Theory of machines notes

2,602 views 102 slides Apr 30, 2021
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About This Presentation

Theory of machines notes for gate, ies and academic purposes


Slide Content

Soumyth

THEORY OF MACHINES

1

CONTENTS
INTRODUCTION ............................................................................................................................................................................. 2
MOTION ANALYSIS ..................................................................................................................................................................... 16
VELOCITY ANALYSIS OF DOUB LE SLIDER CRANK MECHANISM .................................................................................. 22
ACCELERATION ANALYSIS ....................................................................................................................................................... 23
STATIC FORCE ANLAYSIS .......................................................................................................................................................... 29
DYNAMIC FORCE ANALYSIS .................................................................................................................................................... 31
BALANCING OF ROTATING MASS .......................................................................................................................................... 37
BALANCING OF RECIPROCATING MASS ............................................................................................................................... 40
TURNING MOMENT DIAGRAMS .............................................................................................................................................. 48
FLYWHEEL .................................................................................................................................................................................... 50
CAMS .............................................................................................................................................................................................. 51
GEARS ............................................................................................................................................................................................. 57
GEAR TRAINS ............................................................................................................................................................................... 71
GOVERNORS ................................................................................................................................................................................. 75
GYROSCOPE .................................................................................................................................................................................. 82
VIBRATIONS ................................................................................................................................................................................. 86

2

INTRODUCTION
Mechanisms and Machines
If several bodies are assembled in such a way that the motion of one cause constrained and predictable motion
to the other, is called mechanism.
A machine is a mechanism or a combination of mechanisms which, apart from imparting definite motions to the
parts, also transmits and modifies the available mechanical energy into some kind of desired work.

Kinematics deals with the relative motions of different parts of a mechanism without taking into consideration
the forces producing the motions.
Dynamics involves the calculations of forces impressed upon different parts of a mechanism.
Completely constrained motion
When the motion between two elements of a pair is in a definite direction
irrespective of the direction of the force applied, it is known as completely
constrained motion. The constrained motion may be linear or rotary.
Incompletely constrained motion
When the motion between two elements of a pair is possible in more than one
direction of the force applied, it is known as incompletely constrained motion.
Successfully constrained motion
When the motion between two elements of a pair is possible in more than one direction but is
made to have motion in one direction by using some external means, it is successfully
constrained motion.
Rigid and Resistant bodies
A body is said to be rigid if under the action of forces, it does not deform or the distance
between the two points on it remains same.
Resistant bodies are those which are rigid for the purposes they have to serve.
Link
A resistant body or a group of resistant bodies with rigid connections preventing their relative motion is known
as link. A link can also be defined as a member or a combination of members of a mechanism, connecting other
members and having motion relative to them.
Links can be classified into binary, ternary and
quaternary.
Mechanis
m
Machine

3

Kinematic pair
A kinematic pair is a joint of two links having relative motion between them.
Kinematic pairs according to Nature of contact: Lower pair and Higher pair.
Kinematic pairs according to Nature of Mechanical Constraint: Closed pair and Unclosed pair.

Kinematic pairs according to nature of relative motion:
a) Sliding pair, b) turning pair, c) rolling pair, d) screw pair, e) spherical pair.
Types of Joints
Binary joint
If two links are joined at same connection, it is called a binary joint. At B.
Ternary joint
If three links are joined at a connection, it is known as a ternary joint. At T.
Quaternary Joint
If four links are joined at a connection, it is known as quaternary joint. At Q.

4

Kinematic chain
When all the links are connected in such a way that 1
st
link is connected to the last link in order to get the closed
chain and if all the relative motion in these closed chains are constrained then such a chain is known as
kinematic chain.

Degrees of freedom
An unconstrained rigid body moving in space can have translational motion along any three mutually
perpendicular axes and rotational motions about these axes. A rigid body possesses six degrees of freedom.
Degrees of freedom of a pair can be defined as the number of independent relative
motions, both translational and rotational, a pair can have.
Degrees of freedom = 6 – Number of restraints (no. of motion which are not
possible)
Lower pair ⟶ 1 DOF
Higher pair ⟶ 3 DOF
Spherical pair → Degree of freedom – 3
Pair Restrain Degree of Freedom

3T+2R =5 6-5=1

1T=1 6-5=1
Aim: - To find out Degree of Freedom for 2D plane mechanism
�=[3·(�−1)−2�−ℎ]⟶ �������� ��������
L → no. of Links, j → no. of binary joint, h → no. of higher pair
3(�−1)→��.�� ������� ������ �� 2� ������ ���ℎ�����
Note: -
�=[3·(�−1)−2�−ℎ]−��

5

Fr → no. of those motions, which are not the part of mechanism (Dummy motion)
Fr→ s (NOT A PART OF MECHANISM)
Example

6


If F = 0, no relative motion [Frame/structure]

If F < 0, No relative motion [Super
structure/indeterminate structure]

If F = 1, kinematic chain

If F > 1, Unconstrained chain
Degree of freedom is no. of input required to get the constrained output/input in any chain.
An alternate way
1. &#3627408444;&#3627408467; (&#3627408471;+

2
)=(
3&#3627408473;
2
−2) →&#3627408446;&#3627408470;&#3627408475;&#3627408466;&#3627408474;&#3627408462;&#3627408481;&#3627408470;&#3627408464; &#3627408464;ℎ&#3627408462;&#3627408470;&#3627408475;
2. &#3627408444;&#3627408467; (&#3627408471;+

2
)>(
3&#3627408473;
2
−2) →&#3627408441;&#3627408479;&#3627408462;&#3627408474;&#3627408466; ⧸&#3627408480;&#3627408481;&#3627408479;&#3627408482;&#3627408464;&#3627408481;&#3627408482;&#3627408479;&#3627408466; ⧸&#3627408480;&#3627408482;&#3627408477;&#3627408466;&#3627408479;&#3627408480;&#3627408481;&#3627408479;&#3627408482;&#3627408464;&#3627408481;&#3627408482;&#3627408479;&#3627408466;
&#3627408444;&#3627408467; (&#3627408447;&#3627408443;&#3627408454; – &#3627408453;&#3627408443;&#3627408454;)= 0.5→&#3627408441;&#3627408479;&#3627408462;&#3627408474;&#3627408466; &#3627408480;&#3627408481;&#3627408479;&#3627408482;&#3627408464;&#3627408481;&#3627408482;&#3627408479;&#3627408466;
(&#3627408447;&#3627408443;&#3627408454;−&#3627408453;&#3627408443;&#3627408454;)>0.5→&#3627408454;&#3627408482;&#3627408477;&#3627408466;&#3627408479; &#3627408480;&#3627408481;&#3627408479;&#3627408482;&#3627408464;&#3627408481;&#3627408482;&#3627408479;&#3627408466;
3. &#3627408444;&#3627408467; (&#3627408471;+

2
)<(
3&#3627408473;
2
−2) →&#3627408456;&#3627408475;&#3627408464;&#3627408476;&#3627408475;&#3627408480;&#3627408481;&#3627408479;&#3627408462;&#3627408470;&#3627408475;&#3627408466;&#3627408465; &#3627408464;ℎ&#3627408462;&#3627408470;&#3627408475;

7


Spring (links of variable length) as a
link

DOF of an open chain

(One binary joint will restrict 2 motions in 2D)
Grubler’s equation
For those mechanism in which F = 1 & h = 0
Applied Kutzback equation
&#3627408441;=3·(&#3627408473;−1)−2&#3627408471;−ℎ
1=3&#3627408473;−3−2&#3627408471;
3&#3627408473;−2&#3627408471;−4=0→ &#3627408442;&#3627408479;&#3627408482;&#3627408463;&#3627408473;&#3627408466;&#3627408479;

&#3627408480; &#3627408466;&#3627408478;&#3627408482;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475;
l should be even for satisfying Grubler’s equation and lmin = 4 (for lower pairs)
lmin = 4 → First mechanism in Lower pair → Simple mechanism (can’t have a chain with 2 links)
a) Four bar mechanism
b) Single slider crank mechanism
c) Double slider crank mechanism

8

Four bar mechanism (Quadric cycle mechanism)
4 links + 4 turning pair
Best position → Fixed (because it governs both input & output)
Worst position → coupler (because it is just a transmitting body
Input/output
(Having only one option of motion i.e., rotation)
⟶ Complete rotation (360°) → crank
⟶ Partial rotation (<360°) (Oscillation) → Rocker/lever
Inversions
Mechanisms which are obtained by fixing one by one different link.
• Double crank mechanism
• Crank – rocker mechanism
• Double – rocker mechanism
Grashof’s Law
For the continuous relative motion between the number of links in four bar mechanism the summation of longer
of shortest and greatest should not be greater than summation of length of other two links.
For continuous relative motion
(&#3627408454;+&#3627408447;)≤(&#3627408477;+&#3627408478;)
Best position → fixed (because it governs both input and output)
Best link for complete rotation → shortest (s)
Is S+L < p + q (Law satisfied)
1. S → fixed → Double crank
2. S → Adjacent to fix → Crank-Rocker
3. S → couple → Double Rocker
If (S+L) = (p + q) law is satisfied.
a) Not having equal pair or equal link
5, 4, 3, 2
Same as previous
b) Having equal pair or equal link
2, 2, 5, 5
&#3627408480; &#3627408480; &#3627408473; &#3627408473;
i) Parallelogram linkage (same length of
the links)
S – fixed → Double crank
l – fixed → double crank
ii) Detroit linkage
S – fixed → Double crank
l – fixed → double crank
If (S+L) > (p + q) law is not satisfied
Any link fixed → double rocker
If no. of links = l
No. of inversions ≤ l (less when for different fixing relative motion is same)

9

Some practical examples of 4 bar mechanisms
Beam engine mechanism (James Watt)

Coupling Rod of locomotives

Transmission angle (μ)
An angle between the coupler link and the output link in four bar mechanism is known as transmission angle.
&#3627408436;&#3627408438;
2
=&#3627408462;
2
+&#3627408463;
2
−2&#3627408462;&#3627408463;&#3627408464;&#3627408476;&#3627408480;&#3627409155;=&#3627408464;
2
+&#3627408465;
2
−2&#3627408464;&#3627408465;&#3627408464;&#3627408476;&#3627408480;&#3627409159;
Differentiating both sides,
(−2&#3627408462;&#3627408463;)·(−&#3627408480;&#3627408470;&#3627408475;&#3627409155;)·&#3627408465;&#3627409155;=(−2&#3627408464;&#3627408465;)·(−&#3627408480;&#3627408470;&#3627408475;&#3627409159;)·&#3627408465;&#3627409159;
&#3627408465;&#3627409159;
&#3627408465;&#3627409155;
=(
&#3627408462;&#3627408463;
&#3627408464;&#3627408465;

&#3627408480;&#3627408470;&#3627408475;&#3627409155;
&#3627408480;&#3627408470;&#3627408475;&#3627409159;

For μ to be max/min,

&#3627408465;&#3627409159;
&#3627408465;&#3627409155;
=0→(
&#3627408462;&#3627408463;
&#3627408464;&#3627408465;

&#3627408480;&#3627408470;&#3627408475;&#3627409155;
&#3627408480;&#3627408470;&#3627408475;&#3627409159;
=0 →&#3627408480;&#3627408470;&#3627408475;&#3627409155;=0
??????=&#3627409358;°,&#3627409359;??????&#3627409358;°→ &#3627409159;
&#3627408474;??????&#3627408475;=0°,&#3627409159;
&#3627408474;&#3627408462;&#3627408485;=180
James Watt Beam engine couldn’t be used as steam engine, so he
converted one of the turning pair into 4 bar to a sliding pair. (Sliding
pair Mechanism)
Rotation ⟷ Oscillation
Crank ⟷ Rocker
Ex: - Sewing Machine
4 turning pairs
4 links
Ex: - Steam engine

10

Single Slider crank mechanism (Drag-link Mechanism)
I
st
inversion (Cylinder fixed)

Rotation ⟷ Oscillation
Crank ⟷ Piston
Output ⟵ Input (Piston to Crank) ⟶ Reciprocating engine
Input ⟶ Output (Crank to Piston) ⟶ Reciprocating compressor
II
nd
Inversion (Crank fixed)
→ Whitworth quick return motion mechanism
→ Rotary IC engine mechanism (Gnome engine)
III
rd
Inversion (Connecting Rod fixed)
→ Crank and slotted lever quick return mechanism
→ Oscillating cylinder engine mechanism
IV
th
Inversion (Slider/Piston Fixed)
→ Hand Pump (Pendulum pump, Bull engine)

11

Crank and slotted Lever (Quick Return Motion Mechanism)
III
rd
Inversion (Connecting Rod Fixed)

Quick Return Ratio (QRR)
&#3627408452;&#3627408453;&#3627408453;=
&#3627408455;&#3627408470;&#3627408474;&#3627408466;
&#3627408464;&#3627408482;&#3627408481;&#3627408481;??????&#3627408475;??????
&#3627408455;&#3627408470;&#3627408474;&#3627408466;
&#3627408453;&#3627408466;&#3627408481;&#3627408482;&#3627408479;&#3627408475;
=
&#3627409149;
&#3627409148;
>1 (&#3627408514;&#3627408525;&#3627408536;&#3627408514;&#3627408538;&#3627408532;)
Stroke R1R2
&#3627408453;
1&#3627408453;
2=&#3627408438;
1&#3627408438;
2=2×&#3627408438;
1&#3627408448;=2×&#3627408436;&#3627408438;
1×&#3627408464;&#3627408476;&#3627408480;
&#3627409148;
2
=2×&#3627408436;&#3627408438;
1×(
&#3627408450;&#3627408437;
1
&#3627408450;&#3627408436;
)=
2×&#3627408436;&#3627408438;×&#3627408450;&#3627408437;
&#3627408450;&#3627408436;

&#3627408454;&#3627408481;&#3627408479;&#3627408476;&#3627408472;&#3627408466;=
2×&#3627408436;&#3627408438;×&#3627408450;&#3627408437;
&#3627408450;&#3627408436;
=
2×[&#3627408473;&#3627408466;&#3627408475;&#3627408468;&#3627408481;ℎ &#3627408476;&#3627408467; &#3627408480;&#3627408473;&#3627408476;&#3627408481;&#3627408481;&#3627408466;&#3627408465; &#3627408463;&#3627408462;&#3627408479;]×[&#3627408473;&#3627408466;&#3627408475;&#3627408468;&#3627408481;ℎ &#3627408476;&#3627408467; &#3627408464;&#3627408479;&#3627408462;&#3627408475;&#3627408472;]
[&#3627408473;&#3627408466;&#3627408475;&#3627408468;&#3627408481;ℎ &#3627408476;&#3627408467; &#3627408464;&#3627408476;&#3627408475;&#3627408475;&#3627408466;&#3627408464;&#3627408481;&#3627408470;&#3627408475;&#3627408468; &#3627408479;&#3627408476;&#3627408465;]

As α < β, return stroke is quicker than Cutting stroke, so it is called Quick return mechanism.
QRR can never be less than 1.

12

Whitworth Quick Return Motion Mechanism
II
nd
Inversion (crank fixed)
Rotation ⟶ Rotation (Double crank)
Stroke ⟹ R1R2 ⟹ C1C2 ⟹2 (OC)

Oscillating Cylinder Mechanism
III
rd
Mechanism (Connecting Rod fixed)


Rotary IC Engine Mechanism (GNOME Engine)
II
nd
Inversion

13


Hand Pump
IV
th
Inversion (slider fixed)


When combustion takes place inside the
cylinder
Input force comes on piston
This force is transmitted to Connecting Rod
Connecting Rod and piston both rotate
Cylinder block rotates
(Propeller is mounted on Cylinder block)

14

Double slider crank chain
(4 links + 2 Turning Pairs+ 2 Sliding Pairs)

1. Slotted plate is fixed (Elliptical Trammels)

&#3627408464;&#3627408476;&#3627408480;&#3627409155;=
&#3627408485;
&#3627408436;&#3627408451;
&#3627408480;&#3627408470;&#3627408475;&#3627409155;=
&#3627408486;
&#3627408437;&#3627408451;

&#3627408485;
2
&#3627408436;&#3627408451;
2
+
&#3627408486;
2
&#3627408437;&#3627408451;
2
=1→&#3627408440;&#3627408473;&#3627408473;&#3627408470;&#3627408477;&#3627408480;&#3627408466;
Locus of any point ‘P’ on link AB except midpoint is on ellipse.

15

2. If any of the slider is fixed (Switch yoke mechanism)
Rotary to Reciprocatory
Practical use ⟶ Power hex

3. If link connecting slider is fixed (old ham coupling)
Oldham coupling is used to connect shaft having lateral misalignment.

Maximum sliding velocity of this intermediate plate links = rw = (distance between the shaft) × (wdriver)
Mechanical Advantage (M.A)
&#3627408448;&#3627408466;&#3627408464;ℎ&#3627408462;&#3627408475;&#3627408470;&#3627408464;&#3627408462;&#3627408473; &#3627408436;&#3627408465;&#3627408483;&#3627408462;&#3627408475;&#3627408481;&#3627408462;&#3627408468;&#3627408466;=
&#3627408441;
&#3627408476;&#3627408482;&#3627408481;&#3627408477;&#3627408482;&#3627408481;
&#3627408441;
??????&#3627408475;&#3627408477;&#3627408482;&#3627408481;
=
&#3627408455;
&#3627408476;&#3627408482;&#3627408481;&#3627408477;&#3627408482;&#3627408481;
&#3627408455;
??????&#3627408475;&#3627408477;&#3627408482;&#3627408481;

&#3627409154;
&#3627408474;&#3627408466;&#3627408464;ℎ&#3627408462;&#3627408475;??????&#3627408480;&#3627408474;=
&#3627408451;
&#3627408476;&#3627408482;&#3627408481;&#3627408477;&#3627408482;&#3627408481;
&#3627408451;
??????&#3627408475;&#3627408477;&#3627408482;&#3627408481;
=
&#3627408441;
&#3627408476;&#3627408482;&#3627408481;&#3627408477;&#3627408482;&#3627408481;×&#3627408457;
&#3627408476;&#3627408482;&#3627408481;&#3627408477;&#3627408482;&#3627408481;
&#3627408441;
??????&#3627408475;&#3627408477;&#3627408482;&#3627408481;×&#3627408457;
??????&#3627408475;&#3627408477;&#3627408482;&#3627408481;
=
&#3627408455;
&#3627408476;&#3627408482;&#3627408481;&#3627408477;&#3627408482;&#3627408481;×&#3627408484;
&#3627408476;&#3627408482;&#3627408481;&#3627408477;&#3627408482;&#3627408481;
&#3627408455;
??????&#3627408475;&#3627408477;&#3627408482;&#3627408481;×&#3627408484;
??????&#3627408475;&#3627408477;&#3627408482;&#3627408481;

&#3627408448;.&#3627408436;=
&#3627408457;
??????&#3627408475;&#3627408477;&#3627408482;&#3627408481;
&#3627408457;
&#3627408476;&#3627408482;&#3627408481;&#3627408477;&#3627408482;&#3627408481;
×&#3627409154;
&#3627408474;&#3627408466;&#3627408464;ℎ&#3627408462;&#3627408475;??????&#3627408480;&#3627408474;=
&#3627408484;
??????&#3627408475;&#3627408477;&#3627408482;&#3627408481;
&#3627408484;
&#3627408476;&#3627408482;&#3627408481;&#3627408477;&#3627408482;&#3627408481;
×&#3627409154;
&#3627408474;&#3627408466;&#3627408464;ℎ&#3627408462;&#3627408475;??????&#3627408480;&#3627408474;

16

MOTION ANALYSIS
Motion of a link
Let a rigid link OA, of length r, rotate about a fixed-point O with uniform angular velocity ω rad/s in the
counter-clockwise direction. OA turns through a small angle δθ in a small interval of time δt. Then A will travel
along the arc AA′ as shown.

&#3627408457;&#3627408466;&#3627408473;&#3627408476;&#3627408464;&#3627408470;&#3627408481;&#3627408486; &#3627408476;&#3627408467; &#3627408436; &#3627408479;&#3627408466;&#3627408473;&#3627408462;&#3627408481;&#3627408470;&#3627408483;&#3627408466; &#3627408481;&#3627408476; &#3627408450;=
&#3627408436;&#3627408479;&#3627408464; &#3627408436;&#3627408436;

&#3627409151;&#3627408481;
⟹ &#3627408483;
&#3627408462;&#3627408476;=
&#3627408479;·&#3627409151;&#3627409155;
&#3627409151;&#3627408481;

&#3627408484;ℎ&#3627408466;&#3627408475; &#3627409151;&#3627408481;→0,&#3627408483;
&#3627408462;&#3627408476;=&#3627408479;
&#3627408465;&#3627409155;
&#3627408465;&#3627408481;
=&#3627408479;&#3627409172;
The velocity of A is ωr and is perpendicular to OA. It’s represented by a vector oa.
Consider a point B on the link OA.
&#3627408457;&#3627408466;&#3627408473;&#3627408476;&#3627408464;&#3627408470;&#3627408481;&#3627408486; &#3627408476;&#3627408467; &#3627408437; = &#3627409172;×&#3627408450;&#3627408437; ⊥&#3627408481;&#3627408476; &#3627408450;&#3627408437;
&#3627408424;&#3627408411;
&#3627408424;&#3627408410;
=
&#3627409172;·&#3627408450;&#3627408437;
&#3627409172;·&#3627408450;&#3627408436;
=
&#3627408450;&#3627408437;
&#3627408450;&#3627408436;

Magnitude of instantaneous linear velocity of a point on a rotating body is proportional to its distance from the
axis of rotation.
Four link Mechanism

AB is the driver rotating at an angular speed of ω rad/s in the clockwise direction if it is a crank or moving at
this angular velocity at this instant if it’s a rocker.
It is required to find the absolute velocity of C.
&#3627408483;
&#3627408464;&#3627408462;=&#3627408483;
&#3627408464;&#3627408463;+&#3627408483;
&#3627408463;&#3627408462;
Velocity of any point on the fixed link AD is always zero.
Therefore, the velocity of C relative to A is the same as velocity of C relative to D.
&#3627408483;
&#3627408464;&#3627408465;=&#3627408483;
&#3627408463;&#3627408462;+&#3627408483;
&#3627408464;&#3627408463; (&#3627408476;&#3627408479;) &#3627408413;&#3627408412;=&#3627408410;&#3627408411;+&#3627408411;&#3627408412;
&#3627408535;
&#3627408515;&#3627408514; &#3627408470;&#3627408480; &#3627408472;&#3627408475;&#3627408476;&#3627408484;&#3627408475; &#3627408484;ℎ&#3627408470;&#3627408464;ℎ &#3627408470;&#3627408480; ??????·&#3627408488;&#3627408489;,&#3627408462;&#3627408475;&#3627408465; &#3627408535;
&#3627408516;&#3627408515; & &#3627408535;
&#3627408517;&#3627408516; &#3627408462;&#3627408479;&#3627408466; &#3627408482;&#3627408475;&#3627408472;&#3627408475;&#3627408476;&#3627408484;&#3627408475;.

17

Velocity diagram is constructed as follows,
1. Take the first vector ab, as it is completely known.
2. To add vector bc to ab, raw a line ⊥ BC through b, of any length. Since the direction-sense of bc is
unknown, it can lie in either side of b. A convenient length of the line can be taken on both sides of b.
3. Through d, draw a line ⊥ DC to locate the vector dc. The intersection of this line with the line of vector
bc locates the point c.
4. Mark arrowheads on the vectors bc and dc to give the proper sense. Then dc is the magnitude and
represents the direction of the velocity of C relative to A (or D). it is also the absolute velocity of the
point C (A & D being fixed points).
5. Remember that the arrowheads on vector bc can be put in any direction because both ends of the link
BC are movable. If the arrowhead is put from c to b, then the vector is read as cb. The above equation is
modified as
→&#3627408413;&#3627408412;=&#3627408410;&#3627408411;−&#3627408412;&#3627408411; ⟹ &#3627408413;&#3627408412;+&#3627408412;&#3627408411;=&#3627408410;&#3627408411;
The velocity of an intermediate point on any of the links can be found easily by dividing the corresponding
velocity vector in the same ratio as the point divides the link. For point E in the link BC,
&#3627408411;&#3627408414;
&#3627408411;&#3627408412;
=
&#3627408437;&#3627408440;
&#3627408437;&#3627408438;

ae represents the absolute velocity of E.
Angular velocity of Links
&#3627408483;
&#3627408464;&#3627408463;=&#3627408411;&#3627408412; &#3627408483;
&#3627408463;&#3627408464;=&#3627408412;&#3627408411;
&#3627408483;
&#3627408464;&#3627408463;=&#3627409172;
&#3627408464;&#3627408463;×&#3627408437;&#3627408438;=&#3627409172;
&#3627408464;&#3627408463;×&#3627408438;&#3627408437;→ &#3627409172;
&#3627408464;&#3627408463;=
&#3627408483;
&#3627408464;&#3627408463;
&#3627408438;&#3627408437;
&#3627409172;
&#3627408463;&#3627408464;=
&#3627408483;
&#3627408463;&#3627408464;
&#3627408438;&#3627408437;
;
The magnitude of ωcb = ωbc as vcb = vbc and the direction of rotation is the same.
&#3627408483;
&#3627408464;&#3627408465;=&#3627408413;&#3627408412;
&#3627409172;
&#3627408464;&#3627408465;=
&#3627408483;
&#3627408464;&#3627408465;
&#3627408438;&#3627408439;


Slider-crank Mechanism
Figure shows OA is the crank moving with uniform angular velocity ω rad/s in the clockwise direction. At point
B, a slider moves on the fixed guide G. AB is the coupler joining
A and B. it is required to find the velocity of the slider at B.
&#3627408483;
&#3627408463;&#3627408476;=&#3627408483;
&#3627408463;&#3627408462;+&#3627408483;
&#3627408462;&#3627408476; (&#3627408476;&#3627408479;)&#3627408483;
&#3627408463;??????=&#3627408483;
&#3627408462;&#3627408476;+&#3627408483;
&#3627408463;&#3627408462;
??????&#3627408411;=&#3627408424;&#3627408410;+&#3627408410;&#3627408411;
Take the vector vao which is completely known.
Vba is ⊥AB, draw a line ⊥AB through a;
Through g (or a), draw a line parallel to the motion of B.
The intersection of the two lines locates the point b.
gb (or ob) indicates the velocity of the slider B relative to the
guide G. this is also the absolute velocity of the slider (G is fixed).
The slider moves towards the right as indicated by gb. When the crank assumes the position OA′ while rotating,
it will be found that the vector gb lies on the left of g indicating that B moves towards left.
For the given configuration, the coupler AB has angular velocity in the counter-clockwise direction, the
magnitude being
&#3627408483;
&#3627408463;&#3627408462;
&#3627408437;&#3627408436;(&#3627408476;&#3627408479; &#3627408436;&#3627408437;)

18

Crank and Slotted lever Mechanism
A crank and slotted lever mechanism is a form of quick return mechanism used for slotting and shaping
machines.

OP is the crank rotating at an angular velocity of ω rad/s in the clockwise direction about the center O. at the
end of the crank, a slider P is pivoted which moves on an oscillating link AR.
In such problems, it is convenient if a point Q on the link AR immediately below P is assumed to exist (P & Q are
known as coincident points). As the crank rotates, there is relative movement of the points P and Q along AR.
&#3627408483;
&#3627408478;&#3627408476;=&#3627408483;
&#3627408478;&#3627408477;+&#3627408483;
&#3627408477;&#3627408476; (&#3627408476;&#3627408479;)&#3627408483;
&#3627408478;&#3627408462;=&#3627408483;
&#3627408477;&#3627408476;+&#3627408483;
&#3627408478;&#3627408477;
&#3627408410;&#3627408426;=&#3627408424;&#3627408425;+&#3627408425;&#3627408426;
Take the vector vpo which is completely known.
Vqa is ⊥AR, draw a line ⊥AR through a;
Vpq is ∥ AR, draw a line ∥AR through p.
The intersection locates the point q. observe that the velocity diagrams obtained in the two cases are the same
expect that the direction of vpq is the reverse of that of vqp
As the vectors oq and qp are perpendicular to each other, the vector vpo may be assumed to have two
components, one perpendicular to AR and the other parallel to AR.
The component of velocity along AR, ie., qp indicates the relative velocity between Q & P or the velocity of
sliding of the block on link AR.
Now, the velocity of R is perpendicular to AR. As the velocity of Q perpendicular to AR is known, the point r will
lie on the vector aq produced such that ar/aq = AR/AQ
To find the velocity of ram S, write the velocity vector equation,
&#3627408483;
&#3627408480;&#3627408476;=&#3627408483;
&#3627408480;&#3627408479;+&#3627408483;
&#3627408479;&#3627408476; (&#3627408476;&#3627408479;)&#3627408483;
&#3627408480;??????=&#3627408483;
&#3627408479;&#3627408476;+&#3627408483;
&#3627408480;&#3627408479;
??????&#3627408428;=&#3627408424;&#3627408427;+&#3627408427;&#3627408428;

19

vro is already there in the diagram. Draw a line through r perpendicular to RS for the vector vsr and a line
through g, parallel to the line of motion of the slider S on the guide G, for the vector vsg. In this way the point s is
located.
The velocity of the ram S = os (or gs) towards right for the given position of the crank.
&#3627408436;&#3627408473;&#3627408480;&#3627408476;,&#3627409172;
&#3627408479;&#3627408480;=
&#3627408483;
&#3627408479;&#3627408480;
&#3627408453;&#3627408454;
&#3627408438;&#3627408473;&#3627408476;&#3627408464;&#3627408472;&#3627408484;&#3627408470;&#3627408480;&#3627408466;
Usually, the coupler RS is long and its obliquity is neglected. Then or ≈ os.
&#3627408455;&#3627408470;&#3627408474;&#3627408466; &#3627408476;&#3627408467; &#3627408464;&#3627408482;&#3627408481;&#3627408481;&#3627408470;&#3627408475;&#3627408468;
&#3627408455;&#3627408444;&#3627408474;&#3627408466; &#3627408476;&#3627408467; &#3627408479;&#3627408466;&#3627408481;&#3627408482;&#3627408479;&#3627408475;
=
&#3627409155;
&#3627409149;

When the crank assumes the position OP’ during the cutting stroke, the component of velocity along AR (i.e, pq)
is zero and oq is maximum (=op)
&#3627408479;→&#3627408473;&#3627408466;&#3627408475;&#3627408468;&#3627408481;ℎ &#3627408476;&#3627408467; &#3627408464;&#3627408479;&#3627408462;&#3627408475;&#3627408472;(&#3627408450;&#3627408451;), &#3627408473;→&#3627408473;&#3627408466;&#3627408475;&#3627408468;&#3627408481;ℎ &#3627408476;&#3627408467; &#3627408480;&#3627408473;&#3627408476;&#3627408481;&#3627408481;&#3627408466;&#3627408465; &#3627408473;&#3627408466;&#3627408483;&#3627408466;&#3627408479;(&#3627408436;&#3627408453;), &#3627408464;→&#3627408465;&#3627408470;&#3627408480;&#3627408481;&#3627408462;&#3627408475;&#3627408464;&#3627408466; &#3627408463;&#3627408466;&#3627408481;&#3627408484;&#3627408466;&#3627408466;&#3627408475; &#3627408467;&#3627408470;&#3627408485;&#3627408466;&#3627408465; &#3627408464;&#3627408466;&#3627408475;&#3627408481;&#3627408479;&#3627408466;&#3627408480;(&#3627408436;&#3627408450;)
&#3627408439;&#3627408482;&#3627408479;&#3627408470;&#3627408475;&#3627408468; &#3627408481;ℎ&#3627408466; &#3627408464;&#3627408482;&#3627408481;&#3627408481;&#3627408470;&#3627408475;&#3627408468; &#3627408480;&#3627408481;&#3627408479;&#3627408476;&#3627408472;&#3627408466;, &#3627408483;
&#3627408480; &#3627408474;&#3627408462;&#3627408485;=&#3627409172;×&#3627408450;&#3627408451;

×
&#3627408436;&#3627408453;
&#3627408436;&#3627408452;
=&#3627409172;&#3627408479;×
&#3627408473;
&#3627408464;+&#3627408479;

This is by neglecting the obliquity of the link RS, i.e., assuming the velocity of S equal to that of R.
Similarly, during the return stroke,
&#3627408483;
&#3627408480; &#3627408474;&#3627408462;&#3627408485;=&#3627409172;×&#3627408450;&#3627408451;
′′
×
&#3627408436;&#3627408453;
&#3627408436;&#3627408452;
′′
=&#3627409172;&#3627408479;×
&#3627408473;
&#3627408464;−&#3627408479;

&#3627408483;
&#3627408480; &#3627408474;&#3627408462;&#3627408485; (&#3627408464;&#3627408482;&#3627408481;&#3627408481;&#3627408470;&#3627408475;&#3627408468;)
&#3627408483;
&#3627408480; &#3627408474;&#3627408462;&#3627408485; (&#3627408479;&#3627408466;&#3627408481;&#3627408482;&#3627408479;&#3627408475;)
=
&#3627409172;&#3627408479;×
&#3627408473;
&#3627408464;+&#3627408479;
&#3627409172;&#3627408479;×
&#3627408473;
&#3627408464;−&#3627408479;
=
&#3627408464;−&#3627408479;
&#3627408464;+&#3627408479;

20

Velocity analysis (Instantaneous center method approach)

&#3627409172;
&#3627408436;&#3627408437;=
&#3627408483;
&#3627408436;
&#3627408436;&#3627408444;
=
&#3627408483;
&#3627408437;
&#3627408437;&#3627408444;
=
&#3627408483;
&#3627408438;
&#3627408438;&#3627408444;
=
&#3627408483;
&#3627408439;
&#3627408439;&#3627408444;
=
&#3627408483;
&#3627408440;
&#3627408440;&#3627408444;
=
&#3627408483;
&#3627408441;
&#3627408441;&#3627408444;
=⋯
I⟶ defined for the relative motion between two links
I24 ⟶ Instantaneous center for the relative motion between link 2 and link 4.
In general, when the link moves, its relative motion IC keeps on changing.
Locus of I-center for the relative motion between the links ⟹ centrode.
Locus of I-center of rotation for the relative motion between the links ⟹ Axode.
Motions Centrode Axode
General Motion Curve Curved surface
Pure Translation Straight line Plane surface
Pure rotation Point Straight line
In general, the motion of a link in a mechanism is neither pure translation nor pure rotation.
It is a combination of translation and rotation which we normally say the link is in general motion.
But any link at any instant can be assumed to be in pure rotation with respect to the point in the space
known as instantaneous center of rotation. this center is also known as virtual center.
&#3627408449;&#3627408476;.&#3627408476;&#3627408467; &#3627408444;&#3627408475;&#3627408480;&#3627408481;&#3627408462;&#3627408475;&#3627408481;&#3627408462;&#3627408475;&#3627408466;&#3627408476;&#3627408482;&#3627408480; &#3627408464;&#3627408466;&#3627408475;&#3627408481;&#3627408466;&#3627408479;&#3627408480; &#3627408470;&#3627408475; &#3627408462; &#3627408474;&#3627408466;&#3627408464;ℎ&#3627408462;&#3627408475;&#3627408470;&#3627408480;&#3627408474;=
&#3627408525;·(&#3627408525;−&#3627409359;)
&#3627409360;
&#3627408447;⟶ &#3627408475;&#3627408476;.&#3627408476;&#3627408467; &#3627408473;&#3627408470;&#3627408475;&#3627408472;&#3627408480;

Sir Arnold, Kennedy
In reality, AA1 & BB1 ⟶ 0 (≈0).
AA1 and BB1 are very small (negligible).
The link AB at this instant is in General motion.

21

Basics of I-center for a mechanism
Turning pair

Rolling pair

Sliding pair


&#3627408441;&#3627408476;&#3627408479; &#3627408473;=6,&#3627408449;&#3627408476;.&#3627408476;&#3627408467; &#3627408444;&#3627408475;&#3627408480;&#3627408481;&#3627408462;&#3627408475;&#3627408481;&#3627408462;&#3627408475;&#3627408466;&#3627408476;&#3627408482;&#3627408480; &#3627408464;&#3627408466;&#3627408475;&#3627408481;&#3627408466;&#3627408479;&#3627408480; =
&#3627408525;·(&#3627408525;−&#3627409359;)
&#3627409360;
=15
I12 I13 I14 I15 I16
I23 I24 I25 I26
I34 I35 I36
I45 I46
I56

22

Kennedy’s theorem
For the relative motion between the no. of links in a mechanism any three links, their three I·C must lie in
straight line.
Theorem of angular velocities
Any I.C Imn can be treated as on link m or its on link n.
&#3627408457;
&#3627408444;&#3627408474;&#3627408475;
=&#3627409172;
&#3627408474;&#3627408475;·(&#3627408444;
&#3627408474;&#3627408475;&#3627408444;
1&#3627408474;
̅̅̅̅̅̅̅̅̅)=&#3627409172;
&#3627408475;·(&#3627408444;
&#3627408474;&#3627408475;&#3627408444;
1&#3627408475;
̅̅̅̅̅̅̅̅)
This theorem is applied at Imn.
Total I.C in use
&#3627408444;
&#3627408474;&#3627408475;
&#3627408444;
1&#3627408474;
&#3627408444;
1&#3627408475;
}
&#3627408473;&#3627408470;&#3627408475;&#3627408472; 1
&#3627408473;&#3627408470;&#3627408475;&#3627408472; &#3627408474;
&#3627408473;&#3627408470;&#3627408475;&#3627408472; &#3627408475;

If I1m, I1n lies on same side of Imn ⟶ same direction, otherwise opposite direction.
Relative velocity method

VELOCITY ANALYSIS OF DOUBLE SLIDER CRANK
MECHANISM
Links 2 & 4 are relatively translating i.e., there is no orientation change b/w
links.
&#3627408457;
2=&#3627408457;
&#3627408444;23
= &#3627409172;
3∗&#3627408444;
23∗&#3627408444;
13
̅̅̅̅̅̅̅̅̅̅
&#3627409172;
3=
&#3627408457;
2
&#3627408444;
23∗&#3627408444;
13
̅̅̅̅̅̅̅̅̅̅
=
&#3627408457;
2
&#3627408447;
2&#3627408480;&#3627408470;&#3627408475;&#3627409155;

&#3627408457;
4=&#3627408457;
&#3627408444;34
= &#3627409172;
3∗&#3627408444;
34∗&#3627408444;
13
̅̅̅̅̅̅̅̅̅̅
&#3627408457;
4=
&#3627408457;
2
&#3627408447;
2&#3627408480;&#3627408470;&#3627408475;&#3627409155;
∗&#3627408447;
2&#3627408464;&#3627408476;&#3627408480;&#3627409155;
&#3627408457;
4=
&#3627408457;
2
&#3627408481;&#3627408462;&#3627408475;&#3627409155;



The velocity of point B w.r.t point A will
be in the direction perpendicular to the
link AB
Intersection of 12, 14 & 23, 34 is at ∞.
So, I24 is at ∞.

