Velocity in Mechanisms

velocity in mechanisms relative velocity method and velocity in mechanisms instantaneous centre method, velocity in mechanism by relative velocity method pdf free download
HartJohnson Profile Pic
HartJohnson,United States,Professional
Published Date:02-08-2017
Your Website URL(Optional)
Comment
 Features Features Features Features Features Velocity in Velocity in Velocity in Velocity in Velocity in            Mechanisms Mechanisms      Mechanisms Mechanisms Mechanisms      " (Instantaneous Centre Method) (Instantaneous Centre Method) (Instantaneous Centre Method) (Instantaneous Centre Method) (Instantaneous Centre Method)       6.1. 6.1. 6.1. 6.1. 6.1. Introduction Introduction Introduction Introduction Introduction % &"              ' (           )             +,-(     . . /(    0                                   +        +  +        "                                                 +  + ′         +        +      "   +          +  +′  +          +′   119 120 Theory of Machines A B . Such a motion of link AB to A 1 1 1 B is an example of combined motion 1 of rotation and translation, it being immaterial whether the motion of rotation takes first, or the motion of translation. In actual practice, the motion of link AB is so gradual that it is difficult to see the two separate motions. But we see the two separate motions, though the point B moves faster than the point A. Thus, this Mechanisms on a steam automobile engine. combined motion of rotation and translation of the link AB may be assumed to be a motion of pure rotation about some centre I, known as the instantaneous centre of rotation (also called centro or virtual centre). The position of instantaneous centre may be located as discussed below: Since the points A and B of the link has moved to A and B 1 1 respectively under the motion of rotation (as assumed above), there- fore the position of the centre of rotation must lie on the intersection of the right bisectors of chords A A and B B . Let these bisectors intersect 1 1 at I as shown in Fig. 6.2, which is the instantaneous centre of rotation or virtual centre of the link AB. From above, we see that the position of the link AB goes on changing, therefore the centre about which the motion is assumed to take place (i.e. the instantaneous centre of rotation) also goes on chang- ing. Thus the instantaneous centre of a moving body may be defined as that centre which goes on changing from one instant to another. The locus of all such instantaneous centres is known as centrode. A line Fig. 6.2. Instantaneous centre of rotation. drawn through an instantaneous centre and perpendicular to the plane of motion is called instantaneous axis. The locus of this axis is known as axode. 6.2. Space and Body Centrodes A rigid body in plane motion relative to a second rigid body, supposed fixed in space, may be assumed to be rotating about an instantaneous centre at that particular moment. In other words, the instantaneous centre is a point in the body which may be considered fixed at any particular moment. The locus of the instantaneous centre in space during a definite motion of the body is called the space centrode and the locus of the instantaneous centre relative to the body itself is called the body centrode. These two centrodes have the instantaneous centre as a common point at any instant and during the motion of the body, the body centrode rolls without slipping over the space centrode. Fig. 6.3. Space and body centrode. Let I and I be the instantaneous centres for the 1 2 two different positions A B and A B of the link A B 1 1 2 2 1 1 after executing a plane motion as shown in Fig. 6.3. Similarly, if the number of positions of the link A B are considered and a curve is drawn passing through these instantaneous centres (I , I ....), then 1 1 1 2 the curve so obtained is called the space centrode. Chapter 6 : Velocity in Mechanisms 121 Now consider a point C to be attached to the body or link A B and moves with it in such 1 1 1 a way that C coincides with I when the body is in position A B . Let C be the position of the 1 1 1 1 2 point C when the link A B occupies the position A B . A little consideration will show that the 1 1 1 2 2 point C will coincide with I (when the link is in position A B ) only if triangles A B C and 2 2 2 2 1 1 1 A B C are identical. 2 2 2 ∴ A C = A I and B C = B I 1 2 2 2 1 2 2 2 In the similar way, the number of positions of the point C can be obtained for different 1 positions of the link A B . The curve drawn through these points (C , C ....) is called the body 1 1 1 2 centrode. 