TOM Theory

THEORY OF MACHINE

THEORY OF MACHINE (Theory)

CONTENTS

Unit Chapters Page No.
1A Theory of Machines 2-6
1B Velocity and Acceleration Analysis 7 - 11
2A Gears 12 - 23
2B Gear Trains 24 - 29
3 Dynamics ForceAnalysis and Flywheel 30 - 38
4 Vibrations 39 - 50

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1 A THEORY OF MACHINES

Analysis includes kinematic and dynamic analysis
• KinematicAnalysis : Deals with the geometry of motion without considering the force that cause the
motion.
• Dynamic Analysis : Includes kinematics and the effect of forces that cause the motion
• Mechanism : A system consists of links and joints and converts one form of motion to another form
(or)Asystem of links and joints that converts the available form of motion to the desired form.
• Planar Machanism :Amechanism that is constrained to move in a single palne or in parallel planes is
referred as a planar mechanism or plane mechanism.
• Degrees of Freedom : Number of independent co-ordinates that are required to specify the system
completely.
• DOF of a rigid body in spatial motion are 6, consisting of 3 translatory freedoms and 3 rotational
freedoms
• DOF of a rigid body in spatial motion are 3, consisting of 2 translatory freedoms and one rotational
freedoms.
• Link/Kinematic Element : A rigid body or a resistant body that forms the part of a mechanism.
• Kinematic Pair : Two links or elements connected with a joint that allows the relative motion between
the links.
CLASSIFICATION OF LINKS :

The links or the kinematic elements are the basic building block. They are classified as follows.
Binary link : It connects with two other links

o oo o o o
(
Ternary link: It connects with three other links in a system. ( ((

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o( (o (

Quatenary Link : It connects with four other links.

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o o (
(

CLASSIFICATION OF KINEMATIC PAIRS :
Based on Degrees of Freedom :Akinematic pair allows few degrees of freedom and constraints some of
them. When two bodies are joined together one of the body (base) has all the DOF, whereas the other body
loses some DOF and has few DOF relative to the base body. Depending on the allowed degrees of

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THEORY OF MACHINE

freedom and constrained degree of freedom they are classified Class I, Class II ........., Class-n kinematic
pair allows n degree of freedom for the pair and constraints 6-n (3-n) degrees of freedom. A pair that
constrains all the degrees of freedom of the second link relative to the first link is not considered as a
kinematic pair. It is a rigid joint.
• Based on Nature of relative motion : Based on the relative motion that exists between the two links
the pairs can be classified as Rotary/Revolute pairs. Sliding/Prismatic pairs, Cylindrical pairs, special pairs
and so on.
• Based on Nature of contact : Based on the nature of contact between the two links the kinematic
paris are classified as lower pairs and higher pairs. When the two bodies have surface to surface contact
they are referred as lower pairs. When the contact between the bodies is a point or line contact they are
referred as higher pairs.
• Based on type of closure : Closure means the way the two dodies are held together to have continu-
ous contact. Two types of joint closures exists they are form closure and force closure.
• Form Closure : The two links are held together by the links and they cannot be detached easily.
• Force Closure : The contact is maintained by an external force either the gravity force or spring force
and the two bodies can be separated easily.
KINEMATIC CHAIN :

A kinematic chain is formed by connecting number of links with kinematic pairs so that there exist
de3finite relation between the motion of various links. They can be of two types closed kinematic chains and
open kinematic chains.
MECHANISM :Amechanism is obtained by fixing any one link in a kinematic chain.
Degrees of Freedom of a Kinematic chain :Akinematic chain is formed by connecting number of links
with number of pairs. Let ‘n’ be the no. of links and Jn is the number of pairs of class n. Then as for grublers
criterion the DOF of a spatial kinematic chain is given by

dof = 6n - 5J1 - 4J2 - 3J3 - 2J4 - J5
For a planar kinematic chain,

DOF = 3n - 2J1 - J2
Degrees of Freedom of a Mechanism : As one link is fixed in a kinematic chain to get mechanism,

Grublers equation for the DOF of a mechanism is as follows for spatial mechanism
DOF = 6(n-1) - 2J - J

12

Note : A mechanism has six (three) degrees of freedom less compared to that of the kinematic chain from
which it is obtained.
Classification based on degrees of greedom:
Zero degrees of freedom : Structure
Negative degrees of freedom : Super structure/Preloaded structure
Positive degrees of Freedom : Mechanism
FOUR BAR CHAIN/QUADRIC CYCLE CHAIN :

It is the basic chain that consists of four links and four turning pairs. It is the basic from which many one
DOF mechanisms can be derived. The necessary condition to form a four bar chain based on Their lengths
is l < s + p + q.
When l is the length to the longest link, s is that of the shortest link and p,q are the lengths of the remaining

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MECHANICAL ENGINEERING

two links. Though a chain is formed by satisfying the above condition it may not result in useful mechanisms

if one barely satisfy the condition.

Grashoffs Condition : Grashoft’s condition checks the link proportions and classifies the chains/mecha-

nisms

If l + s < p + q Grashoffs or Class - I

If l + s > p + q Non Grashoffs or Class - II

If l + s = p + q Special Grashoffs or Class - III

Inversion :

By fixing one link in a kinematic chain a mechanism is obtained. By fixing different links, different

mechanisms are obtained. Inversion is the process of obtaining different mechanism by fixing different links

in a kinematic chain.

Nomenclature of four mechanisms : C
Afour bar mechanism is as shown in figure ;AD, the link one is known

as the fixed link, AB, the link 2 acts as input link. The link 3 CD

is the coupler and the link 4 DC is the output link. The input and

output links can be interchanged. If the input/output link can have

complete rotation about its centre it is known as crank. If it has only

a partial revolution it is known as a rocker or an oscillatory link.

Based on this the mechanisms can be classified as C-C, C-R, R-C, R-R mechanisms.

Inversions of Grashoff’s 4-bar chain (l + s < p + q) : The mechanisms obtained from the Grashoffs

Kinematic chain are based on the position of the shortest link

• Shortest link fixed-Double crank mechanism.

• Link adjacent to shortest link is fixed crank-Rocker mechanism

• Link opposite to the Shortest link is fixed : Rocker-Rocker mechanism

Inversions of Non-Grashoffs 4-bar chain : l + s > p + q

By fixing any link it results in Rocker-Rocker mechanism.

Inversion of Special Grashoff’s Chain :

Parallelogram on anti paralleogram connection will result in duble crank or drag link mechanism Deltoid

connection will result in crank-Rocker mehanism.

In parallelogram connection both long links and short links are opposite to each other. In deltoid connection

both long links and short links are side by side.

Equivalent linkage : By replacing any pair in a kinematic chain with its equivalent (from the same class).

An equivalent chain can be obtained, by replacing any turning pair in a four bar chain with a siliding pair a

slider crank chain can be obtained. By replacing any two turning pairs with sliding pairs a double slider crank

chain can be obtained.

Inversions of a single slider crank chain.

I Inversion :An I.C engine mechanism/compressor mechanism

II Inversion : wit-worth Quick return motion mechanism and rotary engine.

III Inversion : Crank and slotted lever type quick return motion mechanism and oscillating

cylinder engine

IV Inversion : Hand pump

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THEORY OF MACHINE

Inversions of Double Slider Crank chain :

I inversion : ScothYoke-mechanism. Useful for generating trigonometic functions

II inversion : Elliptical trammel. Useful for tracing the elliptical curves

III inversion : Oldham’s Coupling. Useful for connecting two parallel shafts with little offset.

Position Analysis of 4-bar mechanism

Position analysis deals with the determination of the positions of all other links by knowing the position of

input link. It also involves the determination of extreme positions and the transmission angle.

