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- 1. CONTENTSCONTENTS174 l Theory of MachinesFeatures eatur tures Warping Machine1. Introduction. Acceleration 82. Acceleration Diagram for a Link.3. Acceleration of a Point on a in Mechanisms Link.4. Acceleration in the Slider 8.1. Introduction Introduction Crank Mechanism. We have discussed in the previous chapter the5. Coriolis Component of Acceleration. velocities of various points in the mechanisms. Now we shall 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. 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 an angular velocity of ω rad/s and let α rad/s2 be the angular acceleration of the link AB. (a) Link. (b) Acceleration diagram. Fig. 8.1. Acceleration for a link. 174CONTENTSCONTENTS
- 2. Chapter 8 : Acceleration in Mechanisms l 175 We have already discussed that acceleration of a particle whose velocity changes both inmagnitude and direction at any instant has the following two components : 1. The centripetal or radial component, which is perpendicular to the velocity of theparticle at the given instant. 2. The tangential component, which is parallel to the velocity of the particle at the giveninstant. Thus for a link A B, the velocity of point B with respect to A (i.e. vBA) is perpendicular to thelink A B as shown in Fig. 8.1 (a). Since the point B moves with respect to A with an angular velocityof ω rad/s, therefore centripetal or radial component of the acceleration of B with respect to A , v aBA = ω2 × Length of link AB = ω2 × AB = vBA / AB r 2 ... 3 ω = BA AB This radial component of acceleration acts perpendicular to the velocity v BA, In other words,it acts parallel to the link AB. We know that tangential component of the acceleration of B with respect to A , aBA = α × Length of the link AB = α × AB t This tangential component of acceleration acts parallel to the velocity v BA. In other words,it acts perpendicular to the link A B. In order to draw the acceleration diagram for a link A B, as shown in Fig. 8.1 (b), from anypoint b, draw vector bx parallel to BA to represent the radial component of acceleration of B with rrespect to A i.e. aBA and from point x draw vector xa perpendicular to B A to represent the tangential tcomponent of acceleration of B with respect to A i.e. aBA . Join b a. The vector b a (known asacceleration image of the link A B) represents the total acceleration of B with respect to A (i.e. aBA) r tand it is the vector sum of radial component (aBA ) and tangential component (aBA ) of acceleration.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 accelerationof the point A i.e. aA is known in magnitude and direction and the direction of path of B is given. Theacceleration of the point B is determined in magnitude and direction by drawing the accelerationdiagram as discussed below. 1. From any point o, draw vector oa parallel to the direction of absolute acceleration atpoint A i.e. aA , to some suitable scale, as shown in Fig. 8.2 (b).
- 3. 176 l Theory of Machines 2. We know that the acceleration of B withrespect to A i.e. a BA has the following twocomponents: (i) Radial component of the acceleration rof B with respect to A i.e. aBA , and (ii) Tangential component of the tacceleration B with respect to A i.e. aBA . These twocomponents are mutually perpendicular. 3. Draw vector ax parallel to the link A B(because radial component of the acceleration of Bwith respect to A will pass through AB), such that vector a′x = aBA = vBA / AB r 2where vBA = Velocity of B with respect to A .Note: The value of v BA may be obtained by drawing thevelocity diagram as discussed in the previous chapter. 4. From point x , draw vector xbperpendicular to A B or vector ax (because tangential tcomponent of B with respect to A i.e. aBA , is r A refracting telescope uses mechanisms toperpendicular to radial component aBA ) and change directions.through o draw a line parallel to the path of B to Note : This picture is given as additionalrepresent the absolute acceleration of B i.e. aB. The information and is not a direct example of thevectors xb and o b intersect at b. Now the values current chapter. tof aB and aBA may be measured, to the scale. 5. By joining the points a and b we may determine the total acceleration of B with respectto A i.e. aBA. The vector a b is known as acceleration image of the link A B. 6. For any other point C on the link, draw triangle a b c similar to triangle ABC. Nowvector b c represents the acceleration of C with respect to B i.e. aCB, and vector a c represents theacceleration of C with respect to A i.e. aCA. As discussed above, aCB and aCA will each have twocomponents as follows : r t (i) aCB has two components; aCB and aCB as shown by triangle b zc in Fig. 8.2 (b), in which b z is parallel to BC and zc is perpendicular to b z or BC. r t (ii) aCA has two components ; aCA and aCA as shown by triangle a yc in Fig. 8.2 (b), in which a y is parallel to A C and yc is perpendicular to a y or A C. 7. The angular acceleration of the link AB is obtained by dividing the tangential components tof the acceleration of B with respect to A (aBA ) to the length of the link. Mathematically, angularacceleration of the link A B, α AB = aBA / AB t8.4. 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 withthe inner dead centre (I.D.C) and rotates in a clockwise direction about the fixed point O withuniform angular velocity ωBO rad/s. ∴ Velocity of B with respect to O or velocity of B (because O is a fixed point), vBO = vB = ωBO × OB , acting tangentially at B .
- 4. Chapter 8 : Acceleration in Mechanisms l 177 We know that centripetal or radial acceleration of B with respect to O or acceleration of B(because O is a fixed point), v2 aBO = aB = ωBO × OB = BO r 2 OBNote : A point at the end of a link which moves with constant angular velocity has no tangential componentof 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: 1. Draw vector o b parallel to BO and set off equal in magnitude of aBO = aB , to some rsuitable scale. 2. From point b, draw vector bx parallel to B A. The vector bx represents the radial componentof the acceleration of A with respect to B whose magnitude is given by : aAB = vAB / BA r 2 Since the point B moves with constant angular velocity, therefore there will be no tangentialcomponent of the acceleration. 3. From point x, draw vector xa perpendicular to bx (or A B). The vector xa represents the ttangential component of the acceleration of A with respect to B i.e. aAB .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 A O, therefore the acceleration must be parallel tovelocity. Therefore from o, draw o a parallel to A O, intersecting the vector xa at a. t Now the acceleration of the piston or the slider A (aA) and aAB may be measured to the scale. 5. The vector b a, which is the sum of the vectors b x and x a, represents the total accelerationof A with respect to B i.e. aAB. The vector b a represents the acceleration of the connecting rod A B. 6. The acceleration of any other point on A B such as E may be obtained by dividing the vectorb a at e in the same ratio as E divides A B in Fig. 8.3 (a). In other words a e / a b = AE / AB 7. The angular acceleration of the connecting rod A B may be obtained by dividing thetangential component of the acceleration of A with respect to B ( aAB ) to the length of A B. In other twords, angular acceleration of A B, α AB = aAB / AB (Clockwise about B) t Example 8.1. The crank of a slider crank mechanism rotates clockwise at a constant speedof 300 r.p.m. The crank is 150 mm and the connecting rod is 600 mm long. Determine : 1. linearvelocity and acceleration of the midpoint of the connecting rod, and 2. angular velocity and angularacceleration of the connecting rod, at a crank angle of 45° from inner dead centre position.
- 5. 178 l Theory of Machines Solution. Given : N BO = 300 r.p.m. or ωBO = 2 π × 300/60 = 31.42 rad/s; OB = 150 mm =0.15 m ; B A = 600 mm = 0.6 m We know that linear velocity of B with respect to O or velocity of B, vBO = v B = ωBO × OB = 31.42 × 0.15 = 4.713 m/s ...(Perpendicular to BO) (a) Space diagram. (b) Velocity diagram. (c) Acceleration diagram. Fig. 8.4 Ram moves Ram moves outwards Load moves inwards outwards Oil pressure on lower side of piston Oil pressure on 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 thevelocity 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 ofB with respect to O or simply velocity of B i.e. vBO or vB, such that vector ob = v BO = v B = 4.713 m/s 2. From point b, draw vector ba perpendicular to BA to represent the velocity of A withrespect to B i.e. vAB , and from point o draw vector oa parallel to the motion of A (which is along A O)to represent the velocity of A i.e. vA. The vectors ba and oa intersect at a.
