Dynamics of Machinery PDF

COURSE MATERIAL II Year B. Tech II- Semester MECHANICAL ENGINEERING AY: 2023-24 DYNAMICS OF MACHINERY R22A0310 Prepared by: Mr. C.Daksheeswara Reddy Assistant Professor M...

COURSE MATERIAL II Year B. Tech II- Semester MECHANICAL ENGINEERING AY: 2023-24 DYNAMICS OF MACHINERY R22A0310 Prepared by: Mr. C.Daksheeswara Reddy Assistant Professor MALLA REDDY COLLEGE OF ENGINEERING & TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING (Autonomous Institution-UGC, Govt. of India) Secunderabad-500100,Telangana State, India. www.mrcet.ac.in MALLA REDDY COLLEGE OF ENGINEERING & TECHNOLOGY (Autonomous Institution – UGC, Govt. of India) DEPARTMENT OF MECHANICAL ENGINEERING CONTENTS 1. Vision, Mission & Quality Policy 2. Pos, PSOs & PEOs 3. Blooms Taxonomy 4. Course Syllabus 5. Lecture Notes (Unit wise) a. Objectives and outcomes b. Notes c. Presentation Material (PPT Slides/ Videos) d. Industry applications relevant to the concepts covered e. Question Bank for Assignments f. Tutorial Questions 6. Previous Question Papers www.mrcet.ac.in MALLA REDDY COLLEGE OF ENGINEERING & TECHNOLOGY (Autonomous Institution – UGC, Govt. of India) VISION  To establish a pedestal for the integral innovation, team spirit, originality and competence in the students, expose them to face the global challenges and become technology leaders of Indian vision of modern society. MISSION  To become a model institution in the fields of Engineering, Technology and Management.  To impart holistic education to the students to render them as industry ready engineers.  To ensure synchronization of MRCET ideologies with challenging demands of International Pioneering Organizations. QUALITY POLICY  To implement best practices in Teaching and Learning process for both UG and PG courses meticulously.  To provide state of art infrastructure and expertise to impart quality education.  To groom the students to become intellectually creative and professionally competitive.  To channelize the activities and tune them in heights of commitment and sincerity, the requisites to claim the never - ending ladder of SUCCESS year after year. For more information: www.mrcet.ac.in MALLA REDDY COLLEGE OF ENGINEERING & TECHNOLOGY (Autonomous Institution – UGC, Govt. of India) www.mrcet.ac.in Department of Mechanical Engineering VISION To become an innovative knowledge center in mechanical engineering through state-of- the-art teaching-learning and research practices, promoting creative thinking professionals. MISSION The Department of Mechanical Engineering is dedicated for transforming the students into highly competent Mechanical engineers to meet the needs of the industry, in a changing and challenging technical environment, by strongly focusing in the fundamentals of engineering sciences for achieving excellent results in their professional pursuits. Quality Policy  To pursuit global Standards of excellence in all our endeavors namely teaching, research and continuing education and to remain accountable in our core and support functions, through processes of self-evaluation and continuous improvement.  To create a midst of excellence for imparting state of art education, industry- oriented training research in the field of technical education. MALLA REDDY COLLEGE OF ENGINEERING & TECHNOLOGY (Autonomous Institution – UGC, Govt. of India) www.mrcet.ac.in Department of Mechanical Engineering PROGRAM OUTCOMES Engineering Graduates will be able to: 1. Engineering knowledge: Apply the knowledge of mathematics, science, engineering fundamentals, and an engineering specialization to the solution of complex engineering problems. 2. Problem analysis: Identify, formulate, review research literature, and analyze complex engineering problems reaching substantiated conclusions using first principles of mathematics, natural sciences, and engineering sciences. 3. Design/development of solutions: Design solutions for complex engineering problems and design system components or processes that meet the specified needs with appropriate consideration for the public health and safety, and the cultural, societal, and environmental considerations. 4. Conduct investigations of complex problems: Use research-based knowledge and research methods including design of experiments, analysis and interpretation of data, and synthesis of the information to provide valid conclusions. 5. Modern tool usage: Create, select, and apply appropriate techniques, resources, and modern engineering and IT tools including prediction and modeling to complex engineering activities with an understanding of the limitations. 6. The engineer and society: Apply reasoning informed by the contextual knowledge to assess societal, health, safety, legal and cultural issues and the consequent responsibilities relevant to the professional engineering practice. 7. Environment and sustainability: Understand the impact of the professional engineering solutions in societal and environmental contexts, and demonstrate the knowledge of, and need for sustainable development. 8. Ethics: Apply ethical principles and commit to professional ethics and responsibilities and norms of the engineering practice. 9. Individual and teamwork: Function effectively as an individual, and as a member or leader in diverse teams, and in multidisciplinary settings. 10.Communication: Communicate effectively on complex engineering activities with the engineering community and with society at large, such as, being able to comprehend and write effective reports and design documentation, make effective presentations, and give and receive clear instructions. 11.Project management and finance: Demonstrate knowledge and understanding of the engineering and management principles and apply these to one’s own work, as a member and leader in a team, to manage projects and in multidisciplinary environments. MALLA REDDY COLLEGE OF ENGINEERING & TECHNOLOGY (Autonomous Institution – UGC, Govt. of India) www.mrcet.ac.in Department of Mechanical Engineering 12.Life-long learning: Recognize the need for and have the preparation and ability to engage in independent and life-long learning in the broadest context of technological change. PROGRAM SPECIFIC OUTCOMES (PSOs) PSO1 Ability to analyze, design and develop Mechanical systems to solve the Engineering problems by integrating thermal, design and manufacturing Domains. PSO2 Ability to succeed in competitive examinations or to pursue higher studies or research. PSO3 Ability to apply the learned Mechanical Engineering knowledge for the Development of society and self. Program Educational Objectives (PEOs) The Program Educational Objectives of the program offered by the department are broadly listed below: PEO1: PREPARATION To provide sound foundation in mathematical, scientific and engineering fundamentals necessary to analyze, formulate and solve engineering problems. PEO2: CORE COMPETANCE To provide thorough knowledge in Mechanical Engineering subjects including theoretical knowledge and practical training for preparing physical models pertaining to Thermodynamics, Hydraulics, Heat and Mass Transfer, Dynamics of Machinery, Jet Propulsion, Automobile Engineering, Element Analysis, Production Technology, Mechatronics etc. PEO3: INVENTION, INNOVATION AND CREATIVITY To make the students to design, experiment, analyze, interpret in the core field with the help of other inter disciplinary concepts wherever applicable. PEO4: CAREER DEVELOPMENT To inculcate the habit of lifelong learning for career development through successful completion of advanced degrees, professional development courses, industrial training etc. MALLA REDDY COLLEGE OF ENGINEERING & TECHNOLOGY (Autonomous Institution – UGC, Govt. of India) www.mrcet.ac.in Department of Mechanical Engineering PEO5: PROFESSIONALISM To impart technical knowledge, ethical values for professional development of the student to solve complex problems and to work in multi-disciplinary ambience, whose solutions lead to significant societal benefits. MALLA REDDY COLLEGE OF ENGINEERING & TECHNOLOGY (Autonomous Institution – UGC, Govt. of India) www.mrcet.ac.in Department of Mechanical Engineering Blooms Taxonomy Bloom’s Taxonomy is a classification of the different objectives and skills that educators set for their students (learning objectives). The terminology has been updated to include the following six levels of learning. These 6 levels can be used to structure the learning objectives, lessons, and assessments of a course. 1. Remembering: Retrieving, recognizing, and recalling relevant knowledge from long‐ term memory. 2. Understanding: Constructing meaning from oral, written, and graphic messages through interpreting, exemplifying, classifying, summarizing, inferring, comparing, and explaining. 3. Applying: Carrying out or using a procedure for executing or implementing. 4. Analyzing: Breaking material into constituent parts, determining how the parts relate to one another and to an overall structure or purpose through differentiating, organizing, and attributing. 5. Evaluating: Making judgments based on criteria and standard through checking and critiquing. 6. Creating: Putting elements together to form a coherent or functional whole; reorganizing elements into a new pattern or structure through generating, planning, or producing. MALLA REDDY COLLEGE OF ENGINEERING & TECHNOLOGY (Autonomous Institution – UGC, Govt. of India) www.mrcet.ac.in Department of Mechanical Engineering MALLA REDDY COLLEGE OF ENGINEERING & TECHNOLOGY II Year B. Tech, ME-II Sem L T/P/D C 2 1 3 (R22A0310) DYNAMICS OF MACHINERY Course Objectives: 1. To study about gyroscope and its effects during precession motion of moving vehicles. 2. To understand the force-motion relationship in components subjected to external forces and analysis of standard mechanisms. 3. Able to learn about the working of Clutches, Brakes, Dynamometers and Fly wheel. 4. To study about the balancing, unbalancing of rotating masses and the effect of Dynamics of undesirable vibrations. 