23

ACCELERATION ANAL YSIS
The rate of change of velocity w.r.t time is known as acceleration and it acts in the direction of change in
velocity. It’s a vector quantity.
Let a link OA, of length r, rotate in circular path in the clockwise direction. It has an instantaneous angular
velocity ω and an angular acceleration α in the same direction, i.e., the angular velocity increases in the
clockwise direction.

&#3627408455;&#3627408462;&#3627408475;&#3627408468;&#3627408466;&#3627408475;&#3627408481;&#3627408470;&#3627408462;&#3627408473; &#3627408483;&#3627408466;&#3627408473;&#3627408476;&#3627408464;&#3627408470;&#3627408481;&#3627408486; &#3627408476;&#3627408467; &#3627408436;,&#3627408483;
&#3627408462;=&#3627409172;&#3627408479;
&#3627408502;&#3627408488; &#3627408470;&#3627408480; &#3627408479;&#3627408476;&#3627408481;&#3627408462;&#3627408481;&#3627408466;&#3627408465; &#3627408463;&#3627408486; &#3627409209;??????,&#3627408481;&#3627408476; &#3627408502;&#3627408488;’,&#3627408470;&#3627408475; &#3627408462; &#3627408480;&#3627408477;&#3627408462;&#3627408475; &#3627408476;&#3627408467; &#3627409209;&#3627408533;.
&#3627408436;&#3627408475;&#3627408468;&#3627408482;&#3627408473;&#3627408462;&#3627408479; &#3627408483;&#3627408466;&#3627408473;&#3627408476;&#3627408464;&#3627408470;&#3627408481;&#3627408486; &#3627408476;&#3627408467; &#3627408450;&#3627408436;

,&#3627409172;
&#3627408462;

=&#3627409172;+&#3627409148;·&#3627409151;&#3627408481;
&#3627408455;&#3627408462;&#3627408475;&#3627408468;&#3627408466;&#3627408475;&#3627408481;&#3627408470;&#3627408462;&#3627408473; &#3627408483;&#3627408466;&#3627408473;&#3627408476;&#3627408464;&#3627408470;&#3627408481;&#3627408486; &#3627408476;&#3627408467; &#3627408436;

,&#3627408483;
&#3627408462;

=(&#3627409172;+&#3627409148;·&#3627409151;&#3627408481;)·&#3627408479;
The tangential velocity of A’ (&#3627408535;
&#3627408514;

) have two components of velocity, one parallel and other perpendicular to OA.
&#3627408488;&#3627408516;&#3627408516;&#3627408518;&#3627408525;&#3627408518;&#3627408531;&#3627408514;&#3627408533;&#3627408522;&#3627408528;&#3627408527; &#3627408528;&#3627408519; &#3627408436;⊥&#3627408533;&#3627408528; &#3627408450;&#3627408436;=
&#3627408483;
&#3627408462;

·&#3627408464;&#3627408476;&#3627408480;&#3627409151;&#3627409155;−&#3627408483;
&#3627408462;
&#3627409151;&#3627408481;
=
((&#3627409172;+&#3627409148;·&#3627409151;&#3627408481;)·&#3627408479;·&#3627408464;&#3627408476;&#3627408480;&#3627409151;&#3627409155;)−(&#3627409172;&#3627408479;)
&#3627409151;&#3627408481;

&#3627408436;&#3627408480; &#3627409151;&#3627408481;→0,&#3627408464;&#3627408476;&#3627408480;&#3627409151;&#3627409155;→1 ⟹ &#3627408436;&#3627408464;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408467; &#3627408488;⊥&#3627408481;&#3627408476; &#3627408502;&#3627408488;=
((&#3627409172;+&#3627409148;·&#3627409151;&#3627408481;)·&#3627408479;·1)−(&#3627409172;&#3627408479;)
&#3627409151;&#3627408481;

&#3627408507;&#3627408514;&#3627408527;&#3627408520;&#3627408518;&#3627408527;&#3627408533;&#3627408522;&#3627408514;&#3627408525; &#3627408514;&#3627408516;&#3627408516;&#3627408518;&#3627408525;&#3627408518;&#3627408531;&#3627408514;&#3627408533;&#3627408522;&#3627408528;&#3627408527;→ &#3627408519;
&#3627408514;&#3627408528;
&#3627408533;
=??????·&#3627408531;=(
&#3627408465;&#3627409172;
&#3627408465;&#3627408481;
)·&#3627408479;=
&#3627408465;&#3627408483;
&#3627408465;&#3627408481;

&#3627408488;&#3627408516;&#3627408516;&#3627408518;&#3627408525;&#3627408518;&#3627408531;&#3627408514;&#3627408533;&#3627408522;&#3627408528;&#3627408527; &#3627408528;&#3627408519; &#3627408436;∥&#3627408533;&#3627408528; &#3627408450;&#3627408436;=
&#3627408483;
&#3627408462;

·&#3627408480;&#3627408470;&#3627408475;&#3627409151;&#3627409155;−0
&#3627409151;&#3627408481;
=
((&#3627409172;+&#3627409148;·&#3627409151;&#3627408481;)·&#3627408479;·&#3627408480;&#3627408470;&#3627408475;&#3627409151;&#3627409155;)
&#3627409151;&#3627408481;

&#3627408436;&#3627408480; &#3627409151;&#3627408481;→0,&#3627408480;&#3627408470;&#3627408475;&#3627409151;&#3627409155;→&#3627409151;&#3627409155; ⟹ &#3627408436;&#3627408464;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408467; &#3627408488;⊥&#3627408481;&#3627408476; &#3627408502;&#3627408488;=
((&#3627409172;+&#3627409148;·&#3627409151;&#3627408481;)·&#3627408479;·&#3627409151;&#3627409155;)
&#3627409151;&#3627408481;
=&#3627409172;·&#3627408479;·
&#3627409151;&#3627409155;
&#3627409151;&#3627408481;
=&#3627409172;&#3627408479;·&#3627409172;
&#3627408505;&#3627408514;&#3627408517;&#3627408522;&#3627408514;&#3627408525; &#3627408528;&#3627408531; &#3627408490;&#3627408518;&#3627408527;&#3627408533;&#3627408531;&#3627408522;&#3627408529;&#3627408518;&#3627408533;&#3627408514;&#3627408525; &#3627408514;&#3627408516;&#3627408516;&#3627408518;&#3627408525;&#3627408518;&#3627408531;&#3627408514;&#3627408533;&#3627408522;&#3627408528;&#3627408527;→ &#3627408519;
&#3627408514;&#3627408528;
&#3627408516;
=&#3627409172;
2
·&#3627408479;=
&#3627408483;
2
&#3627408479;

When α=0, Tangential acceleration = 0, only Centripetal acceleration will be present.
When ω =0, Centripetal acceleration will be zero, A has linear motion.
When α is negative, tangential acceleration will be in negative direction.

24

Four-link mechanism

Acceleration diagram Construction: -
a) Select the pole point a1 or d1.
b) Take the first vector from the above table,
i.e., take a1ba to a convenient scale in the
proper direction and sense.
c) Add the second vector to the first and then
the third vector to the second.
d) For the addition of the fourth vector, draw a
line perpendicular to BC through the head cb
of the third vector. The magnitude of the
fourth vector is unknown and cc can lie on
either side of cb.
e) Take the fifth vector from d1.
f) For the addition of sixth vector to the fifth,
draw a line perpendicular to DC through the
head cd of the fifth vector.
The intersection of this line with line drawn
in the step (d) locates the point c1.
Total acceleration of B=a1b1
Total acceleration of C relative to B = b1c1
Total acceleration of C = d1c1
S. no Vector Magnitude Direction Sense
1 &#3627408467;
&#3627408463;&#3627408462;
&#3627408464;
&#3627408476;&#3627408479; &#3627408514;
&#3627409359;&#3627408515;
&#3627408514;
(&#3627408514;&#3627408515;)
2
&#3627408436;&#3627408437;
∥ AB ⟶ A
2 &#3627408467;
&#3627408463;&#3627408462;
&#3627408481;
&#3627408476;&#3627408479; &#3627408515;
&#3627408514;&#3627408515;
&#3627409359; α × AB ⊥ AB or a1ba or ∥ ab ⟶ b
3
&#3627408467;
&#3627408464;&#3627408463;
&#3627408464;
&#3627408476;&#3627408479; &#3627408515;
&#3627409359;&#3627408516;
&#3627408515; (&#3627408515;&#3627408516;)
2
&#3627408437;&#3627408438;
∥ AB ⟶ B
4 &#3627408467;
&#3627408464;&#3627408463;
&#3627408481;
&#3627408476;&#3627408479; &#3627408516;
&#3627408515;&#3627408516;
&#3627409359; - ⊥ BC or b1cb -
5
&#3627408467;
&#3627408464;&#3627408465;
&#3627408464;
&#3627408476;&#3627408479; &#3627408517;
&#3627409359;&#3627408516;
&#3627408517; (&#3627408515;&#3627408516;)
2
&#3627408439;&#3627408438;
∥ DC ⟶ D
6 &#3627408467;
&#3627408464;&#3627408465;
&#3627408481;
&#3627408476;&#3627408479; &#3627408516;
&#3627408517;&#3627408516;
&#3627409359; - ⊥ DC or d1cd -
Acceleration of intermediate points on the links can be obtained by dividing the acceleration vectors in the same
ratio as the points divide the links. For point E on the link BC,
&#3627408437;&#3627408440;
&#3627408437;&#3627408438;
=
&#3627408463;
1&#3627408466;
1
&#3627408463;
1&#3627408464;
1

a1e1 gives the total acceleration of the point E.

25

Slider crank Mechanism


Acceleration of B relative to O = Acceleration of B relative to A + Acceleration of A relative to O.
&#3627408467;
&#3627408463;&#3627408476;=&#3627408467;
&#3627408463;&#3627408462;+&#3627408467;
&#3627408462;&#3627408476;
&#3627408467;
&#3627408463;??????=&#3627408467;
&#3627408462;&#3627408476;+&#3627408467;
&#3627408463;&#3627408462;=&#3627408467;
&#3627408462;&#3627408476;+(&#3627408467;
&#3627408463;&#3627408462;
&#3627408464;
+&#3627408467;
&#3627408463;&#3627408462;
&#3627408481;
)
&#3627408520;
&#3627409359;&#3627408515;
&#3627409359;=&#3627408528;
&#3627409359;&#3627408514;
&#3627409359;+&#3627408514;
&#3627409359;&#3627408515;
&#3627408514;+&#3627408515;
&#3627408514;&#3627408515;
&#3627409359;
Crank OA rotates at a uniform velocity. So, the acceleration of A relative to O has only the centripetal
component.
Slider moves in a linear direction and has no centripetal component.
S. no Vector Magnitude Direction Sense
1 &#3627408519;
&#3627408514;&#3627408528;&#3627408528;&#3627408531; &#3627408528;
&#3627409359;&#3627408514;
&#3627409359;
(&#3627408528;&#3627408514;)
2
&#3627408450;&#3627408436;
∥ AB ⟶ O
2 &#3627408519;
&#3627408515;&#3627408514;
&#3627408516;
&#3627408528;&#3627408531; &#3627408514;
&#3627409359;&#3627408515;
&#3627408514;
(&#3627408514;&#3627408515;)
2
&#3627408436;&#3627408437;
⊥ AB or a1ba or ∥ ab ⟶ A
3 &#3627408519;
&#3627408515;&#3627408514;
&#3627408533;
&#3627408528;&#3627408531; &#3627408515;
&#3627408514; &#3627408515;
&#3627409359; ─ ∥ AB ─
4 &#3627408519;
&#3627408515;&#3627408520; &#3627408528;&#3627408531; &#3627408520;
&#3627409359; &#3627408515;
&#3627409359; ─ ⊥ BC or b1cb ─
Construction
1. Take the first vector fao
2. Add the second vector to the first
3. For the third vector, draw a line ⊥ to AB through the head ba of the second vector
4. For the fourth vector, draw a line through g1 parallel to the line of motion of the slider
Acceleration of the slider B = o1b1 (or g1b1)
Total acceleration of B relative to A = a1b1
The direction of slider is opposite to that of velocity. Therefore, the acceleration is negative, or the slider
decelerates while moving.

26

Coriolis Acceleration Component
It is seen that the acceleration of a moving point relative to a fixed body may have two components of
acceleration: the centripetal and tangential.
However, in some cases, the point may have its motion relative to a
moving body system, for example, motion of a slider on a rotating link.
Following analysis is made to investigate the acceleration at that point
P.
Let a link AR rotate about a fixed-point A on it. P is a point on a slider
on the link.
ω = angular velocity of link, α = angular acceleration of the link,
v = linear velocity of the slider on the link,
f = linear acceleration of the slider on the link,
r = radial distance of point P on the slider.
In a short interval of time δt, let δθ be the angular displacement of the
link and δr be the radial displacement of the slider in the outward
direction.
After the short interval of time δt, let
&#3627409172;

=&#3627409172;+&#3627409148;·&#3627409151;&#3627408481;&#3627408483;

=&#3627408483;+&#3627408467;·&#3627409151;&#3627408481;&#3627408479;

=&#3627408479;+&#3627409151;&#3627408479;
Acceleration of P parallel to AR
&#3627408444;&#3627408475;&#3627408470;&#3627408481;&#3627408470;&#3627408462;&#3627408473; &#3627408483;&#3627408466;&#3627408473;&#3627408476;&#3627408464;&#3627408470;&#3627408481;&#3627408486; &#3627408476;&#3627408467; &#3627408503; &#3627408462;&#3627408473;&#3627408476;&#3627408475;&#3627408468; &#3627408488;&#3627408505;=&#3627408535;=&#3627408535;
&#3627408529;&#3627408530;
&#3627408441;&#3627408470;&#3627408475;&#3627408462;&#3627408473; &#3627408483;&#3627408466;&#3627408473;&#3627408476;&#3627408464;&#3627408470;&#3627408481;&#3627408486; &#3627408476;&#3627408467; &#3627408503; &#3627408462;&#3627408473;&#3627408476;&#3627408475;&#3627408468; &#3627408488;&#3627408505;=(&#3627408483;

·&#3627408464;&#3627408476;&#3627408480;&#3627409151;&#3627409155;−&#3627409172;

·&#3627408479;

·&#3627408480;&#3627408470;&#3627408475;&#3627409151;&#3627409155;)
&#3627408436;&#3627408464;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408467; &#3627408503; &#3627408462;&#3627408473;&#3627408476;&#3627408475;&#3627408468; &#3627408488;&#3627408505;=
(&#3627408483;

·&#3627408464;&#3627408476;&#3627408480;&#3627409151;&#3627409155;−&#3627409172;

·&#3627408479;

·&#3627408480;&#3627408470;&#3627408475;&#3627409151;&#3627409155;)−&#3627408483;
&#3627409151;&#3627408481;
=
((&#3627408483;+&#3627408467;·&#3627409151;&#3627408481;)·&#3627408464;&#3627408476;&#3627408480;&#3627409151;&#3627409155;−(&#3627409172;+&#3627409148;·&#3627409151;&#3627408481;)·(&#3627408479;+&#3627409151;&#3627408479;)·&#3627408480;&#3627408470;&#3627408475;&#3627409151;&#3627409155;)−&#3627408483;
&#3627409151;&#3627408481;

In the limit, as δt ⟶ 0, cos δθ ⟶ 1, sin δθ ⟶ 0.&#3627409151;&#3627409155;
&#3627408436;&#3627408464;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408467; &#3627408503; &#3627408462;&#3627408473;&#3627408476;&#3627408475;&#3627408468; &#3627408488;&#3627408505;=&#3627408467;−&#3627409172;&#3627408479;
&#3627408465;&#3627409155;
&#3627408465;&#3627408481;
=&#3627408467;−&#3627409172;&#3627408479;&#3627409172;=&#3627408519;−??????
&#3627409360;
&#3627408531;
Acceleration of P along AR = (Acceleration of slider) ─ (Centripetal acceleration)
This is the acceleration of f along AR 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 outward direction.
Acceleration of P perpendicular to AR,
&#3627408436;&#3627408464;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408467; &#3627408451; ⊥&#3627408436;&#3627408453;=
(&#3627408483;

·&#3627408480;&#3627408470;&#3627408475;&#3627409151;&#3627409155;+&#3627409172;

&#3627408479;

·&#3627408464;&#3627408476;&#3627408480;&#3627409151;&#3627409155;)−&#3627408483;
&#3627409151;&#3627408481;
=
((&#3627408483;+&#3627408467;·&#3627409151;&#3627408481;)·&#3627408464;&#3627408476;&#3627408480;&#3627409151;&#3627409155;−(&#3627409172;+&#3627409148;·&#3627409151;&#3627408481;)·(&#3627408479;+&#3627409151;&#3627408479;)·&#3627408464;&#3627408476;&#3627408480;&#3627409151;&#3627409155;)−&#3627409172;&#3627408479;
&#3627409151;&#3627408481;

In the limit, as δt ⟶ 0, cos δθ ⟶ 1, sin δθ ⟶ 0.
&#3627408436;&#3627408464;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408467; &#3627408503; ⊥&#3627408488;&#3627408505;=&#3627408483;
&#3627408465;&#3627409155;
&#3627408465;&#3627408481;
+&#3627409172;
&#3627408465;&#3627408479;
&#3627408465;&#3627408481;
+&#3627409148;·&#3627408479;=&#3627408483;·&#3627409172;+&#3627409172;·&#3627408483;+&#3627409148;·&#3627408479;=&#3627409360;??????&#3627408535;+??????&#3627408531;
Acceleration of P ⊥ AR = 2ωv + Tangential acceleration
The component 2ωv is known as the Coriolis acceleration component.
It is positive if both ω and v are either positive or negative.
The Coriolis component is positive if the link AR rotates clockwise and the slider moves radially outwards or
link rotates counter-clockwise and the slider moves radially inwards.

27

The direction of the Coriolis acceleration component is obtained by rotating the radial velocity vector v through
90° in the direction of rotation of the link.
Let Q be a point on the link AR immediately beneath the point P at the instant. Then
&#3627408436;&#3627408464;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408467; &#3627408503; = &#3627408462;&#3627408464;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408467; &#3627408503; ∥&#3627408481;&#3627408476; &#3627408488;&#3627408505;+&#3627408462;&#3627408464;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408467; &#3627408503; ⊥&#3627408481;&#3627408476; &#3627408488;&#3627408505;
&#3627408467;
&#3627408477;&#3627408462;=(&#3627408467;−&#3627409172;
2
&#3627408479;)+(2&#3627409172;&#3627408483;+&#3627409148;&#3627408479;)=&#3627408467;+(&#3627409148;&#3627408479;−&#3627409172;
2
&#3627408479;)+2&#3627409172;&#3627408483;
&#3627408467;
&#3627408477;&#3627408462;=&#3627408462;&#3627408464;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408467; &#3627408503; &#3627408479;&#3627408466;&#3627408473;&#3627408462;&#3627408481;&#3627408470;&#3627408483;&#3627408466; &#3627408481;&#3627408476; &#3627408504;+&#3627408462;&#3627408464;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408467; &#3627408504; &#3627408479;&#3627408466;&#3627408473;&#3627408462;&#3627408481;&#3627408470;&#3627408483;&#3627408466; &#3627408481;&#3627408476; &#3627408488;+&#3627408490;&#3627408528;&#3627408531;&#3627408522;&#3627408528;&#3627408525;&#3627408522;&#3627408532; &#3627408514;&#3627408516;&#3627408516;&#3627408518;&#3627408525;&#3627408518;&#3627408531;&#3627408514;&#3627408533;&#3627408522;&#3627408528;&#3627408527; &#3627408516;&#3627408528;&#3627408526;&#3627408529;&#3627408528;&#3627408527;&#3627408518;&#3627408527;&#3627408533;
&#3627408519;
&#3627408529;&#3627408514;=&#3627408467;
&#3627408477;&#3627408478;

+&#3627408467;
&#3627408478;&#3627408462;+&#3627408519;
&#3627408516;&#3627408531;

&#3627408467;
&#3627408477;&#3627408478;

→&#3627408462;&#3627408464;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408484;ℎ&#3627408470;&#3627408464;ℎ &#3627408462;&#3627408475; &#3627408476;&#3627408463;&#3627408480;&#3627408466;&#3627408479;&#3627408483;&#3627408466;&#3627408479; &#3627408480;&#3627408481;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475;&#3627408466;&#3627408465; &#3627408476;&#3627408475; &#3627408473;&#3627408470;&#3627408475;&#3627408472; &#3627408488;&#3627408505; &#3627408484;&#3627408476;&#3627408482;&#3627408473;&#3627408465; &#3627408476;&#3627408463;&#3627408480;&#3627408466;&#3627408479;&#3627408483;&#3627408466; &#3627408467;&#3627408476;&#3627408479; &#3627408481;ℎ&#3627408466; &#3627408480;&#3627408473;&#3627408470;&#3627408465;&#3627408466;&#3627408479;.
Remember Coriolis component of acceleration exists only if there are two coincident points which has
• Linear relative velocity of sliding
• Angular motion about fixed finite centres of rotation.
&#3627408519;
&#3627408529;&#3627408514;=&#3627408467;
&#3627408477;&#3627408478;

+&#3627408467;
&#3627408478;&#3627408462;+&#3627408519;
&#3627408516;&#3627408531;
=(&#3627408467;
&#3627408477;&#3627408478;

+&#3627408519;
&#3627408516;&#3627408531;
)+&#3627408467;
&#3627408478;&#3627408462;=&#3627408519;
&#3627408529;&#3627408530;+&#3627408519;
&#3627408530;&#3627408514;
Crank and Slotted-Lever Mechanism
Crank OP rotates at uniform angular velocity of ω rad/s clockwise.


&#3627408467;
&#3627408477;&#3627408462;=&#3627408467;
&#3627408477;&#3627408478;+&#3627408467;
&#3627408478;&#3627408462; (&#3627408476;&#3627408479;) &#3627408467;
&#3627408477;&#3627408476;=&#3627408467;
&#3627408478;&#3627408462;+&#3627408467;
&#3627408477;&#3627408478;
&#3627408528;
&#3627409359;&#3627408529;
&#3627409359;=&#3627408514;
&#3627409359;&#3627408530;
&#3627408514;+&#3627408530;
&#3627408514;&#3627408530;
&#3627409359;+&#3627408530;
&#3627409359;&#3627408529;
&#3627408530;+&#3627408529;
&#3627408530;&#3627408529;
&#3627409359;

28

Construction of Acceleration diagram:
1. Take the first vector fpo which is completely known.
2. Take the second vector from the point a1 (or o1). This vector is also known.
3. Only the direction of the third vector &#3627408519;
&#3627408530;&#3627408514;
&#3627408533;
is known. Draw a line ⊥ to AQ through the head qa of the
second vector.
4. As the head of the third vector is not available, the fourth vector cannot be added to it.
Take the last vector &#3627408519;
&#3627408529;&#3627408530;
&#3627408516;&#3627408531;
which is completely known. Place this vector in the proper direction and sense
so that p1 becomes the head of the vector.
pq can’t lie on the right side of p1 because then the vector would become p1pq and not pqp1.
5. For the fourth vector, draw a line parallel to AR through the point pq of the fifth vector.
the intersection of this line with line drawn in the step 3 locates the point q1.
6. Total acceleration of P relative to Q, fpq = q1p1
total acceleration of Q relative to A, fqa = a1q1
the acceleration of R relative to A is given on a1q1 produced such that
&#3627408462;
1&#3627408479;
1
&#3627408462;
1&#3627408478;
1
=
&#3627408436;&#3627408453;
&#3627408436;&#3627408452;


S. no Vector Magnitude Direction Sense
1 &#3627408519;
&#3627408529;&#3627408528;&#3627408528;&#3627408531; &#3627408528;
&#3627409359;&#3627408529;
&#3627409359; ω × OP ∥ OP ⟶ O
2 &#3627408519;
&#3627408530;&#3627408514;
&#3627408516;
&#3627408528;&#3627408531; &#3627408514;
&#3627409359;&#3627408530;
&#3627408514;
(&#3627408514;&#3627408530;)
2
&#3627408436;&#3627408452;
∥ AQ ⟶ A
3 &#3627408519;
&#3627408530;&#3627408514;
&#3627408533;
&#3627408528;&#3627408531; &#3627408530;
&#3627408514; &#3627408530;
&#3627409359; ─ ⊥ AQ or a1qa ─
4 &#3627408519;
&#3627408529;&#3627408530;
&#3627408532;
&#3627408528;&#3627408531; &#3627408530;
&#3627409359;&#3627408529;
&#3627408530; ─ ∥ AR ─
5 &#3627408519;
&#3627408529;&#3627408530;
&#3627408516;&#3627408531;
&#3627408528;&#3627408531; &#3627408529;
&#3627408530;&#3627408529;
&#3627409359; Coriolis component* ⊥ AR Refer*

&#3627408467;
&#3627408477;&#3627408478;
&#3627408464;&#3627408479;
=2&#3627409172;
1&#3627408483;
&#3627408477;&#3627408478; (&#3627409172;
1=&#3627408462;&#3627408475;&#3627408468;&#3627408482;&#3627408473;&#3627408462;&#3627408479; &#3627408483;&#3627408466;&#3627408473;&#3627408470;&#3627408464;&#3627408481;&#3627408486; &#3627408476;&#3627408467; &#3627408436;&#3627408453;)=2(
&#3627408514;&#3627408530;
&#3627408436;&#3627408452;
)&#3627408478;&#3627408477;

29

STATIC FORCE ANLAYSIS
There are 2 types of forces on a mechanism
1. Constraint forces- A pair of action and reaction forces which constrain two bodies to behave in a
manner depending upon the nature of connection are known as constraint forces.
2. Applied forces- Forces acting from outside on a system of bodies are called applied forces.

A body is in static equilibrium if it remains in its state of rest or motion.
• The vector sum of all forces acting on the body is zero. (∑&#3627408441;⃗=0)
• The vector sum of all moments about the arbitrary point is zero. (∑&#3627408448;⃗⃗⃗=0)
2-FORCE SYSTEM 3-FORCE SYSTEM

A member under the action of 2 force system will be in equilibrium if
• The forces are of same magnitude
• The forces act along the same line
• The forces are in opposite directions.
A member under the action of 3 force system will be in equilibrium if
• The resultant of action of 3 forces is zero.
• The lines of action of the forces intersect at a point.
A member under the action of two applied forces and an applied torque will be in equilibrium if
• The forces are in equal in magnitude, parallel in direction and opposite in sense
• The forces form a couple which is equal and opposite to applied torque.

30

Equilibrium of four force members
First look for the forces completely known and combine them into a single force using vector addition method,
then we can use three force method.
FREE BODY DIAGRAM

SUPERPOSITION
In linear systems, if a number of loads act on a system of forces, the net effect is equal to superposition of the
effects of the individual loads taken at a time.
PRINCIPLE OF VIRTUAL WORK
The work done during a virtual displacement from the equiibrium is equal to zero.

According to the principle of virtual work,
&#3627408458;=&#3627408455;&#3627409151;&#3627409155;+&#3627408441;&#3627409151;&#3627408485;=0
An virtual displacement must take place during the same interval δt,
&#3627408455;
&#3627408465;&#3627409155;
&#3627408465;&#3627408481;
+&#3627408441;
&#3627408465;&#3627408485;
&#3627408465;&#3627408481;
=0
&#3627408455;&#3627409172;+&#3627408441;&#3627408483;=0
&#3627408455;=−
&#3627408441;
&#3627409172;
&#3627408483;
(ω→angular velocity, v→linear velocity)

31

DYNAMIC FORCE ANALYSIS
D’ALEMBERT’S PRINCIPLE
Inertia forces and couples, and the external forces and torques on a body together give static equilibrium.
Inertia is a property of matter by virtue of which a body resists any change in velocity.
&#3627408444;&#3627408475;&#3627408466;&#3627408479;&#3627408481;&#3627408470;&#3627408462; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;&#3627408441;
?????? =−&#3627408474;&#3627408467;
??????
(&#3627408467;
??????→&#3627408462;&#3627408462;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408467; &#3627408464;&#3627408466;&#3627408475;&#3627408481;&#3627408479;&#3627408466; &#3627408476;&#3627408467; &#3627408474;&#3627408462;&#3627408480;&#3627408480; &#3627408476;&#3627408467; &#3627408463;&#3627408476;&#3627408465;&#3627408486;)
Inertia force acts in opposite direction to that of acceleration.
&#3627408444;&#3627408475;&#3627408466;&#3627408479;&#3627408481;&#3627408470;&#3627408462; &#3627408464;&#3627408476;&#3627408482;&#3627408477;&#3627408473;&#3627408466; &#3627408438;
??????=−&#3627408444;
??????.&#3627409148;
(&#3627408444;
??????→&#3627408474;&#3627408476;&#3627408474;&#3627408466;&#3627408475;&#3627408481; &#3627408476;&#3627408467; &#3627408444;&#3627408475;&#3627408466;&#3627408479;&#3627408481;&#3627408470;&#3627408462; &#3627408462;&#3627408463;&#3627408476;&#3627408482;&#3627408481; &#3627408462;&#3627408475; &#3627408462;&#3627408485;&#3627408470;&#3627408480; &#3627408477;&#3627408462;&#3627408480;&#3627408480;&#3627408470;&#3627408475;&#3627408468; &#3627408481;ℎ&#3627408479;&#3627408476;&#3627408482;&#3627408468;ℎ &#3627408464;&#3627408466;&#3627408475;&#3627408481;&#3627408479;&#3627408466; &#3627408476;&#3627408467; &#3627408474;&#3627408462;&#3627408480;&#3627408480; &#3627408442;)
According to D’Alembert’s principle
∑&#3627408441;+&#3627408441;
??????=0
∑&#3627408455;+&#3627408438;
??????=0
EQUIVALENT OFFSET INERTIA FORCE
If the body is acted upon by the forces such that resultant force does not pass through the centre of mass, a
couple is acting on the system.
It is possible to replace inertia force and inertia couple by an equivalent offset force, this is done by displacing
the line of action of inertia force from the centre of mass.
ℎ=
&#3627408438;
??????
&#3627408441;
??????
=
&#3627408472;
2
&#3627409148;
&#3627408467;
??????
(&#3627408438;
??????=&#3627408441;
??????.ℎ)
(&#3627408438;
??????=−&#3627408444;
??????.&#3627409148;)(&#3627408441;
?????? =−&#3627408474;&#3627408467;
??????)
DYNAMIC ANALYSIS OF FOUR LINK MECHANISM
For dynamic analysis of four-link mechanisms, the following procedure is followed.
1. Draw the velocity and acceleration diagram of the mechanism from the configuration diagram by usual
methods.
2. Determine the linear acceleration of the centers of masses of various links, and the angular
accelerations of the links
3. Calculate the inertia forces and inertia couples from the relations &#3627408441;
?????? =−&#3627408474;&#3627408467;
?????? and &#3627408438;
??????=−&#3627408444;
??????.&#3627409148; .
4. Replace Fi with equivalent offset inertia force to consider Fi as well as Ci.
5. Assume equivalent offset inertia forces on the links as static forces and analyze the mechanism by any of
the methods.

32

STATIC ANALYSIS OF SLIDER CRANK MECHANISMS
Velocity and acceleration of a piston

&#3627408485;=&#3627408437;
1&#3627408437;=&#3627408437;&#3627408450;−&#3627408437;
1&#3627408450;=&#3627408437;&#3627408450;−(&#3627408437;
1&#3627408436;
1+&#3627408436;
1&#3627408450;)
&#3627408485;=(&#3627408479;+&#3627408473;)−(&#3627408479;&#3627408464;&#3627408476;&#3627408480;&#3627409155;+&#3627408473;&#3627408464;&#3627408476;&#3627408480;&#3627409149;)
(&#3627408525;=&#3627408531;&#3627408527;)
&#3627408485;=&#3627408479;[(&#3627408475;+1)−(&#3627408475;&#3627408464;&#3627408476;&#3627408480;&#3627409149;+&#3627408479;&#3627408464;&#3627408476;&#3627408480;&#3627409155;)]
&#3627408464;&#3627408476;&#3627408480;&#3627409149;=
1
&#3627408475;
√&#3627408475;
2
−&#3627408480;&#3627408470;&#3627408475;
2
&#3627409155;
&#3627408485;=&#3627408479;[(&#3627408475;+1)−(√&#3627408475;
2
−&#3627408480;&#3627408470;&#3627408475;
2
&#3627409155;+&#3627408479;&#3627408464;&#3627408476;&#3627408480;&#3627409155;)]
&#3627408485;=&#3627408479;[(1−&#3627408479;&#3627408464;&#3627408476;&#3627408480;&#3627409155;)+(&#3627408475;−√&#3627408475;
2
−&#3627408480;&#3627408470;&#3627408475;
2
&#3627409155;)]
If (l>>>>n) n
2
is very large and (n
2
-sin
2
  n
2
)
&#3627408485;=&#3627408479;(1−&#3627408479;&#3627408464;&#3627408476;&#3627408480;&#3627409155;)

Velocity of the piston
&#3627408483;=
&#3627408465;&#3627408485;
&#3627408465;&#3627409155;
×
&#3627408465;&#3627409155;
&#3627408465;&#3627408481;

&#3627408483;=&#3627408479;&#3627409172;[&#3627408480;&#3627408470;&#3627408475;&#3627409155;+
&#3627408480;&#3627408470;&#3627408475;2&#3627409155;
2√&#3627408475;
2
−&#3627408480;&#3627408470;&#3627408475;
2
&#3627409155;
]

&#3627408483;=&#3627408479;&#3627409172;[&#3627408480;&#3627408470;&#3627408475;&#3627409155;+
&#3627408480;&#3627408470;&#3627408475;2&#3627409155;
2&#3627408475;
]
Acceleration of the piston
&#3627408467;=
&#3627408465;&#3627408483;
&#3627408465;&#3627409155;
×
&#3627408465;&#3627409155;
&#3627408465;&#3627408481;

&#3627408467;=&#3627408479;&#3627409172;
2
[&#3627408464;&#3627408476;&#3627408480;&#3627409155;+
&#3627408464;&#3627408476;&#3627408480;2&#3627409155;
&#3627408475;
]
&#3627408467;=&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;&#3627409155; (if n>>>cos 2)
Velocity of the piston will be maximum at crank angle , to find 
&#3627408465;&#3627408483;
&#3627408465;&#3627409155;
=0
(n
2
>> sin
2
)

33

&#3627408465;
&#3627408465;&#3627409155;
[&#3627408480;&#3627408470;&#3627408475;&#3627409155;+
&#3627408480;&#3627408470;&#3627408475;2&#3627409155;
2&#3627408475;
]=0 (&#3627408475;=3.5)
&#3627409155;=75.5°,284.5°
&#3627408436;&#3627408481; =75.5°,&#3627408475;=3.5
&#3627408483;
&#3627408480; = 1.037&#3627408479;&#3627409172;.
&#3627408436;&#3627408481; =284.5°,&#3627408475;=3.5
&#3627408483;
&#3627408480; = −1.037&#3627408479;&#3627409172;.
Acceleration of the piston will be maximum at crank angle , to find 
&#3627408465;&#3627408462;
&#3627408480;
&#3627408465;&#3627409155;
=0
&#3627408465;
&#3627408465;??????
[&#3627408464;&#3627408476;&#3627408480;&#3627409155;+
&#3627408464;&#3627408476;&#3627408480;2??????
&#3627408475;
]=0 (n=3.5)
=151°, 209°
&#3627408436;&#3627408481; =151°,=209°,
&#3627408462;
&#3627408480; = −0.723·&#3627408479;·&#3627409172;
2
.
&#3627408436;&#3627408481;  = 0°,
&#3627408462;
&#3627408480; = −1.3·&#3627408479;·&#3627409172;.
Angular velocity and angular acceleration of connecting rod
&#3627408480;&#3627408470;&#3627408475;&#3627409149;=
&#3627408480;&#3627408470;&#3627408475;&#3627409155;
&#3627408475;

Differentiating with respecting to time, we get
&#3627408436;&#3627408475;&#3627408468;&#3627408482;&#3627408473;&#3627408462;&#3627408479; &#3627408483;&#3627408466;&#3627408473;&#3627408476;&#3627408464;&#3627408470;&#3627408481;&#3627408486; &#3627408476;&#3627408467; &#3627408464;&#3627408476;&#3627408475;&#3627408475;&#3627408466;&#3627408464;&#3627408481;&#3627408470;&#3627408475;&#3627408468; &#3627408479;&#3627408476;&#3627408465;=&#3627409172;
&#3627408464;=
&#3627408465;&#3627409149;
&#3627408465;&#3627408481;

&#3627409172;
&#3627408464;=&#3627409172;
&#3627408464;&#3627408476;&#3627408480;&#3627409155;
√&#3627408475;
2
−&#3627408480;&#3627408470;&#3627408475;
2
&#3627409155;

(ω-angular velocity of rod)
&#3627409148;
&#3627408464; = &#3627408462;&#3627408475;&#3627408468;&#3627408482;&#3627408473;&#3627408462;&#3627408479; &#3627408462;&#3627408464;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408467; &#3627408464;&#3627408476;&#3627408475;&#3627408475;&#3627408466;&#3627408464;&#3627408481;&#3627408470;&#3627408475;&#3627408468; &#3627408479;&#3627408476;&#3627408465;=
&#3627408465;&#3627409172;
&#3627408465;&#3627408481;

&#3627409148;
&#3627408464;=−&#3627409172;
2
&#3627408480;&#3627408470;&#3627408475;&#3627409155;[
&#3627408475;
2
−1
(&#3627408475;
2
−&#3627408480;&#3627408470;&#3627408475;
2
&#3627409155;)
32⁄
]

34

DYNAMIC ANALYSIS OF SLIDER CRANK MECHANISM
(neglecting the effect of the weights and the inertia effect of the connecting rod)

Piston Effort
It is the net or effective force applied on the piston.
Force on piston due to gas pressure = Fp
&#3627408444;&#3627408475;&#3627408466;&#3627408479;&#3627408481;&#3627408470;&#3627408462; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;=&#3627408474;&#3627408467;=&#3627408474;&#3627408479;&#3627409172;
2
(&#3627408464;&#3627408476;&#3627408480;&#3627409155;+
&#3627408464;&#3627408476;&#3627408480;2&#3627409155;
&#3627408475;
)
F = Fp - Fb - Ff
In case of vertical engines, the weight of the piston or reciprocating parts also acts as force and thus,
F=Fp + mg - Fb - Ff

Force (thrust) along the connecting rod
Fc = Force in the connecting rod

&#3627408441;
&#3627408464;×&#3627408464;&#3627408476;&#3627408480;&#3627409149;=&#3627408441;
&#3627408441;
&#3627408464;=
&#3627408441;
&#3627408464;&#3627408476;&#3627408480;&#3627409149;

Thrust on the sides of cylinder
It is the normal reaction on the cylinder walls
&#3627408441;
&#3627408475;=&#3627408441;
&#3627408464;&#3627408480;&#3627408470;&#3627408475;&#3627409149;=&#3627408467;&#3627408481;&#3627408462;&#3627408475;&#3627409149;
Crank Effort
Force is exerted on the crankpin because of the force on the piston.