6.3. Methods for Determining the Velocity of a Point on a Link Though there are many methods for determining the velocity of any point on a link in a mechanism whose direction of motion (i.e. path) and velocity of some other point on the same link is known in magnitude and direction, yet the following two methods are important from the subject point of view. 1. Instantaneous centre method, and 2. Relative velocity method. The instantaneous centre method is convenient and easy to apply in simple mechanisms, whereas the relative velocity method may be used to any configuration diagram. We shall discuss the relative velocity method in the next chapter. 6.4. Velocity of a Point on a Link by Instantaneous Centre Method The instantaneous centre method of analysing the motion in a mechanism is based upon the concept (as discussed in Art. 6.1) that any displacement of a body (or a rigid link) having motion in one plane, can be considered as a pure rotational motion of a rigid link as a whole about some centre, known as instantaneous centre or virtual centre of rotation. Consider two points A and B on a rigid link. Let v and Fig. 6.4. Velocity of a point on A v be the velocities of points A and B, whose directions are given a link. B by angles α and β as shown in Fig. 6.4. If v is known in A magnitude and direction and v in B direction only, then the magnitude of v may be determined by the B instantaneous centre method as discussed below : Draw AI and BI perpendicu- lars to the directions v and v respec- A B tively. Let these lines intersect at I, which is known as instantaneous cen- tre or virtual centre of the link. The complete rigid link is to rotate or turn about the centre I. Since A and B are the points on a rigid link, therefore there cannot be any relative motion between them along the line AB. Robots use various mechanisms to perform jobs. 122 Theory of Machines Now resolving the velocities along AB, v cos α = v cos β A B v cosβ° sin (90 –β) A = or = ...(i) v cosα° sin (90 –α) B Applying Lami’s theorem to triangle ABI, AI BI = sin (90°β – ) sin (90° –α) AI sin (90°β – ) = or ...(ii) BI sin (90°α – ) From equation (i) and (ii), vAI A vv AB = ==ω or ...(iii) vBI AI BI B where ω = Angular velocity of the rigid link. If C is any other point on the link, then vv v AB C == ...(iv) AI BI CI From the above equation, we see that 1. If v is known in magnitude and direction and v in direction only, then velocity of point A B B or any other point C lying on the same link may be determined in magnitude and direction. 2. The magnitude of velocities of the points on a rigid link is inversely proportional to the distances from the points to the instantaneous centre and is perpendicular to the line joining the point to the instantaneous centre. 6.5. Properties of the Instantaneous Centre The following properties of the instantaneous centre are important from the subject point of view : 1. A rigid link rotates instantaneously relative to another link at the instantaneous centre for the configuration of the mechanism considered. 2. The two rigid links have no linear velocity relative to each other at the instantaneous centre. At this point (i.e. instantaneous centre), the two rigid links have the same linear velocity relative to the third rigid link. In other words, the velocity of the instantaneous centre 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. 6.6. Number of Instantaneous Centres in a Mechanism Bar 2 The number of instantaneous centres in a constrained kinematic chain is equal to the number of possible combina- Bar 3 tions of two links. The number of pairs of links or the number 3 2 Bar 1 of instantaneous centres is the number of combinations of n Revolutes links taken two at a time. Mathematically, number of instanta- 4 neous centres, nn(–1) Ground 1 Ground 2 N = , where n = Number of links. Base 2 Four bar mechanisms. Chapter 6 : Velocity in Mechanisms 123 6.7. Types of Instantaneous Centres The instantaneous centres for a mechanism are of the following three types : 1. Fixed instantaneous centres, 2. Permanent instantaneous centres, and 3. Neither fixed nor per- manent instantaneous centres. The first two types i.e. fixed and permanent instantaneous centres are together known as primary instantaneous centres and the third type is known as secondary instantaneous centres. Consider a four bar mechanism ABCD as shown in Fig. 6.5. The number of instantaneous cen- Fig. 6.5. Types of instantaneous centres. tres (N) in a four bar mechanism is given by nn ( – 1) 4 (4 – 1) N== = 6 ... (∵ n = 4) 22 The instantaneous centres I and I are called the fixed instantaneous centres as they re- 12 14 main in the same place for all configurations of the mechanism. The instantaneous centres I and I 23 34 are the permanent instantaneous centres as they move when the mechanism moves, but the joints are of permanent nature. The instantaneous centres I and I are neither fixed nor permanent 13 24 instantaneous centres as they vary with the configuration of the mechanism. Note: The instantaneous centre of two links such as link 1 and link 2 is usually denoted by I and so on. It is 12 read as I one two and not I twelve. 