Theory of machines notes for gate, ies and academic purposes

1. Soumyth
THEORY OF MACHINES

2. 1
CONTENTS
INTRODUCTION.............................................................................................................................................................................2
MOTION ANALYSIS.....................................................................................................................................................................16
VELOCITY ANALYSIS OF DOUBLE 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

3. 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

4. 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.

5. 4
Kinematic chain
When all the links are connected in such a way that 1st
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
𝐹 = [3 · (𝐿 − 1) − 2𝑗 − ℎ] ⟶ 𝐾𝑢𝑡𝑧𝑏𝑎𝑐𝑘 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛
L → no. of Links, j → no. of binary joint, h → no. of higher pair
3(𝐿 − 1) → 𝑛𝑜. 𝑜𝑓 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑚𝑜𝑡𝑖𝑜𝑛 𝑖𝑛 2𝐷 𝑝𝑙𝑎𝑛𝑎𝑟 𝑚𝑒𝑐ℎ𝑎𝑛𝑖𝑠𝑚
Note: -
𝐹 = [3 · (𝐿 − 1) − 2𝑗 − ℎ] − 𝐹𝑟

6. 5
Fr → no. of those motions, which are not the part of mechanism (Dummy motion)
Fr→ s (NOT A PART OF MECHANISM)
Example

7. 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. 𝐼𝑓 (𝑗 +
ℎ
2
) = (
3𝑙
2
− 2) → 𝐾𝑖𝑛𝑒𝑚𝑎𝑡𝑖𝑐 𝑐ℎ𝑎𝑖𝑛
2. 𝐼𝑓 (𝑗 +
ℎ
2
) > (
3𝑙
2
− 2) → 𝐹𝑟𝑎𝑚𝑒 ⧸𝑠𝑡𝑟𝑢𝑐𝑡𝑢𝑟𝑒 ⧸𝑠𝑢𝑝𝑒𝑟𝑠𝑡𝑟𝑢𝑐𝑡𝑢𝑟𝑒
𝐼𝑓 (𝐿𝐻𝑆 – 𝑅𝐻𝑆) = 0.5 → 𝐹𝑟𝑎𝑚𝑒 𝑠𝑡𝑟𝑢𝑐𝑡𝑢𝑟𝑒
(𝐿𝐻𝑆 − 𝑅𝐻𝑆) > 0.5 → 𝑆𝑢𝑝𝑒𝑟 𝑠𝑡𝑟𝑢𝑐𝑡𝑢𝑟𝑒
3. 𝐼𝑓 (𝑗 +
ℎ
2
) < (
3𝑙
2
− 2) → 𝑈𝑛𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑒𝑑 𝑐ℎ𝑎𝑖𝑛

8. 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
𝐹 = 3 · (𝑙 − 1) − 2𝑗 − ℎ
1 = 3𝑙 − 3 − 2𝑗
3𝑙 − 2𝑗 − 4 = 0 → 𝐺𝑟𝑢𝑏𝑙𝑒𝑟′
𝑠 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛
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

9. 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
(𝑆 + 𝐿) ≤ (𝑝 + 𝑞)
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
𝑠 𝑠 𝑙 𝑙
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)

10. 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.
𝐴𝐶2
= 𝑎2
+ 𝑏2
− 2𝑎𝑏 𝑐𝑜𝑠 𝜃 = 𝑐2
+ 𝑑2
− 2𝑐𝑑 𝑐𝑜𝑠 𝜇
Differentiating both sides,
(−2𝑎𝑏) · (− 𝑠𝑖𝑛 𝜃) · 𝑑𝜃 = (−2𝑐𝑑) · (− 𝑠𝑖𝑛 𝜇) · 𝑑𝜇
𝑑𝜇
𝑑𝜃
= (
𝑎𝑏
𝑐𝑑
) ·
𝑠𝑖𝑛 𝜃
𝑠𝑖𝑛 𝜇
For μ to be max/min,
𝑑𝜇
𝑑𝜃
= 0 → (
𝑎𝑏
𝑐𝑑
) ·
𝑠𝑖𝑛 𝜃
𝑠𝑖𝑛 𝜇
= 0 → 𝑠𝑖𝑛 𝜃 = 0
𝜽 = 𝟎°, 𝟏𝟖𝟎° → 𝜇𝑚𝑖𝑛 = 0°, 𝜇𝑚𝑎𝑥 = 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