5. To understand the working principles of different type governors and its characteristics. UNIT-I Precession: Gyroscopes, effect of precession motion on the stability of moving vehicles such as motor car, motor cycle, aero planes and ships. UNIT-II Static and Dynamic Force Analysis of Planar Mechanisms: Introduction -Free Body Diagrams – Conditions for equilibrium – Two, Three and Four Members – Inertia forces and D‘Alembert‘s Principle – planar rotation about a fixed centre. Friction in Machine Elements: Inclined plane-Friction of screw and nuts – Pivot and collars uniform pressure, uniform wear-friction circle and friction axis: lubricated surfaces boundary friction-film lubrication. UNIT–III Clutches: Friction clutches- Single Disc or plate clutch, Multiple Disc Clutch, Cone Clutch, Centrifugal Clutch. Brakes and Dynamometers: Simple block brakes, internal expanding brake, band brake of vehicle. Dynamometers – absorption and transmission types. General description and methods of operations. Turning Moment Diagram and Fly Wheels: Turning moment – Inertia Torque connecting rod angular velocity and acceleration, crank effort and torque diagrams – Fluctuation of energy – Fly wheels and their design. UNIT–IV Balancing: Balancing of rotating masses Single and multiple – single and different planes. Balancing of Reciprocating Masses- Primary, Secondary, and higher balancing of reciprocating masses. Analytical and graphical methods. Vibration: Free Vibration of mass attached to vertical spring – Simple problems on forced damped vibration, Vibration Isolation & Transmissibility Whirling of shafts. DEPARTMENT OF MECHANICAL ENGINEERING UNIT–V Governors: Watt, Porter and Proell governors. Spring loaded governors – Hartnell and hartung with auxiliary springs. Sensitiveness, isochronism and hunting. TEXT BOOKS: 1. Theory of Machines / Thomas Bevan / CBS Publishers 2. Theory of Machines / Jagadish Lal &J.M.Shah / Metropolitan. 3. Theory of machines / Khurmi/S.Chand Publications. REFERENCE BOOKS: 1. Theory of Machines / Shiegly / MGH Publishers. 2. Mechanism and Machine Theory / JS Rao and RV Dukkipati / New Age International Publishers 3. Theory of Machines / S.S Ratan/ Mc. Graw Hill Publishers. Course Outcomes: 1. Knowledge acquired about Gyroscope and its precession motion. 2. Able to predict the force analysis in mechanical system and able to solve the problem. 3. The student will learn about the kinematics and dynamic analysis of machine elements. 4. Ability to understand the importance of balancing and implications of computed results in dynamics to improve the design of a mechanism. 5. Student gets the exposure of different governors and its working principle. DEPARTMENT OF MECHANICAL ENGINEERING  MALLA REDDY COLLEGE OF ENGINEERING & TECHNOLOGY (Autonomous Institution – UGC, Govt. of India) DYNAMICS OF MACHINARY DEPARTMENT OF MECHANICAL(R22A0310) ENGINEERING COURSE OBJECTIVES UNIT - 1 CO1: To study about gyroscope and its effects during precession motion of moving vehicles. UNIT - 2 CO2: To understand the force-motion relationship in components subjected to external forces and analysis of standard mechanisms UNIT - 3 CO3: Able to learn about the working of Clutches, Brakes, Dynamometers and Fly wheel. UNIT - 4 CO4: To study about the balancing, unbalancing of rotating masses and the effect of Dynamics of undesirable vibrations. UNIT - 5 CO5: To understand the working principles of different type governors and its characteristics s  COURSE OUTLINE UNIT – 1 NO OF LECTURE HOURS: 10 LECTURE LECTURE TOPIC KEY ELEMENTS LEARNING OBJECTIVES (2 to 3 objectives) 1. Introduction-Gyroscopic Definition, Precessional Angular Understand the concept of Gyroscopic (B2). Motion 2. Gyroscopic Couple Derivation, Understand the concept of Gyroscopic Couple(B2) Gyroscopic Couple Understand the how to calculate the moments(B2) Apply the formulas for couple (B3) 3. Gyroscopic Couple Effect of the Gyroscopic Couple Understand the concept of Gyroscopic Couple on an aero plane. on aero plane(B2) Apply the formulas for couple (B3) Evaluate the concept (B5) 4. Gyroscopic Couple Effect of Gyroscopic Couple on a Understand the concept of Gyroscopic Couple Naval Ship during Steering on aero plane(B2) Apply the formulas for couple (B3) Evaluate the concept (B5) 5. Gyroscopic Couple Effect of Gyroscopic Couple on a Understand the concept of Gyroscopic Couple Naval Ship during Rolling on aero plane(B2) Apply the formulas for couple (B3)  Evaluate the concept (B5) 6. Gyroscopic Couple Stability of a Four Wheel Drive Understand the concept of Gyroscopic Couple Moving in a Curved Path on aero plane(B2) Apply the formulas for couple (B3) Evaluate the concept (B5) 7. Problems Problems on Gyroscopic Couple Understand the concept of Gyroscopic Couple on an aero plane on aero plane(B2) Apply the formulas for couple (B3) Evaluate the concept (B5) 8. Problems Problems on Gyroscopic Naval Apply the formulas for couple (B3) Ship during Steering 9. Problems Problems on Gyroscopic Naval Apply the formulas for couple (B3) Ship during rolling 10. Practice Practice on problems Apply the formulas for couple (B3)  UNIT – 2 NO OF LECTURE HOURS: 10 LECTURE LECTURE TOPIC KEY ELEMENTS LEARNING OBJECTIVES (2 to 3 objectives) 1. Static and Dynamic force analysis Definitions , Inertia force, Understand the concept static and Dynamic Resultant Effect of a System of forces (B2) Forces Acting on a Rigid Body 2. Free Body Diagrams & Inertia forces Two, Three and Four Members, Understand the concept of inertia forces(B3) D‘Alembert‘s Principle 3. Friction in Machine Elements Torque Required to Lifting and Understand the concept of Gyroscopic Couple lowering the Load by a Screw on aero plane(B2) Jack, Apply the formulas for couple (B3) 4. Friction in Journal Bearing-Friction Circle Flat Pivot Bearing , Conical Pivot Understand the concept of Friction in Journal Bearing, Bearing-Friction Circle (B2) 5. Friction in Journal Bearing-Friction Circle Pivot and collars uniform Understand the concept of Friction in Journal pressure, uniform wear Bearing-Friction Circle (B2) 6. problems problems Apply the formulas (B3) 7. problems problems Apply the formulas (B3) 8. problems problems Apply the formulas (B3)  MALLA REDDY COLLEGE OF ENGINEERING & TECHNOLOGY (Autonomous Institution – UGC, Govt. of India) DEPARTMENT OF MECHANICAL ENGINEERING UNIT – 3 NO OF LECTURE HOURS: 17 LECTURE LECTURE TOPIC KEY ELEMENTS LEARNING OBJECTIVES (2 to 3 objectives) 1. Introduction of Clutches Definition, Single Disc or plate Remember the standard design formulas(B1) clutch Understand the concept (B2) 2. Problems Problems on Single Disc or plate Apply the formulas(B3) clutch. Remember the standard design formulas(B1) Apply the formulas(B3) 3. CONE CLUTCH Definition, Derivation Remember the standard design formulas(B1) Understand the concept (B2) Apply the formulas(B3) 4. Problems Problems on CONE CLUTCH Remember the standard design formulas(B1) Apply the formulas(B3) 5. centrifugal Clutch Definition, Derivation. Remember the standard design formulas(B1) Understand the concept (B2) Apply the formulas(B3) 6. Problems Problems on centrifugal Clutch Remember the standard design formulas(B1) Apply the formulas(B3) 7. Brakes and Dynamometers Introduction and Types Understand the concept (B2) Remember the standard design formulas(B1) 8. Problems brakes Remember the standard design formulas(B1) Apply the formulas(B3) 9. Brakes Double Block or Shoe Brake Understand the concept (B2) Remember the standard design formulas(B1) 10. Brakes Internal Expanding Brake Understand the concept (B2) Remember the standard design formulas(B1) 11. Problems Double Block or Shoe Brake Remember the standard design formulas(B1) Apply the formulas(B3) 12. Problems Internal Expanding Brake Remember the standard design formulas(B1) Apply the formulas(B3) 13. Dynamometers – absorption and transmission types Rope Brake Dynamometer, Understand the concept (B2) Classification of Transmission Dynamometers 14. Turning Moment and Flywheel Turning moment diagram for a Understand the concept (B2) single cylinder, double acting steam engine, multi cylinder 15. Problems Turning moment diagram Remember the standard design formulas(B1) Apply the formulas(B3) Analyze the Data(B4) 16. Design of flywheel Dimensions of the Flywheel Rim Understand the concept (B2) 17. Problems Dimensions of the Flywheel Rim Remember the standard design formulas(B1) Apply the formulas(B3)  MALLA REDDY COLLEGE OF ENGINEERING & TECHNOLOGY (Autonomous Institution – UGC, Govt. of India) DEPARTMENT OF MECHANICAL ENGINEERING UNIT – 4 NO OF LECTURE HOURS:13 LECTURE LECTURE TOPIC KEY ELEMENTS LEARNING OBJECTIVES (2 to 3 objectives) 1. Balancing Introduction, Balancing of a single Remember the standard design formulas(B1) rotating mass by a single mass Understand the concept (B2) rotating in the same plane 2. Balancing Balancing of a single rotating Remember the standard design formulas(B1) mass by two masses rotating in Understand the concept (B2) different planes. 3. Balancing Balancing of different masses Remember the standard design formulas(B1) rotating in the same plane. Understand the concept (B2) Balancing of different masses rotating in different planes. 4. Balancing of Coupled Locomotives Balancing of Primary Forces of Remember the standard design formulas(B1) Multi-cylinder In-line Engines Apply the formulas(B3) 5. Balancing of Coupled Locomotives Balancing of Secondary Forces of Remember the standard design formulas(B1) Multi-cylinder In-line Engines Apply the formulas(B3) 6. Problems Problems Remember the standard design formulas(B1) Apply the formulas(B3) 7. Problems Problems Remember the standard design formulas(B1) Apply the formulas(B3) 8. Problems Problems Remember the standard design formulas(B1) Apply the formulas(B3) 9. Vibrations Terms Used in Vibratory Motion, Remember the standard design formulas(B1) Types of Free Vibrations, Understand the concept (B2) Derivations. 10. Free Torsional Vibrations of a Two Rotor System Concept and Derivation Understand the concept (B2) Remember the standard formulas(B1) 11. Free Torsional Vibrations of a Three Rotor System Concept and Derivation Understand the concept (B2) Remember the standard formulas(B1 12. Problems Free torsional vibrations of a two Remember the standard design formulas(B1) rotor system Apply the formulas(B3) 13. Problems Free Torsional Vibrations of a Remember the standard design formulas(B1) Three Rotor System Apply the formulas(B3)  UNIT – 5 NO OF LECTURE HOURS: 08 LECTUR E LECTURE TOPIC KEY ELEMENTS LEARNING OBJECTIVES (2 to 3 objectives) 1. Introduction Governors, Classification, Watt, Understand the concept (B2) Porter 2. Method of resolution of forces Method of resolution of forces, Understand the concept (B2) Instantaneous centre method 3. Types of governors Proell governors Remember the standard design formulas(B1) Understand the concept (B2) 4. Types of governors Spring loaded governors – Remember the standard design formulas(B1) Hartnell and hartung with Understand the concept (B2) auxiliary springs Derivations 5. Types of governors Sensitiveness, isochronism and Remember the standard design formulas(B1) hunting. Understand the concept (B2) 6. problems Problems on all types of Analyze the data(B4) governors Remember the standard design formulas(B1) Apply the formulas(B3) 7. problems Problems on all types of Analyze the data(B4) governors Remember the standard design formulas(B1) Apply the formulas(B3) 8. problems Problems on all types of Analyze the data(B4) governors Remember the standard design formulas(B1) Apply the formulas(B3)  