6.8. Location of Instantaneous Centres The following rules may be used in locating the instantaneous centres in a mechanism : 1. When the two links are connected by a pin joint (or pivot joint), the instantaneous centre Arm moves to a track to retrive information stored there Track selector mechanism The read/write head is guided by informa- tion stored on the disk itself The hard disk is coated with a magnetic materials Computer disk drive mechanisms. Note : This picture is given as additional information and is not a direct example of the current chapter. 124 Theory of Machines lies on the centre of the pin as shown in Fig. 6.6 (a). Such a instantaneous centre is of permanent nature, but if one of the links is fixed, the instantaneous centre will be of fixed type. 2. When the two links have a pure rolling contact (i.e. link 2 rolls without slipping upon the fixed link 1 which may be straight or curved), the instantaneous centre lies on their point of contact, as shown in Fig. 6.6 (b). The velocity of any point A on the link 2 relative to fixed link 1 will be perpendicular to I A and is proportional to I A . In other words 12 12 vIA A12 = vIB B12 3. When the two links have a sliding contact, the instantaneous centre lies on the common normal at the point of contact. We shall consider the following three cases : (a) When the link 2 (slider) moves on fixed link 1 having straight surface as shown in Fig. 6.6 (c), the instantaneous centre lies at infinity and each point on the slider have the same velocity. (b) When the link 2 (slider) moves on fixed link 1 having curved surface as shown in Fig. 6.6 (d),the instantaneous centre lies on the centre of curvature of the curvilinear path in the configuration at that instant. (c) When the link 2 (slider) moves on fixed link 1 having constant radius of curvature as shown in Fig. 6.6 (e), the instantaneous centre lies at the centre of curvature i.e. the centre of the circle, for all configuration of the links. Fig. 6.6. Location of instantaneous centres. 6.9. Aronhold Kennedy (or Three Centres in Line) Theorem The Aronhold Kennedy’s theorem states that if three bodies move relatively to each other, they have three instantaneous centres and lie on a straight line. Consider three kinematic links A, B and C having relative plane motion. The number of instantaneous centres (N) is given by nn ( – 1) 3(3 – 1) N== = 3 22 where n = Number of links = 3 The two instantaneous centres at the pin joints of B with A, and C with A (i.e. I and I ) are the permanent instantaneous centres. ab ac According to Aronhold Kennedy’s theorem, the third instantaneous Fig. 6.7. Aronhold Kennedy’s centre I must lie on the line joining I and I . In order to prove this, bc ab ac theorem. Chapter 6 : Velocity in Mechanisms 125 let us consider that the instantaneous centre I lies outside the line joining I and I as shown in Fig. 6.7. bc ab ac The point I belongs to both the links B and C. Let us consider the point I on the link B. Its velocity bc bc v must be perpendicular to the line joining I and I . Now consider the point I on the link C. Its BC ab bc bc velocity v must be perpendicular to the line joining I and I . BC ac bc We have already discussed in Art. 6.5, that the velocity of the instantaneous centre is same whether it is regarded as a point on the first link or as a point on the second link. Therefore, the velocity of the point I cannot be perpendicular to both lines I I and I I unless the point I lies on the line bc ab bc ac bc bc joining the points I and I . Thus the three instantaneous centres (I , I and I ) must lie on the same ab ac ab ac bc straight line. The exact location of I on line I I depends upon the directions and magnitudes of the bc ab ac angular velocities of B and C relative to A. Drawing Pencil Winding handle to operate the device Central ring Ellipses drawn by the ellipsograph The above picture shows ellipsograph which is used to draw ellipses. Note : This picture is given as additional information and is not a direct example of the current chapter. 6.10. Method of Locating Instantaneous Centres in a Mechanism Consider a pin jointed four bar mechanism as shown in Fig. 6.8 (a). The following procedure is adopted for locating instantaneous centres. 1. First of all, determine the number of instantaneous centres (N) by using the relation nn(–1) N = , where n = Number of links. 2 4(4 – 1) N== 6 In the present case, ...