For 2014 (IES, GATE & PSUs) Theory of Machines Contents Chapter – 1: Mechanisms Chapter - 2 : Flywheel Chapter - 3 : Governor Chapter - 4 : CAM Chapter - 5 : Balancing of Rigid Rotors and Field Balancing Chapter - 6: Balancing of single and multi- cylinder engines Chapter - 7: Linear Vibration Analysis of Mechanical Systems Chapter - 8: Critical speeds or Whirling of Shafts Chapter - 9: Gear Train Chapter - 10: Miscellaneous S K Mondal IES Officer (Railway), GATE topper, NTPC ET-2003 batch, 12 years teaching experienced, Author of Hydro Power Familiarization (NTPC Ltd) Note “Asked Objective Questions” is the total collection of questions from:- 22 yrs IES (2009-1992) [Engineering Service Examination] 22 yrs. GATE (2010-1992) [Mechanical Engineering] and 14 yrs. IAS (Prelim.) [Civil Service Preliminary] Copyright © 2007 S K Mondal Every effort has been made to see that there are no errors (typographical or otherwise) in the material presented. However, it is still possible that there may be a few errors (serious or otherwise). I would be thankful to the readers if they are brought to my attention at the following e-mail address:
[email protected] S K Mondal Guidelines (a) How to use this material for preparation of GATE i. Attend all classes (Coaching) and meticulously read class note OR read book by Ghosh-Mallick ii. Topic list (which are to be covered for GATE) a. Mechanism (VIMP) b. Linear Vibration Analysis of Mechanical Systems (VIMP) c. Gear train (VIMP) d. Flywheel (Coefficient of Fluctuation of speed, Coefficient of Fluctuation of energy), mass calculation, e. Critical Speed of Shafts iii. Practice all solved examples from the book of ONLY the topics mentioned in the above topic list. iv. Then solve my question set (GATE +IES+IAS) on your own and cross check with my explanations of ONLY the topics which are to be covered for GATE.(mentioned above) (b) How to use this material for preparation of IES v. Attend all classes (Coaching) and meticulously read class note OR read book by Ghosh-Mallick vi. All Topics should be covered which is given in this booklet ( Question set) vii. Practice all solved examples from the book of all the topics given in this booklet. viii. Then solve my question set (GATE +IES+IAS) on your own and cross check with my explanations. GATE & PSU) What is TOM The subject theory of machine may be defined as that branch of engineering science which deals with the study of relative motion both the various parts of m/c and forces which act on them. 3. which can be classified according on the primary goal of the mechanism: . It should have relative motion.g. with a constant or variable velocity ratio. and 2. A link should have the following two characteristics: 1. Continuous rotation into oscillation or reciprocation (or the reverse). spring. It must be a resistant body. Mechanism Theory at a glance (IES. belt. Resistance Body: Resistant bodies are those which do not suffer appreciable distortion or change in physical form by the force acting on them e. while acting upon the m/c part in motion. Stated more specifically linkages may be used to convert: 1. oscillating. Dynamics: It deals with the force and their effects. Functions of Linkages The function of a link mechanism is to produce rotating.. Continuous rotation into continuous rotation. Kinematic Link Element: A resistant body which is a part of an m/c and has motion relative to the other connected parts is term as link. Linkages have many different functions. or reciprocating motion from the rotation of a crank or vice versa. with a constant or variable velocity ratio. Oscillation into oscillation. 2. with a constant or variable angular velocity ratio. or reciprocation into reciprocation. Kinemics: It deals with the relative motion between the various parts of the machine 2. Mechanism S K Mondal’s Chapter 1 1. A link may consist of one or more resistant bodies. The theory of m/c may be sub divided into the following branches: 1. Thus a link may consist of a number of parts connected in such away that they form one unit and have no relative motion to each other. The kinematic pairs are divided into lower pairs and higher pairs. Kinematic Pair Two element or links which are connected together in such a way that their relative motion is completely or successfully constrained form a kinematic pair. Higher Pair: When two elements have point or line of contact while in motion. 2. Fluid Link: It formed by having the motion which is transmitted through the fluid by pressure. 3. Flexible Link: Partly deformed while transmitting motion–spring. Types 1. which can be expressed by a single coordinate angle ' ' . Its 6 types are: Revolute Pair Prismatic Pair Screw Pair Cylindrical Pair Spherical Pair Planar Pair.e. i. hydraulic press. hydraulic brakes. depending on how the two bodies are in contact. etc. Revolute Pair A revolute allows only a relative rotation between elements 1 and 2. Mechanism S K Mondal’s Chapter 1 Function generation: the relative motion between the links connected to the frame. belts. e. Lower Pairs A pair is said to be a lower pair when the connection between two elements is through the area of contact. The term kinematic pairs actually refer to kinematic constraints between rigid bodies. . Rigid Link: It is one which does not undergo any deformation while transmitting motion–C. Lower Pair: When two elements have surface contact while in motion.Thus a revolute pair has a single degree of freedom. Path generation: the path of a tracer point.R. g. or Motion generation: the motion of the coupler link. Mechanism S K Mondal’s Chapter 1 Y A F=1 X REVOLUTE (R) Prismatic Pair A prismatic pair allows only a relative translation between elements 1 and 2. which can be expressed by a single coordinate angle ' ' or ‗x‘. Y X Screw Pair A screw pair allows only a relative movement between elements 1 and 2. . Thus a prismatic pair has a single degree of freedom. Example-lead screw and nut of lathe. Thus a screw pair has a single degree of freedom. which can be expressed by a single coordinate 'x'. screw jack. Z X f=3 SPHERICAL (G) . Thus a cylindrical pair has two degrees of freedom.g. which can be expressed as two independent coordinate angle ' ' and ‗x‘. S f=2 CYLINDRICAL (C) Spherical Pair A spherical pair allows three degrees of freedom since the complete description of relative movement between the connected elements needs three independent coordinates. e. Two of the coordinates ' ' and ' ' are required to specify the position of the axis OA and the third coordinate ' ' describes the rotation about the axis OA. – Mirror attachment of motor cycle. Mechanism S K Mondal’s Chapter 1 Cylindrical Pair A cylindrical pair allows both rotation and translation between elements 1 and 2. ram and its guides in shaper. The degree of freedom of a kinetic pair is given by the number independent coordinates required to completely specify the relative movement. between teeth of most of the gears and in cam-follower motion. The piston and cylinder. Two coordinates x and y describe the relative translation in the xy-plane and the third ' ' describes the relative rotation about the z-axis. chains. Wrapping Pairs Wrapping Pairs comprise belts. A point contact takes place when spheres rest on plane or curved surfaces (ball bearings) or between teeth of a skew-helical gears. A cylinder and a hole of equal radius and with axis parallel make contact along a surface. lathe . Higher Pairs A higher pair is defined as one in which the connection between two elements has only a point or line of contact. the pair is known as a sliding pair. cross-head and guides of a reciprocating steam engine. In roller bearings. the pair is known as turning pair. A little consideration will show that a sliding pair has a completely constrained motion. are the examples of a sliding pair. Two cylinders with unequal radius and with axis parallel make contact along a line. A shaft with collars at both ends fitted into a circular hole. tail stock on the lathe bed etc. the crankshaft in a journal bearing in an engine. Turning pair When the two elements of a pair are connected in such a way that one can only turn or revolve about a fixed axis of another link. Mechanism S K Mondal’s Chapter 1 Planar Pair A planar pair allows three degrees of freedom. and other such devices Sliding Pair When the two elements of a pair are connected in such a way that one can only slide relative to the other. 5.2 Square bar in a square hole Fig.4. e.e. e. is an incompletely constrained motion as it may either rotate or slide in a hole . Mechanism S K Mondal’s Chapter 1 spindle supported in head stock. it will only reciprocate) relative to the cylinder irrespective of the direction of motion of the crank.2. as shown in Fig. cycle wheels turning over their axles etc. Kinematic constraints are constraints between rigid bodies that result in the decrease of the degrees of freedom of rigid body system. 2. as shown in Fig. 5. cam and follower. .1 Square hole Collar Squre bar Shaft Fig. When motion between a pair is limited to a definite direction irrespective of the direction of force applied. and the motion of a shaft with collars at each in a circular hole. For example. are the examples of a turning pair.g. A turning pair also has a completely constrained motion. then the motion is said to be a completely constrained motion.3 Shaft with collar in a circular hole. Kinematic Constraints Two or more rigid bodies in space are collectively called a rigid body system. The change in the direction of impressed force may alter the direction of relative motion between the pair.5. 2.. Closed Pair: When two elements of a pair are held together mechanically. Rolling pair: When the two elements of a pair are connected in such a way that one rolls over another fixed link. as shown in Fig. Incompletely constrained motion: When the motion between a pair can take place in more than one direction. Completely constrained motion. the pair is known as rolling pair. the piston and cylinder (in a steam engine) form a pair and the motion of the piston is limited to a definite direction (i.. Types of Constrained Motions Following are the three types of constrained motions: 1. A circular bar or shaft in a circular hole..5. Ball and roller bearings are examples of rolling pair. are also examples of completely constrained motion. According to mechanical constraint between the elements: 1. all lower pair and some of higher pair. 5. as shown in Fig.g.3. We can hinder the motion of these independent rigid bodies with kinematic constraints. Unclosed Pair (Open Pair): When two elements of a pair are not held together mechanically. 5. then the motion is called an incomplete constrained motion. These both motions have no relationship with the other. The motion of a square bar in a square hole. it is called ternary link. forming a pair. Another relation between the number of links (l) and the number of joints (j) which constitute a kinematic chain is given by the expression: 3 j = l-2 2 Where. If each link is assumed to form two pairs with two adjacent links. is such that the constrained motion is not completed by itself. A chain which consists of only binary links is called simple chain. If it is connected to three other links. The motion of an I. 3. of links . But if the load is placed on the shaft to prevent axial upward movement of the shaft. Successfully constrained motion: When the motion between the elements. 5. Mechanism S K Mondal’s Chapter 1 Load Shaft Round hole Shaft Foot step bearing. then relation between the No. The chain is said to be closed chain if every link is connected to at least two other links. then the motion is said to be successfully constrained motion.5. The shaft may rotate in a bearing or it may rotate in a bearing or it may upwards. Fig. Consider a shaft in a foot-step bearing as shown in Fig 5. But if the pair is said to be successfully constrained motion.C. l = no.4 Shaft in a circular hole. but by some other means. therefore there will be as many links as the number of pairs. 5. If it is connected to two other links. and so on. Kinematic chain A kinematic chain is a series of links connected by kinematic pairs. of pairs (p) formatting a kinematic chain and the number of links (l) may be expressed in the from of an equation: l = 2p – 4 Since in a kinematic chain each link forms a part of two pairs. Fig. This is a case of incompletely constrained motion.4 Shaft in a f oot step bearing. otherwise it is called an open chain. A link which is connected to only one other link is known as singular link. then the motion. engine valve (these are kept on their seat by a spring) and the piston reciprocating inside an engine cylinder are also the examples of successfully constrained motion. it is called binary link. g. . The number of nodes defines the type of link. Binary Joint 2.H. Structure L.H.S. We can look at the illustration above and determine the degrees of freedom of each. Links.S. Binary link . of pairs j = no. = R.One link with four nodes DOF=+1 DOF=0 DOF=-1 Now that we have defined degrees of freedom. of binary joints.S.H.H. Mechanism S K Mondal’s Chapter 1 p = no.One link with two nodes Ternary link . > R.H.One link with three nodes Quaternary link .S.H. < R. B C C 3 3 B 4 B 2 3 4 D 2 2 5 A C A D 1 1 A E Structure Constrained chain 1 Unconstrained chain Types of Joins 1. Unconstrained chain. If L. Joints and Kinematic Chains Binary Ternary Quaternary Link Link Link Every link must have nodes.S. Quaternary Joint. Constrained chain L.S. Ternary Joint 3. e. Structure: An assemblage of resistant bodies having no relative motion between them and meant for carrying load having straining action called structure.o. A kinematic chain is said to be movable when its d. If the different link of the same kinematic chain is fixed. An assembly with negative degrees of freedom is a structure with residual stresses. Machine: When a mechanism is required to transmit power or to do some particular kind of work it is known as a machine. Mechanism S K Mondal’s Chapter 1 Joint Type DOF Description A First order pin joint 1 two binary links joined at a common point B Second order pin joint 2 three binary links joined at a common point C Half Joint 1 or 2 Rolling or sliding or both Mechanism Mechanism: When one of the link of a kinematic chain is fixed.f. A structure is an assembly that has zero degrees of freedom. In a two dimensional plane such as this . Degrees of freedom Degree of Freedom: It is the number of independent variables that must be specified to define completely the condition of the system. Mechanisms and Structures A B C A mechanism is defined by the number of positive degrees of freedom. 