Mechanisms A mechanism is a combination of rigid or restraining bodies so shaped and connected that they move upon each other with a definite relative motion. A simple example of this is the slider crank mechanism used in an internal combustion or reciprocating air compressor. Machine A machine is a mechanism or a collection of mechanisms which transmits force from the source of power to the resistance to be overcome,and thus perform a mechanical work. Plane and Spatial Mechanisms
If all the points of a mechanism move in parallel planes, then it is defined as a plane mechanism. If all the points do not move in parallel planes then it is called spatial mechanism. Kinematic Pairs
A mechanism has been defined as a combination so connected that each moves with respect to each other.A clue to the behaviour lies in in the nature of connections,known as kinetic pairs. The degree of freedom of a kinetic pair is given by the number independent coordinates required to completely specify the relative movement. Lower Pairs
A pair is said to be a lower pair when the connection between two elementsis through the area of contact.Its 6 types are :
Revolute Pair Prismatic Pair Screw Pair Cylindrical Pair Spherical Pair Planar Pair
Revolute Pair A revolute allows only a relative rotation between elements 1 and 2, which can be expressed by a single coordinate angle 'theta' .Thus a revolute pair has a single degree of freedom.
Prismatic Pair A prismatic pair allows only a relative translation between elements 1 and 2, which can be expressed by a single coordinate 'S'.Thus a prismatic pair has a single degree of freedom.
Screw Pair A screw pair allows only a relative movement between elements 1 and 2, which can be expressed by a single coordinate angle 'theta' or 'S' .Thus a screw pair has a single degree of freedom.These two coordinates are related as : 'theta/2'pi'=S/L
Cylindrical Pair A cylindrical pair allows both rotation and translation between elements 1 and 2, which can be expressed as two independent coordinates angle 'theta' and 'S' .Thus a cylinderical pair has two degrees of freedom.
Spherical Pair A spherical pair allows three degrees of freedom since the complete description of relative movement between the connected elements needs three independent cooordinates.Two of the coordinates 'alpha' and 'beta' are required to specify the position of the axis OA and the third coordinate 'theta' describes the rotatio about the axis OA.
Planar Pair A planar pair allows three degrees of freedom.Two coordinates x and y describe the relative translation in the xy-plane and the third 'theta' describes the relative rotation about the zaxis.
Higher Pairs
A higher pair is defined as one in which the connection between two elements has only a point or line of contact. A cylinder and a hole of equal radius and with axis parallel make contact along a surface. Two cylinders with unequal radius and with axis parallel make contact along a line. A point contact takes place when spheres rest on plane or curved surfaces (ball bearings) or between teeth of a skew-helical gears. in roller bearings, between teeth of most of the gears and in cam-follower motion. The degree of freedom of a kinetic pair is given by the number independent coordinates required to completely specify the relative movement. Wrapping Pairs
Wrapping Pairs comprise belts, chains, and other such devices. To define a mechanism we define the basic elements as follows : Link
A material body which is common to two or more kinematic pairs is called a link. Kinematic Chain
A kinematic chain is a series of links connected by kinematic pairs. The chain is said to be closed chain if every u link is connected to atleast two other links, otherwise it is calledan open chain. A link which is connected to only one other link is known as singular link.If it is connected to two other links, it is called binary link.If it is connected to three other links, it is called ternary link, and so on. A chain which consists of only binary links is called simple chain. A type of kinematic chain is one with constrained motion, which means that a definite motion of any link produces unique motion of all other links. Thus motion of any point on one link defines the relative position of any point on any other link.So it has one degree of freedom.
The process of fixing different links of a kinematic chain one at a time to produce distinct mechanisms is called kinematic inversion.Here the relative motions of the links of the mechanisms remain unchanged. First, let us consider the simplest kinematic chain,i.e., achain consisting of four binary links and four revolute pairs. The four different mechanisms can be obtained by four different inversions of the chain. Slider Crank mechanism It has four binary links, three revolute pairs, one prismatic pair.By fixing links 1, 2, 3 in turn we get various inversions.
Double Slider Crank mechanism It has four binary links, two revolute pairs, two sliding pairs.Its various types are : Scotch Yoke mechanism: Here the constant rotation of the crank produces harmonic translation of the yoke.Its four binary links are : 1. 2. 3. 4. Fixed Link Crank Sliding Block Yoke
The four kinematic pairs are : 1. 2. 3. 4. revolute pair (between 1 & 2) revolute pair (between 2 & 3) prismatic pair (between 3 & 4) prismatic pair (between 4 & 1)
Oldhams Coupling:
It is used for transmitting anbgular velocity between two parallel but eccentric shafts
Elliptical Trammel: Here link 4 is fixed. Any point on the link 2 describes an ellipse as it moves.The mid-point of the link 2 will obiviously describe a circle. Very often a mechanism with higher pair can be replaced by an equivalent mechanism with lower pair.This equivalence is valid for studying only the instantaneous characteristics.The equivalent lower-pairmechanism facilitates analysis as a certain amount of sliding takes place between connecting links in a higher-pair mechanism. Another example of an equivalent lower-pair mechanism for a cam-follower system is shown.The sliding block is the additional link and thebhigher pair is replaced by two lower pairs, one revoluteand other prismatic. C is the center of curvature of the cam surface at the point of contact between the cam and the follower.