35

Ft = crank effort
&#3627408441;
&#3627408481;×&#3627408479;= &#3627408441;
&#3627408464;&#3627408479;&#3627408480;&#3627408470;&#3627408475;&#3627409149;
&#3627408441;
&#3627408481;=&#3627408441;
&#3627408464;&#3627408480;&#3627408470;&#3627408475;&#3627409149;=
&#3627408441;
&#3627408464;&#3627408476;&#3627408480;&#3627409149;
&#3627408480;&#3627408470;&#3627408475;(&#3627409155;+&#3627409149;)
Thrust on the Bearings
&#3627408441;
&#3627408479;=&#3627408441;
&#3627408464;&#3627408464;&#3627408476;&#3627408480;&#3627409149;=
&#3627408441;
&#3627408464;&#3627408476;&#3627408480;&#3627409149;
&#3627408464;&#3627408476;&#3627408480;(&#3627409155;+&#3627409149;)
Turning Moment on Crankshaft
&#3627408455;=&#3627408441;
&#3627408481;×&#3627408479;=
&#3627408441;
&#3627408464;&#3627408476;&#3627408480;&#3627409149;
&#3627408480;&#3627408470;&#3627408475;(&#3627409155;+&#3627409149;)×&#3627408479;
&#3627408455;=&#3627408441;&#3627408479;(&#3627408480;&#3627408470;&#3627408475;&#3627409155;+
&#3627408480;&#3627408470;&#3627408475;2&#3627409155;
2√&#3627408475;
2
−&#3627408480;&#3627408470;&#3627408475;
2
&#3627409155;
)

&#3627408455;=&#3627408441;
&#3627408481;×&#3627408479;=&#3627408441;×&#3627408450;&#3627408439;
Dynamically equivalent 2-point mass system (connecting rod)

&#3627408444;
&#3627408464;&#3627408479;
&#3627408442;
= &#3627408474;
&#3627408464;&#3627408479;&#3627408446;
&#3627408442;
&#3627408464;&#3627408479;

&#3627408446;
&#3627408442;=&#3627408479;&#3627408462;&#3627408465;&#3627408470;&#3627408482;&#3627408480; &#3627408476;&#3627408467; &#3627408468;&#3627408486;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408462;&#3627408463;&#3627408476;&#3627408482;&#3627408481; &#3627408442;.
&#3627408463;+&#3627408465;=&#3627408447;
For complete dynamic equivalence b/w actual Connecting Rod & 2-point mass system
1. mb + ms = mcr
2. mb. b = ms. d
3. &#3627408474;
&#3627408463;&#3627408463;
2
+&#3627408474;
&#3627408480;&#3627408465;
2
=&#3627408474;
&#3627408464;&#3627408479;&#3627408446;
??????
2


Solving for 1 & 2 & 3

&#3627408474;
&#3627408463;=
&#3627408465;
??????
&#3627408474;
&#3627408464;&#3627408479; &#3627408474;
&#3627408480;=
&#3627408463;
??????
&#3627408474;
&#3627408464;&#3627408479;
&#3627408446;
&#3627408442;
2
=&#3627408463;.&#3627408465;
It is the condition on b & d so that 2-point mass system is completely dynamically equivalent.
&#3627408444;=&#3627408474;
&#3627408463;&#3627408463;
2
+&#3627408474;
&#3627408480;&#3627408465;
2
=&#3627408474;
&#3627408464;&#3627408479;.&#3627408463;.&#3627408465;
Result can be compared with equivalent length of simple pendulum

36


The equivalent length of simple pendulum is

Generally, ??????
&#3627408494;
&#3627409360;
≤&#3627408515;.&#3627408517;
(Inertia couple) actual connecting rod < inertia couple of equivalent system (taken)
On the 2-point mass system, a correction couple is applied.
&#3627408455;
&#3627408464;=&#3627408474;
&#3627408464;&#3627408479;&#3627408446;
&#3627408442;
2
&#3627409148;−&#3627408474;
&#3627408464;&#3627408479;&#3627408463;&#3627408465;&#3627409148;
&#3627408455;
&#3627408464;=−&#3627408474;
&#3627408464;&#3627408479;&#3627409148;(&#3627408463;&#3627408465;−&#3627408446;
&#3627408442;
2
)
Correction couple must be applied in the direction of angular acceleration
&#3627408441;
&#3627408464;=
&#3627408455;
&#3627408464;
&#3627408447;

37

BALANCING OF ROTATING MASS
Often an unbalance of forces is produced in rotary or reciprocating machinery due to inertia forces (ex-
centrifugal force in rotating mass) associated with the moving masses.
Balancing is the process of designing or modifying machinery so that unbalance is reduced to an acceptable
level and if possible is eliminated entirely.

Static Balancing
A system of rotating masses is said to be in static balancing if the combined mass center of the system lies on the
axis of rotation.

&#3627408441;= &#3627408474;
1&#3627408531;
&#3627409359;&#3627409172;
2
+&#3627408474;
2&#3627408531;
&#3627409360;&#3627409172;
2
+&#3627408474;
3&#3627408531;
&#3627409361;&#3627409172;
2

&#3627408474;
1&#3627408531;
&#3627409359;&#3627409172;
2
+&#3627408474;
2&#3627408531;
&#3627409360;&#3627409172;
2
+&#3627408474;
3&#3627408531;
&#3627409361;&#3627409172;
2
+&#3627408474;
&#3627408464;&#3627408531;
&#3627408516;&#3627409172;
2
=0
&#3627408474;
1&#3627408531;
&#3627409359;+&#3627408474;
2&#3627408531;
&#3627409360;+&#3627408474;
3&#3627408531;
&#3627409361;+&#3627408474;
&#3627408464;&#3627408531;
&#3627408516;=0
∑&#3627408474;&#3627408479;&#3627408464;&#3627408476;&#3627408480;&#3627409155;+&#3627408474;
&#3627408464;&#3627408479;
&#3627408464;&#3627408464;&#3627408476;&#3627408480;&#3627409155;
&#3627408464;=0
∑&#3627408474;&#3627408479;&#3627408480;&#3627408470;&#3627408475;&#3627409155;+&#3627408474;
&#3627408464;&#3627408479;
&#3627408464;&#3627408480;&#3627408470;&#3627408475;&#3627409155;
&#3627408464;=0
&#3627408474;
&#3627408464;&#3627408479;
&#3627408464;=√(??????&#3627408474;&#3627408479;&#3627408464;&#3627408476;&#3627408480;&#3627409155;)
2
+(??????&#3627408474;&#3627408479;&#3627408480;&#3627408470;&#3627408475;&#3627409155;)
2

&#3627408481;&#3627408462;&#3627408475;&#3627409155;
&#3627408464;=
??????&#3627408474;&#3627408479;&#3627408464;&#3627408476;&#3627408480;&#3627409155;
??????&#3627408474;&#3627408479;&#3627408480;&#3627408470;&#3627408475;&#3627409155;

38

Dynamic Balancing
When several masses rotate in different planes, the centrifugal forces, in addition to being out of balance, also
form couples.
A system of rotating masses is in dynamic balance when there does not exist any centrifugal force as well as
resultant couple.
Transferring Force from one plane to another plane-
Force of mass m will be replaced by Force F1 and as a
result a couple will be acting at O along OA.






Balancing of Several Masses in Different Planes

For complete balancing of the rotor, the resultant force, and the resultant couple both should be zero.
If resultant force and couple are not zero, then mass placed in reference plane may satisfy force equation, but
for couple equation to be balanced, two forces in different transverse planes are required.
&#3627408474;
1&#3627408531;
1&#3627409172;
2
+&#3627408474;
2&#3627408531;
2&#3627409172;
2
+&#3627408474;
3&#3627408531;
3&#3627409172;
2
+&#3627408474;
&#3627408464;1&#3627408531;
&#3627408464;1&#3627409172;
2
+&#3627408474;
&#3627408464;2&#3627408531;
&#3627408464;2&#3627409172;
2
=0
&#3627408474;
1&#3627408531;
1+&#3627408474;
2&#3627408531;
2+&#3627408474;
3&#3627408531;
3+&#3627408474;
&#3627408464;1&#3627408531;
&#3627408464;1+&#3627408474;
&#3627408464;2&#3627408531;
&#3627408464;2= 0
??????&#3627408474;&#3627408531;+&#3627408474;
&#3627408464;1&#3627408531;
&#3627408464;1+&#3627408474;
&#3627408464;2&#3627408531;
&#3627408464;2=0
Let the counter masses be placed in transverse planes at axial locations at O & Q.
Taking moments about O,
&#3627408474;
1&#3627408531;
1&#3627408473;
1&#3627409172;
2
+&#3627408474;
2&#3627408531;
2&#3627408473;
2&#3627409172;
2
+&#3627408474;
3&#3627408531;
3&#3627408473;
3&#3627409172;
2
+&#3627408474;
&#3627408464;2&#3627408531;
&#3627408464;2&#3627408473;
&#3627408464;2&#3627409172;
2
=0
&#3627408474;
1&#3627408531;
1&#3627408473;
1+&#3627408474;
2&#3627408531;
2&#3627408473;
2+&#3627408474;
3&#3627408531;
3&#3627408473;
3+&#3627408474;
&#3627408464;2&#3627408531;
&#3627408464;2&#3627408473;
&#3627408464;2=0
??????&#3627408474;&#3627408531;&#3627408473;
1+&#3627408474;
&#3627408464;2&#3627408531;
&#3627408464;2&#3627408473;
&#3627408464;2=0

39


This can be also solved analytically,
??????&#3627408474;&#3627408531;&#3627408473;
1&#3627408464;&#3627408476;&#3627408480;&#3627409155;+&#3627408474;
&#3627408464;2&#3627408531;
&#3627408464;2&#3627408473;
&#3627408464;2&#3627408464;&#3627408476;&#3627408480;&#3627409155;
&#3627408438;2=0
??????&#3627408474;&#3627408531;&#3627408473;
1&#3627408480;&#3627408470;&#3627408475;&#3627409155;+&#3627408474;
&#3627408464;2&#3627408531;
&#3627408464;2&#3627408473;
&#3627408464;2&#3627408480;&#3627408470;&#3627408475;&#3627409155;
&#3627408438;2=0
??????&#3627408474;&#3627408531;&#3627408473;&#3627408464;&#3627408476;&#3627408480;&#3627409155;=−&#3627408474;
&#3627408464;2&#3627408531;
&#3627408464;2&#3627408473;
&#3627408464;2&#3627408464;&#3627408476;&#3627408480;&#3627409155;
&#3627408438;2
??????&#3627408474;&#3627408531;&#3627408473;&#3627408480;&#3627408470;&#3627408475;&#3627409155;=−&#3627408474;
&#3627408464;2&#3627408531;
&#3627408464;2&#3627408473;
&#3627408464;2&#3627408480;&#3627408470;&#3627408475;&#3627409155;
&#3627408438;2
&#3627408474;
&#3627408464;2&#3627408531;
&#3627408464;2&#3627408473;
&#3627408464;2=√(??????&#3627408474;&#3627408531;&#3627408473;&#3627408464;&#3627408476;&#3627408480;&#3627409155;)
2
+(??????&#3627408474;&#3627408531;&#3627408473;&#3627408480;&#3627408470;&#3627408475;&#3627409155;)
2

&#3627408481;&#3627408462;&#3627408475;&#3627409155;
&#3627408438;2=
−??????&#3627408474;&#3627408531;&#3627408473;
1&#3627408480;&#3627408470;&#3627408475;&#3627409155;
−??????&#3627408474;&#3627408531;&#3627408473;
1&#3627408464;&#3627408476;&#3627408480;&#3627409155;


Substituting the value of m2 & C2 in above equations, we get
&#3627408474;
&#3627408464;1&#3627408531;
&#3627408464;1=√(??????&#3627408474;&#3627408531;&#3627408464;&#3627408476;&#3627408480;&#3627409155;+&#3627408474;
&#3627408464;2&#3627408531;
&#3627408464;2&#3627408464;&#3627408476;&#3627408480;&#3627409155;
&#3627408438;2)
2
+(??????&#3627408474;&#3627408531;&#3627408480;&#3627408470;&#3627408475;&#3627409155;+&#3627408474;
&#3627408464;2&#3627408531;
&#3627408464;2&#3627408480;&#3627408470;&#3627408475;&#3627409155;
&#3627408438;2)
2

&#3627408481;&#3627408462;&#3627408475;&#3627409155;
&#3627408438;1=
−(??????&#3627408474;&#3627408531;&#3627408480;&#3627408470;&#3627408475;&#3627409155;+&#3627408474;
&#3627408464;2&#3627408531;
&#3627408464;2&#3627408480;&#3627408470;&#3627408475;&#3627409155;
&#3627408438;2)
−(??????&#3627408474;&#3627408531;&#3627408464;&#3627408476;&#3627408480;&#3627409155;+&#3627408474;
&#3627408464;2&#3627408531;
&#3627408464;2&#3627408464;&#3627408476;&#3627408480;&#3627409155;
&#3627408438;2)

40

BALANCING OF RECIPROCATING MASS
&#3627408467;=&#3627408479;&#3627409172;
2
(&#3627408464;&#3627408476;&#3627408480;&#3627409155;+
&#3627408464;&#3627408476;&#3627408480;2&#3627409155;
&#3627408475;
)
&#3627408441;
&#3627408453;=&#3627408474;&#3627408467;=&#3627408474;&#3627408479;&#3627409172;
2
(&#3627408464;&#3627408476;&#3627408480;&#3627409155;+
&#3627408464;&#3627408476;&#3627408480;2&#3627409155;
&#3627408475;
)
&#3627408441;
&#3627408453;=&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;&#3627409155;+&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;2&#3627409155;
&#3627408475;

FR=Force required to accelerate the
reciprocating parts.
FI = Inertia force due to reciprocating parts
FN = Force on the sides of the cylinder walls
or normal force acting on the crosshead
guides
FB = Force acting on the crankshaft bearing
or main bearing.

FI & FR are balanced and FBH is unbalanced and acting along OA (FBH = FU) (FU = unbalanced force = FI=FR)
There will be an unbalanced force & unbalanced couple caused by FN & FBV (unbalanced couple = &#3627408441;
??????×&#3627408485; =
&#3627408441;
&#3627408437;??????×&#3627408485;)
Both FU and unbalanced couple vary in magnitude while rotating and causes serious vibration.
→ &#3627408526;&#3627408531;??????
&#3627409360;
&#3627408516;&#3627408528;&#3627408532;?????? is called primary unbalancing force and &#3627408526;&#3627408531;??????
&#3627409360;
&#3627408516;&#3627408528;&#3627408532;&#3627409360;??????
&#3627408527;
is called secondary unbalancing force.
Partial Balancing of Unbalanced Primary Force in a Reciprocating Engine
The primary unbalanced force (m⋅ω
2
⋅rcosθ) may be considered as the component of the centrifugal force
produced by a rotating mass m placed at the crank radius r.

B= mass of balancing force
b = distance of balancing force



We placed of mass of B, at b distance.
&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;&#3627409155;=&#3627408437;&#3627409172;
2
&#3627408463;&#3627408464;&#3627408476;&#3627408480;&#3627409155;
&#3627408526;&#3627408531;=&#3627408489;&#3627408515;
But still vertical force of mass B is not balanced (Bω
2
b sin) and there will be to-fro motion of system.
So, there will be only partial balancing of the system (B will be c.m & b=r)

41


Unbalanced force along the line of stroke =&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;&#3627409155;−&#3627408516;.&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;&#3627409155;
=(1−&#3627408516;)&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;&#3627409155;
Unbalanced force along the perpendicular to the line of stroke=c.mrω
2
sin.
Resultant unbalanced force at any instant=√((1−&#3627408516;)&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;&#3627409155;)
2
+(&#3627408516;.&#3627408474;&#3627408479;&#3627409172;
2
&#3627408480;&#3627408470;&#3627408475;&#3627409155;.)
2
Effect of Partial Balancing of Reciprocating Parts of Two Cylinder Locomotives
The effect of an unbalanced primary force along the line of stroke is to produce.
1. Variation in tractive force along the line of stroke
2. Swaying couple.
3. Hammer blow

A single or uncoupled locomotive is one, in which the effort is transmitted to one pair of the wheels only;
whereas in coupled locomotives, the driving wheels are connected to the leading and trailing wheel by an
outside coupling rod.

Hammer Blow
The effect of an unbalanced primary force perpendicular to the line of stroke is to produce variation in pressure
on the rails, which results in hammering action on the rails.
The maximum magnitude of the unbalanced force along the perpendicular to the line of stroke is known as a
hammer blow. Its value is mrω
2
.

Variation of Tractive force
The resultant unbalanced force due to the two cylinders, along the line of stroke, is known as tractive force.
Since the crank for the second cylinder is at right angle to the first crank, therefore the angle of inclination for
the second crank will be (90° + θ).
We know that unbalanced force along cylinder 1 =(1−&#3627408464;)&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;&#3627409155;
Unbalanced force along cylinder 2=(1−&#3627408464;)&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;(90+&#3627409155;)= −(1−&#3627408464;)&#3627408474;&#3627408479;&#3627409172;
2
&#3627408480;&#3627408470;&#3627408475;&#3627409155;

&#3627408441;
&#3627408455;=&#3627408479;&#3627408466;&#3627408480;&#3627408482;&#3627408473;&#3627408481;&#3627408462;&#3627408475;&#3627408481; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466; &#3627408462;&#3627408473;&#3627408476;&#3627408475;&#3627408468; &#3627408481;ℎ&#3627408466; &#3627408473;&#3627408470;&#3627408475;&#3627408466;= (1−&#3627408464;)&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;&#3627409155;−(1−&#3627408464;)&#3627408474;&#3627408479;&#3627409172;
2
&#3627408480;&#3627408470;&#3627408475;&#3627409155;=(&#3627409359;−&#3627408516;)&#3627408526;&#3627408531;??????
&#3627409360;
(&#3627408516;&#3627408528;&#3627408532;−&#3627408532;&#3627408522;&#3627408527;??????)
&#3627408448;&#3627408462;&#3627408485;&#3627408470;&#3627408474;&#3627408482;&#3627408474; &#3627408462;&#3627408475;&#3627408465; &#3627408448;&#3627408470;&#3627408475;&#3627408470;&#3627408474;&#3627408482;&#3627408474; &#3627408481;&#3627408479;&#3627408462;&#3627408464;&#3627408481;&#3627408470;&#3627408483;&#3627408466; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;= ±√&#3627409360;(&#3627409359;−&#3627408516;)&#3627408526;&#3627408531;??????
&#3627409360;

42

Swaying Couple
The unbalanced forces along the line of stroke for the two cylinders constitute a couple about the centre YY
between the cylinders. This couple has swaying effect about a vertical axis, and tends to sway the engine
alternately in clockwise and anticlockwise directions. Hence the couple is known as swaying couple.

&#3627408454;&#3627408484;&#3627408462;&#3627408486;&#3627408470;&#3627408475;&#3627408468; &#3627408438;&#3627408476;&#3627408482;&#3627408477;&#3627408473;&#3627408466;=(1−&#3627408464;)&#3627408474;&#3627409172;
2
&#3627408479;&#3627408464;&#3627408476;&#3627408480;&#3627409155;×
&#3627408462;
2
− (1−&#3627408464;)&#3627408474;&#3627409172;
2
&#3627408479;&#3627408464;&#3627408476;&#3627408480;(90+&#3627409155;)×
&#3627408462;
2

&#3627408454;&#3627408484;&#3627408462;&#3627408486;&#3627408470;&#3627408475;&#3627408468; &#3627408438;&#3627408476;&#3627408482;&#3627408477;&#3627408473;&#3627408466;=(&#3627409359;−&#3627408516;)&#3627408526;??????
&#3627409360;
&#3627408531;(&#3627408516;&#3627408528;&#3627408532;??????+&#3627408532;&#3627408522;&#3627408527;??????)×
&#3627408514;
&#3627409360;

&#3627408448;&#3627408462;&#3627408485;&#3627408470;&#3627408474;&#3627408482;&#3627408474; &#3627408462;&#3627408475;&#3627408465; &#3627408448;&#3627408470;&#3627408475;&#3627408470;&#3627408474;&#3627408482;&#3627408474; &#3627408454;&#3627408484;&#3627408462;&#3627408486;&#3627408470;&#3627408475;&#3627408468; &#3627408438;&#3627408476;&#3627408482;&#3627408477;&#3627408473;&#3627408466;= ±
&#3627408514;
√&#3627409360;
(&#3627409359;−&#3627408516;)&#3627408526;??????
&#3627409360;


Secondary Balancing
&#3627408454;&#3627408466;&#3627408464;&#3627408476;&#3627408475;&#3627408465;&#3627408462;&#3627408479;&#3627408486; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466; =&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;2&#3627409155;
&#3627408475;

&#3627408444;&#3627408481; &#3627408464;&#3627408462;&#3627408475; &#3627408462;&#3627408473;&#3627408480;&#3627408476; &#3627408463;&#3627408466; &#3627408484;&#3627408479;&#3627408470;&#3627408481;&#3627408481;&#3627408466;&#3627408475; &#3627408462;&#3627408480; =&#3627408474;(
&#3627408479;
4&#3627408475;
)(2&#3627409172;)
2
&#3627408464;&#3627408476;&#3627408480;2&#3627409155;
The effect of secondary forces is equivalent to an imaginary crank of length ‘r/4n’ rotating at twice the angular
speed.
It is equal to component of primary force along the length of stroke.


Balancing of Inline Engines
The following two conditions must be satisfied to give the primary balance of the reciprocating parts of a multi-
cylinder engine,
1. The algebraic sum of the primary forces must be equal to zero. In other words, the primary force polygon
must close.
2. The algebraic sum of the couples about any point in the plane of the primary forces must be equal to zero. In
other words, the primary couple polygon must close.
The reciprocating mass is transferred to crank pin to give the primary balance of the
reciprocating engine, which is along the line of stroke and treated as revolving masses.

43


&#3627408451;&#3627408479;&#3627408470;&#3627408474;&#3627408462;&#3627408479;&#3627408486; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;= ∑&#3627408474;&#3627409172;
2
&#3627408479;&#3627408464;&#3627408476;&#3627408480;&#3627409155;
&#3627408451;&#3627408479;&#3627408470;&#3627408474;&#3627408462;&#3627408479;&#3627408486; &#3627408464;&#3627408476;&#3627408482;&#3627408477;&#3627408473;&#3627408466;= ∑&#3627408474;&#3627409172;
2
&#3627408479;&#3627408473;&#3627408464;&#3627408476;&#3627408480;&#3627409155;
&#3627408454;&#3627408466;&#3627408464;&#3627408476;&#3627408475;&#3627408465;&#3627408462;&#3627408479;&#3627408486; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;=∑&#3627408474;(
&#3627408479;
4&#3627408475;
)(2&#3627409172;)
2
&#3627408464;&#3627408476;&#3627408480;2&#3627409155;
&#3627408454;&#3627408466;&#3627408464;&#3627408476;&#3627408475;&#3627408465;&#3627408462;&#3627408479;&#3627408486; &#3627408438;&#3627408476;&#3627408482;&#3627408477;&#3627408473;&#3627408466;
= ∑&#3627408474;(
&#3627408479;
4&#3627408475;
)(2&#3627409172;)
2
&#3627408473;&#3627408464;&#3627408476;&#3627408480;2&#3627409155;


Three-cylinder inline engine with crank offset of 120˚
&#3627408451;&#3627408479;&#3627408470;&#3627408474;&#3627408462;&#3627408479;&#3627408486; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;= ∑&#3627408474;&#3627409172;
2
&#3627408479;&#3627408464;&#3627408476;&#3627408480;&#3627409155;=&#3627408474;&#3627409172;
2
&#3627408479;&#3627408464;&#3627408476;&#3627408480;&#3627409155;+&#3627408474;&#3627409172;
2
&#3627408479;&#3627408464;&#3627408476;&#3627408480;(120+&#3627409155;)+&#3627408474;&#3627409172;
2
&#3627408479;&#3627408464;&#3627408476;&#3627408480;(240+&#3627409155;)=0
&#3627408454;&#3627408466;&#3627408464;&#3627408476;&#3627408475;&#3627408465;&#3627408462;&#3627408479;&#3627408486; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;=∑&#3627408474;(
&#3627408479;
4&#3627408475;
)(2&#3627409172;)
2
&#3627408464;&#3627408476;&#3627408480;2&#3627409155;=&#3627408474;(
&#3627408479;
4&#3627408475;
)(2&#3627409172;)
2
(&#3627408464;&#3627408476;&#3627408480;2&#3627409155;+&#3627408464;&#3627408476;&#3627408480;(240+2)+&#3627408464;&#3627408476;&#3627408480;(480+2&#3627409155;))=0
&#3627408448;
&#3627408477;=&#3627408451;&#3627408479;&#3627408470;&#3627408474;&#3627408462;&#3627408479;&#3627408486; &#3627408464;&#3627408476;&#3627408482;&#3627408477;&#3627408473;&#3627408466;=&#3627408474;&#3627408479;&#3627409172;
2
&#3627408473;&#3627408464;&#3627408476;&#3627408480;(240+&#3627409155;)−&#3627408474;&#3627408479;&#3627409172;
2
&#3627408473;&#3627408464;&#3627408476;&#3627408480;&#3627409155;= &#3627408474;&#3627408479;&#3627409172;
2
&#3627408473;(&#3627408464;&#3627408476;&#3627408480;(240+&#3627409155;)−&#3627408464;&#3627408476;&#3627408480;&#3627409155;)
&#3627408448;
&#3627408477; &#3627408526;&#3627408514;&#3627408537;= &#3627408474;&#3627408479;&#3627409172;
2
&#3627408473;(2&#3627408464;&#3627408476;&#3627408480;30)= √3&#3627408474;&#3627408479;&#3627409172;
2
&#3627408473;
&#3627408454;&#3627408466;&#3627408464;&#3627408476;&#3627408475;&#3627408465;&#3627408462;&#3627408479;&#3627408486; &#3627408438;&#3627408476;&#3627408482;&#3627408477;&#3627408473;&#3627408466;= &#3627408448;
&#3627408480;=
&#3627408474;&#3627408479;&#3627409172;
2
&#3627408473;
&#3627408475;
(&#3627408464;&#3627408476;&#3627408480;(480+2&#3627409155;)−&#3627408464;&#3627408476;&#3627408480;2&#3627409155;)
&#3627408448;
&#3627408480; &#3627408526;&#3627408514;&#3627408537;=
√3&#3627408474;&#3627408479;&#3627409172;
2
&#3627408473;
&#3627408475;


Inline 2-cylinder engine

Cranks are 180⁰ apart and have equal reciprocating masses.
&#3627408451;&#3627408479;&#3627408470;&#3627408474;&#3627408462;&#3627408479;&#3627408486; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;= &#3627408474;&#3627408479;&#3627409172;
2
[&#3627408464;&#3627408476;&#3627408480;&#3627409155;+ &#3627408464;&#3627408476;&#3627408480;(180°+&#3627409155;)]=0
&#3627408451;&#3627408479;&#3627408470;&#3627408474;&#3627408462;&#3627408479;&#3627408486; &#3627408438;&#3627408476;&#3627408482;&#3627408477;&#3627408473;&#3627408466;=&#3627408448;
&#3627408477;=&#3627408474;&#3627408479;&#3627409172;
2
[
&#3627408473;
2
&#3627408464;&#3627408476;&#3627408480;&#3627409155;+(−
&#3627408473;
2
)&#3627408464;&#3627408476;&#3627408480;(180
0
+&#3627409155;)]= &#3627408474;&#3627408479;&#3627409172;
2
&#3627408473;&#3627408464;&#3627408476;&#3627408480;&#3627409155;
&#3627408448;
&#3627408477; &#3627408526;&#3627408514;&#3627408537;=&#3627408526;&#3627408531;??????
&#3627409360;
&#3627408525; &#3627408462;&#3627408481; &#3627409155;=0° & 180°

44

&#3627408454;&#3627408466;&#3627408464;&#3627408476;&#3627408475;&#3627408465;&#3627408462;&#3627408479;&#3627408486; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;=&#3627408441;
&#3627408480;=
&#3627408474;&#3627408479;&#3627409172;
2
&#3627408475;
[&#3627408464;&#3627408476;&#3627408480;2&#3627409155;+&#3627408464;&#3627408476;&#3627408480;(360°+2&#3627409155;)]=
2&#3627408474;&#3627408479;&#3627409172;
2
&#3627408475;
&#3627408464;&#3627408476;&#3627408480;2&#3627409155;
&#3627408441;
&#3627408480; &#3627408526;&#3627408514;&#3627408537;=
2&#3627408474;&#3627408479;&#3627409172;
2
&#3627408475;
&#3627408484;ℎ&#3627408466;&#3627408475; &#3627409155;=0°,90°,180°,270°.
&#3627408454;&#3627408466;&#3627408464;&#3627408476;&#3627408475;&#3627408465;&#3627408462;&#3627408479;&#3627408486; &#3627408438;&#3627408476;&#3627408482;&#3627408477;&#3627408473;&#3627408466;= &#3627408448;
&#3627408480;= &#3627408474;&#3627408479;&#3627409172;
2
[
&#3627408473;
2
&#3627408464;&#3627408476;&#3627408480;&#3627409155;+(−
&#3627408473;
2
)&#3627408464;&#3627408476;&#3627408480;(360
0
+&#3627409155;)]=0
Inline four-cylinder Four Stroke Engine
&#3627408451;&#3627408479;&#3627408470;&#3627408474;&#3627408462;&#3627408479;&#3627408486; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;=&#3627408474;&#3627408479;&#3627409172;
2
[&#3627408464;&#3627408476;&#3627408480;&#3627409155;+&#3627408464;&#3627408476;&#3627408480;(180+&#3627409155;)+&#3627408464;&#3627408476;&#3627408480;(180+&#3627409155;)+&#3627408464;&#3627408476;&#3627408480;&#3627409155;]=0
&#3627408451;&#3627408479;&#3627408470;&#3627408474;&#3627408462;&#3627408479;&#3627408486; &#3627408438;&#3627408476;&#3627408482;&#3627408477;&#3627408473;&#3627408466;=&#3627408474;&#3627408479;&#3627409172;
2
[
3&#3627408473;
2
&#3627408464;&#3627408476;&#3627408480;&#3627409155;+
&#3627408473;
2
&#3627408464;&#3627408476;&#3627408480;(180+&#3627409155;)+(−
3&#3627408473;
2
&#3627408464;&#3627408476;&#3627408480;&#3627409155;)+(−
&#3627408473;
2
&#3627408464;&#3627408476;&#3627408480;(180+&#3627409155;))]=0
&#3627408454;&#3627408466;&#3627408464;&#3627408476;&#3627408475;&#3627408465;&#3627408462;&#3627408479;&#3627408486; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;=&#3627408441;
&#3627408480;=
&#3627408474;&#3627408479;&#3627409172;
2
&#3627408475;
[&#3627408464;&#3627408476;&#3627408480;2&#3627409155;+&#3627408464;&#3627408476;&#3627408480;2(180+&#3627409155;)+&#3627408464;&#3627408476;&#3627408480;2(180+&#3627409155;)+&#3627408464;&#3627408476;&#3627408480;2&#3627409155;]=
4&#3627408474;&#3627408479;&#3627409172;
2
&#3627408475;
&#3627408464;&#3627408476;&#3627408480;2&#3627409155;
&#3627408441;
&#3627408480; &#3627408526;&#3627408514;&#3627408537;=
4&#3627408474;&#3627408479;&#3627409172;
2
&#3627408475;
&#3627408462;&#3627408481;  =0˚,90˚,180˚,270˚
&#3627408454;&#3627408466;&#3627408464;&#3627408476;&#3627408475;&#3627408465;&#3627408462;&#3627408479;&#3627408486; &#3627408438;&#3627408476;&#3627408482;&#3627408477;&#3627408473;&#3627408466;=&#3627408474;&#3627408479;&#3627409172;
2
[
3&#3627408473;
2
&#3627408464;&#3627408476;&#3627408480;&#3627409155;+
&#3627408473;
2
&#3627408464;&#3627408476;&#3627408480;2(180+&#3627409155;)+(−
3&#3627408473;
2
&#3627408464;&#3627408476;&#3627408480;2&#3627409155;)+(−
&#3627408473;
2
&#3627408464;&#3627408476;&#3627408480;2(180+&#3627409155;))]=0

45

Six-cylinder four stroke Engine

46

Balancing of radial engines
The method of direct and reverse cranks is used in balancing of radial or V-engines, in which
the connecting rods are connected to a common crank.
The indirect or reverse crank OC′ is the image of the
direct crank OC, when seen through the mirror placed
at the line of stroke. When the direct crank revolves in
a clockwise direction, the reverse crank will revolve in
the anticlockwise direction.

Primary forces
Now let us suppose that the mass (m) of the
reciprocating parts is divided into two parts,
each equal to m / 2.



&#3627408464;&#3627408476;&#3627408474;&#3627408477;&#3627408476;&#3627408475;&#3627408466;&#3627408475;&#3627408481; &#3627408476;&#3627408467; &#3627408464;&#3627408466;&#3627408475;&#3627408481;&#3627408479;&#3627408470;&#3627408467;&#3627408482;&#3627408468;&#3627408462;&#3627408473; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466; &#3627408462;&#3627408473;&#3627408476;&#3627408475;&#3627408468; &#3627408502;&#3627408503; &#3627408467;&#3627408476;&#3627408479; &#3627408477;&#3627408479;&#3627408470;&#3627408474;&#3627408462;&#3627408479;&#3627408486; &#3627408465;&#3627408470;&#3627408479;&#3627408466;&#3627408464;&#3627408481; &#3627408464;&#3627408479;&#3627408462;&#3627408475;&#3627408472; =
&#3627408474;
2
&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;&#3627409155;
&#3627408464;&#3627408476;&#3627408474;&#3627408477;&#3627408476;&#3627408475;&#3627408466;&#3627408475;&#3627408481; &#3627408476;&#3627408467; &#3627408464;&#3627408466;&#3627408475;&#3627408481;&#3627408479;&#3627408470;&#3627408467;&#3627408482;&#3627408468;&#3627408462;&#3627408473; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466; &#3627408462;&#3627408473;&#3627408476;&#3627408475;&#3627408468; &#3627408502;&#3627408503; &#3627408467;&#3627408476;&#3627408479; &#3627408477;&#3627408479;&#3627408470;&#3627408474;&#3627408462;&#3627408479;&#3627408486; &#3627408479;&#3627408466;&#3627408483;&#3627408466;&#3627408479;&#3627408480;&#3627408466; &#3627408464;&#3627408479;&#3627408462;&#3627408475;&#3627408472; =
&#3627408474;
2
&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;&#3627409155;
&#3627408455;&#3627408476;&#3627408481;&#3627408462;&#3627408473; &#3627408464;&#3627408466;&#3627408475;&#3627408481;&#3627408479;&#3627408470;&#3627408467;&#3627408482;&#3627408468;&#3627408462;&#3627408473; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466; &#3627408462;&#3627408473;&#3627408476;&#3627408475;&#3627408468; &#3627408502;&#3627408503;=&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;&#3627409155;
Component of centrifugal force perpendicular to OP are balanced.
Secondary forces
&#3627408454;&#3627408466;&#3627408464;&#3627408476;&#3627408475;&#3627408465;&#3627408462;&#3627408479;&#3627408486; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;&#3627408480;=&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;2??????
&#3627408475;
=&#3627408474;(2&#3627409172;)
2
&#3627408479;
4&#3627408475;
&#3627408464;&#3627408476;&#3627408480;2&#3627409155;
Similar to primary balancing, masses are assumed
to be m/2 at D and D’.
Secondary direct crank and rotates at 2ω rad/s in
the clockwise direction, while the crank OD′ is the
secondary reverse crank and rotates at 2ω rad/s
in the anticlockwise direction

m/2
m/2

47

Balancing of V-type engines

&#3627408451;&#3627408479;&#3627408470;&#3627408474;&#3627408462;&#3627408479;&#3627408486; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;&#3627408480; &#3627408462;&#3627408473;&#3627408476;&#3627408475;&#3627408468; &#3627408473;&#3627408470;&#3627408475;&#3627408466; &#3627408476;&#3627408467; &#3627408480;&#3627408481;&#3627408479;&#3627408476;&#3627408472;&#3627408466; &#3627408467;&#3627408476;&#3627408479; &#3627408464;&#3627408486;&#3627408473;&#3627408470;&#3627408475;&#3627408465;&#3627408466;&#3627408479; 1=&#3627408441;
??????1=&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;(&#3627409148;−&#3627409155;)
&#3627408464;&#3627408476;&#3627408474;&#3627408477;&#3627408476;&#3627408475;&#3627408466;&#3627408475;&#3627408481; &#3627408476;&#3627408467; &#3627408441;
??????1 &#3627408462;&#3627408473;&#3627408476;&#3627408475;&#3627408468; &#3627408502;&#3627408512;=&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;(&#3627409148;−&#3627409155;)&#3627408464;&#3627408476;&#3627408480;&#3627409148;
&#3627408464;&#3627408476;&#3627408474;&#3627408477;&#3627408476;&#3627408475;&#3627408466;&#3627408475;&#3627408481; &#3627408476;&#3627408467; &#3627408441;
??????1 &#3627408462;&#3627408473;&#3627408476;&#3627408475;&#3627408468; &#3627408502;&#3627408511;=&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;(&#3627409148;−&#3627409155;)&#3627408480;&#3627408470;&#3627408475;&#3627409148;
&#3627408451;&#3627408479;&#3627408470;&#3627408474;&#3627408462;&#3627408479;&#3627408486; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;&#3627408480; &#3627408462;&#3627408473;&#3627408476;&#3627408475;&#3627408468; &#3627408473;&#3627408470;&#3627408475;&#3627408466; &#3627408476;&#3627408467; &#3627408480;&#3627408481;&#3627408479;&#3627408476;&#3627408472;&#3627408466; &#3627408467;&#3627408476;&#3627408479; &#3627408464;&#3627408486;&#3627408473;&#3627408470;&#3627408475;&#3627408465;&#3627408466;&#3627408479; 2=&#3627408441;
??????2=&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;(&#3627409148;+&#3627409155;)
&#3627408464;&#3627408476;&#3627408474;&#3627408477;&#3627408476;&#3627408475;&#3627408466;&#3627408475;&#3627408481; &#3627408476;&#3627408467; &#3627408441;
??????2 &#3627408462;&#3627408473;&#3627408476;&#3627408475;&#3627408468; &#3627408502;&#3627408512;=&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;(&#3627409148;+&#3627409155;)&#3627408464;&#3627408476;&#3627408480;&#3627409148;
&#3627408464;&#3627408476;&#3627408474;&#3627408477;&#3627408476;&#3627408475;&#3627408466;&#3627408475;&#3627408481; &#3627408476;&#3627408467; &#3627408441;
??????2 &#3627408462;&#3627408473;&#3627408476;&#3627408475;&#3627408468; &#3627408502;&#3627408511;′=&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;(&#3627409148;+&#3627409155;)&#3627408480;&#3627408470;&#3627408475;&#3627409148;
&#3627408507;&#3627408528;&#3627408533;&#3627408514;&#3627408525; &#3627408503;&#3627408531;&#3627408522;&#3627408526;&#3627408514;&#3627408531;&#3627408538; &#3627408519;&#3627408528;&#3627408531;&#3627408516;&#3627408518;&#3627408532; &#3627408514;&#3627408525;&#3627408528;&#3627408527;&#3627408520; &#3627408502;&#3627408512;= &#3627408493;
&#3627408503;??????=&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;(&#3627409148;−&#3627409155;)&#3627408464;&#3627408476;&#3627408480;&#3627409148;+&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;(&#3627409148;+&#3627409155;)&#3627408464;&#3627408476;&#3627408480;&#3627409148;
=&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;&#3627409148;(&#3627408464;&#3627408476;&#3627408480;(&#3627409148;−&#3627409155;)+&#3627408464;&#3627408476;&#3627408480;(&#3627409148;+&#3627409155;))
=&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;&#3627409148;(2×&#3627408464;&#3627408476;&#3627408480;&#3627409148;×&#3627408464;&#3627408476;&#3627408480;&#3627409155;)
=2&#3627408474;&#3627408479;&#3627409172;
2
.&#3627408464;&#3627408476;&#3627408480;
2
&#3627409148;.&#3627408464;&#3627408476;&#3627408480;&#3627409155;
&#3627408507;&#3627408528;&#3627408533;&#3627408514;&#3627408525; &#3627408503;&#3627408531;&#3627408522;&#3627408526;&#3627408514;&#3627408531;&#3627408538; &#3627408519;&#3627408528;&#3627408531;&#3627408516;&#3627408518;&#3627408532; &#3627408514;&#3627408525;&#3627408528;&#3627408527;&#3627408520; &#3627408502;&#3627408511;=&#3627408493;
&#3627408503;&#3627408495;=&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;(&#3627409148;−&#3627409155;)&#3627408480;&#3627408470;&#3627408475;&#3627409148;−&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;(&#3627409148;+&#3627409155;)&#3627408480;&#3627408470;&#3627408475;&#3627409148;
=&#3627408474;&#3627408479;&#3627409172;
2
&#3627408480;&#3627408470;&#3627408475;&#3627409148;(&#3627408464;&#3627408476;&#3627408480;(&#3627409148;−&#3627409155;)−&#3627408464;&#3627408476;&#3627408480;(&#3627409148;+&#3627409155;))
=&#3627408474;&#3627408479;&#3627409172;
2
&#3627408480;&#3627408470;&#3627408475;&#3627409148;(2×&#3627408480;&#3627408470;&#3627408475;&#3627409148;&#3627408480;&#3627408470;&#3627408475; )
=2.&#3627408474;&#3627408479;&#3627409172;
2
.&#3627408480;&#3627408470;&#3627408475;
2
&#3627409148;.&#3627408480;&#3627408470;&#3627408475;&#3627409155;
&#3627408505;&#3627408518;&#3627408532;&#3627408534;&#3627408525;&#3627408533;&#3627408514;&#3627408527;&#3627408533; &#3627408503;&#3627408531;&#3627408522;&#3627408526;&#3627408514;&#3627408531;&#3627408538; &#3627408519;&#3627408528;&#3627408531;&#3627408516;&#3627408518;= √(&#3627408441;
????????????)
2
+(&#3627408441;
??????&#3627408443;)
2
=2.&#3627408474;&#3627408479;&#3627409172;
2
√(&#3627408464;&#3627408476;&#3627408480;
2
&#3627409148;.&#3627408464;&#3627408476;&#3627408480;&#3627409155;)
2
+(&#3627408480;&#3627408470;&#3627408475;
2
&#3627409148;.&#3627408480;&#3627408470;&#3627408475;&#3627409155;)
2