B C
γ(

Ao o
C
D

o

VelocityAnalysis in 4-bar Mechanism

Consider the mechanism shown in figure. The pre requisite for the velocity analysis is the knowledge of

position of all the links which is available from the position analysis or from configuration diagram. If the link
2 rotates with ω2 rad/sec in counter clock wise direction. The velocity of other links are obtained as follows.

Construction procedure for velocity polygon.

A and D are fixed points having zero velocity mark a, d at a convenient location they act as reference

for tha velocity polygon.
Velocity of B relative toAwill be l2 ω2. perpendicular toAB in direction of ω2 so draw ab ⊥ toAB with

a length l2 ω2. Velocity of C relative to B will be ⊥ to BC but sense is not known hence draw a line ⊥ to BC
passing through b. Velicity of C relative to D will be ⊥ ’lr to DC sense is not known. So draw a line ⊥ lr to

DC through d. These two lines will intersect at C that completes the velocity polygon.
In the velicty digram the vector bc indicates the velocity of C relative to B and ω3l3 = bc gives ω3.

similarly dc. = ω4l4 from which ω4 can be obtained.

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Note:

Capital letters deals with the configuration diagram is AB, CD are respective link positions, lower case

letters indicate the points on the velocity diagram

Special cases:

1. When the link AB and BC are paralel to each other

Velocity of polygon will be a strainght line

Velocity of B is equal to the velocity of C.

∴ l ω2 and ω3 =0

2

If bothAB and DC are on the same side ofAB both have the anular velocity in the same sense. If they

are on opposite side i.e. BC crosses AD; AD and DC will have velocities in the opposite sense
2. When AB and BC are parallel i.e. they are in the same line, ω4 = 0 and ω2ω1 =ω3l3.
3. When BC and CD are parallel i.e. they are in same line, ω2 = 0 and ω3 l3 = ω4 l4

Consider the configuration shown in fig. Let the crank of lenght ‘r’rotates in CCW sense at ω2 red/sec.
The velocity of B relative toAis rω2 ⊥lr to AB draw ab ⊥lr to AB. Velocity of C relative to B will be ⊥lr to
BC draw a line passing through b and ⊥ to BC. Velocity of C relative to G is parallel to G. So draw a line
through g parallel to G. The intersecting point is C.

Now the vector bc gives velocity of C relative to B and lω2 = bc. Vector gc ac gives the velocity of
slider.
Special case :
1. Slider at TDC

Velocity of slider = 0
rω2 = lω3 or ω3. (r/l) = ω2/n, ω3 and ω2 is same sense
2. Slider at BDC
Velocity of slider = 0
rω2 = lω2 and ω3 and ω2 are in opposite directions.
3. Crank perpendicular to line of stroke
Velocity of slider = tω2 and ω3 = 0

“Great works are performed not by strength
but by perseverance”

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1 B “VELOCITY AND ACCELERATION ANALYSIS”

VELOCITY OF LINK

• Velocity of A relative to O = Arc AA’

δt dθ
dt
Vao = r.δθ = rω
δt = r

Vao A

B A` A
r
r
ω δθ
Ο

)

Ο
as δt → o, AA` will be perpendicular to OA, Thus the velocity of A is ωr and is perpendicular to OA.

Note :

• The velocity of any point relative to any other point on a fixed link is always zero.

• The velocity of an intermediate point on any of links can be found easily by dividing the corre-

sponding velocity vector in the same ratio as the point divides link.

• The angular velocity of a link about one extremity is the same as the angular velocity about the

other.

VELOCITY OF RUBBING

The velocity of rubbing of the two surface will depend upon the angular velocity of a link, relative to the

other.

• For pin at A

•C

D•
ω

••
AB

Pin at ‘A’joins linksAD andAB.AD being fixed the velocity of rubbing will depend only upon angular

velocity ofAB.
∴ Velocity of rubbing

= ra.ωAB

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ra = Radius of pin ‘A’

• Pin at D
Velocity of rubbing = rd.ωcd

• Pin at B

Both linkAB and BC is moving

ω = ω = clockwise
ab
ω ω
oc = = anticlockwise

∴ Velocity of rubbing = rb (ωab + ω )
bc

• Pin at C

Velocity of rubbing = rc (ωbc + ω )
dc

INSTANTANEOUS CENTRE (I-CENTRE)

Instantaneous centre of rotation or virtual centre

Let plane body ‘P’having non-linear motion relative to another plane Q.At any instant, the linear velocity
of the point ‘A’and ‘B’on the body ‘P’ are ‘Va’and ‘Vb’respectively.
If a line is drawn perpendicular to the direction of Va at ‘A’, the body can be imagined to rotate about
some point on this line. Similarly for point B. If the intersection of the two lines is at ‘I’ the body ‘P’ will
be rotating about I at the instant.
This point ‘I’is known as instantaneous centre of velocity.
Note : If the direction V and V are parallel than the I-centre of body lies at infinity.

ab

Centrode
• As we know, in general the position of instantaneous centre changes throughout the whole motion.
The locus of all these instantaneous centre for a particular link is known as “Centrode”. It is a curve.
Axode
• The line passing through instantaneous centre and perpendicular to the plane of motion is known as
instantaneous axis. The locus of instantaneous axis for a link during the whole motion is known as
“Axode”. It is a surface.
• Number of instantaneous centre

I= n (n-1)
2

here n= Number of Link

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KENNEDY’S THEOREM
• If three plane bodies have relative motion among themselves, their I-center must lie on a straight line.

This is known as Kennedy’s theorem.

pq

r

• Consider three plane bodies ‘p’, ‘q’ and ‘r’; ‘r’being a fixed body. ‘p’ and ‘q’ rotate about centre
pr and qr respectively relative to the body ‘r’. Thus, pr is the I-centre of bodies ‘p’and ‘r’whereas qr
isthe I-centre of bodies ‘q’and ‘r’. Assume the I-centre of the bodies ‘p’ and ‘q’at the point pq.
• If the point pq is considered on the body p, its velocity vp is perpendicular to the line joining pq and
pr. If the point pq is considered on the body q, its velocity vq is perpendicular to the line joining pq and qr.
• The two velocities of the I-centre pq are in different directions which is impossible.
• The velocities vp and vq of the I-centre will be same only if this centre lies on the line joining pr and
qr.

l13

Angular Velocity Ratio Theorem

B l23 l34 C

4

l24 l12 2 l14

A1 D

It is used to find angular velocity of a link if angular velocity of another link is known.
The angular velocity ratio of two links relative to third link is inversely proportional to the distances of
their common I-centre from their respective centre of rotation.

ω = l24 - l12
4 l24 - l14

ω
2

ACCELERATION ANALYSIS

The rate of change of velocity with respect to time is known as acceleration and it acts in the direction

of the change in velocity.

Tangential Acceleration

• The rate of change of velocity in the tangential direction of the motion is known as tangential

acceleration.

dv
at = dt

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The tangential component of acceleration occurs due to the angular acceleration of link.

Centripetal Acceleration

• The rate of change of velocity towards the centre of rotation is known as centripetal or radial

acceleration. v2
ac = r

Note : The acceleration of intermediate points on the links can be obtained by dividing the accel-

eration vectors in the same ratio as the points divide the links.

Coriolis Acceleration Component

• It comes into picture when a body has its motion relative to a moving body, for example, motion of a

slider on a rotating link.
Let , ω =Angular velocity of the link

α =Angular acceleration of the link

v = Linear velocity of the slider on the link V V`sin δθ
f = Linear acceleration of the slider on the link

r = Radial distance of point P on the slider. v`cos δθ v`
ω‘r’sin δθδθ

)

R

R

• Acceleration of P along AR =

Acceleration of slider - Centripetal acceleration
= f - ω2r

This is the acceleration of P alongAR in the radially outward direction. f will be negative if the slider has

deceleration while moving in the outward direction or has acceleration while moving in the inward direc-
tion.
Acceleration of P perpendicular to AR

= 2 ωV + Tangential acceleration
= 2 ωV + rα
Note -
• The component 2 ωv is known as the Coriolis acceleration component.

