounted on the output shaft
Designation Main dimensions Load capacity, kN The load factor
d, mm D, mm T, mm b, mm с, mm Α° Cr C0r Y
7315 75 130 24 23 19 12 97,6 84,5 1.547

On the side of the output end of the shaft, an end cap with a hole for the lip seal is placed, selected depending on the diameter of the outer ring of the bearing [4], [3].
On the other hand, a blind end cap is placed, selected depending on the diameter of the outer ring of the bearing [4], [3].
Section IV – section for installing the wheel. The diameter is determined by the formula

                   where   – is the torque on the shaft in question, Nm;
   – reduced permissible torsion stresses, MPa, at the wheel landing points are assumed to be equal to   MPa;;

5.3. Pre-selection of rolling bearings
The choice of the most rational type of bearing for the given conditions of operation is very complex and depends on a number of factors: transmitted power reducer, a transmission type, the ratio of forces engaged, the frequency of rotation of the inner ring of the bearing, the required period of service, the acceptable cost, the installation scheme.
Choosing bearings for shafts [1, p. 111]. On low-speed and high-speed shafts, we install bearings of the radial conical single-row type. Installation scheme — with one fixing support. The series is average. According to the size of the diameters d2 and d4 choose bearings [1, с.410]:
for high speed shaft 7212;
for the low-speed shaft 7215.

  1. CALCULATION OF THE MAIN ELEMENTS OF THE HOUSING
    The gearbox housing is designed to accommodate the transmission parts, to perceive the forces that arise during operation, and to protect the transmission parts from damage and contamination
    General-purpose gearboxes are designed to be split for easy Assembly and disassembly. The plane of the connector passes, as a rule, through the axes of the shafts parallel to the plane of the base. In this case, each gear shaft with all the parts located on it is an independent Assembly unit, which is assembled and controlled in advance independently of the other shafts and then mounted in the housing.
    The dimensions and shape of the gearbox are determined by the number and size of the gears enclosed in the housing, the position of the connector plane and the location of the shafts.
    An inspection window is provided in the housing cover for oil filling, Assembly control and gearbox inspection during operation. It is located in places that are convenient for inspection of the engagement. The size of the window should provide a good view of the engagement. The shape of the holes can be rectangular, round, or oval.
    In the lower part of the housing base, there is an oil drain hole closed with a threaded plug, and an opening for installing an oil indicator. Hooks, eyelets or eye bolts are provided for lifting and transporting the gearbox.
  2. the wall Thickness of a single-stage worm gear is determined by the formula:

where – is the thickness of the walls of the gearbox base, mm;
– thickness of the walls of the gearbox cover, mm;
– center distance, mm;

  1. The depth of the gearbox housing must provide the required volume of oil to be filled V=(0.4-0.8) liter / KW (crankcase grease)
    H=230 mm
  2. The dimensions of the interfaces are selected depending on the wall thickness [1]:
    a) distance from the wall –
    b) distance from the flange –
    c) the radius of curvature –
  3. The diameters of the bolts::
    a) a) Foundation:

b) connecting the housing cover to the gearbox base:
1) for bearings

2) other

c) bearings that attach the cover to the housing, determined based on the size of the cover [1]
d) fixing the inspection cover

  1. The number of Foundation bolts is determined by the formula
    .
    where M and N – are the dimensions of the housing base,
  2. Dimensions of flanges shall be determined depending on the diameter of the bolts

Table 12
Dimensions of flange elements
Flange elements The diameter of the bolt
М8 М10 М12 М16 М20 М24
Flange width К, mm 24 28 33 39 48 54
Distance from the bolt axis to the wall С, мм 13 15 18 21 25 27
Bore d0, мм 9 11 13 17 22 26
Layout diameter D0, мм 17 20 26 32 38 45
Rounding radius R, мм 3 3 4 5 5 8

