10/29/2013. Chapter 9. Mechanisms with Lower Pairs. Dr. Mohammad Abuhiba, PE

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1 Chapter 9 Mechanisms with Lower Pairs 1

2 Introduction When the two elements of a pair have a surface contact and a relative motion takes place, the surface of one element slides over the surface of the other, the pair formed is known as lower pair.

3 Pantograph A pantograph is an instrument used to reproduce to an enlarged or a reduced scale and as exactly as possible the path described by a given point. Bars BA & BC are extended to O & E respectively, such that:

4 Pantograph For all relative positions of the bars, triangles OAD & OBE are similar and points O, D and E are in one straight line. Point E traces out same path as described by D From similar triangles OAD and OBE,

5 Exact Straight Line Motion Mechanisms Made up of Turning Pairs O = a point on circumference of a circle of diameter OP OA = any chord B = a point on OA, such that OA OB = constant Locus of a point B will be a straight line perpendicular to diameter OP Draw BQ perpendicular to OP Triangles OAP & OQB are similar

6 Exact Straight Line Motion Mechanisms Made up of Turning Pairs OP is constant If OA OB is constant, then OQ will be constant. Point B moves along straight path BQ which is perpendicular to OP.

7 Exact Straight Line Motion Mechanisms Made up of Turning Pairs Peaucellier mechanism Pin at A is constrained to move along circumference of a circle with fixed diameter OP, by means of link O 1 A. AC = CB = BD = DA; OC = OD ; and OO 1 = O 1 A Product OA OB remains constant, when link O 1 A rotates.

8 Exact Straight Line Motion Mechanisms Made up of Turning Pairs Peaucellier mechanism OC & BC are of constant length OB OA remains constant B traces a straight path perpendicular to OP

9 Exact Straight Line Motion Mechanisms Made up of Turning Pairs Hart s mechanism. FC = DE & CD = EF O, A, B divide links FC, CD, EF in the same ratio BOCE is a trapezium and OA & OB are respectively parallel to FD & CE.

10 Exact Straight Line Motion Mechanisms Made up of Turning Pairs Hart s mechanism

11 Exact Straight Line Motion Mechanisms Made up of Turning Pairs Hart s mechanism From point E, draw EM parallel to CF & EN perpendicular to FD Point B will trace a straight line perpendicular to the diameter OP produced

12 Exact Straight Line Motion Consisting of One Sliding Pair - Scott Russell s Mechanism OA = AP = AQ

13 Approximate Straight Line Motion Mechanisms - Watt s mechanism

14 Approximate Straight Line Motion Mechanisms - Tchebicheff s mechanism OA = O 1 B P, mid of AB traces out an approximately straight line parallel to OO 1 P is exactly above O or O 1 in the extreme positions (when BA lies along OA or when BA lies along BO 1 ) P will lie on a straight line parallel to OO 1, in the two extreme positions and in the mid position, if the lengths of the links are in proportions AB:OO 1 :OA = 1:2:2.5

15 Approximate Straight Line Motion Mechanisms - Roberts mechanism a four bar chain mechanism OA = O 1 B A bar PQ is rigidly attached to link AB at its middle point P. Q will trace out an approximately straight line.

16 Steering Gear Mechanism Used for changing direction of two or more of the wheel axles with reference to the chassis. In automobiles, front wheels are placed over the front axles, which are pivoted at points A and B (Fig. 9.15). These points are fixed to the chassis. Back wheels are placed over the back axle, at the two ends of the differential tube.

17 Steering Gear Mechanism When the vehicle takes a turn, the front wheels along with the respective axles turn about the respective pivoted points.

18 Steering Gear Mechanism To avoid skidding (slipping of wheels sideways), the two front wheels must turn about the same instantaneous center I which lies on the axis of the back wheels. If the instantaneous center of the two front wheels do not coincide with the instantaneous center of the back wheels, the skidding on the front or back wheels will definitely take place.

19 Steering Gear Mechanism The condition for correct steering is that all the four wheels must turn about the same instantaneous center. The axis of the inner wheel makes a larger turning angle than the angle subtended by the axis of outer wheel.

20 Steering Gear Mechanism a = Wheel track b = Wheel base c = Distance between pivots A and B From triangle IBP, From triangle IAP, Fundamental equation for correct steering

21 9.9. Davis Steering Gear 21 Fig Exact steering gear mechanism

22 Davis Steering Gear Slotted links AM & BH are attached to front wheel axle, which turn on pivots A & B respectively. Rod CD is constrained to move in direction of its length, by sliding members at P & Q. These constraints are connected to slotted link AM & BH by a sliding and a turning pair at each end. Steering is affected by moving CD to right or left.