- 6. Chapter 8 : Acceleration in Mechanisms l 179 By measurement, we find that velocity of A with respect to B, vAB = vector ba = 3.4 m / sand Velocity of A, vA = vector oa = 4 m / s 3. In order to find the velocity of the midpoint D of the connecting rod A B, divide the vectorba at d in the same ratio as D divides A B, in the space diagram. In other words, bd / ba = BD/BANote: Since D is the midpoint of A B, therefore d is also midpoint of vector ba. 4. Join od. Now the vector od represents the velocity of the midpoint D of the connectingrod i.e. vD. By measurement, we find that vD = vector od = 4.1 m/s Ans.Acceleration of the midpoint of the connecting rod We know that the radial component of the acceleration of B with respect to O or theacceleration of B, 2 vBO (4.713)2 aBO = aB = r = = 148.1 m/s 2 OB 0.15and the radial component of the acceleraiton of A with respect to B, 2 vAB (3.4)2 aAB = r = = 19.3 m/s 2 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 rof the acceleration of B with respect to O or simply acceleration of B i.e. aBO or aB , such that vector o′ b′ = aBO = aB = 148.1 m/s 2 rNote: Since the crank OB rotates at a constant speed, therefore there will be no tangential component of theacceleration 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. aAB , and t (b) The tangential component of the acceleration of A with respect to B i.e. aAB . These two components are mutually perpendicular. Therefore from point b, draw vector b x parallel to A B to represent aAB = 19.3 m/s 2 and rfrom 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 A O) torepresent the acceleration of A i.e. aA . The vectors xa and o a intersect at a. Join a b. 4. In order to find the acceleration of the midpoint D of the connecting rod A B, divide thevector a b at d in the same ratio as D divides A B. In other words b′d ′ / b′a ′ = BD / BANote: Since D is the midpoint of A B, 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 rodi.e. aD. By measurement, we find that aD = vector o d = 117 m/s2 Ans.
- 7. 180 l Theory of Machines2. Angular velocity of the connecting rod We know that angular velocity of the connecting rod A B, vAB 3.4 ωAB = = = 5.67 rad/s2 (Anticlockwise about B ) Ans. BA 0.6Angular acceleration of the connecting rod From the acceleration diagram, we find that aAB = 103 m/s 2 t ...(By measurement) We know that angular acceleration of the connecting rod A B, t aAB 103 α AB = = = 171.67 rad/s 2 (Clockwise about B ) Ans. BA 0.6 Example 8.2. An engine mechanism is shown in Fig. 8.5. The crank CB = 100 mm and theconnecting rod BA = 300 mm with centre of gravity G, 100 mm from B. In the position shown, thecrankshaft has a speed of 75 rad/s and an angular acceleration of 1200 rad/s2. Find:1. velocity ofG and angular velocity of AB, and 2. acceleration of G and angular acceleration of AB. Fig. 8.5 Solution. Given : ωBC = 75 rad/s ; αBC = 1200 rad/s2, CB = 100 mm = 0.1 m; B A = 300 mm= 0.3 m We know that velocity of B with respect to C or velocity of B, vBC = vB = ωBC × CB = 75 × 0.1 = 7.5 m/s ...(Perpendicular to BC) Since the angular acceleration of the crankshaft, αBC = 1200 rad/s2, therefore tangentialcomponent of the acceleration of B with respect to C, aBC = αBC × CB = 1200 × 0.1 = 120 m/s2 tNote: 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 thevelocity 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 ofB with respect to C or velocity of B (i.e. vBC or vB), such that vector cb = vBC = vB = 7.5 m/s 2. From point b, draw vector ba perpendicular to B A to represent the velocity of A withrespect to B i.e. vAB , and from point c, draw vector ca parallel to the path of motion of A (which isalong A C) to represent the velocity of A i.e. vA.The vectors ba and ca intersect at a. 3. Since the point G lies on A B, therefore divide vector ab at g in the same ratio as G dividesA B 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 G = vector cg = 6.8 m/s Ans.
- 8. Chapter 8 : Acceleration in Mechanisms l 181 From velocity diagram, we find that velocity of A with respect to B, vAB = vector ba = 4 m/s We know that angular velocity of A B, vAB 4 ωAB = = = 13.3 rad/s (Clockwise) Ans. BA 0.3 (a) Space diagram. (b) Velocity diagram. Fig. 8.62. Acceleration of G and angular acceleration of AB We know that radial component of the acceleration of B withrespect to C, 2 2 * a r = vBC = (7.5) = 562.5 m/s2 BC CB 0.1and radial component of the acceleration of A with respect to B, 2 vAB 42 aAB = r = = 53.3 m/s 2 BA 0.3 Now the acceleration diagram, as shown in Fig. 8.6 (c), is drawnas 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 ri.e. aBC , such that vector c ′ b′′ = aBC = 562.5 m/s 2 r 2. From point b, draw vector b b perpendicular to vector c b or CB to represent the ttangential component of the acceleration of B with respect to C i.e. aBC , such that vector b′′ b′ = aBC = 120 m/s 2 t ... (Given) 3. Join c b. The vector c b represents the total acceleration of B with respect to C i.e. aBC. 4. From point b, draw vector b x parallel to B A to represent radial component of the racceleration of A with respect to B i.e. aAB such that vector b′x = aAB = 53.3 m/s 2 r 5. From point x, draw vector xa perpendicular to vector bx or B A to represent tangential tcomponent of the acceleration of A with respect to B i.e. aAB , whose magnitude is not yet known. 6. Now draw vector c a parallel to the path of motion of A (which is along A C) to representthe acceleration of A i.e. aA.The vectors xa and ca intersect at a. Join b a. The vector b arepresents the acceleration of A with respect to B i.e. aAB. t r* When angular acceleration of the crank is not given, then there is no aBC . In that case, aBC = aBC = aB , as discussed in the previous example.
- 9. 182 l Theory of Machines 7. In order to find the acceleratio of G, divide vector a b in g in the same ratio as G dividesB A 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, aG = vector c g = 414 m/s2 Ans. From acceleration diagram, we find that tangential component of the acceleration of A withrespect to B, aAB = vector xa′ = 546 m/s 2 t ...(By measurement) ∴ Angular acceleration of A B, t aAB 546 α AB = = = 1820 rad/s 2 (Clockwise) Ans. BA 0.3 Example 8.3. In the mechanism shown in Fig. 8.7, the slider C ismoving to the right with a velocity of 1 m/s and an acceleration of 2.5 m/s2. The dimensions of various links are AB = 3 m inclined at 45° with thevertical and BC = 1.5 m inclined at 45° with the horizontal. Determine: 1. themagnitude of vertical and horizontal component of the acceleration of thepoint B, and 2. the angular acceleration of the links AB and BC. Solution. Given : v C = 1 m/s ; aC = 2.5 m/s2; A B = 3 m ; BC = 1.5 m First of all, draw the space diagram, as shown in Fig. 8.8 (a), to some Fig. 8.7suitable scale. Now the velocity diagram, as shown in Fig. 8.8 (b), is drawn asdiscussed below: 1. Since the points A and D are fixed points, therefore they lie at one place in the velocitydiagram. Draw vector dc parallel to DC, to some suitable scale, which represents the velocity ofslider C with respect to D or simply velocity of C, such that vector dc = v CD = v C = 1 m/s 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 A B to represent the velocity of B withrespect to A , i.e. vBA and from point c draw vector cb perpendicular to CB to represent the velocityof B with respect to C i.e. vBC .The vectors ab and cb intersect at b. (a) Space diagram. (b) Velocity diagram. (c) Acceleration diagram. Fig. 8.8 By measurement, we find that velocity of B with respect to A , vBA = vector ab = 0.72 m/sand velocity of B with respect to C, vBC = vector cb = 0.72 m/s
- 10. Chapter 8 : Acceleration in Mechanisms l 183 We know that radial component of acceleration of B with respect to C, 2 vBC (0.72) 2 aBC = r = = 0.346 m/s2 CB 1.5and radial component of acceleration of B with respect to A , 2 vBA (0.72)2 aBA = r = = 0.173 m/s 2 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 accelerationdiagram. Draw vector d c parallel to DC, to some suitable scale, to represent the acceleration of Cwith respect to D or simply acceleration of C i.e. aCD or aC such that vector d ′ c′ = aCD = aC = 2.5 m/s 2 2. The acceleration of B with respect to C will have two components, i.e. one radial componentof B with respect to C (aBC ) and r the other tangential component of B with respect toC ( aBC ). Therefore from point c, draw vector c x parallel to CB to represent aBC such that t r vector c ′x = aBC = 0.346 m/s 2 r 3. Now from point x, draw vector xb perpendicular to vector c x or CB to represent atBCwhose magnitude is yet unknown. 4. The acceleration of B with respect to A will also have two components, i.e. one radialcomponent of B with respect to A (arBA) and other tangential component of B with respect to A (at BA).Therefore from point a draw vector a y parallel to A B to represent arBA, such that vector a y = arBA = 0.173 m/s2 t 5. From point y, draw vector yb perpendicular to vector ay or AB to represent aBA . Thevector yb intersect the vector xb at b. Join a b and c b. The vector a b represents the accelerationof point B (aB) and the vector c b represents the acceleration of B with respect to C.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 accelerationof the point B and a b is the horizontal component of the acceleration of the point B. By measurement, vector b b = 1.13 m/s2 and vector a b = 0.9 m/s2 Ans.2. Angular acceleration of AB and BC By measurement from acceleration diagram, we find that tangential component of accelerationof the point B with respect to A , aBA = vector yb′ = 1.41 m/s 2 tand tangential component of acceleration of the point B with respect to C, aBC = vector xb′ = 1.94 m/s 2 t* 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.
