Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com CONTENTS CONTENTS C H A P T E R Introduction Definition Classifications of Machine Design General Considerations in Machine Design General Procedure in Machine Design Fundamental Units Derived Units System of Units S.I Units (Inter national System of Units) Metre 10 Kilogram 11 Second 12 Presentation of Units and their values 13 Rules for S.I Units 14 Mass and Weight 15 Inertia 16 Laws of Motion 17 Force 18 Absolute and Gravitational Units of Force 19 Moment of a Force 20 Couple 21 Mass Density 22 Mass Moment of Inertia 23 Angular Momentum 24 Torque 25 Work 26 Power 27 Energy 1.1 Definition The subject Machine Design is the creation of new and better machines and improving the existing ones A new or better machine is one which is more economical in the overall cost of production and operation The process of design is a long and time consuming one From the study of existing ideas, a new idea has to be conceived The idea is then studied keeping in mind its commercial success and given shape and form in the form of drawings In the preparation of these drawings, care must be taken of the availability of resources in money, in men and in materials required for the successful completion of the new idea into an actual reality In designing a machine component, it is necessary to have a good knowledge of many subjects such as Mathematics, Engineering Mechanics, Strength of Materials, Theory of Machines, Workshop Processes and Engineering Drawing CONTENTS CONTENTS Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com n 1.2 A Textbook of Machine Design Classifications of Machine Design The machine design may be classified as follows : Adaptive design In most cases, the designer’s work is concerned with adaptation of existing designs This type of design needs no special knowledge or skill and can be attempted by designers of ordinary technical training The designer only makes minor alternation or modification in the existing designs of the product Development design This type of design needs considerable scientific training and design ability in order to modify the existing designs into a new idea by adopting a new material or different method of manufacture In this case, though the designer starts from the existing design, but the final product may differ quite markedly from the original product New design This type of design needs lot of research, technical ability and creative thinking Only those designers who have personal qualities of a sufficiently high order can take up the work of a new design The designs, depending upon the methods used, may be classified as follows : (a) Rational design This type of design depends upon mathematical formulae of principle of mechanics (b) Empirical design This type of design depends upon empirical formulae based on the practice and past experience (c) Industrial design This type of design depends upon the production aspects to manufacture any machine component in the industry (d) Optimum design It is the best design for the given objective function under the specified constraints It may be achieved by minimising the undesirable effects (e) System design It is the design of any complex mechanical system like a motor car (f) Element design It is the design of any element of the mechanical system like piston, crankshaft, connecting rod, etc (g) Computer aided design This type of design depends upon the use of computer systems to assist in the creation, modification, analysis and optimisation of a design 1.3 General Considerations in Machine Design Following are the general considerations in designing a machine component : Type of load and stresses caused by the load The load, on a machine component, may act in several ways due to which the internal stresses are set up The various types of load and stresses are discussed in chapters and Motion of the parts or kinematics of the machine The successful operation of any machine depends largely upon the simplest arrangement of the parts which will give the motion required The motion of the parts may be : (a) Rectilinear motion which includes unidirectional and reciprocating motions (b) Curvilinear motion which includes rotary, oscillatory and simple harmonic (c) Constant velocity (d) Constant or variable acceleration Selection of materials It is essential that a designer should have a thorough knowledge of the properties of the materials and their behaviour under working conditions Some of the important characteristics of materials are : strength, durability, flexibility, weight, resistance to heat and corrosion, ability to cast, welded or hardened, machinability, electrical conductivity, etc The various types of engineering materials and their properties are discussed in chapter Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com Introduction n Form and size of the parts The form and size are based on judgement The smallest practicable cross-section may be used, but it may be checked that the stresses induced in the designed cross-section are reasonably safe In order to design any machine part for form and size, it is necessary to know the forces which the part must sustain It is also important to anticipate any suddenly applied or impact load which may cause failure Frictional resistance and lubrication There is always a loss of power due to frictional resistance and it should be noted that the friction of starting is higher than that of running friction It is, therefore, essential that a careful attention must be given to the matter of lubrication of all surfaces which move in contact with others, whether in rotating, sliding, or rolling bearings Convenient and economical features In designing, the operating features of the machine should be carefully studied The starting, controlling and stopping levers should be located on the basis of convenient handling The adjustment for wear must be provided employing the various takeup devices and arranging them so that the alignment of parts is preserved If parts are to be changed for different products or replaced on account of wear or breakage, easy access should be provided and the necessity of removing other parts to accomplish this should be avoided if possible The economical operation of a machine which is to be used for production, or for the processing of material should be studied, in order to learn whether it has the maximum capacity consistent with the production of good work Use of standard parts The use of standard parts is closely related to cost, because the cost of standard or stock parts is only a fraction of the cost