(∵ n = 4) 2 2. Make a list of all the instantaneous centres in a mechanism. Since for a four bar mecha- nism, there are six instantaneous centres, therefore these centres are listed as shown in the following table (known as book-keeping table). Links 1 2 3 4 Instantaneous 12 23 34 – centres 13 24 (6 in number) 14 126 Theory of Machines 3. Locate the fixed and permanent instantaneous centres by inspection. In Fig. 6.8 (a), I 12 and I are fixed instantaneous centres and I and I are permanent instantaneous centres. 14 23 34 Note. The four bar mechanism has four turning pairs, therefore there are four primary (i.e. fixed and permanent) instantaneous centres and are located at the centres of the pin joints. Fig. 6.8. Method of locating instantaneous centres. 4. Locate the remaining neither fixed nor permanent instantaneous centres (or secondary centres) by Kennedy’s theorem. This is done by circle diagram as shown in Fig. 6.8 (b). Mark points on a circle equal to the number of links in a mechanism. In the present case, mark 1, 2, 3, and 4 on the circle. 5. Join the points by solid lines to show that these centres are already found. In the circle diagram Fig. 6.8 (b) these lines are 12, 23, 34 and 14 to indicate the centres I , I , I and I . 12 23 34 14 6. In order to find the other two instantaneous centres, 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. In Fig. 6.8 (b), join 1 and 3 to form the triangles 123 and 341 and the instantaneous centre I will lie on the intersection of I 13 12 I and I I , produced if necessary, on the mechanism. Thus the instantaneous centre I is located. 23 14 34 13 Join 1 and 3 by a dotted line on the circle diagram and mark number 5 on it. Similarly the instanta- neous centre I will lie on the intersection of I I and I I , produced if necessary, on the mecha- 24 12 14 23 34 nism. Thus I is located. Join 2 and 4 by a dotted line on the circle diagram and mark 6 on it. Hence 24 all the six instantaneous centres are located. Note: Since some of the neither fixed nor permanent instantaneous centres are not required in solving problems, therefore they may be omitted. Example 6.1. In a pin jointed four bar mecha- nism, as shown in Fig. 6.9, AB = 300 mm, BC = CD = 360 mm, and AD = 600 mm. The angle BAD = 60°. The crank AB rotates uniformly at 100 r.p.m. Locate all the instanta- neous centres and find the angular velocity of the link BC. Solution. Given : N = 100 r.p.m or AB ω = 2 π × 100/60 = 10.47 rad/s AB Fig. 6.9 Since the length of crank AB = 300 mm = 0.3 m, therefore velocity of point B on link AB, We may also say as follows: Considering links 1, 2 and 3, the instantaneous centres will be I , I and I . 12 23 13 The centres I and I have already been located. Similarly considering links 1, 3 and 4, the instantaneous 12 23 centres will be I , I and I , from which I and I have already been located. Thus we see that the centre 13 34 14 14 34 I lies on the intersection of the lines joining the points I I and I I . 13 12 23 14 34 Chapter 6 : Velocity in Mechanisms 127 v = ω × AB = 10.47 × 0.3 = 3.141 m/s B AB Location of instantaneous centres The instantaneous centres are located as discussed below: 1. Since the mechanism consists of four links (i.e. n = 4 ), therefore number of instantaneous centres, nn ( – 1) 4 (4 – 1) N== = 6 22 2. For a four bar mechanism, the book keeping table may be drawn as discussed in Art. 6.10. 3. Locate the fixed and permanent instantaneous centres by inspection. These centres are I , 12 I , I and I , as shown in Fig. 6.10. 23 34 14 4. Locate the remaining neither fixed nor permanent instantaneous centres by Aronhold Kennedy’s theorem. This is done by circle diagram as shown in Fig. 6.11. Mark four points (equal to the number of links in a mechanism) 1, 2, 3, and 4 on the circle. Fig. 6.10 5. Join points 1 to 2, 2 to 3, 3 to 4 and 4 to 1 to indicate the instantaneous centres already located i.e. I , I , I and I . 12 23 34 14 6. Join 1 to 3 to form two triangles 1 2 3 and 3 4 1. The side 13, common to both triangles, is responsible for completing the two triangles. Therefore the instanta- neous centre I lies on the intersection of the lines joining the points I 13 12 I and I I as shown in Fig. 6.10. Thus centre I is located. Mark 23 34 14 13 number 5 (because four instantaneous centres have already been located) on the dotted line 1 3. 