# KINEMATICS OF MACHINES (ME 232) UNIT-1 SIMPLE MECHANISM. TWO MARKS :-

Mechanics

Nov 14, 2013 (5 years and 1 month ago)

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KINEMATICS OF MACHINES (ME 232)

UNIT
-
1

SIMPLE MECHANISM.

TWO MARKS :
-

Define 1)Kinematic Link 2)Kinematic pair 3)Kinematic chain 4)Klein’s equation for
joints 5)Degree of freedom of mechanism.6)Kutzbach’s relation 7)Grashoff’s law
8)Inversion of me
chanism 9)Mechanical advantage of mechanism 10)Transmission
angle?

It is a resistive body which go to make a part of a machine having relative
motion between them.

2)Kinematic pair.

When two links are in contact
with each other it is known as a pair.If the pair
makes constrain motion it is known as kinematic pair.

3)Kinematic chain.

When a number of links connected in space make relative motion of any point on
a link with respect to any other point on

kinematic chain.

4)Klein’s equation for joints.

h
-
Higher pair joint

l
-

j
-
Lower pair joint

5)Degree of freedom. Of mechanism.

It is defined as the minimum number of input parameters which must be
independently controlled inorder to bring the mechanism into a useful engineering
purpose.

6)Kutzbach’s relation.

n
-
Degree of freedom.

l
-

h
-
Higher pair joint

j
-
Lower pair joint.

7)Grashoff’s law.

Sum of shortest link length and sum of longest link length is not greater than the

8)Inversion of mechanism.

The method of obtaining
different mechanism by fixing different links in a
kinematic chain is known as inversion of mechanism.

It is defined as the ratio of output torque to the input torque also defined as the
ort.

10)Transmission angle.

The acute angle between follower and coupler is known as transmission angle.

11)Toggle position.

If the driver and coupler lie in the same straight line at this point mechanical
his condition the mechanism is known as toggle position.

12)List out few types of rocking mechanism?

Pendulam motion is called rocking mechanism.

1.Quick return motion mechanism.

2.Crank and rocker m
echanism.

3.Cam and follower mechanism.

13)Define pantograph?

It is device which is used to reproduce a displacement exactly in a enlarged
scale.

14)Name the application of crank and slotted lever quick return motion
mechanism?

1.Shamping machines.

2.Siotting mechanism.

3.Rotary internal combustion engine.

15)Define structure?

It is an assemblage of a number of resistant bodies having no relative motion between
themand meant for carrying loads havin
g straining action.

16)What is simple mechanism?

A mechanism with four link is known as simple mechanism.

17)Define mechanism?

When one of the link of a kinematic chain is fixed,the chain is known as a mechanism.

18)Define equivalent mechanism?

The mechanism, that obtained has the same number of the degree of freedom,as the
original mechanism called equivalent mechanism.

19)Define single slider crank chain mechanism?

A single slider crank chain is a modification of the basic four
bar chain. It consist of
one sliding pair and three turning pair.

20)Define double slider crank chain mechanism?

A kinematic chain which consist of two turning pair and two sliding pair is
known as double slider crank mechanism.

16

MARKS :
-

1) Explain

(B) 1)Indexing mechanism 2)Snap action mechanism 3)Motion

(a)Classification based on relative

1)Sliding pair.

In a sliding pair minimum number of degree of freedom is only one.

2)Turning pair.

In a turning pair also degree of freedom is one.when two links are connected such

another link it forms a turning pair.

3)Cylindrical pair.

In a cylindrical pair degree of freedom is two.If one link turns and slides along
another link it forms a cylindrical pair.

4)Rolling pair.

In a rolling pair degree of freedom is two.
The object moves both linearly and
angularly.

5)Spherical pair.

In a spherical pair degree of freedom is three.It can both move left and right,up and
down,and rotate along the same point.

(b)Based on nature of contact.

1)Lower pair.

If

contact between two links is surface contact also having degree of freedom one,
then the pair is known as lower pair.

Example: Sliding pair.

2)Higher pair.

If contact between two links is either point contact or line contact then the pair is
kno
wn as higher pair.

Example: Point contact
-
Rolling pair.

Line contact
-
Cylindrical pair.

3)Mechanical pair.

(a)Open pair.

In this pair everything is open to the admosphere.

(b)Closed pair.

In this pair ever
ything is closed from the admosphere.

(B)Indexing mechanism.

This type of mechanism is used in automatic lathe’s etc.

Assume one revolution of the driver.In 360 degrees,270 degrees makes locking of
follower. Remaining 90 degrees is used to make

rotation of the follower.

2)Snap action mechanism.

It is used in calling bells, bicycle bells etc.

The mechanism used to adjust or modify any one of the links in a mechanism is
anism. Differential screw used in bench vice, pipe
wrench, Lathe chuck and screw jack are some of the examples of motion adjustment
mechanism.

4)Scott Russel mechanism.

This is one of the mechanism to produce straight line motion mechanism.The mechanism
in which the straight line is copied from a existing straight line constrain is known as
Scott Russel mechanism.

2) In a crank and slotted lever quick return motion mechanism,the length of the fixed
link is 300mm and that of the crank is 150mm.Determine

the maximum angle the slotted
lever will make the fixed link.Also determine the ratio of the time of cutting and return
stroke.If the length of slotted bar is 700mm, What would be the length of the
stroke,assuming that the line of stroke passes through th
e positions of the free end of the
slotted lever?