UNIT 1 PRECESSION COURSE OBJECTIVES To study about gyroscope and its effects during precession motion of moving vehicles. COURSE OUTCOMES Knowledge acquired about Gyroscope and its precession motion. DEPARTMENT OF MECHANICAL ENGINEERING Introduction When a body moves along a curved path with a uniform linear velocity, a force in the direction of centripetal acceleration (known as centripetal force) has to be applied externally over the body, so that it moves along the required curved path. This external force applied is known as active force. When a body, itself, is moving with uniform linear velocity along a circular path, it is subjected to the centrifugal force* radially outwards. This centrifugal force is called reactive force. The action of the reactive or centrifugal force is to tilt or move the body along radially outward direction. Precessional Angular Motion We have already discussed that the angular acceleration is the rate of change of angular velocity with respect to time. It is a vector quantity and may be represented by drawing a vector diagram with the help of right hand screw rule. Consider a disc, as shown in Fig (a), revolving or spinning about the axis OX (known as axis of spin) in anticlockwise when seen from the front, with an angular velocity ω in a plane at right angles to the paper. After a short interval of time δt, let the disc be spinning about the new axis of spin OX ′(at an angle δθ) with an angular velocity (ω+ δω). Using the right hand screw rule, initial angular velocity of the disc (ω) is represented by vector ox; and the final angular velocity of the disc (ω+ δω) is represented by vector ox′ as shown in Fig. 14.1 (b). The vector xx′ represents the change of angular velocity in time δt i.e. the angular acceleration of the disc. This may be resolved into two components. One parallel to ox and the other perpendicular to ox. DEPARTMENT OF MECHANICAL ENGINEERING Where dθ/dt is the angular velocity of the axis of spin about a certain axis, which is perpendicular to the plane in which the axis of spin is going to rotate. This angular velocity of the axis of spin (i.e.dθ/dt) is known as angular velocity of precession and is denoted by ωP. The axis, about which the axis of spin is to turn, is known as axis of precession. The angular motion of the axis of spin about the axis of precession is known as precessional angular motion. Gyroscopic Couple Consider a disc spinning with an angular velocity ω rad/s about the axis of spin OX, in anticlockwise direction when seen from the front, as shown in Fig. 14.2 (a). Since the plane in which the disc is rotating is parallel to the plane YOZ, therefore it is called plane of spinning. The plane XOZ is a horizontal plane and the axis of spin rotates in a plane parallel to the horizontal plane about an axis OY. In other words, the axis of spin is said to be rotating or processing about an axis OY. In other words, the axis of spin is said to be rotating or processing about an axis OY (which is perpendicular to both the axes OX and OZ) at an angular velocity _P rap/s. This horizontal plane XOZ is called plane of precession and OY is the axis of precession. Let I = Mass moment of inertia of the disc about OX, and ω= Angular velocity of the disc. Angular momentum of the disc = I. ω Since the angular momentum is a vector quantity, therefore it may be represented by the vector ox′ , as shown in Fig. 14.2 (b). The axis of spin OX is also rotating anticlockwise when seen from the top about the axis OY. Let the axis OX is turned in the plane XOZ through a small angle δθ radians to the position OX′ , in DEPARTMENT OF MECHANICAL ENGINEERING time δt seconds. Assuming the angular velocity ω to be constant, the angular momentum will now be represented by vector ox′. Change in angular momentum and rate of change of angular momentum Since the rate of change of angular momentum will result by the application of a couple to the disc, Therefore the couple applied to the disc causing precession, Where ωP = Angular velocity of precession of the axis of spin or the speed of rotation of the axis of spin about the axis of precession OY. The couple I.ω.ωp, in the direction of the vector xx′ (representing the change in angular momentum) is the active gyroscopic couple, which has to be applied over the disc when the axis of spin is made to rotate with angular velocity ωP about the axis of precession. The vector xx′ lies in the plane XOZ or the horizontal plane. In case of a very small displacement δθ, the vector xx′ will be perpendicular to the vertical plane XOY. Therefore the couple causing this change in the angular momentum will lie in the plane XOY. The vector xx′, as shown in Fig(b), represents an anticlockwise couple in the plane XOY. Therefore, the plane XOY is called the plane of active gyroscopic couple and the axis OZ perpendicular to the plane XOY, about which the couple acts, is called the axis of active gyroscopic couple. When the axis of spin itself moves with angular velocity ωP, the disc is subjected to reactive couple whose magnitude is same (i.e. I.ω.ωP) but opposite in direction to that of active couple. This reactive couple to which the disc is subjected when the axis of spin DEPARTMENT OF MECHANICAL ENGINEERING rotates about the axis of precession is known as reactive gyroscopic couple. The axis of the reactive gyroscopic couple is represented by OZ′ in Fig(a). The gyroscopic couple is usually applied through the bearings which support the shaft. The bearings will resist equal and opposite couple. The gyroscopic principle is used in an instrument or toy known as gyroscope. The gyroscopes are installed in ships in order to minimize the rolling and pitching effects of waves. They are also used in aero planes, monorail cars, gyrocompasses etc. Effect of the Gyroscopic Couple on an aero plane: The top and front view of an aero plane is shown in Fig (a). Let engine or propeller rotates in the clockwise direction when seen from the rear or tail end and the aero plane takes a turn to the left. Let ω= Angular velocity of the engine in rad/sec, m = Mass of the engine and the propeller in kg, k = Its radius of gyration in metre I = Mass moment of inertia of the engine and the propeller in kg-m2 = mk2 v = Linear velocity of the aero plane in m/s, R = Radius of curvature in metres, and ωp = Angular velocity of precession=V/R DEPARTMENT OF MECHANICAL ENGINEERING Before taking the left turn, the angular momentum vector is represented by ox. When it takes left turn, the active gyroscopic couple will change the direction of the angular momentum vector from ox to ox′ as shown in Fig(a). The vector xx′, in the limit, represents the change of angular momentum or the active gyroscopic couple and is perpendicular to ox. Thus the plane of active gyroscopic couple XOY will be perpendicular to xx′, i.e. vertical in this case, as shown in Fig (b). By applying right hand screw rule to vector xx′, we find that the direction of active gyroscopic couple is clockwise as shown in the front view of Fig (a). In other words, for left hand turning, the active gyroscopic couple on the aero plane in the axis OZ will be clockwise as shown in Fig (b).The reactive gyroscopic couple.(equal in magnitude of active gyroscopic couple) will act in the opposite direction (i.e. in the anticlockwise direction) and the effect of this couple is, therefore, to raise the nose and dip the tail of the aero plane. Terms Used in a Naval Ship. The top and front views of a naval ship are shown in Fig 14.7. The fore end of the ship is called bow and the rear end is known as stern or aft. The left hand and right hand sides of the ship, when viewed from the stern are called port and star-board respectively. We shall now discuss the effect of gyroscopic couple on the naval ship in the following three cases: Steering, Pitching, Rolling. DEPARTMENT OF MECHANICAL ENGINEERING Effect of Gyroscopic Couple on a Naval Ship during Steering Steering is the turning of a complete ship in a curve towards left or right, while it moves forward. Consider the ship taking a left turn, and rotor rotates in the clockwise direction when viewed from the stern, as shown in Fig. The effect of gyroscopic couple on a naval ship during steering taking left or right turn may be obtained in the similar way as for an aeroplane. When the rotor of the ship rotates in the clockwise direction when viewed from the stern, it will have its angular momentum vector in the direction ox as shown in Fig(a). As the ship steers to the left, the active gyroscopic couple will change the angular momentum vector from ox to ox′. The vector xx′ now represents the active gyroscopic couple and is perpendicular to ox. Thus the plane of active gyroscopic couple is perpendicular to xx′ and its direction in the axis OZ for left hand turn is clockwise as shown in Fig. The reactive gyroscopic couple of the same magnitude will act in the opposite direction (i.e. in anticlockwise direction). The effect of this reactive gyroscopic couple is to raise the bow and lower the stern. DEPARTMENT OF MECHANICAL ENGINEERING Effect of Gyroscopic Couple on a Naval Ship during Pitching Pitching is the movement of a complete ship up and down in a vertical plane about transverse axis, as shown in Fig(a). In this case, the transverse axis is the axis of precession. The pitching of the ship is assumed to take place with simple harmonic motion i.e. the motion of the axis of spin about transverse axis is simple harmonic. Let I = Moment of inertia of the rotor in kg-m2, and ω= Angular velocity of the rotor inrad/s. Minimum gyroscopic couple, When the pitching is upward, the effect of the reactive gyroscopic couple, as shown in Fig.