11. 10
Single Slider crank mechanism (Drag-link Mechanism)
Ist
inversion (Cylinder fixed)
Rotation ⟷ Oscillation
Crank ⟷ Piston
Output ⟵ Input (Piston to Crank) ⟶ Reciprocating engine
Input ⟶ Output (Crank to Piston) ⟶ Reciprocating compressor
IInd
Inversion (Crank fixed)
→ Whitworth quick return motion mechanism
→ Rotary IC engine mechanism (Gnome engine)
IIIrd
Inversion (Connecting Rod fixed)
→ Crank and slotted lever quick return mechanism
→ Oscillating cylinder engine mechanism
IVth
Inversion (Slider/Piston Fixed)
→ Hand Pump (Pendulum pump, Bull engine)

12. 11
Crank and slotted Lever (Quick Return Motion Mechanism)
IIIrd
Inversion (Connecting Rod Fixed)
Quick Return Ratio (QRR)
𝑄𝑅𝑅 =
𝑇𝑖𝑚𝑒𝑐𝑢𝑡𝑡𝑖𝑛𝑔
𝑇𝑖𝑚𝑒𝑅𝑒𝑡𝑢𝑟𝑛
=
𝛽
𝛼
> 1 (𝒂𝒍𝒘𝒂𝒚𝒔)
Stroke R1R2
𝑅1𝑅2 = 𝐶1𝐶2 = 2 × 𝐶1𝑀 = 2 × 𝐴𝐶1 × 𝑐𝑜𝑠
𝛼
2
= 2 × 𝐴𝐶1 × (
𝑂𝐵1
𝑂𝐴
) =
2 × 𝐴𝐶 × 𝑂𝐵
𝑂𝐴
𝑆𝑡𝑟𝑜𝑘𝑒 =
2 × 𝐴𝐶 × 𝑂𝐵
𝑂𝐴
=
2 × [𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑙𝑜𝑡𝑡𝑒𝑑 𝑏𝑎𝑟] × [𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑟𝑎𝑛𝑘]
[𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑖𝑛𝑔 𝑟𝑜𝑑]
As α < β, return stroke is quicker than Cutting stroke, so it is called Quick return mechanism.
QRR can never be less than 1.

13. 12
Whitworth Quick Return Motion Mechanism
IInd
Inversion (crank fixed)
Rotation ⟶ Rotation (Double crank)
Stroke ⟹ R1R2 ⟹ C1C2 ⟹2 (OC)
Oscillating Cylinder Mechanism
IIIrd
Mechanism (Connecting Rod fixed)
Rotary IC Engine Mechanism (GNOME Engine)
IInd
Inversion

14. 13
Hand Pump
IVth
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)

15. 14
Double slider crank chain
(4 links + 2 Turning Pairs+ 2 Sliding Pairs)
1. Slotted plate is fixed (Elliptical Trammels)
𝑐𝑜𝑠 𝜃 =
𝑥
𝐴𝑃
𝑠𝑖𝑛 𝜃 =
𝑦
𝐵𝑃
𝑥2
𝐴𝑃2
+
𝑦2
𝐵𝑃2
= 1 → 𝐸𝑙𝑙𝑖𝑝𝑠𝑒
Locus of any point ‘P’ on link AB except midpoint is on ellipse.