Given data:

AB=300mm=0.3m

AE=150mm=0.15m

BP1=700mm=0.7m

Inclination of the slotted bar with fixed link:

Let

ABE=
inclination of the slotted bar with

the vertical axis
.

sin

ABE=sin(90
0
-
ß
/2)

=AE/AB

=0.15/0.3=0.5

ABE=(90
-
ß
/2)

=30
0

Time ratio of cutting stroke to the return stroke:

We know that,

(90
0
-
ß
/2)=30

ß
=120
0

Time for cutting stroke=
α

Time for return stroke=
ß

=
360
0
-
120
0

120
0

Length of stroke=p
1
p
2

=2(BP
1
)sin(90
0
-
ß
/2)

=2*0.7sin(90
0
-
60
0
)

L=450mm.

3) Explain the different types of quick return motion mechanism?

Quick return motion are of two types.They are,

1)Crank and slotted lever quick return motion mechanism.

2)Whitw
orth quick return motion mechanism

Crank and slotted lever quick return motion mechanism.

In this mechanism,the link AC forming the turning pair is fixed.The driving crank CB
revolves with uniform angular speed about the fixed
center C.A sliding block attached to
the crank pin at B slides along the slotted bar AP and thus causes AP to oscillate about
the pivorted point A.A short link PR transmits the motion from AP to ram which carries
the tool and reciprocates along the line of

stroke R
1
R
2
.In the extreme positions,AP
1

and
AP
2

are tangential to the circle and the cutting tool is at the end of the stroke.The forward
or cutting stroke occurs when the crank rotates from the position CB
1

to CB
2

in the
clockwise direction.The return s
troke occurs when the crank rotates from the position
CB
2

to CB
1

in the clockwise direction.Since the crank has uniform angular
speed,therefore

Time of cutting stroke

Time of return stroke

=

Since the tool travels a distance of R
1
R
2
during cutting and return ,therefore
length of stroke =R
1
R
2
=P
1
P
2
=2AP

Whitworth quick return motio
n mechanism.

In this mechanism,the link CD forming the turning pair is fixed.The driving crank CA
rotates at a uniform angular speed.The slider attached to the crank pin at A slides along
the slotted bar PA which oscillates at a pivoted point D.The

connecting rod PR carries the
ram at R to which a cutting tool is fixed.The motion of the tool is constrained along the
line RD produced along a line passing through D and perpendicular to CD.When the
driving crank CA moves from the position CA
1

to CA
2

th
rough an angle in the
clockwise direction,the tool moves from left to right through a distance 2PD.Now when
the driving crank moves from the position CA
1

to CA
2

through an angle in the
clockwise direction,the tool moves back from right to left han
d end.Since CA rotates at
uniform angular velocity therefore time taken for return stroke is less than time taken for
cutting stroke.then ratio between time taken for cutting and return stroke is,

Time of cutting stroke

Time o
f return stroke

Explain 1)Elliptical trammels 2)Skotch yoke mechanism 3)Offset mechanism 4)Ratchet
mechanism 5)Escapement mechanism?

1)Elliptical trammels.

It is one of the inversion of double slider crank chain mechanism.It is an

instrument used to draw ellipse.This inversion is obtained by fixing the slotted bar.Link 4
has two straight grooves cut in it, at right angle to each other.the link 1 and 3,are known
as sliders and forms sliding pair with link4.the link AB is a bar which

forms turning pair
with link 1 and 3.When the links 1 and 3 slide along their respective grooves,such as P
traces out a ellipse on the surface of link 4.

This is the equation of a circle of radius AP.

4) Explain

(i) the different types of joints

(ii
) Inversion of four bar kinetic chain

(ii) Inversion of Basic Single Slider Crank Chain

(i)
The different types of joints are

a. Binary Joint

b. Ternary Joint

c. Quaternary Joint

a) Binary Joint

If two links are connected at the same junction it is calle
d binary joint.

Illustration:

In the above figure ( kinetic Chain)

Number of binary joints j = 4

L = 2/3 (j+2)

L = 4

4=2/3 (6)

4=4

D

A

L.H.S. = R.H.S.

It is a kinetic chain based on Kline’s equation.

b) Ternary Joint

(i)

If three are connected to th
e same junction, then is known ternary
joint

(ii)

One ternary joint is equivalent to two binary joints.

Illustration:

In the fig:

No of binary joints = A+B+C

= 3

Number of Ternary joints = C+E

Equivalent binary joints = 2+2

= 4

Hence total

number of binary joints = 3 + 4 = 7

Based on Kline’s Equation

L = 2/3 (J+2)

6 = 2/3 (9)

6=6

Hence is a Kinematic Chain.

D

c) Quaternary Joint

If four links are connected to the same joint then it is a Quaternary joint.
One quaternary joint =
Three binary joints

In the figure Number of binary joint C = 1

Number of ternary joints = A + B + E + F

Equivalent binary joints = 8

Number of quaternary joints = D, G

Equivalent binary joints = 6,

Total Number of Equivalent

binary joints

J = 15

Based on Kline’s equation

L = 2/3 (I+2)

Lin
k (5)

A

B

D

C

E

F

G

11 = 2/3 (17)

L.H.S.

R.H.S

Hence it is not a kinematics chain

In the figure if the link (DG) is deleted then it would be a kinematic chain.