7. Now join 2 to 4 to complete two triangles 2 3 4 and 1 2 4. The side 2 4, common to both triangles, is responsible for completing the two triangles. Therefore centre I lies on the intersection of the lines 24 joining the points I I and I I as shown in Fig. 6.10. Thus centre I 23 34 12 14 24 is located. Mark number 6 on the dotted line 2 4. Thus all the six instan- Fig. 6.11 taneous centres are located. Angular velocity of the link BC Let ω = Angular velocity of the link BC. BC Since B is also a point on link BC, therefore velocity of point B on link BC, v = ω × I B B BC 13 128 Theory of Machines By measurement, we find that I B = 500 mm = 0.5 m 13 v 3.141 B == 6.282 rad/s ∴ ω = Ans. BC IB 0.5 13 Example 6.2. Locate all the instantaneous centres of the slider crank mechanism as shown in Fig. 6.12. The lengths of crank OB and connecting rod AB are 100 mm and 400 mm respectively. If the crank rotates clockwise with an angular velocity of 10 rad/s, find: 1. Velocity of the slider A, and 2. Angular velocity of the connecting rod AB. Fig. 6.12 Solution. Given : ω = 10 rad/ s; OB = 100 mm = 0.1 m OB We know that linear velocity of the crank OB, v = v = ω × OB = 10 × 0.1 = 1 m/s OB B OB Location of instantaneous centres The instantaneous centres in a slider crank mechanism are located as discussed below: 1. Since there are four links (i.e. n = 4), therefore the number of instantaneous centres, nn ( – 1) 4 (4 – 1) N== = 6 22 Bearing block Pin Slider Connecting rod Crank Slider crank mechanism. 2. For a four link mechanism, the book keeping table may be drawn as discussed in Art. 6.10. 3. Locate the fixed and permanent instantaneous centres by inspection. These centres are I , 12 I and I as shown in Fig. 6.13. Since the slider (link 4) moves on a straight surface (link 1), there- 23 34 fore the instantaneous centre I will be at infinity. 14 Note: Since the slider crank mechanism has three turning pairs and one sliding pair, therefore there will be three primary (i.e. fixed and permanent) instantaneous centres. Chapter 6 : Velocity in Mechanisms 129 4. Locate the other two remaining neither fixed nor permanent instantaneous centres, by Aronhold Kennedy’s theorem. This is done by circle diagram as shown in Fig. 6.14. Mark four points 1, 2, 3 and 4 (equal to the number of links in a mechanism) on the circle to indicate I , I , I and I . 12 23 34 14 Fig. 6.13 Fig. 6.14 5. Join 1 to 3 to form two triangles 1 2 3 and 3 4 1 in the circle diagram. The side 1 3, common to both triangles, is responsible for completing the two triangles. Therefore the centre I 13 will lie on the intersection of I I and I I , produced if necessary. Thus centre I is located. Join 12 23 14 34 13 1 to 3 by a dotted line and mark number 5 on it. 6. Join 2 to 4 by a dotted line to form two triangles 2 3 4 and 1 2 4. The side 2 4, common to both triangles, is responsible for completing the two triangles. Therefore the centre I lies on the 24 intersection of I I and I I . Join 2 to 4 by a dotted line on the circle diagram and mark number 6 23 34 12 14 on it. Thus all the six instantaneous centres are located. By measurement, we find that I A = 460 mm = 0.46 m ; and I B = 560 mm = 0.56 m 13 13 1. Velocity of the slider A Let v = Velocity of the slider A. A vv AB = We know that IA I B 13 13 IA 0.46 13 or vv=× =10 × =.82m/s Ans. AB IB 0.56 13 2. Angular velocity of the connecting rod AB Let ω = Angular velocity of the connecting rod AB. AB vv AB ==ω We know that AB IA I B 13 13 130 Theory of Machines Exhaust waste heat Engine Hydraulic rams Load The above picture shows a digging machine. Note : This picture is given as additional information and is not a direct example of the current chapter. v 1 B ∴ ω= = = 1.78 rad/s Ans. AB IB 0.56 13 Note: The velocity of the slider A and angular velocity of the connecting rod AB may also be determined as follows : From similar triangles I I I and I I I , 13 23 34 12 23 24 II I I 12 23 23 24 = ...(i) II I I 13 23 23 34 II I I 13 34 12 24 = and ...(ii) II II 34 23 23 24 vO ω×B BOB ω= = AB We know that ...(∵ v = ω × OB) B OB IB I B 13 13 II I I 12 23 23 24 =ω × = ω × OB OB ...From equation (i) ...(iii) II I I 13 23 23 34 II 23 24 vI =ω ×A= ω × ×II . AAB 13 OB 1334 Also ...From equation (iii) II 23 34 = ω × I I = ω × OD ...From equation (ii) OB 12 24 OB Example 6.3. A mechanism, as shown in Fig. 6.15, has the following dimensions: OA = 200 mm; AB = 1.5 m; BC = 600 mm; CD = 500 mm and BE = 400 mm. Locate all the instantaneous centres. If crank OA rotates uniformly at 120 r.p.m. clockwise, find 1. the velocity of B, C and D, 2. the angular velocity of the links AB, BC and CD. Chapter 6 : Velocity in Mechanisms 131 Solution. Given : N = 120 r.p.m. or ω = 2 π × 120/60 = 12.57 rad/s OA OA Since the length of crank OA = 200 mm = 0.2 m, therefore linear velocity of crank OA, v = v = ω × OA = 12.57 × 0.2 = 2.514 m/s OA A OA Fig. 6.15 Location of instantaneous centres The instantaneous centres are located as discussed below: 1. Since the mechanism consists of six links (i.e. n = 6), therefore the number of instanta- neous centres, nn ( – 1) 6 (6 – 1) N== = 15 22 2. Make a list of all the instantaneous centres in a mechanism. Since the mechanism has 15 instantaneous centres, therefore these centres are listed in the following book keeping table. Links 1 2 3 4 5 6 Instantaneous 12 23 34 45 56 centres 13 24 35 46 (15 in number) 14 25 36 15 26 16 Fig. 6.16 132 Theory of Machines 3. Locate the fixed and permanent instantaneous cen- tres by inspection. These centres are I I , I , I , I , I and 12 23 34 45 56 16 I as shown in Fig. 6.16. 