Velocity and Acceleration in mechanism

how the coriolis acceleration in mechanism and acceleration analysis of complex mechanism, acceleration diagram quick return mechanism velocity acceleration mechanism
HartJohnson Profile Pic
HartJohnson,United States,Professional
Published Date:02-08-2017
Your Website URL(Optional)
Comment
 174 Theory of Machines 8 Warping Machine F F F F Fea ea ea ea eatur tur tur tur tures es es es es Acceleration Acceleration Acceleration Acceleration Acceleration 1. Introduction. 2. Acceleration Diagram for a Link. in Mechanisms in Mechanisms in Mechanisms in Mechanisms in Mechanisms 3. Acceleration of a Point on a Link. 4. Acceleration in the Slider 8.1. 8.1. 8.1. Intr Intr Introduction oduction oduction 8.1. 8.1. Intr Introduction oduction Crank Mechanism. We have discussed in the previous chapter the 5. Coriolis Component of velocities of various points in the mechanisms. Now we shall Acceleration. discuss the acceleration of points in the mechanisms. The acceleration analysis plays a very important role in the development of machines and mechanisms. 8.2. 8.2. 8.2. 8.2. 8.2. Acceleration Diagram for a Link Acceleration Diagram for a Link Acceleration Diagram for a Link Acceleration Diagram for a Link Acceleration Diagram for a Link Consider two points A and B on a rigid link as shown in Fig. 8.1 (a). Let the point B moves with respect to A, with 2 an angular velocity of ω rad/s and let α rad/s be the angular acceleration of the link AB. (a) Link. (b) Acceleration diagram. Fig. 8.1. Acceleration for a link. 174 Chapter 8 : Acceleration in Mechanisms 175 We have already discussed that acceleration of a particle whose velocity changes both in magnitude and direction at any instant has the following two components : 1. The centripetal or radial component, which is perpendicular to the velocity of the particle at the given instant. 2. The tangential component, which is parallel to the velocity of the particle at the given instant. Thus for a link AB, the velocity of point B with respect to A (i.e. v ) is perpendicular to the BA link AB as shown in Fig. 8.1 (a). Since the point B moves with respect to A with an angular velocity of ω rad/s, therefore centripetal or radial component of the acceleration of B with respect to A, v  BA r222 ... ω=  aA =ω × Length of linkB = ω ×AB =v /AB BA BA  AB This radial component of acceleration acts perpendicular to the velocity v , In other words, BA it acts parallel to the link AB. We know that tangential component of the acceleration of B with respect to A, t aA =α× Length of the linkB = α×AB BA This tangential component of acceleration acts parallel to the velocity v . In other words, BA it acts perpendicular to the link AB. In order to draw the acceleration diagram for a link AB, as shown in Fig. 8.1 (b), from any point b', draw vector b'x parallel to BA to represent the radial component of acceleration of B with r respect to A i.e. and from point x draw vector xa' perpendicular to BA to represent the tangential a BA t a . component of acceleration of B with respect to A i.e. Join b' a'. The vector b' a' (known as BA acceleration image of the link AB) represents the total acceleration of B with respect to A (i.e. a ) BA r t and it is the vector sum of radial component () and tangential component () of acceleration. a a BA BA 8.3. Acceleration of a Point on a Link (a) Points on a Link. (b) Acceleration diagram. Fig. 8.2. Acceleration of a point on a link. Consider two points A and B on the rigid link, as shown in Fig. 8.2 (a). Let the acceleration of the point A i.e. a is known in magnitude and direction and the direction of path of B is given. The A 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 acceleration at point A i.e. a , to some suitable scale, as shown in Fig. 8.2 (b). A  176 Theory of Machines 2. We know that the acceleration of B with respect to A i.e. a has the following two BA components: (i) Radial component of the acceleration r of B with respect to A i.e. , and a BA (ii) Tangential component of the t acceleration B with respect to A i.e. a . These two BA components are mutually perpendicular. 3. Draw vector a'x parallel to the link AB (because radial component of the acceleration of B with respect to A will pass through AB), such that r 2 vector ′ ax== a v / AB BA BA wherevB = Velocity of with respect toA. BA Note: The value of v may be obtained by drawing the BA velocity diagram as discussed in the previous chapter. 