1 otherwise it will be locked. Figure4-1 shows a rigid body in a plane. is 1 the chain is said to be constrained. it will be a mechanism. If the d. Inversions: Mechanism is one in which one of the link of kinematic chain is fixed. To determine the DOF this body we must consider how many distinct ways the bar can be moved. the result is a different mechanism.f. Different mechanism are formed by fixing different link of the same kinematic chain are known as inversions of each other.o. The primary function of a mechanism is to transmit or modify motion. If the assembly has zero or negative degrees of freedom it is a structure. Fig. there are 3 DOF. The bar can be translated along the x axis. .f. two independent variables will be required to fully define its position. y and z axes respectively. Y O X Figure 4-1 Degrees of freedom of a rigid body in a plane Consider a pencil on a table. y and z axes and three rotary motions around the x. No single variable by itself can never fully define its position.o. Therefore the system has two degrees of freedom. and rotated about its central. Degree of freedom of a Rigid Body in Space An unrestrained rigid body in space has six degrees of freedom: three translating motions Along the x. Either an X-Y coordinate of an endpoint and an angle or two X coordinates or two Y coordinates. Y X Z Figure 4-2 Degrees of freedom of a rigid body in space Unconstrained rigid body in space possesses 6 d. If the corner of the table was used as a reference point. Mechanism S K Mondal’s Chapter 1 computer screen. translated along the y axis. It has one degree of freedom. turning around point A. The Kutzbach criterion calculates the mobility. j. In two dimensions. Figure shows several cases of a rigid body constrained by different kinds of pairs. The only way the rigid body can move is to rotate about the fixed point A. a rigid body is constrained by a higher pair. Pair 2 θ. In Figure-c. the number of independent input motions must equal the number of degrees of freedom of the mechanism. S Spherical Pair 3 θ. Variable Pin Joint 1 θ Sliding Joint 1 S Screw Pair 1 θ or S Revolute Pair Cost. The two lost degrees of freedom are translational movements along the x and y axes. Now let us consider a plane mechanism with / number of links.F. The mobility is the number of input parameters (usually pair variables) that must be independently controlled to bring the device into a particular position. and it cannot move along the y axis. In this example. ψ Planar Pair 3 xy θ Kutzbach criterion The number of degrees of freedom of a mechanism is also called the mobility of the device. a rigid body is constrained by a prismatic pair which allows only translational motion. Since in a mechanism. one of the links is to be fixed. a rigid body is constrained by a revolute pair which allows only rotational movement around an axis. Mechanism S K Mondal’s Chapter 1 Joint/Pair D. In Figure -b. it has one degree of freedom. translating along the x axis. In order to control a mechanism. For example. therefore the number of movable links will be (l-1) and thus the total . Rigid bodies constrained by different kinds of planar pairs In Figure-a. Y Y A A X X O c a b Figure. the body has lost the ability to rotate about any axis.O. It has two degrees of freedom: translating along the curved surface and turning about the instantaneous contact point. then h = 0.e.e. In general. Figure 4-11 shows the three kinds of pairs in planar mechanisms. Let's calculate its degree of freedom. b). Mechanism S K Mondal’s Chapter 1 number of degrees of freedom will be 3 (l-1) before they are connected to any other link. higher pairs). n = degree of freedom l = no. Substituting h= 0 in equation (i). of higher pair. of joints/no.e. we have n 3(l 1) 2j Where. n n Turning pair Prismatic pair Higher pair Two DOF lost Two DOF lost One DOF lost a b c Figure. Similarly. two degree of freedom pairs ). If we create a lower pair (Figure 4-11a. the degrees of freedom are reduced to 2. If there are no two degree of freedom pairs (i. the degrees of freedom are reduced to 1. In general. These pairs reduce the number of the degrees of freedom. single degree of freedom pairs ) and h number of higher pairs (i. a mechanism with l number of links connected by j number of binary joints or lower pairs (i. then the number of degrees of freedom of a mechanism is given by n 3(l 1) 2j h This equation is called Kutzbach criterion for the movability of a mechanism having plane motion. if we create a higher pair (Figure 4-11c). of lower pair h = no. Kinematic Pairs in Planar Mechanisms Example 1 Look at the transom above the door in Figure 4-13a. of link j = no. a rigid body in a plane has three degrees of freedom. The opening and closing mechanism is shown in Figure 4-13b. Kinematic pairs are constraints on rigid bodies that reduce the degrees of freedom of a mechanism. . C. . 3 and frame 4). D). B. h = 0 F = 3 (4 – 1) – 2 4 – 1 0 =1 (4-2) Note: D and E function as a same prismatic pair. D). Transom mechanism n = 4 (link 1. C C B B A D D A b Figure. Dump truck n = 4. h = 0 F = 3 (4 – 1) – 2 4 – 1 0 =1 (4-3) Example 3 Calculate the degrees of freedom of the mechanisms shown in Figure. C. B. l = 4 (at A. Mechanism S K Mondal’s Chapter 1 4 4 E C 1 C 2 3 A 1 B D B 3 A 4 D a b Figure. l = 4 (at A. 3. Figure 4-14a is an application of the mechanism. Example 2 Calculate the degrees of freedom of the mechanisms shown in Figure 4-14b. so they only count as one lower pair. Hence. as shown in Fig.1 or less. h = 2 F = 3 (4 – 1) – 2 4 – 1 2 =1 Note: The rotation of the roller does not influence the relationship of the input and output motion of the mechanism. it is called a passive or redundant degree of freedom. Degrees of freedom calculation For the mechanism in Figure-(a) n = 6. n = . Imagine that the roller is welded to link 2 when counting the degrees of freedom for the mechanism. Mechanism S K Mondal’s Chapter 1 6 B A 1 B 1 2 A 2 C C 3 3 D E 5 D 6 6 (a) (b) Figure. Redundant link When a link is move without disturbing other links that links is treated as redundant link. then there are redundant constraints in the chain and it forms a statically indeterminate structure. l = 3. R 1 n =0 When. 3 2 4 1 1 n = 3(l 1) 2j h R l 4. l = 7. the freedom of the roller will not be considered. .(e). j 4. h = 0 F = 3 (6 – 1) – 2 7 – 1 0 =1 For the mechanism in Figure-(b) n = 4. L 4 6 P1 11 by counting or P1 N L 1 11 4 3 F 3(N 1) 2P1 8 5 3(8 1) 2 11 1 7 or F N 2L 1 8 2 4 1 1 2 The linkage has negative degree of freedom and thus is a SUPER STRUCTURE. However. The degree of such a linkage will be F 3(8 1) 2 10 =1 N b 4. 7 and 8 constitute a double pair so that the total number of pairs is again 10.N 0 0.N 8. It has 8 links but only three ternary links. the linkage has a constrained motion when one of the seven moving links is 1 1 1 driven by an external source.e. 1 N b 4..N t 4.N t 4. . L 3 5 P1 10 by counting 6 or P1 N L 1 10 F N 2L 1 8 2 3 1 1 3 2 4 or F 3 N 1 2P1 8 7 3(8 1) 2 10 1 i. consider the kinematic chain shown in Fig. Mechanism S K Mondal’s Chapter 1 D 3 C 6 4 2 5 A 1 B (e) Six bar mechanism Now.N 0 0. the links 6.N 8. As the number 4 of links is not to be increased by more than one. 8 1 2 12 1 11 3 4 5 6 7 8 9 10 It has 5 loops and 12 links. 8 links and 10 3 joints. the number of links 5 must be 10 to obtain one degree of freedom.20 (b) shows one of the possible solutions.N t 2. Three excess joints can be 4 11 formed by 6 2 3 6 ternary links or 4 ternary links and 1 Quaternary link or 5 2 ternary links. Referring Table 1. or 8 9 a combination of ternary and quaternary links with double joints. 1 Figure 1. . it 7 10 has 2 degrees of freedom. There are 4 loops and 8 links. the number of joints 13.N 11 4 7 6 L 5. 6 F N 2L 8 4 2 1 1 7 It is a superstructure.2. It has 1 degree of freedom and thus is a mechanism. the linkage is a structure. the number 2 of loops has to be decreased. With 4 loops. P1 15 10 8 F N 2L 1 11 2 5 1 0 9 11 2 Therefore. One of the many solutions is shown in Fig. With 4 loops and 1 degree of freedom. 5 The linkage has 4 loops and 11 links. Mechanism S K Mondal’s Chapter 1 3 N b 7. the required linkage can be designed.N 0 2. With 3 loops. Grashof ’s law In the range of planar mechanisms.) =10 (At the slider. Total number of links = 7 (Fig. Grubler Criterion Grubler’s Criterion for Plane Mechanisms The Grubler‘s criterion applies to mechanisms with only single degree of freedom joints where the overall movability of the mechanism is unity. one sliding pair and two turning pairs) F 3 N 1 2P1 P2 3(8 1) 2 10 0 1 The mechanism has a can pair. Therefore. A little consideration will show that a plane mechanism with a movability of 1 and only single degree of freedom joints can not have odd number of links. its degree of freedom must be found from Gruebler‘s criterion. we have 1 = 3(l-1)-2j or 3l-2j – 4 =0 This equation is known as the Grubler‘s criterion for plane mechanisms with constrained motion. its degree of freedom must be found from Gruebler‘s criterion.) Number of pairs with 1 degree of freedom = 8 Number of pairs with 2 degrees of freedom=1 F = 3 (N-1)-2P1-P2 =3(7-1) – 2 8 – 1 =1 Thus. Substituting n = 1 and h = 0 in Kutzbach equation. Therefore. The simplest possible mechanisms of this type are a four bar mechanism and a slider – crank mechanism in which l = 4 and j = 4. Total number of links = 8 (Fig. it is a mechanism with one degree of freedom. Mechanism S K Mondal’s Chapter 1 The mechanism has a sliding pair. A four bar linkage comprises four bar-shaped links and four turning pairs as shown in Figure 5-8. the simplest groups of lower pair mechanisms are four bar linkages. . In a four-bar linkage. if both of the side links revolve. and the links which are hinged to the frame are called side links. it is called a double-crank mechanism. if both of the side links rock. Some important concepts in link mechanisms are: 1. we need to introduce some basic nomenclature. 3. four bar linkage The link opposite the frame is called the coupler link. Double-rocker mechanism: In a four bar linkage. Double-crank mechanism: In a four bar linkage. Crank: A side link which revolves relative to the frame is called a crank. we refer to the line segment between hinges on a given link as a bar where: s = length of shortest bar l = length of longest bar p. Crank-rocker mechanism: In a four bar linkage.e. All four-bar mechanisms fall into one of the four categories listed in Table 5-1: Case l + s vers. 2. Mechanism S K Mondal’s Chapter 1 Fig. Classification Before classifying four-bar linkages. A link which is free to rotate through 360 degree with respect to a second link will be said to revolve relative to the second link (not necessarily a frame). it is called a double-rocker mechanism. oscillates). Rocker: Any link which does not revolve is called a rocker. q = lengths of intermediate bar Grashof's theorem states that a four-bar mechanism has at least one revolving link if s+l p+q and all three mobile links will rock if s+l>p+q The inequality 5-1 is Grashof's criterion. p + q Shortest Bar Type 1 < Frame Double-crank 2 < Side Rocker-crank 3 < Coupler Double rocker 4 = Any Change point . such a state is called a change point. if the shorter side link revolves and the other one rocks (i. 5. it is called a crank-rocker mechanism.. 4. If it is possible for all four bars to become simultaneously aligned. the slider-crank mechanism shown in Figure 5-14a can be inverted into the mechanisms shown in Figure 5-14b. For example. Single slider crank chain. and d. The following three types of kinematic chains with four lower pairs are important from the subject point of view: 1. taking a different link as the fixed link. c. the mechanism of the pump device in Figure 5-15 is the same as that in Figure 5-14b. When the shortest link is a side link. Inversion is a term used in kinematics for a reversal or interchanges of form or function as applied to kinematic chains and mechanisms. Double slider crank chain. However. the mechanism is a double- crank mechanism. the mechanism is a crank-rocker mechanism. each pair being a sliding pair or a turning pair. For example. Mechanism S K Mondal’s Chapter 1 5 > Any Double-rocker Table: Classification of Four-Bar Mechanisms From Table we can see that for a mechanism to have a crank. When the shortest link is the coupler link. The shortest link is the crank in the mechanism. 2. When the shortest link is the frame of the mechanism. Inversion of Mechanism Method of obtaining different mechanisms by fixing different links in a kinematic chain. 2. and 3. Types of Kinematic Chains The most important kinematic chains are those which consist of four lower pairs. is known as inversion of the mechanism. Four bar chain or quadric cyclic chain. the sum of the length of its shortest and longest links must be less than or equal to the sum of the length of the other two links. this condition is necessary but not sufficient. Mechanisms satisfying this condition fall into the following three categories: 1. 3. the mechanism is a double-rocker mechanism. Different examples can be found in the application of these mechanisms. . Inversion of a kinematic chain has no effect on the relative motion of its links. Mechanism S K Mondal’s Chapter 1 Figure: Inversions of the crank-slide mechanism Figure. The motion of links in a kinematic chain relative to some other links is a property of the chain and is not that of the mechanism. 2. For L number of links in a mechanism. 3 4 2 1 (c) Double lever mechanism (Ackermann steering) 3rd inversion. 4 Oscillator 3 Coupler . A pump device Keep in mind that the inversion of a mechanism does not change the motions of its links relative to each other but does change their absolute motions. the number of possible inversions is equal to L. Inversion of four bar chain (a) Crank and lever mechanism/Beam engine (1st inversion). 4 (lever) 3 1 2 (b) Double crank mechanism (Locomotive mechanism) 2nd inversion. 1. For example. Inversion of the Slider-Crank Mechanism Inversion is a term used in kinematics for a reversal or interchanges of form or function as applied to kinematic chains and mechanisms. .e. A pump device Keep in mind that the inversion of a mechanism does not change the motions of its links relative to each other but does change their absolute motions. Inversions of the crank-slide mechanism Figure. the mechanism of the pump device in Figure is the same as that in Figure (b). sliding pair). Pendulum pump or Bull engine: In this mechanism the inversion is obtained by fixing cylinder or link4 (i. Figure. (c). as shown in figure below. 1. taking a different link as the fixed link. Mechanism S K Mondal’s Chapter 1 1 2 4 3 2. and d. For example. 4 oscillator 3 couplier Watts mechine 2. Different examples can be found in the application of these mechanisms. the slider-crank mechanism shown in Figure (a) can be inverted into the mechanisms shown in Figure (b). Oscillating cylinder engine: It is used to convert reciprocating motion into rotary motion. Oscillating cylinder engine 3. Rotary internal combustion engine or Gnome engine: Fig. slotting machines and in rotary internal combustion engines. 2. Mechanism S K Mondal’s Chapter 1 The duplex pump which is used to supply feed water to boilers uses this mechanism. Quick return motion mechanism 4. Rotary internal combustion engine Sometimes back. Fig. But now-a-days gas turbines are used in its place. rotary internal combustion engines were used in aviation. . Crank and slotted lever quick return motion mechanism: This mechanism is mostly used in shaping machines. Fig.e. The motion of the tool is constrained along the line RD produced. 5. therefore the return stroke is completed within shorter time. as shown in Figure. The connecting rod PR carries the ram at R to which a cutting tool is fixed. Thus it is called quick return motion mechanism. In this mechanism. The link 2 corresponds to a crank in a reciprocating steam engine. Since the crank rotates with uniform angular speed. Whitworth quick return motion mechanism: This mechanism is mostly used in shaping and slotting machines. The driving crank CA (link 3) rotates at a uniform angular speed. the link CD (link 2) formatting the turning pair is fixed. along a line passing through D perpendicular to CD. The slider (link 4) attached to the crank pin at A slides along the slotted bar PA (link 1) which oscillates at a pivoted point D. i. Whitworth quick return motion mechanism . Mechanism S K Mondal’s Chapter 1 Time of cutting stroke 360o or Time of return stroke 360o Length of stroke CB 2 AP AC Note: We see that the angle made by the forward or cutting stroke is greater than the angle described by the return stroke. to prevent over winding. The device in this modified form was used in watches.. From this application it received the name Geneva stop. Figure. coming in contact with the corresponding hollow circular parts of the driven wheel. The number of slots or interval units in B depends upon the desired number of turns for the spring shaft. and B turns on the axis of the spring barrel. as in the following picture. music boxes. mark P1 R1 = P2 P2 = PR. working in the slots. In this case the driven wheel. A can only move through part of the revolution in either direction before pin a strikes the closed slot and thus stop the motion. b. retains it in position when the pin or tooth a is out of action. Mechanism S K Mondal’s Chapter 1 o Time of cutting stroke 360 o or Time of return stroke 360 Note: In order to find the length of effective stroke R1 R2. An example of this mechanism has been made in Sim Design. wheel A is secured to the spring shaft. The length of effective stroke is also equal to 2 PD The Geneva Wheel An interesting example of intermittent gearing is the Geneva Wheel shown in Figure 8-4. Arranged as a stop. The circular portion of the driver. A. a. The wheel A is cut away near the pin a as shown. . B. causing the motion of B. etc. Geneva wheel If one of the slots is closed. to provide clearance for wheel B in its motion. the pin. makes one fourth of a turn for one turn of the driver. are known as sliders and form sliding pairs with link 4. This can be proved as follows: . Inversion of Double slider crank chain It has four binary links. two revolute pairs. Mechanism S K Mondal’s Chapter 1 Geneva mechanism is used to transfer components from one station to the other in a rotary transfer machine Geneva mechanism produces intermittent rotary motion from continuous rotary motion. it is possible for link 1 to reciprocate along a vertical straight line. The link 1 and link 3. At the same time link 2 will rotate and link 3 will oscillate about the pin. Its various types are: Elliptical Trammel Scotch Yoke mechanism Oldham‘s coupling. Elliptical trammels It is an instrument used for drawing ellipses. The fixed plate or link 4 has two straight grooves cut in it. as shown in Figure (a). two sliding pairs. at right angles to each other. any point on the link 2 such as P traces out an ellipse on the surface of link 4. This inversion is obtained by fixing the slotted plate (link 4). in order to get the desired motion. as shown in Figure. A little consideration will show that AP and BP are the semi-major axis and semi-major axis of the ellipse respectively. The link AB (link 2) is a bar which forms turning pair with links 1 and 3. Hand Pump: Here also the slotted link shape is given to the slider and vice-versa. C 2 B 1 4 3 A Here the slider (link 4) is fixed and hence. When the links 1 and 3 slide along their respective grooves. prismatic pair (between 3 & 4) 4. Oldham's coupling is the inversion of double slider crank mechanism. Scotch yoke mechanism Here the constant rotation of the crank produces harmonic translation of the yoke. prismatic pair (between 4 & 1) This mechanism is used for converting rotary motion into a reciprocating motion. Oldham’s coupling It is used for transmitting angular velocity between two parallel but eccentric shafts. Mechanism S K Mondal’s Chapter 1 Note: If P is the mid-point of link BA. revolute pair (between 1 & 2) 2. . revolute pair (between 2 & 3) 3. An Oldham‘s coupling is used for connecting two parallel shafts whose axes are at a small distance apart. Crank 3. then AP = BP. The shafts are coupled in such a way that if one shaft rotates the other shaft also rotates at the same speed. Yoke The four kinematic pairs are: 1. it will trace a circle.