&#3627408454;&#3627408466;&#3627408464;&#3627408476;&#3627408475;&#3627408465;&#3627408462;&#3627408479;&#3627408486; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;&#3627408480; &#3627408462;&#3627408473;&#3627408476;&#3627408475;&#3627408468; &#3627408473;&#3627408470;&#3627408475;&#3627408466; &#3627408476;&#3627408467; &#3627408480;&#3627408481;&#3627408479;&#3627408476;&#3627408472;&#3627408466; &#3627408467;&#3627408476;&#3627408479; &#3627408464;&#3627408486;&#3627408473;&#3627408470;&#3627408475;&#3627408465;&#3627408466;&#3627408479; 1=&#3627408441;
&#3627408454;1=&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;2(&#3627409148;−&#3627409155;)
&#3627408475;

&#3627408438;&#3627408476;&#3627408474;&#3627408477;&#3627408476;&#3627408475;&#3627408466;&#3627408475;&#3627408481; &#3627408476;&#3627408467; &#3627408441;
&#3627408454;1 &#3627408462;&#3627408473;&#3627408476;&#3627408475;&#3627408468; &#3627408502;&#3627408512;=&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;2(&#3627409148;−&#3627409155;)
&#3627408475;
&#3627408464;&#3627408476;&#3627408480;&#3627409148;
&#3627408438;&#3627408476;&#3627408474;&#3627408477;&#3627408476;&#3627408475;&#3627408466;&#3627408475;&#3627408481; &#3627408476;&#3627408467; &#3627408441;
??????1 &#3627408462;&#3627408473;&#3627408476;&#3627408475;&#3627408468; &#3627408502;&#3627408511;=&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;2(&#3627409148;−&#3627409155;)
&#3627408475;
&#3627408480;&#3627408470;&#3627408475;&#3627409148;
&#3627408454;&#3627408466;&#3627408464;&#3627408476;&#3627408475;&#3627408465;&#3627408462;&#3627408479;&#3627408486; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;&#3627408480; &#3627408462;&#3627408473;&#3627408476;&#3627408475;&#3627408468; &#3627408473;&#3627408470;&#3627408475;&#3627408466; &#3627408476;&#3627408467; &#3627408480;&#3627408481;&#3627408479;&#3627408476;&#3627408472;&#3627408466; &#3627408467;&#3627408476;&#3627408479; &#3627408464;&#3627408486;&#3627408473;&#3627408470;&#3627408475;&#3627408465;&#3627408466;&#3627408479; 2=&#3627408441;
&#3627408454;2=&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;2(&#3627409148;+&#3627409155;)
&#3627408475;

&#3627408464;&#3627408476;&#3627408474;&#3627408477;&#3627408476;&#3627408475;&#3627408466;&#3627408475;&#3627408481; &#3627408476;&#3627408467; &#3627408441;
&#3627408454;2 &#3627408462;&#3627408473;&#3627408476;&#3627408475;&#3627408468; &#3627408502;&#3627408512;=&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;2(&#3627409148;+&#3627409155;)
&#3627408475;
&#3627408464;&#3627408476;&#3627408480;&#3627409148;
&#3627408464;&#3627408476;&#3627408474;&#3627408477;&#3627408476;&#3627408475;&#3627408466;&#3627408475;&#3627408481; &#3627408476;&#3627408467; &#3627408441;
&#3627408454;2 &#3627408462;&#3627408473;&#3627408476;&#3627408475;&#3627408468; &#3627408502;&#3627408511;′=&#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;2(&#3627409148;+&#3627409155;)
&#3627408475;
&#3627408480;&#3627408470;&#3627408475;&#3627409148;
&#3627408507;&#3627408528;&#3627408533;&#3627408514;&#3627408525; &#3627408506;&#3627408518;&#3627408516;&#3627408528;&#3627408527;&#3627408517;&#3627408514;&#3627408531;&#3627408538; &#3627408519;&#3627408528;&#3627408531;&#3627408516;&#3627408518;&#3627408532; &#3627408514;&#3627408525;&#3627408528;&#3627408527;&#3627408520; &#3627408502;&#3627408512;= &#3627408493;
&#3627408506;??????= &#3627408474;&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;&#3627409148;(
&#3627408464;&#3627408476;&#3627408480;2(&#3627409148;−&#3627409155;)
&#3627408475;
+
&#3627408464;&#3627408476;&#3627408480;2(&#3627409148;+&#3627409155;)
&#3627408475;
)=
2&#3627408474;
&#3627408475;
&#3627408479;&#3627409172;
2
&#3627408464;&#3627408476;&#3627408480;&#3627409148;&#3627408464;&#3627408476;&#3627408480;2&#3627409148;&#3627408464;&#3627408476;&#3627408480;2&#3627409155;
&#3627408507;&#3627408528;&#3627408533;&#3627408514;&#3627408525; &#3627408506;&#3627408518;&#3627408516;&#3627408528;&#3627408527;&#3627408517;&#3627408514;&#3627408531;&#3627408538; &#3627408519;&#3627408528;&#3627408531;&#3627408516;&#3627408518;&#3627408532; &#3627408514;&#3627408525;&#3627408528;&#3627408527;&#3627408520; &#3627408502;&#3627408511;= &#3627408493;
&#3627408506;&#3627408495;= &#3627408474;&#3627408479;&#3627409172;
2
&#3627408480;&#3627408470;&#3627408475;&#3627409148;(
&#3627408464;&#3627408476;&#3627408480;2(&#3627409148;−&#3627409155;)
&#3627408475;

&#3627408464;&#3627408476;&#3627408480;2(&#3627409148;+&#3627409155;)
&#3627408475;
)=
2&#3627408474;
&#3627408475;
&#3627408479;&#3627409172;
2
&#3627408480;&#3627408470;&#3627408475;&#3627409148;&#3627408480;&#3627408470;&#3627408475;2&#3627409148;&#3627408480;&#3627408470;&#3627408475;2&#3627409155;
&#3627408505;&#3627408518;&#3627408532;&#3627408534;&#3627408525;&#3627408533;&#3627408514;&#3627408527;&#3627408533; &#3627408506;&#3627408518;&#3627408516;&#3627408528;&#3627408527;&#3627408517;&#3627408514;&#3627408531;&#3627408538; &#3627408519;&#3627408528;&#3627408531;&#3627408516;&#3627408518;= √(&#3627408441;
&#3627408454;??????)
2
+(&#3627408441;
&#3627408454;&#3627408443;)
2
=
2&#3627408474;
&#3627408475;
&#3627408479;&#3627409172;
2
√(&#3627408464;&#3627408476;&#3627408480;&#3627409148;&#3627408464;&#3627408476;&#3627408480;2&#3627409148;&#3627408464;&#3627408476;&#3627408480;2&#3627409155;)
2
+(&#3627408480;&#3627408470;&#3627408475;&#3627409148;&#3627408480;&#3627408470;&#3627408475;2&#3627409148;&#3627408480;&#3627408470;&#3627408475;2&#3627409155;)
2

48

TURNING MOMENT DIAGRAMS

During 1 revolution of crank shaft, &#3627408455;=&#3627408441;&#3627408479;(&#3627408480;&#3627408470;&#3627408475;&#3627409155;+
&#3627408480;??????&#3627408475;2??????
2√&#3627408475;
2
−&#3627408480;??????&#3627408475;
2
??????
)
&#3627408455;&#3627408482;&#3627408479;&#3627408475;&#3627408470;&#3627408475;&#3627408468; &#3627408448;&#3627408476;&#3627408474;&#3627408466;&#3627408475;&#3627408481;(&#3627408455;)=&#3627408467;(&#3627409155;)=&#3627408467;&#3627408482;&#3627408475;&#3627408464;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408467; &#3627408464;&#3627408479;&#3627408462;&#3627408475;&#3627408472; &#3627408462;&#3627408475;&#3627408468;&#3627408473;&#3627408466;
F is the net piston effort, r is the crank radius,  is the crank angle.
T  constant, but we want constant ω.
&#3627408455;=&#3627408444;∝

&#3627408455;&#3627408476;&#3627408481;&#3627408462;&#3627408473; &#3627408484;&#3627408476;&#3627408479;&#3627408472; &#3627408477;&#3627408479;&#3627408476;&#3627408465;&#3627408482;&#3627408464;&#3627408466;&#3627408465;= ∫ &#3627408455;
4??????2??????⁄
0
&#3627408465;&#3627409155;
Average work produced =
Tmean×4π(2π)
&#3627408455;
&#3627408474;&#3627408466;&#3627408462;&#3627408475;=
∫ &#3627408455;
4??????2??????⁄
0
&#3627408465;&#3627409155;
4&#3627409163;(2&#3627409163;)

Tmean = mean resisting torque



The area of the turning moment diagram represents the work done per revolution. In actual practice, the engine
is assumed to work against the mean resisting torque.
➢ If (T –Tmean) is positive, the flywheel accelerates and if (T – Tmean) is negative, then the flywheel retards.

49

Fluctuation of energy
The fluctuation of energy may be determined by the turning moment diagram for one complete cycle of
operation.
The variations of energy above and below the mean resisting torque line are called fluctuations of energy. The
areas BbC, CcD, DdE, etc. represent fluctuations of energy.
The difference between the maximum and the minimum energies is known as maximum fluctuation of energy.

Maximum energy in flywheel
= E + a1
Minimum energy in the flywheel
= E + a1 – a2 + a3 – a4





Maximum fluctuation of energy,
Δ E = Maximum energy – Minimum energy
= (E + a1) – (E + a1 – a2 + a3 – a4) = a2 – a3 + a4
Coefficient of Fluctuation of Energy
It may be defined as the ratio of the maximum fluctuation of energy to the work done per cycle.
&#3627408438;
&#3627408466;=
&#3627408448;&#3627408462;&#3627408485;&#3627408470;&#3627408474;&#3627408482;&#3627408474; &#3627408467;&#3627408473;&#3627408482;&#3627408464;&#3627408481;&#3627408482;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408467; &#3627408466;&#3627408475;&#3627408466;&#3627408479;&#3627408468;&#3627408486;
&#3627408458;&#3627408476;&#3627408479;&#3627408472; &#3627408465;&#3627408476;&#3627408475;&#3627408466; &#3627408463;&#3627408486; &#3627408464;&#3627408486;&#3627408464;&#3627408473;&#3627408466;

Work done per cycle = Tmean × 
&#3627408455;
&#3627408474;&#3627408466;&#3627408462;&#3627408475;=
??????
??????
=
??????×60
2????????????
N = Speed in r.p.m

&#3627408458;&#3627408476;&#3627408479;&#3627408472; &#3627408465;&#3627408476;&#3627408475;&#3627408466; &#3627408477;&#3627408466;&#3627408479; &#3627408464;&#3627408486;&#3627408464;&#3627408473;&#3627408466;=
??????×60
&#3627408475;
n= no. of working strokes per minute

50

FLYWHEEL
➢ A flywheel used in machines serves as a reservoir, which stores energy during the period when the supply of
energy is more than the requirement and releases it during the period when the requirement of energy is
more than the supply.
➢ It is used to store the energy when the demand of energy of energy is less and deliver it when the demand of
energy is high.
➢ The excess energy developed during power stroke is absorbed by the flywheel and releases it to the crankshaft
during other strokes in which no energy is developed, thus rotating the crankshaft at a uniform speed.
➢ Hence a flywheel does not maintain a constant speed, it simply reduces the fluctuation of speed. In other
words, a flywheel controls the speed variations caused by the fluctuation of the engine turning moment
during each cycle of operation.
I= moment of inertia of flywheel, ω1= maximum speed, ω2= minimum speed, ω= mean speed,
E= kinetic energy of the flywheel at mean speed, e= maximum fluctuation of energy,
&#3627408446;=&#3627408464;&#3627408476;&#3627408466;&#3627408467;&#3627408467;&#3627408470;&#3627408464;&#3627408470;&#3627408466;&#3627408475;&#3627408481; &#3627408476;&#3627408467; &#3627408467;&#3627408473;&#3627408482;&#3627408464;&#3627408481;&#3627408482;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408467; &#3627408480;&#3627408477;&#3627408466;&#3627408466;&#3627408465;=
&#3627409172;
1−&#3627409172;
2
&#3627409172;

&#3627408466;=
1
2
&#3627408444;&#3627409172;
1
2

1
2
&#3627408444;&#3627409172;
2
2
=
1
2
&#3627408444;(&#3627409172;
1
2
−&#3627409172;
2
2
)=&#3627408444;
(&#3627409172;
1+&#3627409172;
2)
2
(&#3627409172;
1−&#3627409172;
2)=&#3627408444;&#3627409172;(&#3627409172;
1−&#3627409172;
2)=&#3627408444;&#3627409172;
2
(&#3627409172;
1−&#3627409172;
2)
&#3627409172;
=&#3627408444;&#3627409172;
2
&#3627408446;
&#3627408466;= &#3627408444;&#3627409172;
2
&#3627408446;⇒ &#3627408446;=
&#3627408466;
&#3627408444;&#3627409172;
2
=
&#3627408466;

&#3627408444;&#3627409172;
2
2
=
&#3627408466;
2&#3627408440;

&#3627408466;=2&#3627408440;&#3627408446;
Dimensions of Flywheel Rims
&#3627408438;&#3627408466;&#3627408475;&#3627408481;&#3627408479;&#3627408470;&#3627408467;&#3627408482;&#3627408468;&#3627408462;&#3627408473; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466; &#3627408476;&#3627408475; &#3627408466;&#3627408473;&#3627408466;&#3627408474;&#3627408466;&#3627408475;&#3627408481; / &#3627408482;&#3627408475;&#3627408470;&#3627408481; &#3627408473;&#3627408466;&#3627408475;&#3627408468;&#3627408481;ℎ =[&#3627409164;·(&#3627408479;.&#3627408465;)&#3627408481;].&#3627408479;&#3627409172;
2

&#3627408455;&#3627408476;&#3627408481;&#3627408462;&#3627408473; &#3627408483;&#3627408466;&#3627408479;&#3627408481;&#3627408470;&#3627408464;&#3627408462;&#3627408473; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;/ &#3627408482;&#3627408475;&#3627408470;&#3627408481; &#3627408473;&#3627408466;&#3627408475;&#3627408468;&#3627408481;ℎ =2·&#3627409164;.&#3627408479;
2
.&#3627408481;.&#3627409172;
2
For equilibrium
??????(2&#3627408481;).1=2&#3627409164;.&#3627408479;
2
.&#3627408481;.&#3627409172;
2

??????=&#3627409164;.&#3627408479;
2
.&#3627409172;
2
=&#3627409164;&#3627408483;
2

&#3627408483;=2&#3627409163;&#3627408449;60⁄
&#3627408474;=&#3627409164;.&#3627409163;.&#3627408463;.&#3627408465;.&#3627408481;
Punching press
Here flywheel is used to reduce the fluctuation of speed, when torque is
constant, and load is varying.
d - diameter of punch, t – thickness of hole, r- radius of crank shaft
Energy required to punch the plate/unit shear area = E
Total energy required for punching 1 hole = E ·(π.d.t)
Time required for 1 punching cycle = T
Avg. time required per second = power of motor(P) =
&#3627408440;??????&#3627408465;&#3627408481;
&#3627408455;

&#3627408451; = &#3627408455;&#3627408476;&#3627408479;&#3627408478;&#3627408482;&#3627408466;
&#3627408474;&#3627408466;&#3627408462;&#3627408475; × &#3627409172;
&#3627408474;&#3627408466;&#3627408462;&#3627408475;
Actual punching time = Tp
Energy given by motor during punching=
&#3627408440;??????&#3627408465;&#3627408481;
&#3627408455;
×&#3627408455;
&#3627408477;

??????&#3627408440;=&#3627408440;&#3627409163;&#3627408465;&#3627408481;−
&#3627408440;&#3627409163;&#3627408465;&#3627408481;
&#3627408455;
×&#3627408455;
&#3627408477;
??????&#3627408440;=&#3627408440;&#3627409163;&#3627408465;&#3627408481;[1−
&#3627408481;
4&#3627408479;
]

51

CAMS
A cam is a rotating machine element which gives reciprocating or oscillating motion to another element
known as follower.
Classification of Cams & Followers
Classification of Followers
Surface in Contact Motion of the follower Path of motion of follower
Knife edge follower Reciprocating or translating follower Radial follower
Roller follower Oscillating or rotating follower Offset follower
Mushroom follower
Spherical faced follower
Classification of Cams
Shape Follower Movement Manner of Constraint of Follower
Wedge and Flat cams Rise – Return – Rise Pre – loaded Spring cam
Radial of Disc Cams Dwell – Rise – Return – Dwell Positive – drive Cam
Spiral cams Dwell – Rise – Dwell – Return – Return Gravity cam
Cylindrical Cams
Conjugate Cams
Globodial Cams
Cam Nomenclature

• By giving offset to line of follower w.r.t cam center, pressure angle can be reduced there by reducing side
thrust.
• When base circle size increases, pressure angle reduces.
• When dwell period increases, pressure angle reduces. (dwell period– follower remains at rest)

52

Follower motion programming
&#3627408454;=&#3627408470;&#3627408475;&#3627408480;&#3627408481;&#3627408462;&#3627408475;&#3627408481;&#3627408462;&#3627408475;&#3627408466;&#3627408476;&#3627408482;&#3627408480; &#3627408467;&#3627408476;&#3627408473;&#3627408473;&#3627408476;&#3627408484;&#3627408466;&#3627408479; &#3627408465;&#3627408470;&#3627408480;&#3627408477;&#3627408473;&#3627408462;&#3627408464;&#3627408466;&#3627408474;&#3627408466;&#3627408475;&#3627408481;=&#3627408467;() ( -> cam rotation angle)
&#3627408483;=
&#3627408465;&#3627408480;
&#3627408465;&#3627408481;
=
&#3627408465;&#3627408480;
&#3627408465;
×
&#3627408465;
&#3627408465;&#3627408481;

&#3627408465;&#3627408480;
&#3627408465;&#3627408481;
=&#3627408477;ℎ&#3627408486;&#3627408480;&#3627408470;&#3627408464;&#3627408462;&#3627408473; &#3627408481;&#3627408470;&#3627408474;&#3627408466; &#3627408465;&#3627408466;&#3627408479;&#3627408470;&#3627408483;&#3627408462;&#3627408481;&#3627408470;&#3627408483;&#3627408466;,
&#3627408465;&#3627408480;
&#3627408465;
=&#3627408472;&#3627408470;&#3627408475;&#3627408466;&#3627408474;&#3627408462;&#3627408481;&#3627408470;&#3627408464; &#3627408465;&#3627408466;&#3627408479;&#3627408470;&#3627408483;&#3627408462;&#3627408481;&#3627408470;&#3627408483;&#3627408466;,
&#3627408465;
&#3627408465;&#3627408481;
=&#3627409172;=&#3627408462;&#3627408475;&#3627408468;.&#3627408483;&#3627408466;&#3627408473;.&#3627408476;&#3627408467; &#3627408464;&#3627408462;&#3627408474;
&#3627408462;=
&#3627408465;&#3627408483;
&#3627408465;&#3627408481;
=
&#3627408465;&#3627408483;
&#3627408465;
×
&#3627408465;
&#3627408465;&#3627408481;
=
&#3627408465;
2
&#3627408480;
&#3627408465;&#3627409155;
2
(
&#3627408465;&#3627409155;
&#3627408465;&#3627408481;
)
2

&#3627408471;&#3627408466;&#3627408479;&#3627408472; (&#3627408523;)=&#3627409172;
3
&#3627408465;
3
&#3627408480;
&#3627408465;&#3627409155;
3
=
&#3627408465;&#3627408462;
&#3627408465;&#3627408481;

Follower motion
There is Rise, Return, Dwell, Fall of Follower.

Since the follower moves with uniform velocity during its rise and return stroke, therefore the slope of the
displacement curves must be constant.
&#3627408454;=
&#3627408483;&#3627409155;
&#3627409172;
=&#3627409172;&#3627408481; &#3627408480;=&#3627408483;.&#3627408481;
ℎ=
&#3627408483;
&#3627409172;

&#3627408462;

&#3627408483;
&#3627408462;=
ℎ&#3627409172;

&#3627408462;

a=0
j=0
&#3627408483;
&#3627408465;=
ℎ&#3627409172;

&#3627408465;

In order to have the acceleration and retardation within the finite limits.
This may be done by rounding off the sharp corners of the displacement diagram at the beginning and at the end
of each stroke.

53


Simple Harmonic motion of Follower
Construction
s= follower displacement, h= maximum follower displacement, v= velocity of the follower,
f= acceleration of the follower,  = cam rotation angle,
 =cam rotation angle for maximum follower displacement, β= angle on the harmonic circle
&#3627408436;&#3627408481; &#3627408462;&#3627408475;&#3627408486; &#3627408470;&#3627408475;&#3627408480;&#3627408481;&#3627408462;&#3627408475;&#3627408481;,&#3627408465;&#3627408470;&#3627408480;&#3627408477;&#3627408473;&#3627408462;&#3627408464;&#3627408466;&#3627408474;&#3627408466;&#3627408475;&#3627408481; &#3627408532;=

2


2
&#3627408464;&#3627408476;&#3627408480;&#3627409149;
&#3627409149;=&#3627409163;
&#3627409155;
??????

&#3627408532;=

2


2
&#3627408464;&#3627408476;&#3627408480;&#3627409163;
&#3627409155;
??????
=

2
(1−&#3627408464;&#3627408476;&#3627408480;&#3627409163;
&#3627409155;
??????
)
=

2
(1−&#3627408464;&#3627408476;&#3627408480;
&#3627409163;&#3627409172;&#3627408481;
??????
)
&#3627408535;=
&#3627408465;&#3627408480;
&#3627408465;&#3627408481;
=

2
&#3627409163;&#3627409172;
??????
&#3627408480;&#3627408470;&#3627408475;
&#3627409163;&#3627409172;&#3627408481;
??????
=

2
&#3627409163;&#3627409172;
??????
&#3627408480;&#3627408470;&#3627408475;
&#3627409163;
??????

&#3627408535;
&#3627408474;&#3627408462;&#3627408485;=

2
&#3627409163;&#3627409172;
??????
&#3627408462;&#3627408481; =
??????
2

&#3627408519;=
&#3627408465;&#3627408535;
&#3627408465;&#3627408533;
=

2
(
&#3627409163;&#3627409172;
??????
)
2
&#3627408464;&#3627408476;&#3627408480;
&#3627409163;
??????

&#3627408519;
&#3627408474;&#3627408462;&#3627408485;=

2
(
&#3627409163;&#3627409172;
??????
)
2
&#3627408462;&#3627408481; =0˚
o = angle of ascent, R = angle of descent

Here acceleration is abruptly increasing from zero to maximum, which results in infinite jerk, vibration and
noise.

54

Constant acceleration and deceleration (Parabolic)
Here, there is acceleration in the first half and deceleration in the second half and the displacement curve is
parabolic.


&#3627408532;=&#3627408483;
&#3627408476;&#3627408481;+
1
2
&#3627408467;&#3627408481;
2

&#3627408532;=
1
2
&#3627408467;&#3627408481;
2
&#3627408462;&#3627408480; &#3627408483;
&#3627408476;=0
&#3627408467;=
2&#3627408480;
&#3627408481;
2
=&#3627408464;&#3627408476;&#3627408475;&#3627408480;&#3627408481;&#3627408462;&#3627408475;&#3627408481;
&#3627408532;=

2
& &#3627409172;&#3627408481;=
??????
2
,&#3627408481;=
??????
2&#3627409172;

&#3627408519;=
4ℎ&#3627409172;
2
??????
2
=&#3627408464;&#3627408476;&#3627408475;&#3627408480;&#3627408481;&#3627408462;&#3627408475;&#3627408481;
&#3627408535;=&#3627408467;&#3627408481;=
4ℎ&#3627409172;
2
??????
2
×
&#3627409155;
&#3627409172;
=
4ℎ&#3627409172;
??????
2
&#3627409155;
&#3627408535;
&#3627408526;&#3627408514;&#3627408537;=
4ℎ&#3627409172;
??????
2
×
??????
2
=
2ℎ&#3627409172;
??????



Here acceleration is abruptly increasing from maximum to minimum, which results in infinite jerk, vibration
and noise.

55

Constant Velocity
Constant velocity of follower implies the displacement of follower is proportional to cam rotation.
&#3627408483;=0→
&#3627408465;&#3627408532;
&#3627408465;&#3627408533;
=0→&#3627408480;∝ &#3627409155;

&#3627408480;=ℎ
&#3627409155;
??????
=ℎ
&#3627409172;&#3627408481;
??????

&#3627408483;=
&#3627408465;&#3627408532;
&#3627408465;&#3627408533;
=
ℎ&#3627409172;
??????
= &#3627408464;&#3627408476;&#3627408475;&#3627408480;&#3627408481;&#3627408462;&#3627408475;&#3627408481;
&#3627408467;=
&#3627408465;&#3627408535;
&#3627408465;&#3627408533;
=0

There is an abrupt increase and decrease in velocity which results in
infinite inertia forces and not suitable for practical use.



Modified constant velocity program

56

Cycloid
A cycloid is locus of point on a circle rotating on a straight line.


&#3627408480;=

&#3627409163;
(
&#3627409163;&#3627409155;
??????

1
2
&#3627408480;&#3627408470;&#3627408475;
2&#3627409163;&#3627409155;
??????
)
&#3627408483;=
&#3627408465;&#3627408532;
&#3627408465;&#3627408533;
=
&#3627408465;&#3627408532;
&#3627408465;??????
×
&#3627408465;??????
&#3627408465;&#3627408533;
= [

??????


??????
&#3627408464;&#3627408476;&#3627408480;
2&#3627409163;
??????
]&#3627409172;
&#3627408483;=[
ℎ&#3627409172;
??????

ℎ&#3627409172;
??????
&#3627408464;&#3627408476;&#3627408480;
2&#3627409163;
??????
]=
ℎ&#3627409172;
??????
(1−&#3627408464;&#3627408476;&#3627408480;
2&#3627409163;
??????
)
&#3627408483;
&#3627408474;&#3627408462;&#3627408485;=
ℎ&#3627409172;
??????
&#3627408462;&#3627408481; &#3627409155;=
??????
2

&#3627408467;=
&#3627408465;&#3627408535;
&#3627408465;&#3627408533;
=
&#3627408465;&#3627408535;
&#3627408465;??????
×
&#3627408465;??????
&#3627408465;&#3627408533;
= [
2ℎ&#3627409163;&#3627409172;
2
??????
2
&#3627408480;&#3627408470;&#3627408475;
2&#3627409163;
??????
]
&#3627408467;
&#3627408474;&#3627408462;&#3627408485;=
2ℎ&#3627409163;&#3627409172;
2
??????
2
&#3627408462;&#3627408481; &#3627409155;=
&#3627409163;
4



There are no abrupt changes in velocity and acceleration.
So, this is the most ideal one to use.

57


GEARS
Concept of friction wheels

Toothed wheel
Motion and power transfer was primarily achieved by using
friction discs/wheels in contact. Due to friction force between the
wheel, motion and power are transferred from one axis to
another axis.
There is a limitation for maximum value for maximum value of
power transfer due to limiting static friction force. Hence beyond
certain input torque there will be slip between discs.
To overcome this problem, toothed wheels (GEARS) are used in
place of friction wheels to create a positive drive, improving
torque transmission capability
&#3627408483;
&#3627408477;=&#3627409172;
1&#3627408479;
1=&#3627409172;
2&#3627408479;
2
&#3627408483;
&#3627408477;=2&#3627409163;&#3627408449;
1&#3627408479;
1= 2&#3627409163;&#3627408449;
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23

To ensure constant angular velocity in case of toothed wheels in
mesh, the Instantaneous centre of wheels shall be static as, meshing
progresses.
Classification of Gears
Parallel shafts
Depending upon the teeth of equivalent cylinders i.e., straight or helical, following are the main types of gears to
join parallel shafts.
Spur Gears
They have straight teeth parallel to the axes.
They have a line contact, which results in the high impact stresses and excessive noise at high speeds.
Spur Rack and Pinion
Spur rack is a special case of a spur gear where it is made of infinite diameter so that pitch surface is a plane.
It converts rotary motion into translatory motion.
Helical spur gears
In helical gears, the teeth are curved, each being helical in shape.
At the beginning of engagement, contact occurs only at the point of leading edge of curved teeth, as gear rotates,
the contact extends along a diagonal line across the teeth.
Load application is gradual which results in low impact stresses and reduction in noise.

58

Double-Helical and Herringbone gears
It is equivalent to a pair of helical gears secured together, one having right-hand helix and other having left-
hand helix.
Axial thrust which occurs in case of single-helical gears is eliminated in double-helical gears.
It can run at high speed with less noise and vibrations.

Intersecting shafts
The motion between 2 intersecting shafts is equivalent to the rolling if 2 cones, assuming no slipping. The gears,
in general are known as bevel gears.
When the teeth formed on cones are straight, the gears are known as straight bevel and when inclined, they are
known as helical bevel gears.
Straight bevel gear
The teeth are straight, radial to the point of intersection of the shaft axes and vary in cross section throughout
their length.
Shafts are connected at right angles and gears are of the same size.
Spiral bevel gear
Teeth of bevel gear are inclined at an angle to the face of bevel.
They are smother and quieter in action than straight bevel gears because of low impact stresses and gradual
application of load.
Zero bevel gear
Spiral bevel gear with curved teeth but with a zero-angle spiral angle.
They are quieter in action than the spiral bevel gear.

59

Skew Shafts
The two non-intersecting and non-parallel i.e. non-coplanar shaft connected by gears. This type of gearing also
has a line contact, the rotation of which about the axes generates the two pitch surfaces known as hyperboloids.


Crossed Helical Gear
By using a suitable choice of helix angle for the mating gears, two shafts can be set at any angle.

Worm gear
Worm gear is a special case of spiral gear in which the larger wheel, usually has a hollow shape (gear spacing
for rotating) such that other gear’s teeth is fitted partially. The smaller wheel is called worm and has large
spiral angle.

Non-throated -The contact between the teeth is concentrated at a point.
Single-throated- Gear teeth are curved to envelop the worm. There is a line contact between the teeth.
Double-throated- There is an area contact between the teeth.

60

Gear Nomenclature

Pitch circle- It is an imaginary circle which by pure rolling action, would give the same motion as the actual
gear.
Pitch diameter- Diameter of a pitch circle.
Pitch point- Point of contact of two pitch circles is known as the pitch point.
Line of centers- A line through the centres of rotation of the mating gears
Pinion- It is the smaller gear and usually driving gear.
Rack- It is a part of a gear wheel of infinite diameter.

Circular pitch- It is the distance measured along the circumference of a pitch circle from a point on one tooth
to the corresponding point on the adjacent tooth.
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Diametral pitch- The number of teeth per unit length of pitch circle diameter in inches.
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Module- It is the ratio of pitch diameter (in mm) to the number of teeth.
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Gear Ratio- It is the number of teeth on the gear to that of the pinion.
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61

Velocity Ratio- The ratio of angular velocity of the follower to angular velocity of driving gear.
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Addendum circle- It is a circle passing through the tips of teeth.
Addendum- It is the radial height of a tooth above the pitch circle.
Dedendum or root circle- It is the circle passing through the roots of the teeth.
Dedendum- it is the radial depth of a tooth below the pitch circle.
Clearance- Radial difference between the addendum and dedendum of a tooth.
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Backlash- Space Width ― Tooth thickness
Line of action- The force, the driving force exerts on the driven tooth, is
along a line from the pitch point to the point of contact of the two teeth. The
line is also common at the point of contact of the mating gears.
Pressure angle- The angle between the pressure line and the common
tangent to the pitch circles is known as the pressure angle.
Path of contact- It is the path traced by the point of contact of two teeth
from the beginning to the end of engagement.
CP→ Path of approach
PD→ Path of recess
Arc of contact- It is the path traced by a point on the pitch circle from the
beginning to the end of engagement of a given pair of teeth. The arc of
contact consists of two parts.
Arc of approach (AP/EP) - It is the portion of the path of contact from the beginning of the engagement to the
pitch point.
Arc of recess(PB/PF) -It is the portion of the path of contact from the pitch point to the end of the engagement of
a pair of teeth.
Angle of Action (δ) – It is the angle turned by a gear from the beginning of engagement to end of engagement
of a pair of teeth.
Angle of approach (δ) = angle of approach (α) + angle of recess(β)
Contact Ratio- It is angle of action divided by pitch angle.
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Law of Gearing
It states the condition which must be satisfied by the gear tooth profiles to maintain a constant angular velocity
ratio between 2 gears.
(πd1N1=πd2N2)

62


Point C on gear 1 is in contact with point D on gear 2, they have a common normal n-n.
If the curved surfaces are to remain in contact, one surface may slide relative to other along the common
tangent t-t.
The relative motion between the surfaces along the n-n must be zero to avoid the separation.
vc = velocity of C (on 1) perpendicular to AC = ω1.AC
vd = velocity of D (on 2) perpendicular to BD = ω1.BD &#3627408438;&#3627408476;&#3627408474;&#3627408477;&#3627408476;&#3627408475;&#3627408466;&#3627408475;&#3627408481; &#3627408476;&#3627408467; &#3627408483;
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For constant angular velocity ratio two gears, the common normal at the point of contact of two mating teeth
must pass through the pitch point. We see that the angular velocity ratio is inversely proportional to the ratio of
the distances of the point P from the centres A & B.
&#3627409172;
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Velocity of Sliding
The velocity of sliding is the velocity of one tooth relative to its mating tooth along the common tangent at the
point of contact. If the curved surfaces of the two teeth of the gears 1 & 2 are to remain in contact, one can have
sliding motion relative to other along the common tangent t-t.
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Velocity of Sliding = Sum of angular velocities × distance between the pitch point and the point of contact.
Forms of teeth
Common forms of teeth that can satisfy the law of gearing
1. Cycloidal profile teeth
2. Involute profile teeth
Cycloidal profile teeth
In this type, the faces are epicycloids and flanks are hypocycloids.
Hypocycloid- Curve traced by a point on the circumference of a circle which is rolling on the interior of
another circle.
P is the point dividing AB by n-n.

63

Epicycloid- Curve traced by a point on the circumference of a circle rolling on the exterior of another circle.

Here the circle H rotates inside, along the circumference of pitch circle upto Point P which forms flank (only
small portion of curve is taken) and similarly circle E rotates outside till P forming face of flank.
Cycloid is always perpendicular(normal) to the line(CD) joining point of contact(D) and point on cycloid(C).
Law of gearing is satisfied as common normal at any point on cycloid always passes through the pitch point.

Involute Profile
An involute is the locus of a point on straight line which rolls without slipping on the circumference of a circle. It
is also the path traced by the end of cord(wire) being unwound from the
circumference of the circle.
As the line rolls on circle, the path traced by A is involute (AFBC)
A short length EF of the involute drawn can be utilized to make the profile of an
involute tooth.


Common tangent to base circle passes through pitch point.
Common tangent to base circles is generatrix line for involute profile.
Any point on common tangent traces involute profiles when
generatrix line rolls without slipping on base circles.
The tangent CE is normal to involute GC or tangent t-t and CF to DC
or t-t.
As both CE & CF both are normal to t-t and have a common point,
EPF is a straight line.
As wheel 1 rotates, GC pushes DH along the common tangent of base
circles, hence the path of contact is along the common tangent of
base circles.
This common tangent (ECH) is also common normal to involutes which passes through the pitch point.