14 4. Locate the remaining neither fixed nor permanent instantaneous centres by Aronhold Kennedy’s theorem. Draw a circle and mark points equal to the number of links such as 1, 2, 3, 4, 5 and 6 as shown in Fig. 6.17. Join the points 12, 23, 34, 45, 56, 61 and 14 to indicate the centres I , I , I , I , I , I 12 23 34 45 56 16 and I respectively. 14 5. Join point 2 to 4 by a dotted line to form the triangles 1 2 4 and 2 3 4. The side 2 4, common to both triangles, is Fig. 6.17 responsible for completing the two triangles. Therefore the in- stantaneous centre I lies on the intersection of I I and I I produced if necessary. Thus centre 24 12 14 23 34 I is located. Mark number 8 on the dotted line 24 (because seven centres have already been lo- 24 cated). 6. Now join point 1 to 5 by a dotted line to form the triangles 1 4 5 and 1 5 6. The side 1 5, common to both triangles, is responsible for completing the two triangles. Therefore the instantaneous centre I lies on the intersection of I I and I I produced if necessary. Thus centre I is located. 15 14 45 56 16 15 Mark number 9 on the dotted line 1 5. 7. Join point 1 to 3 by a dotted line to form the triangles 1 2 3 and 1 3 4. The side 1 3, common to both triangles, is responsible for completing the two triangles. Therefore the instanta- neous centre I lies on the intersection I I and I I produced if necessary. Thus centre I is 13 12 23 34 14 13 located. Mark number 10 on the dotted line 1 3. 8. Join point 4 to 6 by a dotted line to form the triangles 4 5 6 and 1 4 6. The side 4 6, common to both triangles, is responsible for completing the two triangles. Therefore, centre I lies 46 on the intersection of I I and I I . Thus centre I is located. Mark number 11 on the dotted line 45 56 14 16 46 4 6. 9. Join point 2 to 6 by a dotted line to form the triangles 1 2 6 and 2 4 6. The side 2 6, common to both triangles, is responsible for completing the two triangles. Therefore, centre I lies 26 on the intersection of lines joining the points I I and I I . Thus centre I is located. Mark 12 16 24 46 26 number 12 on the dotted line 2 6. 10. In the similar way the thirteenth, fourteenth and fifteenth instantaneous centre (i.e. I , I 35 25 and I ) may be located by joining the point 3 to 5, 2 to 5 and 3 to 6 respectively. 36 By measurement, we find that I A = 840 mm = 0.84 m ; I B = 1070 mm = 1.07 m ; I B = 400 mm = 0.4 m ; 13 13 14 I C = 200 mm = 0.2 m ; I C = 740 mm = 0.74 m ; I D = 500 mm = 0.5 m 14 15 15 1. Velocity of points B, C and D Let v , v and v = Velocity of the points B, C and D respectively. B C D vv AB We know that = ...(Considering centre I ) 13 IA I B 13 13 v 2.514 A ∴ vI =×B= × 1.07= 3.2 m/s Ans. B13 IA 0.84 13 v v B C = Again, ...(Considering centre I ) 14 IB I C 14 14 Chapter 6 : Velocity in Mechanisms 133 v 3.2 B ∴ vI =×C= × 0.2= 1.6m/s Ans. C14 IB 0.4 14 v v CD = Similarly, ...(Considering centre I ) 15 IC I D 15 15 v 1.6 C ∴ vI =×D= × 0.5= 1.08 m/s Ans. D5 0.74 IC 15 2. Angular velocity of the links AB, BC and CD Let ω , ω and ω = Angular velocity of the links AB, BC and CD respectively. AB BC CD v 2.514 A We know that ω= = = 2.99 rad/s Ans. AB IA 0.84 13 v 3.2 B ω= = = 8rad/s Ans. BC IB 0.4 14 v 1.6 C ω= = = 2.16 rad/s and Ans. CD IC 0.74 15 Example 6.4. The mechanism of a wrapping machine, as shown in Fig. 6.18, has the follow- ing dimensions : O A = 100 mm; AC = 700 mm; BC = 200 mm; O C = 200 mm; O E = 400 mm; 1 3 2 O D = 200 mm and BD = 150 mm. 2 The crank O A rotates at a uniform speed of 100 rad/s. Find the velocity of the point E of the 1 bell crank lever by instantaneous centre method. Fig. 6.18 Solution. Given : ω = 100 rad/s ; O A = 100 mm = 0.1 m O1A 1 We know that the linear velocity of crank O A, 1 v = v = ω × O A = 100 × 0.1 = 10 m/s O1A A O1A 1 Now let us locate the required instantaneous centres as discussed below : 1. Since the mechanism consists of six links (i.e. n = 6), therefore number of instantaneous centres, nn ( – 1) 6 (6 – 1) N== = 15 22 2. Since the mechanism has 15 instantaneous centres, therefore these centres may be listed in the book keeping table, as discussed in Example 6.3. 134 Theory of Machines Fig. 6.19 Fig. 6.20 3. Locate the fixed and the permanent instantaneous centres by inspection. These centres are I , I , I , I , I , I and I as shown in Fig. 6.19. 12 23 34 35 14 56 16 4. Locate the remaining neither fixed nor permanent instantaneous centres by Aronhold Kennedy’s theorem. This is done by circle diagram as shown in Fig. 6.20. Mark six points on the circle (i.e. equal to the number of links in a mechanism), and join 1 to 2, 2 to 3, 3 to 4, 3 to 5, 4 to 1, 5 to 6, and 6 to 1, to indicate the fixed and permanent instantaneous centres i.e. I , I , I , I , 12 23 34 35 I , I , and I respectively. 