‘ Hence if P is the midpoint of link BA. Sliding Block 4. Its four binary links are: 1. Fixed Link 2. . The intermediate piece (link 4) which is a circular disc. The link 4 can slide or reciprocate in the slots in the flanges. the centre of intermediate piece will describe a circle of radius equal to the distance between the axes of the two shafts. :. Let the distance between the axes of the shafts is constant. These flanges have diametrical slots cut in their inner faces. It also used for shafts with angular misalignment where flexible coupling does not serve the purpose. Thus Hooke‘s Joint connecting two rotating shafts whose axes lies in one plane.e. v= . Mechanism S K Mondal’s Chapter 1 Driven Driving shaft shaft 4 2 3 1 The link 1 and link 3 form turning pairs with link 2. The tongues on the link 4 closely fit into the slots in the two flanges (link 1 and link 3). and r = Distance between the axes of the shafts in metres. Maximum sliding' speed of each tongue (in m/s). diametrical projections) T1 and T2 on each face at right angles to each other.r Where = Angular velocity of each shaft in rad/s. have two tongues (i. Then the maximum sliding speed of each tongue along its slot is equal to the peripheral velocity of the centre of the disc along its circular path. Hooke’s Joint (Universal Coupling) This joint is used to connect two non-parallel intersecting shafts. (a) Single Hooke‘s Joint (b) Double Hooke‘s Joint One disadvantage of single Hooke‘s joint is that the velocity ratio is not constant during rotation. Mechanism S K Mondal’s Chapter 1 Figure Universal (Cardan) joint Figure 8-6 General form for a universal joint There are two types of Hooke’s joints in use. But this can be overcome by using double Hooke‘s joint. Let = input angular displacement = output angular displacement = shaft angle tan cos tan tan = cos tan Let 1 = angular velocity of driving shaft 2 = angular velocity of driven shaft d 1 = dt d 2 = dt Now different above equation d d sec2 = cos sec2 dt dt sec2 1 = cos sec2 2 2 sec2 = …(a) 1 cos sec2 now sec2 1 tan2 tan 2 tan =1 cos2 tan cos sin 2 = 1 cos2 cos2 cos2 cos2 sin 2 = cos2 cos2 Equation (a) will be 2 sec2 = 1 cos2 cos2 sin 2 cos cos2 cos2 . . etc. . when = 0. Mechanism S K Mondal’s Chapter 1 cos = cos cos2 sin 2 2 cos cos = = cos [1 sin ] sin 1 cos2 sin 2 2 2 2 1 cos 2 1 cos2 sin2 Special Cases: For a given shaft angle . 1 (iii) For equal speed: 2 cos =1= 1 1 cos2 sin 2 1 cos = 1 cos 2 Thus is unity at given by above equation i. two times in one revolution of driving shaft. Velocity ratio: 2 is maximum when 0. 1 Now 1 cos2 = 1 cos 1 cos sin 2 1 1 cos 1 cos tan2 cos tan2 cos . 1 2 cos 1 2 1 max 1 sin cos 2 Thus is maximum at 0 or 180.. 270 . . i.R: 2 cos 1 min 2 is minimum at 90. when = .e.e. i. .e. four times in one revolution of driving shaft. the expresses given above is maximum when cos 1 i. 2 2 (i) Max. two times in one revolution.e.e. etc.. And will be minimum when cos 0 3 i. 1 (ii) Minimum V. Peaculier mechansim 1. and 2.bar chain mechanisms. Robert ' s mechanism 3. 4. 1 2max 2min 1 cos cos 2 1 = w1tan sin Since is a small angle sin tan Average equation of driven shaft: d2 d2 d d 12 cos sin 2 sin2 . These mechanisms are of the following two types: 1. Hart ' s mechansim 2. 1 2 = dt d dt d (1 cos2 sin2 )2 2 Condition for maximum acceleration d2 2sin 2 0 cos 2 d 2 sin 2 Straight Line Mechanisms One of the most common forms of the constraint mechanisms is that it permits only relative motion of an oscillatory nature along a straight line. In which one sliding pairs are used. Mechanism S K Mondal’s Chapter 1 (iv) Maximum variation of the velocity of driven shaft Variation of velocity of driven shaft 2 2min = max 2m ean But 2m ean = 1 . The mechanisms used for this purpose are called straight line mechanisms. Grass Hooper mech. Approximate Straight Line Motion Mechanisms The approximate straight line motion mechanisms are the modifications of the four. S cot t Russel mechansim 3. . Exact straight line mechanism Approximate striaght line mechanis 1. In which only turning pairs are used. Modified s cot t Russel. because both shafts complete one revolution during the same interval of time 1 1 cos maximum variation of velocity of w2 = cos 1 maximum variation of 2 sin tan 2max 2min 1 tan sin Maximum fluctuation of speed of driven shaft. Watts mechanism 2. and c = Distance between the pivots A and B of the front axle. 2. The difference between the Ackerman and Davis steering gears are: 1. The whole mechanism of the Ackerman steering gear is on the back of the front wheels. The Ackerman steering gear consists of turning pairs.cot = c / b Where a = wheel track. Mechanism S K Mondal’s Chapter 1 Steering Gear Mechanism Fig. Davis Steering Gear Ackerman Steering Gear The Ackerman steering gear mechanism is much simpler than Davis gear. so as move the automobile in any desired path. Steering Gear Mechanism The steering gear mechanism is used for changing the direction of two or more of the wheel axles with reference to the chassis. Velocity and Acceleration . b = wheel base. Usually the two back wheels have-a common axis. cot . whereas Davis steering gear consists of sliding members. it is in front of the wheels. whereas in Davis steering gear. which is fixed in direction with reference to the chassis and the steering is done by means of the front wheels. and higher. Mechanism S K Mondal’s Chapter 1 The concept of velocity and acceleration images is used extensively in the kinematic analysis of mechanisms having ternary. with the help of images the velocity and acceleration of any other point on the link can be easily determined. then. the radius of curvature of it is at infinity and hence instantaneous centre of this ties at infinite.order links. An example is 1. Hence for the body which having straight line motion. Relative Velocity Method Velocity by Instantaneous Centre Method Instantaneous centre is one point about which the body has pure rotation. Special cases of ICR I12 at 2 2 A B I12 VA VB 1 1 2 2 B VA VB 2 VA C 2 A 1 B A VC VB I12 I12 1 Types of ICR: . Instantaneous Centre Method 2. quaternary. If the velocities and accelerations of any two points on a link are known. Also the velocity of P as a point on link 3 must have the direction VP3 . Then the velocity of P as a point on link 2 must have the direction VP2 . Let us suppose this is not true and I23 is located at the point P. Mechanism S K Mondal’s Chapter 1 I13 3 I34 I23 4 2 I24 I12 I14 1 1 (i) Fixed ICR: I12 . except the position of P chosen on the straight line passing through I12 and I13 . The point P chosen therefore. VP VP 2 3 P 2 3 I13 A B I12 1 1 The Theorem can be proved by contradiction. to AP. I34 (iii) Neither Fixed nor Permanent I. I23 must all lie on the same straight line on the line connecting two pins. The direction is inconsistent with the definition that an instantaneous centre must have equal absolute velocity as a part of either link. their relative instantaneous centre lies on straight line‖. I14 (ii) Permanent ICR: I23 . This same contradiction in the direction of VP2 and VP3 occurs for any location chosen for point P. to BP. I24 Three-Centre-in-line Theorem (Kennedy’s Theorem) Kennedy Theorem states that ―If three links have relative motion with respect to each other. The Kennedy Theorem states that the three IC I12 . . cannot be the IC I23 . I13 .C: I13 . This justify the Kennedy Theorem. Mechanism S K Mondal’s Chapter 1 Properties of the IC: 1. of IC. Locate the fixed and permanent instantaneous centre by inspection.Cs and I23 and I34 are permanent instantaneous centre locate the remaining neither fixed nor permanent IC by Kennedy‘s Theorem. 1. I13 1 2 I34 3 4 I23 2 I14 I24 3 4 1 1. Method of locating instantaneous centre in mechanism Consider a pin jointed four bar mechanism as shown in fig. The following procedure is adopted for locating instantaneous centre. In fig I12 and I14 are fixed I. A rigid link rotates instantaneously relative to another link at the instantaneously centre for the configuration of the mechanism considered. Number of I.C in a mechanism: n(n 1) N 2 N = no. In other words. The two rigid links have no linear velocity relative to each other at the instantaneous centre. Make a list at all the instantaneous centre in a mechanism. n(4 1) 4(4 1) N= 6 3 2 2. 2. The instantaneous centre changes with alteration of configuration of mechanism.C. n = no. Each configuration of the link has one centre. This is done by circle diagram . of links. of I. the velocity of the IC relative to any third rigid link will be same whether the instantaneous centre is regarded as a point on the first rigid link or on the second rigid link. determine the no. First of all. Links 1 2 3 4 12 23 34 IC 13 24 14 3. Join the points by solid line to show these centres are already found. In present case 4 links. Fig shows a typical position of four bar linkage in toggle. Mechanism S K Mondal’s Chapter 1 as shown mark the points on a circle equal to the no. 23. 5. I23 and I14 I34 . 4 I12 I24 I34 3 I23 4 2 4 2 I24 I12 I14 1 1 Let T2 represent the input torque T4 represent the output torque. 2 I13 I23 3 I12 I23 2 I14 I24 4 I12 I24 Indices of Merit (Mechanical Advantage) From previous concept are know that 2 I14 I24 as per angular velocity ratio Theorem. join two such points that the line joining them forms two adjacent triangles in the circle diagram. and 14 to indicate the ICs I12 . Also consider that there is no friction or inertia force. Angular Velocity Ratio Theorem According to this Theorem “the ratio of angular velocity of any two links moving in a constrained system is inversely proportional to the ratio of distance of their common instantaneous centre from their centre of rotation”. 34. In order to find the other two IC. T4 2 I14 I24 T2 4 I12 I24 The mechanical advantage of a mechanism is the instantaneous ratio of the output force (torque) to the input force (torque). similarly IC I24 is located. . Then T22 T4 4 ve sign indicates that power is applied to link 2 which is negative of the power applied to link 4 by load. I34 and I14 . In the circle diagram these lines are 12. In fig join 1 and 3 to form triangle 123 and 341 and the instantaneous centre I13 will lie on the intersection of I12 . of links in mechanism. 4. From above equation we know that mechanical advantage is the reciprocal of the velocity ratio. The line which is responsible for completing two triangles should be a common side to the two triangles. I23 . where link 2 and 3 are on the same straight line. Let the absolute velocity of the point A i. the distance I24 I24 is zero.e. 3. draw a line parallel to VB intersecting the line of VBA at b. Take some convenient point o.e. I 24 I14 At this position. Through o. This line will represent the velocity of B with respect to A. which gives the required velocity of point B to the scale. draw oa parallel and equal to VA. 5. VB is known in direction only. VA is known in magnitude and direction and the absolute velocity of the point B i. draw a line perpendicular to AB. 2. I12 and I24 is coincident at A and hence. 1. Through a. Relative Velocity and Acceleration: Relative velocity of coincident points in two kinematic links: . Mechanism S K Mondal’s Chapter 1 C I34 3 4 B 4 2 2 A D I12. Through o.e. Consider two points A and B on a link. Measure ob. Then the velocity of B may be determined by drawing the velocity diagram as shown. 4 I12 I24 0 =0 2 I14 . 4. Known as the pole. to some convenient scale. I24 I14 I24 4 0 T4 2 Mechanical advantage T2 4 Hence the mechanical advantage for the toggle position is infinity The relative velocity method is based upon the velocity of the various points of the link. i. 1. 2/3 = relative angular velocity between link 2 and 3. r = radius of rotation of a point on link . + for opposite rotation. Relative Acceleration Method: f fc ft f = total acceleration f c = Centripetal acceleration f t = Tangential acceleration V2 f c = r2 f t = r r Where.2/3 2/3 = 2 3 . Mechanism S K Mondal’s Chapter 1 VP Q 3 VP VQ P. 3 Crank pin B 2 rub rb . Q 2 O 4 O 1 P on link 2 and 3 Q on link 4 Rubbing Velocity: Let rb = radius of pin B. Corioli’s Component of Acceleration: Fig. A pantograph is mostly used for the reproduction of plane areas and figures such as maps. plans etc.. It is also used to guide cutting tools.v Where = Angular velocity of the link OA. and V = Velocity of slider B with respect to coincident point C.C engine: . It is. Direction of coriolis component of acceleration This tangential component of acceleration of the slider B with respect to the coincident point C on link is known as coriolis component of acceleration and is always perpendicular to the link. acBC aBC t 2. Mechanism S K Mondal’s Chapter 1 = Angular velocity of rotation V = linear velocity of a point on link = Angular acceleration Direction of f c is along radius of rotation and towards centre. used as an indicator rig in order to reproduce to a small scale the displacement of the crosshead and therefore of the piston of a reciprocating steam engine. sometimes. Pantograph A pantograph is an instrument used to reproduce to an enlarged or a reduced scale and as exactly as possible the path described by a given point. Coriolis component of the acceleration of B with respect to C. Direction of f t is perpendicular to radius of rotation. Velocity and Acceleration by Analytical method Kinematic analysis of piston in I. on enlarged or reduced scales. r = length of crank l = length of connecting rod n = obliquity ratio l = r = angular velocity of the crank = inclination of the crank to i. r P1 P odc (outer dead centre) idc R O xp Let. C.d. d d VP = x P [r (1 cos n n2 sin2 )] dt dt d d = [r (1 cos n n2 sin2 )] . Mechanism S K Mondal’s Chapter 1 C l Top D. When the crank rotates through angle from its inner dead centre position the piston. x P = displacement of piston VP = velocity of the piston f P = acceleration of the piston. d dt . Displacement x P = P1 P = P1O PO = (l r) (l cos r cos ) x P = r(1 cos ) 1(1 cos ) Now from figure CR = r sin l sin r sin sin = sin l n 2 sin cos = 1 sin = 1 2 n n2 sin2 n 2 sin 2 = = n2 n n sin 2 2 x P r(1 cos ) l 1 n 1 = r(1 cos ) (n n2 sin2 ) n = r [1 cos n n 2 sin 2 ] x P r[1 cos n n 2 sin 2 ] Now differentiating above equation with respect to„t‟. C. D. receives displacement x P . = inclination of connecting rod to the line of stroke. B.c. 5. r .R.r. r . d dt n d dt cos cos . sin .R. r = angular velocity of the C. r .r. While the crank is approaching the inner dead center and the connecting rod is normally to the crank. Solution: Stroke length = 200 mm = 2r n 4. n2 2 sin r = n Q.C.R now CR = r sin l sin sin = sin = n now Differentiating above w. Mechanism S K Mondal’s Chapter 1 d = r. clockwise. r = angular velocity of the C. n cos r = n cos Also from previous section n2 sin2 cos = n2 cos cos r = 2 2 n sin n n. engine: Let. There find VP.5. N = 400 rpm. fP .R.to t cos 2 fP r2 cos n Analysis of connecting rod in I. to‗t‘ d 1 d sin sin dt n dt d d 1 d d sin . The crank rotates uniformly at 400 rpm clockwise. Find (i) Velocity of piston and angular velocity of the C. In a slider crank mechanism the stroke of the slider is 200 mm and the obliquity ratio is 4. [1 cos n n2 sin2 ] d sin 2 sin 2 VP r sin r sin 2 n2 sin2 2n Again differentiating above equation w. (ii) Acceleration of the piston and angular acceleration of the C. clockwise l tan 1 = 77. 