• The Coriolis component is positive if link AR rotates clockwise and slider moves radially

outwards the link rotates counter clockwise and slider moves radially inwards. Otherwise

Coriolis component will be negative.

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• The direction of the coriolis acceleration component is obtained by rotating the radial velocity
vector‘v’ through 90o in the direction of rotation of the link.

KLEIN’S CONSTRUCTION
In Klein’s construction, the velocity and the acceleration diagrams are made on the configuration diagram
itself. The line that represents the crank in the configuration diagram also represents the velocity and
acceleration of its moving end in the velocity and acceleration diagram respectively.

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“To be a champ, you have to believe in yourself
when nobody else will”

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2 A GEARS

INTRODUCTION

Gears uses no intermediate link or connector and tranmit the motion by direct contact. The two

bodies have either a rolling or a sliding motion along the tangent at the point of contact. No motion is

possible along the common normal as that will either break the contact or one body will tend to penetrate

into other. VP

21

- - -o- - - - - - - - - - -o- - - -
(
(

Point P can be assumed on gear 2 or gear 1.

Vp =ω2r2 = ω1r1 ω= 2µN
60

ω r NT
1 = 2 = 1 = 2

ω 2 r1 N2 T1

Symbols has usual meaning.

CLASSIFICATION OF GEARS BASED ON THE ARRANGEMENT OF SHAFT
1. Parallel shaft
A. Spur Gears : They have straight teeth parallel to the axes and thus are not subjected to axial thrust due
to tooth load.

( ((

(

Line contact

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( ( ((

Note : At the time of engagement of the two gears, the contact extends across the entire width on a
line parallel to the axis of rotation. This results in sudden application of the load, high impact stresses
and excessive noise at high speeds.

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B. Spur Rack and Pinion : Spur rack is a special case of spur gear where it is made of infinite diameter so
that the pitch surface is plane. The spur rack and pinion combination converts rotary motion into translatory
motion or vice-versa. It is used in lathe in which the rack transmits motion to the saddle.

C. Helical gears or Helical spur gears :- In helical gears, the teeth are curved. Two mating gears have
the same helix angle, but have teeth of opposite lands.

Note:-
• At the beginning of engagement, contact occure only at the point of leading edge of the curved teeth.
Thus the load application is gradual which results in low impact stresses.
• The helical gears can be used at higher velocities than the spur gears and have greater load - carrying
capacity.
• Helical gears have the disadvantage of having end thrust as there is a force component along the gear
axis.
D. Double - helical and Herring bone Gears :A dauble - helical gear is equivalent to a pair of helical
gears secured together, one having a right hand helix and other a left hand helix.
Note :
• No axial thrust is present.
• If the left and the right inclinations of a double - helical gear meet at a common apex and there is no
groove in between the gear is known as herringbone gear.
2. Intersecting Shaft

The motion between two intersecting shafts is equivalent to the rolling of two cones assuming no
slipping.
A. Straight bevel Gears : The teeth are straight, radial to the point of inter-section of the shaft axis
and vary in cross-section throughout their length.
Note: Gears of the same size and connecting two shafts at right angle to each other are known as mitre
gears.

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B. Spiral bevel Gears : When the teeth of a bevel gear are inclined at angle to the face of the bevel, they
are known as spiral bevel or helical bevels.
Note :
• There is gradual load application and low impact stresses.
• These are used for the drive to the differential of an automobile.

• Zerol bevel Gears : Spiral bevel gears with curved teeth but with a zero degree spiral angle are
known as zerol bevel gears.
3. Skew Shaft
• In case of skew (non-intersecting) shafts, a uniform rotary motion is not possible by pure rolling con-
tact.
• If if is desired that the two hyperboloids touch each other no the entire length as they roll, they must
have some sliding motion parallel to the line of contact.
• If the two hyperboloids rotate on their respective axes, the motion between them would be a combina-
tion of rolling and sliding action.

A. Crossed helical gears
• The use of crossed - helical gears or spiral gears is limited to light loads. These gears are used to drive
feed mechanism on machine tools, camshafts and oil pumps in I.C. engine.

B. Worm Gears
• It is a special case of a spiral gear in which the larger wheel usually has a hallow or concave shape.
• The smaller of the two wheels is called the worm which also has a large spiral angle.
• The sliding velocity of a worm gear is higher as compared to other types of gears.
• In Non-throated worm gear the contact between the teeth is concentrated at a point.

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• In single throated worm gear, teeth are curved to envelope the worm gear, gear teeth are curved to
envelope the worm. There is line contact between the teeth.
• In double throated worm gear, there is area contact between the teeth.
• Aworm may be cut with a single or multiple thread cutter.

C. Hypoid Gear
• A hypoid pinion is larger and stronger than a spiral bevel pinion.
• A hypoid pair has a quite and smooth action.
• There is continuous pitch line contact of the mating hypoid gears while in action and they have larger
number of teeth in contact than straight tooth bevel gears.

GEAR TERMINOLOGY

• Pitch Circle : It is an imaginary circle drawn in such a way that a pure rolling motion on this circle
gives the motion which is exactly similar to the gear motion.

Note:-
• These pitch circle always touch each other for the correct power transmission.
• It is not fundamental characteristics of gear.
• Pitch Point : It is a point where the two pitch circle of the mating gear touch each others.
• Pressure angle (φ) : It is the angle between common normal at a point of contact tangent at a pitch
point.

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The standard value of pressure angle are 14¼o ,20o , 25o.
• Module (m) : It is defined as the ratio of pitch circle diameter in mm to the number of teeth.

C = πD
T

• Addendum Circle : A circle drawn from top of tooth and concentric to pitch circle is known as

addendum circle.Addendum is Radial distance between pitch circle and addendum circle and it is equal to

1 module.

Note : Clearance = 0.157 m

• Dedendum Circle :Acircle drawn from botton of the teeth and concentric with pitch circle. Dedendum

is Radial distance between pitch circle and dedendum circle and it is equal to 1.157 module.

• Circular Pitch (C) : It is a distance along a pitch circle from one point on a tooth to the corresponding

point on the next tooth.

Pd = T
Dmm

D = Pitch circle-diameter

T = Number of Teeth

• Diametral Pitch : It is the ratio of number of teeth to the pitch circle diameter but diameter should be

in mm.

• Relation between circular pitch (C) diametral pitch (Pd)
Circular pitch × Diametral pitch = π

• Tooth Thickness : It is the thickness of tooth measured along pitch circle.
• Tooth Space : The space between the consecutive teeth measured along the pitch circle.
• Backlash: It is difference between tooth pace and tooth thickness, which is generally provided to
avoid jamming due to thermal expansion.
• Face : The portion of tooth profile above the pich durface.
• Flank : The portion of tooth profile below the pitch surface.
• Profile : The curvature contained by face and flank.
• Path of contact (POC): It is the path travelled by point of contact from the starting of engagement to
the end of engagement.

POC = Path of approch + Path of Recess.
• Arc of Contact (AOC): It is the path traced by a point on the pich circle during starting of engage-
ment to the end of engagement.

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• Gear Ratio (G)
T

G= t
Here,

T = Number of teeth of gear.
t = Number of teeth of pinion (small gear).
• Velocity Ratio (VR)

VR = 1
Gear ratio

• Angle of Action:Angle turned by gear from the beginning of engagement to the end of engagement
of a pair of teeth.

= Angle of approch +Angle of recess
• Contact Ratio
= Angle of action = Arc of contact

Pitch angle Circular pitch
LAW OF GEARING
• The law of gearings states the condition which must be fulfilled by the gear tooth profiles to maintain a
constant angular velocity ratio between two gears.