MOBILITY AND RANGE OF MOVEMENT It is important to note that an arbitary number of links connected by an arbitary number of kinematic pairs do nat result in a mechanism. Some conditions have to be satisfied for a ystem of interconnected links to serve as a useful mecahnism. The degrees of freedom, is equal to the number independent coordinates required specify its configuration, i.e., the relative positions of all the links.Let n be the no. of links in a mechanism out of which, one is fixed, and let j be the no. of simple hinges(ie, those connect two links.) Now, as the (n-1) links move in a plane, in the absence of any connections, each has 3 degree of freedom; 2 coordinates are required to specify the location of any reference point on the link and 1 to specify the orientation of the link. Once we connect the linmks there cannot be anyrelative translation betweenthem and only one coordinate is necessary to specify their relative orientation.Thus, 2 degrees of freedom (translation) are lost, and only one degree of freedom (rotational) is left. So, no. of degrees of freedom is: F=3(n-1)-2j Most mechanisms are constrained, ie F=1. Therefore the above relation becomes, 2j-3n+4=0 ,this is called Grubler's Criterion. Failure of Grubler's criterion A higher pair has 2 degrees of freedom .Following the same argument as before, The degrees of freedom of a mechanism having higher pairs can be written as, F=3(n-1)-2j-h Often some mechanisms have a redundant degree of freedom. If a link can move without causing any movement in the rest of the mechanism, then the link is said to have a redundant degree of freedom. Example of redundant degree of freedom
Displacement analysis of plane mechanisms.The objective of kinematic analysis is to determine the kinematic quantities such as displacements, velocities, and accelerations of the elements of a mechanism when the input
motion is given. It establishes the relationship between the motions of various components of the linkage. When the kinematic dimensions and the configurations of the input link of a mechanism are prescribed, the configurations of all the other links are determined by displacement analysis. 1. Graphical Method 2. Analytical Method
Graphical MethodIn a graphical method of displacement analysis, the mechanism is drawn to a convenient scale and the desired unknown quantities are determined through suitable geometrical constructions and calculations. 1. The configurations of a rigid body in plane motion are completely defined by the locations of any two points on it. 2. Two intersecting circles have two points of intersection and one has to be careful, when necessary, to choose the correct point for the purpose in hand. 3. The use of tracing paper, as an overlay, is very convenient and very often provides an unambiguous and quick solution. 4. The graphical method fails if no closed loop with four links exists in the mechanism.
Analytical MethodAn analytical method of displacement analysis, is preferred whenever 1. high level of accuracy is required 2. a large number of configurations have to be solved 3. The graphical method fails. In this method every link is represented by two dimensional are represented by two dimensional vectors expressed through complex notation. Considering each closed loop in the mechanism, a vector equation is established. Separating the real and imaginary parts , sufficient number of nonlinear algebraic equations are obtained to solve for the unknown quantities. Let us consider a 4R linkage of given link lengths, viz., i=1, 2, 3, and 4. The configuration of the input link (2) is also prescribed by the angle 2, and we have to determine the configurations of the other two links, namely, the coupler and the follower, expressed by the angles 3 and 4. Referring to Figure, all links are denoted as vectors, viz., l1, l2, l3 and l4. All angles are measured CCW from the X-axis which is along the fixed vector l1, rendering 1=0. Considering the closed loop O2O4BAO2, we can write Using complex exponential notation with 1=0 can be written as Equating the real and imaginary parts of this equation separately to zero, we get ........(1a) ........(1b)
Thus, the two unknowns, namely, 3 and 4 can be solved from the two equations (1a) and (1b) as now explained. Rearranging (1a) and (1b), we get
Squaring both sides of these two equations and adding, we obtain Or Where , ...............(2)
It may be noted that with the prescribed data, the coefficients a, b, and c of (2) are known. To solve for 4, from (2) without ambiguity of quadrant, it is better to substitute
In (2) to yield Thus, for a given position of the input link, two different values of 4 are obtained as follows:
These two values correspond to the two different ways in which the 4R mechanism can be formed for any given value of 2, as explained in Figure where the same problem has been solved by a graphical method. To solve for the coupler orientation 3, we can eliminate 4 from (1a) and (1b) to get
, Where
TRANSMISSION ANGLE
For a 4R linkage, the transmission angle ( ) is defined as the acute angle between the coupler (AB) and the follower ).4B), as indicated in Fig. 2.11. If ( - ABO 4) is acute (Fig.2.11), then =- ABO4. On the other hand, if - ABO4 is obtuse, then =-ABO4. As explained in this figure, if = /2, then the entire coupler force is utilized to drive the follower. For good transmission quality, the minimum value of (min)>300. For a crank-rocker mechanism, the minimum value of occurs when the crank becomes collinear with the frame, i.e., . If the swing angle ( ) of the rocker is increased maintaining the same quick-return ratio, then the maximum possible value of min decreases. If the forward and return strokes of the rocker take equal time, then (min)max is restricted to (see Problem 2.6). Therefore,