16. 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)
𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝐴𝑑𝑣𝑎𝑛𝑡𝑎𝑔𝑒 =
𝐹𝑜𝑢𝑡𝑝𝑢𝑡
𝐹𝑖𝑛𝑝𝑢𝑡
=
𝑇𝑜𝑢𝑡𝑝𝑢𝑡
𝑇𝑖𝑛𝑝𝑢𝑡
𝜂𝑚𝑒𝑐ℎ𝑎𝑛𝑖𝑠𝑚 =
𝑃𝑜𝑢𝑡𝑝𝑢𝑡
𝑃𝑖𝑛𝑝𝑢𝑡
=
𝐹𝑜𝑢𝑡𝑝𝑢𝑡 × 𝑉𝑜𝑢𝑡𝑝𝑢𝑡
𝐹𝑖𝑛𝑝𝑢𝑡 × 𝑉𝑖𝑛𝑝𝑢𝑡
=
𝑇𝑜𝑢𝑡𝑝𝑢𝑡 × 𝑤𝑜𝑢𝑡𝑝𝑢𝑡
𝑇𝑖𝑛𝑝𝑢𝑡 × 𝑤𝑖𝑛𝑝𝑢𝑡
𝑀. 𝐴 =
𝑉𝑖𝑛𝑝𝑢𝑡
𝑉𝑜𝑢𝑡𝑝𝑢𝑡
× 𝜂𝑚𝑒𝑐ℎ𝑎𝑛𝑖𝑠𝑚 =
𝑤𝑖𝑛𝑝𝑢𝑡
𝑤𝑜𝑢𝑡𝑝𝑢𝑡
× 𝜂𝑚𝑒𝑐ℎ𝑎𝑛𝑖𝑠𝑚

17. 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.
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝐴 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑡𝑜 𝑂 =
𝐴𝑟𝑐 𝐴𝐴′
𝛿𝑡
⟹ 𝑣𝑎𝑜 =
𝑟 · 𝛿𝜃
𝛿𝑡
𝑤ℎ𝑒𝑛 𝛿𝑡 → 0, 𝑣𝑎𝑜 = 𝑟
𝑑𝜃
𝑑𝑡
= 𝑟𝜔
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.
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝐵 = 𝜔 × 𝑂𝐵 ⊥ 𝑡𝑜 𝑂𝐵
𝐨𝐛
𝐨𝐚
=
𝜔 · 𝑂𝐵
𝜔 · 𝑂𝐴
=
𝑂𝐵
𝑂𝐴
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.
𝑣𝑐𝑎 = 𝑣𝑐𝑏 + 𝑣𝑏𝑎
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.
𝑣𝑐𝑑 = 𝑣𝑏𝑎 + 𝑣𝑐𝑏 (𝑜𝑟) 𝐝𝐜 = 𝐚𝐛 + 𝐛𝐜
𝒗𝒃𝒂 𝑖𝑠 𝑘𝑛𝑜𝑤𝑛 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝝎 · 𝑨𝑩, 𝑎𝑛𝑑 𝒗𝒄𝒃 & 𝒗𝒅𝒄 𝑎𝑟𝑒 𝑢𝑛𝑘𝑛𝑜𝑤𝑛.

18. 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
→ 𝐝𝐜 = 𝐚𝐛 − 𝐜𝐛 ⟹ 𝐝𝐜 + 𝐜𝐛 = 𝐚𝐛
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,
𝐛𝐞
𝐛𝐜
=
𝐵𝐸
𝐵𝐶
ae represents the absolute velocity of E.
Angular velocity of Links
𝑣𝑐𝑏 = 𝐛𝐜 𝑣𝑏𝑐 = 𝐜𝐛
𝑣𝑐𝑏 = 𝜔𝑐𝑏 × 𝐵𝐶 = 𝜔𝑐𝑏 × 𝐶𝐵 → 𝜔𝑐𝑏 =
𝑣𝑐𝑏
𝐶𝐵
𝜔𝑏𝑐 =
𝑣𝑏𝑐
𝐶𝐵
;
The magnitude of ωcb = ωbc as vcb = vbc and the direction of rotation is the same.
𝑣𝑐𝑑 = 𝐝𝐜
𝜔𝑐𝑑 =
𝑣𝑐𝑑
𝐶𝐷
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.
𝑣𝑏𝑜 = 𝑣𝑏𝑎 + 𝑣𝑎𝑜 (𝑜𝑟) 𝑣𝑏𝑔 = 𝑣𝑎𝑜 + 𝑣𝑏𝑎
𝐠𝐛 = 𝐨𝐚 + 𝐚𝐛
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
𝑣𝑏𝑎
𝐵𝐴(𝑜𝑟 𝐴𝐵)