1-n Solution: .C engine mechanism having obliquity ratio n.53 4.53 9 = –4286.29 m / sec (direction away from the crank) (ii) Acceleration cos2 f P = r2 cos n cos565.47° r = 360 77.63 rad / sec 2 .53° sin 2 (i) VP = r sin 2n sin 565.75 m/sec2 (direction towards the crank) cos (iii) r = = 2.17 mm/sec = 4.88 sin 282.88 rad/sec. show that for uniform engine speed the ratio for piston acceleration at the beginning of stroke and 1+ n end of the stroke is given by . n (Direction of r is opposite to that of ) 2sin (iv) r = = 380. Q. Mechanism S K Mondal’s Chapter 1 1 DC B l r Now r = 100 mm n 4.05 = 100 41. In an I. n (Direction of r is same as that of ).5 = 2.47 = 282.05 = 100 41.88 2 cos 282.5 l = 450 mm = 41.02 rad/sec. Firstly draw the configuration diagram of slider crank mechanism to the scale 1: K. Mechanism S K Mondal’s Chapter 1 P P1 P2 P1 is beginning of stroke P2 is end of stroke at P1 .c — Angular velocity of crank. fP2 1n Velocity and Acceleration by Klein’s Construction: K x M C Q X x P O N OC — crank CP — connecting rod — Angle made by crank with i. 0 and at P1 180 cos 2 fP r2 cos n 1 1 n fP1 r2 1 = r2 n n 1 1 n fP2 r2 1 = r2 n n fP1 1n . . now draw a line through ‗O‘ to the line of stroke OP. Procedure to draw velocity diagram: 1. After getting configuration diagram OCP. 2.d. K Acceleration Analysis: 1. . 6. Special Cases of Klein’s Construction: When the crank is at i. With C as centre and with CM as radius draw a circle. OC represents VC . Mechanism S K Mondal’s Chapter 1 3.c. As discussed earlier OCM is velocity polygon of slider crank mechanism to scale ‗kW‘. CM represents VP / C And OM represents VP C VC VP C O P VP VC OC. 5. 4. Extend KL to meet the line of stroke at N.K VP/C OM. 4. Then OCM represent the acceleration polygon. OCM Represent the velocity polygon of slider crank mechanism to the scale ‗KW‘.K Velocity of any point x lying on connecting rod: Cx Cx Cx Cx = CM CP CM CP VX Ox . Then KL represents the common chord of these two circles.K VP OM. Extend the connecting rod length PC. to meet this name the intersection point as M. In OCM. 3.d. 2. Draw another circle with diameter as length PC. Also KL is intersecting to PC at Q. M Q. P /C = =0 PC it means angular velocity of connecting rod is maximum. Mechanism S K Mondal’s Chapter 1 K O. C. M. Here QN = 0 fP/t C NQ kW2 = 0 fPt / C i. VP/C CM k [ CM OC VC VP / C Acceleration polygon is OCQN. Q P O N OCM Velocity triangle OCQN Acceleration polygon.e.. VP OM k 0 VC OC k. VC OC k VP/C CM k 0 VP / C WP / C =0 PC at = 90º WP/C = 0 VP PM k VP VC fPt / C = QC k2 = 0 fC = CO × k 2 . N C L OM = 0. fC CO k2 fP NO k2 fP/C QC k2 (ii) When the crank is at 90° from idc. In polygon. CO represents fCC fC .e. VP (OM) k 0 VC (OC) k VP/C CM k VC VP/ C fPt / C = NQ k2 = 0 P / C = 0 i. it means angular velocity of C. k 2 direction N to O fPt / C = NQ. K Q C O N N L OCM Velocity triangle OCQN Acceleration polygon. K 2 direction Q to C fPC/ C = QC. M Q C P N O OCM velocity triangle OCQN Acceleration polygon. k 2 direction C to O f P = No. No represents fPt fP .c.R..d. is maximum fC = (CO) k 2 fP (NO) k2 Retardation fPe / C = (QC) k 2 (v) When crank and connecting rod are mutual perpendicular. Mechanism S K Mondal’s Chapter 1 f P = NO k case of retardation 2 fPt / C = NQ k2 When the crank is at 180° from i. K 2 direction Q to C . t NQ represents f P/C and NC represents fP/C fC = CO. CN CP CN CP f X Ox1 . FR = Radial load on the crank shaft bearing. K 2 direction N to C. Mechanism S K Mondal’s Chapter 1 f P / C = NC. Acceleration of any point X lying on connecting rod: Cx Cx1 Cx Cx1 = .C engine Mechanism: FT FQ C ( + ) FQ l FR r P FP O FN FP = Net axial force on the piston or piston effort. k2 Force analysis in I. r *Net axial force (FP ) : During acceleration F g M o ti on F F P I FP Fg FI 2 cos 2 FP d Pg mr2 cos 4 n During retardation: F g F I F P M ot ion FP Fg FI d Diameter of the piston (m) Pg = intensity of pressure in the cylinder (N/m2) . T = Turning moment or Torque on the crank T FT . FT = Tangential force at crank pin or force perpendicular to the crank. FQ = Force acting along connecting rod FN = Normal reaction on the side of the cylinder or piston side thrust. 25 m n = 4. FI = inertia force = mfP *In case of vertical engine. The mass of reciprocating parts of a steam engine is 225 kg. the difference in the pressure of the two sides of the piston is 5 bar. length of the stroke is 500 mm and the ratio of length of connecting rod to crank is 4.4)2 5 105 225 0. FP Fg FI Ff WR Q. At what speed must the engine run so that the thrust in the connecting rod in this piston is equal to 5200 N? Solution: Data m 225 kg .7 rpm.4 m Stroke length = 500 mm = 2r r = 250 mm = 0. If the speed of the engine crank shaft is 250 rpm clockwise find the turning moment on the crank shaft. N = 274. the wt of the reciprocating parts (WR ) and friction force (Ff ). Mechanism S K Mondal’s Chapter 1 or Different in the intensity of pressure on two sides of the piston.166 T =? . When the crank is at inner dead center. A single cylinder two stroke vertical engine a bore of 30 cm and a stroke of 40 cm with a connecting rod of 80 cm long. When the piston is at quarter stroke and moving down.25 2 1 4 4.2 = 28. diagram of the cylinder is 400 mm. Q.25 stroke length Pg 70 N / cm2 N = 250 rpm (clockwise).767 rad/sec. The mass of the reciprocating parts is 120 kg.2 0 g = 5 bar FQ 5200 N =? Here FQ FP Now FP Fg FI 2 cos 2 = d Pg mr2 cos 4 n 1 5200 = (0. Neglect the mass and inertia effects on connecting rods and crank.2. Solution: Data d 30 cm Stroke = 40 cm = 2r r = 20 cm n=4 l 80 cm m 120 kg x P 0. d 400 mm 0. = 26. the pressure on it is 70 N/cm2. 62 = 120 0.7 N .25 cos2 9 cos 9 cos 20.31º Fg Motion FI sin Now sin n = 11.71° Now FP FG FI 2 FG = d .31 4 = 8274.7 N W = mg = 1177.5 cos 16 sin2 20.1662 cos54. Pg 4 = 302 70 4 = 49455 N cos 2 FI mr2 cos n cos108.25 5.7 + 1177.25 16 sin2 cos2 = 5.2 N FP = 49455 8274.25 cos 9 54.2 = 42357. Mechanism S K Mondal’s Chapter 1 Now x P r (1 cos n n 2 sin 2 ) 0.20 26.25 × 2r = r (1 cos 4 16 sin 2 ) 1 5 cos 16 sin2 2 16 sin 2 = 4. (b) IES-15. (a) Two link system shown in the above figure has 2 degrees of freedom. Mechanism S K Mondal’s Chapter 1 V1 2 12m s 2N 2 120 60 60 12. IES-5. IEA-13. (d) 4 links and 4 turning pairs satisfy the equation L = (j + 2). (a) IES-8. following six degrees of freedom are arrested (two line movements along y-axis.566rad / s Corioli‘s component of acceleration 2V1 2 2 12 12. (c) 3 IES-4. Ans. Ans. (c) When two elements or links are connected in such a way that their relative motion is constrained they form a kinematic pair. Ans. the joint is equivalent to (l . Ans. Ans. (d) IES-12.566 302m s2 GATE-27. (d) Sliding pair piston and cylinder Revolute pair Crank shaft in a journal bearing in an engine Rolling A road roller rolling over the ground Spherical pair Ball and socket joint IES-2. Ans. Ans. Ans. 3 links and 3 turning pairs form rigid frame. two rotational movements each along x- axis and z-axis. (d) When supported on three points. (b) IES-10. Ans. IES-14. Ans. Ans. Ans. (c) IES-11. Ans. (d) IES-16. Ans. IES-20. (a) IES-7. (a) Elements of higher pairs must be force closed to provide completely constrained motion. Foot step bearing results in successful constraint and 5 links and 5 turning pairs provide incomplete constraint. (a) IES-18. (d) IES-19.) . The relative motion of a kinematic pair may be completely. Ans.1) binary joints. (a) IES-9. (b) GATE-28. Ans. Ans. Ans. Ans. incompletely or successfully constrained IES-3. Ans. (c) When ‗l‘ number of links are joined at the same connection. (d) Previous 20-Years IES Answers IES-1. It is case of 2 complete constraint. if the centre of mass of the system does not lie on the axis of rotation. For the system to be statically balanced. (i) The resultant of all the dynamic forces acting on the system during rotation must be zero. The effects due to resultant inertia forces and couples acting on the machine parts. 2. Dynamic Balancing The system is said to be dynamically (complete) balanced. (ii) Non-homogeneity of material (iii) Unsymmetrical shape of the rotors due to functional requirement. Static Force: The static forces acting on the machine components are due to the weight of components. then the system is unbalanced. These m/c parts are subjected to different forces. if it satisfies following two condition. The motion of moving parts may be of rotary or reciprocating type. GATE & PSU) Balancing A machine consists of a number of moving part. Balancing of Rigid Rotors and field Balancing S K Mondal’s Chapter 5 5. Balancing is the process of correcting or eliminating. . Balancing of Rotating Masses In any rotating system. Static balancing The system is said to be statically balanced if the centre of mass (C. The unbalanced in rotating system is mainly due to the following factors. (i) Errors and tolerances in manufacturing and assembly. about any plane must be zero. Balancing of Rigid Rotors and Field Balancing Theory at a glance (IES. either partially or completely. having one or more rotating masses. 1. Of the system of masses lies on the axis of rotation. In comparison with the static forces. the dynamic forces are very large in magnitude. Cx. the resultant of all the dynamic forces (centrifugal forces) acting on the system during rotation must be zero. (ii) The resultant couples due to all the dynamic forces acting on the system during rotation. Inertia or Dynamic force: The inertia force is due to acceleration of various components or members of the m/c. such that centrifugal force due to the two masses is equal and opposite. . Case of Dynamic Unbalance Fig. Balancing of Rigid Rotors and field Balancing S K Mondal’s Chapter 5 Fig. (i) Balancing by single mass rotating in same plane (Internal balancing): To balance the single rotating mass. Illustration of Static Unbalance Balancing of mass rotating in single plane 1. Thin Rotor on a knife edge.G of mass m from the axis of rotation. Balancing of single rotating mass mrw 2 Bearing r Shaft Consider a single mass m attached to a shaft which is rotating with an angular velocity ‗ ‘. During the rotation of shaft a dynamic force equal to mrw2 acts in radial outward direction. This dynamic force can be balanced by either of the following two methods. a counter mass m b is placed in the plane of rotation at a radius rb and exactly opposite to it. Let r be the distance of C. Case I: Balancing masses are placed in two different planes on opposite side of the plane of rotation of disturbing mass. The balancing masses may be arranged in one of the two ways.w 2 mb 2 m2 . r1 . In order to achieve the complete balancing of the system. r2 . Balancing of Rigid Rotors and field Balancing S K Mondal’s Chapter 5 mrw 2 m r rb mb mrw 2 mr2 = m brb2 or. mr m b rb (ii) Balancing by two masses rotating in different planes (External Balancing): If the balancing mass cannot be placed in the plane of rotation of the disturbing mass then it is not possible to balance the disturbing mass by single balancing mass. mrw 2 m Plane B Plane l1 r l2 A rb 1 rb 2 mb 1 m1 . If a single balancing mass is placed in a plane parallel to the plane of rotation of disturbing mass. while it will produce an unbalanced couple. the dynamic force can be balanced. at least two balancing masses are required to be placed in two planes parallel to the plane of disturbing mass.w 2 . mb1 rb1 l1 mb2 rb2 l 2 .w 2 The balancing mass are m b1 and mb2 at radius rb1 and rb2 from shaft axis and distance. mb1 rb1 2l1 = mb2 rb2 2l2 mb1 rb1 l1 mb2 rb2 l 2 Case II: Balancing mass is placed opposite to each other in two difference planes. rb . Balancing of Rigid Rotors and field Balancing S K Mondal’s Chapter 5 Balancing mass m b1 are at radius rb1 from shaft axis at plane B from distance l1 to disturbing mass. Thus taking moments about plane A. mr mb2 rb2 m b1 rb1 (ii) mb1 rb1 2l1 = mb2 rb2 2l2 Or. mrw 2 mb . r1 . In order to achieve the complete balancing of the system the following two conditions must satisfied.w 2 2 2 m rb 2 r l1 B A l1 C rb 1 mb 1 m1 . l1 And l 2 from plane A. 2 mr = mb1 rb1 2 mb2 rb2 2 Or mr m b1 rb1 m b2 rb2 (ii) The sum of moments due to centrifugal forces about any point must be zero. (i) mr2 mb2 rb2 2 = mb1 rb1 2 Or. To achieve complete balancing following two conditions must satisfy (i) Centrifugal force of disturbing mass ‗m‘ must be equal to the sum of the centrifugal forces of the balancing masses m b1 and mb2 . and balancing mass mb2 at radius rb2 from shaft axis at plane C from distance to disturbing mass. kg-m2. m2 . 1. and 4 with horizontal line OX. r3 and r4 from the axis of rotation of shaft. Such type of balancing can be done by the following two methods. . r2 . Balancing of several masses rotating in same plane Consider the number of masses (say four) m1 . the magnitude of centrifugal force for each mass is proportion to the product of mass and radius of rotation. Graphical Method m3 Fc m2 r m1 r2 r3 3 2 r1 4 1 x r4 b rb mb m4 Fc = m4r4 4 Fc = mbrb b 4 Fc 3 Fc c 4 3 b b d Fc 2 2 a Fc 1 1 O Force polygon (i) First draw configuration diagram with given position. kg-m and the couple due to centrifugal force is taken as term ‗mrl‘. the centrifugal force is taken as term ‗mr‘. Balancing of Rigid Rotors and field Balancing S K Mondal’s Chapter 5 2. 3 . The angular position is measured in anticlockwise direction. (ii) Calculate centrifugal force by individual masses on the rotating shaft. 2 . The masses are at angular position 1 . *Since 2 is same for all masses. m3 and m4 attached to a shaft at a distance of r1 . Hence in subsequent section. FH = m1r1 cos 1 m2 r2 cos 2 m2 r3 cos 3 m4 r4 cos 4 and FV = m1r1 sin 1 m2 r2 sin 2 m3 r3 sin 3 m4 r4 sin 4 (ii) Calculate the magnitude of resultant centrifugal force. m2 . The relative angular position shown in fig. FCr = ( FH )2 ( FV )2 (iii) Calculate the angle made by the resultant centrifugal force with horizontal OX. – (iv) The closing side of the force polygon do represents the balancing centrifugal force. (v) Determine the magnitude of balancing mass m b at a given radius of rotation ' rb ' such — that balancing centrifugal force = m b rb = do scale 2. m3 and m4 revolving in planes A. B. C and D respectively. Balancing of Rigid Rotors and field Balancing S K Mondal’s Chapter 5 FC1 = m1r1 . FC4 = m4 r4 — (iii) Draw the force polygon such that o a represent the centrifugal force exerted by the — — — mass m1 m given direction with some suitable scale. Analytical Method Refer configuration diagram. (i) Resolve the centrifugal force horizontally and vertically and find their sums. bc . FV tan r = FH (iv) The balancing centrifugal force FCb should be equal to magnitude of the FCr but opposite in direction ( b ). FC2 = m2 r2 FC3 = m3 r3 . (v) Determine the magnitude and radius of rotation of balance mass by using relation m b rb = FCr Balancing of several masses rotating in different plane Consider mass m1 . . and cd . Similar draw ab. P +Ve m3 mL C mM m4 L m2 M D m1 B A rL r2 r3 rM r4 r1 l1 l2 l3 lM l4 Such system is balanced by two masses m L and m M which are put in planes L and M respectively. Balancing of Rigid Rotors and field Balancing S K Mondal’s Chapter 5 –ve R. Say ‗L‘ as reference plane (R. Position of Planes of Masses m3 r3 m2 r2 3 2 L 1 m1 r1 4 M rL rm mL r4 mb mM m4 The complete balancing of such system can be done by following two methods: 1. (ii) Tabulate the centrifugal forces and couples due to centrifugal forces. .P are taken as negative. Graphical Method (i) Take one of the planes from balancing plane.P).P are taken as positive and the left of R. The distance of other planes to the right of R. Analytical Method Refer previous table . 