4. From point x, draw vector xb' perpendicular to AB or vector a'x (because tangential t component of B with respect to A i.e. is a , BA r A refracting telescope uses mechanisms to perpendicular to radial component ) and a BA change directions. through o' draw a line parallel to the path of B to Note : This picture is given as additional represent the absolute acceleration of B i.e. a . The B information and is not a direct example of the vectors xb' and o' b' intersect at b'. Now the values current chapter. t of a and may be measured, to the scale. a B BA 5. By joining the points a' and b' we may determine the total acceleration of B with respect to A i.e. a . The vector a' b' is known as acceleration image of the link AB. BA 6. For any other point C on the link, draw triangle a' b' c' similar to triangle ABC. Now vector b' c' represents the acceleration of C with respect to B i.e. a , and vector a' c' represents the CB acceleration of C with respect to A i.e. a . As discussed above, a and a will each have two CA CB CA components as follows : rt (i) a has two components; aa and as shown by triangle b' zc' in Fig. 8.2 (b), in CB CB CB which b' z is parallel to BC and zc' is perpendicular to b' z or BC. rt (ii) a has two components ; aa and as shown by triangle a' yc' in Fig. 8.2 (b), in CA CA CA which a' y is parallel to AC and yc' is perpendicular to a' y or AC. 7. The angular acceleration of the link AB is obtained by dividing the tangential components t of the acceleration of B with respect to A to the length of the link. Mathematically, angular () a BA acceleration of the link AB, t α=aA /B AB BA 8.4. Acceleration in the Slider Crank Mechanism A slider crank mechanism is shown in Fig. 8.3 (a). Let the crank OB makes an angle θ with the inner dead centre (I.D.C) and rotates in a clockwise direction about the fixed point O with uniform angular velocity ω rad/s. BO ∴ Velocity of B with respect to O or velocity of B (because O is a fixed point), vv==ω ×OB, acting tangentially atB. BO B BO Chapter 8 : Acceleration in Mechanisms 177 We know that centripetal or radial acceleration of B with respect to O or acceleration of B (because O is a fixed point), 2 v r 2 BO aa==ω ×OB= BO B BO OB Note : A point at the end of a link which moves with constant angular velocity has no tangential component of acceleration. (a) Slider crank mechanism. (b) Acceleration diagram. Fig. 8.3. Acceleration in the slider crank mechanism. The acceleration diagram, as shown in Fig. 8.3 (b), may now be drawn as discussed below: r 1. Draw vector o' b' parallel to BO and set off equal in magnitude of , to some aa = BO B suitable scale. 2. From point b', draw vector b'x parallel to BA. The vector b'x represents the radial component of the acceleration of A with respect to B whose magnitude is given by : r 2 = / av BA AB AB Since the point B moves with constant angular velocity, therefore there will be no tangential component of the acceleration. 3. From point x, draw vector xa' perpendicular to b'x (or AB). The vector xa' represents the t tangential component of the acceleration of A with respect to B i.e. a . AB Note: When a point moves along a straight line, it has no centripetal or radial component of the acceleration. 4. Since the point A reciprocates along AO, therefore the acceleration must be parallel to velocity. Therefore from o', draw o' a' parallel to AO, intersecting the vector xa' at a'. t Now the acceleration of the piston or the slider A (a ) and may be measured to the scale. a A AB 5. The vector b' a', which is the sum of the vectors b' x and x a', represents the total acceleration of A with respect to B i.e. a . The vector b' a' represents the acceleration of the connecting rod AB. AB 6. The acceleration of any other point on AB such as E may be obtained by dividing the vector b' a' at e' in the same ratio as E divides AB in Fig. 8.3 (a). In other words a' e' / a' b' = AE / AB 7. The angular acceleration of the connecting rod AB may be obtained by dividing the t tangential component of the acceleration of A with respect to B() a to the length of AB. In other AB words, angular acceleration of AB, t / (Clockwise about ) α=aAB B AB AB Example 8.1. The crank of a slider crank mechanism rotates clockwise at a constant speed of 300 r.p.m. The crank is 150 mm and the connecting rod is 600 mm long. Determine : 1. linear velocity and acceleration of the midpoint of the connecting rod, and 2. angular velocity and angular acceleration of the connecting rod, at a crank angle of 45° from inner dead centre position. 178 Theory of Machines Solution. Given : N = 300 r.p.m. or ω = 2 π × 300/60 = 31.42 rad/s; OB = 150 mm = BO BO 0.15 m ; BA = 600 mm = 0.6 m We know that linear velocity of B with respect to O or velocity of B, v = v = ω × OB = 31.42 × 0.15 = 4.713 m/s BO B BO ...(Perpendicular to BO) (a) Space diagram. (b) Velocity diagram. (c) Acceleration diagram. Fig. 8.4 Ram moves Ram moves Load moves outwards inwards outwards Oil pressure on lower side of Oil pressure on piston upper side of piston Load moves inwards Pushing with fluids Note : This picture is given as additional information and is not a direct example of the current chapter. 