Lower Pairs - Theory Of Machines.pdf

8/14/2019 Lower Pairs - Theory Of Machines.pdf

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Theory Of Machines - Part 1Recommend 2.2k

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ChaptersHomeTopics

Chapter 1 : Basic Concepts of MachinesChapter 2 : Lower Pairs

Chapter 3 : Belts, Ropes andChainsChapter 4 : Cams

Chapter 5 : Friction Devices

Chapter 6 : Flywheels

Chapter 7 : Governors

Chapter 8 : Balancing

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Chapter 2 : Lower Pairs

Q. 2.1. Explain universal joint briefly.

Ans. This joint, also called Hookes joint or Cardon joint, is used .to connect twointersecting shafts. This joint can transmit power from one shaft to the other not onlywhen their axes are at an angle but also when the angle between the axes is changing.Its most common use is to connect the horizontal output shaft of gear box to the inclinedpropeller shaft in an automobile. The angle between these shafts goes on changing whenthe rear axes moves over bumps.

Universal joint, as shown in Fig. 2.1 consists of two forks F1 & F2 formed at the ends of shafts s and 2 respectively. The vertical arms of the cross are free to turn in the twobearings provided in F1. Similarly due to the two bearings in F2, the F2 can turn aboutthe horizontal arm of cross for the position shown in Fig. 2.1. The angle between the axesof the two shafts is ct.When the driving shaft takes one revolution, the driven shafts2 also gets rotated throughone revolution. But within one revolution, the angular speed of s2 will become maximumand minimum twice although the shaft s1 is rotating at uniform angular speed w1.It can be proved mathematically that the maximum and the minimum speed of drivenshaft are as under:

Thus greater the angle between the axes of the shafts, the more is the variation in speedof driven shaft.

Q. 2.2 Derive expression, for angular velocity of driven shaft in case of Universal joint Ans. x1 Axis of driving shaft in top view.Axis of driven shafta = angle between the axes. AD = Cross arm in the fork of driving shaft. Let it be horizontal in the beginning.Cl) = Cross arm in the fork of driven shaft. Let itbe vertical in the beginning. -The ends A, B, C and D move along circular pathwhen the shafts rotate. -But when seen in the direction of x1, the path of A & B appears to be circle ADBC and thepath of 1) & C appears to be an ellipse. A bis the top view of path of A&BThe dotted line at angle a to ab in the top view is The path of C&l)When the driving shaft rotates through angle 0, the cross arm AB takes the position AB



and the crossarm CD appears to take position CD when seen in direction of x1. d1 is the top view of Din this position

CD has turned, we take ad2 ad1 and locate 2 at the vertical projection through d2 and

horizontal projection through The point 1)2 will be on the circle ABCD



Q. 2.3. Draw a polar diagram to show the velocity of driving shaft and drivenshaft connected by universal joint.

where speed of driving shafta = Angle between the axes of the shafts.= speed of driven shaft ot the instant when the driving shafthas turned through 0 from the position the cross arm in its fork was horizontal.

This equation gives four values of 0 for a given value of a.Thus the polar diagram is as shown in fig. 2.3



Q. 2.4. Derive an expression for the coefficient of variation of speed of drivenshaft in case of universal joint.

Q. 2.5. Speed of driving shaft is 240rpm, fluctuation of driven shaft is 24 rpm.The shafts are connected by universal joint. What is the angle between them?

Q. 2.6. Speed of driving shaft is 1000 rpm. Fluctuation of speed of driven shaftIs 150 rpm. Determine(i) Angle between the shafts(ii) Maxm. speed of driven shaft.(iii) Minm. speed of driven shaft.



Q. 2.7 Derive an expression for the fluctuation in torque of driven shaftconnected to driving shaft with universal joint.

Q. 2.8. Derive an expression for the acceleration of driven shaft of universal joint.

Q. 2.9 The angle between the axes of the two shafts joined by universal joint is25 Driving shaft rotates of uniform speed of 180 rpm The driven shaft carries



a steady load of 75 kW Calculate the mass of the flywheel of driven shaft if itsradius of gyration is 150 mm and the output torque of driven shaft does notvary more than by 15% input shaft torque

Q. 2.10. Explain why two Hookes joints are used to transmit motion from theengine to the differential of an automobile.

(1) Gear box output shaft.(2) Propeller shaft the intermediate shaft.(3) Differential input shaft.By using two universal joints the speed of (3) is uniform although speed of (2) becomesmaxm. of minm. twice in one rotation provided) a1 = a2 (ii) The fork of (1) & the fork of (3) are in one plane, i.e. when the fork of (1)is vertical, the fork of (3) should also be vertical.



