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Theory of machines by rs. khurmi_ solution manual _ chapter 7Darawan Wahid

Velocity and acceleration of mechanismsHareesha N Gowda, Dayananda Sagar College of Engg, Bangalore

Module 2 instantenous center methodtaruian

Torsional vibrationsM.D.Raj Kamal

GyroscopeHareesha N Gowda, Dayananda Sagar College of Engg, Bangalore

8.acceleration analysisvarun teja G.V.V

- 1. 0 Theory of machines _ Khurmi Solution of Exercises . Ch.7(Relative velocity method Solved & prepared By : Bakhtiar Essa , Ac.year : 2016 – 2017 Edition one UNIVERSITY OF SALAHADDIN COLLEGE OF ENGINEERING MECHANICAL DEPARTMENT
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- 3. 2 EXERCISES Chapter 7 : Velocity in Mechanisms _Page : 169 Given : N = 600 rpm (c.w) , Va =? , Vg = ? , Wab = ? Wob = = 62.83 rad/s Vob = Wob * OB = 62.83 * 0.125 = 7.85 rad /s v. d.scale [ 7.85 m /s = 45 mm ] VA = 37 * = 6.45 m / s VG = 38.7 * = 6.75 m/s VAB = ab * = 5.4 m/s WAB = = 10.8 rad /s 1. In a slider crank mechanism, the length of crank OB and connecting rod AB are 125 mm and 500 mmrespectively. The centre of gravity E of the connecting rod is 275 mm from the slider A. The crankspeed is 600 r.p.m. clockwise. When the crank has turned 45° from the inner dead centre position,determine: 1. velocity of the slider A, 2. velocity of the point E, and 3. angular velocity of theconnecting rod AB[Ans. 6.45 m/s ; 6.75 m/s ; 10.8 rad/s]
- 4. 3 Solution : WOA = WOB = = 4.188 rad / s . VA= WOA * OA , VA = 4.18 * 50 = 209.4 mm /s b [209.4 mm/s =60mm] a velocity diagrams scale . Vg = fg * c g o d Vg = 19.5 * = 68mm/s f e 2. In the mechanism, as shown in Fig. 7.32, OA and OB are two equal cranks at right angles rotating about O at a speed of 40 r.p.m. anticlockwise. The dimensions of the various links are as follows :OA = OB = 50 mm ; AC = BD = 175 mm ; DE = CE = 75 mm ; FG = 115 mm and EF = FC.Draw velocity diagram for the given configuration of the mechanism and find velocity of the slider G. [Ans. 68 mm/s]
- 5. 4 We can find ( dc ) on the S.DIA. by this relationship . = . b de = dc * = 36.8 mm f e VF = af * VF = 46.45 * = 250mm/s o , d c 3. The dimensions of various links in a mechanism, as shown in Fig. 7.33, are as follows :AB = 60 mm ; BC = 400 mm ; CD = 150 mm ; DE = 115 mm ; and EF = 225 mm. Find the velocity of the slider F when the crank AB rotates uniformly in clockwise direction at a speedof 60 r.p.m. [Ans. 250 mm/s]
- 6. 5 Given : N = 150 rpm , d = 50 mm . R = 0.25 cm Space diagram scale 1 : 10 it means 150 mm ( given at the statement ) 15 mm (on the space diagram) WBA = = 15.7 rad/s VBA = WBA* AB VBA= 15.7 * 150 mm = 2355 mm/s Chose this it as the velocity diagram scale . So 2355m m/ sec = 45 mm ( in v.daigram ) 4. In a link work, as shown , the crank AB rotates about A at a uniform speed of 150 r.p.m.The lever DC oscillates about the fixed point D, being connected to AB by the connecting link BC. The block F moves, in horizontal guides being driven by the link EF, when the crank AB is at 30°. The dimensions of the various links are : AB = 150 mm ; BC = 450 mm ; CE = 300 mm ; DE = 150 mm ; and EF = 350 mm. Find, for the given configuration, 1. velocity of slider F, 2. angular velocity of DC, and 3. Rubbing speed at pin C which is 50 mm in diameter. [Ans. 500 mm/s ; 3.5 rad/s ; 2.4 cm/s]
- 7. 6 Since the point E lies on DC, therefore divide vector dc in e in the same ratio as E dividesCD In other words Ce = cd * = 30 * = 20 mm VF = fd * = 9.56 mm * VF = 500 mm/s VDC = dc * = 1575 mm/s when [ dc = 30.1mm ] WDC = = = 3.5 rad/s VCB = cb * = 2669 mm /s when [ c b = 51mm ] WCB = = = 6 rad /s V.R© = ( WCB + WDC ) * R V.R © = ( 6 + 3.5 ) * 0.25 = 2.38 cm