- 11. 184 l Theory of Machines We know that angular acceleration of A B, t aBA 1.41 α AB = = = 0.47 rad/s 2 Ans. AB 3and angular acceleration of BC, t aBA 1.94 α BC = = = 1.3 rad/s 2 Ans. 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/sclockwise. Draw the velocity and acceleration diagram when angle QPS = 60° and Q and R lie onthe same side of PS. Find the angular velocity and angular acceleration of links QR and RS. Solution. Given : ωQP = 10 rad/s; PQ = 62.5 mm = 0.0625 m ; QR = 175 mm = 0.175 m ;R S = 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 QP = v Q = ωQP × PQ = 10 × 0.0625 = 0.625 m/s ...(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 inFig. 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 withrespect to P or velocity of Q i.e. v QP or v Q such that vector pq = v QP = v Q = 0.625 m/s 2. From point q, draw vector qr perpendicular to QR to represent the velocity of R withrespect to Q (i.e. vRQ) and from point s, draw vector sr perpendicular to S R to represent the velocityof R with respect to S or velocity of R (i.e. vRS or v R). The vectors qr and sr intersect at r. Bymeasurement, we find that vRQ = vector qr = 0.333 m/s, and v RS = v R = vector sr = 0.426 m/s We know that angular velocity of link QR, vRQ 0.333 ωQR = = = 1.9 rad/s (Anticlockwise) Ans. RQ 0.175
- 12. Chapter 8 : Acceleration in Mechanisms l 185and angular velocity of link R S, vRS 0.426 ωRS = = = 3.78 rad/s (Clockwise) A Ans. SR 0.1125Angular acceleration of links QR and RS Since the angular acceleration of the crank PQ is not given, therefore there will be no tangentialcomponent 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 accelerationof Q), 2 vQP (0.625) 2 aQP = aQP = aQ = r = = 6.25 m/s 2 PQ 0.0625 Radial component of the acceleration of R with respect to Q, 2 vRQ (0.333)2 aRQ = r = = 0.634 m/s2 QR 0.175and radial component of the acceleration of R with respect to S (or the acceleration of R), 2 vRS (0.426)2 aRS = aRS = aR = r = = 1.613 m/s 2 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 accelerationdiagram. Draw vector pq parallel to PQ, to some suitable scale, to represent the radial component rof acceleration of Q with respect to P or acceleration of Q i.e aQP or aQ such that vector p′q′ = aQP = aQ = 6.25 m/s 2 r 2. From point q, draw vector q x parallel to QR to represent the radial component of racceleration of R with respect to Q i.e. aRQ such that vector q ′x = aRQ = 0.634 m/s 2 r 3. From point x, draw vector xr perpendicular to QR to represent the tangential component tof acceleration of R with respect to Q i.e aRQ whose magnitude is not yet known. 4. Now from point s, draw vector sy parallel to S R to represent the radial component of the racceleration of R with respect to S i.e. aRS such that vector s ′y = aRS = 1.613 m/s 2 r 5. From point y, draw vector yr perpendicular to S R to represent the tangential component tof acceleration of R with respect to S i.e. aRS . 6. The vectors xr and yr intersect at r. Join pr and q r. By measurement, we find that aRQ = vector xr ′ = 4.1 m/s 2 and aRS = vector yr ′ = 5.3 m/s 2 t t We know that angular acceleration of link QR, t aRQ 4.1 α QR = = = 23.43 rad/s 2 (Anticlockwise) Ans. QR 0.175and angular acceleration of link R S, t aRS 5.3 α RS = = = 47.1 rad/s 2 (Anticlockwise) Ans. SR 0.1125
- 13. 186 l Theory of Machines Example 8.5. The dimensions andconfiguration of the four bar mechanism, shown inFig. 8.10, are as follows : P1A = 300 mm; P2B = 360 mm; AB = 360mm, and P1P2 = 600 mm. The angle AP1P2 = 60°. The crank P1A hasan angular velocity of 10 rad/s and an angularacceleration of 30 rad/s 2 , both clockwise.Determine the angular velocities and angularaccelerations of P2B, and AB and the velocity and Fig. 8.10acceleration of the joint B. Solution. Given : ωAP1 = 10 rad/s ; αAP1 = 30 rad/s2; P1A = 300 mm = 0.3 m ; P2B = A B =360 mm = 0.36 m We know that the velocity of A with respect to P1 or velocity of A , vAP1 = vA = ωAP1 × P1A = 10 × 0.3 = 3 m/sVelocity of B and angular velocitites of P2B and AB First of all, draw the space diagram, to some suitable scale, as shown in Fig. 8.11 (a). Nowthe velocity diagram, as shown in Fig. 8.11 (b), is drawn as discussed below: 1. Since P1 and P2 are fixed points, therefore these points lie at one place in velocity diagram.Draw vector p1 a perpendicular to P1A , to some suitable scale, to represent the velocity of A withrespect to P1 or velocity of A i.e. vAP1 or vA, such that vector p1a = v A P1 = v A = 3 m/s 2. From point a, draw vector ab perpendicular to AB to represent velocity of B with respectto A (i.e. vBA) and from point p2 draw vector p2b perpendicular to P2B to represent the velocity of Bwith respect to P2 or velocity of B i.e. vBP2 or v B. The vectors ab and p2b intersect at b. By measurement, we find that vBP2 = v B = vector p2b = 2.2 m/s Ans.and vBA = vector ab = 2.05 m/s We know that angular velocity of P2B, vBP 2 2.2 ωP2B = = = 6.1 rad/s (Clockwise) Ans. P2 B 0.36and angular velocity of A B, vBA 2.05 ωAB = = = 5.7 rad/s (Anticlockwise) Ans. AB 0.36Acceleration of B and angular acceleration of P2B and AB We know that tangential component of the acceleration of A with respect to P1, atA P1 = α A P1 × P A = 30 × 0.3 = 9 m/s 2 1 Radial component of the acceleration of A with respect to P1, 2 vA P1 aAP1 = r = ωAP1 × P A = 102 × 0.3 = 30 m/s2 2 1 PA 1
- 14. Chapter 8 : Acceleration in Mechanisms l 187 Radial component of the acceleration of B with respect to A . 2 vBA (2.05) 2 aBA = r = = 11.67 m/s 2 AB 0.36and radial component of the acceleration of B with respect to P2, 2 vBP 2 (2.2)2 aBP 2 = r = = 13.44 m/s2 P2 B 0.36 (a) Space diagram. (b) Velocity diagram. Fig. 8.11 The acceleration diagram, as shown in Fig. 8.11 (c), isdrawn as follows: 1. Since P1 and P2 are fixed points, therefore these pointswill lie at one place, in the acceleration diagram. Draw vectorp1 x parallel to P1A , to some suitable scale, to represent theradial component of the acceleration of A with respect to P1,such that vector p1′ x = aA P = 30 m/s 2 r 1 2. From point x, draw vector xa perpendicular to P1A torepresent the tangential component of the acceleration of A withrespect to P1, such that vector xa ′ = aA P1 = 9 m/s 2 t 3. Join p1 a. The vector p1 a represents the acceleration (c) Acceleration diagramof A . By measurement, we find that the acceleration of A , Fig. 8.11 aA = aAP1 = 31.6 m/s2 4. From point a, draw vector a y parallel to A B to represent the radial component of theacceleration of B with respect to A , such that vector a′y = aBA = 11.67 m/s 2 r 5. From point y, draw vector yb perpendicular to A B to represent the tangential component tof the acceleration of B with respect to A (i.e. aBA ) whose magnitude is yet unknown. 6. Now from point p ′ , draw vector p ′ z parallel to P B to represent the radial component 2 2 2of the acceleration B with respect to P2, such that vector p2′ z = aBP 2 = 13.44 m/s 2 r