The chain is represented as

5) Inversion of fo
ur Bar Kinematic Chain

The inversions of four bar kinematic chain are as follows

a) Beam Engine:

i) This is also known as crank and lever mechanism.

r arrangement

ii) In this mechanism one link oscillated while the other rotates about fixed

Piston

Crank

Connecting rod

iii) It is used to convert the rotary motion into reciprocating motion.

b) Coupling Rod of Locomotive Engine Wheel

i) Thi
s is also known as Double Crank mechanism. Since both cranks rotate about
the points in the fixed link.

ii) It consists of four links

iii) The opposite links are equal in length,

iv) Links (1) and (3) work as two cranks

v) This motion is also known as rota
ry

rotary converter.

c) Watt’s Indicator Mechanism:

i)

This mechanism was invented by Watt for his steam Engine to guide the
position rod.

ii)

It is also known as simple indicator.

iii)

It is also known as dou
ble lever mechanism

A

C

B

D

B
1

D
1

E
1

F
1

F

E

C
1

Piston

Cylinder

Gas

Pressure

Lever 1

Lever 2

AB

(3))

A

D

B

C

Crank (1)

Crank (2)

iv)

Links BC and DEF work as levers whose displacement is directly
proportional to steam or gas pressure.

iii) Inversion of Basic Single Slider Crank Chain

Single Slider Crank Chain

a) Pendulum Of Bull En
gine

i)

This mechanism is obtained by fixing link (4)

Piston Piston rod

(Slider)

Guide way

s

Fly wheel

B

O

A

ii)

iii)

This is used to supply feed water to bo
iler pans.

b) Oscillating Cylinder Engine

1)

This is used to convert reciprocally motion into rotary motion.

2)

c) Int
ernal Rotary Combustion Engine.

(i) In the engine, the slider is replaced by a piston and link (1) by a cylinder
pivoted at O.

(ii) Odd numbers of cylinder are symmetrically placed at regular interval in same
plane.

(iii) The Crank is fixed and common to a
ll cylinders.

(iv) The reciprocating motion is converted into rotary motion.

Piston

Cylinder

Crank

B

A

O

MODULE

2

INSTANTANEOUS CENTRE, VELOCITY & ACCELARATION DIAGRAMS

TWO MARKS :
-

1.What are the components of acceleration?

i

ii

Ta
ngential component of acceleration

2.Write an expression for find number of instantaneous centers in a mechanism?

N=n(n
-
1)/2,n
-

3.What is expression for Cariolis componenet of acceleration?

a
BC
=2r

Where

=Angular velocity of ‘OA’

V=Linear velocity of ‘B’

4.What are the expression for radial and tangential component of acceleration?

i

a
r
OB
=

OB
*
OB

ii

Tangential component

a
r
OB
=

OB
*OB

Where

OB

OB

5.How can we represent the dir
ection of linear velocity of any point on a link with

respect to another point on the same line?

The direction is perpendicular to the line joining the points.

6.Define instantaneous center axis?

Instantaneous axis is a line drawn through an instantaneous center

and perpendicular to the plane of motion.

7.What are the names of instantaneous center?

i.

Virtual center.

ii.

Centro.

iii.

Rotopole.

8.How can we apply instantaneous center method to determine v
elocity?

Consider three points A,B,C on a rigid link.I being instantaneous

Center.Let V
A
,V
B
,V
C

be the points A,B&C.

Then we have

V
A
/
I
A
=V
B
/
I
B
=V
C
/
I
C

9.What is the objective of Kinematic analysis?

The objective of

Kinematic analysis is to determine the Kinematic

quantities such as displacement,velocity and acceleration of the element in a

mechanism.

10.Write any two rules to locate Instantaneous center?

a)

When two links are connected by a pin joint the inst
antaneous

center lies on the center of the pin.

b)

When two links have a sliding contact,the instantaneous center

lies at infinity in a direction perpendicular to the path of motion

of slide.

16 MARKS :
-

1) .PQRS is a four bar chain with link PS fixed

PQ=62.5mm,QR=175mm,RS=112.5mm,PS=200mm.If the crank

PQ rotates at 10 rad/sec clockwise direction.Draw the velocity and

acceleration diagram when angle QPS=60
0

and Q and R lie on the

same side of PS.Find the angular ve
locity and angular acceleration

Space diagram: Scale

1cm=30mm

Velocity diagram: Scale

1cm=0.1m/s

W
PQ

V
QP
=V
Q
=0.0625*10

=0.625m/s

W
QR
=0.36
/
/
s

W
RS
=0.43
/
/
s

acceleration diagram:

Scale

1cm=1m/s
2

a
r
PQ
=V
2
Q
/
QP=0.625
2
/
0.0625

=6.25m/s

a
r
RS
=V
2
/RS=0.43
2
/
0.1125=1.644m/s
2

a
r
QR
=V
2
QR/QR=.36
2
/0.175=0.74m/s
2

Angular ac
celeration of QR,

QR
=a
t
QR
/QR

=3.6
/
0.175

/
sec
2

Angular acceleration of RS,

RS=a
t
RS
/
RS

=5.3
/
0.1
125

2

RESULT:

i

/
sec

ii

/
sec

iii

/
sec2

iv

2

2.The crank of a slider
crank mechanism rotates clockwise at a constant

speed of 300r.p.m.The crank is 150mm and the connecting rod is 600mm

long.Determine

i

Linear velocity and acceleration of the midpoint of the

connecting rod.

