Chapter 9
Mechanisms with
Lower Pairs
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Dr. Mohammad Abuhiba, PE 1
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 as lower pair.
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9.2. 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:
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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 OAD
and 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 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
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9.4. Exact Straight Line Motion
Mechanisms Made up of Turning Pairs
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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.
9.4. 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 O1A.
AC = CB = BD = DA; OC = OD ; and OO1 = O1A
Product OA×OB remains constant, when link O1A rotates.
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9.4. Exact Straight Line Motion
Mechanisms Made up of Turning Pairs
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OC & BC are of constant
length
OB×OA remains constant
B traces a straight path
perpendicular to OP
Peaucellier mechanism
9.4. 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.
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9.4. Exact Straight Line Motion
Mechanisms Made up of Turning Pairs
Hart’s mechanism
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9.4. Exact Straight Line Motion
Mechanisms Made up of Turning Pairs
Hart’s mechanism
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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
9.5. Exact Straight Line Motion Consisting of
One Sliding Pair - Scott Russell’s Mechanism
OA = AP = AQ
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9.6. Approximate Straight Line Motion
Mechanisms - Watt’s mechanism
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9.6. Approximate Straight Line Motion
Mechanisms - Tchebicheff’s 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 positions
and in the mid position, if the
lengths of the links are in
proportions AB:OO1:OA = 1:2:2.5
9.6. Approximate Straight Line Motion
Mechanisms - Roberts mechanism
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a four bar chain
mechanism
OA = O1B
A bar PQ is rigidly
attached to link AB
at its middle point
P.
Q will trace out an
approximately
straight line.
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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9.8. Steering Gear Mechanism
When the vehicle takes a turn, the front wheels along
with the respective axles turn about the respective
pivoted points.
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9.8. 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.
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9.8. 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.
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9.8. 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
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9.9. Davis Steering Gear
Fig. 9.16
Exact steering
gear mechanism
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9.9. 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.
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9.9. 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’,
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9.9. Davis Steering Gear
From triangle A A’C,
From triangle BB’D’,
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9.9. Davis Steering Gear
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 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.
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9.10. Ackerman Steering Gear
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9.10. 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. 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 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.
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9.12. Ratio of Shafts Velocities
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9.13. Max & Min Speeds of Driven Shaft
w1 will be max for a given value of a when denominator
of above equation is 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
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 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.
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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.
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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.
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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.
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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 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 = 45°.
2. At what value of a will 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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