4 c l m 3r 3 3 r l4 m4 4 3 d b M l rM M mM 2 m2r2l2 a m1r1l1 o Couple polygon 4 c l m4r4l4 m 3r 3 3 3 d M mMrMlM b e L r2l2 mLrLlL 2 O a o m1r1 Force polygon — (iv) Now draw force polygon.m) R. eo is balancing force.m 2 ) (m) A m1 r1 m1r1 l1 m1r1l1 L(R. from Couple 2 (kg) (m) (kg. taking some suitable scale. the centrifugal forces and centrifugal couple are taken as terms (mr) and (mr l) respectively.F 2 Dist. Balancing of Rigid Rotors and field Balancing S K Mondal’s Chapter 5 Plane Mass Radius C. 2. Closing vector do represent the couple mM rM l M which is called balancing couple.P) ML rL m L rL O O B m2 r2 m 2r2 l2 m 2r2l 2 C m3 r3 m3r3 l3 m3r3l 3 M mM rM m M rM lM m M rM l M D m4 r4 m 4 r4 l4 m 4 r4 l 4 Since 2 is same for all masses. (iii) Draw the couple polygon. P. l (kg. respectively. P and R are the centers of gravity of the cantilever part and the counterweight respectively. GATE) Previous 20-Years GATE Questions GATE-1. A rotating disc of 1 m diameter has two eccentric masses of 0.5 kg each at radii of 50 mm and 60 mm at angular positions of 0°and 150°.e. for static balance.1 kg is to be used to balance the rotor. A cantilever type gate hinged at Q is shown in the figure. (m r l) = 0 Now resolve the couple horizontally and vertically and find their summation. and mL rL . What is the radial position of the balancing mass? [GATE-2005] (a) 50 mm (b) 120 mm (c) 150 mm (d) 280 mm . (m r l)H = 0 And find mM rM l M cos M … (i) (m r l)V = 0 and find mM rM l M sin M … (ii) by the use of (i) and (ii) m M rM Can be finding and also M can be find. IAS. The mass of the counterweight. L can be find. Objective Questions (IES. is [GATE-2008] Balancing of a single rotating mass by a single mass rotating in a same plane GATE-2. Similarly above process can be done for centrifugal forces. The mass of the cantilever part is 75 kg. a balancing mass of 0. Balancing of Rigid Rotors and field Balancing S K Mondal’s Chapter 5 (i) (couples) = 0 i. Ans.12 0. but positioned exactly 1800 apart.2 0. (d) IES-5. then the system will be balanced (a) statically (b) dynamically [IAS-1999] (c) statically as well as dynamically (d) neither statically nor dynamically Balancing of several masses rotating in a same plane Balancing of several masses rotating in different planes IAS-12. Ans.8076 mm Along y-axis. Ans. (c) IES-6. (d) IES-4. Ans. Ans. (b) IES-7.32 mm GATE-3. Ans. Two rotors are mounted on a shaft.4kgm2 2 I2 10 0. (a) IES-2. Ans. Ans. Balancing of Rigid Rotors and field Balancing S K Mondal’s Chapter 5 IAS-10. (d) Taking Moment about ‗Q‘ 75 2.5(-60 10-3 cos30o + 50 10-3)2 = 0.5 or R 300 kg GATE-2. the minimum number of balancing weights to be introduced in different planes is [IAS-2001] (a) 1 (b) 2 (c) 3 (d) 4 Answers with Explanation (Objective) Previous 20-Years GATE Answers GATE-1. (c) A system in dynamic balance implies that the system is also statically balanced. If the unbalanced force due to one rotor is equal in magnitude to the unbalanced force due to the other rotor. IES-3(i). Ans. 0. The balancing weights are introduced in planes parallel to the plane of rotation of the disturbing mass.12y y = 150 mm r= x2 y2 150. (b) IES-3. To obtain complete dynamic balance. 0.5(60 10-3 sin30o) )2 = 0. Ans. (c) Moment about A .6kg m2 2 I2 I1 %Increase 100 50% I1 Previous 20-Years IES Answers IES-1.12 x 10-3 x = -9. (b) I1 10 0.2 20 0.0 R 0. Ans. (c) Along x-axis. 15. GATE & PSU) Balancing of Reciprocating Masses In application like IC engine. n n . reciprocating compressor and reciprocating pumps the reciprocating parts are subjected to continuous acceleration and retardation. cos 2 cos 2 Fv = m.8 that the acceleration. We have already discussed in Art. Unbalanced force.r cos n Inertia force due to reciprocating parts or force required to accelerate the reciprocating parts.r cos + m.2. This force is an unbalanced one and is denoted by Fv.FBH) is equal opposite to inertia force exerted and opposite to inertia force (Ft). of the reciprocating parts is approximately given by the expression.2. the inertia force acts on the reciprocating parts which are in a direction opposite to the direction of acceleration. Balancing of single and multi-cylinder engines Theory at a glance (IES. Let m = Mass of the reciprocating parts. = Angular speed of the crank. This inertia force is unbalanced dynamic force acting on the reciprocating parts. cos 2 Fl = FR = Mass Acceleration = m. Due to this accelerate and retardation. l = length of the connecting rod PC r = radius of the crank OC.r cos = m. cos 2 a R 2 .r .e.2. Primary and Secondary Unbalanced Forces of Reciprocating masses Consider a reciprocating engine mechanism as shown in Figure. = Angle of inclination of the crank with the line of stroke PO.r cos n We have discussed in the previous article that the horizontal component of the force exerted on the crank shaft bearing (i. n =Ratio of length of the connecting rod to the crank radius = lIr.2. Balancing of single and multi-cylinder engines S K Mondal’s Chapter 6 6. The maximum secondary unbalanced force is given by r FS = m.The primary unbalanced force is maximum. the primary force is maximum twice in one revolution of the crank.2 n Balancing of Reciprocating Mass in single cylinder engine C l r P idc f O FI Let m = mass of reciprocating parts = angular speed of the crank r = radius of crank l = length of connecting rod n = obliquity ratio l/r = angle made by crank with i. d. FI = inertia force due to reciprocating mass Now cos 2 f = r2 cos n 2 cos 2 FI = mr cos n Hence unbalanced force 2 cos2 FU = FI = mr cos n cos2 = mr 2 cos mr2 n = FP FS Where FP = primary unbalanced force = mr2 cos FS = secondary unbalanced force . Thus. f = acceleration of the reciprocating mass.r 2. Primary unbalanced force.90oor 180o and 360o .2.2 .2 . FP = m. when = 0oor 180o. The maximum Primary unbalanced force is given by FP = m.r n Note: 1. Fs = m.Thus.r is n called secondary unbalanced force.r cos ) is known as primary unbalanced force and m. Balancing of single and multi-cylinder engines S K Mondal’s Chapter 6 cos 2 The expression (m.2. c. the secondary force is maximum four times in one revolution of the crank.r cos cos2 And secondary unbalanced force. The secondary unbalanced force is maximum. when = 0o .2. Hence.e. . The unbalanced force due to reciprocating mass varies in magnitude but constant in direction.e. This is the unbalanced force due to balancing of primary force (FP ). (ii) mbrb2 sin Perpendicular to line of stroke mbrb2 cos is responsible for balancing the primary unbalanced force. twice in one rotation of the crank. Therefore a single rotating mass cannot be used to balance a reciprocating mass completely. however a single rotating mass can be used to partially balance the reciprocating mass. in single cylinder reciprocating engines. 90°. four times in one rotating of the crank. the secondary unbalanced force is small and is generally neglected. while the unbalanced force due to rotating mass is constant in magnitude but varies in direction. 180° and 270 i. the magnitude of the secondary unbalanced force is times the of the n primary unbalanced force. When = 0°. But in case high speed engine the secondary unbalanced force is taken into account. The primary unbalanced force acting on the reciprocating engine is given by FP = mr2 cos 2 Mr FP mb r bcos 2 2 mr cos m 2 r sin b b m 2 r sin b b The primary unbalanced force may be treated as the component of centrifugal force along the line of stroke produced by an imaginary rotating mass m placed at crank pin ‗C‘. for complete balance only forces have to be balanced and the is no unbalanced couple. All reciprocating and rotating masses are in the single plane. 2 mr cos = mbrb cos 2 mr m b rb mb2rb sin is perpendicular to the line of stroke. It is important to note that. Due to balancing mass m b two component of force (i) mbrb2 cos Along the line of stroke. Now balancing can be done by attaching a balancing mass m1 at radius rb diametrically opposite to the cram. Balancing of single and multi-cylinder engines S K Mondal’s Chapter 6 cos 2 = mr2 n The primary unbalanced force is maximum when = 0° or 180° i. The secondary unbalanced force is maximum. However. the frequency of secondary unbalanced force is twice as that of the primary unbalanced 1 force. Therefore in case of low or moderate speed engines. Thus. Consider the multi cylinder inline engine having two inner cranks and two outer cranks. l l l l Crank 1 Crank 4 (outer) (outer) Crank 2 l2 l3 Crank 23 (inner) (inner) l1 l4 –ve R. The unbalancing force along the line of stroke FH = m2r cos c mr2 cos = (1 c) mr2 cos The unbalanced force perpendicular to line of stroke FV = cmr2 sin Resultant unbalanced force FR = FH2 FV2 FR mrw2 (1 c)2 cos2 c2 sin2 If the balancing mass m b has to balance the rotating mass m r rotating at radius rr as well as give partial balancing of reciprocating part. is known as incline engine. Thus the effect of the above method of balancing is to change the direction of maximum unbalanced force. mbrb2 cos = cmr2 cos m b rb cmr This is known as partial primary balancing.P +Ve . m b rb m r rr cmr Balancing of reciprocating masses in multi cylinder inline engine The multi cylinder engines having the axis of the entire cylinder in the same plane and on the same side of the axis of crank shaft. In practice. a compromise is achieved by balancing only a portion of the primary unbalanced force. Now. Balancing of single and multi-cylinder engines S K Mondal’s Chapter 6 In such case the maximum unbalanced force perpendicular to the line of stroke is m brb2 which is same as the maximum primary unbalanced force. unbalanced primary force due to partial balancing. The balancing mass m b is placed diametrically opposite to the crank pin such that. Secondary Balancing cos2 (i) mr 2 n =0 cos 2 (ii) mrw2l n = 0 The above condition can be written as r (i) m(2)2 4n cos 2 = 0 r (ii) m(2)2 4n l cos 2 = 0 This condition are equivalent to the condition of primary balancing for an imaginary crank of r length . for multi cylinder inline with four or more cylinders. rotating at speed 2 and inclined at an angle 2 to idc. The above condition are modified as mr (i) n =0 mrl (ii) n Analytical Method For complete secondary balancing: mr (i) n cos 2 = 0 . To ensure that the above two conditions are satisfied for all angular positions of the crank shaft. Balancing of single and multi-cylinder engines S K Mondal’s Chapter 6 Thus unbalanced force due to reciprocating mass of each cylinder FP = mr2 cos cos FS = mr2 n Unbalanced couple CP = mr2l cos cos2 CS mr2l n Analytical Method: For complete primary balancing (i) mr cos = 0 (ii) mr sin = 0 (iii) mrl cos = 0 (iv) mrl sin = 0 Up to three cylinder inline engines. complete primary balancing can be achieved by suitably arranging the cranks. complete primary balancing is not possible. This imaginary crank is 4n known as secondary crank. However. (i) mr cos 2 = 0 (ii) mr sin 2 = 0 (iii) mrl cos2 (iv) mrl sin 2 = 0 Balancing of V-engine V-engine is two cylinder radial engine in which the connecting rod are fixed to common crank. FP2 = mr2 cos( ) Total primary force along vertical axis OY is FPV = FP1 cos FP2 cos = 2mr2 cos2 cos Total primary force along horizontal line OX is FPH = FP1 sin FP2 sin FPH = 2mr2 sin2 sin Resultant primary force FP = (FPV )2 (FPH )2 = 2mrw2 (cos2 cos )2 (sin2 sin )2 . Y r2 i nde cylinder 1 cy l FP v Q P FS v FP C 2 FP FS FS 1 Line of stroke 2 1 of cylinder 1 X O FS H FP H Line of stroke of cylinder 2 2 V angle Primary Forces: The primary unbalanced force acting along the line of stroke of cylinder 1 is FP1 = mr2 cos( ) Similarly. Balancing of single and multi-cylinder engines S K Mondal’s Chapter 6 mr (ii) n sin 2 = 0 mrl (iii) n cos 2 = 0 mrl (iv) n sin 2 = 0 If n is same for all cylinder. Hence there is no couple. The Primary force FP = mr2 cos can be imagining as. mr sin 2 2 m r 2 prim ary direct crank 2 FP = mr cos m r 2 2 m 2 Primary 2 r sin reverse crank c Secondary Force cos2 FS = mr2 can be assumed as n . Balancing of single and multi-cylinder engines S K Mondal’s Chapter 6 The angle made by resultant force FP with vertical axis OY (measured in clockwise direction) F P = tan 1 PH FPV Secondary Forces: cos2( ) FS1 = mr2 n cos 2( ) FS2 = mr2 n FSV = FS1 cos FS2 cos 2 = mr2 cos cos2 cos2 n FSH = FS1 sin FS2 sin 2 = mr2 sin sin 2 sin 2 n Resultant secondary forces FS = (FSH )2 (FS V )2 and angle 1 FSH S = tan FS V Direct and Reverse Crank In a radial engine and V-engine all the connecting rods are connected to a common crank and this crank revolves in one plane. Unbalanced force along the line of stroke for cylinder 1 = (1 C) mr2 cos Similarly for 2nd cylinder = (1 C) mr2 cos(90 ) Resultant unbalanced force along the line of stroke (FT ) = (1 C) mr2 cos (1 C) mr2 cos(90 ) FT (1 C)mr2 (cos sin ) FT will be maximum or minimum at d FT = 0 FTmax = 2 (1 C)mrw2 . Therefore the angle of inclination for the second crank will be (90 + ). Since the crank for the second cylinder is at right angle to the first crank. . C = Fraction of the reciprocating parts to be balanced. Previous are see that the reciprocating parts are only partially balanced. (ii) Swaying couple The maximum magnitude of the unbalanced force along the perpendicular to the line of stroke is known as hammer blow. The effect of an unbalanced primary force along the line of stroke is to produce. along the line of stroke is known as tractive force. (i) Variation in tractive force. Let m = Mass of reciprocating parts per cylinder. (i) Variation of tractive force The resultant unbalanced force due to the two cylinders. Let the crank of the for the first cylinder be inclined at an angle with the line of stroke. Due to this partial balancing of reciprocating parts there is an unbalanced primary force along the line of stroke and also an unbalanced primary force perpendicular to the line of stroke. Balancing of single and multi-cylinder engines S K Mondal’s Chapter 6 2 r ) 4n m (2 ar y 2 Sec ond a nk cr dire ct 2 2 F5 –2 2 2 r Secondary m (2 ) n reverse crank 2 4 Partial balancing of locomotive: The locomotive usually have two cylinder with crank placed at right angles to each other. d Maximum or minimum FT at = 135° or 315° (ii) Swaying Couple The unbalanced force along the line of stroke for two cylinder constitute a couple about the centre line y y between the cylinder. This is known as swaying couple. D-Alembert’s Principle Thus D-Alembert‘s principle states that the resultant force acting on a body together with the reversed effective force (or inertia force). engine) may be determined by graphical method or analytical method. This principle is used to reduce a dynamic problem into an equivalent static problem. Klien‘s construction .Ritterhaus‘s construction. may be determined by one of the following constructions: 1. Klien’s Construction .2. 1 Max value = (1 C) mr 2a 2 a 90 + Hammer Blow We have already discussed that the maximum magnitude of the unbalanced force along the perpendicular to the line of stroke is known as hammer blow. by graphical method. Velocity and Acceleration of the Reciprocating Parts in Engines The velocity and acceleration of the reciprocating parts of the steam engine or internal combustion engine (briefly called as I. are in equilibrium. Bennett‘s construction. and 3.C. The velocity and acceleration. Balancing of single and multi-cylinder engines S K Mondal’s Chapter 6 2 (1– c)mr cos a 2 Y Y a 2 2 (1– c)mrcos(90 + ) a a Swaying couple = (1 C) mr2 cos (1 C) mr2 cos ( 90) 2 2 a = (1 C) mr2 (cos sin ) 2 It will be maximum at v = 45° or 225°. 3 cm and 0. The respective eccentricity of each masses measured from axis of rotation is 0. 170 kg and 85 kg respectively. The bearing of a shaft A and B are 5 m apart. D and E. as shown in Figure. The shaft carries three eccentric masses C. Let the crank marks an angle with the line of stroke PO and rotates with uniform Angular velocity rad /s in a clockwise direction.5 cm. Balancing of single and multi-cylinder engines S K Mondal’s Chapter 6 Let OC be crank and PC the connecting rod of a reciprocating stream engine. [ IES-2002] Solution: C D E R. Which are 160 kg. a D 2 OD2 Q. 0. 3m and 4m respectively. Determine angular position of each mass with respect to C so that no dynamic force is exerted at B and also find dynamic force at A for this arrangement when the shaft runs at 100 rpm.3 m.6 cm and distance from A is 1. The Klien‘s velocity and acceleration diagrams are drawn discussed below: aPC 2 CN Velocity of D. VD OD1 Acceleration of D.P A B l1 l2 l3 . 6 cm 1 = 0 = 10.88 N 13. Balancing of single and multi-cylinder engines S K Mondal’s Chapter 6 D 3 2 C 1 = 0 = 10. Mass of reciprocating part = 40 kg Mass of revolving parts = 30 kg at crank pin .47 r/sec Couple balancing X-comp m1r1l1 cos 1 m2r2l 2 cos 2 m3r3l 3 cos 2 0 = 0 104 153 cos 2 204 cos 3 = 0 …(i) Y-comp m1r1l1 sin 1 m2r2l 2 sin 2 m3r3l 3 sin 3 = 0 153 sin 2 204 sin 3 = 0 … (ii) Squaring and adding cos ( 3 2 ) = 0.2° … (iii) Putting in (ii) 101.3° (with C) Q.15 3 = 226. The following data relate to a single cylinder reciprocating engine.6° 24.868 3 2 = 150.6 F = 63.5 cm r2 = 0.47 r/sec E m1 = 160 kg m2 = 170 kg m3 = 85 kg r1 = 0.6 tan = 62.3 cm r3 = 0.41 y -comp Fsin m1r1 sin 12 m2r22 sin 2 m3r32 sin 3 = 0 F sin = 13.41 = 192.85° Let the dynamic force at A is F Force balance F cos m1 r1 cos 12 m2r22 cos 2 m3r32 cos 3 = 0 F cos = 62.18 tan 2 = 2 = 76. r = 175 mm (i) Mass to be balanced at crank pin M = 0. rb = m. (ii) The variation in tractive effort (iii) The maximum swaying couple. 