KINEMATICS
OF MACHINES
UNIT-1
SIMPLE MECHANISM.
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
mechanism 9)Mechanical advantage of mechanism 10)Transmission angle?
1)Kinematic link
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 the other link follow a definite
law it is known as kinematic chain.
4)Klein’s
equation for joints.
h-Higher pair joint
l-Number of links
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-Number of links.
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 sum of remaining link length.
8)Inversion of
mechanism.
The method of obtaining different
mechanism by fixing different links in a kinematic chain is known as inversion
of mechanism.
9)Mechanical
advantage of mechanism.
It is defined as the ratio of output
torque to the input torque also defined as the ratio of load to effort.
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 advantage is maximum.Under this
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 mechanism.
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 having 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 (A)Classification
of kinematic link?
(B) 1)Indexing mechanism 2)Snap action mechanism 3)Motion adjustment mechanism 4)Scott
Russel mechanism.
(A)Classification
of kinematic link
(a)Classification
based on relative motion between links.
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 that one link revolves around
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 known 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 everything 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.
3)Motion
adjustment mechanism.
The mechanism used to adjust or
modify any one of the links in a
mechanism is known as motion adjustment mechanism. 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 the 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(900-ß/2)
=AE/AB
=0.15/0.3=0.5
∟ABE=(90-ß/2)
=300
Time ratio of
cutting stroke to the return stroke:
We know that,
(900-ß/2)=30
ß=1200
Time for cutting stroke=α
Time for return
stroke=ß
=3600-1200
1200
Answer=2
Length of
stroke=p1p2
=2(BP1)sin(900-ß/2)
=2*0.7sin(900-600)
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)Whitworth 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 R1R2.In
the extreme positions,AP1 and AP2 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 CB1 to CB2
in the clockwise direction.The return stroke occurs when the crank rotates from
the position CB2 to CB1 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 R1R2 during cutting and return ,therefore
length of stroke =R1R2=P1P2=2AP
Whitworth quick
return motion 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 CA1 to CA2 through 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 CA1 to CA2 through an angle in the clockwise direction,the tool moves
back from right to left hand 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 of 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 called 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
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 the 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.
c)
Quaternary Joint
|
 |
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)
11 = 2/3 (17)
L.H.S.¹ 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 four 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.
Link (1) Frame
Link (2) Crank
Link (3)
Connecting rod
Link (4) lever arrangement
Link (5) Piston
ii) In this mechanism one link oscillated while the other
rotates about fixed link.
iii) It is used to convert the rotary motion into
reciprocating motion.
b) Coupling Rod of Locomotive Engine Wheel
i) This 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 rotary – 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 double lever
mechanism
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 Engine
 |
i)
This mechanism is obtained by
fixing link (4)
ii)
The crank (Link (2)) rotates while
link (3) Oscillates about a pin pivoted to the fixed link (4) at A end link (1)
reciprocates
iii)
This is used to supply feed water
to boiler pans.
b)
Oscillating Cylinder Engine
1)
This is used to convert
reciprocally motion into rotary motion.
2)
Link (3) is fixed link (3) rotates,
the link (1) reciprocates, link (4) Oscillates about a pin pivoted to the link
(A)
c)
Internal 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 all cylinders.
(iv)
The reciprocating motion is converted into rotary motion.
MODULE –2
INSTANTANEOUS CENTRE,
VELOCITY & ACCELARATION
DIAGRAMS
TWO MARKS :-
1.What are the components of acceleration?
i
Radial component of
acceleration
ii
Tangential component of
acceleration
2.Write an expression for find number of instantaneous centers in a
mechanism?
N=n(n-1)/2,n-no of links
3.What is expression for Cariolis componenet of acceleration?
aBC=2rw
Where w=Angular velocity of ‘OA’
V=Linear velocity of ‘B’
4.What are the expression for radial and tangential component of
acceleration?
i
Radial component
arOB=wOB *OB
ii
Tangential component
arOB=aOB *OB
Where wOB =Angular
velocity of link OB
aOB =Angular
acceleration of link OB
5.How can we represent the direction 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 velocity?
Consider
three points A,B,C on a rigid link.I being instantaneous
Then we have
VA/IA=VB/IB=VC/IC
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 instantaneous
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 the length of the
links
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 velocity and angular acceleration
of links QR and RS.
Space diagram:
Scale
1cm=30mm
Velocity diagram:
Scale
1cm=0.1m/s
WPQ=10 rad/sec
VQP=VQ=0.0625*10
=0.625m/s
WQR=0.36/0.175=2rad/s
WRS=0.43/0.1125=3.8rad/s
acceleration diagram: Scale
1cm=1m/s2
arPQ=V2Q/QP=0.6252/0.0625
=6.25m/s
arRS=V2/RS=0.432/0.1125=1.644m/s2
arQR=V2QR/QR=.362/0.175=0.74m/s2
Angular acceleration of QR,aQR=atQR/QR
=3.6/0.175
=20.57rad/sec2
Angular acceleration of RS,aRS=atRS/RS
=5.3/0.1125
=47.1rad/sec2
RESULT:
i
Angular velocity of QR=2rad/sec
ii
Angular velocity of RS=3.8rad/sec
iii
Angular acceleration of
QR=20.57rad/sec2
iv
Angular acceleration of
RS=47.1rad/sec2
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
accelerationof the connecting
rod,at a crank angle of
450 from inner dead center
position.
