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 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
adjustment mechanism 4)Scott Russel mechanism.
(A)Classification of kinematic link
(a)Classification based on relative motion between links.
1)Sliding pair.
3)Motion
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:
D
Link (3)
Link (4)
A
Link (2)
Link (1)
B
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
C
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:
D
Link(4)
E
Link (3)
Link (5)
Link (6)
C
Link (2)
A Link (1) B
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
If four links are connected to the same joint then it is a Quaternary joint.
One quaternary joint = Three binary joints
C
Link (3)
E
Link (4)
Link (7)
D
Link (5)
F
Link (8)
Link (11)
Link (6)
G
Link (10)
Link (9)
A
B
Link (1)
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)
Link (2)
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 four Bar Kinematic Chain
The inversions of four bar kinematic chain are as follows
a) Beam Engine:
Connecting rod
Link(3)
Piston
Crank
Link (2)
Fixed frame (link (1))
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
D
C
Link (4)
Crank (1)
Link (3)
Crank (2)
A
Link (1)
B
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:
B1
D1
E1
C1
B
1
F
E
D
F
Link (4)
A
C
Gas
Pressure
i)
Lever 1 ac (link (4))
Lever 2 DFE (link(3))
AB  Connecting rod (link
(3))
Piston
Cylinder
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
Link (4)
Guide way
s
Link (2) Crank
Fly wheel
Piston
Piston rod
Link (1)
(Slider)
a) Pendulum Of Bull Engine
B
Connecting rod ( link (3))
A
O
Crank (link (2))
Piston ( link (4))
Cylinder (link (1))
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
A
Link (2)
Crank
O
B
Link (4)
Piston
Link (3) Connecting rod
Cylinder
(link (1))
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=2r
Where =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=OB*OB
ii Tangential component
arOB=OB*OB
Where OB=Angular velocity of link OB
OB=Angular acceleration of link OB
OB=Length 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
Center.Let VA,VB,VC be the points A,B&C.
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,QR=atQR/QR
=3.6/0.175
=20.57rad/sec2
Angular acceleration of RS,RS=atRS/RS
=5.3/0.1125
=47.1rad/sec2
RESULT:
i
ii
Angular velocity of QR=2rad/sec
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**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
Angular acceleration,BC=atBC/BC
=102.5/0.6
=170.8rad/sec2
RESULT:
i
Linear velocity=4m/s
ii
Linear acceleration=117.5m/s2
iii Angular Velocity=5.83rad/sec2
iv Angular 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:
WCB=75rad/sec
VBC=VB=CB*WCB
=0.1*75
=7.5 m/s
Scale
1cm=2m/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)
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
Cam profile (axial /radial )(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 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)
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 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 .
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. 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= /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 ().
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
Flat belts
V-belts
Ropes
BELT MATERIALS
Leather, canvas, cotton & rubber.
Rubberized fabric & rubber.
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.