23 Davis Steering Gear a = Vertical distance between AB & CD b = Wheel base d = Horizontal distance between AC & BD c = Distance between pivots A & B of front axle x = Distance moved by AC to AC = CC = DD a = Angle of inclination of links AC & BD, to vertical From triangle A A C,

24 Davis Steering Gear From triangle A A C, From triangle BB D,

25 Davis Steering Gear For correct steering,

26 26 Example 9.1 In a Davis steering gear, the distance between the pivots of the front axle is 1.2m and the wheel base is 2.7m. Find the inclination of the track arm to the longitudinal axis of the car, when it is moving along a straight path.

27 Ackerman Steering Gear The difference between Ackerman and Davis steering gears are : 1. Whole mechanism of Ackerman steering gear is on back of front wheels; whereas in Davis steering gear, it is in front of wheels. 2. Ackerman steering gear consists of turning pairs, whereas Davis steering gear consists of sliding members.

28 Ackerman Steering Gear

29 Ackerman Steering Gear Mechanism ABCD is a four bar crank chain BC = AD & AB CD The following are positions for correct steering: 1. When vehicle moves along a straight path, links AB & CD are parallel and shorter links BC & AD are equally inclined to longitudinal axis of vehicle. 2. When vehicle is steering to left, position of gear is shown by dotted lines in Fig In this position, lines of front wheel axle intersect on back wheel axle at I, for correct steering. To satisfy the fundamental equation for correct steering, links AD & DC are suitably proportioned.

Arms of cross are perpendicular to each other.

30 Universal or Hooke s Joint Used to connect two shafts, which are intersecting at a small angle End of each shaft is forked to U-type and each fork provides two bearings for arms of a cross. Arms of cross are perpendicular to each other. Motion is transmitted from driving shaft to driven shaft through a cross. Inclination of the two shafts may be constant, but in actual practice it varies, when the motion is transmitted.

31 Ratio of Shafts Velocities

32 Max & Min Speeds of Driven Shaft w 1 will be max for a given value of a when denominator of above equation is min. This will happen, when w 1 is min when denominator of above equation is max. This will happen when

33 Max & Min Speeds of Driven Shaft

34 Condition for Equal Speeds of the Driving and Driven Shafts

35 Angular Acceleration of the Driven Shaft For angular acceleration to be maximum

36 Max Fluctuation of Speed Max fluctuation of speed of driven shaft approximately varies as square of angle between the two shafts

37 9.17 Double Hooke s Joint In order to have a constant velocity ratio of driving and driven shafts, an intermediate shaft with a Hooke s joint at each end is used.

37 Double Hooke s Joint In order to have a constant velocity ratio of driving and driven shafts, an intermediate shaft with a Hooke s joint at each end is used. This joint gives a velocity ratio equal to unity, if 1. Axes of driving & driven shafts are in same plane, and 2. Driving & driven shafts make equal angles with the intermediate shaft.

38 38 Example. 9.2 Two shafts with an included angle of 160 are connected by a Hooke s joint. The driving shaft runs at a uniform speed of 1500 rpm. The driven shaft carries a flywheel of mass 12 kg and 100 mm radius of gyration. Find the max angular acceleration of the driven shaft and the max torque required.

39 39 Example 9.3 The angle between the axes of two shafts connected by Hooke s joint is 18. Determine the angle turned through by the driving shaft when the velocity ratio is maximum and unity.

40 40 Example 9.4 Two shafts are connected by a Hooke s joint. The driving shaft revolves uniformly at 500 rpm. If the total permissible variation in speed of the driven shaft is not to exceed ± 6% of the mean speed, find the greatest permissible angle between the center lines of the shafts.

41 41 Example 9.5 Two shafts are connected by a universal joint. The driving shaft rotates at a uniform speed of 1200 rpm. Determine the greatest permissible angle between the shaft axes so that the total fluctuation of speed does not exceed 100 rpm. Also calculate the maximum and minimum speeds of the driven shaft.

42 42 Example 9.6 The driving shaft of a Hooke s joint runs at a uniform speed of 240 rpm and the angle between the shafts is 20. The driven shaft with attached masses has a mass of 55 kg at a radius of gyration of 150 mm. 1. If a steady torque of 200 N.m resists rotation of the driven shaft, find the torque required at the driving shaft, when q = At what value of a will the total fluctuation of speed of the driven shaft be limited to 24 rpm?

43 43 Example 9.7 A double universal joint is used to connect two shafts in the same plane. The intermediate shaft is inclined at an angle of 20 to the driving shaft as well as the driven shaft. Find the maximum and minimum speed of the intermediate shaft and the driven shaft if the driving shaft has a constant speed of 500 rpm.

2. a) What is pantograph? What are its uses? b) Prove that the peaucellier mechanism generates a straight-line motion. (5M+10M)

2. a) What is pantograph? What are its uses? b) Prove that the peaucellier mechanism generates a straight-line motion. (5M+10M) Code No: R22032 R10 SET - 1 1. a) Define the following terms? i) Link ii) Kinematic pair iii) Degrees of freedom b) What are the inversions of double slider crank chain? Describe any two with neat sketches.