Mechanisms With Lower Pairs

8/9/2019 Mechanisms With Lower Pairs

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Chapter 9

Mechanisms with

Lower Pairs

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Dr. Mohammad Abuhiba, PE1


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9.1. 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 aslower pair

.

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Dr. Mohammad Abuhiba, PE

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9.2. Pantograph

A pantograph is aninstrument used to

reproduce to an enlarged or

a reduced scale and as

exactly as possible the pathdescribed by a given point.

Bars BA & BC are extended

to O & E respectively, suchthat:

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9.2. 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 OADand OBE,

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9.4. 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 thatOAOB = 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

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OP is constant

If OAOB is constant,

then OQ will beconstant.

Point B moves along

straight path BQ

which is

perpendicular to OP.


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eaucellier mechanism

Pin atA is constrained to

move alongcircumference of a circlewith fixed diameter OP,by means of link O1A.

AC = CB = BD = DA; OC= OD ; and OO1= O1A

Product OAOB remainsconstant, when link O1Arotates.

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OC & BC are of constantlength

OBOA remains constant

B traces a straight path

perpendicular to OP

eaucellier mechanism


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arts 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 respectivelyparallel toFD & CE.

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arts mechanism

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9.5. Exact Straight Line Motion Consisting of

One Sliding Pair -

Scott Russells Mechanism

OA =AP =AQ

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9.6. Approximate Straight Line Motion

Mechanisms - atts

me h nism

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9.6. Approximate Straight Line Motion

Mechanisms -

Tchebicheffs

mechanism

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OA = O1B

P, mid of AB traces out an

approximately straight line parallel to

OO1 P is exactly above O or O1 in the

extreme positions (when BA lies

along OA or when BA lies along BO1)

P will lie on a straight line parallel to

OO1, in the two extreme positionsand in the mid position, if the

lengths of the links are in

proportionsAB:OO1:OA = 1:2:2.5


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9.8. 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.

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When the vehicle takes a turn, the front wheels along

with the respective axles turn about the respective

pivoted points.

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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 thefront or back wheels will definitely take place.

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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.

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a = Wheel track

b = Wheel base

c = Distance between pivotsA and B

From triangle IBP,

From triangle IAP,

Fundamental equation for correct steering

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9.9. Davis Steering Gear

Fig. 9.16

Exact steering

gear mechanism

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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.

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a = Vertical distance betweenAB & CD

b = Wheel base

d = Horizontal distance betweenAC & BD

c = Distance between pivotsA & B of front axle

x = Distance moved byAC toAC = CC = DD

a = Angle of inclination of links AC & BD, tovertical

From triangleA AC,

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From triangleAAC,

From triangle BBD,

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For correct steering,

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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.

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9.10. Ackerman Steering Gear

The difference between Ackerman andDavis steering gears are :

1. Whole mechanism of Ackerman steering

gear is on back of front wheels; whereas inDavis steering gear, it is in front of wheels.

2. Ackerman steering gear consists of turning

pairs, whereas Davis steering gear consists

of sliding members.

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MechanismABCD is a four bar crank chainBC =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. 9.17. 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.

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9 11 Universal or Hookes Joint

Used to connect two shafts, whichare intersecting at a small angle

End of each shaft is forked to U-typeand each fork provides two bearingsfor arms of a cross.

Arms of cross are perpendicular toeach 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 itvaries, when the motion istransmitted.

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9.12. Ratio of Shafts Velocities

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9.13. Max Min Speeds of Driven Shaft

w1will be max for a given value of awhen denominator

of above equationis min. This will happen, when

w1

is min when denominator of above equation is max.

This will happen when

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9.13. Max Min Speeds of Driven Shaft

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9.14. Condition for Equal Speeds

of the Driving and Driven Shafts

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9.15. Angular Acceleration of the

Driven Shaft

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For angular acceleration to be maximum


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9.16. Max Fluctuation of Speed

Max fluctuation of speed of driven shaft approximately

varies as square of angle between the two shafts

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9.17

Double Hookes

Joint

In order to have a constant velocity ratio of driving and driven

shafts, an intermediate shaft with a Hookes 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.

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Example. 9.2

Two shafts with an included angle of 160 are

connected by a Hookes 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.

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Example 9.3

The angle between the axes of two shafts

connected by Hookesjoint is 18. Determine the

angle turned through by the driving shaft when

the velocity ratio is maximum and unity.

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Example 9.4

Two shafts are connected by a Hookesjoint. 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.

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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.

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Example 9.6

The driving shaft of a Hookes 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 = 45.

2. At what value of awill the total fluctuation of

speed of the driven shaft be limited to 24 rpm ?

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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.

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Mechanisms With Lower Pairs