1. Linear velocity of the midpoint of the connecting rod First of all draw the space diagram, to some suitable scale; as shown in Fig. 8.4 (a). Now the velocity diagram, as shown in Fig. 8.4 (b), is drawn as discussed below: 1. Draw vector ob perpendicular to BO, to some suitable scale, to represent the velocity of B with respect to O or simply velocity of B i.e. v or v , such that BO B vector ob = v = v = 4.713 m/s BO B 2. From point b, draw vector ba perpendicular to BA to represent the velocity of A with respect to B i.e. v , and from point o draw vector oa parallel to the motion of A (which is along AO) AB to represent the velocity of A i.e. v . The vectors ba and oa intersect at a. A Chapter 8 : Acceleration in Mechanisms 179 By measurement, we find that velocity of A with respect to B, vb== vectora 3.4 m / s AB and Velocity of Av ,== vectoroa 4 m / s A 3. In order to find the velocity of the midpoint D of the connecting rod AB, divide the vector ba at d in the same ratio as D divides AB, in the space diagram. In other words, bd / ba = BD/BA Note: Since D is the midpoint of AB, therefore d is also midpoint of vector ba. 4. Join od. Now the vector od represents the velocity of the midpoint D of the connecting rod i.e. v . D By measurement, we find that v = vector od = 4.1 m/s Ans. D Acceleration of the midpoint of the connecting rod We know that the radial component of the acceleration of B with respect to O or the acceleration of B, 2 2 v (4.713) r 2 BO aa== = = 148.1 m/s BO B 0.15 OB and the radial component of the acceleraiton of A with respect to B, 2 2 v (3.4) r 2 AB a == = 19.3 m/s AB BA 0.6 Now the acceleration diagram, as shown in Fig. 8.4 (c) is drawn as discussed below: 1. Draw vector o' b' parallel to BO, to some suitable scale, to represent the radial component r of the acceleration of B with respect to O or simply acceleration of B i.e. such that aa or , BO B r 2 ′′ vector ob== a a= 148.1 m/s BO B Note: Since the crank OB rotates at a constant speed, therefore there will be no tangential component of the acceleration of B with respect to O. 2. The acceleration of A with respect to B has the following two components: r (a) The radial component of the acceleration of A with respect to B i.e. a , and AB t (b) The tangential component of the acceleration of A with respect to B i.e. These two a . AB components are mutually perpendicular. r 2 Therefore from point b', draw vector b' x parallel to AB to represent and a = 19.3 m/s AB from point x draw vector xa' perpendicular to vector b' x whose magnitude is yet unknown. 3. Now from o', draw vector o' a' parallel to the path of motion of A (which is along AO) to represent the acceleration of A i.e. a . The vectors xa' and o' a' intersect at a'. Join a' b'. A 4. In order to find the acceleration of the midpoint D of the connecting rod AB, divide the vector a' b' at d' in the same ratio as D divides AB. In other words ′′ ′ ′ bd// ba = BD BA Note: Since D is the midpoint of AB, therefore d' is also midpoint of vector b' a'. 5. Join o' d'. The vector o' d' represents the acceleration of midpoint D of the connecting rod i.e. a . D By measurement, we find that 2 a = vector o' d' = 117 m/s Ans. D 180 Theory of Machines 2. Angular velocity of the connecting rod We know that angular velocity of the connecting rod AB, v 3.4 2 AB ω= = = 5.67 rad/s (Anticlockwise about B) AB Ans. BA 0.6 Angular acceleration of the connecting rod From the acceleration diagram, we find that t 2 ...(By measurement) a = 103 m/s AB We know that angular acceleration of the connecting rod AB, t a 103 AB 2 α= = = 171.67 rad/s (Clockwise about B) Ans. AB BA 0.6 Example 8.2. An engine mechanism is shown in Fig. 8.5. The crank CB = 100 mm and the connecting rod BA = 300 mm with centre of gravity G, 100 mm from B. In the position shown, the 2 crankshaft has a speed of 75 rad/s and an angular acceleration of 1200 rad/s . Find:1. velocity of G and angular velocity of AB, and 2. acceleration of G and angular acceleration of AB. Fig. 8.5 2 Solution. Given : ω = 75 rad/s ; α = 1200 rad/s , CB = 100 mm = 0.1 m; BA = 300 mm BC BC = 0.3 m We know that velocity of B with respect to C or velocity of B, vv==ω ×CB= 75× 0.1= 7.5 m/s ...(Perpendicular to BC) BC B BC 2 Since the angular acceleration of the crankshaft, α = 1200 rad/s , therefore tangential BC component of the acceleration of B with respect to C, t 2 =α × = 1200× 0.1= 120 m/s aCB BC BC Note: When the angular acceleration is not given, then there will be no tangential component of the acceleration. 1. Velocity of G and angular velocity of AB First of all, draw the space diagram, to some suitable scale, as shown in Fig. 8.6 (a). Now the velocity diagram, as shown in Fig. 8.6 (b), is drawn as discussed below: 1. Draw vector cb perpendicular to CB, to some suitable scale, to represent the velocity of B with respect to C or velocity of B (i.e. v or v ), such that BC B vector cb== v v= 7.5 m/s BC B 2. From point b, draw vector ba perpendicular to BA to represent the velocity of A with respect to B i.e. v , and from point c, draw vector ca parallel to the path of motion of A (which is AB along AC) to represent the velocity of A i.e. v .The vectors ba and ca intersect at a. A 3. Since the point G lies on AB, therefore divide vector ab at g in the same ratio as G divides AB in the space diagram. In other words, ag// ab = AG AB The vector cg represents the velocity of G. By measurement, we find that velocity of G, v = vector cg = 6.8 m/s Ans. G Chapter 8 : Acceleration in Mechanisms 181 From velocity diagram, we find that velocity of A with respect to B, v = vector ba = 4 m/s AB We know that angular velocity of AB, v 4 AB ω= = = 13.3 rad/s (Clockwise) Ans. AB BA 0.3 (a) Space