- 8. 7 WOB = = 9.42 rad /s . VOB = WOB * OB VOB = 9.42 * 0. 225 = 2.12 m /s . v.d Scale : [VOB = 9.42m/s = 50mm ] = oa = ob * = 33. 3 mm VC = oc = 30mm * = 1.2 m/s a b VAC = ac = 10 mm * = 0.424 m/s WAC = = = 1.6 rad / s . o c b b b b b b b b b b b 5. The oscillating link OAB of a mechanism,as shown , is pivoted at O and is moving at 90 r.p.m. anticlockwise. If OA= 150 mm ; AB = 75 mm, and AC = 250 mm ,calculate 1. the velocity of the block C; 2. the angular velocity of the link AC; and 3. the rubbing velocities of the pins at O, A and C, assuming that these pins are of equal diameters of 20 mm [Ans. 1.2 m/s ; 1.6 rad/s clockwise ; 200 mm/s, 782 mm/s, 160 mm/s]
- 9. 8 WOB = = 9.42 rad /s . = WOA { because on the one link} WAC = = = 1.6 rad / s . R = = 10 mm . Ru.O = ( WOA ) * R Ru.O = ( 9.42 ) * 10 = 94 .2 mm /s Ru.A = ( WOB + WAC ) * R Ru.A = ( 9 . 42 – 1.6 ) * 10 = 78 . 2 mm /s Ru.C = ( WAC ) * R Ru.C = ( 1.6 ) * 10 mm = 16 mm /s
- 10. 9 WBA = = 4 rad /s VBA = 4 *30 = 377 mm /s [ V.d scale 377mm /s = 50 mm ] VEA = ae =16 * VBC = bc = 44 * WBC = = = 4 .14 rad /s . VCD = dc = 24 * = 180 . 96 mm /s WCD = = = 2. 2 rad /s VCE = ce = 20 . 5 * = 154 . 5 mm /s WCE = = = 1.28 rad /s 6. The dimensions of the various links of a mechanism, as shown , : AB = 30 mm ; BC = 80 mm ; CD = 45 mm ; and CE = 120 mm. The crank AB rotates uniformly in the clockwise direction at 120 r.p.m. Draw the velocity diagram for the given configuration of the mechanism and determine the velocity of the slider E and angular velocities of the links BC, CD and CE. [Ans. 120 mm/s ; 2.8 rad/s ; 5.8 rad/s ; 2 rad/s ]
- 11. 10 N = 120 rpm WOA = = 4 VOA = WOA * OA VOA = VOA = 2.51 mm /s = 70 mm]V.D.S : [ Vp = op = 17 .6 * = 0.63 m /s VQB = qb = 26 .4 * = 0 . 946 m /s o,q p WQB = = = 6 .31 rad /s . b a 9. In a Whitworth quick return motion mechanism, as shown , the dimensions of various links are as follows : OQ = 100 mm ; OA = 200 mm ; BQ = 150 mm and BP = 500 mm. If the crank OA turns at 120 r.p.m. in clockwise direction and makes an angle of 120° with OQ, Find : 1. velocity of the block P, and 2. angular velocity of the slotted link BQ. [Ans. 0.63 m/s ; 6.3 rad/s ].
- 12. 11 Given : N = 60 r.p.m Tp = 115 N.m = 60 % = 0.6 Ts = ? , Fs = ? Solution : WP = WP = = rad /s VP = WP * OP s q , o VP = * 50 m = 0 .3141 m /s V.D Scale : [ 0.3141 m /s = 60 mm ] p r VS = os * , when{ os = 6mm} VS = 0 .03141 m /sec 10. A toggle press mechanism, as shown, has the dimensions of various links as follows :OP = 50 mm ; RQ = RS = 200 mm ; PR = 300 mm. Find the velocity of S when the crank OP rotates at 60 r.p.m. inthe anticlockwise direction. If the torque on P is 115 N-m, what pressure will be exerted at S when the overall efficiency is 60 percent. [Ans. 400 m/s ; 3.9 kN]
- 13. 12 TP * WP = in put Fs * VS = out put = = , FS = FS = = 13802 .6 N = 13 .8 KN .
- 14. 13 WOA = = 2 rad /s VOA = WOA * OA VOA = 2 * 150 * = 0.942 m /s Velocity diagram Scale [ 0.942m/s = 70 mm ] VCB = ob = 67 * = 0.9 m /s VD = od = 39* = 0.52 m /s 0.5 m /s . a b VAB = ab = 67 * = 0.9 m /s VBD = bd = 66 * = 0.88 m /s WBD = = 1.76 rad /s d o,c 11. in the figure shows a toggle mechanism in which link D is constained to move inzhorizontal direction.For the given configuration, find out : 1.velocities of points band D ; and 2. angular velocities of links AB, BC, and BD.The rank OA rotates at 60 r.p.m. in anticlockwise direction. [Ans. 0.9 m/s; 0.5 m/s; 0.0016 rad/s (anticlockwise) 0.0075 rad/s (anti-clockwise),0.0044 rad/s (anticlockwise)] . 35 degree