of similar parts made to order The standard or stock parts should be used whenever possible ; parts for which patterns are already in existence such as gears, pulleys and bearings and parts which may be selected from regular shop stock such as screws, nuts and pins Bolts and studs should be as few as possible to Design considerations play important role in the successful avoid the delay caused by changing production of machines drills, reamers and taps and also to decrease the number of wrenches required Safety of operation Some machines are dangerous to operate, especially those which are speeded up to insure production at a maximum rate Therefore, any moving part of a machine which is within the zone of a worker is considered an accident hazard and may be the cause of an injury It is, therefore, necessary that a designer should always provide safety devices for the safety of the operator The safety appliances should in no way interfere with operation of the machine Workshop facilities A design engineer should be familiar with the limitations of his employer’s workshop, in order to avoid the necessity of having work done in some other workshop It is sometimes necessary to plan and supervise the workshop operations and to draft methods for casting, handling and machining special parts 10 Number of machines to be manufactured The number of articles or machines to be manufactured affects the design in a number of ways The engineering and shop costs which are called fixed charges or overhead expenses are distributed over the number of articles to be manufactured If only a few articles are to be made, extra expenses are not justified unless the machine is large or of some special design An order calling for small number of the product will not permit any undue Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com n A Textbook of Machine Design expense in the workshop processes, so that the designer should restrict his specification to standard parts as much as possible 11 Cost of construction The cost of construction of an article is the most important consideration involved in design In some cases, it is quite possible that the high cost of an article may immediately bar it from further considerations If an article has been invented and tests of hand made samples have shown that it has commercial value, it is then possible to justify the expenditure of a considerable sum of money in the design and development of automatic machines to produce the article, especially if it can be sold in large numbers The aim of design engineer under all conditions, should be to reduce the manufacturing cost to the minimum 12 Assembling Every machine or structure must be assembled as a unit before it can function Large units must often be assembled in the shop, tested and then taken to be transported to their place of service The final location of any machine is important and the design engineer must anticipate the Car assembly line exact location and the local facilities for erection 1.4 Procedur ocedure General Procedure in Machine Design In designing a machine component, there is no rigid rule The problem may be attempted in several ways However, the general procedure to solve a design problem is as follows : Recognition of need First of all, make a complete statement of the problem, indicating the need, aim or purpose for which the machine is to be designed Synthesis (Mechanisms) Select the possible mechanism or group of mechanisms which will give the desired motion Analysis of forces Find the forces acting on each member of the machine and the energy transmitted by each member Material selection Select the material best suited for each member of the machine Design of elements (Size and Stresses) Find the size of each member of the machine by considering the force acting on the member and the permissible stresses for the material used It should be kept in mind that each member should not deflect or deform than the permissible limit Modification Modify the size of the member to agree with Fig 1.1 General procedure in Machine Design the past experience and judgment to facilitate manufacture The modification may also be necessary by consideration of manufacturing to reduce overall cost Detailed drawing Draw the detailed drawing of each component and the assembly of the machine with complete specification for the manufacturing processes suggested Production The component, as per the drawing, is manufactured in the workshop The flow chart for the general procedure in machine design is shown in Fig 1.1 Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com Introduction n Note : When there are number of components in the market having the same qualities of efficiency, durability and cost, then the customer will naturally attract towards the most appealing product The aesthetic and ergonomics are very important features which gives grace and lustre to product and dominates the market 1.5 Fundamental Units The measurement of physical quantities is one of the most important operations in engineering Every quantity is measured in terms of some arbitrary, but internationally accepted units, called fundamental units 1.6 Derived Units Some units are expressed in terms of other units, which are derived from fundamental units, are known as derived units e.g the unit of area, velocity, acceleration, pressure, etc 1.7 System of Units There are only four systems of units, which are commonly used and universally recognised These are known as : C.G.S units, F.P.S units, M.K.S units, and S.I units Since the present course of studies are conducted in S.I system of units, therefore, we shall discuss this system of unit only (Interna national 1.8 S.I Units (International System of Units) The 11th General Conference* of Weights and Measures have recommended a unified and systematically constituted system of fundamental and derived units for international use This system is now being used in many countries In India, the standards of Weights and Measures Act 1956 (vide which we switched over to M.K.S units) has been revised to recognise all the S.I units in industry and commerce In this system of units, there are seven fundamental units and two supplementary units, which cover the entire field of science and engineering These units are shown in Table 1.1 supplementary units Table 1.1 Fundamental and supplementar y units S.No Physical quantity Unit Fundamental units Length (l) Metre (m) Mass (m) Kilogram (kg) Time (t) Second (s) Temperature (T) Kelvin (K) Electric current (I) Ampere (A) Luminous intensity(Iv) Candela (cd) Amount of substance (n) Mole (mol) Plane angle (α, β, θ, φ ) Radian (rad) Solid angle (Ω) Steradian (sr) Supplementary units * It is known as General Conference of Weights and Measures (G.C.W.M) It is an international organisation of which most of the advanced and developing countries (including India) are members The conference has been entrusted with the task of prescribing definitions for various units of weights and measures, which are the very basics of science and technology today Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com n A Textbook of Machine Design The derived units, which will be commonly used in this book, are given in Table 1.2 Deriv ived units Table 1.2 Derived units S.No Quantity Symbol Units Linear velocity V m/s Linear acceleration a m/s2 Angular velocity ω rad/s Angular acceleration α rad/s2 Mass density ρ kg/m3 Force, Weight F, W Pressure Work, Energy, Enthalpy Power 10 Absolute or dynamic viscosity µ N-s/m2 11 Kinematic viscosity v m2/s 12 Frequency f Hz ; 1Hz = 1cycle/s 13 Gas constant R J/kg K 14 Thermal conductance h W/m2 K 15 Thermal conductivity k W/m K P W, E, H P N ; 1N = 1kg-m/s2 N/m2 J ; 1J = 1N-m W ; 1W = 1J/s 16 1.9 Specific heat c J/kg K 17 Molar mass or Molecular mass M kg/mol Metre Metre The metre is defined as the length equal to 650 763.73 wavelengths in vacuum of the radiation corresponding to the transition between the levels p10 and d5 of the Krypton– 86 atom 1.10 Kilogram The kilogram is defined as the mass of international prototype (standard block of platinumiridium alloy) of the kilogram, kept at the International Bureau of Weights and Measures at Sevres near Paris 1.11 Second The second is defined as the duration of 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium – 133 atom Presenta esentation 1.12 Presentation of Units and their Values The frequent changes in the present day life are facilitated by an international body known as International Standard Organisation (ISO) which makes recommendations regarding international standard procedures The implementation of lSO recommendations, in a country, is assisted by its organisation appointed for the purpose In India, Bureau of Indian Standards (BIS), has been created for this purpose We have already discussed that the fundamental units in S.I units for length, mass and time is metre, kilogram and second respectively But in actual practice, it is not necessary to express all lengths in metres, all masses in kilograms and all times in seconds We shall, sometimes, use the convenient units, which are multiples or divisions of our basic units in tens As a typical example, although the metre is the unit of length, yet a smaller length of one-thousandth of a metre proves to be more convenient unit, especially in the dimensioning of drawings Such convenient units Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com Introduction n are formed by using a prefix in the basic units to indicate the multiplier The full list of these prefixes is given in the following table : Pref xes efi units Table 1.3 Prefi xes used in basic units Factor by which the unit is multiplied Standard form Prefix Abbreviation 1012 tera T 000 000 000 109 giga G 000 000 106 mega M 1000 103 kilo K 100 102 hecto* h 10 101 deca* da 0.1 10–1 deci* d 0.01 10–2 centi* c 0.001 10–3 milli m 0.000 001 10–6 micro µ 0.000 000 001 10–9 nano n 10–12 pico p 000 000 000 000 0.000 000 000 001 1.13 Rules for S.I Units The eleventh General Conference of Weights and Measures recommended only the fundamental and derived units of S.I units But it did not elaborate the rules for the usage of the units Later on many scientists and engineers held a number of meetings for the style and usage of S.I units Some of the decisions of the meeting are : For numbers having five or more digits, the digits should be placed in groups of three separated by spaces (instead of commas)** counting both to the left and right of the decimal point In a four*** digit number, the space is not required unless the four digit number is used in a column of numbers with five or more digits A dash is to be used to separate units that are multiplied together For example, newton × metre is written as N-m It should not be confused with mN, which stands for milli newton Plurals are never used with symbols For example, metre or metres are written as m All symbols are written in small letters except the symbol derived from the proper names For example, N for newton and W for watt The units with names of the scientists should not start with capital letter when written in full For example, 90 newton and not 90 Newton At the time of writing this book, the authors sought the advice of various international authorities, regarding the use of units and their values Keeping in view the international reputation of the authors, as well as international popularity of their books, it was decided to present **** units and * These prefixes are generally becoming obsolete, probably due to possible confusion Moreover it is becoming a conventional practice to use only those power of ten which conform to 103x, where x is a positive or negative whole number ** In certain countries, comma is still used as the decimal mark *** In certain countries, a space is used even in a four digit number **** In some of the question papers of the universities and other examining bodies standard values are not used The authors have tried to avoid such questions in the text of the book However, at certain places the questions with sub-standard values have to be included, keeping in view the merits of the question from the reader’s angle Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com n A Textbook of Machine Design their values as per recommendations of ISO and BIS It was decided to use : 4500 not 500 or 4,500 75 890 000 not 75890000 or 7,58,90,000 0.012 55 not 0.01255 or 01255 30 × 106 not 3,00,00,000 or × 107 The above mentioned figures are meant for numerical values only Now let us discuss about the units We know that the fundamental units in S.I system of units for length, mass and time are metre, kilogram and second respectively While expressing these quantities, we find it time consuming to write the units such as metres, kilograms and seconds, in full, every time we use them As a result of this, we find it quite convenient to use some standard abbreviations : We shall use : m for metre or metres km for kilometre or kilometres kg for kilogram or kilograms t for tonne or tonnes s for second or seconds for minute or minutes N-m for netwon × metres (e.g work done) kN-m for kilonewton × metres rev for revolution or revolutions rad for radian or