ii

Angular velocity and Angular accele
rationof the connecting

rod,at a crank angle of 45
0

Space diagram: Scale

1cm=75mm

Velocity diagram: Scale

1cm=1m/s

N
AB
=300r.p.m

W
AB
=2*

*300
/
/
sec

V
AB
=V
B
=AB*W
AB

=0.15*31.41

=4.7m/s

Acceleration diagram: Scale

1c
m=50m/sec
2

a
r
BA
=a
B
=V
2
BA
/BA

=4.72/0.15

=147.27m
/
sec
2

Linear Velocity,V
d
=4m/s

Linear acceleration,a
d
=117.5m/s

Angular velocity,W
BC
=V
BC
/BC

=3.5/0.6

2

Angular

acceleration,

BC
=a
t
BC
/BC

=102.5/0.6

2

RESULT:

i

Linear velocity=4m
/
s

ii

Linear acceleration=117.5m
/
s
2

iii

/
sec
2

iv

/
sec
2

3.An engine mechanism is shown in the following figure the crank CB=100mm

and the connecting rod BA=300mm with a center of gravity G 100mm from

point B in the position shown the crank shaft has a speed of 75 rad
/
sec and

an angular accelera
/
sec
2
.Find

i

Velocity of G and angular Velocity of AB

ii

Acceleration of G and angular aceeleration of AB

Space diagram: Scale

1cm=50mm

Velocity diagram: Scale

1cm=2m/s

W
CB
/
sec

V
BC
=V
B
=CB*WCB

=0.1*75

=7.5 m
/
s

MODULE
-
3

KINAMATICS OF CAMS

TWO MARKS :
-

1.What is cam ?

Cam is a rotating mechanical member used for transmitting desired motion to a

follower by direct contact

2.Classification of cam?

(i)

accor
ding to cam shape

(ii)

according to follower movement

(iii)

according to manner of constraint of the follower

3.Classify cam based on a shape ?

(i)

wedge cam

(ii)

(iii)

spiral cams

(iv)

drum cams

(v)

spherical cams

4.classification of follower ?

(i) According to fo
llower shape

(ii)according to motion of follower

5.What is roller follower?

In place of a knife edge roller is provided at the contacting end of the follower

6.Spherical follower ?

In the contacting end of the follower is of spherical shape .

7.Angle of ascend ?

The angle of rotation of the cam from the position when the follower begins to rise
till it reaches its highest points . it is d
enoted by θ

8. Angle of descend?

The angle through which the cam rotates during the time the follower returns to the
initial position . It is denoted by θr.

9.Angle of dwell?

It is the angle through which the cam rotates while the follower rema
ins stationary at
the highest or the lowest .

10. Angle of action ?

The total angle moved by the cam during its rotation between the beginning of rise
and the end of return of the follower

11.What is radial or disc cams?

lower reciprocates or oscillates in a direction perpendicular to the
cam axis . The cams are all radial rams. In actual practice, radial cams are widely used
due to their simplicity and compactness.

12.What is dwell?

The zero displacement or the abse
nce of motion of the follower during the motion of the
cam is called dwell.

13. What is classification of followers according to follower shape?

(i)

Knife edge follower

(ii)

Roller follower

(iii)

Mushroom or flat faced follower and

(iv)

Spherical faced or curved shoe follo
wer

14.What is classification of follower according to the motion of the follower?

(i)

Reciprocating or translating follower

(ii)

Oscillating or rotating follower

15.What is classification of followers according to the path of motion ?

(i)

(ii)

Off
set follower

16.What are the motion of the follower ?

The follower can have any of the following four types of motions

(i)

Uniform velocity

(ii)

Simple harmonic motion

(iii)

uniform acceleration and retardation

(iv)

cycloidal motion.

17.What
is the application of cam?

Closing and opening of inlet and exit value operating in IC engine .

18.What are the necessary elements of a cam mechanism?

(i)

Cam
-
The driving member is known as the cam

(ii)

Follower
-
The driven member is known as the follower.

(iii)

Frame
-
It supports the cam and guider the follower.

19.What is translating angle?

The wedge is replaced by a flat plate with a groove . The plate cam moves back and
forth imparting a translatory motion to the follower. Thus these cams are also known as
tra
nslating cams.

20.Write the formula for maximum velocity?

Vo (max) =
2
ωs

θo

Vr (max)=

2ωs

θo

16 MARKS

:
-

(1) A cam rotating clockwise at a uniform speed of 1000 r.p.m Is required to
give a roller followe
r the motion defined below :

1.Follower to move outwards through 50mm during 120 of cam rotation

2. Follower to dwell for next 60 of cam rotation

3. Follower to return to its standing position during next 90 of cam rotation

4. Follower to dwell for the

next of the cam rotation

The minimum radius of the cam is 50 mm and the diameter of roller is
comm. . The line of stroke of the follower is offset by 20mm from the
axis of the cam shaft . If the displacement of the follower takes place
with uniform and
equal acceleration and retardation of both the outward
and return stoke draw profile of the cam and find the maximum velocity
and acceleration during outstroke and return stroke.(16)

Displacement diagram(8)

Draw a rectangular block of length 18
cm breath 50cm

Divided the block for forward dwell of return stroke

Divide forward and return stroke to equal halves.