(b), will try to move the ship toward star-board. On the other hand, if the pitching is downward, the effect of the reactive gyroscopic couple, as shown in Fig(c), is to turn the ship towards port side. Effect of Gyroscopic Couple on a Naval Ship during Rolling We know that, for the effect of gyroscopic couple to occur, the axis of precession should always be perpendicular to the axis of spin. If, however, the axis of precession becomes parallel to the axis of spin, there will be no effect of the gyroscopic couple acting on the body of the ship. In case of rolling of a ship, the axis of precession (i.e. longitudinal axis) is always parallel to the axis of spin for all positions. Hence, there is no effect of the gyroscopic couple acting on the body of a ship. Stability of a Four Wheel Drive Moving in a Curved Path DEPARTMENT OF MECHANICAL ENGINEERING Consider the four wheels A, B, C and D of an automobile locomotive taking a turn towards left as shown in Fig. The wheels A and C are inner wheels, whereas B and D are outer wheels. The centre of gravity (C.G.) of the vehicle lies vertically above the road surface. Let m = Mass of the vehicle inkg, W = Weight of the vehicle in newtons = m.g, rW = Radius of the wheels in metres, R = Radius of curvature in metres (R >rW), h = Distance of centre of gravity, vertically above the roadsurfacein metres, x = Width of track in metres, IW = Mass moment of inertia of one of the wheels in kg-m2, ωW= Angular velocity of the wheels or velocity of spin in rad/s, IE = Mass moment of inertia of the rotating parts of the engine in kg-m2 ωE = Angular velocity of the rotating parts of the engine in rad/s, G = Gear ratio =ωE/ωw v = Linear velocity of the vehicle in m/s = ωW.rW A little consideration will show that the weight of the vehicle (W) will be equally distributed over the four wheels which will act downwards. The reaction between each wheel and the road surface of the same magnitude will act upwards. DEPARTMENT OF MECHANICAL ENGINEERING Therefore Road reaction over each wheel = W/4 = m.g/4 newtons Let us now consider the effect of the gyroscopic couple and centrifugal couple on the vehicle. Effect of the gyroscopic couple Since the vehicle takes a turn towards left due to the precession and other rotating parts, therefore a gyroscopic couple will act. We know that velocity of precession, ωP = v/R Gyroscopic couple due to 4 wheels, CW = 4 IW.ωW.ωP and gyroscopic couple due to the rotating parts of the engine, CE = IE.ωE.ωP = IE.G.ωW.ωP Net gyroscopic couple, C = CW ± CE = 4 IW.ωW.ωP ± IE.G.ωW.ωP DEPARTMENT OF MECHANICAL ENGINEERING = ωW.ωP (4 IW ± G.IE) The positive sign is used when the wheels and rotating parts of the engine rotate in the same direction. If the rotating parts of the engine revolve in opposite direction, then negative sign is used. Due to the gyroscopic couple, vertical reaction on the road surface will be produced. The reaction will be vertically upwards on the outer wheels and vertically downwards on the inner wheels. Let the magnitude of this reaction at the two outer or inner wheels be P new tons. Then P × x = C or P = C/x Vertical reaction at each of the outer or inner wheels, P /2 = C/2x Effect of the centrifugal couple Since the vehicle moves along a curved path, therefore centrifugal force will act outwardly at the centre of gravity of the vehicle. The effect of this centrifugal force is also to overturn the vehicle. We know that centrifugal force, The couple tending to overturn the vehicle or overturning couple, This overturning couple is balanced by vertical reactions, which are vertically up wards on the outer wheels and vertically downwards on the inner wheels. Let the magnitude of this reaction at the two outer or inner wheels be Q. Then Vertical reaction at each of the outer or inner wheels, Total vertical reaction at each of the outer wheel, DEPARTMENT OF MECHANICAL ENGINEERING Total vertical reaction at each of the inner wheel A little consideration will show that when the vehicle is running at high speeds, PI may bezero or even negative. This will cause the inner wheels to leave the ground thus tending to overturn the automobile. In order to have the contact between the inner wheels and the ground, the sum of P/2 and Q/2 must be less than W/4. Stability of a Two Wheel Vehicle Taking a Turn Consider a two wheel vehicle (say a scooter or motor cycle) taking a right turn as shown in fig. Let m = Mass of the vehicle and its rider in kg, W = Weight of the vehicle and its rider in Newton’s = m.g, h = Height of the centre of gravity of the vehicle and rider, rW = Radius of the wheels, R = Radius of track or curvature, IW = Mass moment of inertia of each wheel, IE = Mass moment of inertia of the rotating parts of the v engine, ωW = Angular velocity of the wheels, ωE= Angular velocity of the engine, G = Gear ratio = ωE / ωW, DEPARTMENT OF MECHANICAL ENGINEERING v = Linear velocity of the vehicle = ωW × rW, θ=Angle of wheel. It is inclination of the vehicle to the vertical for equilibrium. Let us now consider the effect of the gyroscopic couple and centrifugal couple on the vehicle, Effect of gyroscopic couple We know that v = ωW× rW or ωW = v /rW Velocity of precession, ωP = v /R A little consideration will show that when the wheels move over the curved path, the vehicle is always inclined at an angle θ with the vertical plane as shown in Fig (b). This angle is known as angle of heel. In other words, the axis of spin is inclined to the horizontal at an angle θ, as shown in Fig (c). Thus the angular Momentum vector Iω due to spin is represented by OA inclined to OX at an angle θ. But the precession axis is vertical. Therefore the spin vector is resolved along OX. Effect of centrifugal couple We know that centrifugal force, This force acts horizontally through the centre of gravity (C.G.) along the outward direction. DEPARTMENT OF MECHANICAL ENGINEERING Centrifugal couple, Since the centrifugal couple has a tendency to overturn the vehicle, therefore Total overturning couple, We know that balancing couple = m.g.h sinθ The balancing couple acts in clockwise direction when seen from the front of the vehicle. Therefore for stability, the overturning couple must be equal to the balancing couple,i.e. From this expression, the value of the angle of heel (θ) may be determined, so that the vehicle does not skid. PROBLEMS Example 1. A uniform disc of diameter 300 mm and of mass 5 kg is mounted on one end of an arm of length 600 mm. The other end of the arm is free to rotate in a universal bearing. If the disc rotates about the arm with a speed of 300 r.p.m. clockwise, looking from the front, with what speed will it process about the vertical axis? Solution. Given: d = 300 mm or r = 150 mm = 0.15 m ;m = 5 kg ; l = 600 mm = 0.6 m N= 300 r.p.m. or ω= 2π× 300/60 = 31.42 rad/s We know that the mass moment of inertia of the disc, about an axis through its centre of gravity and perpendicular to the plane of disc, I = m.r2/2 = 5(0.15)2/2 = 0.056 kg-m2 couple due to mass of disc, C = m.g.l = 5 × 9.81 × 0.6 = 29.43 N-m Let ωP = Speed of precession. We know that couple (C), DEPARTMENT OF MECHANICAL ENGINEERING 29.43 = I.ω.ωP = 0.056 × 31.42 × ωP = 1.76 ωP ωP= 29.43/1.76 = 16.7 rad/s Example 2. An aeroplane makes a complete half circle of 50 metres radius, towards left, when flying at 200 km per hr. The rotary engine and the propeller of the plane has a mass of 400 kg and a radius of gyration of 0.3 m. The engine rotates at 2400 r.p.m. clockwise when viewed from the rear. Find the gyroscopic couple on the aircraft and state its effect on it. Solution: Given :R = 50 m ; v = 200 km/hr = 55.6 m/s ; m = 400 kg ; k = 0.3 m ; N = 2400 r.p.m. or ω= 2π× 2400/60 = 251 rad/s We know that mass moment of inertia of the engine and the propeller, I = m.k2 = 400(0.3)2 = 36 kg-m2 Angular velocity of precession, ωP = v/R= 55.6/50 = 1.11 rad/s We know that gyroscopic couple acting on the aircraft, C = I. ω. ωP = 36 × 251.4 × 1.11 = 100 46 N-m=10.046 kN-m When the aeroplane turns towards left, the effect of the gyroscopic couple is to lift the nose upwards and tail downwards. Example 3 :The turbine rotor of a ship has a mass of 8 tonnes and a radius of gyration0.6 m.It rotates at 1800 r.p.m. clockwise, when looking from the stern. Determine the gyroscopic couple, if the ship travels at 100 km/hr and steer to the left in a curve of 75 m radius. Solution.Given: m = 8 t = 8000 kg ;k = 0.6 m ; N = 1800 r.p.m. or ω = 2π× 1800/60= 188.5 rad/s ; v = 100 km/h = 27.8 m/s ; R = 75 m We know that mass moment of inertia of the rotor, I = m.k2 = 8000 (0.6)2 = 2880 kg-m2 Angular velocity of precession, ω = v / R = 27.8 / 75 = 0.37 rad/s We know that gyroscopic p Couple, C = I.ω.ωP = 2880 × 188.5 × 0.37 = 200 866 N-m = 200.866 kN-m DEPARTMENT OF MECHANICAL ENGINEERING When the rotor rotates in clockwise direction when looking from the stern and the ship steers to the left, the effect of the reactive gyroscopic couple is to raise the bow and lower the stern. DEPARTMENT OF MECHANICAL ENGINEERING  INDUSTRIAL APPLICATIONS OF GYROSCOPIC 1. Motor car 2. Motor cycle 3. Aero planes 4. Ships 5. Applications of gyroscopes include inertial navigation systems, such as in the Hubble Telescope, or inside the steel hull of a submerged submarine. Due to their precision, gyroscopes are also used in gyrotheodolites to maintain direction in tunnel mining. DEPARTMENT OF MECHANICAL ENGINEERING DEPARTMENT OF MECHANICAL ENGINEERING  TUTORIAL QUESTIONS 1. What do you understand by gyroscopic couple? Derive a formula for its magnitude. 2. Explain the effect of the gyroscopic couple on the reaction of the four wheels of a vehicle negotiating a curve. 3. Discuss the effect of the gyroscopic couple on a two wheeled vehicle when taking a turn. 4. What will be the effect of the gyroscopic couple on a disc fixed at a certain angle to a rotating shaft? 5. A flywheel of mass 10 kg and radius of gyration 200 mm is spinning about its axis, which is horizontal and is suspended at a point distant 150 mm from the plane of rotation of the flywheel. Determine the angular velocity of precession of the flywheel. The spin speed of flywheel is 900 r.p.m. 6. Find the angle of inclination with respect to the vertical of a two wheeler negotiating a turn. Given : combined mass of the vehicle with its rider 250 kg ; moment of inertia of the engine flywheel 0.3 kg-m2 ; moment of inertia of each road wheel 1 kg-m2 ; speed of engine flywheel 5 times that of road wheels and in the same direction ; height of centre of gravity of rider with vehicle 0.6 m ; two wheeler speed 90 km/h ; wheel radius 300 mm ; radius of turn 50 m. 7. A pair of locomotive driving wheels with the axle, have a moment of inertia of 180 kg-m2. The diameter of the wheel treads is 1.8 m and the distance between wheel centres is 1.5 m. When the locomotive is travelling on a level track at 95 km/h, defective ballasting causes one wheel to fall 6 mm and to rise again in a total time of 0.1 s. If the displacement of the wheel takes place with simple harmonic motion, find: 1. The gyroscopic couple set up, and 2. The reaction between the wheel and rail due to this couple. 8. A four-wheeled trolley car of total mass 2000 kg running on rails of 1.6 m gauge, rounds a curve of 30 m radius at 54 km/h. The track is banked at 8°. The wheels have an external diameter of 0.7 m and each pair with axle has a mass of 200 kg. The radius of gyration for each pair is 0.3 m. The height of centre of gravity of the car above the wheel base is 1 m. Determine, allowing for centrifugal force and gyroscopic couple actions, the pressure on each rail. 9. A horizontal axle AB, 1 m long, is pivoted at the midpoint C. It carries a weight of 20 N at A and a wheel weighing 50 N at B. The wheel is made to spin at a speed of 600 r.p.m in a clockwise direction looking from its front. Assuming that the weight of the flywheel is uniformly distributed around the rim whose mean diameter is 0.6 m; calculate the angular velocity of precession of the system around the vertical axis through C. DEPARTMENT OF MECHANICAL ENGINEERING  ASSIGNMENT QUESTIONS 1. Explain the effect of the gyroscopic couple on the reaction of the four wheels of a vehicle negotiating a curve 2. Discuss the effect of the gyroscopic couple on a two wheeled vehicle when taking a turn. 3. What will be the effect of the gyroscopic couple on a disc fixed at a certain angle to a rotating shaft? 