radians 1.14 Mass and Weight Sometimes much confusion and misunderstanding is created, while using the various systems of units in the measurements of force and mass This happens because of the lack of clear understanding of the difference between the mass and weight The following definitions of mass and weight should be clearly understood : Mass It is the amount of matter contained in a given body and does not vary with the change in its position on the earth’s surface The mass of a body is measured by direct comparison with a standard mass by using a lever balance Weight It is the amount of pull, which the earth exerts upon a given body Since the pull varies with the distance of the body from the centre of the earth, therefore, the weight of the body will vary with its position on the earth’s surface (say latitude and elevation) It is thus obvious, that the weight is a force The pointer of this spring gauge shows the tension in the hook as the brick is pulled along Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com Introduction n The earth’s pull in metric units at sea level and 45° latitude has been adopted as one force unit and named as one kilogram of force Thus, it is a definite amount of force But, unfortunately, has the same name as the unit of mass The weight of a body is measured by the use of a spring balance, which indicates the varying tension in the spring as the body is moved from place to place Note : The confusion in the units of mass and weight is eliminated to a great extent, in S.I units In this system, the mass is taken in kg and the weight in newtons The relation between mass (m) and weight (W) of a body is W = m.g or m = W / g where W is in newtons, m in kg and g is the acceleration due to gravity in m/s2 1.15 Inertia It is that property of a matter, by virtue of which a body cannot move of itself nor change the motion imparted to it 1.16 Laws of Motion Newton has formulated three laws of motion, which are the basic postulates or assumptions on which the whole system of dynamics is based Like other scientific laws, these are also justified as the results, so obtained, agree with the actual observations Following are the three laws of motion : Newton’s First Law of Motion It states, “Every body continues in its state of rest or of uniform motion in a straight line, unless acted upon by some external force” This is also known as Law of Inertia Newton’s Second Law of Motion It states, “The rate of change of momentum is directly proportional to the impressed force and takes place in the same direction in which the force acts” Newton’s Third Law of Motion It states, “To every action, there is always an equal and opposite reaction” Force 1.17 Force It is an important factor in the field of Engineering science, which may be defined as an agent, which produces or tends to produce, destroy or tends to destroy motion According to Newton’s Second Law of Motion, the applied force or impressed force is directly proportional to the rate of change of momentum We know that Momentum = Mass × Velocity Let m = Mass of the body, u = Initial velocity of the body, v = Final velocity of the body, a = Constant acceleration, and t = Time required to change velocity from u to v ∴ Change of momentum = mv – mu and rate of change of momentum  v −u  mv − mu m(v − u) = = m.a = ∴ t = a    t t or Force, F ∝ ma or F =kma where k is a constant of proportionality For the sake of convenience, the unit of force adopted is such that it produces a unit acceleration to a body of unit mass ∴ F = m.a = Mass × Acceleration Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com 10 n A Textbook of Machine Design In S.I system of units, the unit of force is called newton (briefly written as N) A newton may be defined as the force, while acting upon a mass of one kg, produces an acceleration of m/s2 in the direction in which it acts Thus 1N = 1kg × m/s2 = 1kg-m/s2 Exhaust jet (backwards) Acceleration proportional to mass Far away from Earth’s gravity and its frictional forces, a spacecraft shows Newton’s three laws of motion at work Gravitational Force 1.18 Absolute and Gravitational Units of Force We have already discussed, that when a body of mass kg is moving with an acceleration of m/s2, the force acting on the body is one newton (briefly written as N) Therefore, when the same body is moving with an acceleration of 9.81 m/s2, the force acting on the body is 9.81N But we denote kg mass, attracted towards the earth with an acceleration of 9.81 m/s2 as kilogram force (briefly written as kgf) or kilogram weight (briefly written as kg-wt) It is thus obvious that 1kgf = 1kg × 9.81 m/s2 = 9.81 kg-m/s2 = 9.81 N (∵ 1N = 1kg-m/s2) The above unit of force i.e kilogram force (kgf) is called gravitational or engineer’s unit of force, whereas netwon is the absolute or scientific or S.I unit of force It is thus obvious, that the gravitational units are ‘g’ times the unit of force in the absolute or S I units It will be interesting to know that the mass of a body in absolute units is numerically equal to the weight of the same body in gravitational units For example, consider a body whose mass, m = 100 kg ∴ The force, with which it will be attracted towards the centre of the earth, F = m.a = m.g = 100 × 9.81 = 981 N Now, as per definition, we know that the weight of a body is the force, by which it is attracted towards the centre of the earth ∴ Weight of the body, 981 = 100 kgf W = 981 N = (∵ l kgf = 9.81 N) 9.81 In brief, the weight of a body of mass m kg at a place where gravitational acceleration is ‘g’ m/s2 is m.g newtons 1.19 Moment of Force Force It is the turning effect produced by a force, on the body, on which it acts The moment of a force is equal to the product of the force and the perpendicular distance of the point, about which the moment is required, and the line of action of the force Mathematically, Moment of a force = F × l where F = Force acting on the body, and l = Perpendicular distance of the point and the line of action of the force (F) as shown in Fig 1.2 Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com 1200 n A Textbook of Machine Design Let dc = Core diameter of the tappet screw We know that the load on the tappet screw (Fe), π π 2 2460 = (dc ) σc = (dc ) 50 = 39.3 (dc ) 4 ∴ (dc)2 = 2460 / 39.3 = 62.6 or dc = 7.9 say mm and outer or nominal diameter of the screw, dc = = 9.52 say 10 mm Ans d = 0.84 0.84 We shall use 10 mm stud and it is provided with a lock nut The diameter of the circular end of the arm (D3) and its depth (t4) is taken as twice the diameter of stud ∴ D3 = × 10 = 20 mm Ans and t4 = × 10 = 20 mm Ans Design for valve spring First of all, let us find the total load on the valve spring We know that initial load on the spring, W1 = Initial spring force (Fs) = 96.6 N (Already calculated) and load at full lift, W2 = Full valve lift × Stiffiness of spring (s) (Assuming s = 10 N/mm) = 25 × 10 = 250 N ∴ Total load on the spring, W = W1 + W2 = 96.6 + 250 = 346.6 N Now let us find the various dimensions for the valve spring, as discussed below: (a) Mean diameter of spring coil Let D = Mean diameter of the spring coil, and d = Diameter of the spring wire We know that Wahl’s stress factor, K = 4C − 0.615 × − 0.615 + = + = 1.184 4C − C 4×8− (Assuming C = D/d = 8) and maximum shear stress (τ), 420 = K × WC πd = 1.184 × × 346.6 × πd = 8360 d2 .