- 15. 188 l Theory of Machines 7. From point z, draw vector zb perpendicular to P2B to represent the tangential component tof the acceleration of B with respect to P2 i.e. aBP 2 . 8. The vectors yb and zb intersect at b. Now the vector p2 b represents the acceleration ofB with respect to P2 or the acceleration of B i.e. aBP2 or aB. By measurement, we find that aBP2 = aB = vector p2 b = 29.6 m/s2 Ans. Also vector yb′ = aBA = 13.6 m/s 2 , and vector zb′ = aBP 2 = 26.6 m/s 2 t t We know that angular acceleration of P2B, t aBP 2 26.6 α = = = 73.8 rad/s 2 (Anticlockwise) Ans. P2B P2 B 0.36 t aBA 13.6and angular acceleration of A B, α AB = = = 37.8 rad/s 2 (Anticlockwise) Ans. 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 linksare 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. angularvelocity of CD, 3. linear acceleration of D, and 4. angular acceleration of CD. Solution. Given : NAO = 20 r.p.m. or ωAO = 2 π × 20/60 = 2.1 rad/s ; OA = 300 mm = 0.3 m ;A B = 1200 mm = 1.2 m ; BC = CD = 450 mm = 0.45 m
- 16. Chapter 8 : Acceleration in Mechanisms l 189 We know that linear velocity of A with respect to O or velocity of A , vAO = v A = ωAO × O A = 2.1 × 0.3 = 0.63 m/s ...(Perpendicular to O A)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). Nowthe 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 O A, to some suitable scale, to represent the velocity ofA with respect to O (or simply velocity of A ), such that vector oa = v AO = v A = 0.63 m/s 2. From point a, draw vector ab perpendicular to A B to represent the velocity of B withrespect to A (i.e. vBA) and from point o draw vector ob parallel to path of motion B (which is alongBO) to represent the velocity of B with respect to O (or simply velocity of B). The vectors ab and obintersect at b. 3. Divide vector ab at c in the same ratio as C divides A B in the space diagram. In otherwords, BC/CA = bc/ca 4. Now from point c, draw vector cd perpendicular to CD to represent the velocity of D withrespect to C (i.e. vDC) and from point o draw vector od parallel to the path of motion of D (whichalong the vertical direction) to represent the velocity of D. By measurement, we find that velocity of sliding at B, vB = vector ob = 0.4 m/s Ans.and velocity of sliding at D, v D = vector od = 0.24 m/s Ans.2. Angular velocity of CD By measurement from velocity diagram, we find that velocity of D with respect to C, vDC = vector cd = 0.37 m/s
- 17. 190 l Theory of Machines ∴ Angular velocity of CD, vDC 0.37 ωCD = = = 0.82 rad/s (Anticlockwise). Ans. CD 0.453. Linear acceleration of D We know that the radial component of the acceleration of A with respect to O or accelerationof A, 2 vAO aAO = aA = r = ωAO × OA = (2.1)2 × 0.3 = 1.323 m/s2 2 OA Radial component of the acceleration of B with respect to A , 2 vBA (0.54)2 aBA = r = = 0.243 m/s 2 AB 1.2 ...(By measurement, v BA = 0.54 m/s) Radial component of the acceleration of D with respect to C, 2 vDC (0.37)2 aDC = r = = 0.304 m/s 2 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 O A, to some suitable scale, to represent the radial componentof the acceleration of A with respect to O or simply the acceleration of A , such that vector o′a′ = aAO = aA = 1.323 m/s 2 r 2. From point a, draw vector a x parallel to A B to represent the radial component of theacceleration of B with respect to A , such that vector a ′x = aBA = 0.243 m/s 2 r 3. From point x, draw vector xb perpendicular to A B to represent the tangential component tof the acceleration of B with respect to A (i.e. aBA ) whose magnitude is not yet known. 4. From point o, draw vector o b parallel to the path of motion of B (which is along BO) torepresent the acceleration of B (aB). The vectors xb and o b intersect at b. Join a b. The vectora b represents the acceleration of B with respect to A . 5. Divide vector a b at c in the same ratio as C divides A B in the space diagram. In otherwords, BC / B A = b c/b a 6. From point c, draw vector cy parallel to CD to represent the radial component of theacceleration of D with respect to C, such that vector c ′y = aDC = 0.304 m/s 2 r 7. From point y, draw yd perpendicular to CD to represent the tangential component of tacceleration of D with respect to C (i.e. aDC ) whose magnitude is not yet known. 8. From point o, draw vector o d parallel to the path of motion of D (which is along thevertical direction) to represent the acceleration of D (aD). The vectors yd and o d intersect at d. By measurement, we find that linear acceleration of D, aD = vector o d = 0.16 m/s2 Ans.4. Angular acceleration of CD From the acceleration diagram, we find that the tangential component of the acceleration ofD with respect to C, aDC = vector yd ′ = 1.28 m/s 2 t ...(By measurement)
- 18. Chapter 8 : Acceleration in Mechanisms l 191 ∴ Angular acceleration of CD, t aDC 1.28 α CD = = = 2.84 rad/s2 (Clockwise) Ans. CD 0.45 Example 8.7. Find out the acceleration of the slider D andthe angular acceleration of link CD for the engine mechanism shownin Fig. 8.14. The crank OA rotates uniformly at 180 r.p.m. in clockwisedirection. The various lengths are: OA = 150 mm ; AB = 450 mm;PB = 240 mm ; BC = 210 mm ; CD = 660 mm. Solution. Given: N AO = 180 r.p.m., or ωAO = 2π × 180/60 =18.85 rad/s ; O A = 150 mm = 0.15 m ; A B = 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 velocityof A , v AO = v A = ωAO × O A = 18.85 × 0.15 = 2.83 m/s All dimensions in mm. ...(Perpendicular to O A) Fig. 8.14 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 discussedbelow: (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 velocitydiagram. Draw vector oa perpendicular to O A, to some suitable scale, to represent the velocity of Awith respect to O or velocity of A (i.e. vAO or vA), such that vector oa = v AO = v A = 2.83 m/s 2. Since the point B moves with respect to A and also with respect to P, therefore drawvector ab perpendicular to A B to represent the velocity of B with respect to A i.e. vBA ,and from pointp draw vector pb perpendicular to PB to represent the velocity of B with respect to P or velocity ofB (i.e. vBP or vB). The vectors ab and pb intersect at b. 3. Since the point C lies on PB produced, therefore divide vector pb at c in the same ratio asC divides PB in the space diagram. In other words, pb/pc = PB/PC.