4. A uniform disc of 150 mm diameter has a mass of 5 kg. It is mounted centrally in bearings which maintain its axle in a horizontal plane. The disc spins about it axle with a constant speed of 1000 r.p.m. while the axle precesses uniformly about the vertical at 60 r.p.m. The directions of rotation are as shown in Fig. 14.3. If the distance between the bearings is 100 mm, find the resultant reaction at each bearing due to the mass and gyroscopic effects. 5. An aeroplane makes a complete half circle of 50 metres radius, towards left, when flying at 200 km per hr. The rotary engine and the propeller of the plane has a mass of 400 kg and a radius of gyration of 0.3 m. The engine rotates at 2400 r.p.m. clockwise when viewed from the rear. Find the gyroscopic couple on the aircraft and state its effect on it. 6. The turbine rotor of a ship has a mass of 8 tonnes and a radius of gyration 0.6 m. It rotates at 1800 r.p.m. clockwise, when looking from the stern. Determine the gyroscopic couple, if the ship travels at 100 km/hr and steer to the left in a curve of 75 m radius. 7. The heavy turbine rotor of a sea vessel rotates at 1500 r.p.m. clockwise looking from the stern, its mass being 750 kg. The vessel pitches with an angular velocity of 1 rad/s. Determine the gyroscopic couple transmitted to the hull when bow is rising, if the radius of gyration for the rotor is 250 mm. Also show in what direction the couple acts on the hull? 8. The turbine rotor of a ship has a mass of 3500 kg. It has a radius of gyration of 0.45 m and a speed of 3000 r.p.m. clockwise when looking from stern. Determine the gyroscopic couple and its effect upon the ship: 1. when the ship is steering to the left on a curve of 100 m radius at a speed of 36 km/h. 2. when the ship is pitching in a simple harmonic motion, the bow falling with its maximum velocity. The period of pitching is 40 seconds and the total angular displacement between the two extreme positions of pitching is 12 degrees. 9. A ship propelled by a turbine rotor which has a mass of 5 tonnes and a speed of 2100 r.p.m. The rotor has a radius of gyration of 0.5 m and rotates in a clockwise direction when viewed from the stern. Find the gyroscopic effects in the following conditions: DEPARTMENT OF MECHANICAL ENGINEERING  1. The ship sails at a speed of 30 km/h and steers to the left in a curve having 60 m radius. 2. The ship pitches 6 degree above and 6 degree below the horizontal position. The bow is descending with its maximum velocity. The motion due to pitching is simple harmonic and the periodic time is 20 seconds. 3. The ship rolls and at a certain instant it has an angular velocity of 0.03 rad/s clockwise when viewed from stern. Determine also the maximum angular acceleration during pitching. Explain how the direction of motion due to gyroscopic effect is determined in each case. DEPARTMENT OF MECHANICAL ENGINEERING  UNIT 2 STATIC AND DYNAMICS FORCE ANALYSIS OF PLANNER MECHANISMS /FRICTION IN MACHINE MECHANISMS DEPARTMENT OF MECHANICAL ENGINEERING Course Objectives To understand the force-motion relationship in components subjected to external forces and analysis of standard mechanisms. Course Outcomes Able to predict the force analysis in mechanical system and able to solve the problem. DEPARTMENT OF MECHANICAL ENGINEERING Introduction: Then the system can be treated as static, which permits application of. Techniques of static force analysis. Dynamic force analysis is the evaluation of input forces or torques and joint forces. Considering motion of members. Evaluation of the inertia force /torque is explained first. The inertia force is an imaginary force, which when acts upon a rigid body, brings it in an equilibrium position. It is numerically equal to the accelerating force in magnitude, but opposite in direction. Mathematically, Inertia force = – Accelerating force = – m.a Where m = Mass of the body, and a = Linear acceleration of the centre of gravity of the body. Similarly, the inertia torque is an imaginary torque, which when applied upon the rigid body, brings it in equilibrium position. It is equal to the accelerating couple in magnitude but opposite in direction. Resultant Effect of a System of Forces Acting on a Rigid Body: Consider a rigid body acted upon by a system of forces. These forces may be reduced to a single resultant force F whose line of action is at a distance h from the centre of gravity G. Now let us assume two equal and opposite forces (of magnitude F ) acting through G, and parallel to the resultant force, without influencing the effect of the resultant force F, as shown in Fig. 15.1. A little consideration will show that the body is now subjected to a couple (equal to F × h) and a force; equal and parallel to the resultant force F passing through G. The force F through G causes linear acceleration of the c.g. and the moment of the couple (F × h) causes angular acceleration of the body about an axis passing through G and perpendicular to the point in which the couple acts. α = Angular acceleration of the rigid body due to couple, h = Perpendicular distance between the force and centre of gravity of the body, m = Mass of the body, k = Least radius of gyration about an axis through G, and I = Moment of inertia of the body about an axis passing through its centre of gravity and perpendicular to the point in which the couple acts = m.k2 We know that Force, F = Mass × Acceleration = m.a...(i) and F.h = m.k2.α = I.α D-Alembert’s Principle DEPARTMENT OF MECHANICAL ENGINEERING Consider a rigid body acted upon by a system of forces. The system may be reduced to a single resultant force acting on the body whose magnitude is given by the product of the mass of the body and the linear acceleration of the centre of mass of the body. According to Newton’s second law of motion. F = m.a...(i) Where F = Resultant force acting on the body, m = Mass of the body, and a = Linear acceleration of the centre of mass of the body The equation (i) may also be written as: F – m.a = 0 A little consideration will show, that if the quantity – m.a be treated as a force, equal, opposite and with the same line of action as the resultant force F, and include this force with the system of forces of which F is the resultant, then the complete system of forces will be in equilibrium. This principle is known as D-Alembert’s principle. The equal and opposite force – m.a is known as reversed effective force or the inertia force (briefly written as FI). The equation (ii) may be written as F + FI = 0... (iii) Thus, D-Alembert’s principle states that the resultant force acting on a body together with the reversed effective force (or inertia force), are in equilibrium. This principle is used to reduce a dynamic problem into an equivalent static problem. Friction in Machine Elements Screw Friction The screws, bolts, studs, nuts etc. are widely used in various machines and structures for temporary fastenings. These fastenings have screw threads, which are made by cutting a continuous helical groove on a cylindrical surface. If the threads are cut on the outer surface of a solid rod, these are known as external threads. But if the threads are cut on the internal surface of a hollow rod, these are known as internal threads. The screw threads are mainly of two types i.e. V-threads and square threads. The V- threads are stronger and offer more frictional resistance to motion than square threads. Moreover, the V-threads have an advantage of preventing the nut from slackening. In general, the V threads are used for the purpose of tightening pieces together e.g. bolts and nuts etc. But the square threads are used in screw jacks, vice screws etc. The following terms are important for the study of screw Helix. It is the curve traced by a particle, while describing a circular path at a uniform speed and advancing in the axial direction at a uniform rate. In other words, it is the curve traced by a particle while moving along a screw thread. Pitch. It is the distance from a point of a screw to a corresponding point on the next thread, measured parallel to the axis of the screw DEPARTMENT OF MECHANICAL ENGINEERING Lead. It is the distance; a screw thread advances axially in one turn. Depth of thread. It is the distance between the top and bottom surfaces of a thread (also known as crest and root of a thread). Single-threaded screw. If the lead of a screw is equal to its pitch. It is known as single threaded screw. Lead = Pitch × Number of threads Helix angle. It is the slope or inclination of the thread with the horizontal. The screw jack is a device, for lifting heavy loads, by applying a comparatively smaller effort at its handle. The principle, on which a screw jack works, is similar to that of an inclined plane. Fig (a) shows a common form of a screw jack, which consists of a square threaded rod (also called screw rod or simply screw) which fits into the inner threads of the nut. The load, to be raised or lowered, is placed on the head of the square threaded rod which is rotated by the application of an effort at the end of the lever for lifting or lowering the load. Torque Required to Lifting the Load by a Screw Jack : If one complete turn of a screw thread by imagined to be unwound, from the body of the screw and developed, it will form an inclined plane as shown in Fig (a). Let p = Pitch of the screw, d = Mean diameter of the screw, α= Helix angle, P = Effort applied at the circumference of the screw to lift the load, W = Load to be lifted, and μ = Coefficient of friction, between the screw and nut = tan ø, Where φ is the friction angle. DEPARTMENT OF MECHANICAL ENGINEERING From the geometry of the Fig(a), we find that Since the principle on which a screw jack works is similar to that of an inclined plane, therefore the force applied on the lever of a screw jack may be considered to be horizontal as shown in Fig(b). Since the load is being lifted, therefore the force of friction (F = μ.RN) will act downwards. All the forces acting on the screw are shown in Fig(b). Resolving the forces along the plane, P cos α = W sin α+ F = W sinα+μ.RN (i) and resolving the forces perpendicular to the plane, RN =P sin α+ W cosα (ii) Substituting this value of RN in equation(i), P cos α = W sin α + μ (P sin α + W cos α) = W sin α + μ P sin α + μ W cos α P cos α – μ P sin α = W sin α + μ W cos α P (cos α – μ sin α) = W (sin α + μ cos α) Substituting the value of μ = tan φ in the above equation, we get Multiplying the numerator and denominator by cosφ, = W tan (α+ φ) Torque required to overcoming friction between the screw and nut, When the axial load is taken up by a thrust collar or a flat surface, as shown in Fig (b),so that the load does not rotate with the screw, then the torque required to overcome friction at the collar, DEPARTMENT OF MECHANICAL ENGINEERING R1 and R2 = Outside and inside radii of the collar, R = Mean radius of the collar, and μ1 = Coefficient of friction for the collar. Total torque required to overcome friction (i.e. to rotate the screw), If an effort P1 is applied at the end of a lever of arm length l, then the total torque required to overcome friction must be equal to the torque applied at the end of the lever, i.e. Friction of a V-thread The normal reaction in case of a square threaded screw is RN = W cosα, where α= Helix angle. But in case of V-thread (or acme or trapezoidal threads), the normal reaction between the screw and nut is increased because the axial component of this normal reaction must be equal to the axial load. W, as shown in Fig. Let 2= Angle of the V-thread, and DEPARTMENT OF MECHANICAL ENGINEERING = Semi-angle of the V-thread. Friction in Journal Bearing-Friction Circle A journal bearing forms a turning pair as shown in Fig (a). The fixed outer element of a turning pair is called a bearing and that portion of the inner element (i.e. shaft) which fits in the bearing is called a journal. The journal is slightly less in diameter than the bearing, in order to permit the free movement of the journal in a bearing. When the bearing is not lubricated (or the journal is stationary), then there is a line contact between the two elements as shown in Fig (a). The load W on the journal and normal reaction RN (equal to W) of the bearing acts through the centre. The reaction RN acts vertically upwards at point A. This point A is known as seat or point of pressure. Now consider a shaft rotating inside a bearing in clockwise direction as shown in Fig(b). The lubricant between the journal and bearing forms a thin layer which gives rise to a greasy friction. Therefore, the reaction R does not act vertically upward, but acts at another point of pressure B. This is due to the fact that when shaft rotates, a frictional force F = μ RN acts at the circumference of the shaft which has a tendency to rotate the shaft in opposite direction of motion and this shifts the point A to point B. In order that the rotation may be maintained, there must be a couple rotating the shaft. Let ø= Angle between R (resultant of F and RN) and RN, μ = Coefficient of friction between the journal and bearing, T = Frictional torque in N-m, and r = Radius of the shaft in meters. For uniform motion, the resultant force acting on the shaft must be zero and the resultant turning moment on the shaft must be zero. In other words, DEPARTMENT OF MECHANICAL ENGINEERING R = W, and T = W × OC = W × OB sinφ= W.rsinφ Since ø is very small, therefore substituting sinφ= tanφ T = W.rtanø=μ.W.r (μ =tanø) If the shaft rotates with angular velocity ω rad/s, then power wasted in friction, P = Tω= T × 2πN/60 watts Where N = Speed of the shaft in r.p.m. Friction of Pivot and Collar Bearing The rotating shafts are frequently subjected to axial thrust. The bearing surfaces such as pivot and collar bearings are used to take this axial thrust of the rotating shaft. The propeller shafts of ships, the shafts of steam turbines, and vertical machine shafts are examples of shafts which carry an axial thrust. The bearing surfaces placed at the end of a shaft to take the axial thrust are known as pivots. The pivot may have a flat surface or conical surface as shown in Fig. 10.16 (a) and (b) respectively. When the cone is truncated, it is then known as truncated or trapezoidal pivot as shown in Fig (c). The collar may have flat bearing surface or conical bearing surface, but the flat surface is most commonly used. There may be a single collar, as shown in Fig (d) or several collars along the length of a shaft, as shown in Fig(e) in order to reduce the intensity of pressure. In modern practice, ball and roller thrust bearings are used when power is being transmitted and when thrusts are large as in case of propeller shafts of ships. A little consideration will show that in a new bearing, the contact between the shaft and bearing may be good over the whole surface. In other words, we can say that the pressure over the rubbing surfaces is uniformly distributed. But when the bearing becomes old, all parts of the rubbing surface will not move with the same velocity, because the velocity of rubbing surface increases with the distance from the axis of the bearing. This means that wear may be different at different radii and this causes to alter the distribution of pressure. Hence, in the study of friction of bearings, it is assumed hat The pressure is uniformly distributed throughout the bearing surface, and The wear is uniform throughout the bearing surface. DEPARTMENT OF MECHANICAL ENGINEERING Flat Pivot Bearing: When a vertical shaft rotates in a flat pivot bearing (known as foot step bearing), as shown in Fig., the sliding friction will be along the surface of contact between the shaft and the bearing. Let W =Load transmitted over the bearing surface, R =Radius of bearing surface, p =Intensity of pressure per unit area of bearing Surface between rubbing surfaces, and μ =Coefficient of friction. We will consider the following two cases: 1. When there is a uniform pressure 2. When there is a uniform wear Considering uniform pressure When the pressure is uniformly distributed over the bearing area, then Consider a ring of radius r and thickness drof the bearing area. Area of bearing surface, A = 2πr.dr Load transmitted to the ring, δW= p × A = p × 2πr.dr (i) Frictional resistance to sliding on the ring acting tangentially at radius r, Fr= μ.δW= μ p × 2π r.dr=2πμ.p.r.dr Frictional torque on the ring, Tr = Fr ×r= 2π μ p r.dr× r = 2 π μ pr2dr (ii) DEPARTMENT OF MECHANICAL ENGINEERING Integrating this equation within the limits from 0 to R for the total frictional torque on the pivot bearing. Considering uniform wear We have already discussed that the rate of wear depends upon the intensity of pressure and the velocity of rubbing surfaces (v). It is assumed that the rate of wear is proportional to the product of intensity of pressure and the velocity of rubbing surfaces (i.e. p.v.). Since the velocity of rubbing surfaces increases with the distance (i.e. radius r) from the axis of the bearing, therefore for uniform Wear DEPARTMENT OF MECHANICAL ENGINEERING PROBLEMS Example 1.A vertical shaft 150 mm in diameter rotating at 100 r.p.m. rests on a flat end foot step bearing. The shaft carries a vertical load of 20 kN. Assuming uniform pressure distribution and coefficient of friction equal to 0.05, estimate power lost in friction. Solution.Given :D = 150 mm or R = 75 mm = 0.075 m ; N = 100 r.p.m or ω= 2 δ × 100/60= 10.47 rad/s ; W = 20 kN = 20 × 103 N ; μ = 0.05 Conical Pivot Bearing DEPARTMENT OF MECHANICAL ENGINEERING The conical pivot bearing supporting a shaft carrying a load W is shown in Fig. Let Pn = Intensity of pressure normal to the cone, α = Semi angle of the cone, μ = Coefficient of friction between the shaft and the bearing, R = Radius of the shaft. Consider a small ring of radius r and thickness dr. Let dl is the length of ring along the cone, such that dl= drcosec α Area of the ring, A = 2πr.dl= 2πr.drcosec α ( dl= drcosecα) Considering uniform pressure We know that normal load acting on the ring, δWn= Normal pressure ×Area = pn × 2πr.drcosec α vertical load acting on the ring, δW= Vertical component of δWn= δWn.sinα Total vertical load transmitted to the bearing, We know that frictional force on the ring acting tangentially at radius r, The vertical load acting on the ring is also given by δW= Vertical component of pn× Area of the ring DEPARTMENT OF MECHANICAL ENGINEERING Integrating the expression within the limits from 0 to R for the total frictional torque on the conical pivot bearing. Total frictional torque: Substituting the value of pnin equation (i), Considering uniform wear In Fig. let pr be the normal intensity of pressure at a distance r from the central axis. We know that, in case of uniform wear, the intensity of pressure varies inversely with the distance. pr.r = C (a constant) or pr= C/r The load transmitted to the ring, Total load transmitted to the bearing, We know that frictional torque acting on the ring, DEPARTMENT OF MECHANICAL ENGINEERING Total frictional torque acting on the bearing, Substituting the value of C, we have Trapezoidal or Truncated Conical Pivot Bearing If the pivot bearing is not conical, but a frustum of a cone with r1 and r2, the external and internal radius respectively as shown in Fig, then Considering uniform pressure The total torque acting on the bearing is obtained by integrating the value of Tr, within the limits r1 and r 2. Total torque acting on the bearing, DEPARTMENT OF MECHANICAL ENGINEERING Substituting the value of pn from equation (i), Considering uniform wear the load transmitted to the ring, δW= 2πC.dr Total load transmitted to the ring, We know that the torque acting on the ring, considering uniform wear, is Total torque acting on the bearing, Substituting the value of C from equation (ii), we get PROBLEMS Example 1.A conical pivot supports a load of 20 kN, the cone angle is 120º and the intensity of normal pressure is not to exceed 0.3 N/mm2. The external diameter is twice the internal diameter. Find the outer and inner radii of the bearing surface. If the shaft rotates at 200 r.p.m. and the coefficient of friction is 0.1, find the power absorbed in friction. Assume uniform pressure. Solution: Given: W = 20 kN = 20 × 103 N ; 2 α= 120º or α= 60º ; pn= 0.3 N/mm2 ;N = 200 r.p.m. or ω= 2π× 200/60 = 20.95 rad/s ; μ =0.1 Outer and inner radii of the bearing surface. DEPARTMENT OF MECHANICAL ENGINEERING Let r1 and r2 = Outer and inner radii of the bearing surface, in mm. Since the external diameter is twice the internal diameter, therefore r1 = 2 r2 We know that intensity of normal pressure (pn), Power absorbed in friction We know that total frictional torque (assuming uniform pressure), Power absorbed in friction P = T.ω = 301.76 × 20.95 = 6322 W = 6.322 kW Flat Collar Bearing We have already discussed that collar bearings are used to take the axial thrust of the rotating hafts. There may be a single collar or multiple collar bearings as shown in Fig.