Let ω1 = angular velocity of gear 1 (clockwise)
ω2 = angular vecocity of gear 2 (anticlockwise)

ω1 BP
ω2 AP

Condition :
(i) If it is desired that angular velocities of two gears ramain constant, the common normal at the point of
contact to the two teeth shuld always pass through a fixed point P which divides the line of centres in the
inverse ratio of angular velcities of two gears.
(ii) For constant angular velocity ratio of the two gears, the common normal at the point of contact of the
two mating teeth must pass through the pitch point.
• Velocity of Sliding

If the curved surfaces of the two teeth of the gears are to remain in contact one can have a sliding
motion relative to the other along the common tangent.
Velocity of sliding = (ω1 + ω2) PC

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= Sum of angular velocities × distance between the pich point and point of contact.
TYPE OF TEETH PROFILE
(i) Cycloidal Profile Teeth

A cycloid is the locus of a point on the circumference of a circle that rolls without slipping on the
circumference of another circle. In this type, the faces of the teeth are epicyloids and flans the hypocyc-
loids.
• An epicycloid is the locus of a point on the circumference of a circle that rolls without slipping outside
circumference of another circle.
• An hypocycloid is the locus of a point on the circumference of a circle that rolls without slipping inside
the circumference of another circle.
• A property of the hypocycloid is that at any instant, the line joining the generating point with the point of
contact of the two circles is normal to the hypocycloid.
• Meshing of Teeth : During meshing of teeth, the face of a tooth on one gear is to mesh with the flank
of another tooth no the other gear. Thus for proper meshing it is necessary that the diameter of the circle
generating face of a tooth is the same as the diameter of the circle generaing flank of the meshing tooth, the
pitch circle being the same in the two cases.
• In case of cyloidal teeth. The pressure angle varies from the maximum at the beginning of engagement
to zero when the point of contact coincides with pitch point P and then again increases to maximum in the
reverse direction.
• If the centre distance between the two pitch circle varies, the point P is shifted and the speed of the
driven gear would vary.
(ii) Involute Profile

Involute is a curve generated by point on a tangent which rolls on a circle without slipping. The
involute profile on a gear will be generated through a generating circle and this generating circle will be
known as base circle. It is a fundamental property of a gear its radius will not change in change in any
condition for a gear.

Note:
• A normal on any point of involute profile will be tangent to the base circle.
• Tooth profile is always generated from base circle and the profile between root circle and base circle
will not be of involute type.
• If the centre distance between the two pitch circle varies, the point P is shifted and the speed of the
driven gear would vary.
• For a pair of involute gears, velocity ratio is inversely proportional to the pitch circle diameters as well
as base circle diameters.
• Path of Contact

= √R2a- R2 cos2 φ + √r2a - r2 cos2 φ - (R+r) sin φ

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THEORY OF MACHINE

Here,
r = pitch circle radius of pinion
R = pitch circle radius of wheel
ra = addendum circle radius of pinion
Ra = addendum circle radius of wheel

• Arc of contact

Path of contact
= cos φ
• Number of pairs of teeth in contact

Arc of contact
= Circle pitch

• The maximum value of the addendum radius of the wheel to avoid interference.

=R 1+ r r +2 sin2 φ
R R

• Maximum value of addendum of the wheel

=R 1 + r r +2 sin2 φ − 9
R R

• Minimum Number of teeth on the wheel for the given values of the gear ratio the pressure angle and the
addendum coefficient (aw).

T= 2a
1 w

1+ G 1 + 2 sin2 φ − 1
G

Where aw = addendum coefficient
Note -

• Point of contact will always be at a line tangent to the base circle.
• Base circle diameter = pitch circle diametere X cos φ

• For equal number of teeth on pinion and the wheel, G = 1

Tmin = 2
1 + 3sin2 φ − 1

• To avoid interference, between rack and pinion, the maximum addendum of the rack can be increased
in such a way that C coincides with E

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MECHANICAL ENGINEERING

Let the adopted value of the addendum of the rack be arm, where ar is the addendum coefficient.
GE = mt sin2 φ
2

To avoid interference, GE ≥ arm

t∴≥ 2ar
sin2 φ

• For 20o pressure angle tmin = 13

CONDITION FOR INTERCHANGEABLE GEARS
The gears are interchangeable if they have
• the same module
• the same pressure angle
• the same addendums and dedendums, and
• the same thickness

UNDERCUTTING

• If the addendum of the mating gear is more than the limiting value; it interferes with the dedendum of the
pinion and the two gears are locked.
• If instead of the gear mating with the pinion, a cutting rack having similar teeth is used to cut the teeth
in the pinion, the, portion of the pinion tooth which would have interfered with the gear will be removed.A

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THEORY OF MACHINE

gear having its material removed in this manner is said to be undercut and the process undercutting.

COMPARISON OF CYCLOIDALAND INVOLUTE PROFILE

Cycloidal Teeth

(a) Pressure angle varies from maximum at the beginning of engagement, reduces to zero at the pitch

point and again increases to maximum at the end of engagement resulting in less smooth running of the gears.

(b) It involves double curve for the teeth, epicycloid and hypocycloid. This complicates the manufac-

ture.

(c) Owing to difficulty of manufacture, these are costlier.

(d) Exact centre distance is required to transmit a constant velocity ratio.

(e) Phenomenon of interference does not occur at all.

(f) The teeth have spreading flanks and thus are stronger.

(g) In this, a convex flank always has contact with a concave face resulting in less wear.

Involute Teeth

(a) Pressure angle is constant throughout the engagement of teeth. This results in smooth running of the

gears.

(b) It involves single curve for the teeth resuling i simplicity of manufacturing and of tools.

(c) These are simple to manufacture and thus are cheaper.

(d)Alittle variation in the centre distance does not affect the velocity ratio.

(e) Interence can occur if the condition of minimum number of teeth on a gear is not followed.

(f) The teeth have radial flanks and thus are weaker as compared to the cycloidal form for the same

pitch.

(g) Two convex surfaces are in contact and thus there is more wear.

HELICALAND SPIRAL GEAR

In helical and spiral gears, the teeth are inclined to the axis of a gear. They can be right handed or left

handed. Ψ1 = helix angle for gear 1
Let, Ψ2 = helix angle for gear 2
θ = Angle between shaft
θ = Ψ1+Ψ2 (for gears of same hand)
θ = Ψ1+Ψ2 (for gears of opposite hand)

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MECHANICAL ENGINEERING

Angle between shafts, θ = Ψ1+Ψ2 for gears of same hand
θ = Ψ1+Ψ2 θ = 0, a case of helical gears joining parallel shafts.
Teminology
• Helix angle (Ψ): It is the angle at which the teeth are inclined to the axis of a gear. It is also known as

spiral angle.

• Normal Circular Pitch (Pn) : Normal circle pitch or simply normal pitch is the shortest distance mea-
sured along the normal to the helix between corresponding points on the adjacent teeth, The normal circle

pitch of two mating gears must be same.
Pn = P cos Ψ

Also, we have, P = πm as for spur gear
P = πm

nn

and Mn = m cosΨ

• mn T1 + T2
Centre distance = 2 cosΨ cosΨ
12

• Efficiency

η= cos(θ + φ) cos(Ψ1 − Ψ2 − φ)
cos(θ + φ) cos(Ψ1 − Ψ2 − φ)

η= cos(θ + φ) + 1
max
cos(θ + φ) + 1

Here,

φ = Pressure angle

Ψ= Helix angle

θ = Angle between two shaft

WORM AND WORM GEAR
• To accomplish large speed reduction in skew shaft, spiral gears with a small driver and a larger follower
are required.
• The load transmitted through these gears is limited.
• Usually worm and worm gears are used to connect two skew shafts at right angle to each other.
• A worm can be a single double or triple start if one, two or three threads are traversed on the worm for
one tooth advncement of the gear wheel.