Q. 2.11. Derive the fundamental equation for correct steering or explain theprinciple of steering gear mechanism.

When the vehicle is moving along a curve, all the four wheels should turn along oneinstantaneous centre I to avoid skidding that is, the angle 0 through which inner wheelaxis turns must be greater than the angle cp through which the outer wheel axis turns.

If this condition is satisfied, there will be pure rolling Qfthe four wheels when the vehiclemoves along a curve. V

Q. 2.12. Explain Davis steering mechanism.

Ans.

The mechanism has the following links: V(i) Front axle AB V(n) Bell crank lever FAQ, pivoted to AB at A PA is stub axle and AQ is called track arm(iii) Bell crank lever RBU(iv) Bar ST which is to AB & can move to left or right m guides G1 & G2(v) One slider pin-jointed to ST at S & free to slide over AQ(vi) Another slider pin-jointed to ST at r & free to slide over BUWhen the vehicle is moving straight, the arms AQ & BF are at angle a with the



When the vehicle is to take left turn:(a) PAQ turns through 8and takes position PAQ (b) RBU turns through and takes position RBU (c) Bar ST slides is position S T such that SS = TT x.

Q. 2.13. In Davis steering mechanism of a vehicle calculate the inclination of track arms to the longitudinal axis when the vehicle is moving straight if thewheel base is 2.8 m and the distance between pivots is 1.4 m.



Q. 2.14. With the help of neat sketch explain Ackermann steering mechanism.

Ans.

Ackermann steering mechanism is based on four bar chain. It has four turning pairs. Thefour links are:(i) Front axle AB(ii) Bell crank lever LAC, pivoted to AB at A. LA is stub axle.(liz) Bell crank lever MBD(iv) Cross bar CDWhen the vehicle moves straight (Fig. 2.7) CD I AB

AC & BD are inclined at angle c with longitudinal axis.When the vehicle takes left turn (Fig. 2.8)LAC turns through 8 and take posn. LAC MD turns through and take posn MBD 0>4)Though 6 > 4), yet the condition for correct steering i.e. cot 4) - cot 0 is satisfied inthis mechanism for one particular val of 6 only. Thus this mechanism gires purerollingonly in three cases: *(a) When the vehicle is moving straight(b) When the vehicle takes left turn of a particular radius.(c) When the vehicle takes right turn of the same particular radius. Although thismechanism does not give pure rolling at curves of all radii even then thismechanism is most commonly used because of its simple construction and easymaintenance.

Q. 2.15. What are the differences between Davis streering Mechanism andAckermann steering mechanism.

Ans.


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Q. 2.16. Explain pantograph with the help of diigram. Ans.

Pantograph is used to reproduce a given map or diagram to reduced or enlarged scale.It consists of four lines OAB, BCE, AD and DC.

AD is pin-jointed to OB at A and to DC at UDC is pin-jointed to BE at C AD is equal and I I to is BCDC is equal and I I to ABThus ABC1) is a parallelogram.

Q. 2.17. Explain engine indicator with the help of diagram.

Ans.


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indicator is used for drawing pressure-volume diagram of Ic engine or steam engineThe pressure of steam in steam engine cylinder acts on the and oritniso because theindicator cylinder is connected to engine cylinder A small change m pressure mengineiirgives rise to a small displacement of DThis small displacement of D makes the end E to move vertically through a largedistance. The displacement of E is iecbrded with a pencil on a paper.

Q. 2.18. With the help of a diagram explain a straight line mechanism.

Perpendicular straight line mechanism is shown in Fig. 2.11. In this mechanism when pinA is moved along circular path with centre Q, the pin B traces a straight lineperpendicular to OQ.This mechanism consist of the following eight links:


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Now as CC, CB & OP are constant, therefore ON is constant.It means B traces a straight line passing through N and perpendicular to OQ.

Q. 2.19. What is Hookes joint?

Ans. It is a joint which is used to connect two shafts whose axes intersect at angle. Itconsists mainly of two forks and one cross. It is also called universal joint.

Q. 2.20. What is the principle of steering gear mechanism?

Ans. The principle states that for pure rolling of the wheels of a vehicle when moving ona curved path, the axes of all the wheels should meet at one point.

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Lower Pairs - Theory Of Machines.pdf