- 19. 192 l Theory of Machines 4. From point c, draw vector cd perpendicular to CD to represent the velocity of D withrespect to C and from point o draw vector od parallel to the path of motion of the slider D (which isvertical), to represent the velocity of D, i.e. v D. By measurement, we find that velocity of the slider D, v D = vector od = 2.36 m/s Velocity of D with respect to C, vDC = vector cd = 1.2 m/s Velocity of B with respect to A , vBA = vector ab = 1.8 m/sand velocity of B with respect to P, v BP = vector pb = 1.5 m/sAcceleration of the slider D We know that radial component of the acceleration of A with respect to O or accelerationof A , aAO = aA = ωAO × AO = (18.85)2 × 0.15 = 53.3 m/s2 r 2 Radial component of the acceleration of B with respect to A , 2 vBA (1.8)2 aBA = r = = 7.2 m/s 2 AB 0.45 Radial component of the acceleration of B with respect to P, 2 vBP (1.5) 2 aBP = r = = 9.4 m/s 2 PB 0.24 Radial component of the acceleration of D with respect to C, 2 vDC (1.2)2 aDC = r = = 2.2 m/s 2 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 accelerationdiagram. Draw vector o a parallel to O A, to some suitable scale, to represent the radial component rof the acceleration of A with respect to O or the acceleration of A (i.e. aAO or aA), such that vector o′a′ = aAO = aA = 53.3 m/s 2 r 2. From point a, draw vector a x parallel to A B to represent the radial component of the racceleration of B with respect to A (i.e. aBA ), such that vector a′x = aBA = 7.2 m/s 2 r 3. From point x, draw vector xb perpendicular to the vector ax or AB to represent the ttangential component of the acceleration of B with respect to A i.e. aBA whose magnitude is yetunknown. 4. Now from point p, draw vector p y parallel to PB to represent the radial component of rthe acceleration of B with respect to P (i.e. aBP ), such that vector p′y = aBP = 9.4 m/s 2 r 5. From point y, draw vector yb perpendicular to vector by or PB to represent the tangential tcomponent of the acceleration of B, i.e. aBP . The vectors xb and yb intersect at b. Join p b. Thevector p b represents the acceleration of B, i.e. aB.
- 20. Chapter 8 : Acceleration in Mechanisms l 193 6. Since the point C lies on PB produced, therefore divide vector pb at c in the same ratioas C divides PB in the space diagram. In other words, pb/pc = PB/PC 7. From point c, draw vector cz parallel to CD to represent the radial component of the racceleration of D with respect to C i.e. aDC , such that vector c′z = aDC = 2.2 m/s 2 r 8. From point z, draw vector zd perpendicular to vector cz or CD to represent the tangential tcomponent of the acceleration of D with respect to C i.e. aDC , whose magnitude is yet unknown. 9. From point o, draw vector o d parallel to the path of motion of D (which is vertical) torepresent the acceleration of D, i.e. aD. The vectors zd and o d intersect at d. Join c d. By measurement, we find that acceleration of D, aD = vector od = 69.6 m/s2 Ans.Angular acceleration of CD From acceleration diagram, we find that tangential component of the acceleration of D withrespect to C, aDC = vector zd ′ = 17.4 m/s 2 t ...(By measurement) We know that angular acceleration of CD, t aDC 17.4 α CD = = = 26.3 rad / s 2 (Anticlockwise) Ans. CD 0.66 Example 8.8. In the toggle mechanism shown in Fig. 8.16, the slider D is constrained tomove on a horizontal path. The crank OA is rotating in the counter-clockwise direction at a speed Fig. 8.16of 180 r.p.m. increasing at the rate of 50 rad/s2. 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, and2. Acceleration of slider D and angular acceleration of BD. Solution. Given : NAO = 180 r.p.m. or ωAO = 2 π × 180/60 = 18.85 rad/s ; O A = 180 mm= 0.18 m ; CB = 240 mm = 0.24 m ; A B = 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 , vAO = v A = ωAO × O A = 18.85 × 0.18 = 3.4 m/s ...(Perpendicular to O A)
- 21. 194 l Theory of Machines1. Velocity of slider D and angular velocity of BD First of all, draw the space diagram to some suitable scale, as shown in Fig. 8.17 (a). Nowthe velocity diagram, as shown in Fig. 8.17 (b), is drawn as discussed below: 1. Since O and C are fixed points, therefore these points lie at one place in the velocitydiagram. Draw vector oa perpendicular to O A, to some suitable scale, to represent the velocity of Awith respect to O or velocity of A i.e. v AO or v A, such that vector oa = v AO = v A = 3.4 m/s (a) Space diagram. (b) Velocity diagram. (c) Acceleration diagram. Fig. 8.17 2. Since B moves with respect to A and also with respect to C, therefore draw vector abperpendicular to A B to represent the velocity of B with respect to A i.e. vBA, and draw vector cbperpendicular to CB to represent the velocity of B with respect to C ie. vBC. The vectors ab and cbintersect at b. 3. From point b, draw vector bd perpendicular to BD to represent the velocity of D withrespect to B i.e. v DB, and from point c draw vector cd parallel to CD (i.e., in the direction of motionof the slider D) to represent the velocity of D i.e. v D. By measurement, we find that velocity of B with respect to A , v BA = vector ab = 0.9 m/s Velocity of B with respect to C, v BC = vector cb = 2.8 m/s Velocity of D with respect to B, v DB = vector bd = 2.4 m/sand velocity of slider D, vD = vector cd = 2.05 m/s Ans.Angular velocity of BD We know that the angular velocity of BD, vDB 2.4 ωBD = = = 4.5 rad/s Ans. BD 0.542. Acceleration of slider D and angular acceleration of BD Since the angular acceleration of OA increases at the rate of 50 rad/s2, i.e. αAO = 50 rad/s2,therefore Tangential component of the acceleration of A with respect to O, aAO = αAO × OA = 50 × 0.18 = 9 m/s 2 t