Hypocycloid Epicycloid

64

Pressure angles in this case remain constant throughout the engagement of teeth.
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=
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For a pair of involute gears velocity ratio of gears in inversely proportional to pitch circle diameters as well as
base circle diameter.
Effect of Altering the Centre Distance on the Velocity Ratio for Involute Teeth
Gears
Any shift in centres of gears will change the centre distance.
If the involutes are still in contact, the common normal of two
involutes at the point of contact will be common tangent for both base
circles and its intersection with the line of centres will be new pitch
point.
Shifting of centres will not alter velocity ratio, but the pressure angle
increases (from φ to φ′) with the increase in the centre distance.



Comparison Between Involute and Cycloidal Gears
Cycloidal profile
Advantages Disadvantages
1
Cycloidal gears are stronger than the involute
gears, for the same pitch
1
Pressure angle is not constant

2
Results in less wear in cycloidal gears as
compared to involute gears
2 Manufacturing is difficult and costly
3 The interference does not occur at all 3 Centre distance cannot be maintained accurately


4
Due to wear and tear, it may not satisfy Law of
Gearing.
Involute profile
Advantages Disadvantages
1
The centre distance for a pair of involute gears can
be varied within limits without changing the
velocity ratio
1 Not suitable for lesser numbers of teeth.
2
The pressure angle, from the start of the
engagement of teeth to the end of the engagement,
remains constant
2
Undercut or interference between the teeth may
occur for this gear in case addendum modifications
are not performed properly
3
The involute teeth are easy to manufacture than
cycloidal teeth.

65

Systems of Gear Teeth
The following four systems of gear teeth are commonly used in practice
1. 14.5 ° Composite system – it is used for general purpose
2. 14.5 ° Full depth involute system – it was developed for use with gear hobs for spur and helical gears.
3. 20° Full depth involute system - The increase of the pressure angle from 14.5 ° to 20° results in a stronger
tooth
4. 20° Stub involute system – it has a strong tooth to take heavy loads
S.no Particulars 14.5 ° Composite system or Full
depth involute system
20° Full depth
involute system
20° Stub involute
system.
1 Addendum 1 m 1 m 0.8 m
2 Dedendum 1.25 m 1.25 m 1 m
3 Working depth 2 m 2 m 1.60 m
4 Minimum total
depth
2.25 m 2.25 m 1.80 m
5 Tooth thickness 1.5708 m 1.5708 m 1.5708 m
6 Minimum clearance
0.25 m 0.25 m 0.2 m
7 Fillet radius at root 0.4 m 0.4 m 0.4 m
Path of Contact
Gear 1 is the driver and wheel 2 is driven counter-clockwise.

Contact of two teeth is made where the addendum circle of wheel meets the line of action EF at C, it is broken
where addendum circle of gear meets line of action EF at D.
Path of contact = Path of access + Path of recess
CD = CP + PD
CD = (CF-PF) + (PF-DF)
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Arc of Contact
Arc of contact is the path traced by a point on the pitch circle from the
beginning to the end of engagement of a given pair of teeth.
P’ P’’ is the arc of contact, P’P is arc of approach and PP’’ is arc of recess.
Let the time to transverse the arc of approach is ta.
Then arc of approach = P’P =Tangential velocity of P’ × time of approach
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&#3627408464;&#3627408476;&#3627408480;??????

Similarly, arc of recess is PP’’ = tang. vel. of P × time of recess

66

&#3627408436;&#3627408479;&#3627408464; &#3627408476;&#3627408467; &#3627408479;&#3627408466;&#3627408464;&#3627408466;&#3627408480;&#3627408480;=
&#3627408503;&#3627408491;
&#3627408464;&#3627408476;&#3627408480;??????

&#3627408436;&#3627408479;&#3627408464; &#3627408476;&#3627408467; &#3627408464;&#3627408476;&#3627408475;&#3627408481;&#3627408462;&#3627408464;&#3627408481;=
&#3627408490;&#3627408503;
&#3627408464;&#3627408476;&#3627408480;??????
+
&#3627408503;&#3627408491;
&#3627408464;&#3627408476;&#3627408480;??????
=
&#3627408490;&#3627408491;
&#3627408464;&#3627408476;&#3627408480;??????

&#3627408436;&#3627408479;&#3627408464; &#3627408476;&#3627408467; &#3627408464;&#3627408476;&#3627408475;&#3627408481;&#3627408462;&#3627408464;&#3627408481;=
&#3627408451;&#3627408462;&#3627408481;ℎ &#3627408476;&#3627408467; &#3627408464;&#3627408476;&#3627408475;&#3627408481;&#3627408462;&#3627408464;&#3627408481;
&#3627408464;&#3627408476;&#3627408480;??????

Number of pairs of teeth in contact (Contact ratio)
All the teeth lying in between the arc of contact will be meshing with the teeth on the other wheel.
&#3627408455;ℎ&#3627408466; &#3627408475;&#3627408482;&#3627408474;&#3627408463;&#3627408466;&#3627408479; &#3627408476;&#3627408467; &#3627408481;&#3627408466;&#3627408466;&#3627408481;ℎ &#3627408470;&#3627408475; &#3627408464;&#3627408476;&#3627408475;&#3627408481;&#3627408462;&#3627408464;&#3627408481; (&#3627408527;)=
&#3627408436;&#3627408479;&#3627408464; &#3627408476;&#3627408467; &#3627408464;&#3627408476;&#3627408475;&#3627408481;&#3627408462;&#3627408464;&#3627408481;
&#3627408438;&#3627408470;&#3627408479;&#3627408464;&#3627408482;&#3627408473;&#3627408462;&#3627408479; &#3627408477;&#3627408470;&#3627408481;&#3627408464;ℎ
=
&#3627408438;&#3627408439;
&#3627408464;&#3627408476;&#3627408480;??????
1
&#3627408477;

For continuous transmission of motion, at least one tooth of one wheel must be in contact with another tooth of
the second wheel. Therefore, n must be greater than unity.
Interference in Involute gears
At any instant, the portions of tooth profiles which are in action must be involutes, so that line of action does not
deviate.
If any of the two surfaces is not an involute, the two surfaces would not touch each other tangentially and the
transmission of the power would not be proper. Mating of two non-involute teeth is known as Interference.

Owing to non-involute profile, the contacting teeth have different velocities which can lock the gears.
If pinion is the driver, the line of action will be along EF which is the common tangent to base circles of two
gears. The teeth on pinion wheel are engaged at C and disengaged at D. Now if the addendum circle radius is
increased, D will shift towards F on PF and D coincides with F if add. radius of pinion is AF.
Any further increase in this value of radius will result in shifting the point of contact inside the base circle of the
wheel.
Since an involute can exist only outside the base circle, therefore, any profile of teeth inside the base circle will
be of involute type.
The profiles in such a case cannot be tangent to each other and tip of the pinion will try to dig out the flank of
the tooth of the wheel. Therefore, interference occurs in the mating of two gears.
If the addendum radius of wheel is greater than BE, the tip of the wheel tooth be in contact with a portion of the
non-involute profile of the teeth for some time of engagement. This causes interference.
To have no interference, addendum circles of the wheel and the pinion must intersect the line of action between
E & F.
The points E & F are called interference points.

67

Minimum Number of Teeth
We saw previously that maximum addendum radius of wheel to prevent interference is BE.
&#3627408437;&#3627408440;
2
=&#3627408437;&#3627408441;
2
+&#3627408441;&#3627408440;
2
=&#3627408437;&#3627408441;
2
+(&#3627408441;&#3627408451;+&#3627408451;&#3627408440;)
2

&#3627408437;&#3627408440;
2
=(&#3627408453;&#3627408464;&#3627408476;&#3627408480;??????)
2
+ (&#3627408453;&#3627408480;&#3627408470;&#3627408475;??????+&#3627408479;&#3627408480;&#3627408470;&#3627408475;??????)
2

&#3627408437;&#3627408440;=&#3627408453;√1+
&#3627408479;
&#3627408453;
(
&#3627408479;
&#3627408453;
+2)&#3627408480;&#3627408470;&#3627408475;
2
??????
Therefore, maximum value of addendum of the wheel is
aw max = BE – pitch radius
&#3627408462;
&#3627408484; &#3627408474;&#3627408462;&#3627408485;=&#3627408453;√1+
&#3627408479;
&#3627408453;
(
&#3627408479;
&#3627408453;
+2)&#3627408480;&#3627408470;&#3627408475;
2
??????−&#3627408453;
= &#3627408453;[√1+
&#3627408479;
&#3627408453;
(
&#3627408479;
&#3627408453;
+2)&#3627408480;&#3627408470;&#3627408475;
2
??????−1]
&#3627408453;=
&#3627408474;&#3627408455;
2
,&#3627408479;=
&#3627408474;&#3627408481;
2
& &#3627408442;&#3627408466;&#3627408462;&#3627408479; &#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476; (&#3627408494;)=
&#3627408455;
&#3627408481;
=
&#3627408453;
&#3627408479;

&#3627408462;
&#3627408484; &#3627408474;&#3627408462;&#3627408485;=
&#3627408474;&#3627408455;
2
[√1+
1
&#3627408442;
(
1
&#3627408442;
+2)&#3627408480;&#3627408470;&#3627408475;
2
??????−1]
Let the adopted value of addendum in some cases be aw × m.
&#3627408454;&#3627408476;,&#3627408462;
&#3627408484;×&#3627408474; ≤
&#3627408474;&#3627408455;
2
[√1+
1
&#3627408442;
(
1
&#3627408442;
+2)&#3627408480;&#3627408470;&#3627408475;
2
??????−1]
&#3627408455; ≥
2&#3627408462;
&#3627408484;
[√1+
1
&#3627408442;
(
1
&#3627408442;
+2)&#3627408480;&#3627408470;&#3627408475;
2
??????−1]

&#3627408455;=
2&#3627408462;
&#3627408484;
[√1+
1
&#3627408442;
(
1
&#3627408442;
+2)&#3627408480;&#3627408470;&#3627408475;
2
??????−1]

This gives the minimum number of teeth on the wheel for the given value of gear ratio, pressure angle and the
addendum coefficient aw.
The minimum no. of teeth on pinion is &#3627408481;=
&#3627408455;
&#3627408442;

68

Undercutting
If interference can’t be avoided by the design, it can be minimized by removing interfering portion of teeth. This
is termed Undercutting.

A form tool of same geometry as that of meshing gear teeth is used to remove material at interfering portion.
Due to undercutting, the strength of teeth is reduced.
Effect of wear & tear
Due to wear and tear, teeth size gets reduced but involute profile remains same as offset of involute profile is
involute.

It satisfies Law of Gearing.
Due to wear, back lash increases


Interference between Rack and Pinion
Here to avoid interference, the maximum value of addendum should be
such that C coincides with E.
It means that maximum addendum value of rack is GE.
Let the adopted value of addendum of the rack be ar×m where ar is the
addendum coefficient.
&#3627408442;&#3627408440;=&#3627408451;&#3627408440;×&#3627408480;&#3627408470;&#3627408475;??????=&#3627408479;&#3627408480;&#3627408470;&#3627408475;??????×&#3627408480;&#3627408470;&#3627408475;??????=&#3627408479;&#3627408480;&#3627408470;&#3627408475;
2
??????=
&#3627408474;&#3627408481;
2
×&#3627408480;&#3627408470;&#3627408475;
2
??????
To avoid interference,
&#3627408442;&#3627408440;≥&#3627408462;
&#3627408479;×&#3627408474; (&#3627408476;&#3627408479;)
&#3627408474;&#3627408481;
2
×&#3627408480;&#3627408470;&#3627408475;
2
??????≥&#3627408462;
&#3627408479;×&#3627408474; (&#3627408476;&#3627408479;)&#3627408481;≥
2&#3627408480;&#3627408470;&#3627408475;
2
??????
&#3627408474;

&#3627408451;&#3627408462;&#3627408481;ℎ &#3627408476;&#3627408467; &#3627408464;&#3627408476;&#3627408475;&#3627408481;&#3627408462;&#3627408464;&#3627408481;=&#3627408438;&#3627408451;+&#3627408451;&#3627408439;=
&#3627408436;&#3627408465;&#3627408465;&#3627408466;&#3627408474;&#3627408465;&#3627408482;&#3627408474; &#3627408476;&#3627408467; &#3627408479;&#3627408462;&#3627408464;&#3627408472;
&#3627408464;&#3627408476;&#3627408480;??????
+√&#3627408479;
&#3627408462;
2
−(&#3627408479;&#3627408464;&#3627408476;&#3627408480;??????)
2
−&#3627408479;&#3627408480;&#3627408470;&#3627408475;??????
&#3627408448;&#3627408462;&#3627408485;&#3627408470;&#3627408474;&#3627408482;&#3627408474; &#3627408477;&#3627408462;&#3627408481;ℎ &#3627408476;&#3627408467; &#3627408464;&#3627408476;&#3627408475;&#3627408481;&#3627408462;&#3627408464;&#3627408481; &#3627408481;&#3627408476; &#3627408462;&#3627408483;&#3627408476;&#3627408470;&#3627408465; &#3627408470;&#3627408475;&#3627408481;&#3627408466;&#3627408479;&#3627408467;&#3627408466;&#3627408479;&#3627408466;&#3627408475;&#3627408464;&#3627408466;=&#3627408439;&#3627408440;=√&#3627408479;
&#3627408462;
2
−(&#3627408479;&#3627408464;&#3627408476;&#3627408480;??????)
2

69

Helical Gears
In helical gears teeth are inclined to the axis of the gear, they can be right-handed or left-handed in which the
helix slopes away from the viewer when a gear is viewed parallel to the axis of the gear.


Here, the helix angle of gear 2 is reduced by a few
degrees so that the helix angle of gear 1 is ψ1 and that
of gear 2 is ψ2. Let the angle turned by it be  which is
the angle between the axes of two gears.
 = ψ1- ψ2

when ψ2 = 0 i.e., the helix angle of gear 2 is zero or
gear 2 is a Straight spur gear.
 = ψ1.


if ψ2  0 i.e., helix angle of gear 2 is negative.
 = ψ1― (―ψ2) = ψ1 + ψ2

From above, we can conclude that angle between shafts is equal to
= ψ1― ψ2, in case of gears of opposite hands (ex- one left and one right hand)
= ψ1+ ψ2, in case of gears of same hand (ex- both left hand or both right hand)
In case of helical gears for parallel shafts, there will be line contact whereas for skew
shafts (non-parallel) there will be point of contact.
Helical gears with line of contact are stronger than spur gears and can transmit heavy
loads.
Pitch line velocities of gear 1 & 2 (for ψ2  0)
The magnitude and direction of v12 represents the sliding velocity of gear 1 with respect to gear 2 parallel to t-t.

Side view
Top view
Here the helix angle is
same ψ1=ψ2.
When ψ1=ψ2, the helix
angle is the same as
before.
Then = ψ1- ψ2 = 0.
It’s a case of helical
gears joining parallel
shafts.

70

Terminology of helical Gears
Helix angle (ψ) – it is the angle at which teeth are inclined to axis of a gear.
Circular pitch (pn) – it is the distance between the corresponding points on adjacent teeth
measured on a pitch circle. It is also called as transverse circular pitch.
Normal Circular pitch (p) – it is the shorts distance measured along the normal to the
helix between corresponding points on the adjacent teeth.
&#3627408477;
&#3627408475;=&#3627408477;×&#3627408464;&#3627408476;&#3627408480;??????
Velocity ratio and center distance of Helical Gears
Velocity ratio
vn=v1×cos(ψ1) = v2×cos(ψ2)
&#3627408483;
2
&#3627408483;
1
=
&#3627408464;&#3627408476;&#3627408480;&#3627409171;
1
&#3627408464;&#3627408476;&#3627408480;&#3627409171;
2

??????&#3627408518;&#3627408525;&#3627408528;&#3627408516;&#3627408522;&#3627408533;&#3627408538; &#3627408505;&#3627408514;&#3627408533;&#3627408522;&#3627408528;=
&#3627409172;
2
&#3627409172;
1
=
&#3627408483;
2&#3627408479;
2⁄
&#3627408483;
1&#3627408479;
1⁄
=
&#3627408483;
2&#3627408465;
2⁄
&#3627408483;
1&#3627408465;
1⁄
=
&#3627408483;
2&#3627408465;
2⁄
&#3627408483;
1&#3627408465;
1⁄
=
&#3627408465;
1
&#3627408465;
2
×
&#3627408483;
2
&#3627408483;
1
=
&#3627408465;
1
&#3627408465;
2
×
&#3627408464;&#3627408476;&#3627408480;&#3627409171;
1
&#3627408464;&#3627408476;&#3627408480;&#3627409171;
2
=
&#3627408474;
1&#3627408455;
1&#3627408464;&#3627408476;&#3627408480;&#3627409171;
1
&#3627408474;
2&#3627408455;
2&#3627408464;&#3627408476;&#3627408480;&#3627409171;
2
=
&#3627408507;
&#3627409359;
&#3627408507;
&#3627409360;


Centre distance
Let C be the distance between two skew shaft axes which is the shortest distance between them.
&#3627408438;=&#3627408479;
1+&#3627408479;
2=
1
2
(&#3627408465;
1+&#3627408465;
2)=
1
2
(&#3627408474;
1&#3627408455;
1+&#3627408474;
2&#3627408455;
2)=
1
2
(
&#3627408474;
&#3627408475;
&#3627408464;&#3627408476;&#3627408480;&#3627409171;
1
&#3627408455;
1+
&#3627408474;
&#3627408475;
&#3627408464;&#3627408476;&#3627408480;&#3627409171;
2
&#3627408455;
2)=
&#3627408474;
&#3627408475;
2
(
&#3627408455;
1
&#3627408464;&#3627408476;&#3627408480;&#3627409171;
1
+
&#3627408455;
2
&#3627408464;&#3627408476;&#3627408480;&#3627409171;
2
)
Helical gear forces and efficiency
The force exerted by a helical gear on its mating gear acts normal to contacting surface if friction is neglected.
Normal force in case of helical gears have 3 components, apart from
tangential and radial, there is axial force.
F
t
n = total normal force
Ft= tangential force
Fa= axial force
Fr= radial force
Fn=normal force in the plane of Ft & Fa
φ=pressure angle
φn=normal pressure angle
ψ=helix angle

&#3627408441;
&#3627408475;= &#3627408441;
&#3627408475;
&#3627408481;
&#3627408464;&#3627408476;&#3627408480;??????
&#3627408475; &#3627408462;&#3627408475;&#3627408465; &#3627408441;
&#3627408479;= &#3627408441;
&#3627408475;
&#3627408481;
&#3627408480;&#3627408470;&#3627408475;??????
&#3627408475;
&#3627408441;
&#3627408481;=&#3627408441;
&#3627408475;&#3627408464;&#3627408476;&#3627408480;&#3627409171; &#3627408462;&#3627408475;&#3627408465;&#3627408441;
&#3627408462;=&#3627408441;
&#3627408475;&#3627408480;&#3627408470;&#3627408475;&#3627409171;
Efficiency of spiral and Helical Gears
In spiral or crossed helical gears, the sliding action between the surfaces acts chiefly along the tangent to the
pitch helix. The friction force is equal to μFn and acts in a direction opposite to the direction of sliding of the gear
surface.
&#3627409154;=
&#3627408464;&#3627408476;&#3627408480;(&#3627409155;+??????)+&#3627408464;&#3627408476;&#3627408480;(&#3627409171;
1−&#3627409171;
2−??????)
&#3627408464;&#3627408476;&#3627408480;(&#3627409155;−??????)+&#3627408464;&#3627408476;&#3627408480;(&#3627409171;
1−&#3627409171;
2−??????)

&#3627409154;
&#3627408474;&#3627408462;&#3627408485;=
&#3627408464;&#3627408476;&#3627408480;(&#3627409155;+??????)+1
&#3627408464;&#3627408476;&#3627408480;(&#3627409155;−??????)−1

71

GEAR TRAINS
A gear train is a combination of gears used to transmit motion from one shaft to other. It becomes necessary
when it is required to bring large speed reductions within a small space.
Types of gear trains
1. Simple gear train
2. Compound gear train
3. Reverted gear train
4. Planetary or epicyclic Gear train.
Simple Gear train
A series of gears capable of receiving and transmitting motion from one gear to another is called simple gear
train.
All odd number gears move in one-direction and even-numbered gears
move in other direction.
Speed ratio is the ratio of the speed of driving gear to driven gear, is
negative when input and output gears rotate in opposite directions and
vice versa.
T= no. of teeth on a gear
N=Speed of a gear in r.p.m
&#3627408449;
2
&#3627408449;
1
=
&#3627408455;
1
&#3627408455;
2

&#3627408449;
3
&#3627408449;
2
=
&#3627408455;
2
&#3627408455;
3
,
&#3627408449;
4
&#3627408449;
3
=
&#3627408455;
3
&#3627408455;
4
&
&#3627408449;
5
&#3627408449;
4
=
&#3627408455;
4
&#3627408455;
5

Multiplying,
&#3627408449;
2
&#3627408449;
1
×
&#3627408449;
3
&#3627408449;
2
×
&#3627408449;
4
&#3627408449;
3
×
&#3627408449;
5
&#3627408449;
4
=
&#3627408455;
1
&#3627408455;
2
×
&#3627408455;
2
&#3627408455;
3
×
&#3627408455;
3
&#3627408455;
4
×
&#3627408455;
4
&#3627408455;
5

&#3627408455;&#3627408479;&#3627408462;&#3627408470;&#3627408475; &#3627408457;&#3627408462;&#3627408473;&#3627408482;&#3627408466;=
&#3627408449;
5
&#3627408449;
1
=
&#3627408455;
1
&#3627408455;
5
=
&#3627408475;&#3627408482;&#3627408474;&#3627408463;&#3627408466;&#3627408479; &#3627408476;&#3627408467; &#3627408481;&#3627408466;&#3627408466;&#3627408481;ℎ &#3627408476;&#3627408475; &#3627408465;&#3627408479;&#3627408470;&#3627408483;&#3627408470;&#3627408475;&#3627408468; &#3627408468;&#3627408466;&#3627408462;&#3627408479;
&#3627408475;&#3627408482;&#3627408474;&#3627408463;&#3627408466;&#3627408479; &#3627408476;&#3627408467; &#3627408481;&#3627408466;&#3627408466;&#3627408481;ℎ &#3627408476;&#3627408475; &#3627408465;&#3627408479;&#3627408470;&#3627408483;&#3627408466;&#3627408475; &#3627408468;&#3627408466;&#3627408462;&#3627408479;

&#3627408454;&#3627408477;&#3627408466;&#3627408466;&#3627408465; &#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;=
1
&#3627408455;&#3627408479;&#3627408462;&#3627408470;&#3627408475; &#3627408483;&#3627408462;&#3627408473;&#3627408482;&#3627408466;
=
&#3627408449;
1
&#3627408449;
5

Compound gear train
When a series of gears are connected in such a way that
two or more gears rotate about an axis with same
angular velocity, it is known as compound gear train.

&#3627408449;
2
&#3627408449;
1
=
&#3627408455;
1
&#3627408455;
2
,
&#3627408449;
4
&#3627408449;
3
=
&#3627408455;
3
&#3627408455;
4
&
&#3627408449;
6
&#3627408449;
5
=
&#3627408455;
5
&#3627408455;
6

&#3627408449;
2
&#3627408449;
1
×
&#3627408449;
4
&#3627408449;
3
×
&#3627408449;
6
&#3627408449;
5
=
&#3627408455;
1
&#3627408455;
2
×
&#3627408455;
3
&#3627408455;
4
×
&#3627408455;
5
&#3627408455;
6

&#3627408449;
6
&#3627408449;
5
=
&#3627408455;
1
&#3627408455;
2
×
&#3627408455;
3
&#3627408455;
4
×
&#3627408455;
5
&#3627408455;
6


&#3627408455;&#3627408479;&#3627408462;&#3627408470;&#3627408475; &#3627408457;&#3627408462;&#3627408473;&#3627408482;&#3627408466;=
&#3627408477;&#3627408479;&#3627408476;&#3627408465;&#3627408482;&#3627408464;&#3627408481; &#3627408476;&#3627408467; &#3627408475;&#3627408482;&#3627408474;&#3627408463;&#3627408466;&#3627408479; &#3627408476;&#3627408467; &#3627408481;&#3627408466;&#3627408466;&#3627408481;ℎ &#3627408476;&#3627408475; &#3627408465;&#3627408479;&#3627408470;&#3627408483;&#3627408470;&#3627408475;&#3627408468; &#3627408468;&#3627408466;&#3627408462;&#3627408479;&#3627408480;
&#3627408477;&#3627408479;&#3627408476;&#3627408465;&#3627408482;&#3627408464;&#3627408481; &#3627408476;&#3627408467; &#3627408475;&#3627408482;&#3627408474;&#3627408463;&#3627408466;&#3627408479; &#3627408476;&#3627408467; &#3627408481;&#3627408466;&#3627408466;&#3627408481;ℎ &#3627408476;&#3627408475; &#3627408465;&#3627408479;&#3627408470;&#3627408483;&#3627408466;&#3627408475; &#3627408468;&#3627408466;&#3627408462;&#3627408479;&#3627408480;

(
&#3627409172;
2
&#3627409172;
1
=
&#3627408449;
2
&#3627408449;
1
)

72

Reverted gear train
If the gears of first and last wheels of gear train coincide, it is called reverted gear train. Such an arrangement is
used in clocks and in simple lathes where back gear is used to give a slow aped to the chuck.
&#3627408449;
4
&#3627408449;
1
=
&#3627408477;&#3627408479;&#3627408476;&#3627408465;&#3627408482;&#3627408464;&#3627408481; &#3627408476;&#3627408467; &#3627408475;&#3627408482;&#3627408474;&#3627408463;&#3627408466;&#3627408479; &#3627408476;&#3627408467; &#3627408481;&#3627408466;&#3627408466;&#3627408481;ℎ &#3627408476;&#3627408475; &#3627408465;&#3627408479;&#3627408470;&#3627408483;&#3627408470;&#3627408475;&#3627408468; &#3627408468;&#3627408466;&#3627408462;&#3627408479;&#3627408480;
&#3627408477;&#3627408479;&#3627408476;&#3627408465;&#3627408482;&#3627408464;&#3627408481; &#3627408476;&#3627408467; &#3627408475;&#3627408482;&#3627408474;&#3627408463;&#3627408466;&#3627408479; &#3627408476;&#3627408467; &#3627408481;&#3627408466;&#3627408466;&#3627408481;ℎ &#3627408476;&#3627408475; &#3627408465;&#3627408479;&#3627408470;&#3627408483;&#3627408466;&#3627408475; &#3627408468;&#3627408466;&#3627408462;&#3627408479;&#3627408480;
=
&#3627408455;
1&#3627408455;
3
&#3627408455;
2&#3627408455;
4

If r is the pitch circle radius of the gear,
r1+r2= r3+r4


Planetary or Epicyclic gear train
A gear train having relative motion of axes is called planetary or epicyclic gear train system. In an epicyclic gear
train, the axis of at least one of the gears also moves relative to the frame.
Consider two gears S & P, the axes of which are connected by an arm a, if the arm a is fixed the
wheels S & P constitute a simple train. However, if the wheel S is foxed so the arm can rotate
about the axis of S, the wheel P would also move around S. therefore it is an epicyclic train.
Large speed reductions are possible if the fixed wheel is annular.
In epicyclic gear, usually one gear is fixed.

Analysis of Epicyclic gear train
Epicyclic gear trains usually have complex motions.
Assume that arm a is fixed. Turn S through x revolutions in the clockwise directions. Assuming
clockwise motion of a wheel as positive and counter-clockwise direction as negative.
Revolutions made by a = 0
Revolutions made by S = x
Revolutions made by P = -(Ts/Tp) × x
Now if the mechanism is locked together and turned through number of revolutions, the relative
motions between a, S & P will not alter. Let the locked system be turned through y revolutions in the clockwise
direction. S is turned through x revolutions. Then
Revolutions made by a = y
Revolutions made by S = y + x
Revolutions made by P = y -(Ts/Tp) × x
This implies that if the arm ‘a’ turns through y revolutions and S through (y+x) revolutions in the same
direction then P will rotate through (y -(Ts/Tp) × x) revolutions in space or relative to the foxed axis of S.
Arm (a) (Na) Gear S (Ns) Gear P (Np)
0 1 - (Ts/Tp)
0 +x - (Ts/Tp) ×x
+y y+x y - (Ts/Tp) ×x
(Arm a is fixed)

73

Epicyclic Gear Train—Sun and Planet Gear
The annulus gear D meshes with the planet P and
sun gear meshes with planet P. When the annulus
gear is fixed, the sun gear provides the drive and
when the sun gear is fixed, the annulus gear
provides the drive. In both cases, the arm acts as a
follower.
Let TS , TP and TD be the teeth and NS, NP and ND be
the speeds for the gears S, P and D respectively.



a S(Ns) P(NP) R(NR)
0 1 - (TS/TP) - (TS/TR)
0 x - (TS/TP) × x - (TS/TR) × x
y x + y y- (TS/TP) × x y -(TS/TR) × x
When the ring gear (annulus) is fixed, Ns = 0 → y―(TS/TR) × x = 0.
Compound Epicyclic Gear Train—Sun and Planet Gear
The annulus gear A meshes with the gear B and the sun gear D meshes with the
gear C. It may be noted that when the annulus gear is fixed, the sun gear
provides the drive and when the sun gear is fixed, the annulus gear provides the
drive. In both cases, the arm acts as a follower.
Torques in epicyclic Gear train
Angular velocities of all elements are considered constant.
System is at equilibrium.
∑??????=0 ; ??????
&#3627408480;+??????
&#3627408477;+??????
&#3627408479;+??????
&#3627408462;=0
Planet is rotating freely, τp=0.
??????
&#3627408480;+??????
&#3627408479;+??????
&#3627408462;=0
Total power of system = constant
∑??????&#3627409172;=0 ⇒∑??????&#3627408449;=0; ??????
&#3627408480;&#3627408449;
&#3627408480;+??????
&#3627408479;&#3627408449;
&#3627408479;+??????
&#3627408462;&#3627408449;
&#3627408462;=0
If the ring gear is fixed,
??????
&#3627408480;&#3627408449;
&#3627408480;+??????
&#3627408462;&#3627408449;
&#3627408462;=0
S Arm(a) Gear D Compound gear B-C Gear A
0 1 ―(TD/TC) ―(TD/TC) × (TB/TA)
0 x ―(TD/TC) × x ―(TD/TC) × (TB/TA) × x
y x+y y―(TD/TC) × x y―(TD/TC) × (TB/TA) × x

74

Differentials

Its function is
(a) to transmit motion from the engine shaft to the rear driving wheels, and
(b) to rotate the rear wheels at different speeds while the automobile is taking a turn.
If the automobile is running on a straight path, the rear wheels are driven directly by the engine and speed of
both the wheels is same. But when the automobile is taking a turn, the outer wheel will run faster than the inner
wheel because at that time the outer rear wheel must cover more distance than the inner rear wheel. This is
achieved by epicyclic gear train with bevel gears.
When the automobile runs on a straight path, the gears C and D must rotate together. These gears are rotated
through the spindle on the gear B. The gears E and F do not rotate on the spindle. But when the automobile is
taking a turn, the inner rear wheel should have lesser speed than the outer rear wheel and due to relative speed
of the inner and outer gears D and C, the gears E and F start rotating about the spindle axis and at the same
time revolve about the axle axis.
Due to this epicyclic effect, the speed of the inner rear wheel decreases by a certain amount and the speed of the
outer rear wheel increases, by the same amount.

75

GOVERNORS
The function of the governor is to maintain the speed of an engine within specified limits whenever there is a
variation of load.
Types of Governors-

Centrifugal governor
The centrifugal governors are based on the balancing of centrifugal force on the rotating balls by an equal and
opposite radial force, known as the controlling force
Inertia governor
In this type, position of balls is affected by the forces set up by an angular acceleration or deceleration of the
given spindle in addition to centrifugal forces on the balls.

76

Watt Governor
Upper link is fixed, but lower links are fixed to a sleeve free to move on the vertical spindle.
When the spindle rotates, the ball takes up a position depending upon the
speed of the spindle.
If it lowers, they move near the axis due to reduction in centrifugal force
(reduction in angular velocity), this movement is further taken to throttle
of the engine to maintain fuel supply.
The height of the governor ‘h’ decreases with increase in speed and vice
versa.
&#3627408536;=&#3627408474;&#3627408468; ,&#3627408516;&#3627408518;&#3627408527;&#3627408533;&#3627408531;&#3627408522;&#3627408519;&#3627408534;&#3627408520;&#3627408514;&#3627408525; &#3627408519;&#3627408528;&#3627408531;&#3627408516;&#3627408518;=&#3627408474;&#3627408479;&#3627409172;
2
, &#3627408507;→&#3627408481;&#3627408466;&#3627408475;&#3627408480;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408475; &#3627408482;&#3627408477;&#3627408477;&#3627408466;&#3627408479; &#3627408473;&#3627408470;&#3627408475;&#3627408472;

&#3627408455;&#3627408464;&#3627408476;&#3627408480;&#3627409155;=&#3627408474;&#3627408468; &#3627408462;&#3627408475;&#3627408465; &#3627408455;&#3627408480;&#3627408470;&#3627408475;&#3627409155;=&#3627408474;&#3627408479;&#3627409172;
2

&#3627408481;&#3627408462;&#3627408475;&#3627409155;=
&#3627408474;&#3627408479;&#3627409172;
2
&#3627408474;&#3627408468;
=
&#3627408479;&#3627409172;
2
&#3627408468;
=
&#3627408479;


ℎ=
&#3627408468;
&#3627409172;
2
=
9.81
(
2&#3627409163;&#3627408449;
60
)
2
=
895
&#3627408449;
2
&#3627408474;
In Watt governor, movement of sleeve is very less at high speeds and thus is unsuitable for these high speeds.
But these can be overcome by means of spring.
Porter Governor
If the sleeve of a Watt governor is loaded with a heavy mass, it becomes a Porter
governor.
The net force on sleeve will be Mg+f if sleeve moves up and Mg-f when sleeve
moves down.
I → Instantaneous center of rotation of the link AB.

Considering left hand side for equilibrium of link AB, taking moments of forces
about the instantaneous center I.
&#3627408474;&#3627408479;&#3627409172;
2
.&#3627408462;=&#3627408474;&#3627408468;.&#3627408464;+
&#3627408448;&#3627408468;±&#3627408467;
2
(&#3627408464;+&#3627408463;) ⇒ &#3627408474;&#3627408479;&#3627409172;
2
=&#3627408474;&#3627408468;.
&#3627408464;
&#3627408462;
+
&#3627408448;&#3627408468;±&#3627408467;
2
(
&#3627408464;
&#3627408462;
+
&#3627408463;
&#3627408462;
)
&#3627408474;&#3627408479;&#3627409172;
2
=&#3627408474;&#3627408468;&#3627408481;&#3627408462;&#3627408475;&#3627409155;+
&#3627408448;&#3627408468;±&#3627408467;
2
(&#3627408481;&#3627408462;&#3627408475;&#3627409155;+&#3627408481;&#3627408462;&#3627408475;&#3627409149;)
&#3627408474;&#3627408479;&#3627409172;
2
=&#3627408481;&#3627408462;&#3627408475;&#3627409155;(&#3627408474;&#3627408468;+
&#3627408448;&#3627408468;±&#3627408467;
2
(1+&#3627408472;)) &#3627408472;=
&#3627408481;&#3627408462;&#3627408475;&#3627409149;
&#3627408481;&#3627408462;&#3627408475;&#3627409155;

&#3627408463;&#3627408486; &#3627408481;&#3627408462;&#3627408472;&#3627408470;&#3627408475;&#3627408468; &#3627408481;&#3627408462;&#3627408475;&#3627409155;=
&#3627408479;

& &#3627409172;=
2&#3627409163;&#3627408449;
60
&#3627408462;&#3627408475;&#3627408465; &#3627408480;&#3627408470;&#3627408474;&#3627408477;&#3627408473;&#3627408470;&#3627408467;&#3627408486;&#3627408470;&#3627408475;&#3627408468;
&#3627408449;
2
=
895

(
2&#3627408474;&#3627408468;+(&#3627408448;&#3627408468;±&#3627408467;)(1+&#3627408472;)
2&#3627408474;&#3627408468;
)
This equation provides two values of N, by taking Mg+f in one eqn. and Mg-f in
another, giving two values for one h.
If sleeve has just moved down and if the speed of the engine increases, the
resistance to motion would be Mg+f and the sleeve does not move up until speed
rises to a level as to overcome this resistance.

77

Proell Governor
A Proell governor is a porter governor with two balls are fixed on the upward extensions of lower links which
are inform of bent links.
centrifugal force =mr’ω
2
.
&#3627408484;&#3627408466;&#3627408470;&#3627408468;ℎ&#3627408481; &#3627408476;&#3627408467; &#3627408481;ℎ&#3627408466; &#3627408480;&#3627408473;&#3627408466;&#3627408466;&#3627408483;&#3627408466;=
1
2
(&#3627408448;&#3627408468;±&#3627408467;)
&#3627408474;&#3627408479;

&#3627409172;
2
&#3627408466;=&#3627408474;&#3627408468;(&#3627408464;+&#3627408479;−&#3627408479;

)+
&#3627408448;&#3627408468;±&#3627408467;
2
(&#3627408464;+&#3627408463;)
AE is vertical, r=r’
&#3627408474;&#3627408479;

&#3627409172;
2
=
&#3627408462;
&#3627408466;
(&#3627408474;&#3627408468;
&#3627408464;
&#3627408462;
+
&#3627408448;&#3627408468;±&#3627408467;
2
(
&#3627408464;
&#3627408462;
+
&#3627408463;
&#3627408462;
))
&#3627408474;&#3627408479;

&#3627409172;
2
=
&#3627408462;
&#3627408466;
(&#3627408474;&#3627408468;&#3627408481;&#3627408462;&#3627408475;&#3627409155;+
&#3627408448;&#3627408468;±&#3627408467;
2
(&#3627408481;&#3627408462;&#3627408475;&#3627409155;+&#3627408481;&#3627408462;&#3627408475;&#3627409149;))
&#3627408472;=
&#3627408481;&#3627408462;&#3627408475;&#3627409149;
&#3627408481;&#3627408462;&#3627408475;&#3627409155;
& &#3627408481;&#3627408462;&#3627408475;&#3627409155;=
&#3627408479;

& &#3627409172;=
2&#3627409163;&#3627408449;
60
&#3627408462;&#3627408475;&#3627408465; &#3627408480;&#3627408470;&#3627408474;&#3627408477;&#3627408473;&#3627408470;&#3627408467;&#3627408486;&#3627408470;&#3627408475;&#3627408468;
&#3627408449;
2
=
895


&#3627408462;
&#3627408466;
(
2&#3627408474;&#3627408468;+(&#3627408448;&#3627408468;±&#3627408467;)(1+&#3627408472;)
2&#3627408474;&#3627408468;
)
Hartnell Governor

As the speed increases and the balls move away from the spindle axis, bell crank lever move on pivot and lifts
the sleeve against the spring force. If the speed decreases the sleeve moves downwards.
&#3627408441;
1=&#3627408474;&#3627408479;&#3627409172;
2

Taking moments about the fulcrum,
&#3627408441;
1&#3627408462;
1=
1
2
(&#3627408448;&#3627408468;+&#3627408441;
&#3627408480;1+&#3627408467;)&#3627408463;
1+&#3627408474;&#3627408468;&#3627408464;
1
&#3627408441;
2&#3627408462;
2=
1
2
(&#3627408448;&#3627408468;+&#3627408441;
&#3627408480;2+&#3627408467;)&#3627408463;
2+&#3627408474;&#3627408468;&#3627408464;
2
&#3627408436;&#3627408480; &#3627409155;≈0 ⇒ &#3627408462;
1=&#3627408462;
2 & &#3627408463;
1=&#3627408463;
2 & &#3627408464;
1=&#3627408464;
2 =0
&#3627408441;
1&#3627408462;=
1
2
(&#3627408448;&#3627408468;+&#3627408441;
&#3627408480;1+&#3627408467;)&#3627408463; (&#3627408522;)
&#3627408441;
2&#3627408462;=
1
2
(&#3627408448;&#3627408468;+&#3627408441;
&#3627408480;2+&#3627408467;)&#3627408463; (&#3627408522;&#3627408522;)
Subtracting (i) from (ii)

78

&#3627408441;
&#3627408480;2− &#3627408441;
&#3627408480;1=
2&#3627408462;
&#3627408463;
(&#3627408441;
2−&#3627408441;
1)
h1 = movement of the sleeve, s = stiffness of spring
&#3627408454;&#3627408477;&#3627408479;&#3627408470;&#3627408475;&#3627408468; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;= &#3627408441;
&#3627408480;2− &#3627408441;
&#3627408480;1= ℎ
1&#3627408480;=
2&#3627408462;
&#3627408463;
(&#3627408441;
2−&#3627408441;
1)

1=&#3627409155;&#3627408463;=
&#3627408479;
1−&#3627408479;
2
&#3627408462;
&#3627408463;
&#3627408480;=2(
&#3627408462;
&#3627408463;
)
2
(
&#3627408441;
2−&#3627408441;
1
&#3627408479;
1−&#3627408479;
2
)
Hartung governor
It is a spring-controlled governor in which the vertical arms of the bell-crank lever are fitted with spring balls.
The spring compresses against the frame of the governor while the rollers at the horizontal arm press the
sleeve.

