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Dynamic force analysis – Inertia force and Inertia torque– D Alembert’s principle –Dynamic Analysis in reciprocating engines – Gas forces – Inertia effect of connecting rod– Bearing loads – Crank shaft torque

- 1. 1 INTRODUCTION TO THEORY OF MACHINES AND FUNDAMENTALS THEORY OF MACHINES INTRODUCTION Theory of Machines may be defined as that branch of engineering science, which deals with the study of relative motion between the various parts of machine, and forces which act on them. The knowledge of this subject is very essential for an engineer in designing the various parts of a machine. Theory of machines and mechanisms is an applied science which is used to understand the relationship between the geometry and motions of the parts of the machine or mechanism and the forces which produces these motions. SUB- DIVISIONS OF THEORY OF MACHINES
- 2. 2 Various structures and machines – bridges, cranes, airplanes, ships, etc. will be found, upon examination, to consist of numerous parts or members connected together in such a way as to perform a useful function and to withstand externally applied loads. • Axial Tension • Axial Compression • Torsion • bending The above said four basic types of loading of a member are frequently encountered in both structural and machine design problems. They may be said to constitute essentially the principal subject matter of Strength of materials. INTRODUCTION
- 3. 3 Analysis and design of any structure or machine involve two major questions: 1. Is the structure strong enough to withstand the loads applied to it? 2. Is it stiff enough to avoid excessive deformations and deflections? In statics, the members of a structure were treated as rigid bodies; but actually all materials are deformable and this property will henceforth be taken into account. Thus Strength of materials may be regarded as the statics of deformable or elastic bodies. Both the strength and stiffness of a structural member are functions of its size and shape and also of certain physical properties of the material from which it is made. These physical properties of materials are largely determined from experimental studies of their behavior in a testing machine. The study of Strength of materials is aimed at predicting just how these geometric and physical properties of a structure will influence its behavior under service conditions. INTRODUCTION
- 4. 4 MECHANICS: is an area of science concerned with the behavior of physical bodies when subjected to forces or displacements, and the subsequent effects of the bodies on their environment.
- 5. 5 Statics: is that branch of theory of machines which deals with the forces and their effects, while the machine parts are rest. Dynamics: is that branch of theory of machines which deals with the forces and their effects, while acting upon the machine parts in motion.
- 6. 6 Kinematics: is that branch of theory of machines which is responsible to study the motion of bodies without reference to the forces which are cause this motion, i.e it’s relate the motion variables (displacement, velocity, acceleration) with the time. Kinetics: is that branch of theory of machines which is responsible to relate the action of forces on bodies to their resulting motion.
- 7. 7 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. All physical quantities, met within this subject, are expressed in terms of the following three fundamental quantities: 1. Length (L or l), 2. Mass (M or m), and 3. Time (t). Derived Units: Some units are expressed in terms of fundamental units known as derived units, e.g., the units of area, velocity, acceleration, pressure, etc. Systems of Units: There are only four systems of units, which are commonly used and universally recognized. These are known as: 1. C.G.S. units, 2. F.P.S units, 3. M.K.S. units and 4. S.I. units. 1. C.G.S. Units: In this system, the fundamental units of length, mass and time are centimeter, gram and second respectively. The C.G.S. units are known as absolute units or physicist's units. 2. F.P.S. Units: In this system, the fundamental units of length, mass and time are foot, pound and second respectively. 3. M.K.S. Units: In this system, the fundamental units of length, mass and time are meter, kilogram and second respectively. The M.K.S. units are known as gravitational units or engineer's units. 4. International System of Units (S.I. Units): In this system of units, the fundamental units are meter (m), kilogram (kg) and second (s) respectively. But there is a slight variation in their derived units. FUNDAMENTAL CONCEPTS
- 8. 8 FORCE A force is a push or pull upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects. When the interaction ceases, the two objects no longer experience the force.