- 22. Chapter 8 : Acceleration in Mechanisms l 195 Radial component of the acceleration of A with respect to O, 2 vAO (3.4)2 aAO = r = = 63.9 m/s 2 OA 0.18 Radial component of the acceleration of B with respect to A, 2 vBA (0.9)2 aBA = r = = 2.25 m/s 2 AB 0.36 Radial component of the acceleration of B with respect to C, 2 vBC (2.8)2 aBC = r = = 32.5 m/s 2 CB 0.24and radial component of the acceleration of D with respect to B, 2 vDB (2.4)2 aDB = r = = 10.8 m/s 2 BD 0.54 Now the acceleration diagram, as shown in Fig. 8.17 (c), is drawn as discussed below: 1. Since O and C are fixed points, thereforethese points lie at one place in the accelerationdiagram. Draw vector ox parallel to O A, to somesuitable scale, to represent the radial componentof the acceleration of A with respect to O i.e. r aAO , such that vector o′x = aAO = 63.9 m/s 2 r 2. From point x , draw vector xaperpendicular to vector ox or O A to represent thetangential component of the acceleration of A with trespect to O i.e. aAO ,such that An experimental IC engine with crank shaft and cylinders. vector x a ′ = t aAO = 9 m/s 2 Note : This picture is given as additional informa- 3. Join oa. The vector oa represents the tion and is not a direct example of the currenttotal acceleration of A with respect to O or chapter.acceleration of A i.e. aAO or aA. 4. Now from point a, draw vector ay parallel to A B to represent the radial component of the racceleration of B with respect to A i.e. aBA , such that vector a′y = aBA = 2.25 m/s 2 r 5. From point y, draw vector yb perpendicular to vector ay or A B to represent the tangential tcomponent of the acceleration of B with respect to A i.e. aBA whose magnitude is yet unknown. 6. Now from point c, draw vector cz parallel to CB to represent the radial component of the racceleration of B with respect to C i.e. aBC , such that vector c ′z = aBC = 32.5 m/s 2 r 7. From point z, draw vector zb perpendicular to vector cz or CB to represent the tangential tcomponent of the acceleration of B with respect to C i.e. aBC . The vectors yb and zb intersect at b.Join c b. The vector c b represents the acceleration of B with respect to C i.e. aBC. 8. Now from point b, draw vector bs parallel to BD to represent the radial component of the racceleration of D with respect to B i.e. aDB , such that vector b′s = aDB = 10.8 m/s 2 r
- 23. 196 l Theory of Machines 9. From point s, draw vector sd perpendicular to vector bs or BD to represent the tangential tcomponent of the acceleration of D with respect to B i.e. aDB whose magnitude is yet unknown. 10. From point c, draw vector cd parallel to the path of motion of D (which is along CD)to represent the acceleration of D i.e. aD. The vectors sd and cd intersect at d. By measurement, we find that acceleration of slider D, aD = vector cd = 13.3 m/s2 Ans.Angular acceleration of BD By measurement, we find that tangential component of the acceleration of D with respectto B, aDB = vector sd ′ = 38.5 m/s 2 t We know that angular acceleration of BD, t aDB 38.5 αBD == = 71.3 rad/s2 (Clockwise) Ans. BD 0.54 Example 8.9. The mechanism of a warping machine, as shown in Fig. 8.18, has thedimensions as follows: O1A = 100 mm; AC = 700 mm ; BC = 200 mm ; BD = 150 mm ; O2D = 200 mm ; O2E = 400mm ; O3C = 200 mm. Fig. 8.18 The crank O1A rotates at a uniform speed of 100 rad/s. For the given configuration,determine: 1. linear velocity of the point E on the bell crank lever, 2. acceleration of the points Eand B, and 3. angular acceleration of the bell crank lever. Solution. Given : ωAO1 = 100 rad/s ; O1A = 100 mm = 0.1 m We know that linear velocity of A with respect to O1, or velocity of A , vAO1 = v A = ω AO1 × O1A = 100 × 0.1 = 10 m/s ...(Perpendicular to O1A )1. Linear velocity of the point E on bell crank lever First of all draw the space diagram, as shown in Fig. 8.19 (a), to some suitable scale. Nowthe velocity diagram, as shown in Fig. 8.19 (b), is drawn as discussed below: 1. Since O1, O2 and O3 are fixed points, therefore these points are marked as one point in thevelocity diagram. From point o1, draw vector o1a perpendicular to O1A to some suitable scale, torepresent the velocity of A with respect to O or velocity of A , such that vector o1a = v AO1 = v A = 10 m/s
- 24. Chapter 8 : Acceleration in Mechanisms l 197 2. From point a, draw vector ac perpendicular to A C to represent the velocity of C withrespect to A (i.e. vCA) and from point o3 draw vector o3c perpendicular to O3C to represent thevelocity of C with respect to O3 or simply velocity of C (i.e. vC). The vectors ac and o3c intersect atpoint c. (a) Space diagram. (b) Velocity diagram. (c) Acceleration diagram. Fig. 8.19 3. Since B lies on A C, therefore dividevector ac at b in the same ratio as B divides A C inthe space diagram. In other words, ab/ac = AB/AC 4. From point b, draw vector bdperpendicular to BD to represent the velocity of Dwith respect to B (i.e. vDB), and from point o2 drawvector o2d perpendicular to O2D to represent thevelocity of D with respect to O2 or simply velocityof D (i.e. v D). The vectors bd and o2d intersect at d. 5. From point o 2 , draw vector o 2 eperpendicular to vector o2d in such a way that o2e/o2d = O2E/O2D By measurement, we find that velocity ofpoint C with respect to A , v CA = vector ac = 7 m/s Velocity of point C with respect to O3, Warping machine uses many mechanisms. vCO3 = v C = vector o3c = 10 m/s Velocity of point D with respect to B, vDB = vector bd = 10.2 m/s
- 25. 198 l Theory of Machines Velocity of point D with respect to O2, v DO2 = v D = vector o2d = 2.8 m/sand velocity of the point E on the bell crank lever, v E = v EO2 = vector o2e = 5.8 m/s Ans.2. Acceleration of the points E and B Radial component of the acceleration of A with respect to O1 (or acceleration of A ), 2 vAO1 102 aAO 2 = aAO1 = aA = r = = 1000 m/s 2 O1 A 0.1 Radial component of the acceleration of C with respect to A , 2 vCA 72 aCA = r = = 70 m/s 2 AC 0.7 Radial component of the acceleration of C with respect to O3, 2 10 2 vCO 3 aCO 3 = r = = 500 m/s 2 O3C 0.2 Radial component of the acceleration of D with respect to B, 2 vDB (10.2)2 aDB = r = = 693.6 m/s 2 BD 0.15 Radial component of the acceleration of D with respect to O2, 2 vD (2.8) 2 aDO 2 = O 2 = r = 39.2 m/s2 O2 D 0.2 Radial component of the acceleration of E with respect to O2, 2 vE O 2 (5.8) 2 aEO 2 = r = = 84.1 m/s 2 O2 E 0.4 Now the acceleration diagram, as shown in Fig. 8.19 (c), is drawn as discussed below: 1. Since O1, O2 and O3 are fixed points, therefore these points are marked as one point in theacceleration diagram. Draw vector o1 a parallel to O1A , to some suitable scale, to represent theradial component of the acceleration of A with respect to O1 (or simply acceleration of A ), such that vector o1′ a ′ = aAO1 = aA = 1000 m/s 2 r 2. From point a, draw ax parallel to AC to represent the radial component of the acceleration rof C with respect to A (i.e. aCA ), such that vector a′x = aCA = 70 m/s 2 r 3. From point x, draw vector xc perpendicular to A C to represent the tangential component tof the acceleration of C with respect to A (i.e. aCA ), the magnitude of which is yet unknown. 4. From point o3, draw vector o3 y parallel to O3C to represent the radial component of the racceleration of C with respect to O3 (i.e. aCO 3 ), such that vector o′ y = aCO3 = 500 m/s2 r 3 5. From point y, draw vector yc perpendicular to O3C to represent the tangential component tof the acceleration of C with respect to O3 (i.e. aCO3 ). The vectors xc and yc intersect at c.