diagram. (b) Velocity diagram. Fig. 8.6 2. Acceleration of G and angular acceleration of AB We know that radial component of the acceleration of B with respect to C, 2 2 v (7.5) r 2 BC a == = 562.5 m/s BC CB 0.1 and radial component of the acceleration of A with respect to B, 2 2 v 4 r AB 2 a == = 53.3 m/s AB BA 0.3 Now the acceleration diagram, as shown in Fig. 8.6 (c), is drawn as discussed below: 1. Draw vector c' b'' parallel to CB, to some suitable scale, to (c) Acceleration diagram. represent the radial component of the acceleration of B with respect to C, Fig. 8.6 r i.e. , such that a BC r 2 ′′′ vector cb== a 562.5 m/s BC 2. From point b'', draw vector b'' b' perpendicular to vector c' b'' or CB to represent the t tangential component of the acceleration of B with respect to C i.e. a , such that BC t 2 vector ′′ ′== 120 m/s ... (Given) bb a BC 3. Join c' b'. The vector c' b' represents the total acceleration of B with respect to C i.e. a . BC 4. From point b', draw vector b' x parallel to BA to represent radial component of the r acceleration of A with respect to B i.e. such that a AB r 2 vector ′== 53.3 m/s bx a AB 5. From point x, draw vector xa' perpendicular to vector b'x or BA to represent tangential t component of the acceleration of A with respect to B i.e. , whose magnitude is not yet known. a AB 6. Now draw vector c' a' parallel to the path of motion of A (which is along AC) to represent the acceleration of A i.e. a .The vectors xa' and c'a' intersect at a'. Join b' a'. The vector b' a' A represents the acceleration of A with respect to B i.e. a . AB t r When angular acceleration of the crank is not given, then there is no a . In that case, aa==a , as BC BC B BC discussed in the previous example. 182 Theory of Machines 7. In order to find the acceleratio of G, divide vector a' b' in g' in the same ratio as G divides BA in Fig. 8.6 (a). Join c' g'. The vector c' g' represents the acceleration of G. By measurement, we find that acceleration of G, 2 a = vector c' g' = 414 m/s Ans. G From acceleration diagram, we find that tangential component of the acceleration of A with respect to B, t 2 == vector ′ 546 m/s ...(By measurement) axa AB ∴ Angular acceleration of AB, t a 546 2 AB α= = = 1820 rad/s (Clockwise) Ans. AB BA 0.3 Example 8.3. In the mechanism shown in Fig. 8.7, the slider C is 2 moving to the right with a velocity of 1 m/s and an acceleration of 2.5 m/s . The dimensions of various links are AB = 3 m inclined at 45° with the vertical and BC = 1.5 m inclined at 45° with the horizontal. Determine: 1. the magnitude of vertical and horizontal component of the acceleration of the point B, and 2. the angular acceleration of the links AB and BC. 2 Solution. Given : v = 1 m/s ; a = 2.5 m/s ; AB = 3 m ; BC = 1.5 m C C First of all, draw the space diagram, as shown in Fig. 8.8 (a), to some Fig. 8.7 suitable scale. Now the velocity diagram, as shown in Fig. 8.8 (b), is drawn as discussed below: 1. Since the points A and D are fixed points, therefore they lie at one place in the velocity diagram. Draw vector dc parallel to DC, to some suitable scale, which represents the velocity of slider C with respect to D or simply velocity of C, such that vector dc = v = v = 1 m/s CD C 2. Since point B has two motions, one with respect to A and the other with respect to C, therefore from point a, draw vector ab perpendicular to AB to represent the velocity of B with respect to A, i.e. v and from point c draw vector cb perpendicular to CB to represent the velocity BA of B with respect to C i.e. v .The vectors ab and cb intersect at b. BC (a) Space diagram. (b) Velocity diagram. (c) Acceleration diagram. Fig. 8.8 By measurement, we find that velocity of B with respect to A, va== vectorb 0.72 m/s BA and velocity of B with respect to C, vc== vectorb 0.72 m/s BC Chapter 8 : Acceleration in Mechanisms 183 We know that radial component of acceleration of B with respect to C, 2 2 v (0.72) r 2 BC a == = 0.346 m/s BC 1.5 CB and radial component of acceleration of B with respect to A, 2 2 v (0.72) r BA 2 == = 0.173 m/s a BA AB 3 Now the acceleration diagram, as shown in Fig. 8.8 (c), is drawn as discussed below: 1. Since the points A and D are fixed points, therefore they lie at one place in the acceleration diagram. Draw vector d' c' parallel to DC, to some suitable scale, to represent the acceleration of C with respect to D or simply acceleration of C i.e. a or a such that CD C 2 vector ′′ 2.5 m/s dc== a a= CD C 2. The acceleration of B with respect to C will have two components, i.e. one radial component r of B with respect to C a and the other tangential component of B with respect to () BC t r C . () a Therefore from point c', draw vector c' x parallel to CB to represent a such that BC BC r 2 ′ vector cx== a 0.346 m/s BC t 3. Now from point x, draw vector xb' perpendicular to vector c' x or CB to represent a BC whose magnitude is yet unknown. 4. The acceleration of B with respect to A will also have two components, i.e. one radial r t component of B with respect to A (a ) and other tangential component of B with respect to A (a ). BA BA r Therefore from point a' draw vector a' y parallel to AB to represent a , such that BA r 2 vector a' y = a = 0.173 m/s BA t 5. From point y, draw vector yb' perpendicular to vector a'y or AB to represent The a . BA vector yb' intersect the vector xb' at b'. Join a' b' and c' b'. The vector a' b' represents the acceleration of point B (a ) and the vector c' b' represents the acceleration of B with respect to C. B 1. Magnitude of vertical and horizontal component of the acceleration of the point B Draw b' b'' perpendicular to a' c'. The vector b' b'' is the vertical component of the acceleration of the point B and a' b'' is the horizontal component of the acceleration of the point B. By measurement, 2 2 vector b' b'' = 1.13 m/s and vector a' b'' = 0.9 m/s Ans. 2. Angular acceleration of AB and BC By measurement from acceleration diagram, we find that tangential component of acceleration of the point B with respect to A, t 2 == vector ′ 1.41 m/s a yb BA and tangential component of acceleration of the point B with respect to C, t 2 ′ ax== vectorb 1.94 m/s BC If the mechanism consists of more than one fixed point, then all these points lie at the same place in the velocity and acceleration diagrams. 184 Theory of Machines We know that angular acceleration of AB, t a 1.41 BA 2 α= = = 0.47 rad/s Ans. AB AB 3 and angular acceleration of BC, t a 1.94 2 BA Ans. α= = = 1.3 rad/s BC CB 1.5 Example 8.4. PQRS is a four bar chain with link PS fixed. The lengths of the links are PQ = 62.5 mm ; QR = 175 mm ; RS = 112.5 mm ; and PS = 200 mm. The crank PQ rotates at 10 rad/s clockwise. Draw the velocity and acceleration diagram when angle QPS = 60° and Q and R lie on the same side of PS. Find the angular velocity and angular acceleration of links QR and RS. Solution. Given : ω = 10 rad/s; PQ = 62.5 mm = 0.0625 m ; QR = 175 mm = 0.175 m ; QP RS = 112.5 mm = 0.1125 m ; PS = 200 mm = 0.2 m We know that velocity of Q with respect to P or velocity of Q, v = v = ω × PQ = 10 × 0.0625 = 0.625 m/s QP Q QP ...(Perpendicular to PQ) Angular velocity of links QR and RS First of all, draw the space diagram of a four bar chain, to some suitable scale, as shown in Fig. 8.9 (a). Now the velocity diagram as shown in Fig. 8.9 (b), is drawn as discussed below: (a) Space diagram. (b) Velocity diagram. (c) Acceleration diagram. Fig. 8.9 1. Since P and S are fixed points, therefore these points lie at one place in velocity diagram. Draw vector pq perpendicular to PQ, to some suitable scale, to represent the velocity of Q with respect to P or velocity of Q i.e. v or v such that QP Q vector pq = v = v = 0.625 m/s QP Q 2. From point q, draw vector qr perpendicular to QR to represent the velocity of R with respect to Q (i.e. v ) and from point s, draw vector sr perpendicular to SR to represent the velocity RQ of R with respect to S or velocity of R (i.e. v or v ). The vectors qr and sr intersect at r. By RS R measurement, we find that v = vector qr = 0.333 m/s, and v = v = vector sr = 0.426 m/s RQ RS R We know that angular velocity of link QR, v 0.333 RQ 1.9 rad/s (Anticlockwise) ω= = = Ans. QR RQ 0.175 Chapter 8 : Acceleration in Mechanisms 185 and angular velocity of link RS, v 0.426 RS ω= = = 3.78 rad/s (Clockwise) A Ans. RS SR 0.1125 Angular acceleration of links QR and RS Since the angular acceleration of the crank PQ is not given, therefore there will be no tangential component of the acceleration of Q with respect to P. We know that radial component of the acceleration of Q with respect to P (or the acceleration of Q), 2 2 v (0.625) QP r 2 6.25 m/s aaa == = = = QP QP Q 0.0625 PQ Radial component of the acceleration of R with respect to Q, 2 2 v (0.333) RQ r 2 a == = 0.634 m/s RQ QR 0.175 and radial component of the acceleration of R with respect to S (or the acceleration of R), 2 2 v (0.426) r RS 2 == = = = 1.613 m/s aa a RS RS R SR 0.1125 The acceleration diagram, as shown in Fig. 8.9 (c) is drawn as follows : 1. Since P and S are fixed points, therefore these points lie at one place in the acceleration diagram. Draw vector p'q' parallel to PQ, to some suitable scale, to represent the radial component r of acceleration of Q with respect to P or acceleration of Q i.e aa or such that QP Q r 2 ′′ vector pq==aa= 6.25 m/s QP Q 2. From point q', draw vector q' x parallel to QR to represent the radial component of r acceleration of R with respect to Q i.e. such that a RQ r 2 vector ′== 0.634 m/s qxa RQ 3. From point x, draw vector xr' perpendicular to QR to represent the tangential component t of acceleration of R with respect to Q i.e a whose magnitude is not yet known. RQ 4. Now from point s', draw vector s'y parallel to SR to represent the radial component of the r acceleration of R with respect to S i.e. such that a RS r 2 vector ′== 1.613 m/s s y a RS 5. From point y, draw vector yr' perpendicular to SR to represent the tangential component t of acceleration of R with respect to S i.e. a . RS 6. The vectors xr' and yr' intersect at r'. Join p'r and q' r'. By measurement, we find that tt22 == vector′′ 4.1 m/s and= vector= 5.3 m/s axr a yr RQ RS We know that angular acceleration of link QR, t a 4.1 RQ 2 α= = = 23.43 rad/s (Anticlockwise) Ans. QR QR 0.175 and angular acceleration of link RS, t a 5.3 2 RS α= = = 47.1 rad/s (Anticlockwise) Ans. RS 0.1125 SR 186 Theory of Machines Example 8.5. The dimensions and configuration of the four bar mechanism, shown in Fig. 8.10, are as follows : P A = 300 mm; P B = 360 mm; AB = 360 1 2 mm, and P P = 600 mm. 1 2 The angle AP P = 60°. The crank P A has 1 2 1 an angular velocity of 10 rad/s and an angular 2 acceleration of 30 rad/s , both clockwise. Determine the angular velocities and angular accelerations of P B, and AB and the velocity and 2 Fig. 8.10 acceleration of the joint B. 2 Solution. Given : ω = 10 rad/s ; α = 30 rad/s ; P A = 300 mm = 0.3 m ; P B = AB = AP1 AP1 1 2 360 mm = 0.36 m We know that the velocity of A with respect to P or velocity of A, 1 v = v = ω × P A = 10 × 0.3 = 3 m/s AP1 A AP1 1 Velocity of B and angular velocitites of P B and AB 2 First of all, draw the space diagram, to some suitable scale, as shown in Fig. 8.11 (a). Now the velocity diagram, as shown in Fig. 8.11 (b), is drawn as discussed below: 1. Since P and P are fixed points, therefore these points lie at one place in velocity diagram. 1 2 Draw vector p a perpendicular to P A, to some suitable scale, to represent the velocity of A with 1 1 respect to P or velocity of A i.e. v or v , such that 1 AP1 A vector p a = v = v = 3 m/s 1 AP1 A 2. From point a, draw vector ab perpendicular to AB to represent velocity of B with respect to A (i.e. v ) and from point p draw vector p b perpendicular to P B to represent the velocity of B BA 2 2 2 with respect to P or velocity of B i.e. v or v . The vectors ab and p b intersect at b. 2 BP2 B 2 By measurement, we find that v = v = vector p b = 2.2 m/s Ans. BP2 B 2 and v = vector ab = 2.05 m/s BA We know that angular velocity of P B, 2 v 2.2 BP 2 ω= = = 6.1 rad/s (Clockwise) Ans. P2B 0.36 PB 2 and angular velocity of AB, v 2.05 BA ω= = = 5.7 rad/s (Anticlockwise) AB Ans. AB 0.36 Acceleration of B and angular acceleration of P B and AB 2 We know that tangential component of the acceleration of A with respect to P , 1 t 2 aP =α ×A = 30× 0.3 = 9 m/s A A1 PP 11 Radial component of the acceleration of A with respect to P , 1 2 v A r P22 2 1 aP ==ω ×A= 10× 0.3= 30 m/s AP AP 1 11 PA 1 Chapter 8 : Acceleration in Mechanisms 187 Radial component of the acceleration of B with respect to A. 2 2 v (2.05) r BA 2 11.67 m/s a == = BA 0.36 AB and radial component of the acceleration of B with respect to P , 2 2 2 v (2.2) BP r 2 2 == = 13.44 m/s a BP 2 0.36 PB 2 (a) Space diagram. (b) Velocity diagram. Fig. 8.11 The acceleration diagram, as shown in Fig. 8.11 (c), is drawn as follows: 1. Since P and P are fixed points, therefore these points 1 2 will lie at one place, in the acceleration diagram. Draw vector p ' x parallel to P A, to some suitable scale, to represent the 1 1 radial component of the acceleration of A with respect to P , 1 such that r 2 ′ vector == 30 m/s pxa 1AP 1 2. From point x, draw vector xa' perpendicular to P A to 1 represent the tangential component of the acceleration of A with respect to P , such that 1 t 2 xa′== a vector 9 m/s AP 1 (c) Acceleration diagram 3. Join p ' a'. The vector p ' a' represents the acceleration 1 1 of A . By measurement, we find that the acceleration of A, Fig. 8.11 2 a = a = 31.6 m/s A AP1 4. From point a', draw vector a' y parallel to AB to represent the radial component of the acceleration of B with respect to A, such that r 2 ′ vector a y== a 11.67 m/s BA 5. From point y, draw vector yb' perpendicular to AB to represent the tangential component t of the acceleration of B with respect to A (i.e. ) whose magnitude is yet unknown. a BA ′ ′ , p pz 6. Now from point draw vector parallel to P B to represent the radial component 2 2 2 of the acceleration B with respect to P , such that 2 r 2 ′ vector pza== 13.44 m/s 2BP 2 188 Theory of Machines 7. From point z, draw vector zb' perpendicular to P B to represent the tangential component 2 t of the acceleration of B with respect to P i.e. a . BP 2 2 8. The vectors yb' and zb' intersect at b'. Now the vector p ' b' represents the acceleration of 2 B with respect to P or the acceleration of B i.e. a or a . By measurement, we find that 2 BP2 B 2 a = a = vector p ' b' = 29.6 m/s Ans. BP2 B 2 tt22 ′′ Also vector yb== a 13.6 m/s , and vector zb= a = 26.6 m/s BA BP 2 We know that angular acceleration of P B, 2 t a BP 26.6 2 2 α= = = 73.8 rad/s (Anticlockwise) Ans. P2B PB 0.36 2 t a 13.6 BA 2 α= = = 37.8 rad/s (Anticlockwise) and angular acceleration of AB, Ans. AB AB 0.36 Bicycle is a common example where simple mechanisms are used. Note : This picture is given as additional information and is not a direct example of the current chapter. Example 8.6. In the mechanism, as shown in Fig. 8.12, the crank OA rotates at 20 r.p.m. anticlockwise and gives motion to the sliding blocks B and D. The dimensions of the various links are OA = 300 mm; AB = 1200 mm; BC = 450 mm and CD = 450 mm. Fig. 8.12 For the given configuration, determine : 1. velocities of sliding at B and D, 2. angular velocity of CD, 3. linear acceleration of D, and 4. angular acceleration of CD. Solution. Given : N = 20 r.p.m. or ω = 2 π × 20/60 = 2.1 rad/s ; OA = 300 mm = 0.3 m ; AO AO AB = 1200 mm = 1.2 m ; BC = CD = 450 mm = 0.45 m Chapter 8 : Acceleration in Mechanisms 189 We know that linear velocity of A with respect to O or velocity of A, v = v = ω × OA = 2.1 × 0.3 = 0.63 m/s ...