- 9. 9 TYPES OF FORCES Contd…
- 10. 10 Balanced and unbalanced forces There are two forces acting upon the book. One force - the Earth's gravitational pull - exerts a downward force. The other force - the push of the table on the book (sometimes referred to as a normal force) - pushes upward on the book. Since these two forces are of equal magnitude and in opposite directions, they balance each other. The book is said to be at equilibrium. There is no unbalanced force acting upon the book and thus the book maintains its state of motion. Now consider a book sliding from left to right across a tabletop. The force of gravity pulling downward and the force of the table pushing upwards on the book are of equal magnitude and opposite directions. These two forces balance each other. Yet there is no force present to balance the force of friction. As the book moves to the right, friction acts to the left to slow the book down. There is an unbalanced force; and as such, the book changes its state of motion. The book is not at equilibrium and subsequently accelerates. Unbalanced forces cause accelerations.
- 11. 11 Applied forces Force is a vector quantity. Vector is a physical quantity described by magnitude (meaning the strength of the force) and direction. Applied force means the force with which an object has been pushed or pulled by another object. Inertia forces A force opposite in direction to an accelerating force acting on a body and equal to the product of the accelerating force and the mass of the body. Frictional forces Frictional force is force that acts between two surfaces that are moving past another. The applied forces act from outside on the mechanism. The inertia forces arise due to the mass of the links of the mechanism and their acceleration. Frictional forces are the outcome of friction in the joints. A pair of action and reaction forces acting on a body is called constraint forces. In the design of mechanisms, the following forces are generally considered
- 13. 13 NEWTON’S LAWS OF MOTION FIRST LAW An every body continuous in its state of rest or of uniform motion in a straight line unless an external forces acts on it. SECOND LAW The rate of change of momentum of a body is directly proportional to the force acting on it and takes place in the direction of force. THIRD LAW To every action there is an equal and opposite reaction.
- 14. 14 Kinematics of motion is the relative motion of bodies without consideration of the forces causing the motion. Types: Plane Motion Rectilinear Motion Curvilinear Motion KINEMATICS OF MOTION Plane Motion When the motion of a body is confined to only one plane, the motion is said to be plane motion. The plane motion may be either rectilinear or curvilinear Rectilinear Motion It is the simplest type of motion and is along a straight line path. Such a motion is also known as translator motion. Curvilinear Motion It is the motion along a curved path. Such a motion, when confined to one plane, is called plane curvilinear motion. Contd…
- 15. 15 Linear Displacement It may be defined as the distance moved by a body with respect to a certain fixed point. The displacement may be along a straight or a curved path. Linear Velocity It may be defined as the rate of change of linear displacement of a body with respect to the time. Linear Acceleration It may be defined as the rate of change of linear velocity of a body with respect to the time. It is also a vector quantity.
- 16. 16 Static force analysis When the inertia effect due to the mass of the machine components are neglected in the analysis of the mechanism Dynamic force analysis When the inertia forces are considered in the analysis of the mechanism FORCE ANALYSIS Inertia force A force equal in magnitude but opposite in direction and collinear with the impressed force producing the acceleration, is known as inertia force. Inertia force = – m x a Inertia torque 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. Inertia Torque = -I x α D-Alembert’s principle 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. ٤F = 0
- 17. 17
- 18. 18 ANALYTICAL METHOD FOR VELOCITY AND ACCELERATION OF THE PISTON Consider the motion of a crank and connecting rod of a reciprocating steam engine as shown in Fig. Let OC be the crank and PC the connecting rod. Let the crank rotates with angular velocity of ω rad/s and the crank turns through an angle θ from the inner dead centre (briefly written as I.D.C). Let x be the displacement of a reciprocating body P from I.D.C. after time t seconds, during which the crank has turned through an angle θ Let l=Length of connecting rod between the centres, r=Radius of crank or crank pin circle, φ=Inclination of connecting rod to the line of stroke PO, And n=Ratio of length of connecting rod to the radius of crank = l/r.