- 26. Chapter 8 : Acceleration in Mechanisms l 199 6. Join a c. The vector a c represents the acceleration of C with respect to A (i.e. aCA). 7. Since B lies on A C, therefore divide vector ac at b in the same ratio as B divides AC inthe space diagram. In other words, ab/ac = AB/AC. Join b o2 which represents the acceleration ofpoint B with respect to O2 or simply acceleration of B. By measurement, we find that Acceleration of point B = vector o2 b = 440 m/s2 Ans. 8. Now from point b, draw vector b z parallel to BD to represent the radial component of rthe acceleration of D with respect to B (i.e. aDB ), such that vector b′z = aDB = 693.6 m/s 2 r 9. From point z, draw vector zd perpendicular to BD to represent the tangential component tof the acceleration of D with respect to B (i.e. aDB ), whose magnitude is yet unknown. 10. From point o2 , draw vector o2 z1 parallel to O2D to represent the radial component of rthe acceleration of D with respect to O2 (i.e. aDO 2 ), such that vector o2′ z1 = aDO 2 = 39.2 m/s 2 r 11. From point z 1, draw vector z 1d perpendicular to O2D to represent the tangential component tof the acceleration of D with respect to O2 (i.e. aDO 2 ). The vectors zd and z 1d intersect at d. 12. Join o2 d. The vector o2d represents the acceleration of D with respect to O2 or simplyacceleration of D (i.e. aDO2 or aD). 13. From point o2, draw vector o2 e perpendicular to o2 d in such a way that o2′e′ / o2′ d ′ = O2 E / O2 D r tNote: The point e may also be obtained drawing aEO 2 and aEO 2 as shown in Fig. 8.19 (c). By measurement, we find that acceleration of point E, aE = aEO2 = vector o 2 e = 1200 m/s2 Ans.3. Angular acceleration of the bell crank lever By measurement, we find that the tangential component of the acceleration of D with respectto O2, aD O 2 = vector z1 d1′ = 610 m/s 2 t ∴ Angular acceleration of the bell crank lever t aDO2 610 = = = 3050 rad/s 2 (Anticlockwise) Ans. O2 D 0.2 Example 8.10. A pump is driven from an enginecrank-shaft by the mechanism as shown in Fig. 8.20. Thepump piston shown at F is 250 mm in diameter and thecrank speed is 100 r.p.m. The dimensions of various linksare as follows: OA = 150 mm ; AB = 600 mm ; BC = 350 mm ;CD = 150 mm; and DE = 500 mm. Determine for the position shown : 1. The velocity ofthe cross-head E, 2. The rubbing velocity of the pins Aand B which are 50 mm diameter. 3. The torque requiredat the crank shaft to overcome a presure of 0.35 N/mm2,and 4. The acceleration of the cross-head E. All dimensions in mm. Fig. 8.20
- 27. 200 l Theory of Machines Solution. Given : NAO = 100 r.p.m. or ωAO = 2 π × 100/60 = 10.47 rad/s; OA = 150 mm = 0.15 m ;A B = 600 mm = 0.6 m ; BC = 350 mm = 0.35 m ; CD = 150 mm = 0.15 m ; DE = 500 mm = 0.5 m We know that velocity of A with respect to O or velocity of A , v AO = v A = ωAO × O A = 10.47 × 0.15 = 1.57 m/s ...(Perpendicular to O A)1. Velocity of the cross-head E First of all, draw the space diagram, to some suitable scale, as shown in Fig. 8.21 (a). Nowthe velocity diagram, as shown in Fig. 8.21 (b), is drawn as discussed below: (a) Space diagram. (b) Velocity diagram. (c) Acceleration diagram. Fig. 8.21 1. Since O and C are fixed points, therefore these points are marked as one point in thevelocity diagram. Now draw vector oa perpendicular to O A, to some suitable scale, to represent thevelocity of A with respect ot O or the velocity of A , such that vector oa = v AO = v A = 1.57 m/s 2. From point a, draw vector ab perpendicular to A B to represent the velocity of B withrespect to A (i.e. vBA), and from point c draw vector cb perpendicular to CB to represent the velocityof B with respect to C (i.e. vBC). The vectors ab and cb intersect at b. By measurement, we find that vBA = vector ab = 1.65 m/sand v BC = v B = vector cb = 0.93 m/s 3. From point c, draw vector cd perpendicular to CD or vector cb to represent the velocity ofD with respect to C or velocity of D, such that vector cd : vector cb = CD: CB or vDC : v BC = CD : CB vDC CD CD 0.15 ∴ = or vDC = vBC × = 0.93 × = 0.4 m/s vBC CB CB 0.35 4. From point d, draw vector de perpendicular to DE to represent the velocity of E withrespect to D (i.e. vED), and from point o draw vector oe parallel to the path of motion of E (which isvertical) to represent the velocity of E or F. The vectors oe and de intersect at e. By measurement, we find that velocity of E with respect to D, vED = vector de = 0.18 m/s
- 28. Chapter 8 : Acceleration in Mechanisms l 201and velocity of the cross-head E, vEO = v E = vector oe = 0.36 m/s Ans.2. Rubbing velocity of the pins at A and B We know that angular velocity of A with respect to O, ωAO = 10.47 rad/s ...(Anticlockwise) Angular velocity of B with respect to A , vBA 1.65 ωBA = = = 2.75 rad/s ...(Anticlockwise) AB 0.6and angular velocity of B with respect to C, vBC 0.93 ωBC = = = 2.66 rad/s ...(clockwise) CB 0.35 We know that diameter of pins at A and B, dA = dB = 50 mm = 0.05 m ...(Given)or Radius, rA = rB = 0.025 m ∴ Rubbing velocity of pin at A = (ωAO – ωBA) rA = (10.47 – 2.75) 0.025 = 0.193 m/s Ans.and rubbing velocity of pin at B = (ωBA + ωBC) rB = (2.75 + 2.66) 0.025 = 0.135 m/s Ans.3. Torque required at the crankshaft Given: Pressure to overcome by the crankshaft, pF = 0.35 N/mm2 Diameter of the pump piston DF = 250 mm ∴ Force at the pump piston at F, π π FF = Pressure × Area = pF × ( DF ) 2 = 0.35 × (250)2 = 17 183 N 4 4 Let FA = Force required at the crankshaft at A . Assuming transmission efficiency as 100 per cent, Work done at A = Work done at F FF × vF 17 183 × 0.36 FA × vA = FF × vF or FA = = = 3940 N vA 1.57 ...(3 vF = vE ) ∴ Torque required at the crankshaft, T A = FA × O A = 3940 × 0.15 = 591 N-m Ans.Acceleration of the crosshead E We know that the radial component of the acceleration of A with respect to O or theacceleration of A , v2 (1.57)2 aAO = aA = AO = r = 16.43 m/s2 OA 0.15
- 29. 202 l Theory of Machines Radial component of the acceleration of B with respect to A , 2 vBA (1.65)2 aBA = r = = 4.54 m / s 2 AB 0.6 Radial component of the acceleration of B with respect to C. 2 vBC (0.93)2 aBC = r = = 2.47 m/s 2 CB 0.35and radial component of the acceleration of E with respect to D, 2 vED (0.18)2 aED = r = = 0.065 m/s 2 DE 0.5 Now the acceleration diagram, as shown in Fig. 8.21 (c), is drawn as discussed below: 1. Since O and C are fixed points, therefore these points are marked as one point in theacceleration diagram. Draw vector oa parallel to O A, to some suitable scale, to represent the radialcomponent of the acceleration of A with respect to O or the acceleration of A , such that vector o′a′ = aAO = aA = 16.43 m/s 2 r 2. From point a, draw vector ax parallel to A B to represent the radial component of the racceleration of B with respect to A (i.e. aBA ), such that vector a ′x = aBA = 4.54 m/s 2 r 3. From point x, draw vector xb perpendicular to A B to represent the tangential component tof the acceleration of B with respect to A (i.e. aBA ) whose magnitude is yet unknown. 4. Now from point c, draw vector c y parallel to CB to represent the radial component of rthe acceleration of B with respect to C (i.e. aBC ), such that vector c′y = aBC = 2.47 m/s 2 r 5. From point y, draw vector yb perpendicular to CB to represent the tangential component tof the acceleration of B with respect to C (i.e. aBC ). The vectors yb and xb intersect at b. Join cband ab. The vector cb represents the acceleration of B with respect to C (i.e. aBC) or the accelerationof B (i.e. aB) and vector ab represents the acceleration of B with respect to A (i.e. aBA). By measurement, we find that aBC = aB = vector cb = 9.2 m/s2and aBA = vector ab = 9 m/s2 6. From point c, draw vector cd perpendicular to CD or vector cb to represent theacceleration of D with respect to C or the acceleration of D (i.e. aDC or aD), such that vector cd : vector cb = CD : CB or aD : aBC = CD : CB aD CD CD 0.15 ∴ = or aD = aBC × = 9.2 × = 3.94 m/s 2 aBC CB CB 0.35 7. Now from point d, draw vector dz parallel to DE to represent the radial component of E rwith respect to D (i.e. aED ), such that vector d ′z = aED = 0.065 m/s 2 rNote: Since the magnitude of arED is very small, therefore the points d and z coincide. 8. From point z, draw vector ze perpendicular to DE to represent the tangential component tof the acceleration of E with respect to D (i.e. aED ) whose magnitude is yet unknown. 9. From point o, draw vector oe parallel to the path of motion of E (which is vertical) torepresent the acceleration of E. The vectors ze and oe intersect at e.