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Teminology

• Axial pitch (P ) is the distance between corresponding points on adjacent teeth measured along the
a

direction of the axis.

• The distance by which a helix advances along the axis of the gear for one turn around is known as Lead

(L).
• Lead angle (λ) is the angle at which the teeth are inclined to the normal to the axis of rotation. The lead

angle is the complement of the helix angle.
Ψ +λ = 90

• Velcocity Ratio (VR)

Angle turned by the gear
= Angle turned by the worm

1
= πd

• Velcocity Ratio C = m (T1 cot λ + T2 )
2

2
1 - sinφ
• Efficiency η = 1 + sinφ
max

BEVEL GEAR

• To have a gear drive between two intersecting shafts, belev gears are used.

Let, γg, γ = pitch angles of gear and pinion respectively.
p

r , r = pitch radii of gear and pinion respectively.
gp

rp = ω
g

rp ω
p

γ sin θ
p
tan = rp + cos θ
rg

γ sin θ
p
tan = ω + cos θ
p
g
ω

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MECHANICAL ENGINEERING

2 B GEAR TRAINS

A Gear Train is a combination of gears used to transmit motion from one shaft to another. It is required to
obtain large speed reduction within a small space.
TYPE OF GEAR TRAIN
(i) Simple gear train : In this each shaft support one gear.
(ii) Compound gear train : In this each shaft support two gear wheels except first and last
(iii) Reverted gear train: In this driving and the driven gear are coaxial or coincident.
(iv) Planetary and epicyclic gear train
• It is also possible that in a gear train, the axes of some of the wheels are not fixed but rotate around the
axes of other wheels with which they mesh. Such trains are known as planetary or epicyclic gear trains.
Epicyclic gear trains are useful to fransmit very high velocoty ratios with gear of smaller sizes in a lesser
space.
SIMPLE GEAR TRAIN
• A series of gears, capable of receiving and transmitting motion from one gear to another is called a
simple gear train. In it, all the axes remain fixed relative to the frame and each gear is on a separate shaft.

• A pair of mated external gear always move in opposite direction.
• All odd numbered gears move in one direction and all even numbered gears in the opposite direction.
• A simple gear train can also have bevel gears.

N2 = T1 , N3 = T2
N1 T2 N2 T3

N4 = T3 , N5 = T4
N3 T4 N4 T5

Train Value = N5 = T1
N1 T5

number of teeth on driving gear
= number of teeth on driven gear

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THEORY OF MACHINE

1
Speed Ratio = train value

N1 N5
N5 = T1
• Intermediate gears have no effect on the speed ratio and therefore, they are known as idlers.

COMPOUND GEAR TRAIN
• When a series of gears are connected in such a way that two or more gears rotate about an axis with
the same angular velocity; it is known as compound gear train.

N2 = TT12, N4 = T3 and N6 =T4T5
N1 N3
N5 T6

{N2 = N3 , N4 = N5}

Train value N6 = T1 T3 T5
N1 T2 T4 T6

Product of number of teeth on driving gears
= Product of number of teeth on driving gears

N TT T
Speed Ratio 1 = 2 4 6

N6 T1 T3 T5

REVERTED GEAR TRAIN
• If the axis of the first and last wheel of a compound gear coincide it is called reverted gear train.
• Such arrangment is used in clock and in simple lathe where ‘back gear’is used to give slow speed to
the chuck.

• Train value = N4 T1 T3
N1 T1 T4

If is the pitch circle radius of a gear,

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MECHANICAL ENGINEERING

EPICYCLE GEAR TRAIN
• When there exists a relative motion of axes in a gear train, it is called an epicyclic gear train. Thus in
an epicyclic train, the axis of at least one of the gears also move relative to the frame.
• Large speed reductions are possible with epicyclic gears and if the fixed wheel is annular a more
compact unit could be obtained.
• Important applications of epicyclic gears are in transmission, computing devices.
• In general gear trains have two degrees of freedom.
Analysis

Assume that the arm a is fixed. Turn S through ‘x’revolution in the clockwise dirction.Assuming clockwise
motion of a wheel as positive and counter-clockwise as negative,
revolutions made by a = 0 (Arm a fixed)
revoluctions made by S = x

revolutions made by P = - TS ×
TL

Now, it the mechanism is locked together and turned through a number of revolutions, the relative motions

between a, S and P will not alter. Let the locked system be turned through ‘y’revolution in the clockwise

direction. Then

revolutions made by a = y

revolution made by S = y + x

revolutions made by P y - TS ×
TL

Above procedure can be written in tabulated form

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THEORY OF MACHINE

Relative Velocity Method
• Angular velocity of s = angular velocity of S relative to ‘a’ + angular velocity of a

Ws = Wsa + Wa
Ns = Nsa + Na
∴ Np = - Npa + Na

Np Ns - Na
T N -N

s pa

Torques
• Let Ns , Na , Np and NA be the speeds and Ts , Ta , Tp and TA the external torque transmitted by s, a, P
and A

T = Ts + Ta + Tp + TA = 0
The planet ‘P’ can rotate on its own pin fixed to a, but is not connected to anything outside therefore it do not
transmit external torque. Hence

T +T +T =0
saA

• If A fixed, TA is usually known as the braking of fixing torque.
If there is no losses in power transmission
ΣTω = 0
ΣTN = 0
Ts Ns + Ta Na + TA NA = 0
If A is fixed NA = 0
and Ts Ns + Ta Na = 0

SUN AND PLANET GEAR
Special case of epicyclic gear train.

• When an annular wheel is added to the epicyclic gear train, the combination is usually, referred to as sun
and planet gear.

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MECHANICAL ENGINEERING

• The annular wheel gears with the wheel P Which can rotate freely on the arm ‘a’. The wheel S and P are

generally called the sun and the planet wheels.

• In general, S, Aand ‘a’are free to rotate independently of each other. It is also possible the either S or

Aare fixed. IfAis fixed, S will be the driving member and if S is fixed.Awill be the driving member. In each

case the driven member is the arm a.

Let,

NS = Speed of the sun wheel S
NA = speed of the annular wheel A
Na = speed of arm a
Ts = number of teeth on S
TA = number of teeth on A

Note :
* If the sun wheel S is fixed , NS = 0

Speed of the arm, Na = 1
N TS / TA+ 1
A

* If the annular wheel A is fixed , NA= 0

Na = TS / TA
NA 1 + TS / TA

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THEORY OF MACHINE

Differential Gear
* When a vehicles takes a turn, the outer wheels must travel farther than the inner wheels. Since, Both rear
wheels are drdiven by the engine through gearing. Therefore some sort of automatic device is necessary so
that the two rear wheels are given at slightly different speeds. This is accomplished by fitting a differential
gear on the rear axle.

*An epicyclic gear having two degrees of freedom has been utilized in the differential gear of an automobiel.
It permits the two wheels to rotate at the same speed when driving straight while allowing the wheels to
rotate at different speeds when taking a turn. Thus, a differential gear is a device which adds or subtracts
angular displacement.

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MECHANICAL ENGINEERING

3 DYNAMICS FORCE ANALYSIS AND FLYWHEEL

* Dynamic forces are associated with accelerating masses.
* D’Alembert’s principle states that the inertia forces and couples, and the external forces and torques

on a body together give statical equilibrium.
* Inertia is a properly of matter by virtue of which a body resists any change in velocity.