Join the diagonal of the forward and return stroke block and mark the
mid points

Then divide

the c4ntre line in to six equal parts

the remain four and in to with divide in and that corresponding points are
marked

Join all the points

Offset type;(8)

Draw a circle of radius of 50mm and a roller of diam
eter of 10mm on
the centre

Now draw another circle join center of the roller

Join them tangentially and then transfer the points from the
displacement diameter to here .

Take the degree for forward dwell and out strok
e and divide the form
and latter to six equal parts.

Draw a roller at each points and joining its ends gives the cam profile.

(2)Draw the profile of the cam when the roller follower moves with cycloid
motion during outstroke and return str
oke as given below:

(i)

Out stroke with max displacement of 31.4mm during 180 of cam
rotation.

(ii)

Return stroke for the next 150of cam rotations.

(iii)

Dwell for the remaining 30 of cam rotation min radius of the cam is
15mm and roller diameter of follower is 10mm . dr
aw the profile for

(iv)

Displacement diagram (4)

Draw a rectangular block of length 18cm breath 50cm

Divided the block for forward dwell of return stroke

Divide forward and return stroke to eq
ual halves.

Join the diagonal of the forward and return stroke block and mark the
mid points

Draw a circle of radius 4.99 mm from the end of forward stroke and
starting of return stroke and join opposite angles parallel to the block .

From the middle of the upper and lower of block parallel to the middle
line .

Plot the points and join them.

Draw the circle radius of 30mm and roller diameter 20mm

Now draw another circle join center
of the roller

Take the degree for forward dwell and out stroke and divide the form
and latter to six equal parts.

Join them tangentially and then transfer the points from the displacement

Join them to the centre of the circl
e trans for the points from the
displacement diagram

Draw a roller at each points and joining its ends gives the cam profile.

Offset type(6):

Draw the circle radius of 30mm roller and 20mm

Now draw another circle join center of

the roller

Offset 10mm.

Join them tangentially and then transfer the points from the displacement
diameter to here .

Take the degree for forward dwell and out stroke and divide the form
and latter to six equal parts.

Draw a roller at each points and joining its ends gives the cam profile.

(3)A cam rotating clock wise at uniform speed pf 1000r.p.m is required to give
a knife edge follower the motion defined below

(a)follower to move outward through 2.5m durin
g 120 of cam rotation

(b)follower to dwell for next 60 of cam rotation

(c) follower to set up to its starting position during next 90 of cam rotation

(d)follower to dwell for the next of the rotation

The minimum radius of the cam is 50mm a
n the line of stroke of the follower
is axial . If the displacement of the follower takes place with uniform and equal
acceleration and retardation on both outward and return stroke dean

the cam profile.(16)

Displacement diagram : (7)

Draw a rectangular block of length 18cmand breath of 25cm.

Divided the block for forward dwell of return stroke

Divide forward and return stroke to equal halves.

Join the diagonal of the forward and return stroke block and
mark the
mid points

Then divide the c4ntre line in to six equal parts

the remain four and in to with divide in and that corresponding points are
marked

Join all the points

ial )(9)

Draw the circle of radius 50mm

Now draw the another circle from centre of the roller

Join them axially and transfer the lengths from the displacement diagram.

Take the degree for forward dwell and out stroke and divide the form and
latter to

six equal parts.

Draw a roller at each points and joining its ends gives the cam profile.

Join the all points.

(4)A cam profile is to give the following motion to a knife edge follower

(i)out stroke during 60 of cam rotation.

Dwell for the next 30 of a
m rotation.

Return stroke dieing next60 of cam rotation .

Dwell for the remaining 210 of cam rotation.

The stroke of follower is 40mm and the minimum radius of the cam is 50mm. The
follower moves with uniform velocity during both the outstroke and retu
rn stroke . Draw
the profile of the outstroke and return stroke. Drawn the profile of the cam when

(a)The axis of the follower passes through the axis of the cm shaft

(b)Axis of the follower id offset by 20mm right hand side from axis of the
cam shaft
. (16)

Displacement diagram (7)

Draw a rectangular block of length 18cmand breath of 40mm.

Divided the block for forward dwell of return stroke

Divide forward and return stroke to equal halves.

Join the start and end of the forward and return stroke.

Draw the circle of radius 50mm .

Take the degree for forward dwell and out stroke and divide the form and
latter to six equal parts.

Draw a roller at ea
ch points and joining its ends gives the cam profile.

Join the all points.

5.For the following data draw the profile of a cam with a flat faced follower line of
motion which passes through the cam centre.

Least radius of the cam = 30mm

Angle of asce4nt a
n descent each = 72

Angle of dwell in lifted position =30

Follower lift =27.5mm

The out ward stroke takes place with s.h.m and the inward strokes with uniform
acceleration and retardation . If the uniform speed of rotation of the cam

is 3000r.p.m

What will be the maximum velocity and acceleration during out ward and inward stroke
of the follower?(16)

Displacement diagram(7)

Draw a rectangular block of length 18cmand breath of 25cm.

Divided the block for forward

dwell of return stroke

Divide forward and return stroke to equal halves.

Join the diagonal of forward stroke with curved fine.

Join the diagonal of return stroke and mark mid points . Divide its centre line
in to six equal parts and jo
in the other and four points to upper and lower ends
of diagonal.

Draw the circle of radius 50mm

Now draw the another circle from centre of the roller

Join them axially and transfer the lengths from the displacement diagram.