19. 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.
𝑣𝑞𝑜 = 𝑣𝑞𝑝 + 𝑣𝑝𝑜 (𝑜𝑟) 𝑣𝑞𝑎 = 𝑣𝑝𝑜 + 𝑣𝑞𝑝
𝐚𝐪 = 𝐨𝐩 + 𝐩𝐪
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,
𝑣𝑠𝑜 = 𝑣𝑠𝑟 + 𝑣𝑟𝑜 (𝑜𝑟) 𝑣𝑠𝑔 = 𝑣𝑟𝑜 + 𝑣𝑠𝑟
𝐠𝐬 = 𝐨𝐫 + 𝐫𝐬

20. 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.
𝐴𝑙𝑠𝑜, 𝜔𝑟𝑠 =
𝑣𝑟𝑠
𝑅𝑆
𝐶𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
Usually, the coupler RS is long and its obliquity is neglected. Then or ≈ os.
𝑇𝑖𝑚𝑒 𝑜𝑓 𝑐𝑢𝑡𝑡𝑖𝑛𝑔
𝑇𝐼𝑚𝑒 𝑜𝑓 𝑟𝑒𝑡𝑢𝑟𝑛
=
𝜃
𝛽
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)
𝑟 → 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑟𝑎𝑛𝑘(𝑂𝑃), 𝑙 → 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑙𝑜𝑡𝑡𝑒𝑑 𝑙𝑒𝑣𝑒𝑟(𝐴𝑅), 𝑐 → 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑓𝑖𝑥𝑒𝑑 𝑐𝑒𝑛𝑡𝑟𝑒𝑠(𝐴𝑂)
𝐷𝑢𝑟𝑖𝑛𝑔 𝑡ℎ𝑒 𝑐𝑢𝑡𝑡𝑖𝑛𝑔 𝑠𝑡𝑟𝑜𝑘𝑒, 𝑣𝑠 𝑚𝑎𝑥 = 𝜔 × 𝑂𝑃′
×
𝐴𝑅
𝐴𝑄
= 𝜔𝑟 ×
𝑙
𝑐 + 𝑟
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,
𝑣𝑠 𝑚𝑎𝑥 = 𝜔 × 𝑂𝑃′′
×
𝐴𝑅
𝐴𝑄′′
= 𝜔𝑟 ×
𝑙
𝑐 − 𝑟
𝑣𝑠 𝑚𝑎𝑥 (𝑐𝑢𝑡𝑡𝑖𝑛𝑔)
𝑣𝑠 𝑚𝑎𝑥 (𝑟𝑒𝑡𝑢𝑟𝑛)
=
𝜔𝑟 ×
𝑙
𝑐 + 𝑟
𝜔𝑟 ×
𝑙
𝑐 − 𝑟
=
𝑐 − 𝑟
𝑐 + 𝑟

21. 20
Velocity analysis (Instantaneous center method approach)
𝜔𝐴𝐵 =
𝑣𝐴
𝐴𝐼
=
𝑣𝐵
𝐵𝐼
=
𝑣𝐶
𝐶𝐼
=
𝑣𝐷
𝐷𝐼
=
𝑣𝐸
𝐸𝐼
=
𝑣𝐹
𝐹𝐼
= ⋯
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.
𝑁𝑜. 𝑜𝑓 𝐼𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑐𝑒𝑛𝑡𝑒𝑟𝑠 𝑖𝑛 𝑎 𝑚𝑒𝑐ℎ𝑎𝑛𝑖𝑠𝑚 =
𝒍 · (𝒍 − 𝟏)
𝟐
𝐿 ⟶ 𝑛𝑜. 𝑜𝑓 𝑙𝑖𝑛𝑘𝑠
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.