- 15. 14 Given : FE = 2.2 KN , = 72 % = ? , FD = ? Solution : To drow space diadram we know that BC = 2AB , if we assume that AB = 40 mm , so BC = 80 mm . aslo we assume the scale of velocity digram is 60 mm = ab . VE = ae = 58 mm and VD = ad= ac = 18 mm . d,c a = = 3 .222 = FD = * FD = 0 .72 * = 5 KN b e,f 12. A riveter, as shown below , is operated by a piston F acting through the links EB, AB and BC. The ram D carries the tool. The piston moves in a line perpendicular to the line of motion of D. The length of link BC is twice the length of link AB. In the position shown, AB makes an angle of 12° with AC and BE is at right angle to AC. Find the velocity ratio of E to D. If, in the same position, the total load on the piston is 2.2 kN, find the thrust exerted by D when the efficiency of the mechanism is 72 per cent, Ans. [3.2 ; 5 kN]
- 16. 15 Q1: Describe the method to find the velocity of a point on a link whose direction (or path) is known and the velocity of some other point on the same link in magnitude and direction is given. ANS 1: by Relative Velocity Method The relative velocity method is based upon the relative velocity of the various points of the link ,Consider two points A and B on a link . Let the absolute velocity of the point A i.e. vA is known in magnitude and direction and the absolute velocity of the point B i.e. vB is known in direction only. Then the velocity of B may be determined by drawing the velocity diagramas . Q2: Explain how the velocities of a slider and the connecting rod are obtained in a slider crankmechanism. ANS,2 : A slider crank mechanism is shown below (a). The slider A is attached to the connecting rod AB. Let the radius of crank OB be r and let it rotates in a clockwise direction, about the point O with uniform angular velocity rad/s. Therefore, the velocity of B i.e. vB is known in magnitude and direction. The slider reciprocates along the line of stroke AO. The velocity of the slider A (i.e. vA) may be determined by relative velocity method as discussed below :
- 17. 16 Q3 : Define rubbing velocity at a pin joint. What will be the rubbing velocity at pin joint when the two links move in the same and opposite directions ? ANS 3 : The rubbing velocity is defined as the algebraic sum between the angular velocities of the two links which are connected by pin joints, multiplied by the radius of the pin.Consider two links OA and OB connected by a pin joint at O as shown Let w1 = Angular velocity of the link OA or the angular velocity of the point A with respect to O.w2 = Angular velocity of the link OB or the angular velocity of the point B with respect to O, and r = Radius of the pin. According to the definition, Rubbing velocity at the pin joint O = (w1 – w2) r, if the links move in the same direction = (w1 + w2) r, if the links move in the opposite direction . Q4 :What is the difference betweenideal mechanical advantage and actual mechanical advantage ? ANS 4 : 1- Ideal mechanical advantage : FA × vA = FB × vB , = 100 % ( neglecting effect of feiction ) If we consider the effect of friction, less resistance will be overcome with the given effort.Therefore the actual mechanical advantage will be less. Let : = Efficiency of the mechanism. 2- Actual mechanical advantage, ≠ 100 %
- 18. 17 Solved & prepared By : Bakhtiar Essa Omar , 2016 – 2017 with Good luck for all
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