320 = 54 175 m b = 29.7)2 sin 45]2 = 880. Balancing of single and multi-cylinder engines S K Mondal’s Chapter 6 Speed = 150 rpm Stroke = 350 mm If 60% of the reciprocating parts and all the revolving parts are to be balanced.5 km/hr = 26. Hammer blow = 46 103 N . r = 0.7 rad/sec. between cylinder centre lines = 0.3 m. D = 1. V = 96.175 15.53 kg. The following data refer to two cylinder locomotive with crank 90°. w = 29. = 1. Let M b = balancing mass placed at each driving wheel at a radius rb . Dead centre.5 km/hr.8 m Dist. (ii) Unbalanced force at = 45° = [(1 C) mr2 cos ]2 [cmr 2 sin ]2 = [(1 0.6 40 30 = 54 kg mb . if the hammer blow is not to exceed 46 × 103 N at 96. determine the (i) Balanced mass required at a radius of 320 mm. between the driving wheel = 1. a = 0.78 r/sec.7 N Q. Solution: m = 300 kg.72 cos 45]2 [0. Reciprocating mass per cylinder = 300 kg Crank radius = 0. (ii) Unbalanced force when the crank has turned 45° from the top.175 15.65 m Dist. .65 m.6 40 0.55 m Determine (i) The fraction of the reciprocating masses to be balanced.8 m/sec. r mb .8 m.6) 40 0. Solution: = 15.3 m Driving wheel dia. Let C = fraction of reciprocating masses to be balanced Mass to be balanced = 300C kg. Idealizing the actual mechanical device. 2.87C … (ii) From (i) and (ii) VIBRATORY MOTIONS in machinery arise when variable forces act on elastic parts. and dampers.65 0. m b rb = 69 C Now (i) Hammer blow = m brb2 . Balancing of single and multi-cylinder engines S K Mondal’s Chapter 6 A D Wheel B C Wheel 0.1 mb rb 1. ANALYSING VIBRATIONS requires the following general procedure: 1. for example) they are deliberately designed into the machine. 5.45 0. 0 . Solving the equations and interpreting the results.55m C B ut 1 = 0° 2 = 1 + 9 0° For couple balance X-comp: m1r1l1 cos 1 m2r2l 2 cos 2 mb rbl b cos = 0 300C. Writing differential equations of motion for the idealized system.45 0 mbrb .45 1. Evaluating amount of friction involved. Usually these motions are undesirable although in some cases (vibratory conveyors.55 cos = 0 mb rb cos = 26. 3. springs. Evaluating masses and elasticity of parts involved.1.1 1.3. replacing it by an approximately equivalent system of masses.55 sin = 0 mb rb sin = 63.13C … (i) Y-comp: m1r1l1 sin 1 m2r2l 2 sin 2 mbrbl b sin = 0 0 300C 0. 0. 4.3 1. 1 f Hz. Fig. DEFINITIONS (i) Free (Natural) Vibrations: Elastic vibrations in which there are no friction and external forces after the initial release of the body are known as free or natural vibrations. Oscillating force F(t) applied to freedom system showing the single-degree-of-freedom the mass static deflection due to weight system Time period: It is the time taken by motion to repeat itself or it is the time require to complete one cycle. Model of a single-degree-of- Fig. 2 tP sec. shaft or beam is displaced from the equilibrium position by application of external forces and then release if commences cyclic motion. the vibrations are said to be force. Frequency: The no. Representation of a Fig. GATE & PSU) Vibrations A body is said to vibrate if has a to-and-fro motion When any elastic body such as spring. Linear Vibration Analysis of Mechanical Systems Theory at a glance (IES. The vibrations gradually cease and the system rests in its equilibrium position. of cycle per unit time is known as frequency. The frequency of the vibrations is that of the applied force and is independent of their own natural frequency of vibrations. (ii) Damped Vibrations: When the energy of a vibrating system is gradually dissipated by friction and other resistances. (iii) Forced Vibrations: When a repeated force continuously acts on a system. t P 2 . the vibrations are said to be damped. Such cyclic motion due to elastic deformation under the action of external forces is known as vibration. Linear Vibration Analysis of Mechanical Systems S K Mondal’s Chapter 7 7. (c)]. a rod. Figure. the vibrations are said ti be transverse. X = Max.M X P x t x = : t O A A 2A 3A 4A 2 Let x = displacement of a point from mean position after time‗t‘. (i) Longitudinal Vibrations: If the shaft is elongated and shortened so that same moves up and down resulting in tensile and compressive stresses in the shaft. The system can execute the following types of vibrations. Such state is known as resonance. e.H.H.. S. The particles of the body move approximately perpendicular to its axis. the vibrations are said to be longitudinal. (ii) Transverse Vibrations: When the shaft is bent alternately [Figure (b)] and tensile and compressive stress due to bending result. the vibrations are known as torsional vibrations. (iii) Torsional Vibrations: When the shaft is twisted and untwisted alternately and torsional shear stresses are induced. The particles of the body in a circle about the axis of the shaft [Figure. shaft or spring.g. Linear Vibration Analysis of Mechanical Systems S K Mondal’s Chapter 7 Stiffness of spring (K): F = K F = force applied on spring N = deflection of spring Damping Coefficient (C): F C= N-sec/m V F = Force applied on damper V = velocity of viscous fluid m/sec. Shows a massless shaft end of which is fixed and the other end carrying a heavy disc.M . the body starts vibrating with excessively large amplitude. Resonance: When the frequency of external excitation force acting on a body is equal to the natural frequency of a vibrating body. TYPES OF VIBRATIONS Consider a vibrating body. The different particles of the body move parallel to the axis of the body [figure (a)]. Displacement of point from mean position x = X sin X sin t x = cos t x = 2 sin t x = 2 x or x 2x 0 Equation of S. mg m1g m2g Ke K1 K2 1 1 1 K e K1 K 2 2. Springs in parallel: K1 Ke K1 K2 m m K2 m mg = m1g m2g and 1 2 K e K11 K 22 K e K1 K 2 Natural Frequency of Free Longitudinal Vibrations The natural frequency of the free longitudinal Vibrations may be determined by the following three methods: 1. Linear Vibration Analysis of Mechanical Systems S K Mondal’s Chapter 7 Equivalent Springs 1. Equilibrium Method 2. Springs in series: K1 Ke K2 m m 1 2 and mg = mg = m2g . Equilibrium Method . Energy method 3. Rayleigh‘smethod 1. -s.) … (i) … (Taking upward force as nega5tive) And Accelerating force = Mass Acceleration d2x =m 2 … (Taking downward force as positive) dt Equating equations (i) and (ii). Linear Vibration Analysis of Mechanical Systems S K Mondal’s Chapter 7 Consider a constraint (i. is balanced by a force of spring. .x = 0 …(iv) dt 2 Comparing equations (iii) and (iv). It is usually expressed in N/m. . m = Mass of the body suspended from the constraint in kg. such that W= s.e. . as shown in Figure(c).s. spring) of negligible mass in an unstrained position. Let s = Stiffiness of constraint.x = -s.) t p 2 m 2 Taking the value of g as 9. W = Weight of the body in newtons = m. we have s = m 2 m Time period.g. Natural frequency of free longitudinal vibrations In the equilibrium position. the gravitational pull W= m. Since the mass is now displaced from its equilibrium position by a distance x.g = s. as shown in Figure (a). Fig.81 m/s2 and in metres. fn …( m.x or m 2 + s.s ( + x) = W – s.x = s. in metres.s. . as shown in Figure (b). and is then released. It is the force required to produce unit displacement in the direction of vibration. and x = Displacement given to the body by the external force.g.x … ( W = s.s. t p =2 s 1 1 s 1 g And natural frequency.x = 0 dt dt 2 dx s + x 0 dt 2 m We know that the fundamental equation of harmonic motion is d2x + 2. the equation of motion of the body of mass m after time is d2x d2x m 2 = . Restoring force =W. = Static deflection of the spring in metres due to weight W newtons. therefore after time t. Linear Vibration Analysis of Mechanical Systems S K Mondal’s Chapter 7 1 9.81 0.4985 fn Hz 2 Note: The value of static deflection may be found out from the given conditions of the problem. For longitudinal vibrations, it may be obtained by the relation, Stress W 1 W.t E or E or Strain A E.A Where = Static deflection i.e. extension of the conatraint, W = load attached to the free end of constraint, l = length of the constraint, E = Young‘s modulus for the constraint, and A = Cross-sectional area of the constraint. 2. Energy method Energy Method: 1 K. E of mass = m x2 2 F Kx 1 mg Kx P. E of spring = Kx 2 2 K g m x Total energy= constant K. E + P. E = constant d (K.E P . E) =0 dt 1 1 d m x 2 Kx 2 2 2 =0 m x Kx 0 dt Torsional Stiffness (K t ) : T G IP Kt = = l G = modulus of rigidity of shaft material N/m2 IP = Polar moment of inertia of shaft 4 = d 32 l Length of the shaft d = diameter of the shaft Linear Vibration Analysis of Mechanical Systems S K Mondal’s Chapter 7 I ·· (Inertia torque) ,· , ·· d l O A A A A Kt (Restoring torque) (Inertia torque + External torque) = 0 I Kt . 0 Q. K1 = 2400 N/m, K 2 = 1600 N/m, K 3 = 3600 N/m K 4 K 5 = 500 N/m. Find mass m such that system have natural frequency of 10 Hz. K1 757.89 1757.89 Ke m. K2 m 1000 K3 m K4 K5 1 Ke fn = 2 m m 0.422 kg Q. Determine natural frequency. Linear Vibration Analysis of Mechanical Systems S K Mondal’s Chapter 7 Io ·· r o ·· mx M T Kx = kr m K Angular displacement of pulley = Angular velocity of pulley = Angular Acceleration of pulley = Linear displacement of mass = x = r Linear velocity of mass = x r Linear Acceleration of mass = x r Equilibrium Method: Consider linear motion of mass ‗m‘. (Inertia force + External force)= 0 mx T 0 T = m x Consider rotary motion of pulley. (Inertia force + External torque)= 0 I0 Kr r T.r 0 I0 Kr 2 m x r 0 I0 Kr 2 mr 2 0 (I0 mr 2 ) K r 2 0 Kr 2 1 =0 I0 Mr 2 I0 mr 2 2 Kr 2 0 Mr 2 2 mr 2 K 0 M m 2 K 2n = M m 2 E of spring = Kx 2 = K r 2 2 2 2 1 2 1 1 Total energy = mr 2 I0 2 Kr 2 2 2 2 2 d(Total energy) 0 dt 1 1 1 mr 2x I0 x Kr 2 2 = 0 2 2 2 mr 2 I0 Kr 2 0 Mr 2 mr 2 Kr 2 0 2 M 2 m K 0 K 0 M m 2 K Wn M m 2 1 K fn 2 M m 2 Q. Linear display of mass = x1 r Difference of spring x 2 = R . E of pulley = I0 2 1 1 P. . Let angular display of pulley = . E of the mass = m x 2 = mr 2 2 2 2 1 2 K. Linear Vibration Analysis of Mechanical Systems S K Mondal’s Chapter 7 K n = M m 2 1 K fn 2 M m 2 Energy Method: 1 1 K. E of rotary pulley = I02 I0 mr2 2 2 2 4 1 1 P. E of the spring = Kx 2 Kr 2 2 2 2 .2 KR2 . 1 1 K . Let angular Display pulley = Linear display of pulley = x = r Linear display of spring = x. E of spring = Kx 22 KR22 2 2 d(Total energy) 0 dt 1 1 1 mr 2 2 MR2 . E.E of linear pulley = m x2 mr 2 2 2 2 1 1 2 1 K.2 0 2 4 2 MR2 mr 2 KR2 0 2 KR2 Wn MR2 mr 2 2 Q. E of pulley = I0 MR 2 2 4 1 1 P. Linear Vibration Analysis of Mechanical Systems S K Mondal’s Chapter 7 x2 K r R M m x1 Now 1 1 2 K. x K r o m x In this case pulley has both linear as well as angular display hence it has 2 K. E of mass = m x12 mr 2 2 2 1 2 1 2 K. Therefore the P. E = Ax g x d(T E) Now =0 dt 2g x x =0 l 2g n l Damped Free Vibration . E of display. Find the natural frequency of oscillation of liquid column when tube is slightly displaced. Linear Vibration Analysis of Mechanical Systems S K Mondal’s Chapter 7 d(T E) Now =0 dt 3 mr 2 2 Kr 0 2 2K 0 3m 2K n 3m Q. P. x x l The liquid column is displaced from equilibrium position through a distance x. 1 K. The liquid of mass Ax is raised through a distance x. Therefore each particle in liquid column has a velocity x at that instant. E of fluid = m x 2 2 1 = Al x 2 2 density A = cross area of column. 2 = rn r 2n2 n2 . eSt x S2 eSt Above equation will be mS2 CS K 0 C C2 4Km S1. Linear Vibration Analysis of Mechanical Systems S K Mondal’s Chapter 7 Kx Cx· K C m m x ·· mx C = damping coefficient N. 2 = 2m 2 C C K = 2m 2m m The general solution to the different equation. x = AeS1t BeS2t A and B are constant.s/m (Intertia force + External force) = 0 m x C x Kx 0 Above equation is a linear differential equation of the second order and its solution can be written in the form x eSt x S . Critical damping coefficient (CC ) : the critical coefficient CC is that value of damping coefficient at which 2 CC K 2m m = 0 2 CC K 2m = m CC K = = n 2m m CC 2mn CC 2 Km Damping Factor or Damping ratio (r): C r CC General solution of differential equation: S1. x0 2 1 ) nt x+ Öx – A e (– X t Be (–x– Öx2 –1) nt 2. under damped system (r < 1) : S1 = [ r i 1 r 2 ]n S2 = [ r i 1 r 2 ]n 1 r2 ) nt 1 r2 ) nt x Ae( r i Be( r i = e rWn t [AeiWd t Be iWd t ] x X e rn t sin( d t ) X. over damped system (r > 1): 2 2 x = Ae(–r r 1 )nt Be( r r 1 )nt The value of A and B can be determined from initial condition At t = 0. . Linear Vibration Analysis of Mechanical Systems S K Mondal’s Chapter 7 = [ r r 2 1] n S1 = ( r r 2 1)n S2 = ( r r 2 1)n 1. Critically damped system (r < 1) : S1 n S2 n nt x = (A Bt)e X – (A + Bt)e nt t 3. are constant. x = Xo t = 0. X X Xn 1 = log e 0 = log e 1 = …………. leading us to the concept of ―logarithmic decrement‖ or It is defined as the logarithm of the ratio of any two success amplitudes on the same side of the mean position. The successive peak amplitudes bear a certain specific relationship involving the damping of the system. log e X1 X2 Xn X X Xn 1 n = log e 0 log 1 ………… log e X1 X2 Xn 1 X log e 0 n Xn Now X1 = Xe rnt1 sin( d t1 ) X 2 = Xe rnt2 sin( d t2 ) = Xe rnt2 sin[ d (t1 tP ) ] = Xe rn (t1 tP ) sin[d t1 d tP ] . Linear Vibration Analysis of Mechanical Systems S K Mondal’s Chapter 7 2A tp tp = a –xn t Xe X x1 x sin x2 d ( 1 r2 )n d = natural frequency of damped vibration r=1 r>1 r=0 r<1 Logarithmic Decrement The decrease in a amplitude from one cycle to the next depends on the extent of damping in the system. Equation of motion for x = 2 x2 1 ]n t x2 1 ]n t x = Ae[ x Be[ x x Ae2. (i) 160 N-S/m (ii) 80 N-S/m Solution: M = 4 kg K = 400 N/m K n = = 10 rad/sec.0215 B = 1. Linear Vibration Analysis of Mechanical Systems S K Mondal’s Chapter 7 2 = Xe rn ( t1 tP ) sin d t1 d Wd rn (t1 t P ) = Xe sin( dt1 ) X = log 1 X2 X1 Xe rn ( t1 ) sin( d t1 ) = X2 Xe rn ( t1 tP ) sin( d t1 ) = erWn tP 2 = rn t P = rn .54 103 .02 at t = 0. A spring mass dashpot system consists of spring of stiffness 400 N/m and the mass of 4 kg.67t 37. m C.32B 0 A = 0.67 Ae2. x = 0 2.32 Be37.32t x = 2.32t Now at t = 0.02 m A + B = 0. If the damping coefficient of the dashpot is. Find the equation of motion of the mass. x = 0.67A 37. The mass is displaced 20 mm beyond the equilibrium position and released. 160 x1 = = =2 CC 2 4 400 C2 x2 = =1 CC 1. Wd 2 = rn n ( 1 r2 ) 2r 1 r2 Damped Free Torsional vibration: I Ct K t 0 All term are torque Q.67t Be37. 2te10t Q.02 A = 0. x = 0.936 CC 2mWn K m C Expression for displacement: Since r > 1 2 r2 1 ]nt x = Ae[ r r 1 ]nt Be[ r = Ae( 10. K = 15 × 103 N/m C = 1500. t = 0.14t Substitution t = 0 and x = 0.01 … (i) Substitution t = 0 and x = 10 10.77 Ae 10. x = 0 0 = 10A B B = 0.67t 1.54 10 3 e 37. Equation of motion for x = 1 x (A Bt)ent = (A Bt)e 10t = Ae10t Bte 10t x = 10Ae 10t 10Bte 10t Be 10t Substituting at.01m and released with a velocity of 10 m/s in the direction of return motion Find (i) An expression for the displacement x of the mass interms of times ‘t’ and 1 (ii) the displacement of mass after sec.01 m At t = 0.14Be 139.02e10t 0.77t 139.14)t x = 10.32t 2. a viscous damper having damping coefficient 1500 N-S/m and a mass of 10 kg.02 And at t = 0.72 rad/sec.14 B 10 . 100 Solution: m = 10 kg. Linear Vibration Analysis of Mechanical Systems S K Mondal’s Chapter 7 x 0. The mass is displaced by 0.2 x 0. has a spring stiffness of 15 kN/m. x = 0.215 e2. m C C r = = 1. x = 10 m / s K Now Wn = = 38.01 A + B = 0.77)t Be( 139. The system shown in fig. N-S/m At t = 0.77 A 139. x xC x P x C = Complementary function x P = Particular integral The complementary function is obtained by considering no forcing function i.0670 B = 0. second order differential equation.077e (139. 2x 1 n = tan 2 1 n . The solution of this equation consist two parts.0770 x 0.14 )t 1 At t = 100 x 0.0670e(10.041 m Negative sign indicates the displacement is on opposite side of mean position. Forced Damped Vibrations: K C Kx Cx· m m m·· x x Fo sin t Fo sin t (Inertia force + External force) = 0 m x c x Kx = Fo sin t The equation is a linear. Then Fo X= K 2 2 2 1 2x n n Fo Where = X st = deflection due to force Fo or zero frequency deflection or static K deflection. Linear Vibration Analysis of Mechanical Systems S K Mondal’s Chapter 7 A = 0.77)t 0.e. Angular frequency of external exciting hammons force rad/sec. m x c x Kx 0 which is same as previous (3) cases Now x P X sin( t ) X = amplitude of steady state vibration = Angle by which the displacement vector lags force vector (phase angle). When there is no. n 2. 