22. 21
Basics of I-center for a mechanism
Turning pair
Rolling pair
Sliding pair
𝐹𝑜𝑟 𝑙 = 6, 𝑁𝑜. 𝑜𝑓 𝐼𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑐𝑒𝑛𝑡𝑒𝑟𝑠 =
𝒍 · (𝒍 − 𝟏)
𝟐
= 15
I12 I13 I14 I15 I16
I23 I24 I25 I26
I34 I35 I36
I45 I46
I56

23. 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.
𝑉𝐼𝑚𝑛
= 𝜔𝑚𝑛 · (𝐼𝑚𝑛𝐼1𝑚
̅̅̅̅̅̅̅̅̅) = 𝜔𝑛 · (𝐼𝑚𝑛𝐼1𝑛
̅̅̅̅̅̅̅̅)
This theorem is applied at Imn.
Total I.C in use
𝐼𝑚𝑛
𝐼1𝑚
𝐼1𝑛
}
𝑙𝑖𝑛𝑘 1
𝑙𝑖𝑛𝑘 𝑚
𝑙𝑖𝑛𝑘 𝑛
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.
𝑉2 = 𝑉𝐼23
= 𝜔3 ∗ 𝐼23 ∗ 𝐼13
̅̅̅̅̅̅̅̅̅̅
𝜔3 =
𝑉2
𝐼23 ∗ 𝐼13
̅̅̅̅̅̅̅̅̅̅
=
𝑉2
𝐿2 𝑠𝑖𝑛 𝜃
𝑉4 = 𝑉𝐼34
= 𝜔3 ∗ 𝐼34 ∗ 𝐼13
̅̅̅̅̅̅̅̅̅̅
𝑉4 =
𝑉2
𝐿2 𝑠𝑖𝑛 𝜃
∗ 𝐿2 𝑐𝑜𝑠 𝜃
𝑉4 =
𝑉2
𝑡𝑎𝑛 𝜃
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 ∞.

24. 23
ACCELERATION ANALYSIS
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.
𝑇𝑎𝑛𝑔𝑒𝑛𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝐴, 𝑣𝑎 = 𝜔𝑟
𝑶𝑨 𝑖𝑠 𝑟𝑜𝑡𝑎𝑡𝑒𝑑 𝑏𝑦 𝜹𝜽, 𝑡𝑜 𝑶𝑨’, 𝑖𝑛 𝑎 𝑠𝑝𝑎𝑛 𝑜𝑓 𝜹𝒕.
𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑂𝐴′
, 𝜔𝑎
′
= 𝜔 + 𝛼 · 𝛿𝑡
𝑇𝑎𝑛𝑔𝑒𝑛𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝐴′
, 𝑣𝑎
′
= (𝜔 + 𝛼 · 𝛿𝑡) · 𝑟
The tangential velocity of A’ (𝒗𝒂
′
) have two components of velocity, one parallel and other perpendicular to OA.
𝑨𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝐴 ⊥ 𝒕𝒐 𝑂𝐴 =
𝑣𝑎
′
· 𝑐𝑜𝑠 𝛿𝜃 − 𝑣𝑎
𝛿𝑡
=
((𝜔 + 𝛼 · 𝛿𝑡) · 𝑟 · 𝑐𝑜𝑠 𝛿𝜃) − (𝜔𝑟)
𝛿𝑡
𝐴𝑠 𝛿𝑡 → 0, 𝑐𝑜𝑠 𝛿𝜃 → 1 ⟹ 𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑨 ⊥ 𝑡𝑜 𝑶𝑨 =
((𝜔 + 𝛼 · 𝛿𝑡) · 𝑟 · 1) − (𝜔𝑟)
𝛿𝑡
𝑻𝒂𝒏𝒈𝒆𝒏𝒕𝒊𝒂𝒍 𝒂𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏 → 𝒇𝒂𝒐
𝒕
= 𝜶 · 𝒓 = (
𝑑𝜔
𝑑𝑡
) · 𝑟 =
𝑑𝑣
𝑑𝑡
𝑨𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝐴 ∥ 𝒕𝒐 𝑂𝐴 =
𝑣𝑎
′
· 𝑠𝑖𝑛 𝛿𝜃 − 0
𝛿𝑡
=
((𝜔 + 𝛼 · 𝛿𝑡) · 𝑟 · 𝑠𝑖𝑛 𝛿𝜃)
𝛿𝑡
𝐴𝑠 𝛿𝑡 → 0, 𝑠𝑖𝑛 𝛿𝜃 → 𝛿𝜃 ⟹ 𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑨 ⊥ 𝑡𝑜 𝑶𝑨 =
((𝜔 + 𝛼 · 𝛿𝑡) · 𝑟 · 𝛿𝜃)
𝛿𝑡
= 𝜔 · 𝑟 ·
𝛿𝜃
𝛿𝑡
= 𝜔𝑟 · 𝜔
𝑹𝒂𝒅𝒊𝒂𝒍 𝒐𝒓 𝑪𝒆𝒏𝒕𝒓𝒊𝒑𝒆𝒕𝒂𝒍 𝒂𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏 → 𝒇𝒂𝒐
𝒄
= 𝜔2
· 𝑟 =
𝑣2
𝑟
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.