Fc = centrifugal force, m=mass of each ball,
S=spring force, s=stiffness of the spring,
M= mass of sleeve, r= radial distance of the ball,
ω =angular velocity of the balls at radius r,
ra=radius at which spring force is zero,
a=length of vertical arm of bell-crank lever,
b= length of horizontal arm of bell-crank lever.
Neglecting the obliquity of the arms and taking moments about the fulcrum A,
&#3627408441;
&#3627408464;.&#3627408462;=&#3627408454;.&#3627408462;+
&#3627408448;&#3627408468;
2
.&#3627408463;
&#3627408474;&#3627408479;&#3627409172;
2
=&#3627408480;(&#3627408479;−&#3627408479;
&#3627408476;).&#3627408462;+
&#3627408448;&#3627408468;
2
.&#3627408463;

79

Wilson-Hartnell Governor
It is a spring-loaded type of governor.

Here two bell crank levers are pivoted at the end of two arms which rotate with the spindle. Two balls are
connected by 2 main springs arranged symmetrically on either side of the sleeve. While rotating, when the ball
radius increases with speed, the spring exerts an inward pull Fs on the balls and the rollers press against the
sleeve which is raised.
An auxiliary spring is provided, as main spring is not adjustable. The auxiliary spring tends to keep the sleeve
downwards so that it assists main spring.
Sa = stiffness of auxiliary spring, s = stiffness of each of main spring, F’s = force applied by the auxiliary spring
&#3627408441;
1&#3627408462;
1− &#3627408441;
&#3627408480;1&#3627408462;
1=
1
2
(&#3627408448;&#3627408468;+&#3627408441;

&#3627408480;1
&#3627408486;
&#3627408485;
+&#3627408467;)&#3627408463;
1+&#3627408474;&#3627408468;&#3627408464;
1
&#3627408441;
2&#3627408462;
2− &#3627408441;
&#3627408480;2&#3627408462;
2=
1
2
(&#3627408448;&#3627408468;+&#3627408441;

&#3627408480;2
&#3627408486;
&#3627408485;
+&#3627408467;)&#3627408463;
2+&#3627408474;&#3627408468;&#3627408464;
2
If obliquity effects are neglected, a1 = a2 = a, b1 =b2 =b & c1 = c2 =0
(&#3627408441;
1− &#3627408441;
&#3627408480;1)&#3627408462;=
1
2
(&#3627408448;&#3627408468;+&#3627408441;

&#3627408480;1
&#3627408486;
&#3627408485;
+&#3627408467;)&#3627408463; (&#3627408522;)
(&#3627408441;
2− &#3627408441;
&#3627408480;2)&#3627408462;=
1
2
(&#3627408448;&#3627408468;+&#3627408441;

&#3627408480;2
&#3627408486;
&#3627408485;
+&#3627408467;)&#3627408463; (&#3627408522;&#3627408522;)
Subtracting (i) from (ii)
(&#3627408441;
2− &#3627408441;
1)&#3627408462;− (&#3627408441;
&#3627408480;2− &#3627408441;
&#3627408480;1)&#3627408462;=(&#3627408441;

&#3627408480;2− &#3627408441;

&#3627408480;1)·
&#3627408486;&#3627408463;
2&#3627408485;

The main spring consists of 2 two springs. Therefore, the force exerted is given by
Fs2 – Fs1 = 2 × Force exerted by each spring = 2 × Stiffness of each spring × Elongation of each spring
Fs2 – Fs1 = 2 × s × 2 × (r2 – r1) =4s × (r2 – r1)
h1 = movement of sleeve, h2 = deflection of auxiliary spring
&#3627408441;′
&#3627408480;2−&#3627408441;′
&#3627408480;1= ℎ
2.&#3627408454;
&#3627408462;=(ℎ
1
&#3627408486;
&#3627408485;
)&#3627408454;
&#3627408462;=(&#3627408479;
2−&#3627408479;
1)
&#3627408463;
&#3627408462;
&#3627408486;
&#3627408485;
&#3627408454;
&#3627408462;
Substituting all values, we get
&#3627408441;
2− &#3627408441;
1
&#3627408479;
2−&#3627408479;
1
=4&#3627408480;+
&#3627408454;
&#3627408462;
2
(
&#3627408463;
&#3627408462;
&#3627408486;
&#3627408485;
)
2

80

Characteristics of governor
Stability
A governor is said to be stable if it brings the speed of the engine to required value and there is not much
hunting.
Controlling force
The inward force acting on the rotating balls is known as controlling force. It is equal and opposite to the
centrifugal reaction.
At equilibrium, Fr = Fc = mrω
2
Stable equilibrium
When the radius increases, Fr and ω increases and sleeve moves up and
constraining the fuel supply, reducing the speed.
Similarly, when the radius decreases, Fr and ω decreases and sleeve down and
increasing fuel supply increasing speed.
This is Stable equilibrium.
&#3627408438;&#3627408476;&#3627408475;&#3627408465;&#3627408470;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408467;&#3627408476;&#3627408479; &#3627408466;&#3627408478;&#3627408482;&#3627408473;&#3627408470;&#3627408463;&#3627408479;&#3627408470;&#3627408482;&#3627408474;= &#3627409149;>&#3627409148;⇒
&#3627408465;&#3627408441;
&#3627408479;
&#3627408465;&#3627408479;
>&#3627408474;&#3627409172;
2

Unstable equilibrium
When the radius increases, Fr and ω decreases and sleeve moves down and
increasing fuel supply increasing speed.
Similarly, when the radius decreases, Fr and ω increases and sleeve moves
up and constraining the fuel supply, reducing the speed.
This is Unstable equilibrium.
&#3627408438;&#3627408476;&#3627408475;&#3627408465;&#3627408470;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408467;&#3627408476;&#3627408479; &#3627408482;&#3627408475;&#3627408480;&#3627408481;&#3627408462;&#3627408463;&#3627408473;&#3627408466; &#3627408466;&#3627408478;&#3627408482;&#3627408473;&#3627408470;&#3627408463;&#3627408479;&#3627408470;&#3627408482;&#3627408474;= &#3627409149;<&#3627409148; ⇒
&#3627408465;&#3627408441;
&#3627408479;
&#3627408465;&#3627408479;
<&#3627408474;&#3627409172;
2

Sensitivity
A governor is said to be sensitive when it readily responds(more) to a small
change of speed.
&#3627408454;&#3627408466;&#3627408475;&#3627408480;&#3627408470;&#3627408481;&#3627408470;&#3627408483;&#3627408470;&#3627408481;&#3627408486; &#3627408476;&#3627408467; &#3627408462; &#3627408468;&#3627408476;&#3627408483;&#3627408466;&#3627408479;&#3627408475;&#3627408476;&#3627408479; =
&#3627408479;&#3627408462;&#3627408475;&#3627408468;&#3627408466; &#3627408476;&#3627408467; &#3627408480;&#3627408477;&#3627408466;&#3627408466;&#3627408465;
&#3627408474;&#3627408466;&#3627408462;&#3627408475; &#3627408480;&#3627408477;&#3627408466;&#3627408466;&#3627408465;

&#3627408454;&#3627408466;&#3627408475;&#3627408480;&#3627408470;&#3627408481;&#3627408470;&#3627408483;&#3627408470;&#3627408481;&#3627408486; &#3627408476;&#3627408467; &#3627408462; &#3627408468;&#3627408476;&#3627408483;&#3627408466;&#3627408479;&#3627408475;&#3627408476;&#3627408479; =
&#3627408449;
1−&#3627408449;
2
&#3627408449;
&#3627408474;&#3627408466;&#3627408462;&#3627408475;
=
&#3627409172;
&#3627408474;&#3627408462;&#3627408485;−&#3627409172;
&#3627408474;??????&#3627408475;
&#3627409172;
&#3627408474;&#3627408466;&#3627408462;&#3627408475;

&#3627408464;&#3627408476;&#3627408466;&#3627408467;&#3627408467;&#3627408470;&#3627408464;&#3627408470;&#3627408466;&#3627408475;&#3627408481; &#3627408476;&#3627408467; &#3627408470;&#3627408475;&#3627408480;&#3627408466;&#3627408475;&#3627408480;&#3627408470;&#3627408481;&#3627408470;&#3627408483;&#3627408466;&#3627408475;&#3627408466;&#3627408480;&#3627408480;=
&#3627408449;
1−&#3627408449;
2
&#3627408449;

Effort of Governor
The effort of a governor is the mean force acting on sleeve to raise or lower it for a given change of speed.
When the speed increases, sleeve raises from an equilibrium position to a new position and the average force
acted on it bring it to new position is Effort of the governor.
&#3627408440;
2
=&#3627408466;&#3627408467;&#3627408467;&#3627408476;&#3627408479;&#3627408481; &#3627408476;&#3627408467; &#3627408462; &#3627408468;&#3627408476;&#3627408483;&#3627408466;&#3627408479;&#3627408475;&#3627408476;&#3627408479;
For a porter governor-
ℎ=
2&#3627408474;&#3627408468;+&#3627408448;&#3627408468;(1+&#3627408472;)
2&#3627408474;&#3627409172;
2
(&#3627408522;)
C → fraction of increase in speed, E is the force applied to prevent it from moving and (F+E) is the force applied
on sleeve.
ℎ=
2&#3627408474;&#3627408468;+(&#3627408448;&#3627408468;+&#3627408440;)(1+&#3627408472;)
2&#3627408474;&#3627408468;(1+&#3627408464;)
2
&#3627409172;
2
(&#3627408522;&#3627408522;)
&#3627408439;&#3627408470;&#3627408483;&#3627408470;&#3627408465;&#3627408470;&#3627408475;&#3627408468; (&#3627408470;) &#3627408463;&#3627408486; (&#3627408470;&#3627408470;) &#3627408462;&#3627408475;&#3627408465; &#3627408480;&#3627408470;&#3627408474;&#3627408477;&#3627408473;&#3627408470;&#3627408467;&#3627408486;&#3627408470;&#3627408475;&#3627408468;,&#3627408484;&#3627408466; &#3627408468;&#3627408466;&#3627408481;
&#3627408440;
2
=
&#3627408464;&#3627408468;
1+&#3627408472;
[2&#3627408474;+&#3627408448;(1+&#3627408472;)]

81

For a Watt governor
&#3627408436;&#3627408480; &#3627408448;=0,&#3627408472;=1
&#3627408440;
2
=&#3627408464;&#3627408474;&#3627408468;

For a Hartnell governor
&#3627408440;
2
=&#3627408464;(&#3627408448;&#3627408468;+&#3627408441;
&#3627408480;)
Power of a governor
The power of a governor is the work done at the sleeve for a given percent change of speed.
&#3627408451;=
&#3627408440;
2
×&#3627408465;&#3627408470;&#3627408480;&#3627408477;&#3627408473;&#3627408462;&#3627408464;&#3627408466;&#3627408474;&#3627408466;&#3627408475;&#3627408481; &#3627408476;&#3627408467; &#3627408481;ℎ&#3627408466; &#3627408480;&#3627408473;&#3627408466;&#3627408466;&#3627408483;&#3627408466;
For a porter governor
&#3627408451;&#3627408476;&#3627408484;&#3627408466;&#3627408479;=
&#3627408440;
2
×(2×&#3627408465;&#3627408470;&#3627408480;&#3627408477;&#3627408473;&#3627408462;&#3627408464;&#3627408466;&#3627408474;&#3627408466;&#3627408475;&#3627408481; &#3627408476;&#3627408467; &#3627408481;ℎ&#3627408466; &#3627408480;&#3627408473;&#3627408466;&#3627408466;&#3627408483;&#3627408466;)
&#3627408465;&#3627408470;&#3627408480;&#3627408477;&#3627408473;&#3627408462;&#3627408464;&#3627408466;&#3627408474;&#3627408466;&#3627408475;&#3627408481; &#3627408476;&#3627408467; &#3627408481;ℎ&#3627408466; &#3627408480;&#3627408473;&#3627408466;&#3627408466;&#3627408483;&#3627408466;=(ℎ
1−ℎ
2) 2ℎ(
2&#3627408464;
1+2&#3627408464;
)
&#3627408451;&#3627408476;&#3627408484;&#3627408466;&#3627408479; &#3627408476;&#3627408467; &#3627408468;&#3627408476;&#3627408483;&#3627408466;&#3627408479;&#3627408475;&#3627408476;&#3627408479;=(&#3627408474;+&#3627408448;)&#3627408464;&#3627408468;×2ℎ(
2&#3627408464;
1+2&#3627408464;
)= (&#3627408474;+&#3627408448;)&#3627408468;ℎ(
4&#3627408464;
2
1+2&#3627408464;
)
Hunting in a governor
Due to high sensitivity of governor, sleeve tends to change its position rapidly between limits and leads to an
oscillatory behavior of speed of governor.
For example, when the load on the engine increases, the engine speed decreases and, if the governor is very
sensitive, the governor sleeve immediately falls to its lowest position. This will result in the opening of the
control valve wide which will supply the
fuel to the engine more than its requirement
so that the engine speed rapidly increases
again, and the governor sleeve rises to its
highest position. Due to this movement of
the sleeve, the control valve will cut off the
fuel supply to the engine and thus the
engine speed begins to fall once again. This
cycle is repeated indefinitely and is known
as Hunting.
Isochronism
A governor is said to be isochronous when the equilibrium speed is constant (i.e. range of speed is zero) for all
radii of rotation of the balls within the working range, neglecting friction. The Isochronism is the stage of
infinite sensitivity.
For isochronism, range of speed should be zero i.e. N2 – N1 = 0 or N2 = N1.
The condition for isochronism is
&#3627408448;&#3627408468;+&#3627408441;
&#3627408480;1
&#3627408448;&#3627408468;+&#3627408441;
&#3627408480;2
=
&#3627408479;
1
&#3627408479;
2

82

GYROSCOPE
Gyroscope→ gyro = to rotate, scope = to maintain.
Gyroscope is a device which maintains the direction of angular momentum.
Angular Velocity
The angular velocity of a rotating body is specified by the magnitude
of velocity, the direction of the axis of rotor, the sense of rotation of
rotor, i.e., clockwise, or anticlockwise.
Angular velocity is represented by a vector and if the rotation is anti-
clockwise, then the vector is towards us and if it clockwise, the vector
is into paper.
Angular acceleration

ω changed to ω+δω, and its direction changed to Ox

.
oa is the direction of ω, ob is the direction of ω+δω, ab is the direction if change of angular velocity.ac is the
direction of change in angular velocity along x axis, cb represents change in angular velocity along y axis.
&#3627408438;ℎ&#3627408462;&#3627408475;&#3627408468;&#3627408466; &#3627408470;&#3627408475; &#3627408462;&#3627408475;&#3627408468;&#3627408482;&#3627408473;&#3627408462;&#3627408479; &#3627408483;&#3627408466;&#3627408473;&#3627408476;&#3627408464;&#3627408470;&#3627408481;&#3627408486;=&#3627408514;&#3627408516;=(&#3627409172;+&#3627409151;&#3627409172;)&#3627408464;&#3627408476;&#3627408480;&#3627409151;&#3627409155;−&#3627409172;
&#3627408453;&#3627408462;&#3627408481;&#3627408466; &#3627408476;&#3627408467; &#3627408464;ℎ&#3627408462;&#3627408475;&#3627408468;&#3627408466; &#3627408476;&#3627408467; &#3627408462;&#3627408475;&#3627408482;&#3627408468;&#3627408482;&#3627408473;&#3627408462;&#3627408479; &#3627408483;&#3627408466;&#3627408473;&#3627408476;&#3627408464;&#3627408470;&#3627408481;&#3627408486;=
(&#3627409172;+&#3627409151;&#3627409172;)&#3627408464;&#3627408476;&#3627408480;&#3627409151;&#3627409155;−&#3627409172;
&#3627409151;&#3627408481;

&#3627408436;&#3627408475;&#3627408468;&#3627408482;&#3627408473;&#3627408462;&#3627408479; &#3627408462;&#3627408464;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475;=&#3627408473;&#3627408470;&#3627408474;
??????&#3627408481;→0
(&#3627409172;+&#3627409151;&#3627409172;)&#3627408464;&#3627408476;&#3627408480;&#3627409151;&#3627409155;−&#3627409172;
&#3627409151;&#3627408481;

As δt⟶ 0, δ ⟶ 0 and cos δ ⟶1
&#3627408436;&#3627408475;&#3627408468;&#3627408482;&#3627408473;&#3627408462;&#3627408479; &#3627408462;&#3627408464;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408462;&#3627408473;&#3627408476;&#3627408475;&#3627408468; &#3627408485;−&#3627408462;&#3627408485;&#3627408470;&#3627408480;=&#3627408473;&#3627408470;&#3627408474;
??????&#3627408481;→0
(&#3627409172;+&#3627409151;&#3627409172;)−&#3627409172;
&#3627409151;&#3627408481;
=
&#3627408465;&#3627409172;
&#3627408465;&#3627408481;

&#3627408453;&#3627408462;&#3627408481;&#3627408466; &#3627408476;&#3627408467; &#3627408464;ℎ&#3627408462;&#3627408475;&#3627408468;&#3627408466; &#3627408476;&#3627408467; &#3627408462;&#3627408475;&#3627408468;&#3627408482;&#3627408473;&#3627408462;&#3627408479; &#3627408462;&#3627408464;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475;,&#3627408516;&#3627408515;=
(&#3627409172;+&#3627409151;&#3627409172;)&#3627408480;&#3627408470;&#3627408475;&#3627409151;&#3627409155;
&#3627409151;&#3627408481;

As δt⟶0, δ ⟶0 and sin(δ) ⟶ δ
&#3627408436;&#3627408475;&#3627408468;&#3627408482;&#3627408473;&#3627408462;&#3627408479; &#3627408462;&#3627408464;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408462;&#3627408473;&#3627408476;&#3627408475;&#3627408468; &#3627408486;−&#3627408462;&#3627408485;&#3627408470;&#3627408480;=&#3627408473;&#3627408470;&#3627408474;
??????&#3627408481;→0
(&#3627409172;+&#3627409151;&#3627409172;)&#3627409151;
&#3627409151;&#3627408481;
=&#3627409172;
&#3627408465;
&#3627408465;&#3627408481;

&#3627408455;&#3627408476;&#3627408481;&#3627408462;&#3627408473; &#3627408436;&#3627408475;&#3627408468;&#3627408482;&#3627408473;&#3627408462;&#3627408479; &#3627408462;&#3627408464;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475;,??????=&#3627408514;&#3627408516;⃗⃗⃗⃗⃗+&#3627408516;&#3627408515;⃗⃗⃗⃗⃗=&#3627408514;&#3627408515;⃗⃗⃗⃗⃗⃗=
&#3627408465;&#3627409172;
&#3627408465;&#3627408481;
+&#3627409172;
&#3627408465;
&#3627408465;&#3627408481;

83

Gyroscopic Torque
Let I be the moment of inertia of rotor and ω is the angular velocity about horizontal axis of Ox. Let this axis of
spin turn through a small angle δ in horizontal plane to a position Ox’ in time δt.
The direction of angular velocity is changed
from ox to ox’, xx’ is the change of angular
velocity due to change of axis of spin of the
rotor. This results in angular acceleration
which is in the same direction of change of
angular velocity (xx’).



&#3627408464;ℎ&#3627408462;&#3627408475;&#3627408468;&#3627408466; &#3627408470;&#3627408475; &#3627408462;&#3627408475;&#3627408468;&#3627408482;&#3627408473;&#3627408462;&#3627408479; &#3627408483;&#3627408466;&#3627408473;&#3627408476;&#3627408464;&#3627408470;&#3627408481;&#3627408486;=&#3627408537;&#3627408537;′=&#3627409172;×&#3627409151;&#3627409155;
&#3627408436;&#3627408475;&#3627408468;&#3627408482;&#3627408473;&#3627408462;&#3627408479; &#3627408462;&#3627408464;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475;=??????=&#3627409172;×
&#3627409151;&#3627409155;
&#3627409151;&#3627408481;

&#3627408436;&#3627408480; &#3627409151;&#3627408481;→0,??????=&#3627409172;×
&#3627408465;&#3627409155;
&#3627408465;&#3627408481;


&#3627408465;&#3627409155;
&#3627408465;&#3627408481;
&#3627408470;&#3627408480; &#3627408481;ℎ&#3627408466; &#3627408483;&#3627408466;&#3627408473;&#3627408476;&#3627408464;&#3627408470;&#3627408481;&#3627408486; &#3627408476;&#3627408467; &#3627408481;ℎ&#3627408466; &#3627408462;&#3627408485;&#3627408470;&#3627408480; &#3627408476;&#3627408467; &#3627408480;&#3627408477;&#3627408470;&#3627408475; &#3627408470;&#3627408480; &#3627408464;&#3627408462;&#3627408473;&#3627408473;&#3627408466;&#3627408465; &#3627408483;&#3627408466;&#3627408473;&#3627408476;&#3627408464;&#3627408470;&#3627408481;&#3627408486; &#3627408476;&#3627408467; &#3627408477;&#3627408479;&#3627408466;&#3627408464;&#3627408470;&#3627408480;&#3627408480;&#3627408470;&#3627408476;&#3627408475; &#3627408462;&#3627408475;&#3627408465; &#3627408470;&#3627408480; &#3627408465;&#3627408466;&#3627408475;&#3627408476;&#3627408481;&#3627408466;&#3627408465; &#3627408463;&#3627408486; &#3627409172;
&#3627408477;.
&#3627408436;&#3627408475;&#3627408468;&#3627408482;&#3627408473;&#3627408462;&#3627408479; &#3627408462;&#3627408464;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475;=&#3627409148;=&#3627409172;×&#3627409172;
&#3627408477;
The torque required to produce this acceleration is known as Gyroscopic torque and couple applied to axis of
spin to rotate it with ωp about axis of precession. The vector xx′ lies in the plane XOZ or the horizontal plane. In
case of a very small displacement δθ, the vector xx′ will be perpendicular to the vertical plane XOY. Therefore,
the couple causing this change in the angular momentum will lie in the plane XOY.


Ox is known as axis of spin, YOZ is plane of spin, Oy is the axis of precession, XOZ is plane of precession, Oz is
the axis of gyroscopic couple, XOY is plane of active gyroscopic couple.
A reactive gyroscopic torque or reactive torque is subjected to the disc which tends to rotate the axis of spin in
opposite direction.

&#3627408442;&#3627408486;&#3627408479;&#3627408476;&#3627408480;&#3627408464;&#3627408476;&#3627408477;&#3627408470;&#3627408464; &#3627408455;&#3627408476;&#3627408479;&#3627408478;&#3627408482;&#3627408466;=&#3627408455;=&#3627408444;×&#3627409148;=&#3627408444;×&#3627409172;×&#3627409172;
&#3627408477;

84

Gyroscopic effect on Aeroplanes
When an aeroplane takes a left turn
Here the propeller rotates in anti-clockwise direction, and to
change the direction an active gyroscopic couple is acted, it acts
in clockwise direction when seen from front view.

Reactive gyroscopic couple, acts in anti-clockwise direction (seen from front view) i.e., couple direction is x’x,
but acts in plane of precession xoy.
This reactive gyroscopic couple make the nose rise and tail go down.
When the aeroplane takes a right turn under similar conditions as discussed above, the effect of the reactive
gyroscopic couple will be to dip the nose and raise the tail of the aeroplane.
Gyroscopic effect on Ships

Just like the aeroplane, when the ship turns left, there will be an anti-clock wise couple acting on the ship and
bow rises and stern lowers and when the ship turns right, there will be clock wise couple acting on the ship and
bow lower and stern rises.

When the pitching is upward, the effect of the reactive gyroscopic couple, as shown in (b), will try to move the
ship toward starboard. On the other hand, if the pitching is downward, the effect of the reactive gyroscopic
couple, as shown in (c), is to turn the ship towards port side.
In case of rolling of a ship, the axis of precession (i.e. longitudinal axis) is always parallel to the axis of spin for
all positions. Hence, there is no effect of the gyroscopic couple acting on the body of a ship.

85

Stability of an Automobile (four-wheel drive)
A little consideration will show, that the weight of the vehicle (W) will be
equally distributed over the four wheels which will act downwards. The
reaction between each wheel and the road surface of the same magnitude
will act upwards
When the vehicle takes a turn towards left due to the precession and other
rotating parts, a gyroscopic couple will act.
&#3627408442;&#3627408486;&#3627408479;&#3627408476;&#3627408480;&#3627408464;&#3627408476;&#3627408477;&#3627408470;&#3627408464; &#3627408464;&#3627408476;&#3627408482;&#3627408477;&#3627408473;&#3627408466; &#3627408465;&#3627408482;&#3627408466; &#3627408481;&#3627408476; 4 &#3627408484;ℎ&#3627408466;&#3627408466;&#3627408473;&#3627408480;=&#3627408438;
&#3627408484;=4&#3627408444;
&#3627408484;&#3627409172;
&#3627408484;&#3627409172;
&#3627408477;
&#3627408442;&#3627408486;&#3627408479;&#3627408476;&#3627408480;&#3627408464;&#3627408476;&#3627408477;&#3627408470;&#3627408464; &#3627408464;&#3627408476;&#3627408482;&#3627408477;&#3627408473;&#3627408466; &#3627408465;&#3627408482;&#3627408466; &#3627408481;&#3627408476; &#3627408479;&#3627408476;&#3627408481;&#3627408462;&#3627408481;&#3627408470;&#3627408475;&#3627408468; &#3627408477;&#3627408462;&#3627408479;&#3627408481;&#3627408480; &#3627408476;&#3627408467; &#3627408466;&#3627408475;&#3627408468;&#3627408470;&#3627408475;&#3627408466;=&#3627408438;
&#3627408440;=&#3627408444;
&#3627408440;.&#3627408442;.&#3627409172;
&#3627408484;&#3627409172;
&#3627408477;
&#3627408449;&#3627408466;&#3627408481; &#3627408442;&#3627408486;&#3627408479;&#3627408476;&#3627408480;&#3627408464;&#3627408476;&#3627408477;&#3627408470;&#3627408464; &#3627408464;&#3627408476;&#3627408482;&#3627408477;&#3627408473;&#3627408466;=&#3627408438;
&#3627408442;= &#3627408438;
&#3627408484;±&#3627408438;
&#3627408440;= 4&#3627408444;
&#3627408484;&#3627409172;
&#3627408484;&#3627409172;
&#3627408477;±&#3627408444;
&#3627408440;.&#3627408442;.&#3627409172;
&#3627408484;&#3627409172;
&#3627408477;
The positive sign is used when the wheels and rotating parts of the engine
rotate in the same direction negative if the rotating parts of the engine
revolves in opposite direction.
Due to the gyroscopic couple, vertical reaction on the road surface will be
produced. The reaction will be vertically upwards on the outer wheels and
vertically downwards on the inner wheels.





&#3627408453;&#3627408466;&#3627408462;&#3627408464;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408475; &#3627408466;&#3627408462;&#3627408464;ℎ &#3627408476;&#3627408467; &#3627408476;&#3627408482;&#3627408481;&#3627408466;&#3627408479; &#3627408484;ℎ&#3627408466;&#3627408466;&#3627408473;=&#3627408453;
&#3627408442;=
&#3627408438;
&#3627408442;
2&#3627408484;
(&#3627408482;&#3627408477;&#3627408484;&#3627408462;&#3627408479;&#3627408465;&#3627408480;)
&#3627408453;&#3627408466;&#3627408462;&#3627408464;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408475; &#3627408466;&#3627408462;&#3627408464;ℎ &#3627408476;&#3627408467; &#3627408470;&#3627408475;&#3627408475;&#3627408466;&#3627408479; &#3627408484;ℎ&#3627408466;&#3627408466;&#3627408473;=&#3627408453;
&#3627408442;=
&#3627408438;
&#3627408442;
2&#3627408484;
(&#3627408465;&#3627408476;&#3627408484;&#3627408475;&#3627408484;&#3627408462;&#3627408479;&#3627408465;&#3627408480;)
Effect of centrifugal couple
Since the vehicle moves along a curved path, therefore centrifugal force will act outwardly at the centre of
gravity of the vehicle. The effect of this centrifugal force is also to overturn the vehicle.
&#3627408441;
&#3627408464;=&#3627408474;&#3627408479;&#3627409172;
2
=
&#3627408474;&#3627408483;
2
&#3627408479;

&#3627408438;&#3627408476;&#3627408482;&#3627408477;&#3627408473;&#3627408466; &#3627408481;&#3627408466;&#3627408475;&#3627408465;&#3627408470;&#3627408475;&#3627408468; &#3627408481;&#3627408476; &#3627408476;&#3627408483;&#3627408466;&#3627408479;&#3627408481;&#3627408482;&#3627408479;&#3627408475; &#3627408481;ℎ&#3627408466; &#3627408483;&#3627408466;ℎ&#3627408470;&#3627408464;&#3627408473;&#3627408466;= &#3627408441;
&#3627408464;×ℎ=
&#3627408474;&#3627408483;
2

&#3627408479;

This couple is balanced by reactions at wheels
&#3627408438;
&#3627408464;=&#3627408453;
&#3627408464;×&#3627408484;=
&#3627408438;
&#3627408442;
2&#3627408484;
×&#3627408484;=&#3627408441;
&#3627408464;×ℎ=
&#3627408474;&#3627408483;
2

&#3627408479;

&#3627408457;&#3627408466;&#3627408479;&#3627408481;&#3627408470;&#3627408464;&#3627408462;&#3627408473; &#3627408479;&#3627408466;&#3627408462;&#3627408464;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408475; &#3627408466;&#3627408462;&#3627408464;ℎ &#3627408476;&#3627408482;&#3627408481;&#3627408466;&#3627408479; &#3627408484;ℎ&#3627408466;&#3627408466;&#3627408473;=&#3627408451;
&#3627408476;=
&#3627408458;
4
+
&#3627408438;
&#3627408442;
2&#3627408484;
+
&#3627408438;
&#3627408464;
2&#3627408484;

&#3627408457;&#3627408466;&#3627408479;&#3627408481;&#3627408470;&#3627408464;&#3627408462;&#3627408473; &#3627408479;&#3627408466;&#3627408462;&#3627408464;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408475; &#3627408466;&#3627408462;&#3627408464;ℎ &#3627408470;&#3627408475;&#3627408475;&#3627408466;&#3627408479; &#3627408484;ℎ&#3627408466;&#3627408466;&#3627408473;=&#3627408451;
??????=
&#3627408458;
4

&#3627408438;
&#3627408442;
2&#3627408484;

&#3627408438;
&#3627408464;
2&#3627408484;

Pi≥0, if Pi is less than zero or negative causes the inner wheels to leave the ground and tends to overturn the
vehicle.

86

VIBRATIONS
When elastic bodies such as spring, beam and a shaft are displaced from the equilibrium position by the
application of external forces, and then released, they execute vibratory motion. Elastic forces present in the
body bring back the body to its original position and doesn’t stop at equilibrium due to the presence of kinetic
energy and converts into strain energy at opposite end and comes back to its original position. In this way
vibration is repeated indefinitely.
Terms used in vibration motion
Free (Natural) vibrations– Elastic vibrations in which there are no friction and external forces after the
initial release of the body are known as free or natural vibrations.
Damped vibrations– When the energy of a vibrating system is gradually dissipated by the friction and other
resistances, the vibrations are said to be damped. The vibrations gradually cease and the system rests in its
equilibrium position.
Forced vibrations– When a repeated force continuously acts on a system, the vibrations are said to be forced.
Period – It is the time taken by a motion and is measured in seconds.
Cycle– It is the motion completed during one-time period.
Frequency – Frequency is the number if cycles of motion completed in one second.
Resonance– When the frequency of the external force is the same as that of natural frequency of the system, a
state of resonance is said to have been reached.
Types of vibrations
Longitudinal vibrations
When the particles of shaft move parallel to axis of the shaft, then
the vibrations are known as longitudinal vibrations.
Transverse vibrations
When the particles of shaft move perpendicular to the axis of the
shaft, then the vibrations are known as transverse vibrations.
Torsional vibrations
When the particles of the shaft move in a circle about the axis of
the shaft, then the vibrations are known as torsional vibrations.

The number of independent co-ordinates required to describe a vibratory system is known as its degree of
freedom.