Inertia force , F1 = - m fg
m = mass of body
fg = acceleration of centre of mass of the body
* Inertia couple resists any change in the angular velocity
inerita force Ci = -Ig . α
Ig = moment of inertia about an axis passing through centre of mas G and perpendicular to plane of
rotation of the body
α = angular acceleration of the body
* Let F1, F2, F3 etc. = external forces on the body
Tg1, Tg2, Tg3 etc. = external torques on the body about centre of mass G
So, F1 + F2 + F3 + ...... + Fi = 0 or ΣF = 0
Tg1 + Tg2 + Tg3 + ...... + Ci = 0 or ΣT = 0
Thus a dynamic analysis problem is reduced to one requiring static analysis.
* The perpendicular displacement of the force from centre of mass is such that the torque so produced is
equal to the inertia couple acting on the body.

η = k2a
f

g

where k = radius of gyration

Velocity and Acceleration of Piston

If the crank OA rotates in the clockwise direction © Copyright : Ascent Gate Academy 30
1 and r = lengths of connecting rod and crank
Let, x = displacement of piston from inner dead centre

θ = angle turned by carnk from inner dead centre
n = l/r

then , x = r [ (1 - cos θ) + (n - √ n2 - sin2θ ]

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THEORY OF MACHINE

then maximum value of sin2θ can be unity, and (n2>>1). for large connecting rod.
So x = r ( 1 - cos θ )

This is the expression for a simple harmonic motion. Thus the Piston executes a simple harmonic motion

when the connecting rod is large.

dx
* Velocity of Piston , v = dt

= rω2 cos θ + cos 2 θ
n

= rω2 cos θ [ if n is very large ]

*Angular velocity andAngularAcceleration of connecting rod :

sin β = sin θ
n

ω = ω √ cos θ θ
c n2 - sin2

*Angular acceleration of connecting rod :

α = - ω2 sin θ n2 - 1
c
( n2 - sin2 θ )3/2

The negative sign indicates that the sense of angular acceleration of the rod is such that it tends to reduce the
angle β.

Piston Effort (Effective Driving Force)

* In reciprocating engines, the reciprocating masses accelerate during the first half of the stroke and the
ineertia force due to the same tends to resists the motion. Thus net force on the piston decreased.
* During the later half of the stroke, the reciprocating masses decelerate and the inertia force acts in the
direction of the inertia force acts in the direction of the applied gas pressure and thus increase the effective
froce on the piston.
* In a vertical engine, the weight of the reciprocating masses assists the piston during the out stroke
(down stroke), thus increasing the piston effort by an amount equal to the weight of the piston.
* During the stroke (upstroke) piston effeort is decreased by the same amount

Let, A1 = area of Piston (cover end)
A2 = area of piston end
p1 = pressure on cover end
P2 = pressure on piston end

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MECHANICAL ENGINEERING

m = mass of reciprocating parts

Force on piston Fp = p1A1 - p2A2
Inertia force ; Fb = mf

= mrω2 cos θ + cos2 θ
n

Net force on piston . F = Fp - Fb
* Crank Effort : Crank effort is the net effort (force) applied at the crank pin perphendicular to the

carnk which gives the required turning moment on the crank shaft.

Let, Ft = crank effort

Fc = force on connecting rod
So, Ft = Fc sin (θ + β)

= F sin (θ + β)
cos β
* Turning Moment on Crankshaft

T = Ft x r

= Fr sin θ + sin2θ θ
2√ n2 - sin2

* Inertia of connecting rod :As the motion of connecting rod is not linear.
The mass of connecting rod can be replaced by two point masses at two point , if it is ensured that the

two masses together have the same dynamical properties as before.
* The two members will be dynamically similar if :

* the sum of the two masses is equal to the total mass
* the combined centre of mass coincides with that of the rod
* the moment of inertia of two point masses about perpendicualr axis through their combined cenre of
mass is equal to thatof the rod.

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THEORY OF MACHINE

Let, m = total mass of rod

mb = mass at B
md = mass at D
mb + md = m
and mb.b = md.d

mb= m d
b+d

and d
md= m b + d

So, l = mb.b2 + md.d2

= mbd

if K is the radius of gyration, then K2 = bd

* Let the masses at a and B

ma = mass at A, distance AG = a
ma + mb = m

So, ma = m. b , mb = b .a
l l

I’ = mab

assume, a > d, I’ > l

So, Increase in Torque = ∆T = mb α (1- L),
c

(b + d) = L

This is much torque must be applied to the two mass system in the opposite direction to that of the

angular velocity to make the system dynamically equivalent to that of actual rod.

Correction torque

TC = mb α (1- L), cosθ
c √ n2 - sin2 θ

* Due to the weight of the mass at A, a torque is exerted on the cranks shaft which is Ta = (mag)r cosθ
* In case of vertical engines, a torque is also exerted on the cranks shaft due to weight of mass at B :

Ta = (mag)r sin θ + sin2θ θ
2√ n2 - sin2

* The net torque on the crank shaft will be the algebraic sum of the torque T,Tc,Ta,Tb.

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MECHANICAL ENGINEERING

FLYWHELL
Function
* A flywheel used in machines serve as a reservoir, which stores energy during the period when supply
of energy is more than the requirement , and release it during the period when the requirement of energy is
more than the supply. Flywheel does not maintain a constant speed, it simply reduce fluctuation of speed.

Turning Moment Diagram
* The turning moment diagram is the graphical representation of the turing moment or crank-effort. The
turning moment is taken as ordinate and crank angle as abscissa.

Turning Moment diagram for single cylinder four stroke engine
* In case of a four-stroke internal combustion engine, the diagram repeats itself after every two revolu-
tions instead of one revolution as for a steam engine.

* The pressure inside the engine cylinder is less than the atmospheric pressure during the suction stroke,
therefore a negative loop is formed.
* During the compression stroke, the work is done on the gases, therefore a higher negative loop is
obtained.
* During the expansion or working stroke, the fuel burns and the gases expand, therefore a large positive
loop is obtained. In this stroke the work is done by the gases.
* During exhaust stroke, the work is done on the gases, therefore a negative loop is formed.

T-M diagram for multi-cylinder engine

* The turning moment diagram for a single cylinder, engine varies considerably and a greater variation of
the same is observed in case of a four stroke, single cylinder engine. For engines with more than one cylinder
, the total crank shaft torque at any instant is given by the sum of the torques developed by each cylinder at
the instant.

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THEORY OF MACHINE

Fluctuation of Energy
* The energies of the flywheel corresponding to position of the crank are as follows :

From the two values of the energies of the flywheel corresponding to the position c, it is concluded taht
a1 - a2 + a3 - a4 + a5 - a6 = 0

* The greatest of these energies is the maximum kinetic energy of the flywheel and for the corresponding
crank position, the speed is maximum.
* The least of these energies is the least kinetic energy of the flywhell and for the corresponding crank
position, the speed is minimum.

Flucation of Speed
* The difference between the maximum and minimum speeds during a cycle is called maximum flucation
of speed.
* The ration of maximum fluctuation of speed of the mean speed is called coefficient of fluctuation of
speed (Cs).