25. 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 𝑓𝑏𝑎
𝑐
𝑜𝑟 𝒂𝟏𝒃𝒂
(𝒂𝒃)2
𝐴𝐵
∥ AB ⟶ A
2 𝑓𝑏𝑎
𝑡
𝑜𝑟 𝒃𝒂𝒃𝟏 α × AB ⊥ AB or a1ba or ∥ ab ⟶ b
3
𝑓𝑐𝑏
𝑐
𝑜𝑟 𝒃𝟏𝒄𝒃 (𝒃𝒄)2
𝐵𝐶
∥ AB ⟶ B
4 𝑓𝑐𝑏
𝑡
𝑜𝑟 𝒄𝒃𝒄𝟏 - ⊥ BC or b1cb -
5
𝑓𝑐𝑑
𝑐
𝑜𝑟 𝒅𝟏𝒄𝒅 (𝒃𝒄)2
𝐷𝐶
∥ DC ⟶ D
6 𝑓𝑐𝑑
𝑡
𝑜𝑟 𝒄𝒅𝒄𝟏 - ⊥ 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,
𝐵𝐸
𝐵𝐶
=
𝑏1𝑒1
𝑏1𝑐1
a1e1 gives the total acceleration of the point E.

26. 25
Slider crank Mechanism
Acceleration of B relative to O = Acceleration of B relative to A + Acceleration of A relative to O.
𝑓𝑏𝑜 = 𝑓𝑏𝑎 + 𝑓𝑎𝑜
𝑓𝑏𝑔 = 𝑓𝑎𝑜 + 𝑓𝑏𝑎 = 𝑓𝑎𝑜 + (𝑓𝑏𝑎
𝑐
+ 𝑓𝑏𝑎
𝑡
)
𝒈𝟏𝒃𝟏 = 𝒐𝟏𝒂𝟏 + 𝒂𝟏𝒃𝒂 + 𝒃𝒂𝒃𝟏
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 𝒇𝒂𝒐𝒐𝒓 𝒐𝟏𝒂𝟏
(𝒐𝒂)2
𝑂𝐴
∥ AB ⟶ O
2 𝒇𝒃𝒂
𝒄
𝒐𝒓 𝒂𝟏𝒃𝒂
(𝒂𝒃)2
𝐴𝐵
⊥ AB or a1ba or ∥ ab ⟶ A
3 𝒇𝒃𝒂
𝒕
𝒐𝒓 𝒃𝒂 𝒃𝟏 ─ ∥ AB ─
4 𝒇𝒃𝒈 𝒐𝒓 𝒈𝟏 𝒃𝟏 ─ ⊥ 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.