- 19. 19 ANALYTICAL METHOD FOR VELOCITY AND ACCELERATION OF THE PISTON
- 20. 20 ANALYTICAL METHOD FOR VELOCITY AND ACCELERATION OF THE PISTON
- 21. 21 ANALYTICAL METHOD FOR VELOCITY AND ACCELERATION OF THE PISTON
- 22. 22 ANALYTICAL METHOD FOR VELOCITY AND ACCELERATION OF THE PISTON
- 23. 23 ANALYTICAL METHOD FOR VELOCITY AND ACCELERATION OF THE PISTON
- 24. 24 ANALYTICAL METHOD FOR VELOCITY AND ACCELERATION OF THE PISTON
- 25. 25 ANALYTICAL METHOD FOR VELOCITY AND ACCELERATION OF THE PISTON
- 26. 26 ANALYTICAL METHOD FOR VELOCITY AND ACCELERATION OF THE PISTON
- 27. 27 ANALYTICAL METHOD FOR VELOCITY AND ACCELERATION OF THE PISTON
- 28. 28 ANALYTICAL METHOD FOR VELOCITY AND ACCELERATION OF THE PISTON
- 29. 29 ANALYTICAL METHOD FOR VELOCITY AND ACCELERATION OF THE PISTON
- 30. 30 ANALYTICAL METHOD FOR VELOCITY AND ACCELERATION OF THE PISTON
- 31. 31 FORMULAE TO BE REMEMBER
- 32. 32 DYNAMIC FORCE ANALYSIS OF RECIPROCATING ENGINE MECHANISM
- 33. 33 DYNAMIC FORCE ANALYSIS OF RECIPROCATING ENGINE MECHANISM
- 34. 34 DYNAMIC FORCE ANALYSIS OF RECIPROCATING ENGINE MECHANISM
- 35. 35 DYNAMIC FORCE ANALYSIS OF RECIPROCATING ENGINE MECHANISM
- 36. 36 DYNAMIC FORCE ANALYSIS OF RECIPROCATING ENGINE MECHANISM
- 37. 37 DYNAMIC FORCE ANALYSIS OF RECIPROCATING ENGINE MECHANISM
- 38. 38 DYNAMIC FORCE ANALYSIS OF RECIPROCATING ENGINE MECHANISM
- 45. Equivalent Dynamical System • In order to determine the motion of a rigid body, under the action of external forces, it is usually convenient to replace the rigid body by two masses placed at a fixed distance apart, in such a way that • 1.the sum of their masses is equal to the total mass of the body • 2. the centre of gravity of the two masses coincides with that of the body ; and • 3. the sum of mass moment of inertia of the masses about their centre of gravity is equal to the mass moment of inertia of the body. • When these three conditions are satisfied, then it is said to be an equivalent dynamical system 45
- 46. 46 This equation gives the essential condition of placing the two masses, so that the system becomes dynamical equivalent. The distance of one of the masses (i.e. either l1 or l2) is arbitrary chosen and the other distance is obtained from equation When the radius of gyration kG is not known, then the position of the second mass may be obtained by considering the body as a compound pendulum.