(Perpendicular to OA) AO A AO 1. Velocities of sliding at B and D First of all, draw the space diagram, to some suitable scale, as shown in Fig. 8.13 (a). Now the velocity diagram, as shown in Fig. 8.13 (b), is drawn as discussed below: (a) Space diagram. (b) Velocity diagram. (c) Acceleration diagram. Fig. 8.13 1. Draw vector oa perpendicular to OA, to some suitable scale, to represent the velocity of A with respect to O (or simply velocity of A), such that vector oa = v = v = 0.63 m/s AO A 2. From point a, draw vector ab perpendicular to AB to represent the velocity of B with respect to A (i.e. v ) and from point o draw vector ob parallel to path of motion B (which is along BA BO) to represent the velocity of B with respect to O (or simply velocity of B). The vectors ab and ob intersect at b. 3. Divide vector ab at c in the same ratio as C divides AB in the space diagram. In other words, BC/CA = bc/ca 4. Now from point c, draw vector cd perpendicular to CD to represent the velocity of D with respect to C (i.e. v ) and from point o draw vector od parallel to the path of motion of D (which DC along the vertical direction) to represent the velocity of D. By measurement, we find that velocity of sliding at B, v = vector ob = 0.4 m/s Ans. B and velocity of sliding at D, v = vector od = 0.24 m/s Ans. D 2. Angular velocity of CD By measurement from velocity diagram, we find that velocity of D with respect to C, v = vector cd = 0.37 m/s DC 190 Theory of Machines ∴ Angular velocity of CD, v 0.37 DC 0.82 rad/s (Anticlockwise). ω= = = Ans. CD CD 0.45 3. Linear acceleration of D We know that the radial component of the acceleration of A with respect to O or acceleration of A, 2 v r AO22 2 aa== =ω ×OA= (2.1)× 0.3= 1.323 m/s AO A AO OA Radial component of the acceleration of B with respect to A, 2 2 v (0.54) r 2 BA a == = 0.243 m/s BA AB 1.2 ...(By measurement, v = 0.54 m/s) BA Radial component of the acceleration of D with respect to C, 2 2 v (0.37) r 2 DC a == = 0.304 m/s DC CD 0.45 Now the acceleration diagram, as shown in Fig. 8.13 (c), is drawn as discussed below: 1. Draw vector o' a' parallel to OA, to some suitable scale, to represent the radial component of the acceleration of A with respect to O or simply the acceleration of A , such that r 2 ′′ vector oa== a a= 1.323 m/s AO A 2. From point a', draw vector a' x parallel to AB to represent the radial component of the acceleration of B with respect to A, such that r 2 vector ′== 0.243 m/s ax a BA 3. From point x, draw vector xb' perpendicular to AB to represent the tangential component t of the acceleration of B with respect to A (i.e. ) whose magnitude is not yet known. a BA 4. From point o', draw vector o' b' parallel to the path of motion of B (which is along BO) to represent the acceleration of B (a ). The vectors xb' and o' b' intersect at b'. Join a' b'. The vector B a' b' represents the acceleration of B with respect to A. 5. Divide vector a' b' at c' in the same ratio as C divides AB in the space diagram. In other words, BC / B A = b' c'/b' a' 6. From point c', draw vector c'y parallel to CD to represent the radial component of the acceleration of D with respect to C, such that r 2 vector ′ 0.304 m/s c y== a DC 7. From point y, draw yd' perpendicular to CD to represent the tangential component of t acceleration of D with respect to C (. ie. a ) whose magnitude is not yet known. DC 8. From point o', draw vector o' d' parallel to the path of motion of D (which is along the vertical direction) to represent the acceleration of D (a ). The vectors yd' and o' d' intersect at d'. D By measurement, we find that linear acceleration of D, 2 a = vector o' d' = 0.16 m/s Ans. D 4. Angular acceleration of CD From the acceleration diagram, we find that the tangential component of the acceleration of D with respect to C, t 2 ′ ...(By measurement) a== vector yd 1.28 m/s DC Chapter 8 : Acceleration in Mechanisms 191 ∴ Angular acceleration of CD, t a 1.28 2 DC Ans. α= = = 2.84 rad/s (Clockwise) CD 0.45 CD Example 8.7. Find out the acceleration of the slider D and the angular acceleration of link CD for the engine mechanism shown in Fig. 8.14. The crank OA rotates uniformly at 180 r.p.m. in clockwise direction. The various lengths are: OA = 150 mm ; AB = 450 mm; PB = 240 mm ; BC = 210 mm ; CD = 660 mm. Solution. Given: N = 180 r.p.m., or ω = 2π × 180/60 = AO AO 18.85 rad/s ; OA = 150 mm = 0.15 m ; AB = 450 mm = 0.45 m ; PB = 240 mm = 0.24 m ; CD = 660 mm = 0.66 m We know that velocity of A with respect to O or velocity of A, v = v = ω × OA AO A AO All dimensions in mm. = 18.85 × 0.15 = 2.83 m/s Fig. 8.14 ...