Take th
e degree for forward dwell and out stroke and divide the form and
latter to six equal parts.

Draw a roller at each points and joining its ends gives the cam profile.

Instead of joining the points make a perpendicular line of joints its edges.

6. Dra
w the radial cam and explain its terminologies? (16)

Diagram : (9)

Cam profile ;(7)

The surface of the cam it comes into contact with the follower

Base circle :

The smallest circle that can be drawn to the cam profile .

Trace point :

The reference point on the follower to trace the cam profile.

Pitch curve:

The locus or path of the tracing point

Prime circle :

The smallest circle drawn tangent to the pitch curve .

Pressure angle φ:

The angle betwe
en the direction of the follower motion and the normal to the
pitch curve.

Pitch circle :

The circle passing through the pitch points and concentric with the base circle .

Pitch point :

Point on the pitch curves at which the pres
sure angle is maximum .

Cam angle :

Angle of rotation the am for a definite displacement of the follower

Left or stroke :

Maximum displacement of the follower from the base circle of the cam.

Note: Inside ( ) indicate marks weightage

MODULE
-
4

GEARS & CLUTHES

TWO MARKS :
-

1.
Define spur gear.

A spur gear is a cylindrical gear whose tooth traces are straight line generation

of the reference cylinder. They are used to transmit rotary motion between

parallel shafts.

2

Addendum is the radial distance of a tooth from the pitch circle to the top of the

toot
h.

Dedendum is the radial distance of a tooth from the pitch circle to the bottom of

the tooth.

3.
Define circular pitch
.

It is the distance measured on the circumference of the pitch circle from a point

of one tooth to the corresponding
point on the next tooth. It is denoted by Pc

Circular pitch Pc=

/DT

Where D = Diameter of pitch circle.

T = Number of teeth on the wheel.

4
. Define I) path of co
ntact. II) Length of path of contact.

Path of contact
: It is the path traced by the point of contact of two teeth from the

beginning to the end of engagement.

Length of path of contact
: It is the length of common normal cut
-

off b
y the

addendum circles of the wheel and pinion.

5.
State the law of gearing.

Law of gearing states that, the common normal at the point of contact between a pair

of teeth must always pass through the pitch point.

6.
Define conju
gate action
.

When the tooth profiles are so shaped so as to produce a constant angular velocity ratio

during Meshing, then the surface are said to de conjugate.

7.
Define angle of approach
.

The angle of approach is defined as the angle th
rough which a gear rotates from the

instant a pair of teeth comes into contact until the teeth are in contact at the pitch point.

8
. List out the characteristics of in volute action.

a) Arc of contact.

b) Length of p
ath of contact.

c) Contact ratio.

9.
Define contact ratio.

Contact ratio is defined as the ratio of the length of arc of contact to the circular pitch

ath
ematically.

Contact ratio =
length of arc of contact

Pc

Where Pc = circular path.

10
. What is the advantage of in volute gear?

The most important advantage of involutes gear is that the center
distance for a pair

of involute gears can be varied within limits without changing the velocity ratio.

11.
What are the conditions to be satisfied for interchangeability of all gears.

For interchangeability of all gears, the set must have t
he same circular pitch, module,

diameter pitch, pressure, angle, addendum and dedendum and tooth thickness must be

one half of the circular pitch.

12.
Define gear tooth system.

A tooth system is a standard which specifies the relat

dedendum, working depth, tooth thickness and pressure angle to attain

interchangeability of gears of tooth numbers but of the same pressure angle and pitch

13.
Define cycloid.

A cycloid is the curve traced

by a point on the circumference of a circle which rolls

without slipping on a fixed straight line.

14.
Define clearance.

The amount by which the dedendum of a gear exceeds the addendum of the mating

gear is called clearanc
e.

15.
When in volute interference occurs
.

If the teeth are of such proportion that the beginning of contact occurs before the

interference point is met then the involute proportion of the driven gear will mate a

non in vo
lute portion of the driving gear and involute interference is said to occur.

16.
What is the principle reason for employing non standard gears?

a) To eliminate the undercutting.

b) To prevent interference.

c) To maintain reasonable con
tact ratio.

17.

a) It transmits exact velocity ratio.

b) It has high efficiency.

a)

The manufacture of gears require special tool and equipment.

b)

The error in cutting teeth may cause vibrations and noise during operation.

18.
Define helix angle (

).

It is the angle betwee
n the line drawn through one of the teeth and the center line of

the shaft on which the gear is maintained.

19.
Define gear ratio.

The quotient of the number of teeth on the wheel divided by the number of threads on

the worm.

20.
De
fine gear train.

A combination of gears that is used for transmitting motion from one shaft to another

shaft is known as gear train.

E.g. spur gear, spiral gear.

21
. Define velocity ratio.

Velocity ratio of a

simple gear train is defined as the ratio of the angular velocity of

the first gear in the train to the angular velocity of the last gear.

22
. Define epicycles gear train.

In a gear train when the axes of shafts over which the gears are mou
nted move

relative to a fixed axis is called epicyclic gear train.

.

23.
List out the function of differential gear used in the rear drive of an automobile.

a) To transmit motion from the engine shaft to the rear driving wheels
.

b ) To rotate the rear wheel of different speeds while the automobile is taking a turn.

24.
Define bevel gears.

The gears which are used to connect shafts whose axes of rotation intersect are called

bevel gears.