- 30. Chapter 8 : Acceleration in Mechanisms l 203 By measurement, we find that acceleration of the crosshead E, aE = vector oe = 3.8 m/s2 Ans. Example 8.11. Fig. 8.22 shows the mechanism of a radial valve gear. The crank OA turnsuniformly at 150 r.p.m and is pinned at A to rod AB. The point C in the rod is guided in the circularpath with D as centre and DC as radius. The dimensions of various links are: OA = 150 mm ; AB = 550 mm ; AC = 450 mm ; DC = 500 mm ; BE = 350 mm. Determine velocity and acceleration of the ram E for the given position of the mechanism. All dimensions in mm. Fig. 8.22 Solution. Given : NAO = 150 r.p.m. or ωAO = 2 π × 150/60 = 15.71 rad/s; OA = 150 mm = 0.15 m;A B = 550 mm = 0.55 m ; AC = 450 mm = 0.45 m ; DC = 500 mm = 0.5 m ; BE = 350 mm = 0.35 m We know that linear velocity of A with respect to O or velocity of A , v AO = v A = ωAO × O A = 15.71 × 0.15 = 2.36 m/s ...(Perpendicular to O A)Velocity of the ram E First of all draw the space diagram, as shown in Fig. 8.23 (a), to some suitable scale. Nowthe velocity diagram, as shown in Fig. 8.23 (b), is drawn as discussed below: 1. Since O and D are fixed points, therefore these points are marked as one point in thevelocity diagram. Draw vector oa perpendicular to O A, to some suitable scale, to represent thevelocity of A with respect to O or simply velocity of A , such that vector oa = vAO = vA = 2.36 m/s 2. From point a, draw vector ac perpendicular to A C to represent the velocity of C withrespect to A (i.e. vCA), and from point d draw vector dc perpendicular to DC to represent the velocityof C with respect to D or simply velocity of C (i.e. vCD or vC). The vectors ac and dc intersect at c. 3. Since the point B lies on A C produced, therefore divide vector ac at b in the same ratio asB divides A C in the space diagram. In other words ac:cb = AC:CB. Join ob. The vector ob representsthe velocity of B (i.e. vB) 4. From point b, draw vector be perpendicular to be to represent the velocity of E withrespect to B (i.e. vEB), and from point o draw vector oe parallel to the path of motion of the ram E(which is horizontal) to represent the velocity of the ram E. The vectors be and oe intersect at e. By measurement, we find that velocity of C with respect to A , v CA = vector ac = 0.53 m/s Velocity of C with respect to D, v CD = v C = vector dc = 1.7 m/s
- 31. 204 l Theory of Machines Velocity of E with respect to B, v EB = vector be = 1.93 m/sand velocity of the ram E, vE = vector oe = 1.05 m/s Ans.Acceleration of the ram E We know that the radial component of the acceleration of A with respect to O or theacceleration of A , 2 vAO (2.36)2 aAO = aA = r = = 37.13 m/s 2 OA 0.15 Radial component of the acceleration of C with respect to A , v2 (0.53) 2 aCA = CA = r = 0.624 m/s 2 OA 0.45 Radial component of the acceleration of C with respect to D, v2 (1.7)2 aCD = CD = r = 5.78 m/s 2 DC 0.5 Radial component of the acceleration of E with respect to B, 2 vEB (1.93)2 aEB = r = = 10.64 m/s 2 BE 0.35 The acceleration diagram, as shown in Fig. 8.23 (c), is drawn as discussed below: (c) Acceleration diagram. Fig. 8.23
- 32. Chapter 8 : Acceleration in Mechanisms l 205 1. Since O and D are fixed points, therefore these points are marked as one point in theacceleration diagram. Draw vector oa parallel to O A, to some suitable scale, to represent the radialcomponent of the acceleration of A with respect to O or simply the acceleration of A, such that vector o′a′ = aAO = aA = 37.13 m/s 2 r 2. From point d, draw vector dx parallel to DC to represent the radial component of theacceleration of C with respect to D, such that vector d ′x = aCD = 5.78 m/s 2 r 3. From point x, draw vector xc perpendicular to DC to represent the tangential component tof the acceleration of C with respect to D (i.e. aCD ) whose magnitude is yet unknown. 4. Now from point a, draw vector ay parallel to A C to represent the radial component ofthe acceleration of C with respect to A , such that vector a ′y = aCA = 0.624 m/s 2 r 5. From point y, draw vector yc perpendicular to AC to represent the tangential component tof acceleration of C with respect to A (i.e. aCA ). The vectors xc and yc intersect at c. 6. Join ac. Thevector ac represents theacceleration of C withrespect to A (i.e. aCA). 7. Since the pointB lies on A C produced,therefore divide vector acat b in the same ratio as Bdivides A C in the spacediagram. In other words, ac : c b = A C : CB. 8. From point b,draw vector b z parallel toBE to represent the radial A lathe is a machine for shaping a piece of metal, by rotating it rapidly alongcomponent of the its axis while pressing against a fixed cutting or abrading tool.acceleration of E with Note : This picture is given as additional information and is not a directrespect to B, such that example of the current chapter. vector b′z = aEB = 10.64 m/s 2 r 9. From point z, draw vector ze perpendicular to BE to represent the tangential component tof the acceleration of E with respect to B (i.e. aEB ) whose magnitude is yet unknown. 10. From point o, draw vector oe parallel to the path of motion of E (which is horizontal)to represent the acceleration of the ram E. The vectors ze and oe intersect at e. By measurement, we find that the acceleration of the ram E, aE = vector o′e′ = 3.1 m/s 2 Ans. Example 8.12. The dimensions of the Andreau differential stroke engine mechanism, asshown in Fig. 8.24, are as follows: AB = 80 mm ; CD = 40 mm ; BE = DE = 150 mm ; and EP = 200 mm. The links AB and CD are geared together. The speed of the smaller wheel is 1140 r.p.m.Determine the velocity and acceleration of the piston P for the given configuration.
- 33. 206 l Theory of Machines Solution. Given: N DC = 1140 r.p.m. or ωDC = 2 π × 1140/60 = 119.4 rad/s ; A B = 80 mm= 0.08 m ; CD = 40 mm = 0.04 m ; BE = DE = 150 mm = 0.15 m ; EP = 200 mm = 0.2 m Fig. 8.24 We know that velocity of D with respect to C or velocity of D, vDC = v D = ωDC × CD = 119.4 × 0.04 = 4.77 m/s ...(Perpendicular to CD) Since the speeds of the gear wheels are inversely proportional to their diameters, therefore Angular speed of larger wheel ω 2CD = BA = Angular speed of smaller wheel ωDC 2 AB ∴ Angular speed of larger wheel, CD 0.04 ωBA = ωDC × = 119.4 × = 59.7 rad/s AB 0.08and velocity of B with respect to A or velocity of B, vBA = vB = ωBA × AB = 59.7 × 0.08 = 4.77 m/s ...(Perpendicular to A B)Velocity of the piston P First of all draw the space diagram, to some suitable scale, as shown in Fig. 8.25 (a). Nowthe velocity diagram, as shown in Fig. 8.25 (b), is drawn as discussed below: 1. Since A and C are fixed points, therefore these points are marked as one point in thevelocity diagram. Draw vector cd perpendicular to CD, to some suitable scale, to represent thevelocity of D with respect to C or velocity of D (i.e. vDC or v D), such that vector cd = v DC = v D = 4.77 m/s 2. Draw vector ab perpendicular to A B to represent the velocity of B with respect to A orvelocity of B (i.e. vBA or v B), such that vector ab = v BA = v B = 4.77 m/s 3. Now from point b, draw vector be perpendicular to BE to represent the velocity of E withrespect to B (i.e. vEB), and from point d draw vector de perpendicular to DE to represent the velocityof E with respect to D (i.e. vED). The vectors be and de intersect at e. 4. From point e, draw vector ep perpendicular to EP to represent the velocity of P withrespect to E (i.e. vPE), and from point a draw vector ap parallel to the path of motion of P (which ishorizontal) to represent the velocity of P. The vectors ep and ap intersect at p.

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