(Perpendicular to OA) First of all draw the space diagram, to some suitable scale, as shown in Fig. 8.15 (a). Now the velocity diagram, as shown in Fig. 8.15 (b), is drawn as discussed below: (a) Space diagram. (b) Velocity diagram. (c) Acceleration diagram. Fig. 8.15 1. Since O and P are fixed points, therefore these points lie at one place in the velocity diagram. Draw vector oa perpendicular to OA, to some suitable scale, to represent the velocity of A with respect to O or velocity of A (i.e. v or v ), such that AO A vector oa = v = v = 2.83 m/s AO A 2. Since the point B moves with respect to A and also with respect to P, therefore draw vector ab perpendicular to AB to represent the velocity of B with respect to A i.e. v ,and from point BA p draw vector pb perpendicular to PB to represent the velocity of B with respect to P or velocity of B (i.e. v or v ). The vectors ab and pb intersect at b. BP B 3. Since the point C lies on PB produced, therefore divide vector pb at c in the same ratio as C divides PB in the space diagram. In other words, pb/pc = PB/PC. 192 Theory of Machines 4. From point c, draw vector cd perpendicular to CD to represent the velocity of D with respect to C and from point o draw vector od parallel to the path of motion of the slider D (which is vertical), to represent the velocity of D, i.e. v . D By measurement, we find that velocity of the slider D, v = vector od = 2.36 m/s D Velocity of D with respect to C, v = vector cd = 1.2 m/s DC Velocity of B with respect to A, v = vector ab = 1.8 m/s BA and velocity of B with respect to P, v = vector pb = 1.5 m/s BP Acceleration of the slider D We know that radial component of the acceleration of A with respect to O or acceleration of A, r 22 2 aa==ω ×AO= (18.85)×0.15= 53.3 m/s AO A AO Radial component of the acceleration of B with respect to A, 2 2 v (1.8) r 2 BA a == = 7.2 m/s BA AB 0.45 Radial component of the acceleration of B with respect to P, 2 2 v (1.5) r BP 2 a == = 9.4 m/s BP PB 0.24 Radial component of the acceleration of D with respect to C, 2 2 v (1.2) r 2 DC a == = 2.2 m/s DC CD 0.66 Now the acceleration diagram, as shown in Fig. 8.15 (c), is drawn as discussed below: 1. Since O and P are fixed points, therefore these points lie at one place in the acceleration diagram. Draw vector o' a' parallel to OA, to some suitable scale, to represent the radial component r of the acceleration of A with respect to O or the acceleration of A (i.e. or a ), such that a AO A r 2 vector ′′== = 53.3 m/s oa a a AO A 2. From point a', draw vector a' x parallel to AB to represent the radial component of the r acceleration of B with respect to A (i.e. ), such that a BA r 2 ′ vector ax== a 7.2 m/s BA 3. From point x, draw vector xb' perpendicular to the vector a'x or AB to represent the t tangential component of the acceleration of B with respect to A i.e. whose magnitude is yet a BA unknown. 4. Now from point p', draw vector p' y parallel to PB to represent the radial component of r the acceleration of B with respect to P (i.e. ), such that a BP r 2 ′ vector py== a 9.4 m/s BP 5. From point y, draw vector yb' perpendicular to vector b'y or PB to represent the tangential t component of the acceleration of B, i.e. a . The vectors xb' and yb' intersect at b'. Join p' b'. The BP vector p' b' represents the acceleration of B, i.e. a . B Chapter 8 : Acceleration in Mechanisms 193 6. Since the point C lies on PB produced, therefore divide vector p'b' at c' in the same ratio as C divides PB in the space diagram. In other words, p'b'/p'c' = PB/PC 7. From point c', draw vector c'z parallel to CD to represent the radial component of the r acceleration of D with respect to C i.e. a , such that DC r 2 ′ vector cz== a 2.2 m/s DC 8. From point z, draw vector zd' perpendicular to vector c'z or CD to represent the tangential t component of the acceleration of D with respect to C i.e. whose magnitude is yet unknown. a , DC 9. From point o', draw vector o' d' parallel to the path of motion of D (which is vertical) to represent the acceleration of D, i.e. a . The vectors zd' and o' d' intersect at d'. Join c' d'. D By measurement, we find that acceleration of D, 2 a = vector o'd' = 69.6 m/s Ans. D Angular acceleration of CD From acceleration diagram, we find that tangential component of the acceleration of D with respect to C, t 2 az== vectord′ 17.4 m/s ...(By measurement) DC We know that angular acceleration of CD, t a 17.4 2 DC Ans. α= = = 26.3 rad / s (Anticlockwise) CD CD 0.66 Example 8.8. In the toggle mechanism shown in Fig. 8.16, the slider D is constrained to move on a horizontal path. The crank OA is rotating in the counter-clockwise direction at a speed Fig. 8.16 2 of 180 r.p.m. increasing at the rate of 50 rad/s . The dimensions of the various links are as follows: OA = 180 mm ; CB = 240 mm ; AB = 360 mm ; and BD = 540 mm. For the given configuration, find 1. Velocity of slider D and angular velocity of BD, and 2. Acceleration of slider D and angular acceleration of BD. Solution. Given : N = 180 r.p.m. or ω = 2 π × 180/60 = 18.85 rad/s ; OA = 180 mm AO AO = 0.18 m ; CB = 240 mm = 0.24 m ; AB = 360 mm = 0.36 m ; BD = 540 mm = 0.54 m We know that velocity of A with respect to O or velocity of A, v = v = ω × OA = 18.85 × 0.18 = 3.4 m/s AO A AO ...(Perpendicular to OA)

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