(Assuming τ = 420 MPa or N/mm2) ∴ d = 8360 / 420 = 19.9 or d = 4.46 mm The standard size of the wire is SWG having diameter ( d ) = 4.47 mm Ans (See Table 22.2) ∴ Mean diameter of the spring coil, D = C · d = × 4.47 = 35.76 mm Ans and outer diameter of the spring coil, Do = D + d = 35.76 + 4.47 = 40.23 mm Ans (b) Number of turns of the coil Let n = Number of active turns of the coil We know that maximum compression of the spring, δ = W · C3 · n G·d or C3 · n δ = W G ·d Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com Internal Combustion Engine Parts n 1201 Power-brake mechanism of an automobile Since the stiffness of the springs, s = W / δ = 10 N/mm, therefore, δ / W = 1/10 Taking G = 84 × 103 MPa or N/mm2, we have × 83 × n 10.9 n = = 10 84 × 10 × 4.47 103 ∴ n = 103 / 10.9 × 10 = 9.17 say 10 For squared and ground ends, the total number of the turns, n′ = n + = 10 + = 12 Ans (c) Free length of the spring Since the compression produced under W2 = 250 N is 25 mm (i.e equal to full valve lift), therefore, maximum compression produced ( δ max) under the maximum load of W = 346.6 N is 25 × 346.6 = 34.66 mm 250 We know that free length of the spring, LF = n′ · d + δmax + 0.15 δmax = 12 × 4.47 + 34.66 + 0.15 × 34.66 = 93.5 mm Ans (d) Pitch of the coil We know that pitch of the coil δmax = Free length 93.5 = = 8.5 mm Ans n' − 12 − Example 32 Design the various components of the valve gear machanism for a horizontal diesel engine for the following data: = Bore = 140 mm ; Stroke = 270 mm ; Power = 8.25 kW ; Speed = 475 r.p.m ; Maximum gas pressure = 3.5 N/mm2 Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com 1202 n A Textbook of Machine Design The valve opens 33º before outer dead cerntre and closes 1º after inner dead centre It opens and closes with constant acceleration and decleration for each half of the lift The length of the rocker arm on either side of the fulcrum is 150 mm and the included angle is 160º The weight of the valve is N Solution Given : D = 140 mm = 0.14 m ; L = 270 mm = 0.27 m ; Power = 8.25 kW = 8250 W ; N = 475 r.p.m ; p = 3.5 N/mm2 ; l = 150 mm = 0.15 m ; θ = 160° ; w = N First of all, let us find out dimensions of the valve as discussed below : Size of the valve port Let dp = Diameter of the valve port, and π ap = Area of the valve port = (d p ) We know that area of the piston, a = π π D = (0.14)2 = 0.0154 m2 4 and mean velocity of the piston, L N × 0.27 × 475 = = 4.275 m / s 60 60 From Table 32.3, let us take the mean velocity of the gas through the port (vp) as 40 m/s We know that ap vp = a.p v = π (d )2 40 = 0.0154 × 4.275 or 31.42 (dp)2 = 0.0658 p ∴ (dp)2 = 0.0658 / 31.42 = 2.09 × 10–3 or dp = 0.045 m = 45 mm Ans Maximum lift of the valve We know that maximum lift of the valve, dp 45 = h = = 15.9 say 16 mm Ans cos α cos 45º (Taking α = 45º) Thickness of the valve head We know that thickness of valve head, t = k · dp 3.5 p = 0.42 × 45 = 4.72 mm Ans σb 56 .(Taking k = 0.42 and σb = 56 MPa) Valve stem diameter We know that valve stem diameter, ds = dp 45 + 6.35 = 11.97 say 12 mm Ans + 6.35 mm = 8 Valve head diameter The projected width of the valve seat, for a seat angle of 45º, may be empirically taken as 0.05 dp to 0.07 dp Let us take width of the valve seat as 0.06 dp i.e 0.06 × 45 = 2.7 mm ∴ Valve head diameter, dv = dp + × 2.7 = 45 + 5.4 = 50.4 say 51 mm Ans Now let us calculate the various forces acting on the rocker arm of exhaust valve Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com Internal Combustion Engine Parts n 1203 We know that gas load on the valve, π π (d v ) pc = (51)2 0.4 = 817 N 4 Total load on the valve, considering the weight of the valve, P = P1 + w = 817 + = 820 N Initial spring force, considering the weight of the valve, P1 = Fs = .(Taking pc = 0.4 N/mm2) π π (d ) ps − w = (51) 0.025 − = 48 N v (Taking ps = 0.025 N/mm2) The force due to acceleration (Fa) may be obtained as discussed below : We know that total angle of crank for which the valve remains open = 33 + 180 + = 214º Since the engine is a four stroke engine, therefore the camshaft angle for which the valve remains open = 214 / = 107º Now, when the camshaft turns through 107 / = 53.5°, the valve fifts by a distance of 16 mm It may be noted that the half of this period is occupied by constant acceleration and half by constant decleration The same process occurs when the value closes Therefore, the period for constant acceleration is equal to camshaft rotation of 53.5 / = 26.75 º and during this time, the valve lifts through a distance of mm We know that speed of camshaft N 475 = = 237.5 r.p.m 2 ∴ Angle turned by the camshaft per second = 237.5 × 360 = 1425 deg / s 60 and time taken by the camshaft for constant aceleration, = 26.75 = 0.0188 s 1425 a = Acceleration of the valve t = Let a.t2 (Equation of motion) = × t + a (0.0188)2 = 1.767 × 10−4 a (∵ u = o) ∴ a = / 1.767 × 10–4 = 45 274 mm / s2 = 45.274 m / s2 and force due to valve acceleration, considering the weight of the valve, We know that s =u.t+ × 45.274 + = 16.84 N (∵ m = w/g) 9.81 We know that the maximum load on the rocker arm for exhaust valve, Fe = P + Fs + Fa = 820 + 48 + 16.84 = 884.84 say 885 N Since the length of the two arms of the rocker are equal, therefore, load at the two ends of the arm are equal, i.e Fe = Fc = 885 N We know that reaction at the fulcrum pin F, Fa = m · a + w = Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com 1204 n A Textbook of Machine Design RF = = ( Fe ) + ( Fc ) − 2Fe × Fc × cos θ (885) + (885) − × 885 × 885 × cos 160º = 1743 N The rocker arm is shown in Fig 32.29 We shall now design the various parts of rocker arm as discussed below: Fig 32.29 Design of fulcrum pin Let d1 = Diameter of the fulcrum pin, and l1 = Length of the fulcrum pin = 1.25 d1 (Assume) Considering the bearing of the fulcrum pin We know that load on the fulcrum pin (RF), 1743 = d1 × l1 × pb = d1 × 1.25 d1 × = 6.25 (d1)2 ∴ (For ordinary lubrication, pb is taken as N/mm2) = 1743 / 6.25 = 279 or d1 = 16.7 say 17 mm l1 = 1.25 d1 = 1.25 × 17 = 21.25 say 22 mm Now let us check the average shear stress induced in the pin Since the pin in double shear, therefore, load on the fulcrum pin (RF), π π 2 1743 = × (d1 ) τ = × (17) τ = 454 τ 4 ∴ τ = 1743 / 454 = 3.84 N/mm2 or MPa This induced shear stress is quite safe Now external diameter of the boss, D1 = 2d1 = × 17 = 34 mm Assuming a phosphor bronze bush of mm thick, the internal diameter of the hole in the lever, dh = d1 + × = 17 + = 23 mm and (d1 )2 Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com Internal Combustion Engine Parts n 1205 Now, let us check the induced bending stress for the section of the boss at the fulcrum which is shown in Fig 32.30 Bending moment at this section, M = Fe × l = 885 × 150 N-mm = 132 750 N-mm Section modulus, ∴ × 22 [(34) − (23)3 ] 12 = 2927 mm3 Z= 34 / Induced bending stress, M 132 750 = = 45.3 N/mm2 or MPa Z 2927 Fig 32.30 The induced bending stress is quite safe Design for forked end Let d2= Diameter of the roller pin, and l2= Length of the roller pin = 1.25 d2 (Assume) Considering bearing of the roller pin We know that load on the roller pin (Fc), 885= d2 × l2 × pb = d2 × 1.25 d2 × = 8.75 (d2)2 (Taking pb = N/mm2) ∴ (d2)2= 885 / 8.75 = 101.14 or d2 = 10.06 say 11 mm Ans and l2= 1.25 d2 = 1.25 × 11 = 13.75 say 14 mm Ans σb = Power transmission gears in an automobile engine Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com 1206 n A Textbook of Machine Design Let us now check the roller pin for induced shearing stress Since the pin is in double shear, therefore, load on the roller pin (Fc ), 885 = × π π (d ) τ = × (11) τ = 190 τ 4 ∴ τ = 885 / 190 = 4.66 N/mm2 or MPa This induced shear stress is quite safe The roller pin is fixed in the eye and thickness of each eye is taken as one-half the length of the roller pin ∴ Thickness of each eye, l2 14 = = mm 2 Let us now check the induced bending stress in the roller pin The pin is neither simply supported in fork nor rigidly fixed at the end Therefore, the common practice is to assume the load distribution as shown in Fig 32.31 The maximum bending moment will occur at Y– Y Neglecting the effect of clearance, we have Maximum bending moment at Y–Y, t2 = Fc  l2 t2  Fc l2 M = 2 + 3− ×   Fc  l2 l2  Fc l2 =  +  − × (∵ t2 = l2/2)   × Fc × l2 = 24 × 885 × 14 = 2581 N-mm = 24 and section modulus of the pin, π π (d )3 = (11)3 = 131 mm3 Z= 32 32 ∴ Bending stress induced in the pin Fig 32.31 M 2581 = = 19.7 N/mm2 or MPa Z 131 The bending stress induced in the pin is within permissible limits Since the radial thickness of eye (t3) is taken as d2 / 2, therefore, overall diameter of the eye, D2= d2 = × 11 = 22 mm The outer diameter of the roller is taken slightly larger (at least mm more) than the outer diameter of the eye In the present case, 28 mm outer diameter of the roller will be sufficient Providing a clearance of 1.5 mm between the roller and the fork on either side of the roller, we have = t2 + × 1.5 = 14 + × + = 24 mm 2 Design for rocker arm cross-section Since the engine is a slow speed engine, therefore, a rectangular section may be selected for the rocker arm The cross-section of the rocker arm is obtained by considering the bending of the sections just near the boss of fulcrum on both sides, such as section A–A and B– B l3 = l2 + × Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com Internal Combustion Engine Parts n 1207 Let t1 = Thickness of the rocker arm which is uniform throughout B = Width or depth of the rocker arm which varies from boss diameter of fulcrum to outside diameter of the eye (for the forked end side) and from boss diameter of fulcrum to thickness t3 (for the tappet or stud end side) Now bending moment on section A – A and B – B, 34   M = 885  150 − = 117 705 N-mm    and section modulus at A – A and B – B, 1 2 Z = × t1 · B = × t1 ( D1 ) = × t1 (34)2 = 193 t1 6 (At sections A–A and B–B, B = D) We know that bending stress (σb), M 117 705 70 = Z = 193 t (Taking σb = 70 MPa or N/mm2) ∴ t1 = 117 705 / 193 × 70 = 8.7 say 10 mm Ans Design for tappet screw The adjustable tappet screw carries a comperssive load of Fe = 885 N Assuming the screw to be made of mild steel for which the compressive stress (σc) may be taken as 50 MPa Let dc = Core diameter of the tappet screw We know that load on the tappet screw (Fe), π π (d c ) σ c = (d c ) 50 = 39.3(d c ) 4 ∴ (dc )2 = 885 / 39.3 = 22.5 or dc = 4.74 say mm Ans and outer or nominal diameter of the screw, dc d = 0.84 = 0.8 = 6.25 say 6.5 mm Ans We shall use 6.5 mm stud and it is provided with a lock nut The diameter of the circular end of the arm (D3) and its depth (t4) is taken as twice the diameter of stud ∴ D3 = × 6.5 = 13 mm Ans and t4 = × 6.5 = 13 mm Ans Design for valve spring First of all, let us find the total load on the valve spring We know that intial load on the spring, W1 = Initial spring force (Fs) = 48 N (Already calculated) and load at full lift, W2 = Full valve lift × Sitffness of spring (s) = 16 × = 128 N (Taking s = N/mm) ∴ Total load on the spring, W = W1 + W2 = 48 + 128 = 176 N Now let us find the various dimensions for the valve spring as discussed below: (a) Mean diameter of the spring coil Let D = Mean diameter of the spring coil, and d = Diameter of the spring wire 885 = Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com 1208 n A Textbook of Machine Design Inside view of an automobile We know that Wahl’s stress factor, K = 4C − 0.615 × − 0.615 + = + = 1.2525 4C − 4×6−4 C (Assuming C = D/d = 6) and maximum shear stress (τ), 420 = K × WC πd = 1.2525 × × 176 × πd = 3368 d2 ∴ d = 3368 / 420 = 8.02 or d = 2.83 mm The standard size of the wire is SWG 11 having a diameter (d ) = 2.946 mm Ans (see Table 22.2) ∴ Mean diameter of spring coil, D = C · d = × 2.946 = 17.676 mm Ans and outer diameter of the spring coil, Do = D + d = 17.676 + 2.946 = 20.622 mm Ans (b) Number of turns of the coil Let n = Number of turns of the coil, We know that maximum compression of the spring δ = W · C3 · n δ C3 ·n = or G·d W G ·d Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com Internal Combustion Engine Parts n 1209 Since the stiffness of the spring, s = W / δ = N / mm, therefore δ / W = / Taking G = 84 × 103 MPa or N/mm2, we have × 63 × n 6.98 n = = 84 × 10 × 2.946 103 ∴ n = 103 / × 6.98 = 17.9 say 18 For squared and ground ends, the total number of turns, n′ = n + = 18 + = 20 Ans (c) Free length of the spring Since the compression produced under W2 = 128 N is 16 mm, therefore, maximum compression