14 56 16 5. Join 1 to 3 by a dotted line to form two triangles 1 2 3 and 1 3 4. The side 1 3, common to both triangles, is responsible for completing the two triangles. Therefore the instantaneous cen- tre I lies on the intersection of the lines joining the points I I and I I produced if necessary. 13 12 23 14 34 Thus centre I is located. Mark number 8 (because seven centres have already been located) on the 13 dotted line 1 3. 6. Join 1 to 5 by a dotted line to form two triangles 1 5 6 and 1 3 5. The side 1 5, common to both triangles, is responsible for completing the two triangles. Therefore the instantaneous centre I 15 lies on the intersection of the lines joining the points I I and I I produced if necessary. Thus 16 56 13 35 centre I is located. Mark number 9 on the dotted line 1 5. 15 Note: For the given example, we do not require other instantaneous centres. By measurement, we find that I A = 910 mm = 0.91 m ; I B = 820 mm = 0.82 m ; I B = 130 mm = 0.13 m ; 13 13 15 I D = 50 mm = 0.05 m ; I D = 200 mm = 0.2 m ; I E = 400 mm = 0.4 m 15 16 16 Velocity of point E on the bell crank lever Let v = Velocity of point E on the bell crank lever, E v = Velocity of point B, and B v = Velocity of point D. D vv AB We know that = ...(Considering centre I ) 13 IA I B 13 13 Chapter 6 : Velocity in Mechanisms 135 v 10 A ∴ Ans. vI =×B= × 0.82= 9.01 m/s B13 IA 0.91 13 vv BD = and ...(Considering centre I ) 15 IB ID 15 15 v 9.01 B ∴ vI =×=D × 0.05= 3.46 m/s Ans. D15 IB 0.13 15 vv DE Similarly, ...(Considering centre I ) = 16 ID IE 16 16 v 3.46 D ∴ Ans. vI =×E= × 0.4= 6.92 m/s E16 ID 0.2 16 Example 6.5. Fig. 6.21 shows a sewing needle bar mechanism O ABO CD wherein the different dimensions are as follows: 1 2 Crank O A = 16 mm; ∠β = 45°; Vertical distance between O and 1 1 O = 40 mm; Horizontal distance between O and O = 13 mm; O B = 23 2 1 2 2 mm; AB = 35 mm; ∠ O BC = 90°; BC = 16 mm; CD = 40 mm. D lies 2 vertically below O . 1 Find the velocity of needle at D for the given configuration. The crank O A rotates at 400 r.p.m. 1 Solution. Given : N = 400 r.p.m or ω = 2π × 400/60 = O1A O1A 41.9 rad/s ; O A = 16 mm = 0.016 m 1 We know that linear velocity of the crank O A, 1 v = v = ω × O A = 41.9 × 0.016 = 0.67 m/s O1A A O1A 1 Now let us locate the required instantaneous centres as discussed Fig. 6.21 below : 1. Since the mechanism consists of six links (i.e. n = 6), therefore number of instantaneous centres, nn(–1) 6(6–1) N== = 15 22 2. Since the mechanism has 15 instantaneous centres, therefore these centres may be listed in the book keeping table, as discussed in Example 6.3. 3. Locate the fixed and permanent instantaneous centres by inspections. These centres are I , I , I , I , I , I and I , as shown in Fig. 6.22. 12 23 34 45 56 16 14 Fig. 6.22 136 Theory of Machines 4. Locate the remaining neither fixed nor permanent in- stantaneous centres by Aronhold Kennedy’s theorem. This is done by circle diagram as shown in Fig. 6.23. Mark six points on the circle (i.e. equal to the number of links in a mechanism) and join 1 to 2, 2 to 3, 3 to 4, 4 to 5, 5 to 6, 6 to 1 and 1 to 4 to indicate the fixed and permanent instantaneous centres i.e. I , I , I , I , I , 12 23 34 45 56 I and I respectively. 16 14 5. Join 1 to 3 by a dotted line to form two triangles 1 2 3 and 1 3 4. The side 1 3, common to both the triangles, is responsible for completing the two triangles. Therefore the instantaneous centre I lies on the intersection of I I and 13 12 23 Fig. 6.23 I I produced if necessary. Thus centre I is located. Mark 14 34 13 number 8 (because seven centres have already been located) on the dotted line 1 3. 6. Join 1 to 5 by a dotted line to form two triangles 1 5 6 and 1 4 5. The side 1 5, common to both the triangles, is responsible for completing the two triangles. Therefore the instantaneous centre I lies on the intersection of I I and I I produced if necessary. Thus centre I is 15 16 56 14 45 15 located. Mark number 9 on the dotted line 1 5. Note: For the given example, we do not require other instantaneous centres. By measurement, we find that I A = 41 mm = 0.041 m ; I B = 50 mm = 0.05 m ; I B = 23 mm = 0.023 m ; 13 13 14 I C= 28 mm = 0.028 m ; I C = 65 mm = 0.065 m ; I D = 62 mm = 0.062 m 14 15 15 Let v = Velocity of point B, B v = Velocity of point C, and C v = Velocity of the needle at D. D vv AB = We know that ...(Considering centre I ) 13 IA I B 13 13 v 0.67 A vI =×B= × 0.05= 0.817 m/s B13 ∴ IA 0.041 13 v v B C = and ...(Considering centre I ) 14 IB I C 14 14 v 0.817 B vI =×C= × 0.028= 0.995 m/s C14 ∴ IB 0.023 14 v v CD = Similarly, ...