Cs = Nmax - Nmin
Nmean

* Coefficient of steadiness : The reciprocal of the coefficient of fluctuation of speed is known as
coefficient of steadiness and is denoted by m.

m= 1 = Nmean
Cs Nmax - Nmin

Maximum fluctuation of speed

So, Maximum fluctuation of energy
∆E = Maximum energy - Minimum energy

We can also write

∆E = Emax - Emin = 1 l (ω2max - ω2min)
2
∆E ω2
= l mean Cs

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MECHANICAL ENGINEERING

Here , l = Mass moment of inertia of the flywhel about its axis of rotation

ω = Maximum angular speed during cycle
max
ω
max = Minimum angular speed during cycle

ω ω +ω min
mean = max during cycle

2

* Coefficient of Flutuation of Energy (CE)

Maximum fluctuation of energy
CE = Work done per cyle

Work done per cycle
Work done per cycle = T x θ

mean

Where , Tmean is mean torque
θ is angle turned is one cycle

= 2π , in case of two stroke engine
= 4π, in case of 4 stroke engine

DIMENSION OF FLYWHEEL RIM

Let, ω = angular velocity

r = mean radius

t = thickness of rim

ρ = density of the material of the rim
σ = circumferential stress induced in rim

Consider an element of the rim,

Centrifugal force on the element/unit length
= [ρ(r.dθ)t]ω2

Total vertical force/unit length

= ∫ π ρr2dθω2sinθ = ρr2ω2∫ πsinθdθ
0 0

= ρr2ω2 (- cosθ)π0 = 2ρr2tω2

Then for equilibrium , σ(2t)1 = 2ρr2tω2
σ = ρr2ω2 = ρv2

Then the diameter can be calculated from the relation,
v = πdN/60

Also, mass = density x volume= density x circumference x cross-sectional area
or µ = ρπdbt

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THEORY OF MACHINE

PUNCHING PRESSURE
Assumption
* As the hole is punched, the shear force decrease uniformly from maximum value to zero.
* One punching operation per revolution.

* The Crank is driven by a motor which supplies constant torque and the punch is the position of the

slider in a slider-crank mechanism. We see that the load acts only during the rotation of the crank from θ =

θ to θ = θ2, when the actual punching takes place and the load is zero for the rest of the cycle. When a
1

flywheel is used.

* The speed of the crankshaft will increase too much during the rotation of crank from θ = θ to θ = 2π
2
or θ = 0 and again from θ = 0 to θ = θ1, because there no load while input energy continuesto be supplied.

Let E, be the energy required for punching a hole. This energy is determined by the size of the hole

punched, the thickness at the material and the physical properties of the material.

Let, d1 = Diameter of the hole punched

t1 = Thickness of the plate, and

τ = Ultimate shear stress for the plate material
u

So, Maximum shear force required for punching

Fs= Area sheared x Ultimate shear stress = πd1.t1τu

So, Work done or Energy required for punching a hole

E1 = 1 x Fs x t
2

The energy supplied to the shaft per revolution should also be equal to E1,

The energy supplied bythe motor to the crankshaft during actual punching operation

E2 = E1 θ - θ
2 1
2π

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MECHANICAL ENGINEERING

So, Balance energy required for punching

E1 = E2 = E1 - E1 θ - θ
2 1
2π

= E1 1 - θ - θ
2 1
2π

This energy is to be supplied by the flywheel by the decrease in its kinetic energy when its speed falls
from maximum.Thus maximum fluctuation of energy.

∆E = E1 = E2 = E1 1 - θ - θ
2 1
2π

The value of θ and θ may be determined only if its crank radius (r), length of connecting rod (l) and the
12

relative position of the job with respect to the crankshaft axis are known.

In the absence of data, we assume

θ - θ t t
2 1 2s 4r
2π = =

where, t = Thickness of the material to be punched
s = Stroke of the punch

= 2 x Crank radius = 2r.

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THEORY OF MACHINE

4 VIBRATIONS

Any motion that exactly repeats itself after a certain interval of time is a periodic motion and is called a
vibrations.
TYPE OF VIBRATION
* Free Vibrations : Elastic vibrations in which there are no friction and external after the initial release of
the body, are known as free or natural vibrations.
* Forced Vibrataions : When a repeated force continuously acts on a system, the vibrations are said to
be forced vibration. The frequency of the vibrations is that of the applied force and is indeendent of there
own natural frequency of vibrations.
* Damped Vibrations : When the energy of a vibrating system is gradually dissipated by friction and
other resistances, the vibrations are said to be dampled vibration.
* Undamped Vibrations (Hypothetical) : When there is no friction or resistance present in system to
counter act vibration then body execute undamped vibration.
* Longitudinal Vibrations : If the shaft is elongated and shortened so that the same moves up and down
resulting in tensile and compressive stresses in the shaft , the vibrations are said to be longitudinal.

* Transverse Vibrations : When the shaft is bent altermately and tensile and compressive stresses due
to bending result, the vibrations are said to be transverse vibration.

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MECHANICAL ENGINEERING

* Torsional Vibrations : When the shaft is twisted and untwisted alternately and torsional shear stresses
are induced, the vibrations are known as torsional vibrations.

BASIC ELEMENT OF VIBRATING SYSTEM
* Inertial element : These are represented by lumped masses for rectilinear motion and by lumped
moment of inertia for angulr motion.
* Restoring Elements : Massless linear or torsional springs represent the restoring elements for rectilinear
and torsional motions respectively.
* Damping Elements : Massless dampers of rigid elements may be considred energy dissipation in a
system.
FREE LONGITUDINALVIBRATION
Different methods for finding natural frequency of a vibrating system
(i) Equilibrium method

* It is based on the principle that whenever a vibratory system is in equillibrium, the algebraic sum of

forces and moments acting on it is zero.

* This is in accordance with ‘D’Alembert’s principle that the sum of the inertial forces and the external

forces on a body in equilibrium must be zero.
Let, ∆ = static deflection

S = stiffness of tmhx.e. spring
Inertia force = (upwards)

Sp..ring force = s x (upwards)
mx + sx = 0

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This is the equation of a simple harmonic motion and is analogous to
x.. + ω2ηx = 0

* The solution of which is given by
x = A sin ωηt + B Cos ωηt

where Aand B are the constants of integration and their values depend upon the manner in which the

vibration starts.

ωη = s
m

* Linear frequency

fn = 1 s
2π m

Time period T = 1 = 2π s
fn m

(ii) Energy Method
* In a conservative system (system with no damping) the total mechanical i.e. the sum of the kinetic and
the potential energies remains constant.

d (K.E. + P.E.) = 0
dt

We know

KE = 1 mx. 2, PE = sx2
2 2

d 1 mx. 2 + sx2 =0
dt 2 2

or

d 1 mx. 2 + sx2 =0
dt 2 2

or .. + sx =0
mx

or

ωη = s

m

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(iii) Rayleigh’s Method
* In this method, the maximum kinetic energy at the mean position is made equal to the maximum poten-
tial energy (or strain energy) of the extreme position.

Let the motion be simple harmonic
Therefore, x = Xsinωηt
Where, X = maximum displacement from the mean position to the extreme position
So, x. = ωη X cosωηt ,

x. max = ωηt

or KE at mean position = PE at extreme position

i.e. , 1 m(ωη X )2 = 1 sX2
2 2

or mω2η = S

or ωη = s

m

* Inertial effect of mass of spring
If we consider mass of spring is ‘m1’ then

fn = 1 s
2π
m+ m1
3

DAMPED LONGITUDINAL VIBRATION
* When an elastic body is set in vibration motion, the vibrations die out after some time due to the internal
moledular friction of the mass of the body and the friction of the medium in which it vibrates. The diminishing
of vibrations with time is called damping.

Assumption
* The damping force is proportional to the velocity of vibration at lower values of speed and proportional
to the square of velocity at higher speeds.
Let, s = stiffness of the spring

c = damping coefficient (damping force per unit velocity)
wn = frequency of natural undamped vibrations

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x = d.isplacement of mass from mean position at time ‘t’
v = x = velocity of the mass at time ‘t’
t = x.. = acceleration of the mass at time ‘t’

As the sum of the inertial force and the external forces on a body in any direction is to be zero

mx.. + cs. + sx = 0

.x. + c x. + s x =0
m m

α = c+ 2 s
1,2 2m m
c
2m

* Damping factor

ζ= c Actual damping coefficient
Cc = Critical damping coefficient

ζ= (c/2m)2 c
s/m = 2 √ sm

Here, Cc = 2 √sm

Thus when,
ζ= 1, the damping is critical
ζ > 1, the system is over damping
ζ < 1, the system is under-damping

* Logarithmic Decrement (δ) : The ratio of two successive oscillations is constant in an underdamped
system. Natural loagarithm of this ratio is called logarthmic decrement.