produced under the maximum load of W = 176 N is 16 × 176 = 22 mm 128 We know that free length of the spring, LF = n’ d + δmax + 0.15 δmax = 20 × 2.946 + 22 + 0.15 × 22 = 84.22 say 85 mm Ans (d) Pitch of the coil We know that pitch of the coil Free length 85 = = = 4.47 mm Ans 20 − n′ − Design of cam The cam is forged as one piece with the camshaft It is designed as discussed below : The diameter of camshaft (D′) is taken empirically as D′ = 0.16 × Cylinder bore + 12.7 mm = 0.16 × 140 + 12.7 = 35.1 say 36 mm The base circle diameter is about mm greater than the camshaft diameter ∴ Base circle diameter = 36 + = 39 say 40 mm The width of cam is taken equal to the width of roller, i.e 14 mm The width of cam (w′) is also taken empirically as W ′ = 0.09 × Cylinder bore + mm = 0.09 × 140 +6 = 18.6 mm Let us take the width of cam as 18 mm Now the *cam is drawn according to the procedure given below : First of all, the displacement diagram, as shown in Fig 32.32, is drawn as discussed in the following steps : Draw a horizontal line ANM such that AN represents the angular displacement when valve opens (i.e 53.5º) to some suitable scale The line NM represents the angular displacement of the cam when valve closes (i.e 53.5º) Divid AN and NM into any number of equal even parts (say six) Draw vertical lines through points 0, 1, 2, etc equal to the lift of valve i.e 16 mm Divide the vertical lines – f and 3′ – f ′ into six equal parts as shown by points a, b, c and a′, b′, c′ in Fig 32.32 Since the valve moves with equal uniform acceleration and decleration for each half of the lift, therefore, valve displeacement diagram for opening and closing of valve consists of double parabola δmax = * For complete details, refer Authors’ popular book on ‘Theory of Machines’ Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com 1210 n A Textbook of Machine Design Fig 32.32 Displacement diagram Join Aa, Ab , Ac intersecting the vertical lines 1, 2, at B, C, D respectively Join the points B, C, D with a smooth curve This is the required parabola for the half of valve opening Similarly other curves may be drawn as shown in Fig 32.32 The curve A, B, C, , G, K, L, M is the required displacement diagram Now the profile of the cam, as shown in Fig 32.32, is drawn as discussed in the following steps: Draw a base circle with centre O and diameter equal 40 mm (radius = 40/2 = 20 mm) Draw a prime circle with centre O and radius, OA= Min radius of cam + Gears keyed to camshafts Diameter of roller = 20 + × 28 = 20 + 14 = 34 mm Draw angle AOG = 53.5º to represent opening of valve and angle GOM = 53.5º to represent closing of valve Divide the angular displacement of the cam during opening and closing of the valve (i.e angle AOG and GOM) into same number of equal even parts as in displacement diagram Join the points 1, 2, 3, etc with the centre O and produce the lines beyond prime circle as shown in Fig 32.33 Set off points 1B, 2C, 3D, etc equal to the displacements from displacement diagram Join the points A, B, C, L, M, A The curve drawn through these points is known as pitch curve From the points A, B, C, K, L, draw circles of radius equal to the radius of the roller Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com Internal Combustion Engine Parts n 1211 Fig 32.33 Join the bottoms of the circle with a smooth curve as shown in Fig 32.33 The is the required profile of cam XER CISES EXER CISES A four stroke internal combustion engine has the following specifications: Brake power = 7.5 kW; Speed = 1000 r.p.m.; Indicated mean effective pressure = 0.35 N/mm2;Maximum gas pressure = 3.5 N/mm2; Mechanical efficiency = 80 % Determine: The dimesions of the cylinder, if the length of stroke is 1.4 times the bore of the cylinder; Wall thickness of the cylinder, if the hoop stress is 35 MPa; Thickness of the cylinder head and the size of studs when the permissible stresses for the cylinder head and stud materials are 45 MPa and 65 MPa respectively Design a cast iron trunk type piston for a single acting four stroke engine developing 75 kW per cylinder when running at 600 r.p.m The other avialable data is as follows: Maximum gas pressure = 4.8 N/mm2; Indicated mean effective pressure = 0.65 N/mm2; Mechanical efficiency = 95%; Radius of crank = 110 mm; Fuel consumption = 0.3 kg/BP/hr; Calorific value of fuel (higher) = 44 × 103kJ/kg; Difference of temperatures at the centre and edges of the piston head = 200ºC; Allowable stress for the material of the piston = 33.5 MPa; Allowable stress for the material of the piston rings and gudgeon pin = 80 MPa; Allowable bearing pressure on the piston barrel = 0.4 N/mm2 and allowable bearing pressure on the gudgeon pin = 17 N/mm2 Design a piston for a four stroke diesel engine consuming 0.3 kg of fuel per kW of power per hour and produces a brake mean effective pressure of the 0.7 N/mm2 The maximum gas pressure inside the cylinder is N/mm2 at a speed of 3500 r.p.m The cylinder diameter is required to be 300 mm with stroke 1.5 times the diameter The piston may have compression rings and an oil ring The following data can be used for design: Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com 1212 n A Textbook of Machine Design Higher calorific value of fuel = 46 × 103kJ/kg; Temperature at the piston centre = 700 K; Tempressure at the piston edge = 475 K; Heat conductivity factor = 46.6 W/m/K; Heat conducted through top = 5% of heat produced; Permissible tensile strength for the material of piston = 27 N/mm2; Pressure between rings and piston = 0.04 N/mm2; Permissible tensile stress in rings = 80 N/mm2; Permissible Pressure on piston barrel = 0.4 N/mm2; Permissible pressure on piston pin = 15 N/mm2; Permissible stress in piston pin = 85 N/mm2 Any other data required for the design may be assumed Determine the dimensions of an I-section connecting rod for a petrol engine from the following data: Diameter of the piston = 110 mm; Mass of the reciprocating parts = kg; Length of the connecting rod from centre to centre = 325 mm; Stroke length = 150 mm; R.P.M = 1500 with possible overspeed of 2500; Compression ratio = : 1; Maximum explosion pressure = 2.5 N/mm2 The following particulars refer to a four stroke cycle diesel engine: Cylinder bore = 150 mm; Stroke = 187.5 mm; R.P.M = 1200; Maximum gas pressure = 5.6 N/mm2; Mass of reciprocating parts = 1.75 kg The dimensions of an I-section connecting rod of forged steel with an elastic limit compressive stress of 350 MPa The ratio of the length of connecting rod to the length of crank is and the factor of safety may be taken as 5; The wrist pin and crankpin dimensions on the basis