25.
Define lim
ited slip differential.

The coupling unit which is sensitive to wheel speed causes most of the torque to be

directed to the slow moving wheel. This combination is called limited slip

differential

.

16 MARKS :
-

1
. In a reverted gear train two shafts A and B are in the straight line and are geared

through an intermediate parallel

shaft C .the gears connecting A and C have a module of 2 and those connecting C and

B have a m
odule of 3.5.

Speed of B is less than 1/10 that of A .if two pinions have each 24 teeth ,find suitable

teeth for gears , the actual velocity ratio and corresponding distance of shaft C from

A.

Given data:

Module

of gears 1, 2 = m
A
=2;

Module of gears 3,4 = m
B

=3.5;

N
B

< 1/10 N
A;

T
1

= T
3

=24;

To find:

1. Suitable teeth for gears.

2. Actual

velocity ratio, and

3. Distance between shafts C and A.

Solution

1. SUITABLE TEETH FOR GEARS:

N

B

< 1/10 N
A

.

Let N
B
= 1/11 N
A

.

We know that for reverted gear train

R1 + R2 =

R3 + R4

Or m
A
T
1
/2 + m
A
T
2
/2 = m
B
T
3
/2 + m
B
T
4
/2

M
A
(T
1
+T
2
) = M
B
(T
3
+T
4
)

2(24 + T
2
) = 3.5(24 + T
4
)

T
2

=18 +1.75T
4

Now speed ratio, N
B
/N
A

=T
1
*T
3

/T
2
*T
4

1/11 =24*24/T
2
*T
4

T
2
*T
4

=6336.

Sub the value of T
2

,we get

(18 +1.7574)T
4

= 6336

T
4
= 56

T
2

=116.

2. ACTUAL VELOCITY RATIO.

N
A
/N
B

= T
2
*T
4
/T
1
*T
3

N
A
/N
B

=11.28.

3. DISTANCE BETWEEN SHAFTS C AND A:

C = d
1

+ d
2
/2

= m
A
(T
1

+ T
2
)/2

= 2(116+24)/2

C=140mm.

MODULE

5

FRICTION

TWO MARKS :
-

1. Define clutch.

Clutch is a transmission device of an automobile which is used to engage and
disengage the power from the engine to the rest of the system.

2. What are the types of friction clutches?

Type
s of friction clutches are:

*Disc or plate clutches.

*Cone clutches.

*Centrifugal clutches.

3. Define centrifugal clutch.

Centrifugal clutch is being increasingly used in automobile and machines
obviously it works on the principle of centrifugal force.

4. What are the types of flat drives?

The types of flat drives are:

*Compound belt drive.

*Stepped or cone pulley drive.

*Fast and loose pulley.

5. Define slip.

Slip is defined as the relative motion between the belt and pulley.

6. Define law of belt
ing.

Law of belting states that the centre line of the belt, as if approaches the pulley lie
in a plane perpendicular to the axis of that pulley or must lie in the plane of the pulley
otherwise the belt will run off the pulley.

7. Rope drive: Utility.

Th
e rope drives are widely used when large power is to be transmitted
continuously from one pulley to another over a considerable distance. One advantage of
rope drives is that a number of separate driver may be from the driving pulley.

8. Belt drive: Utilit
y.

Belt drive is commonly used for transmission of power when exact velocity ratio
is not required. Generally, belt drives are used to transmit power from one pulley to
another, when the two pulleys are not more than 10 meters apart.

9. What are the types

of ropes?

The types of ropes are:

*Fiber ropes.

*Wire ropes.

10. Quarter turn left drive.

The quarter turn left drive is used with shafts arranged at right angles and rotating
in one definite direction.

11. Define the velocity ratio of the belt drive.

The velocity ratio of the belt drive is defined as the ratio between the velocities of
the driver and the follower or the driven.

-
belt.

*Power transmitted is more due to wedging action in the grooved pulleys.

*V
-
belt is more compact
, quite and shock absorbing.

*The V
-
belt drive is positive because of negligible slip between the belt and the
groove.

*High velocity ratio may be obtained.

-
belt.

*It cannot be used with large center distances.

*It is not as dura
ble as flat belt.

*It is a costlier system.

14. Circular belts or ropes.

*Ropes are circular in cross section.

*It is used to transmit more power.

*Distance between two pulleys is more than 8metres.

15. Belt materials.

BELT TYPES

BELT MATERIALS

Flat

belts

Leather, canvas, cotton & rubber.

V
-
belts

Rubberized fabric & rubber.

Ropes

Cotton, hemp & manila.

16. Name the types of friction.

*Static friction.

*Dynamic friction.

17. Define the angle of repose.

If the angle of inclination ‘
α’ of the plane to horizontal is such that the body begin
to move down the plane. Then the angle α is called the angle of repose.

18. What is meant by frictional force?

Force of friction is always acting in the direction opposite to the direction of
moti
on.

19. Why self locking screws have lesser efficiency?

Self locking needs some friction on the thread surface of the screw and hence it
needs higher effort to lift a body and hence automatically the efficiency decreases.

20. What is static friction?

It
is the friction experienced by a body, when at rest.

21. What is dynamic friction?

It is the friction experienced by the body, when in motion. The dynamic friction is
also called as kinematic friction.