27. 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
𝜔′
= 𝜔 + 𝛼 · 𝛿𝑡𝑣′
= 𝑣 + 𝑓 · 𝛿𝑡𝑟′
= 𝑟 + 𝛿𝑟
Acceleration of P parallel to AR
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑷 𝑎𝑙𝑜𝑛𝑔 𝑨𝑹 = 𝒗 = 𝒗𝒑𝒒
𝐹𝑖𝑛𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑷 𝑎𝑙𝑜𝑛𝑔 𝑨𝑹 = (𝑣′
· 𝑐𝑜𝑠 𝛿𝜃 − 𝜔′
· 𝑟′
· 𝑠𝑖𝑛 𝛿𝜃)
𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑷 𝑎𝑙𝑜𝑛𝑔 𝑨𝑹 =
(𝑣′
· 𝑐𝑜𝑠 𝛿𝜃 − 𝜔′
· 𝑟′
· 𝑠𝑖𝑛 𝛿𝜃) − 𝑣
𝛿𝑡
=
((𝑣 + 𝑓 · 𝛿𝑡) · 𝑐𝑜𝑠 𝛿𝜃 − (𝜔 + 𝛼 · 𝛿𝑡) · (𝑟 + 𝛿𝑟) · 𝑠𝑖𝑛 𝛿𝜃) − 𝑣
𝛿𝑡
In the limit, as δt ⟶ 0, cos δθ ⟶ 1, sin δθ ⟶ 0.𝛿𝜃
𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑷 𝑎𝑙𝑜𝑛𝑔 𝑨𝑹 = 𝑓 − 𝜔𝑟
𝑑𝜃
𝑑𝑡
= 𝑓 − 𝜔𝑟𝜔 = 𝒇 − 𝝎𝟐
𝒓
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,
𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑃 ⊥ 𝐴𝑅 =
(𝑣′
· 𝑠𝑖𝑛 𝛿𝜃 + 𝜔′
𝑟′
· 𝑐𝑜𝑠 𝛿𝜃) − 𝑣
𝛿𝑡
=
((𝑣 + 𝑓 · 𝛿𝑡) · 𝑐𝑜𝑠 𝛿𝜃 − (𝜔 + 𝛼 · 𝛿𝑡) · (𝑟 + 𝛿𝑟) · 𝑐𝑜𝑠 𝛿𝜃) − 𝜔𝑟
𝛿𝑡
In the limit, as δt ⟶ 0, cos δθ ⟶ 1, sin δθ ⟶ 0.
𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑷 ⊥ 𝑨𝑹 = 𝑣
𝑑𝜃
𝑑𝑡
+ 𝜔
𝑑𝑟
𝑑𝑡
+ 𝛼 · 𝑟 = 𝑣 · 𝜔 + 𝜔 · 𝑣 + 𝛼 · 𝑟 = 𝟐𝝎𝒗 + 𝜶𝒓
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.

28. 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
𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑷 = 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑷 ∥ 𝑡𝑜 𝑨𝑹 + 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑷 ⊥ 𝑡𝑜 𝑨𝑹
𝑓𝑝𝑎 = (𝑓 − 𝜔2
𝑟) + (2𝜔𝑣 + 𝛼𝑟) = 𝑓 + (𝛼𝑟 − 𝜔2
𝑟) + 2𝜔𝑣
𝑓𝑝𝑎 = 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑷 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑡𝑜 𝑸 + 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑸 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑡𝑜 𝑨 + 𝑪𝒐𝒓𝒊𝒐𝒍𝒊𝒔 𝒂𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕
𝒇𝒑𝒂 = 𝑓𝑝𝑞
′
+ 𝑓𝑞𝑎 + 𝒇𝒄𝒓
𝑓𝑝𝑞
′
→ 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑡𝑖𝑜𝑛 𝑤ℎ𝑖𝑐ℎ 𝑎𝑛 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑟 𝑠𝑡𝑎𝑡𝑖𝑜𝑛𝑒𝑑 𝑜𝑛 𝑙𝑖𝑛𝑘 𝑨𝑹 𝑤𝑜𝑢𝑙𝑑 𝑜𝑏𝑠𝑒𝑟𝑣𝑒 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑠𝑙𝑖𝑑𝑒𝑟.
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.
𝒇𝒑𝒂 = 𝑓𝑝𝑞
′
+ 𝑓𝑞𝑎 + 𝒇𝒄𝒓
= (𝑓𝑝𝑞
′
+ 𝒇𝒄𝒓
) + 𝑓𝑞𝑎 = 𝒇𝒑𝒒 + 𝒇𝒒𝒂
Crank and Slotted-Lever Mechanism
Crank OP rotates at uniform angular velocity of ω rad/s clockwise.
𝑓𝑝𝑎 = 𝑓𝑝𝑞 + 𝑓𝑞𝑎 (𝑜𝑟) 𝑓
𝑝𝑜 = 𝑓𝑞𝑎 + 𝑓𝑝𝑞
𝒐𝟏𝒑𝟏 = 𝒂𝟏𝒒𝒂 + 𝒒𝒂𝒒𝟏 + 𝒒𝟏𝒑𝒒 + 𝒑𝒒𝒑𝟏