5. At zero frequency of excitation (i. At very high frequency of excitation.F is maximum when 1.7 0 7 x= 1 x=2 0 1 2 3 Frequency ratio ( / n) Observation made from frequency response curve: 1. For x more than 0.F is below unity. when = 0) the M.5 1 x= 0 . As the x decreases. X M.F = X St 1 M. the M. Linear Vibration Analysis of Mechanical Systems S K Mondal’s Chapter 7 Solution x = xC x P Fo sin( t ) xP = 2 2 2 k 1 2x n n Magnification Factor: It is defined as the ratio of the amplitude of steady state vibration ‗X‘ to the zero frequency deflection Xst .F . In other words. 3. the M.F reaches infinity at 1. the damping does not have any effect on M. 6.1 25 x=0 2 M.F tends to zero.F = 2 2 2 1 2x n n 4 x=0 3 .707. n 4.F at zero frequency of excitation.e. . the condition is known as resonance.F increases. damping ( x 0). the maximum M. 25 x=0 x = 0. the maximum value of M.F is unity for all value of x . The M. 4. higher the damping factor. 0 .5 x=0 150° 0 .2 5 x= . Solution: m = 80 kg K = 1000 N/m . The phase angle varies from 0° at low frequency ratio to 180° at very high frequency ratio. A spring mass damper system has a mass of 80 kg suspended from a spring having stiffness of 1000 N/m and a viscous damper with a damping coefficient of 80 N-s/m. If the mass is subjected to a periodic disturbing force of 50 N at undamped natural frequency. 2. Q.7 07 x= x=1 120° 2 x= Phase angle ( ) 90° 60° 07 2 1 0 . At frequency ratio less than unity. higher is the phase n angle.7 x=0 x= x= x= 30° 0. If there is no damping ( x 0) the phase angle is either 0° or 180° and at resonance the phase angle suddenly change from 0° to 180°. At resonance frequency ( n ) the phase angle is 90° and damping does not have any effect on phase angle. Linear Vibration Analysis of Mechanical Systems S K Mondal’s Chapter 7 Phase angle ( ) VS frequency ratio n 180° 0 . whereas at frequency greater than unity higher the damping factor lower n is the phase angle. (iv) The phase difference. 3. 5 x x= 25 = 0 x=0 1 2 3 ( /n ) 1. The variation in phase angle is because of damping. (ii) The damped natural frequency (wd ) (iii) The amplitude of forced vibration of mass. determine (i) The undamped natural frequency (wn ) . 53 rad/sec. Linear Vibration Analysis of Mechanical Systems S K Mondal’s Chapter 7 C = 80 N-sec/m Fo = 50 N K n = = 3. m C C x = = = 0.1765 m. = angular velocity of rotation of unbalanced mass.49 rad/sec. Transmissibility: (TR ) The ratio of the force transmitted (FT) to the force applied (F0 is known as the isolation factor or transmissibility ratio of the spring support. Fo X= K [here = n ] 2 2 1 2x n n = 0. Vibration isolation Force transmitted to the foundation Force transmissibility = Force impressed upon the system FT TR Fo . Fig.1416 CC 2mn wd = ( 1 x2 )n = 3. 2x 1 n = tan 2 1 n = 90° Forced variation due to rotating unbalance: In this case Fo m oe2 m o = unbalanced mass E = eccentricity of the unbalanced mass. 12 5 x=0 0 1 2 3 Spring Damping Mass controlled controlled control region region region Frequency ratio ( /n ) 1. the transmissibility is greater than one. When 2. n . n 3. Frequency ratio n 4 x = 0. This means the transmitted n force is always greater than the impressed exciting force. When = 1. the transmissibility is maximum. Linear Vibration Analysis of Mechanical Systems S K Mondal’s Chapter 7 2 1 2x n TR 2 2 2 1 2 x n n X Motion transmissiblity Y X = Absolute amplitude of the mass (body) Y = Amplitude of the base excitation. 25 x=0 . n 2. 2 = x x=1 x =2 1 x=2 x=1 x = 0. The transmissibility tends to zero as the frequency ratio tend to infinity. All the curves start from the unit value of transmissibility and pass through the unit value of transmissibility at frequency ratio = 2. 4.125 3 x=0 2 TR 5 0. Also. Find the natural frequency of vibration for the system shown in fig. n Q. mgl 3 = 3EI Stiffness of cantilever beam is mg 3EI K1 = 3 l Then the system is converted into K1 KC K m m 1 1 1 K e K1 K 3EIK Ke = KL3 3EI 1 Ke 1 3EIK fn = = Hz 2 m 2 m(KL3 3EI) (i) When K = 1 1 1 3EI = K e = K1 = Ke K1 K l3 1 3EI fn = 2 ml 3 (ii) When I = . Neglect the mass of cantilever beam. Linear Vibration Analysis of Mechanical Systems S K Mondal’s Chapter 7 5. In order to have low value of transmissibility. the operation of vibrating system generally kept 2. find the natural frequency of vibration when: [ IAS – 1997] (i) K = (ii) I = EI l K m Solution: The deflection of cantilever beam due to weight mg. Linear Vibration Analysis of Mechanical Systems S K Mondal’s Chapter 7 1 1 1 L3 1 = = Ke K1 K 3E .383 10 3 m. The m/c has rotating unbalanced equal to 250 kg mm. The stiffness of the spring is changed such that the static deflection is still the same with the same viscous damper as in earlier case. Determine the dynamic amplitude.354 105 N / m. Determine the change in the dynamic amplitude. w = 125. Change in dynamic amplitude = X Xnew = 0. [IES-2007] Solution: m = 500 kg = 5 103 mo e = 250 10 3 N = 1200 rpm.4 44. K Ke = K 1 K fn = .81 105 N / m K n = = 44.16 10 3 m . It is supported on helical springs which deflect by 5 mm due to the weight of the m/c.543 10 3 m. m Fo X= K 2 2 2 1 2x n n 250 10 3 125.29 44.29 rad/s. 1700 Fo Xnew = K 2 2 1 2x n n = 0.4.66 2 125. The speed of the m/c is 1200 rpm. 2 m Q.66 rad/sec. x = 0. 33.81 105 = 2 125.66 2 9. Now this m/c is mounted on a larger concrete block of mass 1200 kg. .66 2 1 2 0. The damping factor of the viscous damper is 0.29 = 0. Now if m = 500 + 1200 = 1700 kg. A m/c of mass 500 kg.4 mg Now K = = 9.354 105 n = = 44. K = 33.29 rad/sec. Critical speeds or whirling of Shaft S K Mondal’s Chapter 8 8. Critical speeds or whirling of Shaft Theory at a glance (IES, GATE & PSU) Critical Speed Critical speeds are associated with uncontrolled large deflections, which occur when inertial loading on a slightly deflected shaft exceeds the restorative ability of the shaft to resist. Shafts must operate well away from such speeds. Rayleigh‘s equation e KY 2 m MG (e + y) G O s G O, s Fig. Two Position of the rotor Critical Speed of Shafts Critical speeds or whirling of Shaft S K Mondal’s Chapter 8 The speed at which the shafts starts to vibrate violently in the direction perpendicular to the axis of the shaft is known as critical speed or whirling speed. 1. Critical speed of shaft carrying single rotor (without damping): 2 e C y 2 1 C y = deflection of geometric centre due to C.F e = eccentricity of the rotor C = critical speed of shaft. K g C m 2. Critical speed of shaft carrying single rotor (with damping). 2 e y = C 2 2 2 1 2x C C Let 0 = point of intersection of bearing centre line with the rotor. S = geometric centre of the rotor. G = centre of gravity of the rotor. = phase angle between e and y. 2x 1 C = tan 2 1 C G G s S O O << c < c G ° 90 = G S S O O = c > c Critical speeds or whirling of Shaft S K Mondal’s Chapter 8 S 0.4 >>c Multi Rotor Fig. Typical multi-rotor system 1. Rayleigh’s Method: g mi y i wC = mi y2i 2. Dunkerley’s Method: 1 1 1 1 2 = 2 2 ........ C ( C1 ) ( C2 ) ( C5 )2 Objective Questions (IES, IAS, GATE) Previous 20-Years GATE Questions GATE-1. An automotive engine weighing 240 kg is supported on four springs with linear characteristics. Each of the front two springs have a stiffness of 16 Gear Train and Gear Design S K Mondal’s Chapter 9 Gear Train and Gear Design (Same chapter is added to Machine design booklet) Theory at a glance (GATE. intersecting (axis) or Non-intersecting non parallel shafts. including pinion wire. IAS & PSU) Spur gear Basic Purpose of Use of Gears Gears are widely used in various mechanisms and devices to transmit power And motion positively (without slip) between parallel. There are a number of variations of the basic spur gear. rack and internal gears. stem pinions. A SPUR GEAR is cylindrical in shape. • Without change in the direction of rotation • With change in the direction of rotation • Without change of speed (of rotation) • With change in speed at any desired ratio Often some gearing system (rack – and – pinion) is also used to transform Rotary motion into linear motion and vice-versa. IES. . Fig. with teeth on the outer circumference that are straight and parallel to the axis (hole). Pitch circle .Layout of a pair of meshing spur gears Figure. the smaller gear is called the „pinion‟.Spur gear schematic showing principle terminology For a pair of meshing gears. Gear Train and Gear Design S K Mondal’s Chapter 9 Figure. the larger is called the „gear wheel‟ or simply the „gear‟. (a). Various modules are illustrated in figure. . Its diameter is called the pitch diameter. Backlash. Backlash BACKLASH is the distance (spacing) between two “mating” gears measured at the back of the driver on the pitch circle. This is the radial distance from the pitch circle to the bottom land. d m T Addendum. Module. The module is defined by the ratio of pitch diameter and number of teeth. Gear Train and Gear Design S K Mondal’s Chapter 9 This is a theoretical circle on which calculations are based. and T is the number of teeth. Dedendum. m the module. m is the module (mm). The unit of the module should be millimeters (mm). abnormal wear and excessive heat while providing space for lubrication of the gears. The backlash for spur gears depends upon (i) module and (ii) pitch line velocity. and T the Number of teeth. This is the radial distance from the pitch circle to the outside of the tooth. d = mT Where d is the pitch diameter (mm). Clearance (C) is the amount by which the dedendum in a given gear exceeds the addendum of its mating gear. d p m T Where p is the circular pitch (mm). d the pitch diameter (mm). which is purposely built in. Care must be taken to distinguish the module from the unit symbol for a meter. This is the ratio of the pitch diameter to the number of teeth. Factor affected by changing center distance is backlash. Typically the height of a tooth is about 2. (b). Circular pitch This is the distance from a point on one tooth to the corresponding point on the adjacent tooth measured along the pitch circle. is very important because it helps prevent noise.25 times the module. Gear Train and Gear Design S K Mondal’s Chapter 9 Fig. Figure.Schematic showing the pressure line and pressure angle . Line of Action – Line tangent to both base circles .Schematic of the involute form Pitch Circle and pitch point Fig. Gear Train and Gear Design S K Mondal’s Chapter 9 Figure. Fig. Pitch Circle – Circle with origin at the gear center and passing through the pitch point. . Gear Train and Gear Design S K Mondal’s Chapter 9 Pitch Point – Intersection of the line of centers with the line of action Fig. The pressure angle of a spur gear normally varies from 14° to 20° The value of pressure angle generally used for involute gears are 20° Relationship Between Pitch and Base Circles rb r cos Fig. 1. 14 1 0 Composite system. Pressure angle – Angle between the line normal to the line of centers and the line of action. Gear Train and Gear Design S K Mondal’s Chapter 9 Fig. The following four systems of gear teeth are commonly used in practice. 2 . (d) Double enveloping worm gearing. 4. (f) Spiroid and helicon gearing. racks have their teeth cut into the surface of a straight bar instead of on the surface of a cylindrical blank. Non-parallel. 2. The tooth profile of the 14 1 2 full depth involute system was developed for use with gear hobs for spur and helical gears. non. 1. (g) Face gears (off centre). Unlike the basic spur gear. Gear Train and Gear Design S K Mondal’s Chapter 9 2. 2 3. The increase of the pressure angle 0 from 14 1 2 to 20° results in a stronger tooth. . coplanar gears (intersecting axes): (a) Bevel gears (b) Face gears. because the tooth acting as a beam is wider at the base. It is stronger but has no interchangeability.coplanar gears (nonintersecting axes): (a) Crossed axis helical (b) Cylindrical worm gearing (c) Single enveloping worm gearing. The tooth profile of the 20° full depth involute system may be cut by hobs. RACK RACKS are yet another type of spur gear. The tooth profile of this system has cycloidal curves at the top and bottom and involute curve at the middle 0 portion. Special gear types: (a) Square and rectangular gears. Classification of Gears Gears can be divided into several broad classifications. 200 Full depth involute system 4 200 Stub involutes system. (c) Conical involute gearing. 14 1 0 Full depth involute system. 0 The 14 1 2 composite system is used for general purpose gears. Non-parallel. (b) Elliptical gears. Parallel axis gears: (a) Spur gears (b) Helical gears (c) Internal gears. (e) Hypoid gears. 3. The 20° stub involute system has a strong tooth to take heavy loads. The teeth are produced by formed milling cutters or hobs. In case of single helical gears there is some axial thrust between the teeth.e. which is a disadvantage. Gear Train and Gear Design S K Mondal’s Chapter 9 Fig. herringbone gears) are used. In order to eliminate this axial thrust. double helical gears (i. in which equal and opposite thrusts are provided on each gear and the resulting axial thrust is zero. Herringbone gears Figure-Herringbone gear . Rack Helical gear The helical gears may be of single helical type or double helical type. It is equivalent to two single helical gears. in turn. results in impact conditions and generates noise.Herringbone gear Figure. resulting in a sudden application of the load which. the contact between meshing teeth occurs along the entire face width of the tooth. Gear Train and Gear Design S K Mondal’s Chapter 9 Figure.Crossed axis helical gears In spur gears. . There is a gradual pick-up of load by the tooth. Worm Gear Fig. the contact between meshing teeth begins with a point on the leading edge of the tooth and gradually extends along the diagonal line across the tooth. High ratios are easy to obtain with the hypoid gear system. Hypoid pinions may have as few as five teeth in a high ratio set. . resulting in smooth engagement and silence operation. Fig. Fig. Gear Train and Gear Design S K Mondal’s Chapter 9 In helical gears. Ratios can be obtained with hypoid gears that are not available with bevel gears. Bevel Gears Fig. Hypoid Gears Hypoid gears resemble bevel gears and spiral bevel gears and are used on crossed-axis shafts. The distance between a hypoid pinion axis and the axis of a hypoid gear is called the offset. The geometry of hypoid teeth is defined by the various dimensions used to set up the machines to cut the teeth. Figure. Gear Train and Gear Design S K Mondal’s Chapter 9 Hypoid gears are matched to run together. They differ from spiral-bevel gears in that the axis of the pinion is offset from the axis of the gear. Hypoid gears are similar in appearance to spiral-bevel gears. Fig.Hypoid gear . just as zero or spiral bevel gear sets are matched. Gear Train and Gear Design S K Mondal’s Chapter 9 Figure.Comparison of intersecting and offset-shaft bevel-type gearings Figure-Epicyclic gears Mitres gear . therefore. When equal bevel gears (having equal teeth) connect two shafts whose axes are mutually perpendicular. both gears always have the same number of teeth.Miter gears Minimum Number of Teeth on the Pinion in Order to Avoid Interference The number of teeth on the pinion (Tp) in order to avoid interference may be obtained from the following relation: 2 Aw Tp 11 G 1 2 sin 1 2 G G Where AW = Fraction by which the standard addendum for the wheel should be Multiplied. o 1 In full depth 14 degree involute system. the smallest number of teeth in a pinion which meshes 2 with rack without interference is 32. = Pressure angle or angle of obliquity. Gear Train and Gear Design S K Mondal’s Chapter 9 Miter gears are identical to bevel gears except that in a miter gear set. Their ratio. . (generally AW = 1) G = Gear ratio or velocity ratio = TG / TP = DG / DP. then the bevel gears are known as mitres. is always 1 to 1. miter gears are not used when an application calls for a change of speed. Minimum number of teeth for involute rack and pinion arrangement for pressure angle of 20° is 2 AR 2 1 Tmin 17. Figure. Tmin 18 sin 2 sin 2 20o The minimum number of teeth on the pinion to operate without interference in standard full height involute teeth gear mechanism with 20° pressure angle is 18. As a result.1 as 17 So. Fig. . 