- 47. • This means that the second mass is situated at the centre of oscillation or percussion of the body, which is at a distance of 47
- 48. 1.The connecting rod of a gasoline engine is 300 mm long between its centres. It has a mass of 15 kg and mass moment of inertia of 7000 kg-mm2. Its centre of gravity is at 200 mm from its small end centre. Determine the dynamical equivalent two-mass system of the connecting rod if one of the masses is located at the small end centre. 48 Solution. Given : l = 300 mm ; m = 15 kg; I = 7000 kg-mm2 l1 = 200 mm that mass moment of inertia (I),=m (kG)2 kG = 21.6mm
- 49. 2. A connecting rod is suspended from a point 25 mm above the centre of small end, and 650 mm above its centre of gravity, its mass being 37.5 kg. When permitted to oscillate, the time period is found to be 1.87 seconds. Find the dynamical equivalent system constituted of two masses, one of which is located at the small end centre. 49 Given : h = 650 mm = 0.65 m ; l1 = 650 – 25 = 625 mm = 0.625 m ; m = 37.5 kg ; tp = 1.87 s We know that for a compound pendulum, time period of oscillation (tp), To Find the dynamical equivalent system m1 and m2 Solution
- 50. 50
- 51. Correction Couple to be Applied to Make Two Mass System Dynamically Equivalent • When considering the inertia forces on the connecting rod in a mechanism, we replace the rod by two masses arbitrarily. • A little consideration will show that when two masses are placed arbitrarily*, then the conditions (i) and (ii) will only be satisfied. • But the condition (iii) is not possible to satisfy. This means that the mass moment of inertia of these two masses placed arbitrarily, will differ than that of mass moment of inertia of the rigid body. 51
- 52. 52 The torque required to accelerate the body the torque required to accelerate the two-mass system Difference between the torques required to accelerate the two-mass system and the torque required to accelerate the rigid body The difference of the torques T' is known as correction couple. This couple must be applied, when the masses are placed arbitrarily to make the system dynamical equivalent Correction couple L
- 53. Inertia Forces in a Reciprocating Engine, Considering the Weight of Connecting Rod • Analytical Method for Inertia Torque • mC = Mass of the connecting rod, • l = Length of the connecting rod, • l1 = Length of the centre of gravity of the connecting rod from P. Mass of the connecting rod at P Total equivalent mass of the reciprocating parts acting at P 53
- 54. 54 Total equivalent mass of the reciprocating parts acting at P Total inertia force of the equivalent mass acting at P, Corresponding torque exerted on the crank shaft T1
- 55. 55 To Determine the torque exerted on the crankshaft due to mass m2 placed at the big end centre (C) To Determine the torque exerted on the crankshaft due to Correction Couple T2 The total torque exerted on the crankshaft due to the inertia of the moving parts is the algebraic sum of T1 , T2 and T3. T=T1+T2+T3
- 56. 56 T=T1+T2+T3 To find Torque Exerted on Crank shaft
- 57. 57
- 58. 58
- 59. 59 To Determine the torque exerted on the crankshaft due to Correction Couple T2
- 60. 60 To Determine the torque exerted on the crankshaft due to mass m2 placed at the big end centre (C) The total torque exerted on the crankshaft due to the inertia of the moving parts is the algebraic sum of T1 , T2 and T3. T=T1+T2+T3
- 61. A vertical engine running at 1200 r.p.m. with a stroke of 110 mm, has a connecting rod 250 mm between centres and mass 1.25 kg. The mass centre of the connecting rod is 75 mm from the big end centre and when suspended as a pendulum from the gudgeon pin axis makes 21 complete oscillations in 20 Seconds. 1. Calculate the radius of gyration of the connecting rod about an axis through its mass centre. 2. When the crank is at 40° from the top dead centre and the piston is moving downwards, find analytically, the acceleration of the piston and the angular acceleration of the connecting rod. Hence find the inertia torque exerted on the crankshaft. To make the two-mass system to be dynamically equivalent to the connecting rod, necessary correction torque has to be applied and since the engine is vertical, gravity effects are to be considered. 61
- 62. 62
- 63. 63 T=T1+T2+T3+T4
- 64. 64 To Determine the torque exerted on the Crankshaft due to Correction Couple T2
- 65. 65 We know that for a compound pendulum, frequency of oscillation
- 66. 66 Torque due to the mass at P,T3 T3=0.156 N-m To Determine the torque exerted on the crankshaft due to mass m2 placed at the big end centre (C) T4
- 67. 67 Inertia torque exerted on the crankshaft