87

Free longitudinal vibrations
The natural frequency of a vibrating system can be found by any of the following methods
1) Equilibrium method
m = mass of body suspended
s = Stiffness of the constraint
δ = Static deflection of the spring due to the weight
x = displacement of spring due to external force.
&#3627408458; = &#3627408474;&#3627408468; = &#3627408480;&#3627409151;
The mass is initially in unstrained position, then a mass is added
and it reaches to its equilibrium position, where weight of body
(mg) equates spring force (sδ), then an external force is applied on body downwards which causes a
displacement ‘x’.
Since the mass is now displaced from its equilibrium position by a distance x, and is then released, therefore
after time t,
&#3627408453;&#3627408466;&#3627408480;&#3627408481;&#3627408476;&#3627408479;&#3627408470;&#3627408475;&#3627408468; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;=&#3627408458;−&#3627408480;(&#3627409151;+&#3627408485;)=&#3627408458;−&#3627408480;&#3627409151;−&#3627408480;&#3627408485;= &#3627408480;&#3627409151;−&#3627408480;&#3627409151;−&#3627408480;&#3627408485;=−&#3627408480;&#3627408485;
&#3627408436;&#3627408464;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408475;&#3627408468; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;=&#3627408474;&#3627408462;&#3627408480;&#3627408480;×&#3627408462;&#3627408464;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475;=&#3627408474;×
&#3627408465;&#3627408485;
2
&#3627408465;&#3627408481;
2

Equating 1 & 2
&#3627408474;×
&#3627408465;&#3627408485;
2
&#3627408465;&#3627408481;
2
=−&#3627408480;&#3627408485; ⇒ &#3627408474;×
&#3627408465;&#3627408485;
2
&#3627408465;&#3627408481;
2
+&#3627408480;&#3627408485;=0
&#3627408465;&#3627408485;
2
&#3627408465;&#3627408481;
2
+
&#3627408480;
&#3627408474;
&#3627408485;=0
Fundamental equation of simple harmonic equation,
&#3627408465;&#3627408485;
2
&#3627408465;&#3627408481;
2
+&#3627409172;
2
&#3627408485;=0
Comparing 3 & 4,
&#3627409172;= √
&#3627408480;
&#3627408474;

&#3627408481;&#3627408470;&#3627408474;&#3627408466; &#3627408477;&#3627408466;&#3627408479;&#3627408470;&#3627408476;&#3627408465; (&#3627408481;
&#3627408477;)=
2&#3627409163;
&#3627409172;
=2&#3627409163;√
&#3627408474;
&#3627408480;

&#3627408449;&#3627408462;&#3627408481;&#3627408482;&#3627408479;&#3627408462;&#3627408473; &#3627408467;&#3627408479;&#3627408466;&#3627408478;&#3627408482;&#3627408466;&#3627408475;&#3627408464;&#3627408486; (&#3627408467;
&#3627408475;)=
1
&#3627408481;
&#3627408477;
=
1
2&#3627409163;

&#3627408480;
&#3627408474;
=
1
2&#3627409163;

&#3627408468;
&#3627409151;

&#3627408519;
&#3627408527;=
0.4985
√&#3627409151;

2) Equilibrium method
∑&#3627408441;=&#3627408474;&#3627408462; →&#3627408481;&#3627408479;&#3627408462;&#3627408475;&#3627408480;&#3627408473;&#3627408462;&#3627408481;&#3627408470;&#3627408475;&#3627408468; &#3627408480;&#3627408486;&#3627408480;&#3627408481;&#3627408466;&#3627408474;
∑&#3627408448;=&#3627408444;&#3627409148; →&#3627408479;&#3627408476;&#3627408481;&#3627408462;&#3627408481;&#3627408470;&#3627408475;&#3627408468; &#3627408480;&#3627408486;&#3627408480;&#3627408481;&#3627408466;&#3627408474;


∑&#3627408441;−&#3627408474;&#3627408462;=0
Restoring force + inertia force = 0
∑&#3627408448;−&#3627408444;&#3627409148;=0
Restoring Torque + inertia Torque = 0
1
2
3
4
(
&#3627408480;
&#3627408474;
=
&#3627408468;
&#3627409151;
)

88

Example―1 (Simple spring―mass System)

Inertia force + restoring force = 0
&#3627408526;&#3627408537;̈+&#3627408524;&#3627408537;=&#3627409358;→&#3627408442;&#3627408476;&#3627408483;&#3627408466;&#3627408479;&#3627408475;&#3627408470;&#3627408475;&#3627408468; &#3627408465;&#3627408470;&#3627408467;&#3627408467;&#3627408466;&#3627408479;&#3627408466;&#3627408475;&#3627408481;&#3627408470;&#3627408462;&#3627408473; &#3627408466;&#3627408478;&#3627408482;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408467; &#3627408474;&#3627408476;&#3627408481;&#3627408470;&#3627408476;&#3627408475;
&#3627408455;ℎ&#3627408466; &#3627408462;&#3627408463;&#3627408476;&#3627408483;&#3627408466; &#3627408464;&#3627408462;&#3627408475; &#3627408463;&#3627408466; &#3627408484;&#3627408479;&#3627408470;&#3627408481;&#3627408481;&#3627408466;&#3627408475; &#3627408462;&#3627408480; &#3627408485;̈+
&#3627408472;
&#3627408474;
&#3627408485;=0
&#3627408455;ℎ&#3627408470;&#3627408480; &#3627408470;&#3627408480; &#3627408481;ℎ&#3627408466; &#3627408466;&#3627408478;&#3627408482;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408467; &#3627408480;&#3627408470;&#3627408474;&#3627408477;&#3627408473;&#3627408466; ℎ&#3627408462;&#3627408479;&#3627408474;&#3627408476;&#3627408475;&#3627408470;&#3627408464; &#3627408474;&#3627408476;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408462;&#3627408475;&#3627408465; &#3627408470;&#3627408480; &#3627408462;&#3627408475;&#3627408473;&#3627408476;&#3627408468;&#3627408476;&#3627408482;&#3627408480; &#3627408481;&#3627408476; &#3627408485;̈+&#3627409172;
&#3627408475;
2
&#3627408485;=0
&#3627408442;&#3627408466;&#3627408475;&#3627408466;&#3627408479;&#3627408462;&#3627408473; &#3627408480;&#3627408476;&#3627408473;&#3627408482;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408467; &#3627408485;̈+
&#3627408472;
&#3627408474;
&#3627408485;=0 &#3627408470;&#3627408480; &#3627408485;(&#3627408481;)=&#3627408436;&#3627408480;&#3627408470;&#3627408475;(&#3627409172;&#3627408481;)+&#3627408437;&#3627408464;&#3627408476;&#3627408480;(&#3627409172;&#3627408481;)
&#3627408485;(&#3627408481;)=&#3627408436;&#3627408480;&#3627408470;&#3627408475;(&#3627409172;&#3627408481;)+&#3627408437;&#3627408464;&#3627408476;&#3627408480;(&#3627409172;&#3627408481;)⇒&#3627408485;=&#3627408459;&#3627408480;&#3627408470;&#3627408475;(&#3627409172;
&#3627408475;&#3627408481;+??????)
&#3627408441;&#3627408479;&#3627408476;&#3627408474; &#3627408462;&#3627408463;&#3627408476;&#3627408483;&#3627408466; &#3627408466;&#3627408478;&#3627408482;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475;&#3627408480; &#3627408484;&#3627408466; &#3627408464;&#3627408462;&#3627408475; &#3627408480;&#3627408462;&#3627408486;,&#3627409172;
&#3627408475;= √
&#3627408472;
&#3627408474;

&#3627408449;&#3627408462;&#3627408481;&#3627408482;&#3627408479;&#3627408462;&#3627408473; &#3627408467;&#3627408479;&#3627408466;&#3627408478;&#3627408482;&#3627408466;&#3627408475;&#3627408464;&#3627408486;=&#3627408467;
&#3627408475;=
&#3627409172;
&#3627408475;
2&#3627409163;
=
1
2&#3627409163;

&#3627408472;
&#3627408474;

&#3627408455;&#3627408470;&#3627408474;&#3627408466; &#3627408477;&#3627408466;&#3627408479;&#3627408470;&#3627408476;&#3627408465; (&#3627408481;
&#3627408477;)=
1
&#3627408467;
&#3627408475;
=
2&#3627409163;
&#3627409172;
&#3627408475;
=2&#3627409163;√
&#3627408474;
&#3627408472;

Now let us consider different ways of starting motion
1. If the motion is started by displacing the mass through a distance xo and giving a velocity vo, then
&#3627408485;=
&#3627408483;
&#3627408476;
&#3627409172;
&#3627408475;
&#3627408480;&#3627408470;&#3627408475;&#3627409172;
&#3627408475;&#3627408481;+&#3627408485;
&#3627408476;&#3627408464;&#3627408476;&#3627408480;&#3627409172;
&#3627408475;&#3627408481;
&#3627408485;=√&#3627408485;
&#3627408476;
2
+(
&#3627408483;
&#3627408476;
&#3627409172;
&#3627408475;
)
2
&#3627408480;&#3627408470;&#3627408475;(&#3627409172;
&#3627408475;&#3627408481;+??????)
2. If the motion is started by displacing the mass through a distance xo and then releasing it at t=0, x=xo
and v=0
&#3627408485;=&#3627408485;
&#3627408476;&#3627408464;&#3627408476;&#3627408480;&#3627409172;
&#3627408475;&#3627408481;
3. If the motion is started by providing a velocity of vo at equilibrium then at t=0, x=0
&#3627408485;=
&#3627408483;
&#3627408476;
&#3627409172;
&#3627408475;
&#3627408480;&#3627408470;&#3627408475;&#3627409172;
&#3627408475;&#3627408481;
Example 2 (Simple Pendulum)
&#3627408444;&#3627408475;&#3627408466;&#3627408479;&#3627408481;&#3627408470;&#3627408462; &#3627408481;&#3627408476;&#3627408479;&#3627408478;&#3627408482;&#3627408466;= &#3627408444;
&#3627408436;=&#3627408474;&#3627408447;
2
&#3627409155;̈
&#3627408444;
&#3627408436;=&#3627408474;&#3627408462;&#3627408480;&#3627408480; &#3627408474;&#3627408476;&#3627408474;&#3627408466;&#3627408475;&#3627408481; &#3627408476;&#3627408467; &#3627408470;&#3627408475;&#3627408466;&#3627408479;&#3627408481;&#3627408470;&#3627408462; &#3627408462;&#3627408463;&#3627408476;&#3627408482;&#3627408481; &#3627408477;&#3627408476;&#3627408470;&#3627408475;&#3627408481; &#3627408476;&#3627408467; &#3627408479;&#3627408476;&#3627408481;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475;=&#3627408474;&#3627408447;
2

&#3627408453;&#3627408466;&#3627408480;&#3627408481;&#3627408476;&#3627408479;&#3627408470;&#3627408475;&#3627408468; &#3627408455;&#3627408476;&#3627408479;&#3627408478;&#3627408482;&#3627408466;=&#3627408474;&#3627408468;&#3627408447;&#3627408480;&#3627408470;&#3627408475;&#3627409155;
&#3627408474;&#3627408447;
2
&#3627409155;̈+&#3627408474;&#3627408468;&#3627408447;&#3627408480;&#3627408470;&#3627408475;&#3627409155;=0
&#3627409155;̈+
&#3627408468;
&#3627408447;
&#3627408480;&#3627408470;&#3627408475;&#3627409155;=0
Simple pendulum does not execute Simple harmonic motion.
If  is very small, sin  , then
&#3627409155;̈+
&#3627408468;
&#3627408447;
&#3627409155;=0
&#3627409172;
&#3627408475;=√
&#3627408468;
&#3627408447;
,&#3627408467;
&#3627408475;=
1
2&#3627409163;

&#3627408468;
&#3627408447;
,&#3627408481;
&#3627408475;=
1
&#3627408467;
&#3627408475;
= 2&#3627409163;√
&#3627408447;
&#3627408468;

X & φ are constants & must be found
from initial conditions
??????=&#3627408481;&#3627408462;&#3627408475;
−1
&#3627408485;
&#3627408476;&#3627409172;
&#3627408475;
&#3627408483;
&#3627408476;

89

Energy method
In a conservative system (no damping-free vibrations), the sum of the kinetic energy and the potential energy is
constant, So the change in mechanical energy (K.E +P.E ) will be zero.
&#3627408465;
&#3627408465;&#3627408481;
(&#3627408446;.&#3627408440;+&#3627408451;.&#3627408440;)=0
&#3627408446;.&#3627408440;=
1
2
&#3627408474;&#3627408485;̇
2

&#3627408451;.&#3627408440;=&#3627408474;&#3627408466;&#3627408462;&#3627408475; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466; ×&#3627408465;&#3627408470;&#3627408480;&#3627408477;&#3627408473;&#3627408462;&#3627408464;&#3627408466;&#3627408474;&#3627408466;&#3627408475;&#3627408481;=
0+&#3627408472;&#3627408485;
2
×&#3627408485;=
&#3627408472;&#3627408485;
2
2

&#3627408465;
&#3627408465;&#3627408481;
(
1
2
&#3627408474;&#3627408485;̇
2
+
&#3627408472;&#3627408485;
2
2
)=0
&#3627408474;&#3627408485;̈+&#3627408472;&#3627408485;=0 ⇒ &#3627409172;
&#3627408475;= √
&#3627408472;
&#3627408474;

Rayleigh’s Method
In this method, the maximum kinetic energy at the mean position (where potential energy is zero) is made equal
to the maximum potential energy (or strain energy) at the extreme position (where the kinetic energy is zero).
Let the motion be simple harmonic,
Therefore, &#3627408485;=&#3627408459;&#3627408480;&#3627408470;&#3627408475;(&#3627409172;
&#3627408475;&#3627408481;)
where X = maximum displacement from the mean position to the extreme position.
&#3627408485;̇= &#3627409172;
&#3627408475;&#3627408459;&#3627408464;&#3627408476;&#3627408480;&#3627409172;
&#3627408475;&#3627408481; , &#3627408485;̇
&#3627408474;&#3627408462;&#3627408485;=&#3627409172;
&#3627408475;&#3627408459;
K.E at mean position = P.E at extreme position
1
2
&#3627408474;(&#3627409172;
&#3627408475;&#3627408459;)
2
=
1
2
&#3627408472;&#3627408485;
2

&#3627409172;
&#3627408475;= √
&#3627408472;
&#3627408474;

Natural Frequency of Free Transverse Vibrations
&#3627408436;&#3627408464;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408475;&#3627408468; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;=&#3627408474;&#3627408485;̈
&#3627408453;&#3627408466;&#3627408480;&#3627408481;&#3627408476;&#3627408479;&#3627408470;&#3627408475;&#3627408468; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;=−&#3627408480;&#3627408485;
Accelerating force = Restoring force
&#3627408474;&#3627408485;̈+&#3627408480;&#3627408485;=0
&#3627408485;̈=
&#3627408480;
&#3627408474;
&#3627408485;=0
&#3627409172;
&#3627408475;= √
&#3627408480;
&#3627408474;
⇒&#3627408467;
&#3627408475;=
1
2&#3627409163;

&#3627408480;
&#3627408474;

Inertia effect of the mass of spring
So far, the mass of the spring and thus the effect of inertia have been neglected. Now the inertia effect of spring is
taken into account.
m’ = mass of the spring wire per unit length = m1 / l
v=velocity of the free end of the spring at the instant under consideration
l=total length of the spring wire
m1= mass of the spring
Consider an element of length δy at an length y measured round the coils from the fixed end.
&#3627408446;&#3627408470;&#3627408475;&#3627408466;&#3627408464;&#3627408481;&#3627408470;&#3627408464; &#3627408466;&#3627408475;&#3627408466;&#3627408479;&#3627408468;&#3627408486; &#3627408476;&#3627408467; &#3627408466;&#3627408473;&#3627408466;&#3627408474;&#3627408466;&#3627408475;&#3627408481;=
1
2
×&#3627408474;&#3627408462;&#3627408480;&#3627408480; &#3627408476;&#3627408467; &#3627408466;&#3627408473;&#3627408466;&#3627408474;&#3627408466;&#3627408475;&#3627408481;×(&#3627408483;&#3627408466;&#3627408473;&#3627408476;&#3627408464;&#3627408470;&#3627408481;&#3627408486; &#3627408476;&#3627408467; &#3627408466;&#3627408473;&#3627408466;&#3627408474;&#3627408466;&#3627408475;&#3627408481;)
2
=
1
2
×(&#3627408474;

&#3627408486;)×(
&#3627408486;
&#3627408473;
&#3627408483;)
2

&#3627408446;&#3627408470;&#3627408475;&#3627408466;&#3627408481;&#3627408470;&#3627408464; &#3627408466;&#3627408475;&#3627408466;&#3627408479;&#3627408468;&#3627408486; &#3627408476;&#3627408467; &#3627408481;ℎ&#3627408466; &#3627408480;&#3627408477;&#3627408479;&#3627408470;&#3627408475;&#3627408468;=∫
1
2
×&#3627408474;

&#3627408483;
2
(
&#3627408486;
&#3627408473;
)
2
&#3627408465;&#3627408486;=
1
2
&#3627408474;

&#3627408483;
2
&#3627408473;
2
∫&#3627408486;
2
&#3627408465;&#3627408486;
&#3627408473;
0
=
1
2
&#3627408474;

&#3627408483;
2
&#3627408473;
2
×
&#3627408473;
3
3
&#3627408473;
0
=
1
3
1
2
(&#3627408474;

&#3627408473;)&#3627408483;
2

90

&#3627408446;&#3627408470;&#3627408475;&#3627408466;&#3627408481;&#3627408470;&#3627408464; &#3627408466;&#3627408475;&#3627408466;&#3627408479;&#3627408468;&#3627408486; &#3627408476;&#3627408467; &#3627408481;ℎ&#3627408466; &#3627408480;&#3627408477;&#3627408479;&#3627408470;&#3627408475;&#3627408468;=
1
3
[
1
2
×&#3627408474;&#3627408462;&#3627408480;&#3627408480; &#3627408476;&#3627408467; &#3627408480;&#3627408477;&#3627408479;&#3627408470;&#3627408475;&#3627408468;×(&#3627408483;&#3627408466;&#3627408473;&#3627408476;&#3627408464;&#3627408470;&#3627408481;&#3627408486; &#3627408476;&#3627408467; &#3627408467;&#3627408479;&#3627408466;&#3627408466; &#3627408466;&#3627408475;&#3627408465;)
2
]
&#3627408446;.&#3627408440; &#3627408476;&#3627408467; &#3627408481;ℎ&#3627408466; &#3627408480;&#3627408477;&#3627408479;&#3627408470;&#3627408475;&#3627408468;=
1
3
×&#3627408446;.&#3627408440; &#3627408476;&#3627408467; &#3627408462; &#3627408474;&#3627408462;&#3627408480;&#3627408480; &#3627408466;&#3627408478;&#3627408482;&#3627408462;&#3627408473; &#3627408481;&#3627408476; &#3627408481;ℎ&#3627408462;&#3627408481; &#3627408476;&#3627408467; &#3627408480;&#3627408477;&#3627408479;&#3627408470;&#3627408475;&#3627408468; &#3627408474;&#3627408476;&#3627408483;&#3627408470;&#3627408475;&#3627408468; &#3627408484;&#3627408470;&#3627408481;ℎ &#3627408481;ℎ&#3627408466; &#3627408480;&#3627408462;&#3627408474;&#3627408466; &#3627408483;&#3627408466;&#3627408473;&#3627408476;&#3627408464;&#3627408470;&#3627408481;&#3627408486; &#3627408462;&#3627408480; &#3627408481;ℎ&#3627408466; &#3627408467;&#3627408479;&#3627408466;&#3627408466; &#3627408466;&#3627408475;&#3627408465;.
This shows that inertia effect of spring is equal to that if a mass of the spring, concentrated at its free
end
&#3627408455;ℎ&#3627408482;&#3627408480; &#3627408466;&#3627408478;&#3627408482;&#3627408470;&#3627408483;&#3627408462;&#3627408473;&#3627408466;&#3627408475;&#3627408481; &#3627408474;&#3627408462;&#3627408480;&#3627408480; &#3627408462;&#3627408481; &#3627408481;ℎ&#3627408466; &#3627408467;&#3627408479;&#3627408466;&#3627408466; &#3627408466;&#3627408475;&#3627408465; = &#3627408474;+
&#3627408474;
1
3

&#3627408467;
&#3627408475;=
1
2&#3627409163;

&#3627408480;
&#3627408474;+
&#3627408474;
1
3

Consider a rod of length l suspended vertically, a mass m is suspended at the free end.
&#3627408454;&#3627408481;&#3627408462;&#3627408481;&#3627408470;&#3627408464; &#3627408439;&#3627408466;&#3627408467;&#3627408473;&#3627408466;&#3627408464;&#3627408481;&#3627408470;&#3627408476;&#3627408475;,??????=
&#3627408474;&#3627408468;&#3627408473;
&#3627408436;&#3627408440;

&#3627408467;
&#3627408475;=
1
2&#3627409163;

&#3627408468;
??????
=
1
2&#3627409163;

&#3627408468;&#3627408436;&#3627408440;
&#3627408474;&#3627408468;&#3627408473;
=
1
2&#3627409163;

&#3627408436;&#3627408440;
&#3627408474;&#3627408473;

&#3627408467;
&#3627408475;=
1
2&#3627409163;

&#3627408436;&#3627408440;
(&#3627408474;+
&#3627408474;
1
3
)&#3627408473;

Natural frequency of Transverse vibrations due to a Point Load
acting over a Simply Supported Shaft
Consider a shaft AB of length l, carrying a point load W at C. Load causes deflection and when load is released
causes transverse vibrations. In case of shafts carrying a concentraed mass, the force is proportional to the
deflections of the mass from the equlibrium position and the relation derived for natural frequency of
longitudinal vibrations holds good.
&#3627408467;
&#3627408475;=
1
2&#3627409163;

&#3627408468;
&#3627409151;
=
0.4985
√&#3627409151;


&#3627409151;=
&#3627408458;&#3627408473;
3
3&#3627408440;&#3627408444;


&#3627409151;=
&#3627408458;&#3627408462;
2
&#3627408463;
2
3&#3627408440;&#3627408444;&#3627408473;


&#3627409151;=
&#3627408458;&#3627408462;
3
&#3627408463;
3
3&#3627408440;&#3627408444;&#3627408473;
3

A = Cross-section of rod
l = length of rod
E = Young’s modulus of
rod

91

Natural Frequency of Free Transverse Vibrations Due to Uniformly
Distributed Load Acting Over a Simply Supported Shaft

The shaft makes transverse vibrations due to elastic forces. At any instant, let it be deflected by an amount y at a
distance ‘x’ from the end A. The vibrations being free and due to elastic forces, will be of simple harmonic
motion type.
From the theory of bending of shafts,
&#3627408440;&#3627408444;=
&#3627408465;
4
&#3627408486;
&#3627408465;&#3627408485;
4
=&#3627408465;&#3627408486;&#3627408475;&#3627408462;&#3627408474;&#3627408470;&#3627408464; &#3627408473;&#3627408476;&#3627408462;&#3627408465; &#3627408477;&#3627408466;&#3627408479; &#3627408482;&#3627408475;&#3627408470;&#3627408481; &#3627408473;&#3627408466;&#3627408475;&#3627408468;&#3627408481;ℎ=&#3627408474;&#3627408486;&#3627409172;
2
=&#3627408464;&#3627408466;&#3627408475;&#3627408481;&#3627408479;&#3627408470;&#3627408467;&#3627408482;&#3627408468;&#3627408462;&#3627408473; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466; &#3627408477;&#3627408466;&#3627408479; &#3627408482;&#3627408475;&#3627408470;&#3627408481; &#3627408473;&#3627408466;&#3627408475;&#3627408468;&#3627408481;ℎ
&#3627408465;
4
&#3627408486;
&#3627408465;&#3627408485;
4

&#3627408474;&#3627409172;
2
&#3627408486;
&#3627408440;&#3627408444;
=0
&#3627408465;
4
&#3627408486;
&#3627408465;&#3627408485;
4
−&#3627409158;
4
&#3627408486;=0 &#3627408484;ℎ&#3627408466;&#3627408479;&#3627408466; &#3627409158;
4
=
&#3627408474;&#3627409172;
2
&#3627408440;&#3627408444;

(&#3627408439;
4
−&#3627409158;
4
)&#3627408486;=0
&#3627408455;ℎ&#3627408470;&#3627408480; &#3627408468;&#3627408470;&#3627408483;&#3627408466;&#3627408480; &#3627408439;=±&#3627409158; &#3627408462;&#3627408475;&#3627408465; ±&#3627408470;&#3627409158;
&#3627408455;ℎ&#3627408466; &#3627408480;&#3627408476;&#3627408473;&#3627408482;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408484;&#3627408470;&#3627408473;&#3627408473; &#3627408463;&#3627408466; &#3627408470;&#3627408475; &#3627408481;ℎ&#3627408466; &#3627408467;&#3627408476;&#3627408479;&#3627408474;,&#3627408538;=&#3627408488;&#3627408532;&#3627408522;&#3627408527;??????&#3627408537;+&#3627408489;&#3627408516;&#3627408528;&#3627408532;??????&#3627408537;+&#3627408490;&#3627408532;&#3627408522;&#3627408527;&#3627408521;??????&#3627408537;+&#3627408491;&#3627408516;&#3627408528;&#3627408532;&#3627408521;??????&#3627408537; ①
This is the general expression for the deflections in case of uniformly loaded shafts. Constants A,B,C & D are to
be found from the boundary conditions.
Simple Supported Shaft
The boundary conditions are
(&#3627408462;)&#3627408486;=0 &#3627408462;&#3627408481; &#3627408485;=0 & &#3627408473;
(&#3627408463;)
&#3627408465;
2
&#3627408486;
&#3627408465;&#3627408485;
2
=0 &#3627408462;&#3627408481; &#3627408485;=0 & &#3627408473;
When x=0, y=0; B+D=0 ②
&#3627408458;ℎ&#3627408466;&#3627408475; &#3627408485;=&#3627408473;,&#3627408486;=0⇒ &#3627408436;&#3627408480;&#3627408470;&#3627408475;&#3627409158;&#3627408473;+&#3627408437;&#3627408464;&#3627408476;&#3627408480;&#3627409158;&#3627408473;+&#3627408438;&#3627408480;&#3627408470;&#3627408475;ℎ&#3627409158;&#3627408473;+&#3627408439;&#3627408464;&#3627408476;&#3627408480;ℎ&#3627409158;&#3627408473;=0 ③
Differentiating 1 w.r.t x twice gives
&#3627408465;
2
&#3627408486;
&#3627408465;&#3627408485;
2
= &#3627409158;
2
(−&#3627408436;&#3627408480;&#3627408470;&#3627408475;&#3627409158;&#3627408485;−&#3627408437;&#3627408464;&#3627408476;&#3627408480;&#3627409158;&#3627408485;+&#3627408438;&#3627408480;&#3627408470;&#3627408475;ℎ&#3627409158;&#3627408485;+&#3627408439;&#3627408464;&#3627408476;&#3627408480;ℎ&#3627409158;&#3627408485;)
When x=0,
&#3627408465;
2
&#3627408486;
&#3627408465;&#3627408485;
2
=0=&#3627409158;
2
(−&#3627408437;+&#3627408439;)⇒&#3627408437;=&#3627408439; ④
When x=l,
&#3627408465;
2
&#3627408486;
&#3627408465;&#3627408485;
2
=0=&#3627409158;
2
(−&#3627408436;&#3627408480;&#3627408470;&#3627408475;&#3627409158;&#3627408473;−&#3627408437;&#3627408464;&#3627408476;&#3627408480;&#3627409158;&#3627408473;+&#3627408438;&#3627408480;&#3627408470;&#3627408475;ℎ&#3627409158;&#3627408473;+&#3627408439;&#3627408464;&#3627408476;&#3627408480;ℎ&#3627409158;&#3627408473;) ⑤
From 4 & 2 we get, B=0 & D=0.
From 3 & 4, we get C sinh λl = 0 & A sin λl = 0.
&#3627408444;&#3627408467; &#3627408438;&#3627408480;&#3627408470;&#3627408475;ℎ&#3627409158;&#3627408473;=0,&#3627408462;&#3627408480;&#3627408480;&#3627408482;&#3627408474;&#3627408466; &#3627409158;=0,&#3627408481;ℎ&#3627408466;&#3627408475;
&#3627408474;&#3627409172;
2
&#3627408440;&#3627408444;
=0 ⇒ &#3627408467;
&#3627408475;=0
Which means sysyem does not vibrate, So C = 0.
Equation 1 reduces to y=A sin λx
Now, when A sin λl = 0, A can’t be zero as there will be no vibrations.
So, sin λl = 0, λl = 0,π,2π,3π,…..

92

&#3627409158;=
&#3627409163;
&#3627408473;
,
2&#3627409163;
&#3627408473;
,
3&#3627409163;
&#3627408473;
,….&#3627408476;&#3627408479; (
&#3627408474;&#3627409172;
2
&#3627408440;&#3627408444;
)
14⁄
=
&#3627409163;
&#3627408473;
,
2&#3627409163;
&#3627408473;
,
3&#3627409163;
&#3627408473;
,….
&#3627409172;
12⁄
=
&#3627409163;
&#3627408473;
(
&#3627408440;&#3627408444;
&#3627408474;
)
14⁄
,
2&#3627409163;
&#3627408473;
(
&#3627408440;&#3627408444;
&#3627408474;
)
14⁄
,
3&#3627409163;
&#3627408473;
(
&#3627408440;&#3627408444;
&#3627408474;
)
14⁄
,….
&#3627409172;=(
&#3627409163;
&#3627408473;
)
2

&#3627408440;&#3627408444;
&#3627408474;
,(
2&#3627409163;
&#3627408473;
)
2

&#3627408440;&#3627408444;
&#3627408474;
,(
3&#3627409163;
&#3627408473;
)
2

&#3627408440;&#3627408444;
&#3627408474;

&#3627408467;
&#3627408475;=
&#3627409172;
2&#3627409163;
=
&#3627409163;
2

&#3627408440;&#3627408444;
&#3627408474;&#3627408473;
4
,
4&#3627409163;
2

&#3627408440;&#3627408444;
&#3627408474;&#3627408473;
4
,
9&#3627409163;
2

&#3627408440;&#3627408444;
&#3627408474;&#3627408473;
4

A simply supported shaft carrying a uniformly disturbed mass has maximum deflection at he mid-span.
&#3627409151;=
5&#3627408474;&#3627408468;&#3627408473;
4
384&#3627408440;&#3627408444;

&#3627408492;&#3627408496;
&#3627408526;&#3627408525;
&#3627409362;
=
&#3627409363;&#3627408520;
&#3627409361;??????&#3627409362;&#3627409209;

Taking the lowest frequency of transverse vibrations, this frequency is called the fundamental frequency.
&#3627408519;
&#3627408527;=
??????
&#3627409360;

&#3627409363;&#3627408520;
&#3627409361;??????&#3627409362;&#3627409209;

&#3627408486;=&#3627408436;&#3627408480;&#3627408470;&#3627408475;&#3627409158;&#3627408485;=&#3627408436;&#3627408480;&#3627408470;&#3627408475;
&#3627409163;
&#3627408473;
&#3627408485;=0 → &#3627408467;
&#3627408475;=
&#3627409163;
2

5&#3627408468;
384&#3627409151;


&#3627408436;&#3627408480;&#3627408470;&#3627408475;
2&#3627409163;
&#3627408473;
&#3627408485;=0→ &#3627408467;
&#3627408475;=
4&#3627409163;
2

5&#3627408468;
384&#3627409151;


&#3627408436;&#3627408480;&#3627408470;&#3627408475;
3&#3627409163;
&#3627408473;
&#3627408485;=0→ &#3627408467;
&#3627408475;=
9&#3627409163;
2

5&#3627408468;
384&#3627409151;


Higher the frequency, the more number of nodes.
A simply supported shaft will have an infinite number of frequencies under a uniformly distributed load.
Cantilvers
Boundary conditions
&#3627408486;=0 &#3627408462;&#3627408481; &#3627408485;=0 &
&#3627408465;&#3627408486;
&#3627408465;&#3627408485;
=0 &#3627408462;&#3627408481; &#3627408485;=0 &
&#3627408465;
2
&#3627408486;
&#3627408465;&#3627408485;
2
=0 &#3627408462;&#3627408481; &#3627408485;=&#3627408473; &
&#3627408465;
3
&#3627408486;
&#3627408465;&#3627408485;
3
=0 &#3627408462;&#3627408481; &#3627408485;=&#3627408473;
&#3627409151;=
&#3627408474;&#3627408468;&#3627408473;
4
8&#3627408440;&#3627408444;

Both ends fixed
&#3627408486;=0 &#3627408462;&#3627408481; &#3627408485;=0 & &#3627408473;
&#3627408465;&#3627408486;
&#3627408465;&#3627408485;
=0 &#3627408462;&#3627408481; &#3627408485;=0 & &#3627408473;
&#3627409151;=
&#3627408474;&#3627408468;&#3627408473;
4
384&#3627408440;&#3627408444;

93

Natural Frequency of Free Transverse Vibrations for a Shaft Subjected
to several Point Loads
Dunkerley’s method
Let W1, W2, W3, ……. be the concentrated loads on the shaft due to masses m1, m2, m3…... and δ1, δ2, δ3, …... the
static deflections of this shaft under each load when each load acts alone on the shaft. Let the shaft carry a
uniformly distributed mass of m per unit length over its whole span and the static deflection at mid-span due to
the load of this mass be δs.
fn = frequency of transverse vibration of the whole system
fns = frequency with the distributed load acting along
fn1, fn2, fn3, …... = frequency of transverse vibrations when each of W1, W2, W3, ……. Act alone.
Thus, according to Dunkerley’s empirical formula,
&#3627409359;
&#3627408519;
&#3627408527;
&#3627409360;
=
&#3627409359;
&#3627408519;
&#3627408527;&#3627409359;
&#3627409360;
+
&#3627409359;
&#3627408519;
&#3627408527;&#3627409360;
&#3627409360;
+
&#3627409359;
&#3627408519;
&#3627408527;&#3627409361;
&#3627409360;
+⋯+
&#3627409359;
&#3627408519;
&#3627408527;&#3627408532;
&#3627409360;

&#3627408484;ℎ&#3627408466;&#3627408479;&#3627408466; &#3627408467;
&#3627408475;1=
0.4985
√&#3627409151;
1
,&#3627408467;
&#3627408475;2=
0.4985
√&#3627409151;
2
,&#3627408467;
&#3627408475;3=
0.4985
√&#3627409151;
3
,…,&#3627408467;
&#3627408475;&#3627408480;=
0.5614
√&#3627409151;
&#3627408480;

1
&#3627408467;
&#3627408475;
2
=
1
(0.4985)
2
(&#3627409151;
1+&#3627409151;
2+&#3627409151;
3+⋯)+
1
(0.5614)
2
&#3627409151;
&#3627408480;
&#3627408467;
&#3627408475;=
0.4985
√(&#3627409151;
1+&#3627409151;
2+&#3627409151;
3+⋯)+
&#3627409151;
&#3627408480;
1.27

Energy Method
Consider a shaft of negligible mass m, carrying point loads W1, W2, W3, ……. due to masses m1, m2, m3…... and y1,
y2, y3, …... the deflections of these loads.
In the extreme positions of the shaft, it possesses maximum potential energy and zero Kinetic energy and at
mean positions it possesses maximum Kinetic energy and zero potential energy.
&#3627408448;&#3627408462;&#3627408485;&#3627408470;&#3627408474;&#3627408482;&#3627408474; &#3627408477;&#3627408476;&#3627408481;&#3627408466;&#3627408475;&#3627408481;&#3627408470;&#3627408462;&#3627408473; &#3627408466;&#3627408475;&#3627408466;&#3627408479;&#3627408468;&#3627408486;=
1
2
&#3627408474;
1&#3627408468;&#3627408486;
1+
1
2
&#3627408474;
2&#3627408468;&#3627408486;
2+
1
2
&#3627408474;
3&#3627408468;&#3627408486;
3+⋯=
&#3627408468;
2
(&#3627408474;
1&#3627408486;
1+&#3627408474;
2&#3627408486;
2+&#3627408474;
3&#3627408486;
3)=
&#3627408468;
2
∑&#3627408474;&#3627408486;
&#3627408448;&#3627408462;&#3627408485;&#3627408470;&#3627408474;&#3627408482;&#3627408474; &#3627408446;&#3627408470;&#3627408475;&#3627408466;&#3627408481;&#3627408470;&#3627408464; &#3627408440;&#3627408475;&#3627408466;&#3627408479;&#3627408468;&#3627408486;=
1
2
&#3627408474;
1&#3627408483;
1
2
+
1
2
&#3627408474;
2&#3627408483;
2
2
+
1
2
&#3627408474;
3&#3627408483;
3
2
+⋯=
1
2
&#3627408474;
1(&#3627409172;&#3627408486;
1)
2
+
1
2
&#3627408474;
2(&#3627409172;&#3627408486;
2)
2
+
1
2
&#3627408474;
3(&#3627409172;&#3627408486;
3)
2
+⋯
=
&#3627409172;
2
2
∑&#3627408474;&#3627408486;
2

ω is the circular frequency of vibration.
Equating maximum P.E & K.E,
&#3627408468;
2
∑&#3627408474;&#3627408486;=
&#3627409172;
2
2
∑&#3627408474;&#3627408486;
2

&#3627409172;=√
&#3627408468;∑&#3627408474;&#3627408486;
∑&#3627408474;&#3627408486;
2

&#3627408467;
&#3627408475;=
&#3627409172;
2&#3627409163;
=
1
2&#3627409163;

&#3627408468;∑&#3627408474;&#3627408486;
∑&#3627408474;&#3627408486;
2

94

Whirling of shafts
m = mass of rotor
k = stiffness of shaft
e = initial eccentricity of centre of
mass
y = additional deflection of rotor
due to centrifugal force
ω = angular velocity of shaft




Rotor is rotating along the shaft axis, normally the rotor centre of mass should lie on shaft axis but generally it
lies at a distance e from the axis. When rotating along shaft axis the distance increases further ‘y’ form the shaft
axis. This further increase of eccentricity causes to increase eccentricity more as Centrifugal force increases
with radius (eccentricity ‘y’). This continues till the shaft fails or the eccentricity becomes infinity.
The speed at which the shaft runs so that the additional deflection of the shaft from the axis of rotation becomes
infinite, is known as critical or whirling speed.
Since the shaft is rotating at ω rad/s, therefore centrifugal force acting radially outwards through G causing the
shaft to deflect is given by
&#3627408441;
&#3627408464;=&#3627408474;(&#3627408486;+&#3627408466;)&#3627409172;
2

The shaft behaves like a spring, so force resisting the deflection is
&#3627408441;
&#3627408453;=&#3627408472;&#3627408486;
For the equilibrium position,
&#3627408441;
&#3627408438;=&#3627408441;
&#3627408453;
&#3627408474;(&#3627408486;+&#3627408466;)&#3627409172;
2
=&#3627408472;&#3627408486;
&#3627408474;&#3627408486;&#3627409172;
2
+&#3627408474;&#3627408466;&#3627409172;
2
=&#3627408472;&#3627408486;
&#3627408486;(&#3627408472;−&#3627408474;&#3627409172;
2
)=&#3627408474;&#3627408466;&#3627409172;
2

&#3627408486;=
&#3627408474;&#3627408466;&#3627409172;
2
&#3627408472;−&#3627408474;&#3627409172;
2

&#3627408486;=
&#3627408466;&#3627409172;
2
&#3627408472;
&#3627408474;
−&#3627409172;
2
=
&#3627408466;&#3627409172;
2
&#3627409172;
&#3627408475;
2
−&#3627409172;
2
=
&#3627408466;
(
&#3627409172;
&#3627408475;
&#3627409172;
)
2
−1

Thus, when ωn=ω, the deflection y is indefinitely large and the speed ω is called critical speed(ωc).
&#3627409172;
&#3627408475;=&#3627409172;
&#3627408464;=√
&#3627408472;
&#3627408474;
=√
&#3627408468;
&#3627409151;

If the rotor speed increases more than ωn, then the value of y becomes negative, and shaft bends in opposite
direction, if it reaches (y reaches) -e, then the shaft will become straight and the rotor runs steadily.