(Considering centre I ) 15 IC I D 15 15 v 0.995 C vI =×D= × 0.062= 0.95 m/s ∴ Ans. D15 0.065 IC 15 Chapter 6 : Velocity in Mechanisms 137 Example 6.6. Fig. 6.24 shows a Whitworth quick return motion mechanism. The various dimensions in the mechanism are as follows : OQ = 100 mm ; OA = 200 mm ; QC = 150 mm ; and CD = 500 mm. The crank OA makes an angle of 60° with the vertical and rotates at 120 r.p.m. in the clockwise direction. Locate all the instantaneous centres and find the velocity of ram D. Solution : Given. N = 120 r.p.m. or ω = OA OA Fig. 6.24 2 π × 120 / 60 = 12.57 rad/s Location of instantaneous centres The instantaneous centres are located as discussed below : 1. Since the mechanism consists of six links (i.e. n = 6), therefore the number of instanta- neous centres, nn ( – 1) 6 (6 – 1) N== = 15 22 2. Make a list of all the instantaneous centres in a mechanism as discussed in Example 6.3. 3. Locate the fixed and permanent instantaneous centres by inspection. These centres are I , 12 I , I , I , I , I and I as shown in Fig. 6.25. 23 34 45 56 16 14 Fig. 6.25 4. Locate the remaining neither fixed nor permanent instantaneous centres by Aronhold Kennedy’s theorem. Draw a circle and mark points equal to the number of links such as 1, 2, 3, 4, 5, 138 Theory of Machines and 6 as shown in Fig. 6.26. Join the points 1 2, 2 3, 3 4, 4 5, 5 6, 6 1 and 1 4 to indicate the centres I , I , I , I , I , I and I 12 23 34 45 56 16 14 respectively. 5. Join point 1 to 3 by a dotted line to form two triangles 1 2 3 and 1 3 4. The side 1 3, common to both the triangles, is responsible for completing the two triangles. Therefore the instan- taneous centre I lies on the intersection of I I , and I I pro- 13 12 23 14 34 duced if necessary. Thus centre I is located. Mark number 8 on the 13 dotted line 1 3 (because seven centres have already been located). 6. Join point 1 to 5 by a dotted line to form two triangles 1 4 5 and 1 5 6. The side 1 5, common to both the triangles, is Fig. 6.26 responsible for completing the two triangles. Therefore the instan- taneous centre I lies on the intersection of I I and I I produced if necessary. Thus centre I is 15 14 45 56 16 15 located. Mark number 9 on the dotted line 1 5. 7. Join point 2 to 4 by a dotted line to form two triangles 1 2 4 and 2 3 4. The side 2 4, common to both the triangles, is responsible for completing the two triangles. Therefore the instanta- neous centre I lies on the intersection of I I and I I produced if necessary. Thus centre I is 24 12 14 23 34 24 located. Mark number 10 on the dotted line 2 4. 8. Join point 2 to 5 by a dotted line to form two triangles 1 2 5 and 2 4 5. The side 2 5, common to both the triangles, is responsible for completing the two triangles. Therefore the instanta- neous centre I lies on the intersection of I I and I I produced if necessary. Thus centre I is 25 12 15 24 45 25 located. Mark number 11 on the dotted line 2 5. 9. Join point 2 to 6 by a dotted line to form two triangles 1 2 6 and 2 5 6. The side 2 6 common to both the triangles, is responsible for completing the two triangles. Therefore the instanta- neous centre I lies on the intersection of I I and I I produced if necessary. Thus centre I is 26 12 16 25 56 26 located. Mark number 12 on the dotted line 2 6. 10. Join point 3 to 5 by a dotted line to form two triangles 2 3 5 and 3 4 5. The side 3 5, common to both the triangles, is responsible for completing the two triangles. Therefore the instanta- neous centre I lies on the intersection of I I and I I produced if necessary. Thus centre I is 35 23 25 34 45 35 located. Mark number 13 on the dotted line 3 5. 11. Join point 3 to 6 by a dotted line to form two triangles 1 3 6 and 3 5 6. The side 3 6, common to both the triangles, is responsible for completing the two triangles. Therefore the instanta- neous centre I lies on the intersection of I I and I I produced if necessary. Thus centre I is 36 13 16 35 56 36 located. Mark number 14 on the dotted line 3 6. Note. The centre I may also be obtained by considering the two triangles 2 3 6 and 3 4 6. 36 12. Join point 4 to 6 by a dotted line to form two triangles 1 4 6 and 4 5 6. The side 4 6, common to both the triangles, is responsible for completing the two triangles. Therefore the instanta- neous centre I lies on the intersection of I I and I I produced if necessary. Thus centre I is 46 14 16 45 56 46 located. Mark number 15 on the dotted line 4 6. Velocity of ram D By measurement, we find that I I = 65 mm = 0.065 m 12 26 ∴ Velocity of ram, v = ω × I I = 12.57 × 0.065 = 0.817 m/s Ans. D OA 12 26 EXERCISES 1. Locate all the instantaneous centres for a four bar mechanism as shown in Fig. 6.27. The lengths of various links are : AD = 125 mm ; AB = 62.5 mm ; BC = CD = 75 mm. If the link AB rotates at a uniform speed of 10 r.p.m. in the clockwise direction, find the angular velocity of the links BC and CD. Ans. 0.63 rad/s ; 0.65 rad/s

Advise: Why You Wasting Money in Costly SEO Tools, Use World's Best Free SEO Tool Ubersuggest.