  1. Dimensions of bearing socket elements:
    a) a) the bore Diameter D is assumed to be equal to the outer diameter of the bearing or Cup;
    b) b) length of the bearing:
    c) Number of bolts for mounting the bearing cover:

c) Diameter of bolts:

d) the Depth of screwing:

e) thread Depth:

f) drilling Depth:

  1. CHECKING CALCULATION
    7.1. Determination of reactions in supports and construction of diagrams of bending and torques of the output shaft
    The required data is shown in table 13..
    Forces acting on the shaft and the distance between their application points
    Table 13
    Engagement forces and pressure on the supports Geometric parameters of the shaft Geometric parameters of the shaft-gear
    Oblique cylindrical transmission Straight-toothed front Belt drive
  2. Perform a scheme of loading the shaft with the indication of the current forces and distances between the points of their application (taken from the sketch layout)
    The distance between the loading points

Pic. 5.

  1. Draw up a diagram of the shaft loading in the vertical plane (Fig. 6).
  2. According to the rules of resistance of materials, considering the shaft as a beam lying on pivotally movable supports and loaded with concentrated forces, we determine the reactions in the supports in the vertical plane and plot the bending moments
    a) find the reactions at the supports:
    b) find the bending moments:
  3. A Similar scheme of loading the shaft, determining the reactions of supports and plotting bending moments is performed for the horizontal plane (Fig. 6 )
    a) find reactions in the supports:
    b) find the bending moments:
  4. Construct a plot of the torque (Fig. 6 ):
  5. Determine the total radial reactions in the supports:
  6. Determine the total bending moments

7.2. Determination of reactions in supports and construction of diagrams of bending and torques of the input shaft

  1. Perform a scheme of loading the shaft with the indication of the current forces and distances between the points of their application (taken from the sketch layout).

The distance between the loading points

Pic. 6.

  1. We draw up a diagram of the shaft loading in the vertical plane (Fig. 6 ).
  2. According to the rules of resistance of materials, considering the shaft as a beam lying on pivotally movable supports and loaded with concentrated forces, we determine the reactions in the supports in the vertical plane and plot the bending moments (Fig. 6):
    a) find reactions in the supports:
    b) find the bending moments:
  3. Determine the total radial reactions in the supports::
  4. Determine the total bending moments:

7.3. Calculation check shaft bearings durability
The calculation will be carried out for bearings 7215 GOST 333-71.

  1. According to table 7.6 from [1], we find the coefficient e of the pre-selected bearing 7315 GOST 333-71:.
    е=0,388.
  2. Calculate the axial components of the reactions of the supports from the action of radial forces:.
  3. We Determine the calculated axial loads Ra1 and Ra2 taking into account the location of the bearings vraspor:
  4. Determine the ratio
    where V – is the coefficient of rotation when the inner ring rotates V=1, the outer ring rotates V=1,2, and it is compared with the coefficient e:

Since these ratios are less than the coefficient e, X=1, Y=0.

  1. Determine the equivalent dynamic load using the formula:

where Кσ – is the safety coefficient: at a calm load Кσ = 1;
КТ – temperature coefficient: when the bearing temperature is less than 100 °С КТ = 1.

  1. According to table. 7.2 [1] define the coefficient γ:
    γ=3,77.
  2. Calculate the required dynamic load capacity of bearings using the formula:
    1. Since the data for the second ratio is greater than the coefficient e, X=0.4, from (table 11). Y=1.547. Calculate the required dynamic load capacity of bearings using the formula:

The bearings are suitable for mounting on this shaft.
7.4 test calculation of shaft-gear bearings for durability
The calculation will be carried out for bearings 7212 GOST 333-71.

  1. According to table 7.6 from [1], we find the coefficient e of the pre-selected bearing 7212 GOST 333-71:
    е=0,351.
  2. We calculate the axial components of the reactions of the supports from the action of radial forces:
  3. We Determine the calculated axial loads Ra1 and Ra2 taking into account the location of the bearings vrasp:
  4. Determining the ratio
    where V – when the inner ring rotates V=1, the outer ring rotates – V=1,2, and it is compared with the coefficient е:

Since these ratios are less than the coefficient e, X=1, Y=0.