95

Damped vibrations
When an elastic body is set in vibratory motion, the vibration die out after some time due to the internal
molecular 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. The damping provided by viscous fluid is called
Viscous Damping. Damping can be done externally by dampers or
dashpots.
s = stiffness of spring
&#3627408464;=&#3627408465;&#3627408462;&#3627408474;&#3627408477;&#3627408470;&#3627408475;&#3627408468; &#3627408464;&#3627408476;&#3627408466;&#3627408467;&#3627408467;&#3627408470;&#3627408464;&#3627408470;&#3627408466;&#3627408475;&#3627408481;
&#3627409172;
&#3627408475;=&#3627408467;&#3627408479;&#3627408466;&#3627408478;&#3627408482;&#3627408466;&#3627408475;&#3627408464;&#3627408486; &#3627408476;&#3627408467; &#3627408475;&#3627408462;&#3627408481;&#3627408482;&#3627408479;&#3627408462;&#3627408473; &#3627408482;&#3627408475;&#3627408465;&#3627408462;&#3627408474;&#3627408477;&#3627408466;&#3627408465; &#3627408483;&#3627408470;&#3627408463;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475;&#3627408480;
&#3627408485;=&#3627408465;&#3627408470;&#3627408480;&#3627408477;&#3627408473;&#3627408462;&#3627408464;&#3627408466;&#3627408474;&#3627408466;&#3627408475;&#3627408481; &#3627408476;&#3627408467; &#3627408474;&#3627408462;&#3627408480;&#3627408480; &#3627408467;&#3627408479;&#3627408476;&#3627408474; &#3627408474;&#3627408466;&#3627408462;&#3627408475; &#3627408477;&#3627408476;&#3627408480;&#3627408470;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408462;&#3627408481; &#3627408481;&#3627408470;&#3627408474;&#3627408466; &#3627408533;.
&#3627408483;=&#3627408485;̇=&#3627408483;&#3627408466;&#3627408473;&#3627408476;&#3627408464;&#3627408470;&#3627408481;&#3627408486; &#3627408476;&#3627408467; &#3627408474;&#3627408462;&#3627408480;&#3627408480; &#3627408462;&#3627408481; &#3627408481;&#3627408470;&#3627408474;&#3627408466; &#3627408481;
&#3627408467;=&#3627408485;̈=&#3627408462;&#3627408464;&#3627408464;&#3627408466;&#3627408473;&#3627408466;&#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408467; &#3627408474;&#3627408462;&#3627408480;&#3627408480; &#3627408462;&#3627408481; &#3627408481;&#3627408470;&#3627408474;&#3627408466; &#3627408481;
Now, when the mass moves downwards, the friction force of the
damper acts in the upward direction.
The forces acting on mass m are
&#3627408444;&#3627408475;&#3627408466;&#3627408479;&#3627408481;&#3627408470;&#3627408462;=&#3627408474;&#3627408485;̈ (&#3627408482;&#3627408477;&#3627408484;&#3627408462;&#3627408479;&#3627408465;&#3627408480;)
&#3627408439;&#3627408462;&#3627408474;&#3627408477;&#3627408470;&#3627408475;&#3627408468; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;=&#3627408464;&#3627408485;̇ (&#3627408482;&#3627408477;&#3627408484;&#3627408462;&#3627408479;&#3627408465;&#3627408480;)
&#3627408454;&#3627408477;&#3627408479;&#3627408470;&#3627408475;&#3627408468; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;=&#3627408480;&#3627408485; (&#3627408482;&#3627408477;&#3627408484;&#3627408462;&#3627408479;&#3627408465;&#3627408480;)
The sum of inertia forces and external forces on a body in any direction is to be zero.
&#3627408474;&#3627408485;̈+&#3627408464;&#3627408485;̇+&#3627408480;&#3627408485;=0 ⇒ &#3627408537;̈+
&#3627408516;
&#3627408526;
&#3627408537;̇+
&#3627408532;
&#3627408526;
&#3627408537;=&#3627409358;
It is a second order differential equation, its solution will be of the form
&#3627408485;=&#3627408436;&#3627408466;
??????1&#3627408481;
+&#3627408437;&#3627408466;
??????2&#3627408481;

&#3627409148;
2
+
&#3627408464;
&#3627408474;
&#3627409148;+
&#3627408480;
&#3627408474;
=0
&#3627409148;
1,2=−
&#3627408464;
2&#3627408474;
±√(
&#3627408464;
2&#3627408474;
)
2
−(
&#3627408480;
&#3627408474;
)
&#3627408455;ℎ&#3627408466; &#3627408479;&#3627408462;&#3627408481;&#3627408470;&#3627408476; &#3627408476;&#3627408467; (
&#3627408464;
2&#3627408474;
)
2
&#3627408481;&#3627408476;
&#3627408480;
&#3627408474;
&#3627408479;&#3627408466;&#3627408477;&#3627408479;&#3627408466;&#3627408480;&#3627408466;&#3627408475;&#3627408481;&#3627408480; &#3627408481;ℎ&#3627408466; &#3627408465;&#3627408466;&#3627408468;&#3627408479;&#3627408466;&#3627408466; &#3627408476;&#3627408467; &#3627408465;&#3627408462;&#3627408474;&#3627408477;&#3627408475;&#3627408466;&#3627408480;&#3627408480; &#3627408462;&#3627408475;&#3627408465; &#3627408470;&#3627408481;&#3627408480; &#3627408480;&#3627408478;&#3627408482;&#3627408462;&#3627408479;&#3627408466; &#3627408479;&#3627408476;&#3627408476;&#3627408481; &#3627408479;&#3627408466;&#3627408477;&#3627408479;&#3627408466;&#3627408480;&#3627408466;&#3627408475;&#3627408481;&#3627408480; &#3627408481;ℎ&#3627408466; &#3627408465;&#3627408462;&#3627408474;&#3627408477;&#3627408470;&#3627408475;&#3627408468; &#3627408467;&#3627408462;&#3627408464;&#3627408481;&#3627408476;&#3627408479;(&#3627409211;).
&#3627409153;=√
(&#3627408464;2&#3627408474;⁄)
2
(&#3627408480;&#3627408474;⁄)
=
&#3627408464;
2√&#3627408480;&#3627408474;

&#3627408464;=2&#3627409153;√&#3627408480;&#3627408474;=2&#3627409153;&#3627408474;&#3627409172;
&#3627408475;=2&#3627409153;
&#3627408480;
&#3627409172;
&#3627408475;

when ζ=1, the damping is critical. The corresponding value of damping coefficient is denoted by cc
&#3627408456;&#3627408475;&#3627408465;&#3627408466;&#3627408479; &#3627408438;&#3627408479;&#3627408470;&#3627408481;&#3627408470;&#3627408464;&#3627408462;&#3627408473; &#3627408465;&#3627408462;&#3627408474;&#3627408477;&#3627408470;&#3627408475;&#3627408468; &#3627408464;&#3627408476;&#3627408475;&#3627408465;&#3627408470;&#3627408481;&#3627408470;&#3627408476;&#3627408475;&#3627408480; (&#3627409153;=1),&#3627408464;=&#3627408464;
&#3627408464;=2√&#3627408480;&#3627408474;=2&#3627408474;&#3627409172;
&#3627408475;=2&#3627408480;&#3627409172;
&#3627408475;⁄
&#3627409211;=&#3627408439;&#3627408462;&#3627408474;&#3627408477;&#3627408470;&#3627408475;&#3627408468; &#3627408467;&#3627408462;&#3627408464;&#3627408481;&#3627408476;&#3627408479;=
&#3627408516;
&#3627408516;
&#3627408516;
=
&#3627408462;&#3627408464;&#3627408481;&#3627408482;&#3627408462;&#3627408473; &#3627408465;&#3627408462;&#3627408474;&#3627408477;&#3627408470;&#3627408475;&#3627408468; &#3627408464;&#3627408476;&#3627408466;&#3627408467;&#3627408467;&#3627408470;&#3627408464;&#3627408470;&#3627408466;&#3627408475;&#3627408481;
&#3627408464;&#3627408479;&#3627408470;&#3627408481;&#3627408470;&#3627408464;&#3627408462;&#3627408473; &#3627408465;&#3627408462;&#3627408474;&#3627408477;&#3627408470;&#3627408475;&#3627408468; &#3627408464;&#3627408476;&#3627408466;&#3627408467;&#3627408467;&#3627408470;&#3627408464;&#3627408470;&#3627408466;&#3627408475;&#3627408481;

&#3627408485;̈+
&#3627408464;
&#3627408474;
&#3627408485;̇+
&#3627408480;
&#3627408474;
&#3627408485;=0 &#3627408464;&#3627408462;&#3627408475; &#3627408462;&#3627408473;&#3627408480;&#3627408476; &#3627408463;&#3627408466; &#3627408484;&#3627408479;&#3627408470;&#3627408481;&#3627408481;&#3627408466;&#3627408475; &#3627408462;&#3627408480; &#3627408485;̈+2&#3627409153;&#3627409172;
&#3627408475;&#3627408485;̇+&#3627409172;
&#3627408475;
2
&#3627408485;=0
&#3627409148;
1,2=−&#3627409153;&#3627409172;
&#3627408475;±√&#3627409153;
2
&#3627409172;
&#3627408475;
2
−&#3627409172;
&#3627408475;
2
=&#3627409172;
&#3627408475;(−&#3627409153;±√&#3627409153;
2
−1)
The exact solution of the equation will depend upon whether the roots are real or imaginary,
ζ>1, the system is overdamped
&#3627408443;&#3627408466;&#3627408479;&#3627408466;, (
&#3627408464;
2&#3627408474;
)
2
>(
&#3627408480;
&#3627408474;
) &#3627408480;&#3627408476; &#3627408481;ℎ&#3627408466; &#3627408479;&#3627408476;&#3627408476;&#3627408481;&#3627408480; &#3627408462;&#3627408479;&#3627408466; &#3627408479;&#3627408466;&#3627408462;&#3627408473; &#3627408462;&#3627408475;&#3627408465; &#3627408482;&#3627408475;&#3627408466;&#3627408478;&#3627408482;&#3627408462;&#3627408473;
&#3627408455;ℎ&#3627408466; &#3627408479;&#3627408476;&#3627408476;&#3627408481;&#3627408480; &#3627408484;&#3627408470;&#3627408473;&#3627408473; &#3627408463;&#3627408466; &#3627409148;
1=−
&#3627408464;
2&#3627408474;
+√(
&#3627408464;
2&#3627408474;
)
2
−(
&#3627408480;
&#3627408474;
) & &#3627409148;
2=−
&#3627408464;
2&#3627408474;
−√(
&#3627408464;
2&#3627408474;
)
2
−(
&#3627408480;
&#3627408474;
)
&#3627408455;ℎ&#3627408466; &#3627408479;&#3627408476;&#3627408476;&#3627408481;&#3627408480; &#3627408484;&#3627408470;&#3627408473;&#3627408473; &#3627408463;&#3627408466; &#3627409148;
1=&#3627409172;
&#3627408475;(−&#3627409153;+√&#3627409153;
2
−1) & &#3627409148;
2=&#3627409172;
&#3627408475;(−&#3627409153;−√&#3627409153;
2
−1)
(A & B are constants, α1 & α2 are roots of eqn.)
&#3627409172;
&#3627408475;=√&#3627408480;&#3627408474;⁄→&#3627408480;=&#3627408474;&#3627409172;
&#3627408475;
2

96

&#3627408485;=&#3627408436;&#3627408466;
??????
1&#3627408481;
+&#3627408437;&#3627408466;
??????
2&#3627408481;
=&#3627408436;&#3627408466;
??????
&#3627408475;(−??????+√??????
2
−1)&#3627408481;
+&#3627408437;&#3627408466;
??????
&#3627408475;(−??????−√??????
2
−1)&#3627408481;

The constants A & B can be found from initial conditions, this is the equation of aperiodic motion. The system
cannot vibrate due to over-damping. The magnitude of resultant displacement approaches zero with time.
ζ<1, the system is underdamped
&#3627408443;&#3627408466;&#3627408479;&#3627408466;, (
&#3627408464;
2&#3627408474;
)
2
<(
&#3627408480;
&#3627408474;
) &#3627408480;&#3627408476; &#3627408481;ℎ&#3627408466; &#3627408479;&#3627408476;&#3627408476;&#3627408481;&#3627408480; &#3627408462;&#3627408479;&#3627408466; &#3627408470;&#3627408474;&#3627408462;&#3627408468;&#3627408470;&#3627408475;&#3627408462;&#3627408479;&#3627408486;
&#3627408455;ℎ&#3627408466; &#3627408479;&#3627408476;&#3627408476;&#3627408481;&#3627408480; &#3627408484;&#3627408470;&#3627408473;&#3627408473; &#3627408463;&#3627408466; &#3627409148;
1=−
&#3627408464;
2&#3627408474;
+&#3627408470;√(
&#3627408480;
&#3627408474;
)−(
&#3627408464;
2&#3627408474;
)
2
& &#3627409148;
2=−
&#3627408464;
2&#3627408474;
−&#3627408470;√(
&#3627408480;
&#3627408474;
)−(
&#3627408464;
2&#3627408474;
)
2

&#3627408455;ℎ&#3627408466; &#3627408479;&#3627408476;&#3627408476;&#3627408481;&#3627408480; &#3627408484;&#3627408470;&#3627408473;&#3627408473; &#3627408463;&#3627408466; &#3627409148;
1=&#3627409172;
&#3627408475;(−&#3627409153;+&#3627408470;√1−&#3627409153;
2
) & &#3627409148;
2=&#3627409172;
&#3627408475;(−&#3627409153;−&#3627408470;√1−&#3627409153;
2
)
&#3627408485;=&#3627408436;&#3627408466;
??????1&#3627408481;
+&#3627408437;&#3627408466;
??????2&#3627408481;
=&#3627408436;&#3627408466;
??????&#3627408475;(−??????+??????√1−??????
2
)&#3627408481;
+&#3627408437;&#3627408466;
??????&#3627408475;(−??????−??????√1−??????
2
)&#3627408481;

After simplifying, we get
&#3627408537;=&#3627408511;&#3627408518;
−&#3627409211;??????
&#3627408527;&#3627408533;
&#3627408532;&#3627408522;&#3627408527;(??????
&#3627408517;&#3627408533;+??????)
The solution consists of 3 terms:
• X & φ, which are constants and are to be determined from initial conditions
• e
―ζωnt
, which decreases with time and finally e
―∞
= 0
• sin (ωdt + φ) which represents a repetition of motion
Resultant motion is an oscillatory motion with decreasing amplitudes having a frequency of ωd and ultimately
dies with time.
&#3627408467;
&#3627408465;=
&#3627409172;
&#3627408465;
2&#3627409163;
& &#3627408455;
&#3627408465;=
2&#3627409163;
&#3627409172;
&#3627408465;

ζ=1, the damping is critical
&#3627408443;&#3627408466;&#3627408479;&#3627408466;, (
&#3627408464;
2&#3627408474;
)
2
=(
&#3627408480;
&#3627408474;
) ,&#3627408481;ℎ&#3627408466; &#3627408479;&#3627408476;&#3627408476;&#3627408481;&#3627408480; &#3627408462;&#3627408479;&#3627408466; &#3627408466;&#3627408478;&#3627408482;&#3627408462;&#3627408473;,&#3627408462;&#3627408475;&#3627408465; &#3627408470;&#3627408480; &#3627409172;
&#3627408475;
&#3627408537;=(&#3627408488;+&#3627408489;&#3627408533;)&#3627408518;
−??????
&#3627408527;&#3627408533;

&#3627408466;
−??????&#3627408475;&#3627408481;
→0 &#3627408484;ℎ&#3627408466;&#3627408475; &#3627408481; →∞

Logarithmic decrement
In under-damped case, the amplitude reduces gradually.
&#3627408459;
0=&#3627408465;&#3627408470;&#3627408480;&#3627408477;&#3627408473;&#3627408462;&#3627408464;&#3627408466;&#3627408474;&#3627408466;&#3627408475;&#3627408481; &#3627408462;&#3627408481; &#3627408481;ℎ&#3627408466; &#3627408480;&#3627408481;&#3627408462;&#3627408479;&#3627408481; &#3627408476;&#3627408467; &#3627408474;&#3627408476;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408484;ℎ&#3627408466;&#3627408475; &#3627408481;=0
&#3627408459;
1=&#3627408465;&#3627408470;&#3627408480;&#3627408477;&#3627408473;&#3627408462;&#3627408464;&#3627408466;&#3627408474;&#3627408466;&#3627408475;&#3627408481; &#3627408462;&#3627408481; &#3627408481;ℎ&#3627408466; &#3627408466;&#3627408475;&#3627408465; &#3627408476;&#3627408467; &#3627408467;&#3627408470;&#3627408479;&#3627408480;&#3627408481; &#3627408476;&#3627408480;&#3627408464;&#3627408470;&#3627408473;&#3627408473;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408484;ℎ&#3627408466;&#3627408475; &#3627408481;=&#3627408455;
&#3627408465;
&#3627408459;
1=&#3627408459;&#3627408466;
−????????????&#3627408475;&#3627408455;
??????&#3627408480;&#3627408470;&#3627408475;(&#3627409172;
&#3627408465;&#3627408455;
&#3627408465;+??????)=&#3627408459;&#3627408466;
−????????????&#3627408475;&#3627408455;
??????&#3627408480;&#3627408470;&#3627408475;(&#3627409172;
&#3627408465;
2&#3627409163;
&#3627409172;
&#3627408465;
+??????)=&#3627408459;&#3627408466;
−????????????&#3627408475;&#3627408455;
??????&#3627408480;&#3627408470;&#3627408475;??????
&#3627408459;
2=&#3627408465;&#3627408470;&#3627408480;&#3627408477;&#3627408473;&#3627408462;&#3627408464;&#3627408466;&#3627408474;&#3627408466;&#3627408475;&#3627408481; &#3627408462;&#3627408481; &#3627408481;ℎ&#3627408466; &#3627408466;&#3627408475;&#3627408465; &#3627408476;&#3627408467; &#3627408480;&#3627408466;&#3627408464;&#3627408476;&#3627408475;&#3627408465; &#3627408476;&#3627408480;&#3627408464;&#3627408470;&#3627408473;&#3627408473;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408484;ℎ&#3627408466;&#3627408475; &#3627408481;=2&#3627408455;
&#3627408465;=&#3627408459;&#3627408466;
−2????????????&#3627408475;&#3627408455;
??????&#3627408480;&#3627408470;&#3627408475;??????
&#3627408459;
3=&#3627408465;&#3627408470;&#3627408480;&#3627408477;&#3627408473;&#3627408462;&#3627408464;&#3627408466;&#3627408474;&#3627408466;&#3627408475;&#3627408481; &#3627408462;&#3627408481; &#3627408481;ℎ&#3627408466; &#3627408466;&#3627408475;&#3627408465; &#3627408476;&#3627408467; &#3627408481;ℎ&#3627408470;&#3627408479;&#3627408465; &#3627408476;&#3627408480;&#3627408464;&#3627408470;&#3627408473;&#3627408473;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408484;ℎ&#3627408466;&#3627408475; &#3627408481;=3&#3627408455;
&#3627408465;=&#3627408459;&#3627408466;
−3????????????&#3627408475;&#3627408455;
??????&#3627408480;&#3627408470;&#3627408475;??????
……………………………………………………….
&#3627408459;
&#3627408475;=&#3627408459;&#3627408466;
−&#3627408475;????????????&#3627408475;&#3627408455;
??????&#3627408480;&#3627408470;&#3627408475;??????
&#3627408459;
&#3627408475;+1=&#3627408459;&#3627408466;
−&#3627408475;+1????????????&#3627408475;&#3627408455;
??????&#3627408480;&#3627408470;&#3627408475;??????
&#3627408459;
&#3627408475;
&#3627408459;
&#3627408475;+1
=&#3627408466;
????????????&#3627408475;&#3627408455;
??????=
&#3627408459;
0
&#3627408459;
1
=
&#3627408459;
1
&#3627408459;
2
=
&#3627408459;
2
&#3627408459;
3
=⋯
&#3627409209;=&#3627408525;&#3627408527;(
&#3627408511;
&#3627408527;
&#3627408511;
&#3627408527;+&#3627409359;
)=&#3627409211;??????
&#3627408527;&#3627408507;
&#3627408517;
&#3627409209;=&#3627409211;
??????
&#3627408517;
√&#3627409359;−&#3627409211;
&#3627409360;
&#3627408507;
&#3627408517;
(&#3627408436;+&#3627408437;=&#3627408459;&#3627408480;&#3627408470;&#3627408475;?????? ,&#3627408470;(&#3627408436;−&#3627408437;)=&#3627408459;&#3627408464;&#3627408476;&#3627408480;??????)
(&#3627409172;
&#3627408475;√1−&#3627409153;
2
=&#3627409172;
&#3627408465;)

97

Forced vibrations
Step-input Forcing
Application of a constant force F to the mass of a vibrating system is known as step-input forcing. The equation
of motion will be
&#3627408474;&#3627408485;̈+&#3627408480;&#3627408485;=&#3627408441;
After the application of force F, the system will vibrate about the new equilibrium position, the displacement of
which will be F/s.
Harmonic Forcing
Here the mass is subjected to an oscillating force F = Fo sin ωt.
Equation of motion will be
&#3627408474;&#3627408485;̈+&#3627408480;&#3627408485;=&#3627408441;
&#3627408476;&#3627408480;&#3627408470;&#3627408475;&#3627409172;&#3627408481;
The solution for this equation is
&#3627408485;=&#3627408459;&#3627408480;&#3627408470;&#3627408475;(&#3627409172;
&#3627408475;&#3627408481;+??????)+
&#3627408441;
&#3627408476;&#3627408480;⁄
1−(&#3627409172;&#3627409172;
&#3627408475;⁄)
2
&#3627408480;&#3627408470;&#3627408475;&#3627409172;&#3627408481;
The resultant motion is the sum of two harmonics.
Periodic Forcing
A periodic force is one in which the motion repeats itself in all details after a certain interval of time.
Forced-Damped Vibrations
B-B be the static equilibrium position under the weight of the
mass. Now if the mass is subjected to an oscillating force F=Fo
sin ωt, the forces on acting on mass at any instant will be
&#3627408444;&#3627408474;&#3627408477;&#3627408479;&#3627408466;&#3627408480;&#3627408480;&#3627408466;&#3627408465; &#3627408450;&#3627408480;&#3627408464;&#3627408470;&#3627408473;&#3627408473;&#3627408462;&#3627408481;&#3627408470;&#3627408475;&#3627408468; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466; (&#3627408493;)= &#3627408441;
&#3627408476;&#3627408480;&#3627408470;&#3627408475;&#3627409172;&#3627408481;(&#3627408465;&#3627408476;&#3627408484;&#3627408475;&#3627408484;&#3627408462;&#3627408479;&#3627408465;&#3627408480;)
&#3627408444;&#3627408475;&#3627408466;&#3627408479;&#3627408481;&#3627408470;&#3627408462;=&#3627408474;&#3627408485;̈ (&#3627408482;&#3627408477;&#3627408484;&#3627408462;&#3627408479;&#3627408465;&#3627408480;)
&#3627408439;&#3627408462;&#3627408474;&#3627408477;&#3627408470;&#3627408475;&#3627408468; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;=&#3627408464;&#3627408485;̇ (&#3627408482;&#3627408477;&#3627408484;&#3627408462;&#3627408479;&#3627408465;&#3627408480;)
&#3627408454;&#3627408477;&#3627408479;&#3627408470;&#3627408475;&#3627408468; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;=&#3627408480;&#3627408485; (&#3627408482;&#3627408477;&#3627408484;&#3627408462;&#3627408479;&#3627408465;&#3627408480;)

The equation of motion will be
&#3627408474;&#3627408485;̈+&#3627408464;&#3627408485;̇+&#3627408480;&#3627408485;=&#3627408441;
&#3627408476;&#3627408480;&#3627408470;&#3627408475;&#3627409172;&#3627408481;
&#3627408455;ℎ&#3627408466; &#3627408480;&#3627408476;&#3627408473;&#3627408482;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408470;&#3627408480; &#3627408485;=&#3627408438;&#3627408476;&#3627408474;&#3627408477;&#3627408473;&#3627408470;&#3627408474;&#3627408466;&#3627408475;&#3627408481;&#3627408462;&#3627408479;&#3627408486; &#3627408467;&#3627408482;&#3627408475;&#3627408464;&#3627408481;&#3627408470;&#3627408476;&#3627408475; (&#3627408438;&#3627408441;)+&#3627408451;&#3627408462;&#3627408479;&#3627408481;&#3627408470;&#3627408462;&#3627408473; &#3627408467;&#3627408482;&#3627408475;&#3627408464;&#3627408481;&#3627408470;&#3627408476;&#3627408475; (&#3627408451;&#3627408444;)
&#3627408485;=&#3627408438;&#3627408441;+&#3627408451;&#3627408444;=&#3627408459;&#3627408466;
−????????????&#3627408475;&#3627408481;
&#3627408480;&#3627408470;&#3627408475;(&#3627409172;
&#3627408465;&#3627408481;−??????)+
&#3627408441;
&#3627408476;
√(&#3627408480;−&#3627408474;&#3627409172;
2
)
2
+(&#3627408464;&#3627409172;)
2
&#3627408480;&#3627408470;&#3627408475;&#3627409172;&#3627408481;
The damped free vibrations represented by the first part(CF) becomes negligible with time as e
-∞
=0. Also, in
actual practice, the value of the complementary function x1 at any time t is much smaller as compared to
particular integral x2. The steady state response of the system is then given by the second part PI. The amplitude
of steady state response is given by
&#3627408436;=
&#3627408441;
&#3627408476;&#3627408480;⁄
√(1−(
&#3627409172;
&#3627409172;
&#3627408475;
)
2
)
2
+(2&#3627409153;
&#3627409172;
&#3627409172;
&#3627408475;
)
2

φ is the phase lag for the displacement relative to the velocity vector
&#3627408481;&#3627408462;&#3627408475;??????=
&#3627408462;&#3627409172;
&#3627408463;−&#3627409172;
2
=
2&#3627409153;
&#3627409172;
&#3627409172;
&#3627408475;
1−(
&#3627409172;
&#3627409172;
&#3627408475;
)
2

X & φ are obtained from the
initial conditions.

98

Graphical method
Assume that the displacement(x) of the
vibrating mass under the action if the applied
simple harmonic force Fo sin ωt is also simple
harmonic and lags by an amount φ.





&#3627408485;=&#3627408436;&#3627408480;&#3627408470;&#3627408475;(&#3627409172;&#3627408481;−??????)
&#3627408454;&#3627408482;&#3627408463;&#3627408480;&#3627408481;&#3627408470;&#3627408481;&#3627408482;&#3627408481;&#3627408470;&#3627408475;&#3627408468; &#3627408481;ℎ&#3627408466; &#3627408483;&#3627408462;&#3627408473;&#3627408482;&#3627408466;&#3627408480; &#3627408476;&#3627408467; &#3627408485;,&#3627408485;̇,&#3627408485;̈ &#3627408470;&#3627408475; &#3627408481;ℎ&#3627408466; &#3627408466;&#3627408478;&#3627408482;&#3627408462;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408474;&#3627408485;̈+&#3627408464;&#3627408485;̇+&#3627408480;&#3627408485;=&#3627408441;
&#3627408476;&#3627408480;&#3627408470;&#3627408475;&#3627409172;&#3627408481;,&#3627408484;&#3627408466; &#3627408468;&#3627408466;&#3627408481;
&#3627408441;
&#3627408476;&#3627408480;&#3627408470;&#3627408475;&#3627409172;&#3627408481;+&#3627408474;&#3627409172;
2
&#3627408436;&#3627408480;&#3627408470;&#3627408475;(&#3627409172;&#3627408481;−??????)−&#3627408464;&#3627409172;&#3627408436;&#3627408480;&#3627408470;&#3627408475;[
&#3627409163;
2
+(&#3627409172;&#3627408481;−??????)]−&#3627408480;&#3627408436;&#3627408480;&#3627408470;&#3627408475;(&#3627409172;&#3627408481;−??????)=0
&#3627408441;
&#3627408476;=√(&#3627408480;&#3627408436;−&#3627408474;&#3627409172;
2
&#3627408436;)
2
+(&#3627408464;&#3627409172;&#3627408436;)
2

&#3627408436;=
&#3627408441;
&#3627408476;
√(&#3627408480;−&#3627408474;&#3627409172;
2
)
2
+(&#3627408464;&#3627409172;)
2

&#3627408481;&#3627408462;&#3627408475;??????=
&#3627408464;&#3627409172;
&#3627408480;−&#3627408474;&#3627409172;
2


Magnification factor (MF)
The ratio of amplitude of the steady-state response to the static deflection under the action of force Fo is known
as the magnification factor.
&#3627408448;&#3627408441;=
&#3627408436;&#3627408474;&#3627408477;&#3627408473;&#3627408470;&#3627408481;&#3627408482;&#3627408465;&#3627408466; &#3627408476;&#3627408467; &#3627408481;ℎ&#3627408466; &#3627408480;&#3627408481;&#3627408466;&#3627408462;&#3627408465;&#3627408486;−&#3627408480;&#3627408481;&#3627408462;&#3627408481;&#3627408466;
&#3627408454;&#3627408481;&#3627408462;&#3627408481;&#3627408470;&#3627408464; &#3627408465;&#3627408466;&#3627408467;&#3627408473;&#3627408466;&#3627408464;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408482;&#3627408475;&#3627408465;&#3627408466;&#3627408479; &#3627408481;ℎ&#3627408466; &#3627408462;&#3627408464;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408476;&#3627408467; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466; &#3627408441;
&#3627408476;
=
&#3627408441;
&#3627408476;√(&#3627408480;−&#3627408474;&#3627409172;
2
)
2
+(&#3627408464;&#3627409172;)
2

&#3627408441;
&#3627408476;&#3627408480;⁄

&#3627408448;&#3627408441;=
&#3627408480;
√(&#3627408480;−&#3627408474;&#3627409172;
2
)
2
+(&#3627408464;&#3627409172;)
2
=
1
√(1−(
&#3627409172;
&#3627409172;
&#3627408475;
)
2
)
2
+(2&#3627409153;
&#3627409172;
&#3627409172;
&#3627408475;
)
2

Thus, the magnification factor depends on ω/ωn and ζ.
The magnification factor gives the factor by which the static deflection produced by a force F (i.e. xo=Fo/s) must
be multiplied to obtain the maximum amplitude of the forced vibration or amplitude at steady state by the
harmonic force F cos ω.t
??????&#3627408493;(&#3627408493;&#3627408514;&#3627408516;&#3627408533;&#3627408528;&#3627408531;)×&#3627408506;&#3627408533;&#3627408514;&#3627408533;&#3627408522;&#3627408516; &#3627408517;&#3627408518;&#3627408519;&#3627408525;&#3627408518;&#3627408516;&#3627408533;&#3627408522;&#3627408528;&#3627408527; (&#3627408537;
&#3627408528;=&#3627408493;
&#3627408528;&#3627408532;⁄)= &#3627408488;&#3627408526;&#3627408529;&#3627408525;&#3627408522;&#3627408533;&#3627408534;&#3627408517;&#3627408518; &#3627408514;&#3627408533; &#3627408532;&#3627408533;&#3627408518;&#3627408514;&#3627408517;&#3627408538; &#3627408532;&#3627408533;&#3627408514;&#3627408533;&#3627408518;(&#3627408488;)
A plot of magnification factor(MF)
against the ratio of frequencies (ω/ωn)
for different values of ζ is shown

As damping increases (ζ↑), the
maximum value of the MF decreases
and vice-versa. When there is no
damping (ζ=0), the MF reaches
infinity.
When the frequency of forced vibrations equals frequency of free vibrations, the condition is known as
resonance.

99

Vibration isolation and transmissibility
Vibrations are produced in machines having unbalanced masses. These vibrations will be transmitted to the
foundation upon which the machines are installed. This is undesirable. To diminish the transmitted forces, the
machines are usually mounted in springs and dampers or some other vibration isolation material.
Transmissibility is defined as the ratio of force transmitted (to the foundation) to the force applied. It is the
measure of effectiveness of the vibration isolation material.
We have discussed above that the force transmitted to the foundation consists of the following two forces:
1. Spring force or elastic force which is equal to s. xmax, and
2. Damping force which is equal to c. ω. xmax.
These forces are perpendicular to each other.
&#3627408441;
&#3627408481;=√(&#3627408480;&#3627408436;)
2
+(&#3627408464;&#3627409172;&#3627408436;)
2
=&#3627408436;√(&#3627408480;)
2
+(&#3627408464;&#3627409172;)
2
=
&#3627408441;
&#3627408476;
√(&#3627408480;−&#3627408474;&#3627409172;
2
)
2
+(&#3627408464;&#3627409172;)
2
√(&#3627408480;)
2
+(&#3627408464;&#3627409172;)
2
=
&#3627408441;
&#3627408476;√1+2&#3627409153;
&#3627409172;
&#3627409172;
&#3627408475;
√[1−(
&#3627409172;
&#3627409172;
&#3627408475;
)
2
]
2
+(2&#3627409153;
&#3627409172;
&#3627409172;
&#3627408475;
)
2

&#3627408455;&#3627408479;&#3627408462;&#3627408475;&#3627408480;&#3627408474;&#3627408470;&#3627408480;&#3627408480;&#3627408470;&#3627408463;&#3627408470;&#3627408473;&#3627408470;&#3627408481;&#3627408486;,&#3627409210;=
&#3627408441;
&#3627408481;
&#3627408441;
&#3627408476;
=
√1+2&#3627409153;
&#3627409172;
&#3627409172;
&#3627408475;
√[1−(
&#3627409172;
&#3627409172;
&#3627408475;
)
2
]
2
+(2&#3627409153;
&#3627409172;
&#3627409172;
&#3627408475;
)
2

&#3627408436;&#3627408481; &#3627408453;&#3627408466;&#3627408480;&#3627408476;&#3627408475;&#3627408462;&#3627408475;&#3627408464;&#3627408466;,
&#3627409172;
&#3627409172;
&#3627408475;
=1,&#3627409152;=
√1+(2&#3627409153;)
2
2&#3627409153;

&#3627408458;ℎ&#3627408466;&#3627408475; &#3627408475;&#3627408476; &#3627408465;&#3627408462;&#3627408474;&#3627408477;&#3627408466;&#3627408479; &#3627408470;&#3627408480; &#3627408482;&#3627408480;&#3627408466;&#3627408465;,&#3627409153;=0 &#3627409152;=
1
±[1−(
&#3627409172;
&#3627409172;
&#3627408475;
)
2
]

&#3627408458;ℎ&#3627408466;&#3627408475;
&#3627409172;
&#3627409172;
&#3627408475;
<√2,&#3627409210;>1⇒&#3627408481;&#3627408479;&#3627408462;&#3627408475;&#3627408480;&#3627408474;&#3627408470;&#3627408481;&#3627408481;&#3627408466;&#3627408465; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;(&#3627408441;
&#3627408481;)
>&#3627408462;&#3627408477;&#3627408477;&#3627408473;&#3627408470;&#3627408466;&#3627408465; &#3627408467;&#3627408476;&#3627408479;&#3627408464;&#3627408466;(&#3627408441;
&#3627408476;)
&#3627408458;ℎ&#3627408466;&#3627408475;
&#3627409172;
&#3627409172;
&#3627408475;
>√2,&#3627409210;<1⇒&#3627408441;
&#3627408481;<&#3627408441;
&#3627408476;
&#3627408458;ℎ&#3627408466;&#3627408475;
&#3627409172;
&#3627409172;
&#3627408475;
>√2,&#3627409210;↑→&#3627409153;↑
&#3627408458;ℎ&#3627408466;&#3627408475;
&#3627409172;
&#3627409172;
&#3627408475;
=√2,&#3627409210;=1⇒&#3627408441;
&#3627408481;=&#3627408441;
&#3627408476;
&#3627408458;ℎ&#3627408466;&#3627408475;
&#3627409172;
&#3627409172;
&#3627408475;
=1,&#3627409210;→∞,

If damping is used, the magnitude of transmitted force can be reduced

100

Forcing due to unbalance
All types of rotating machinery always consist of unbalance left in them. The net balance in such machines may
be represented by a mass ‘m’ rotating with its centre of mass at a
distance ‘e’ from the axis of rotation. If mass ‘M’ is the total mass of
vibrating system including the unbalanced mass ‘m’, the centrifugal
force acting outwards of rotation is equal to m.e.ω
2
.
The equation of motion can be written as
&#3627408474;&#3627408485;̈+&#3627408464;&#3627408485;̇+&#3627408480;&#3627408485;=&#3627408474;&#3627408466;&#3627409172;
2
&#3627408480;&#3627408470;&#3627408475;&#3627409172;&#3627408481;
&#3627408485;=
&#3627408474;&#3627408466;&#3627409172;
2
√(&#3627408480;−&#3627408474;&#3627409172;
2
)
2
+(&#3627408464;&#3627409172;)
2
&#3627408480;&#3627408470;&#3627408475;(&#3627409172;&#3627408481;−??????)
&#3627408436;=
&#3627408474;&#3627408466;&#3627409172;
2
&#3627408480;⁄
√(1−(
&#3627409172;
&#3627409172;
&#3627408475;
)
2
)
2
+(2&#3627409153;
&#3627409172;
&#3627409172;
&#3627408475;
)
2

&#3627408436;
&#3627408474;&#3627408466;
&#3627408448;
=
(
&#3627409172;
&#3627409172;
&#3627408475;
)
2
√(1−(
&#3627409172;
&#3627409172;
&#3627408475;
)
2
)
2
+(2&#3627409153;
&#3627409172;
&#3627409172;
&#3627408475;
)
2

The equation provides the steady-state amplitude as a function of damping
factor and frequency ratio.
It shows that at higher values of frequency ratio (ω/ωn), the amplitude can be
reduced by mass(m) and eccentricity of rotating unbalance(e).
Forcing due to Support motion
In case of vehicles (cars, bikes, trucks), the excitation of the system is
through the support or base instead of directly to the mass. Assuming that
the support is excited a harmonic motion, y = Y sin(ωt) and displacement
of x is more compared to y in the considered position.
The equation of motion is
&#3627408474;&#3627408485;̈+&#3627408464;(&#3627408485;̇−&#3627408486;̇)+&#3627408480;(&#3627408485;−&#3627408486;)=0
&#3627408474;&#3627408485;̈+&#3627408464;&#3627408485;̇+&#3627408480;&#3627408485;=&#3627408464;&#3627408486;̇+&#3627408480;&#3627408486;
=&#3627408464;&#3627408460;&#3627409172;&#3627408464;&#3627408476;&#3627408480;&#3627409172;&#3627408481;+&#3627408480;&#3627408460;&#3627408480;&#3627408470;&#3627408475;&#3627409172;&#3627408481;
=&#3627408516;&#3627408512;??????&#3627408480;&#3627408470;&#3627408475;(90+&#3627409172;&#3627408481;)+&#3627408532;&#3627408512;&#3627408480;&#3627408470;&#3627408475;&#3627409172;&#3627408481;

&#3627408441;
&#3627408476;=√(&#3627408480;&#3627408460;)
2
+(&#3627408464;&#3627408460;&#3627409172;)
2
=&#3627408460;√&#3627408480;
2
+(&#3627408464;&#3627409172;)
2




We know that,
&#3627408436;=
&#3627408441;
&#3627408476;
√(&#3627408480;−&#3627408474;&#3627409172;
2
)
2
+(&#3627408464;&#3627409172;)
2
=
&#3627408460;√&#3627408480;
2
+(&#3627408464;&#3627409172;)
2
√(&#3627408480;−&#3627408474;&#3627409172;
2
)
2
+(&#3627408464;&#3627409172;)
2

&#3627408436;
&#3627408460;
=
√&#3627408480;
2
+(&#3627408464;&#3627409172;)
2
√(&#3627408480;−&#3627408474;&#3627409172;
2
)
2
+(&#3627408464;&#3627409172;)
2

&#3627408436;
&#3627408460;
=
√1+(2&#3627409153;
&#3627409172;
&#3627409172;
&#3627408475;
)
2
√(1−(
&#3627409172;
&#3627409172;
&#3627408475;
)
2
)
2
+(2&#3627409153;
&#3627409172;
&#3627409172;
&#3627408475;
)
2

The ratio (A/Y) is called Displacement Transmissibility.
(&#3627408474;&#3627408466;
&#3627409172;
2
&#3627408480;
=&#3627408474;&#3627408466;
&#3627409172;
2
&#3627408448;⁄
&#3627408480;&#3627408448;⁄
=
&#3627408474;&#3627408466;
&#3627408448;
(
&#3627409172;
&#3627409172;
&#3627408475;
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101

Torsional vibrations
Consider a uniform shaft of length ‘l’ rigidly fixed at its upper end and carrying a disc of moment of inertia 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.
&#3627409155;=&#3627408462;&#3627408475;&#3627408468;&#3627408482;&#3627408473;&#3627408462;&#3627408479; &#3627408465;&#3627408470;&#3627408480;&#3627408477;&#3627408473;&#3627408462;&#3627408464;&#3627408466;&#3627408474;&#3627408466;&#3627408475;&#3627408481; &#3627408476;&#3627408467; &#3627408481;ℎ&#3627408466; &#3627408465;&#3627408470;&#3627408480;&#3627408464; &#3627408467;&#3627408479;&#3627408476;&#3627408474; &#3627408470;&#3627408481;&#3627408480; &#3627408466;&#3627408478;&#3627408482;&#3627408473;&#3627408470;&#3627408463;&#3627408479;&#3627408470;&#3627408482;&#3627408474; &#3627408477;&#3627408476;&#3627408480;&#3627408470;&#3627408481;&#3627408470;&#3627408476;&#3627408475; &#3627408462;&#3627408481; &#3627408462;&#3627408475;&#3627408486; &#3627408470;&#3627408475;&#3627408480;&#3627408481;&#3627408462;&#3627408475;&#3627408481;.
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At any instant, the torques acting on the disc are
&#3627408444;&#3627408475;&#3627408466;&#3627408479;&#3627408481;&#3627408470;&#3627408462; &#3627408481;&#3627408476;&#3627408479;&#3627408478;&#3627408482;&#3627408466;=−&#3627408444;&#3627409155;̈
&#3627408453;&#3627408466;&#3627408480;&#3627408481;&#3627408476;&#3627408479;&#3627408470;&#3627408475;&#3627408468; &#3627408481;&#3627408476;&#3627408479;&#3627408478;&#3627408482;&#3627408466;=−&#3627408478;&#3627409155;
&#3627408444;&#3627409155;̈+&#3627408478;&#3627409155;=0
&#3627409155;̈+
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&#3627409155;=0
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&#3627408475;=√
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, &#3627408467;
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1
2&#3627409163;

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