of bearing pressures of 10 N/mm2 and 6.5 N/mm2 of the projected area respectively; and The dimensions of the small and big ends of the connecting rods, including the size of the securing bolts of the crankpin end Assume that the allowable stress in the bolts, is not to exceed 35 N/mm2 Draw dimensioned sketches of the connecting rod showing the provisions for lubrication A connecting rod is required to be designed for a high speed, four stroke I.C engine The following data are available Diameter of piston = 88 mm; Mass of reciprocating parts = 1.6 kg; Length of connecting rod (centre to centre) = 300 mm; Stroke = 125 mm; R.P.M = 2200 (when developing 50 kW); Possible overspeed = 3000 r.p.m.; Compression ratio = 6.8 : (approximately); Probale maximum explosion pressure (assumed shortly after dead centre, say at about 3°) = 3.5 N/mm2 Draw fully dimensioned drawings of the connecting rod showing the provision for the lubrication Design a plain carbon steel centre crankshaft for a single acting four stroke, single cylinder engine for the following data: Piston diameter = 250 mm; Stroke = 400 mm; Maximum combustion pressure = 2.5 N/mm2; Weight of the flywheel = 16 kN; Total belt pull = N; Length of connecting rod = 950 mm When the crank has turned through 30° from top dead centre, the pressure on the piston is N/mm2 and the torque on the crank is maximum Any other data required for the design may be assumed Design a side crankshaft for a 500 mm × 600 mm gas engine The weight of the flywheel is 80 kN and the explosion pressure is 2.5 N/mm2 The gas pressure at maximum torque is 0.9 N/mm2 when the crank angle is 30º The connecting rod is 4.5 times the crank radius Any other data required for the design may be assumed Design a rocker arm of I-section made of cast steel for operating an exhaust valve of a gas engine The effective length of the rocker arm is 250 mm and the angle between the arm is 135° The exhaust valve is 80 mm in diameter and the gas pressure when the valve begins to open is 0.4 N/mm2 The greatest suction pressure is 0.03 N/mm2 below atmospheric The initial load may be assumed as 0.05 N/mm2 of valve area and the valve inertia and friction losses as 120 N The ultimate strength of cast steel is 750 MPa The allowable bearing pressure is N/mm2 and the permissible stress in the material is 72 MPa 10 Design the various components of a valve gear mechanism for a horizontal diesel engine having the following specifications: Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com Internal Combustion Engine Parts n 1213 Brake power = 10 kW; Bore = 140 mm; Stroke = 270 mm; Speed = 500 r.p.m and maximum gas pressure = 3.5 N/mm2 The valve open 30° before top dead centre and closes 2° after bottom dead centre It opens and closes with constant acceleration and deceleration for each half of the lift The length of the rocker arm on either side of the fulcrum is 150 mm and the included angle is 135° The mass of the valve is 0.3 kg UEST STIONS Q UEST IONS Explain the various types of cylinder liners Discuss the design of piston for an internal combustion engine State the function of the following for an internal combustion engine piston: (a) Ribs ; (b) Piston rings ; (c) Piston skirt ; and (d) Piston pin What is the function of a connecting rod of an internal combustion engine? Explain the various stresses induced in the connecting rod Under what force, the big end bolts and caps are designed? Explain the various types of crankshafts At what angle of the crank, the twisting moment is maximum in the crankshaft? What are the methods and materials used in the manufacture of crankshafts? 10 Sketch a valve gear mechanism of an internal combustion engine and label its various parts 11 Discuss the materials commonly used for making the valve of an I C engine 12 Why the area of the inlet valve port is made larger than the area of exhaust valve port? Transmission mechanism in a truck engine Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com 1214 A Textbook of Machine Design n OBJECTIVE YPE UEST STIONS OBJECT IVE T YP E Q UEST IONS The cylinders are usually made of (a) cast iron or cast steel (b) aluminium (c) stainless steel (d) copper The length of the cylinder is usually taken as (a) equal to the length of piston (b) equal to the length of stroke (c) equal to the cylinder bore (d) 1.5 times the length of stroke The skirt of piston (a) is used to withstand the pressure of gas in the cylinder (b) acts as a bearing for the side thrust of the connecting rod (c) is used to seal the cylinder in order to prevent leakage of the gas past the piston (d) none of the above The side thrust on the cylinder liner is usually taken as of the maximum gas load on the piston (a) 1/5 (b) 1/8 (c) 1/10 (d) 1/5 The length of the piston usually varies between (a) D and 1.5 D (b) 1.5 D and D (c) 2D and 2.5 D (d) 2.5 D and D where D = Diameter of the piston In designing a connecting rod, it is considered like for buckling about X-axis (a) both ends fixed (b) both ends hinged (c) one end fixed and the other end hinged (d) one end fixed and the other end free Which of the following statement is wrong for a connecting rod? (a) The connecting rod will be equally strong in buckling about X-axis, if Ixx = Iyy (b) If Ixx > Iyy,,the buckling will occur about X-axis (c) If Ixx < 4Iyy,the buckling will occur about X-axis (d) The most suitable section for the connecting rod is T-section The crankshaft in an internal combustion engine (a) is a disc which reciprocates in a cylinder (b) is used to retain the working fluid and to guide the piston (c) converts reciprocating motion of the piston into rotary motion and vice versa (d) none of the above The rocker arm is used to actuate the inlet and exhaust valves motion as directed by the (a) cam and follower (b) crank (c) crankshaft (d) none of these 10 For high speed engines, a rocker arm of should be used (a) rectangular section (b) I-section (c) T-section (d) circular ANSWER ANSWER S (a) (b) (d) (d) (b) (c) (c) (a) (a) 10 (b) GO To FIRST ... material are physical, chemical and mechanical properties We shall now discuss the physical and mechanical properties of the material in the following articles Aluminium Copper Zinc Iron Lead... (Reaffirmed 1993) Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com A Textbook of Machine Design Simpo PDF Merge and Split Unregistered Version - http://www.simpopdf.com. .. phenol-furfural (Durite), ureaformaldehyde (Plaskon), etc The thermoplastic materials not become hard with Reinforced plastic with fibreglass the application of heat and pressure and no chemical change makes