  1. Determine the equivalent dynamic load by the formula:

where Кσ – is the safety coefficient: at a calm load Кσ = 1;
КТ – temperature coefficient: when the bearing temperature is less than 100 °С КТ = 1.

  1. According to table. 7.2 [1] define the coefficient γ:
    γ=3,77.
  2. Calculate the required dynamic load capacity of bearings using the formula:
    1. Since the data for the second ratio is greater than the coefficient e, X=0.4, from (table 10). Y=1.710. Calculate the required dynamic load capacity of bearings using the formula:

The bearings are suitable for mounting on this shaft.
7.5. Checking calculation dowels
Gears, pulleys, sprockets and other parts are attached to the shafts using keyway or spline connections designed to transmit torques.
.. In General-purpose gearboxes, prismatic key connections are widely used due to their simple design, relatively low cost, and ease of Assembly and disassembly. The key section is selected depending on the shaft diameter according to table. 7.7 in [1]. The length of the key is taken along the length of the hub with rounding down to the standard side.

Keyway connection diagram

Pic. 9
Table 14
Prismatic dowels mounted on the output shaft.
Shaft diameter d The cross section of the dowel The depth of the groove Key length l
b h t1 t2
85 22 11 7,5 4,4 70

After determining the size of the key, we perform a test calculation of the connection by the crumpling stresses:

where Т – is the torque on the shaft, Nmm;
d – shaft diameter, mm;
l – working length of the key, mm;
(h–t1) – is the crushed height of the do, мм;
[GСМ] – permissible crumpling stresses, with a steel hu [GСМ] = 110–190 МПа.
a) key connecting shaft and worm wheel:

The key meets the working conditions and is suitable for mounting on the shaft.
b) key connecting the shaft and gear of a straight-line transmission:

The calculated crumpling stresses exceed the permissible values, so two dowels installed at an angle of 180°are used.

7.6. A check calculation of the shaft on the fatigue strength.
The test calculation is performed on the joint action of bending and torsion by determining the safety margin coefficients in dangerous sections of the shaft and comparing them with the permissible value. Recommended to be taken [S] = 1.5 to 2.5. The factor of safety is determined by the formula:

where Sσ and Sτ are the coefficients of safety margin for bending and torsion, respectively.
safety factors are determined in the following sequence:

  1. Select the shaft material according to the recommendations [1]. The selected material is 40 steel..
  2. Determine the design scheme (Fig. 6) dangerous section of the shaft. Dangerous section – section 3.
  3. we Determine the coefficient of safety margin for bending under the assumption that the stresses change in a symmetrical cycle:

where σ–1 – is the limit of bending strength with a symmetrical cycle, for carbon structural steels

σа – amplitude bending stresses in the considered section of the shaft, with a symmetrical stress cycle

Кσ – the effective coefficient of normal stress concentration is determined by table 7.8 V [1].
Кσ=2,3;
Kd – the scale factor is determined by table 7.9 in [1].
Kd=0,77;
KV – coefficient that takes into account the method of surface hardening for shafts without surface hardening KV=1;

  1. Determine the torsion safety factor for the case of a pulsating cycle as the most commonly used (non-reverse transmission):

where τ–1 – is the limit of torsional endurance with a symmetric cycle, for carbon and alloy steels

τа – the amplitude torsion stress in the considered section of the shaft, with a pulsating cycle

Кτ – the effective torsion concentration coefficient is determined by table 7.8 in [1].
Кτ=2,35,
Ψτ – coefficient that takes into account the cycle asymmetry for carbon steels Ψτ=0,05,

The strength condition is met, so the shaft design is suitable for use.