Space diagram:
Scale
1cm=75mm
Velocity diagram:
Scale
1cm=1m/s
NAB=300r.p.m
WAB=2*P*300/60=31.41rad/sec
VAB=VB=AB*WAB
=0.15*31.41
=4.7m/s
Acceleration diagram:
Scale
1cm=50m/sec2
arBA=aB=V2BA/BA
=4.72/0.15
=147.27m/sec2
Linear Velocity,Vd=4m/s
Linear acceleration,ad=117.5m/s
Angular velocity,WBC=VBC/BC
=3.5/0.6
=5.83rad/sec2
=102.5/0.6
=170.8rad/sec2
RESULT:
i
Linear velocity=4m/s
ii
Linear acceleration=117.5m/s2
iiiAngular Velocity=5.83rad/sec2
ivAngular acceleration=170.8rad/sec2
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 acceleration 0f
1200 rad/sec2.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
WCB=75rad/sec
VBC=VB=CB*WCB
=0.1*75
=7.5 m/s
MODULE -3
KINAMATICS OF CAMS
TWO MARKS :-
1.What is cam ?
follower by direct contact
2.Classification of cam?
(i)
according 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)
radial cams
(iii)
spiral cams
(iv)
drum cams
(v)
spherical cams
4.classification of follower ?
(i) According to follower 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 denoted 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 remains
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?
In radial cams the follower 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 absence 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 follower
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)
Radial follower and
(ii)
Offset 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 translating cams.
20.Write the
formula for maximum velocity?
Vo (max) = 2ωs
θo
Vr (max)= 2ωs
θo
16 MARKS :-
BRIEF ANSWERS
(1) A cam rotating clockwise at a uniform speed of
1000 r.p.m Is required to give a roller follower 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 18cm 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 diameter 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 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.
(2)Draw the profile of the cam when the roller
follower moves with cycloid motion during outstroke and return stroke 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 . draw the profile for
(iv)
(a) Radial type
(b)offset type(16)
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 equal 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.
Radial
type:(6)
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 circle 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 during 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 an 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
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 am 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 return
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.
Radial type
cam profile:(9)
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 each 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
an 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 join the other and
four points to upper and lower ends of diagonal.
Radial type:(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.
Instead of
joining the points make a perpendicular line of joints its edges.
6. Draw the
radial cam and explain its terminologies? (16)
Diagram : (9)
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 between 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 pressure angle is maximum .
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. Define addendum and
dedendum.
Addendum is the radial
distance of a tooth from the pitch circle to the top of the
tooth.
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= P/DT
Where D = Diameter of pitch circle.
T =
Number of teeth on the wheel.
4. Define I) path of contact.
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 by 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 conjugate 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 through 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 path 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
athematically.
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 the 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 relationship between addendum,
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 clearance.
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 volute 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
contact ratio.
17. What are the advantages
and disadvantages of gear drive.
Advantages:
a) It
transmits exact velocity ratio.
b) It has high efficiency.
Disadvantages:
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 (b).
It is the angle between
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. Define 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 mounted 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 limited 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 module 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;
NB < 1/10 NA;
T1 = T3 =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 NA .
Let NB
= 1/11 NA .
We know that for reverted gear train
R1
+ R2 = R3 + R4
Or
mAT1/2 + mAT2/2 = mB
T3 /2 + mBT4/2
MA(T1
+T2) = MB(T3 +T4)
2(24 + T2)
= 3.5(24 + T4)
T2 =18 +1.75T4
Now speed ratio, NB/NA =T1*T3 /T2*T4
1/11 =24*24/T2*T4
T2*T4
=6336.
Sub the value of T2 ,we get
(18
+1.7574)T4 = 6336
T4 =
56
T2
=116.
2. ACTUAL VELOCITY
RATIO.
NA/NB = T2*T4/T1*T3
NA/NB
=11.28.
3. DISTANCE BETWEEN
SHAFTS C AND A:
C = d1
+ d2/2
= mA(T1
+ T2 )/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?
Types 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 belting.
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.
The 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: Utility.
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.
12. Advantages of V-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.
13. Disadvantages of V-belt.
*It cannot be used with large center
distances.
*It is not as durable 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 motion.
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(RN) between the two bodies. It
is generally denoted by μ.
μ=F/RN.
23. Define 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.
24. Square thread vs V-thread.
*V-thread is stronger and often
moves frictional to the motion than square threads.
*A given load may be lifted by
applying lesser force by square thread as compared to V-threads.
*V-threads are capable of taking
more loads as compared to square threads.
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 obtain
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 friction.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 RN 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 P1 is exactly balanced by the
force of friction F1.
Now increase the applied force to P2 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 F2 = P2. Thus every time the efforts increased the force of friction is
also increases, so as to become exactly equal 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
limiting 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 expression 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 mounted radially 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 pulley rim.
N = speed of pulley.
w = angular speed of pulley.
w1 = angular speed at which
the begins.
Centrifugal
force on each shoe Fc = mw²r
Spring force
exerted by each spring Fs = mw1²r
Net force on the
shoe =Fc-Fs
= mw²r-mw1²r
Frictional force
acting on each shoe, f = μ (Fc-Fs)
Frictional
torque Fr = F*R
= μ (Fc-Fs) R
Total frictional
torque transmitted, T = n* μ (Fc-Fs) 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.
Screw jack with square threads.
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 effort
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 plane.
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
T1 =
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 T2 = μ1 W R.
Where,
μ1 = co efficient of friction of
the collar.
R1 = outside radius.
R2 = inside radius.
R = (R1+R2)/2.
Total torque T =
T1+T2 = W tan (α+ φ) d/2+ μ1 W R.
If F is the
horizontal force applied tangentially, then
T = F*L = W tan
(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.
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