22. What is co
-
efficient of friction?

It is defined
as the ratio of the limiting friction(F) to the normal reaction(R
N
)
between the two bodies. It is generally denoted by μ.

μ=F/R
N.

23. Define screw jack.

The screw jack is the device used to lift the heavy loa
ds by applying a
comparatively small effort at its handle. The working principle of screw jack is similar to
that of an inclined plane.

-

*V
-
thread is stronger and often moves frictional to the motion than square

*A
given load may be lifted by applying lesser force by square thread as
compared to V
-

*V
-

25. What is the effort required to lift a 50tonne lorry using screw jack? (
μ=0.3, α=20º)

Q = tanˉ (μ)

= tanˉ (0.3)

=
16.64

W = 50tonne

= 50*10*9.81

=
490.5KN

Effort required to lift the lorry P = W tan (α+Q)

= 490.5 tan
(16.69+20)

=
365.59KN
.

26. Open belt drive.

The open belt drive is used with shaft arranged parallel and rotating in same
direction. The tension in the lower side will be more than in the upper side

belt because
of more tension in the lower side belt, the lower side belt is known as tight side where as
the upper side is known as the slack side.

27. Open belt drive with one idler pulley.

Idler pulleys are provide to obt
ain high velocity ratio and when the required belt
tension cannot be obtained by other means. Idler pulley is also known as jockey pulley.

16 MARKS :
-

1) Explain limiting of fri
ction.Draw a neat sketch of a body over the surface with all the
forces

acting on it.

Consider a body of weight W is lying on a rough horizontal body B as shown in
figure. In this position(a), body A is in equilibrium under the action of its own weight W,

and the normal reaction R
N

of B on A. now small force is applied to the body, it does not
move because of the frictional force which prevent the motion. This shows applied force
P
1

is exactly balanced by the force of friction F
1.

Now increase the applied
force to P
2
as shown in fig (1) it is still found to be in
equilibrium. This means that the force of friction has also increased to a value of F
2

= P
2
.
Thus every time the efforts increased the force of friction is also increases, so as to
become exactly e
qual to the applied force. There is however, a limit beyond which the
force of friction cannot increase as shown in fig (d). After this, any increase in the applied
effort will not lead to any further increase in the force of friction as shown in fig (e),
thus
the body A begins to move in the direction of the applied force. This maximum value of
frictional force, which comes into play, when a body just begins to slide over the surface
of the other body, is known as limiting force of friction or simply limit
ing friction. It may
be noted that when the applied force is less than the limiting friction, the body remains at
rest, and the friction into play is called static friction.

2. Explain with neat sketch the working of centrifugal clutch. Deduce the

express
ion for
the total torque transmitted.

Draw the required figure.

Centrifugal clutch works on the principle of centrifugal force. The driving shaft
carries the shoes and springs. While the drivers shaft is connected to the pulley. Shoes are
ly and the springs keep them away from inner rim of the pulley. When the
centrifugal force is less than the spring force, brake lining cannot make any contact with
the pulley rims.

Let

n = no: of shoes.

m = mass of each shoe.

R = inside radius of the p
ulley rim.

N = speed of pulley.

w = angular speed of pulley.

w
1
= angular speed at which the begins.

Centrifugal force on each shoe F
c
= mw²r

Spring force exerted by each spring F
s
= mw
1
²r

Net force on the shoe =F
c
-
F
s

=

mw²r
-
mw
1
²r

Frictional force acting on each shoe, f = μ (F
c
-
F
s
)

Frictional torque Fr = F*R

= μ (F
c
-
F
s
) R

Total frictional torque transmitted, T = n* μ (F
c
-
F
s
) R

= nFR.

3) Define screw jack and screw jack with square threads.

Screw jack

The screw jack is the device used to lift the heavy loads by applying a
comparatively small effort at its handle. The working principle of screw jack is

similar to
that of an inclined plane.

Draw a figure of screw jack with its spindle having square threads.

The load to be raised or lowered is placed on the square threaded rod. It is rotated
by the application of an eff
ort at the end of the Tommy bar (lever). The motion of nut on
the screw is analogous to sliding along an inclined plane.

Let,

W = load to be lifted.

P = horizontal force.

l = horizontal distance between the central axis of the screw and end E of the bar.

φ= friction angle.

μ = tan φ, co
-
efficient of friction.

α = angle of repose.

P = the pitch of the screw.

d = mean diameter of the screw.

The nut is rotated so that the screw moves against the axial load W. it is treated as
motion upwards the inclined pl
ane. All the forces acting on the screw are considered and
the relation.

P =
W sin (α+ φ)

Cos (α+ φ)

= W tan (α+ φ).

4. Turning moment required to overcome friction in screw jack.

Torque required to overcome friction T
1

= p.d/2

= W tan (α+ φ) d/2.

When the axial load is taken up by a thrust collar, then the torque required to
overcome friction at the collar T
2

= μ
1

W R.

Where,

μ
1
= co efficient of friction of

the collar.

R
1

R
2

R = (R1+R2)/2.

Total torque T = T1+T2 = W tan (α+ φ) d/2+ μ
1

W R.

If F is the horizontal force applied tangentially, then

T = F*L = W t
an (α+ φ) d/2.

In case the nut rotates in the opposite direction, i.e. the load is to be lowered, and
then the effort applied tangentially at the end of the Tommy bar is given by

T =
-
F*L =
-
W tan (α
-
φ) d/2.