  1. GEAR REDUCER
    Lubrication of worm gears and bearings reduces friction, wear and heat losses. According to the method of applying the lubricant to the engagement, there are crankcase and circulation lubricants. Crankcase lubrication is performed by dipping the wheel rims into the oil poured inside the body. This lubricant is used at circumferential speeds V<10 m/s. At a higher speed, the oil is discharged by centrifugal force. When lubricating with dipping, the volume of oil poured into the crankcase is determined at the rate of (0.4–0.8) liters of oil per 1 kW of transmitted power.

It is recommended that the oil level should not be higher than the center of the lower rolling body of the bearing. For better lubrication of the wheel, mudguards are installed on the shaft, throwing oil on the engagement. Lubrication of rolling bearings of General-purpose gearboxes is carried out with liquid oils or plate ointments. The most favorable conditions for bearing operation are provided by liquid oils. Their advantages are high lubrication stability, less resistance to rotation, the ability to remove heat and clean the bearing from wear products. Liquid oil is easier to replace without disassembling the unit. The lack of liquid oils is associated with the need to use complex seals
In practice, bearings tend to be lubricated with the same oil that is used to lubricate the gear parts of the mechanism. In this case, the lubrication of bearings is usually carried out by spraying oil with gears, which results in oil entering the bearing units. The selection of the oil grade begins with determining the required kinematic viscosity of the oil depending on the circumferential velocity according to table 8.3 b [1].
V=1.97 м/с
Then, according to the found viscosity value, select the appropriate oil according to table 8.4 [1].
Aviation oil MS-20 GOST 21743-76
To drain the oil from the gearbox housing, an oil drain hole is provided, located in the lower part of the housing and closed with a threaded plug.
Oil drain plug

Pic. 10
During the operation of the gearbox, the pressure inside the housing increases due to heating of the oil and air. This causes oil to escape from the housing through the seals and joints. To avoid this, the internal cavity of the housing is connected to the external environment by installing vents (usually in the cover of the viewing window):

Pic. 11 Plug-vent

The list of references

  1. «Course design» Dulevich A. F., Novikov S. A., Surus A. I., Tsaruk F. F.-Mn: BSTU, 1997.
  2. «machine Parts and design basics» Skoibeda A. T., Kuzmin A.V., Makeychik N. N.-Mn: Higher school, 2000.
  3. «course design of machine parts» Sheinblit A. E.-M.: Higher school, 1985
  4. «Designing nodes and machine parts» Dunaev P. F., Lelikov O. P.-M: Higher school, 1985.
  5. «Course design of machine parts» Chernavsky S. A.-M.: mechanical engineering, 1979.
  6. «Details of machines» Ivanov M. N., Ivanov V. N.-M.: Higher school, 1975.

THE LIST OF REFERENCES.

  1. Explanatory dictionary of the Uzbek language. Academy Of Sciences Of The Republic Of UzbekistanNavoi Institute of language and literature. A. edited by madvaliev. National encyclopedia of Uzbekistan. State scientific publishing house. Tashkent. – 2011.
  2. English-Uzbek, Uzbek-English dictionary. 70,000 words and phrases. Sh. Butaev, A. Iriskulov. Nauka Publishing House” Academy Of Sciences Of The Republic Of Uzbekistan. Tashkent-2009.
  3. Shobidova.A. machine Parts: a textbook for technical schools. — Tashkent:»encyclopedia of Uzbekistan», 2014. -444 P.
  4. Sobitova.A. parts of the machine. Textbook. — Tashkent: 2004. -120 b.
  5. Shobidova. A., Musaev S. U. Yurid. Construction of belt and chain extensions.- Tashkent: 2000. — 182 b.
  6. Shobidova.A., Musaev S. U. Design of gear and worm gears.- Tashkent: 2005. -80 P.
  7. Shobidova.A. parts of the machine. Auto-publishing house. Tashkent-2009. 428 b. 8. Shobidova.A. Musaev S. O. Lifting, transport means. — T.: «shark», 2007. -192 P.