such a crank rocker will have a poor transmission quality if INSTANTANEOUS CENTRE OF VELOCITY
In the figure a rigid body 2 is shown to be in plane motion with respect to fixed link 1. The velocities of two points A and B of the rigid link are shown by VA and VB ,respectively. Two lines drawn through A and B in directions perpendicular to VA and VB meet at P. Let PA=r1 and PB=r2. The velocity of point B in the direction of AB is VBcos, and that of point A in the same direction is VAcos. As the length of AB is fixed, the component of VBA in the direction of AB is zero. Thus VBcos=VAcos. From the triangle PAB, we have
Thus the velocities of the points A and B are proportional and perpendicular to PA and PB, respectively. So ,instantaneously the rigid body can be thought of as being momentarily on pure rotation about the point P.The velocity of any point C on the body at this instant is given by VC=PC.VB/r2 in a direction perpendicular to PC .The point P is called instantaneous centre of velocity, and its instantaneous velocity is zero. Alternatively instantaneous centre of velocity of velocity can be described as a point which has no velocity with respect to the fixed link. If both links 1 and 2 are in motion, in a similar manner we can define relative instantaneous centre of velocity P12 to be a point on 2 having zero velocity with respect to a coincident point on 1.Consequently the relative motion of link 2 with respect to 1 appears to be in pure rotation about P12. Thus if a mechanism has N links, the number of instantaneous centers possible are N (N1)/2. 1. If two links have a hinged joint, the location of the hinge is the relative instantaneous centre because one link is in pure rotation with respect to the other about that hinge. 2. If relative motion between two links is pure sliding, the relative instantaneous centre lies at infinity on a line perpendicular to the direction of sliding. 3. If one link is rolling over another, the point of contact is the relative instantaneous centre. 4. If a link is sliding over a curved element, the centre of curvature is the relative instantaneous centre. 5. If the relative motion between two links is both rolling and sliding the relative instantaneous centre lies on the common normal to the surfaces of these links passing through the point of contact.
ARNOLD KENNEDY THEORM OF THREE CENTRES VELOCITY AND ACCELARATION ANALYSIS (ANALYTICAL)Once the configuration of the mechanism is known, the velocities and acceleration are linear in the unknown quantities and hence are easy to solve. Consequently when the velocity and the acceleration analysis have to be carried out for a large number of configurations, the analytical method turns out to be more advantageous than the graphical method. Accuracy obtained in analytical method is also high. Referring to the precious figure, assume that the configuration of the mechanism has already been determined, i.e., l1,l2, l3,l4, and 2 are prescribed and 3, 4are solved. The task is to determine the angular velocity and acceleration of the coupler and the follower if those of the crank are given. Towards this end differentiate with respect to time and obtain (2a)
(2b) We should note that the above equations are two simultaneous equations in the two unknown i.e., and which can be solved to yield
The concept of velocity and acceleration images is used extensively in the kinematic analysis of mechanisms having ternary, quaternary, and higher-order links. If the velocities and accelerations of any two points on a link are known, then, with the help of images the velocity and acceleration of any other point on the link can be easily determined. An example is illustrated below:
Once the velocity analysis is complete, (2.34a) and (2.34b) again provide two linear equations in and which are obtained as
VELOCITY AND ACCELERATION IMAGESThe concept of velocity and acceleration images is used extensively in the kinematic analysis of mechanisms having ternary, quaternary, and higher-order links. If the velocities and accelerations of any two points on a link are known, then, with the help of images the velocity and acceleration of any other point on the link can be easily determined. An example is illustrated below: A rigid link BCDE having four hinges is sown in figure. Let the angular velocity and acceleration of this be and . The absolute velocity vectors of the E, B, C and D are shown in the figure as VE, VB, VC, and VD respectively. The velocity difference vectors are And their magnitudes are, respectively,
So,
Hence the velocity diagram bcde is a scale drawing of the link BCDE. The figure bcde is called the velocity image of the link BCDE. The velocity image is rotated through 90o in the direction , as all the velocity difference vectors are perpendicular to the corresponding lined. The scale of the image is determined by and therefore the scale will be different for each link of a mechanism. The letters identifying the end points of the image are in the same sequence as that in the link diagram BCDE. The absolute velocity any point X on the link is obtained by joining the image of X(x) with the pole of the velocity diagram o.