δ = ln Xn = lne(ωnTd) = ζω T
Xn+1 nd

2πζ
= 1 − ζ2

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FORCED VIBRATION
* Step-input force

Constant force

.x. + s.x =0
m

F = constant

* Harmonic force

.. + cs = F0 sin ωt
mx

x = X sin (ωnt + φ) + F0
1- s

sinωt
ω2
ωn

FORCED DAMPED VIBRATIONS

If the mass is subjected to an oscillating force F = F0 sin ωt, the forces acting on the mass at any instatn will
be

* Impressed oscillating foce (downwards)

F = F0 sin ωt

* Inertial force (upward) mx..
* = cx.

Damping force (upward) =

* Spring force (restoring force) upwards = sx

Thus the equation of motion will be

m.x. + cx. + sx - F sin ωt = 0 F0 sin(ωt - φ)
0

m.x. + cx. + sx - F sin ωt
0

x = Xe-ζ ωnt . sin(ωd . t + φ1)t +

Demped Free Re sponse √(s - mω)2 + (cω)2

Steady State Response

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TRANSVERSE VIBRATION
Natural frequency of shaft and beams under different type of load and end condition, shows transverse
vibration.

(i) Single concentric loads
* In case of shaft and beams of neglibile mas carrying a consentrated mass, the force is proportional to
the deflection of the mass from the equilibrium position.

Where, ∆= mg/3 for cantilevers supporting a concentrated mass at free end.
3EI

= mga2b3 for simply supported beams
3EIl

= mga2b3 for beams fixed at both ends
3EIl3

Note : A shaft supported in long bearing is assumed to have both end fixed while one in short bearing
is considered to be simply supported.

(ii) Uniformly loaded Shaft (Simple supported)

* A simply supported shaft carrying a uniformly distributed mass has maximum deflection at the mid-

span.

∆= 5mgl4
384EI

EI 5g
So, ml4 = 384∆

π EI , 4π EI , 9π EI
f=
n 2 ml4 2 ml4 2 ml4

π 5g
fn = 2 384∆

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Note :
* This is the lowest frequency of transyerse vibrations and is called the fundamental frequency. It has a
node at each end.
* The next higher frequency is four times the fundamental frequency. It has three nodes.
* The next higher frequency is nine times the fundamental. It has four nodes.

* Thus a simply supported shall will have an infinite number of frequencies under uniformly distributed
load.
* Similarly, the cases of cantilevers and shafts fixed at both ends can be considered.

(iii) Shaft Carrying several loads

A. Dunkerley’s Method

* It is semi-empirical. This gives approximate results but is simple.

fn = frequency of transverse vibration of the whole system
fns = frequency with the distributed load acting alone
fn1,fn2,fn3 ---- = frequencies of transverse vibrations when each of w1,w2,w3, ------- alone.

Then, according do Dunkerley’s empirical formula

1 1 1 11
= 2+ 2+ 3 +... +
f 2 f f f f3
n n1 n2 n3 ns

where g = √9.61 1 = 0.4985
1 ∆ 2π = √∆ √∆

fn1 = 2π 1 1 1
Similarly ,

f= 0.4985 ; fn3 = 0.4985 , and so on
n2 √∆ √∆

2 3

fn = 0.4985 ∆

∆ + ∆ + ∆ + ...... + s
1 2 3
127

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B. Energy Method
* Which gives accurate results but involves heavy calculations if there are many loads.

Maximum P.E.

= 1 w1y1 + 1 w2y2 + 1 w3y3 + ...
2 2 2

g
= 2 Σmy

Maximum K.E.

= 1 m1v12 + 1 m2v 2 + 1 m3v23 + ...
2 2 2 2

ω2
= 2 Σmy2
ω = circular frequency of vibration

K.E. = P.E.

ω 1 gΣmy
fn = 2π = 2π Σmy2

WHIRLING OF SHAFT
* When a rotor is mounted on a shaft, its centre of mass does not usually coincide with the centre line of
the shaft. Therefore, when the shaft rotates it is subjected to a centrifugal force which makes the shaft bend
in the direction of eccentricity ofthe centre of mass. Thus further increases the eccentricity, and hence cen-
trifugal force. The bending of shaft depends upon the eccentricity of the centre of mass of the rotor as well
as upon the speed at which the shaft rotates.
* Critical or whirling or whipping speed is the speed at which the shaft tends to vibrate violently in the
transverse direction.

TORSIONAL VIBRATION

(i) Free torsional vibratoin (single rotor)
* Consider a uniform shaft of length l rigidly fixed at its upper end and carrying a disc of moment of
inertial I at its lower end. The shaft is assumed to be massless. If the disc is given a twist about its vertical axis
and then released, it will start oscillating about the axis and will perform torsional vibrations.

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.

Let,
θ =Angular displacement of the disc from its equilibrium positon at any instant

q = GJ = Torsional stiffness of t he shaft
l

Here,
G = Modulus of rididly of the shaft material
J = Polar moment of inertia of the shaft cross-section
..
lθ + qθ = 0

ω = q
n l

1q
fn = 2π l

* Inertia effect of mass of shaft

1 q
fn = 2π
1+ l1
3
Here,

l1 = Moment of inertia of shaft

(ii) Free torsional vibration (two rotor)
* If a shaft held in bearings carries a rotor at each end, it can vibrate torsionally such that the two rotors
move in the opposite directions. Thus some length of the shaft is twisted in one direction while the rest is
twisted in the other . The section which does not undergo any twist is called the nodal section.

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Let,

L and L = Lengths of two protions of the shaft
ab

la and lb = Moment of inertia of rotors A and B respectively
qa and qb = Torsional stiffness of lengths la and lb of the shaft respectively
f and f = Natural frequencies of torsional vibrations of rotors Aand B respectively

na nb

Then,

f na = fnb

1 qa 1 qb
So, 2π la = 2π lb

or qa = qb
la lb

GJ GJ
or lala = lblb

or la = la or lala = lblb
lb lb

Note :
* Node divides the length of the shaft in the inverse ratio of moment of inertia of two rotors.
* In general, possible number of node points and frequencies is one less than the number of rotors in
torsional vibrating system.

(iii) Free torsional vibration (Three rotor system)
Consider a three-rotor system in which the two rotors A and B are fixed to the ends of the shaft, and

the rotor C is in between.
* Single Node System : Single node case observed when whether ‘A’and ‘B’rotes is same direction
and ‘C’in opposite direction or ‘B’and ‘C’in same direction and ‘A’in opposite direction.

---------------l-a ----------------
----------
Aa Ab

Ac

---------------l-c--------------------------

or

Ab Ac
A

a

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* Double Node System : Double node case occur when ‘A’and ‘C’are in same direction and ‘B’is in
opposite direction.

la l1 - la l2 - la lc
-----------------------
Aa Ac ---------------------------------

l1 Ab ----------------------- l2

fna = 1 kθA A 111111111111222222222222 B 111111111111222222222222
2π IA

C

fnb = 1 Glp 1 1
2π +
I l1 - la l2 - lc
b

fnc = 1 kθc
2π Ic

If, fna = fnb = fnc
la la = lc lc

1 =1 11
+
l -l I l - l l - l
aa b
1 a 2 c

TORSIONALLY EQUIVALENT SHAFT
* A torsionally equivalent shaft is one which has the same torsional stiffness as that of the stepped shaft so
that it twists to the same extent under a given torque as the stepped shaft would.

d 4 d 4 d 4 d 4
d1 d2 d3 d4
l = l1 + l2 + l3 + l4

********************

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