3. the interference does not occur at all. In cycloidal gears. In cycloidal gears. if a circle rolls without slipping on the inside of a fixed circle. whereas in involute gears. cycloidal teeth of a gear Advantages of cycloidal gears Following are the advantages of cycloidal gears: 1. the convex surfaces are in contact. However the difference in wear is negligible. Due to this reason. On the other hand. then the curve traced by a point on the circumference of a circle is called hypocycloid. which rolls on the circle without slipping or by a point on a taut string which is unwrapped from a reel as shown in figure below. When a circle rolls without slipping on the outside of a fixed circle. 2. the cycloidal teeth are preferred especially for cast teeth. the curve traced by a point on the circumference of a circle is known as epicycloid. Gear Train and Gear Design S K Mondal’s Chapter 9 Forms of teeth Cycloidal teeth A cycloid is the curve traced by a point on the circumference of a circle which rolls without slipping on a fixed straight line. the contact takes place between a convex flank and concave surface. This condition results in less wear in cycloidal gears as compared to involute gears. Though there are advantages of cycloidal gears but they are outweighed by the greater simplicity and flexibility of the involute gears. In connection with toothed wheels. Since the cycloidal teeth have wider flanks. the circle is known as base circle. Involute teeth An involute of a circle is a plane curve generated by a point on a tangent. therefore the cycloidal gears are stronger than the involute gears for the same pitch. reduces to zero at pitch point. starts . Figure-involute teeth The tooth profile most commonly used in gear drives for power transmission is an involute. 2. But in cycloidal gears. It is necessary for smooth running and less wear of gears. The most important advantage of the involute gears is that the centre distance for a pair of involute gears can be varied within limits without changing the velocity ratio. In involute gears. from the start of the engagement of teeth to the end of the engagement. remains constant. Advantages of involute gears Following are the advantages of involute gears: 1. It is due to easy manufacturing. the pressure angle. the pressure angle is maximum at the beginning of engagement. This is not true for cycloidal gears which require exact centre distance to be maintained. Gear Train and Gear Design S K Mondal’s Chapter 9 Fig. This may be avoided by altering the heights of addendum and dedendum of the mating teeth or the angle of obliquity of the teeth.e. 3.25. The zone of action of meshing gear teeth is shown in figure above. As shown. The contact ratio for gears is greater than one. Contact ratio Note: The ratio of the length of arc of contact to the circular pitch is known as contact ratio i. In figure above initial contact occurs at a and final contact at b. The ratio of the length of arc of contact to the circular pitch is known as contact ratio i. Tooth profiles drawn through these points intersect the pitch circle at A and B. We recall that tooth Contact begins and ends at the intersections of the two addendum circles with the pressure line.e. the distance AP is called the arc of approach (qa). epicycloids and hypocycloid) are required for the face and flank respectively. Gear Train and Gear Design S K Mondal’s Chapter 9 increasing and again becomes maximum at the end of engagement. The sum of these is the arc of action (qt). number of pairs of teeth in contact.00 or higher for best results. For maximum smoothness and quietness.e. Note: The only disadvantage of the involute teeth is that the interference occurs with pinions having smaller number of teeth. Contact ratio should be at least 1. length of arc of contact Contact ratio = circular pitch RA2 R 2 cos 2 rA2 r 2 cos 2 ( R r )sin = Pc (cos ) Fig. respectively. High-speed applications should be designed with a face-contact ratio of 2.50 and 2. . In involute system. number of pairs of teeth in contact. the contact ratio should be between 1. and the distance P B. The face and flank of involute teeth are generated by a single curve whereas in cycloidal gears. double curves (i. This results in less smooth running of gears. the basic rack has straight teeth and the same can be cut with simple tools. the arc of recess (qr ).00. Thus the involute teeth are easy to manufacture than cycloidal teeth. Increased centre distance. so that more of the tooth profile becomes involute. Face Width . Contact begins when the tip of the driven tooth contacts the flank of the driving tooth. Interference can also be reduced by using a larger pressure angle. Modified addendum iv. The actual effect is that the involute tip or face of the driven gear tends to dig out the noninvolute flank of the driver. However. and this occurs before the involute portion of the driving tooth comes within range. more teeth can be used only by increasing the pitch diameter. There are several ways to avoid interfering: i. This results in a smaller base circle. contact is occurring below the base circle of gear 2 on the noninvolute portion of the flank. Modified involutes iii. In other words. if the pinion is to transmit a given amount of power. In this case the flank of the driving tooth first makes contact with the driven tooth at point A. Fig. Increase number of gear teeth ii. Gear Train and Gear Design S K Mondal’s Chapter 9 Interference The contact of portions of tooth profiles that are not conjugate is called interference. The demand for smaller pinions with fewer teeth thus favors the use of a 25 0 pressure angle even though the frictional forces and bearing loads are increased and the contact ratio decreased. Interference can be eliminated by using more teeth on the pinion. Gear Train and Gear Design S K Mondal’s Chapter 9 Face width. It is the width of the gear tooth measured parallel to its axis. Face width. We know that face width, b = 10 m Where, m is module. Fig. Fig. Face width of helical gear. Beam Strength of Gear Tooth The beam strength of gear teeth is determined from an equation (known as Lewis equation) and the load carrying ability of the toothed gears as determined by this equation gives satisfactory results. In the investigation, Lewis assumed that as the load is being transmitted from one gear to another, it is all given and taken by one tooth, because it is not always safe to assume that the load is distributed among several teeth, considering each tooth as a cantilever beam. Notes: (i) The Lewis equation am applied only to the weaker of the two wheels (i.e. pinion or gear). Gear Train and Gear Design S K Mondal’s Chapter 9 (ii) When both the pinion and the gear are made of the same material, then pinion is the weaker. (iii) When the pinion and the gear are made of different materials, then the product of w y or o y is the deciding factor. The Lewis equation is used to that wheel for which w y or o y is less. Figure- Tooth of a gear The maximum value of the bending stress (or the permissible working stress): w WT h t / 2 (WT h) 6 b.t 3 / 12 b.t 2 Where M = Maximum bending moment at the critical section BC = WT× h, WT = Tangential load acting at the tooth, h = Length of the tooth, y = Half the thickness of the tooth (t) at critical section BC = t/2, I = Moment of inertia about the centre line of the tooth = b.t3/12, b = Width of gear face. Lewis form factor or tooth form factor WT w b pc y w b m y The quantity y is known as Lewis form factor or tooth form factor and WT (which is the tangential load acting at the tooth) is called the beam strength of the tooth. Lewis form factor or tooth form factor 0.684 y 0.124 ,for 14 1 2 0 composite and full depth involute system. T 0.912 0.154 ,for 200 full depth involute system. T 0.841 0.175 ,for 200 stub system. T Gear Train and Gear Design S K Mondal’s Chapter 9 Example: A spur gear transmits 10 kW at a pitch line velocity of 10 m/s; driving gear has a diameter of 1.0 m. find the tangential force between the driver and the follower, and the transmitted torque respectively. Solution: Power transmitted = Force Velocity 10 103 Force 10 10 103 Force 1000 N / m 10 diameter Torque Transmitted Force 2 1 1000 1000 0.5 2 500N m 0.5 kN m Wear Strength of Gear Tooth Wear strength = bQdpK, 2 Tg Where, Q for external gear Tg Tp 2 Tg for internal gear Tg Tp 2c sin cos 1 1 load - stress factor k 1.4 E1 E2 2 BHN 0.16 100 Gear Lubrication All the major oil companies and lubrication specialty companies provide lubricants for gearing and other applications to meet a very broad range of operating conditions. General gear lubrication consists of high- quality machine oil when there are no temperature extremes or other adverse ambient conditions. Many of the automotive greases and oils are suitable for a broad range of gearing applications. For adverse temperatures, environmental extremes, and high-pressure applications, consult the lubrication specialty companies or the major oil companies to meet your particular requirements or specifications. The following points refer especially to spiral and hypoid bevel gears: (a) Both spiral and hypoid bevel gears have combined rolling and sliding motion between the teeth, the rolling action being beneficial in maintaining a film of oil between the tooth mating surfaces. (b) Due to the increased sliding velocity between the hypoid gear pair, a more complicated lubrication system may be necessary. 1 is angular velocity of gear 1. their pitch circles roll on each other without slippage.Simple gear train Compound gear train . When two gears are in mesh. Gear Train and Gear Design S K Mondal’s Chapter 9 Simple Gear train A gear train is one or more pairs of gears operating together to transmit power. If r1 is pitch radius of gear 1. and 2 is angular velocity of gear 2 then the pitch line velocity is given by V r11 r2 2 The velocity ratio is 1 r 2 2 r1 Figure. r2 is pitch radius of gear 2. Gear Train and Gear Design S K Mondal’s Chapter 9 Figure -Compound gear train The velocity ratio in the case of the compound train of wheels is equal to Product of teeth on the followers Product of teeth on the drivers The velocity ratio of the following gear train is Figure.velocity ratio N T T T F A C E N T T T A B D F Reverted gear train . Gear Train and Gear Design S K Mondal’s Chapter 9 Figure. in addition to the previous ratio equations. to Provide for the in-line condition on the input and output shafts. as shown in Fig above. This requires the distances between the shafts to be the same for both stages of the train. From the figure rA rB rC rD or TA +TB = TC+TD and as NB+NC it must be TB =TD & TA=TC NB N Or = D or NC = NAND NA NC where NB = N c Epicyclic gear train Consider the following Epicyclic gear trai .reverted gear train or a compound reverted gear train It is sometimes desirable for the input shaft and the output shaft of a two-stage compound gear train to be in- line. Replacing All the diameters give T2 / 2P T3 / 2P T4 / 2P T5 / 2P Assuming a constant diametral pitch in both stages. The distance constraint is d d d d 2 3 4 5 2 2 2 2 The diametric pitch relates the diameters and the numbers of teeth. In the compound gear train shown in the figure. P = T/d. gears A and C have equal numbers of teeth and gears B and D have equal numbers of teeth. we have the geometry condition Stated in terms of numbers of teeth: T2 + T3 = T4 + T5 This condition must be exactly satisfied. This configuration is called a compound reverted gear train. so n planet / n arm NS n sum / n arm NP n planet n arm NS n sum n arm NP n planet 200 20 1000 ( 200) 30 20 n planet 800 200 333 rpm 30 The speed of rotation of the planetary gears is 333 rpm counter-clockwise. Gear Train and Gear Design S K Mondal’s Chapter 9 Figure. 30 and 80. The speed and direction of the sun gear is 1000 rpm clockwise and the ring gear is held stationary. determine the speed and direction of the final drive and also the speed and direction of the planetary gears. To determine the speed of the planets use The planets and sun are in mesh. Now make a table for the epicyclic gear arrangement shown in the figure below. . The teeth numbers of the sun. planets and ring gear are 20.Epicyclic gear train For the epicyclic gearbox illustrated in figure. The reduction ratio for the gearbox is Given by nsun/narm = 1000/200 = 5. respectively. Solution nsun 1000 narm 200 rpm 80 / 20 1 5 The speed of the final drive is 200 rpm clockwise. 25m . Spur & Helical 6:1 to 10:1 for high speed helical For high speed spur. × x N3 N5 N3 Formula List for Gears: (a) Spur Gear Name Speed ratio 1. Gear Train and Gear Design S K Mondal’s Chapter 9 Arm 2 3 4 5 -N 2 -N 2 -N 4 N 2 1. Bevel 1:1 to 3:1 3. 0 +x x x × x N3 N3 N5 N3 2. 2. x y. y y y y y N2 N4 N2 y x+y y. Worm 10:1 to 100:1 provided 100 KW SPUR GEAR d (i) Circular pitch (p) = T T (ii) Diametral pitch (P) = d (iii) pP = d 1 (iv) Module (m) = = or d = mT T P ωp Tg (v) Speed ratio (G) = = ωg Tp 1 (vi) centre-to-centre distance = 2 dg + d p 1 = 2 m Tg + Tp (vii) Addendum ha = 1m hf =1. 912 Lewis form factor. Y = 0. usually b = 10m (xii) Beam strength b = mbσ b Y Lewis equation σ ult Where σ b = 3 0. for 20° full depth gear. Gear Train and Gear Design S K Mondal’s Chapter 9 Clearance (C) = 0.Tp σ 2c sin cos 1 1 load .4 E1 E2 2 BHN = 0.154 .stress factor k = + 1. pinion = 11 G 1 + + 2 sin2 1 GG zg p G = Gear ratio = zp g ha A p = fraction of addendum to module = for pinion m h Aw = fraction of addendum to module = f for gear (generally 1) m (xi) Face width 8m <b <12m.25 m 2T (viii) Pt = d Pr = Pt tan Pt PN = cos (ix) Minimum number of teeth to avoid interference 2A w Tmin = For 20° full depth T = 18 to 20 sin2 (x) 2 × Aw Tmin.16 100 (xiv) . z (xiii) Wear strength = bQdpK 2 Tg Where Q for external gear Tg + Tp 2 Tg for internal gear Tg . for ordinary cut gear v < 10m / s Cv 3+V 6 = . Gear Train and Gear Design S K Mondal’s Chapter 9 Cs 3 Peff = Pt where Cv = .6 = . .00 +1.6 + v (xv) Spott's equation. for hobbed generated v > 20m / s 6+v 5. Peff = Cs Pt + Pd en pTp br1r2 where Pd = for steel pinion and steel gear 2530 r12 + r22 e = 16.29 d for all. for precision gear v > 20m / s 5.25 for grade 8 = m + 0. 2 cos T d 9 T = 3 .centre distance. p m 5 pa = = . 1. pa = axial pitch. 8 (13) Addendum ha = mn .mm d 2N tann Pr = Pt = Psinαn cos Pa = Pt tan = Pcosn sin (15) Beam strength Sb = mn bσ b Y bQd p K (16) Wear strength Sw = cos2 (17) . (11) Helix angle varies from 15 to 25. cos mn 8 a = T1 + T2 centre – to . 5.5. d = cos cos2 (10) An imaginary spur gear is considered with a pitch circle diameter of d and module m n is called „formative‟ or „virtual‟ spur gear. m mn = normal module. Mt = N .25mn (14) 2Mt 60 x106 x KW Pt = = Pcosn cos . clearence = 0. 4. (12) Preference value of normal modus mn = 1. P 2 Pn = where Pn = normal diametral pitch. tan = transverse pressure angle. helix angle. d = pitch circle diameter. pn = normal circular pitch. 2. 3. 6. 1. 3 pP = 1 4 mn = mcos P = m = transverse module. dedendum hf = 1.25. TP Tmn 7 d = = zm = . cosα P = transverse diametral pitch.25 mn . Gear Train and Gear Design S K Mondal’s Chapter 9 (b) Helical Gear 1 pn = pcosα where p = transverse circular pitch. tan tan tan n 6 cos = n = normal pressure angle usually 20° . 16.2 cos δ clearance c = 0. (v) l = px × T1 . 2. dz = mT2 (vi) Axial pitch of the worm = circular pitch of the worm wheel d 2 Px = = m [ICS .2 cos δ -1 m f1 h f2 = m 1 + 0. (xiii) F F δ = sin-1 .5. 20.6 en p Tp br1r2 Peff = .6 + v 2530 r12 + r22 Peff = Cs Pt + Pd cosα n cosψ (c) Worm Gear (i) Specified and designated by T1 / T2 / q / m d1 Where: q is diametric quotient = m (ii) The threads of the worm have an involute helicoids profile. Pd = Cv 5. (iii) Axial pitch px = distance between two consecutive teeth-measured along the axis of the worm. 12. 25 (x) Number of starts T1 usually taken 1. Cv = . 10.2m cos δ c = 0.2 m cos δ (xii) F = 2m q +1 effective face width of the root of the worm wheel. l r = d a1 + 2C sin -1 d a + 2C = length of the root of the worm wheel teeth d a + 2C 1 1 .04] T2 l = Px T1 = mT1 T l (vii) Lead angle δ = tan-1 1 = tan-1 q d1 1 1 (viii) centre-to-centre distance (a) = d1 + d2 = m T1 + T2 2 2 (ix) Preferred values of q: 8. a distance that a point on the helical profile will move. Gear Train and Gear Design S K Mondal’s Chapter 9 Cs Pt 5. (iv) The lead (l) = when the worm is rotated one revolution. or 4 (xi) addendum h a1 = m h a2 = m 2 cos δ -1 dedendum h = 2. t 0 A K t .t0 A KW = 1000 1 .μsin δ sin (xiv) P1 t = . Gear Train and Gear Design S K Mondal’s Chapter 9 2mt coscosδ . P1 r = P1 t d1 cossinδ + μcosδ cos sinδ + μcosδ Power output cos .η .μtanδ (xv) Efficiency () = = Power input cos + μtanδ d1n1 (xvi) Rubbing velocity Vs = m / s (remaining 4 cheak) 60000 cosδ (xvii) Thermal consideration Hg = 1000 1 .η × KW Hd = K t . P1 a = P1 t .