VELOCITY ANALYSIS (GRAPHICAL)1. Instantaneous Centre Method 2. Relative Velocity Method
INSTANTANEOUS CENTRE METHOD 1. First determine the number of instantaneous centers (N) by using the relation
2. Make a list of all the instantaneous centers in the mechanism. 3. Locate the fixed and permanent instantaneous centers by inspection. 4. Locate the remaining neither fixed nor permanent instantaneous centers by Kennedys theorem. This can be done by circle diagram. Mark points on a circle equal to the number of links in a mechanism. 5. Join the points by solid lines to show that these circles are already found. In the lines indicate the instantaneous centers corresponding to those particular two points. 6. In order to find the remaining instantaneous centers, join two such points that the line joining them forms two adjacent triangles in the circle diagram. The line which is responsible for completing two triangles should be a common side to the two triangles.
RELATIVE VELOCITY METHODThe relative velocity method is based upon the velocity of the various points of the link. Consider two points A and B on a link. Let the absolute velocity of the point A i.e. VA is known in magnitude and direction and the absolute velocity of the point B i.e. VB is known in direction only. Then the velocity of B may be determined by drawing the velocity diagram as shown.
1. Take some convenient point o, known as the pole. 2. Through o, draw oa parallel and equal to VA, to some convenient scale. 3. Through a, draw a line perpendicular to AB. This line will represent the velocity of B with respect to A, i.e. 4. Through o, draw a line parallel to VB intersecting the line of VBA at b. 5. Measure ob, which gives the required velocity of point B to the scale. ACCELARATION ANALYSIS Consider two points A and B on the rigid link. The acceleration of the point A, i.e. aA is known in magnitude and direction and the direction of path of B is given. The acceleration of the point B is determined in magnitude and direction by drawing the acceleration diagram as discussed below:
1. From any point o , draw vector o a parallel to the direction of absolute at point A i.e. to some suitable scale as shown in figure. 2. We know that the acceleration of B with respect A i.e.aBA has the following two components: Radial component of acceleration B with respect to A i.e.arBA and Tangential component of acceleration of B with respect to A i.e.at BA These two components are mutually perpendicular. 3. Draw vector a x parallel to the link AB such that 4. From point x, draw vector xb perpendicular to AB or vector a x and through o draw a line parallel to the path of the path of to represent the absolute acceleration of B i.e. aB. the vectors xb and a b intersect at b . Now the values of aB and atBA may be measured to the scale.
5. By joining the points a and b we may determine the total acceleration of B with respect A i.e. aB. The vector a b is known acceleration image of the link AB.
CORIOLLIS COMPONENT OF ACCELERATION
In a mechanism, a link is quite often guided along a prescribed path in another moving link. For the velocity and acceleration analyses of such a mechanism, the differences in the velocities and accelerations of two instantaneously coincident points belonging to the two links have to be determined. In this section, we shall derive the expressions for these quantities. Figure shows a rotating rigid link (labeled 2) on which link 3 is moving along a straight line. The configurations at the instants t and (t+dt) are, respectively, shown by the symbols without and with a prime. Further, P2 and P3 represent two points on links 2 and 3, respectively, coincident at the instant t. The displacement of P3 can be written as Where P2P2 represents the displacement of P2 and P2 P3 represents the displacement of P3 with respect to link 2. Dividing both sides of the foregoing equation by t and taking the limit , we get
Where VP3/2 is the velocity of P3 as seen by an observer attached to link 2. The direction of VP3/2 is tangential to the path of P3 in link 2.
From figure we see that the block has moved through an additional transverse distance AP3 because of rotation of link 2 and the radial motion of link 3 with respect to link 2. When From this equation we observe that the additional transverse distance is proportional to the square of the time elapsed. Therefore this displacement must be due to an additional acceleration of P3 in the transverse direction. If the magnitude of this additional is ac then
In vector notation, the above equation can be represented as This extra transverse component of acceleration is known as the Coriollis component. The final expression will then be
It should be noted that the direction ac is obtained by rotating VP3/2 through 90 degrees in the sense of 2. For a straight path of P3 on link 2, the direction of is along the straight line. When link 3 moves along a curvilinear path on the rotating link 2, the above equation can be written in terms of the components of as
It may be noted that the magnitude of is equal to where is the radius of curvature of the path of P3 on link 2. the direction of VP3/2 is obviously tangential to this path. For a straight line path of P on link 2, becomes infinite and .