(a) and (b)respectively. The collar bearings are also known as thrust bearings. The friction in the collar bearings may be found as discussed below: Consider a single flat collar bearing supporting a shaft as shown in Fig(a). DEPARTMENT OF MECHANICAL ENGINEERING Let r1 = External radius of the collar, r2 = Internal radius of the collar. Area of the bearing surface, Considering uniform pressure When the pressure is uniformly distributed over the bearing surface, then the intensity of pressure, The frictional torque on the ring of radius r and thickness dr, Integrating this equation within the limits from r2 to r1 for the total frictional torque on the collar. Total frictional torque, Substituting the value of p from equation (i), Considering uniform wear The load transmitted on the ring, considering uniform wear is, DEPARTMENT OF MECHANICAL ENGINEERING Total load transmitted to the collar, We also know that frictional torque on the ring; we also know that frictional torque on the ring, Substituting the value of C from equation (ii), PROBLEMS Example 1.A thrust shaft of a ship has 6collars of 600 mm external diameter and 300 mm internal diameter. The total thrust from the propeller is 100 kN. If the coefficient of friction is 0.12 and speed of the engine90 r.p.m., find the power absorbed in friction at the thrust block, assuming. 1. Uniform pressure 2. Uniform wear. Solution. Given: n = 6 ; d1 = 600 mm or r1 = 300mm ;d2 = 300 mm or r2 = 150 mm ; W = 100 kN= 100 × 103 N ; μ = 0.12 ; N = 90 r.p.m. or ω= 2 π× 90/60 = 9.426 rad/s Power absorbed in friction, assuming uniform pressure We know that total frictional torque transmitted, Power absorbed in friction, P = Tω= 2800 × 9.426 = 26 400 W = 26.4 kW Power absorbed in friction assuming uniform wear We know that total frictional torque transmitted, Power absorbed in friction, P = T.ω = 2700 × 9.426 = 25 450 W = 25.45 kW DEPARTMENT OF MECHANICAL ENGINEERING Example 2.A shaft has a number of a collars integral with it. The external diameter of the collars is 400 mm and the shaft diameter is 250 mm. If the intensity of pressure is 0.35 N/mm 2(uniform) and the coefficient of friction is 0.05, estimate power absorbed when the shaft runs at 105 r.p.m. carrying a load of 150 kN. Number of collars required. Solution. Given: d1 = 400 mm or r1 = 200 mm ; d2 = 250 mm or r2 = 125 mm ; p = 0.35N/mm2 ; μ = 0.05 ; N = 105 r.p.m or ω= 2 π× 105/60 = 11 rad/s ; W = 150 kN = 150 × 103 N Power absorbed We know that for uniform pressure, total frictional torque transmitted Power absorbed, P = Tω = 1240 × 11 = 13640 W = 13.64 kW Number of collarsrequired Let n = Number of collarsrequired. We know that the intensity of uniform pressure (p), DEPARTMENT OF MECHANICAL ENGINEERING  INDUSTRIAL APPLICATIONS 1. Human body modeled as a system of rigid bodies of geometrical solids. Representative bones were added for better visualization of the walking person. 2. Screw jack. DEPARTMENT OF MECHANICAL ENGINEERING  TUTORIAL QUESTIONS 1. Discuss briefly the various types of friction experienced by a body 2. Derive from first principles an expression for the effort required to raise a load with a screw jack taking friction into consideration. 3. What is meant by the expression ‘friction circle’? Deduce an expression for the radius of friction circle in terms of the radius of the journal and the angle of friction. 4. Derive from first principles an expression for the friction moment of a conical pivot assuming (i) Uniform pressure, and (ii) Uniform wear. 5. Find the force required to move a load of 300 N up a rough plane, the force being applied parallel to the plane. The inclination of the plane is such that a force of 60 N inclined at 30º to a similar smooth plane would keep the same load in equilibrium. The coefficient of friction is 0.3. 6. A square threaded screw of mean diameter 25 mm and pitch of thread 6 mm is utilised to lift a weight of 10 kN by a horizontal force applied at the circumference of the screw. Find the magnitude of the 7. Force if the coefficient of friction between the nut and screw is 0.02. 8. A bolt with a square threaded screw has mean diameter of 25 mm and a pitch of 3 mm. It carries an axial thrust of 10 kN on the bolt head of 25 mm mean radius. If µ = 0.12, find the force required at the end of a spanner 450 mm long, in tightening up the bolt. DEPARTMENT OF MECHANICAL ENGINEERING  ASSIGNMENT QUESTIONS 1. The thrust of a propeller shaft in a marine engine is taken up by a number of collars integral with the shaft which is 300 mm in diameter. The thrust on the shaft is 200 kN and the speed is 75 r.p.m. Taking µ constant and equal to 0.05 and assuming intensity of pressure as uniform and equal to 0.3 N/mm2, find the external diameter of the collars and the number of collars required, if the power lost in friction is not to exceed 16 kW. 2. A shaft has a number of a collars integral with it. The external diameter of the collars is 400 mm and the shaft diameter is 250 mm. If the intensity of pressure is 0.35 N/mm2 (uniform) and the coefficient of friction is 0.05, estimate: 1. power absorbed when the shaft runs at 105 r.p.m. carrying a load of 150 kN and 2. number of collars required. 3. A thrust shaft of a ship has 6 collars of 600 mm external diameter and 300 mm internal diameter. The total thrust from the propeller is 100 kN. If the coefficient of friction is 0.12 and speed of the engine 90 r.p.m., find the power absorbed in friction at the thrust block, assuming l. uniform pressure and 2. Uniform wear. 4. A conical pivot bearing supports a vertical shaft of 200 mm diameter. It is subjected to a load of 30 kN. The angle of the cone is 120º and the coefficient of friction is 0.025. Find the power lost in friction when the speed is 140 r.p.m., assuming 1. uniform pressure ; and 2. uniform wear. 5. A conical pivot supports a load of 20 kN, the cone angle is 120º and the intensity of normal pressure is not to exceed 0.3 N/mm2. The external diameter is twice the internal diameter. Find the outer and inner radii of the bearing surface. If the shaft rotates at 200 r.p.m. and the coefficient of friction is 0.1, find the power absorbed in friction. Assume uniform pressure. 6. A vertical shaft 150 mm in diameter rotating at 100 r.p.m. rests on a flat end footstep bearing. The shaft carries a vertical load of 20 kN. Assuming uniform pressure distribution and coefficient of friction equal to 0.05, estimate power lost in friction. 7. Derive the Resultant Effect of a System of Forces Acting on a Rigid Body ? 8. Explain the D-Alembert’s Principle? 9. Explain Velocity and Acceleration of the Reciprocating Parts in Engines? 10. The crank and connecting rod of a reciprocating engine are 200 mm and 700 mm respectively. The crank is rotating in clockwise direction at 120 rad/s. Find with the help of Klein’s construction: 1. Velocity and acceleration of the piston, 2. Velocity and acceleration of the midpoint of the connecting rod, and 3. Angular velocity and angular acceleration of the connecting rod, at the instant when the crank is at 30° to I.D.C. (inner dead centre). DEPARTMENT OF MECHANICAL ENGINEERING  UNIT 3 CLUTCHES,BRAKE/DYNAMOMETERS/ TURNING MOMENT DIAGRAM & FLY WHEEL DEPARTMENT OF MECHANICAL ENGINEERING Course Objectives Able to learn about the working of Clutches, Brakes, Dynamometers and Fly wheel. Course Outcomes The student will learn about the kinematics and dynamic analysis of machine elements. DEPARTMENT OF MECHANICAL ENGINEERING FRICTION CLUTCHES A clutch is a device used to transmit the rotary motion of one shaft to another when desired. The axes of the two shafts are coincident. In friction clutches, the connection of the engine shaft to the gear box shaft is affected by friction between two or more rotating concentric surfaces. The surfaces can be pressed firmly against one another when engaged and the clutch tends to rotate as a single unit. SINGLE PLATE CLUTCH (DISC CLUTCH) A disc clutch consists of a clutch plate attached to a splined hub which is free to slide axially on splines cut on the driven shaft. The clutch plate is made of steel and has a ring of friction lining on each side. The engine shaft supports a rigidly fixed flywheel. A spring-loaded pressure plate presses the clutch plate firmly against the flywheel when the clutch is engaged. When disengaged, the springs press against a cover attached to the flywheel. Thus, both the flywheel and the pressure plate rotate with the input shaft. The movement of the clutch pedal is transferred to the pressure plate through a thrust bearing. Figure 8.13 shows the pressure plate pulled back by the release levers and the friction linings on the clutch plate are no longer in contact with the pressure plate or the flywheel. The flywheel rotates without driving the clutch plate and thus, the driven shaft. When the foot is taken off the clutch pedal, the pressure on the thrust bearing is released. As a result, the springs become free to move the pressure plate to bring it in contact with the clutch plate. The clutch plate slides on the splined hub and is tightly gripped between the pressure plate and the fly wheel. The friction between the linings on the clutch plate, and the flywheel on one side and the pressure plate on the other, cause the clutch plate and hence, the driven shaft to rotate. In case the resisting torque on the drive shaft exceeds the torque at the clutch, clutch slip will occur. DEPARTMENT OF MECHANICAL ENGINEERING Torque transmitted by plate or disc clutch The following notations are used in the derivation T= Torque transmitted by the clutch P= intensity of axial pressure r1&r2=external and internal radii of friction faces µ= co-efficient of friction Consider an elemental ring of radius r and thickness dr Friction surface = 2πrdr Axial force on the dw= pressure *area = P*2πrdr Frictional force acting on the ring tangentially at radius r Fr= µdw=µ*p*2πrdr Frictional torque acting on the ring Tr=Fr*r=µp*2πr*dr*r=2πµpr2dr Considering uniform pressure When the pressure is uniformly distributed over the entire area of the friction face, then the intensity o pressure, P=W/π[(r2-(r)2] (i) Where W = Axial thrust with which the contact or friction surfaces are held together. We have discussed above that the frictional torque on the elementary ring of radius r and thickness dr is Tr= 2 πμ.p.r2dr Integrating this equation within the limits from r 2 to r1 for the total frictional torque. Therefore total frictional torque acting on the friction surface or on the clutch, DEPARTMENT OF MECHANICAL ENGINEERING Substituting the value of p from equation (i), 2. Considering uniform wear Let p be the normal intensity of pressure at a distance r from the axis of the Clutch. Since the intensity of pressure varies inversely with the distance, therefore p.r.= C (a constant) or p = C/r and the normal force on the ring, We know that the frictional torque acting on the ring, Total frictional torque on the friction surface, R = Mean radius of the friction surface = (r1+ r2)/2 Multiple plate clutches In a multi-plate clutch, the number of frictional linings and the metal plates is Increased which increases the capacity of the clutch to transmit torque. Figure8.14 shows a simplified diagram of a multi-plate DEPARTMENT OF MECHANICAL ENGINEERING clutch. The friction rings are splined on their outer circumference and engage with corresponding splines on the flywheel. They are free to slide axially. The Friction material thus, rotates with the flywheel and the engine shaft. The Number of friction rings depends The driven shaft also supports discs on the splines which rotate with the driven shaft and can slide axially. If the actuating force on the pedal is removed, a spring presses the discs into contact with the friction rings and the torque is transmitted between the engine shaft and the driven shaft. If n is the total number of plates both on the driving and the driven members, the number of active surfaces will be n – 1. Let n1 = Number of discs on the driving shaft, and n2 = Number of discs on the driven shaft. Number of pairs of contact surfaces, n = n1 +n2 – 1 And total frictional torque acting on the friction surfaces or on the clutch, T = n.μ.W.R Where R = Mean radius of the friction surfaces PROBLEMS Example1. Determine the maximum, minimum and average pressure in plate clutch when the axial force is 4 kN. The inside radius of the contact surface is 50 mm and the outside radius is 100 mm. Assume uniform wear. DEPARTMENT OF MECHANICAL ENGINEERING Solution. Given: W = 4 kN = 4 × 103 N, r2 = 50 mm ;r1 = 100mm Maximum pressure Let p max = Maximum pressure. Since the intensity of pressure is maximum at the inner radius (r2), therefore pmax × r2 = C or C = 50 pmax We know that the total force on the contact surface (W), 4 × 103 = 2 π C (r1 – r2) = 2 π× 50 pmax (100 – 50) = 15 710 p max Pmax = 4 × 103/15 710 = 0.2546 N/mm2 Minimum pressure Let p min = Minimum pressure. Since the intensity of pressure is minimum at the outer radius (r1), Therefore P min × r1 = C or C = 100 pmin We know that the total force on the contact surface (W), 4 × 103 = 2 π C (r1 – r2) = 2π × 100 pmin (100 – 50) = 31 420 p min P min = 4 × 103/31 420 = 0.1273 N/mm2 Average pressure We know that average pressure, Example2. A single plate clutch, with both sides effective, has outer and inner diameters 300 mm and 200 mm respectively. The maximum intensity of pressure at any point in the contact surfaceis not to exceed 0.1 N/mm2.Ifthe coefficient of friction is 0.3, determine the power transmitted by a clutch at a speed 2500r.p.m. Solution.Given: d1 = 300 mm or r1 = 150 mm ;d2 = 200 mm or r2 = 100 mm ; p = 0.1 N/mm2; μ = 0.3 ;N= 2500 r.p.m. or ω= 2π× 2500/60 = 261.8 rad/s Since the intensity of pressure ( p) is maximum at the inner radius (r2), therefore for uniform DEPARTMENT OF MECHANICAL ENGINEERING p.r2 = C or C = 0.1 × 100 = 10 N/mm We know that the axial thrust, W = 2 π C (r1 – r2) = 2 π × 10 (150 – 100) = 3142 N Mean radius of the friction surfaces for uniform wear, We know that torque transmitted, T = n.μ.W.R= 2 × 0.3 × 3142 × 0.125 = 235.65 N-m Power transmitted by a clutch, P = T*ω= 235.65 × 261.8 = 61 693 W = 61.693 kW CONE CLUTCH A cone clutch, as shown in Fig. 10.24, was extensively used in automobiles but now-a-days it has been replaced completely by the disc clutch It consists of one pair of friction surface only. In a cone clutch, the driver is keyed to the driving shaft by a sunk key and has an inside conical surface or face which exactly fits into the outside conical surface of the driven. The driven member resting on the feather key in the driven shaft, maybe shifted along the shaft by a forked lever provided at B, in order to engage the clutch by bringing the two conical surfaces in contact. Due to the frictional resistance set up at this contact surface, the torque is transmitted from one shaft to another. In some cases, a spring is placed around the driven shaft in contact with the hub of the driven. This spring holds the clutch faces in contact and maintains the pressure between them, and the forked lever is used only for disengagement of the clutch. The contact surfaces of the clutch may be metal to metal contact, but more often the driven member is lined with some material like wood, leather, cork or asbestos etc. The material of the clutch faces (i.e. contact surfaces) depends upon the allowable normal pressure and the coefficient of friction. DEPARTMENT OF MECHANICAL ENGINEERING Consider a pair of friction surface as shown in Fig. Since the area of contact of a pair of friction surface is a frustum of a cone, therefore the torque transmitted by the cone clutch maybe determined in the similar manner as discussed. Let p n = Intensity of pressure with which the conical friction surfaces are held together (i.e. normal pressure between contact surfaces), r1 and r2 = Outer and inner radius of friction surfaces respectively R = Mean radius of the friction surface=(r1+ r2)/2 α= Semi angle of the cone (also called face angle of the cone) or the angle of the friction surface with the axis of the clutch, μ = Coefficient of friction between contact surfaces, and b = Width of the contact surfaces (also known as face width or clutch face). Consider a small ring of radius r and thickness dr, as shown in Fig. 10.25 (b). Let dl is length of ring of the friction surface, such that dl= dr.cose α Area of the ring= A = 2π r.dl= 2πr.drcosec α We shall consider the following two cases : When there is a uniform pressure and when there is a uniform wear. Considering uniform pressure We know that normal load acting on the ring, δWn = Normal pressure × Area of ring = p n × 2 πr.dr.cosec α The axial load acting on the ring, δW= Horizontal component of δWn (i.e. in the direction of W) = δWn × sin α= p n × 2π r.dr. cosec α× sin α = 2π × p n.r.dr DEPARTMENT OF MECHANICAL ENGINEERING Total axial load transmitted to the clutch or the axial spring force required, We know that frictional force on the ring acting tangentially at radius r, Fr = μ.δWn = μ.pn × 2 π r.dr.cosec α Frictional torque acting on the ring, Integrating this expression within the limits from r2 to r1 for the total frictional torque on the clutch Considering uniform wear In Fig., let prbe the normal intensity of pressure at a distance r from the axis of the clutch. We know that, in case of uniform wear, the intensity of pressure varies inversely with thedistance. P r.r = C (a constant) or pr = C / r We know that the normal load acting on the ring, δWn = Normal pressure × Area of ring = pr× 2πr.drcosecα The axial load acting on the ring , δW= δWn × sin α = pr.2 π r.dr.cosec α.sin α= pr × 2π r.dr We know that frictional force acting on the ring, DEPARTMENT OF MECHANICAL ENGINEERING Fr = μ.δWn = μ.pr × 2πr × drcosecα Frictional torque acting on the ring, PROBLEMS Example 1.An engine developing 45 kW at 1000 r.p.m. is fitted with a cone clutch built inside the flywheel. The cone has a face angle of 12.5º and a maximum mean diameter of 500 mm. The coefficient of friction is 0.2. The normal pressure on the clutch face is not to exceed 0.1 N/mm 2. Determine: 1. the axial spring force necessary to engage to clutch, and 2. the face widthrequired. Solution.Given :P = 45 kW = 45 × 103 W ; N = 1000 r.p.m. or ω= 2π× 1000/60 = 104.7rad/s ; α= 12.5º ; D = 500 mm or R = 250 mm = 0.25 m ; μ = 0.2 ; pn=0.1N/mm2 Axial spring force necessary to engage theclutch First of all, let us find the torque (T ) developed by the clutch and the normal load (W n) acting on the friction surface. We know that power developed by the clutch (P), 45 × 103 = Tω= T × 104.7 or T = 45 × 103/104.7 = 430 N-m We also know that the torque developed by the clutch (T), 430 = μ.Wn.R = 0.2 × W n× 0.25 = 0.05 W n W n = 430/0.05 = 8600 N Axial spring force necessary to engage the clutch, We = Wn (sin α+ μ cosα) = 8600 (sin 12.5º + 0.2 cos 12.5º) = 3540 N Face widthrequired DEPARTMENT OF MECHANICAL ENGINEERING Let b = Face widthrequired We know that normal load acting on the friction surface (W n), 8600 = p n × 2πR.b= 0.1 × 2π× 250 × b = 157 b b = 8600/157 = 54.7 mm Example 2. A conical friction clutch is used to transmit 90 kW at 1500 r.p.m. The semi cone angle is 20º and the coefficient of friction is 0.2. If the mean diameter of the bearing surface is375 mm and the intensity of normal pressure is not to exceed 0.25 N/mm2, find the dimensions of the conical bearing surface and the axial load required. Solution.Given: P = 90 kW = 90 × 103 W ;N = 1500 r.p.m. or ω = 2 π× 1500/60 = 156rad/s ; α = 20º ; μ = 0.2 ; D = 375 mm or R = 187.5 mm ; pn = 0.25 N/mm2 Dimensions of the conical bearing surface Let r1 and r2 = External and internal radii of the bearing surface respectively, b = Width of the bearing surface in mm, and T = Torque transmitted. We know that power transmitted (P), 90 × 103 = Tω = T × 156 T = 90 × 103/156 = 577 N-m = 577 × 103 N-mm The torque transmitted (T), 577 × 103 = 2 π μ pn.R2.b = 2π × 0.2 × 0.25 (187.5)2b = 11 046 b b = 577 × 103/11 046 = 52.2 mm We know that r1 + r2 = 2R = 2 × 187.5 =375 mm…………… i r1 – r2 = b sin α= 52.2 sin 20º =18 mm………………….. ii From equations (i) and (ii), r1 = 196.5 mm, and r2 = 178.5 mm Axial load required Since in case of friction clutch, uniform wear is considered and the intensity of pressure is maximum at the minimum contact surface radius (r2), therefore pn.r2 = C (a constant) or C = 0.25 × 178.5 = 44.6 N/mm We know that the axial load required, W = 2πC (r1 – r2) = 2π × 44.6 (196.5 – 178.5) = 5045 N DEPARTMENT OF MECHANICAL ENGINEERING Centrifugal Clutch The centrifugal clutches are usually incorporated into the motor pulleys. It consists of a number of shoes on the inside of a rim of the pulley, as shown in Fig. 10.28. The outer surface of the shoes is covered with a friction material. These shoes, which can move radially in guides, are held Against the boss (or spider) on the driving shaft by means of springs. The springs exert a radially inward force which is assumed constant. The mass of the shoe, when revolving, causes it to exert a radially outward force (i.e. centrifugal force). The magnitude of this centrifugal force depends upon the speed at which the shoe is revolving. A little consideration will show that when the centrifugal force is less than the spring force, the shoe remains in the same position as when the driving shaft was stationary, but when the centrifugal force is equal to the spring force, the shoe is just floating. When the centrifugal force exceeds the spring force, the shoe moves outward and comes into contact with the driven member and presses against it. The force with which the shoe presses against the driven member is the difference of the centrifugal force and the spring force. The increase of speed causes the shoe to press harder and enables more torque to be transmitted. In order to dete