29. 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 𝒇𝒒𝒂
𝒕
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 𝒇𝒑𝒒
𝒄𝒓
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
𝑎1𝑟1
𝑎1𝑞1
=
𝐴𝑅
𝐴𝑄
S. no Vector Magnitude Direction Sense
1 𝒇𝒑𝒐𝒐𝒓 𝒐𝟏𝒑𝟏 ω × OP ∥ OP ⟶ O
2 𝒇𝒒𝒂
𝒄
𝒐𝒓 𝒂𝟏𝒒𝒂
(𝒂𝒒)2
𝐴𝑄
∥ AQ ⟶ A
3 𝒇𝒒𝒂
𝒕
𝒐𝒓 𝒒𝒂 𝒒𝟏 ─ ⊥ AQ or a1qa ─
4 𝒇𝒑𝒒
𝒔
𝒐𝒓 𝒒𝟏𝒑𝒒 ─ ∥ AR ─
5 𝒇𝒑𝒒
𝒄𝒓
𝒐𝒓 𝒑𝒒𝒑𝟏 Coriolis component* ⊥ AR Refer*
𝑓𝑝𝑞
𝑐𝑟
= 2𝜔1𝑣𝑝𝑞 (𝜔1 = 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑖𝑐𝑡𝑦 𝑜𝑓 𝐴𝑅) = 2 (
𝒂𝒒
𝐴𝑄
) 𝑞𝑝

30. 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. (∑ 𝐹
⃗ = 0)
• The vector sum of all moments about the arbitrary point is zero. (∑ 𝑀
⃗⃗⃗ = 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.

31. 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,
𝑊 = 𝑇𝛿𝜃 + 𝐹𝛿𝑥 = 0
An virtual displacement must take place during the same interval δt,
𝑇
𝑑𝜃
𝑑𝑡
+ 𝐹
𝑑𝑥
𝑑𝑡
= 0
𝑇𝜔 + 𝐹𝑣 = 0
𝑇 = −
𝐹
𝜔
𝑣
(ω→angular velocity, v→linear velocity)

32. 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.
𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝑓𝑜𝑟𝑐𝑒𝐹𝑖 = −𝑚𝑓𝑔
(𝑓𝑔 → 𝑎𝑎𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑐𝑒𝑛𝑡𝑟𝑒 𝑜𝑓 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑏𝑜𝑑𝑦)
Inertia force acts in opposite direction to that of acceleration.
𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝑐𝑜𝑢𝑝𝑙𝑒 𝐶𝑖 = −𝐼𝑔. 𝛼
(𝐼𝑔 → 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝑎𝑏𝑜𝑢𝑡 𝑎𝑛 𝑎𝑥𝑖𝑠 𝑝𝑎𝑠𝑠𝑖𝑛𝑔 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑐𝑒𝑛𝑡𝑟𝑒 𝑜𝑓 𝑚𝑎𝑠𝑠 𝐺)
According to D’Alembert’s principle
∑ 𝐹 + 𝐹𝑖 = 0
∑ 𝑇 + 𝐶𝑖 = 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 =
𝐶𝑖
𝐹𝑖
=
𝑘2
𝛼
𝑓𝑔
(𝐶𝑖 = 𝐹𝑖. ℎ)
(𝐶𝑖 = −𝐼𝑔. 𝛼)(𝐹𝑖 = −𝑚𝑓𝑔)
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 𝐹𝑖 = −𝑚𝑓𝑔 and 𝐶𝑖 = −𝐼𝑔. 𝛼 .
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.

33. 32
STATIC ANALYSIS OF SLIDER CRANK MECHANISMS
Velocity and acceleration of a piston
𝑥 = 𝐵1𝐵 = 𝐵𝑂 − 𝐵1𝑂 = 𝐵𝑂 − (𝐵1𝐴1 + 𝐴1𝑂)
𝑥 = (𝑟 + 𝑙) − (𝑟 𝑐𝑜𝑠 𝜃 + 𝑙 𝑐𝑜𝑠 𝛽)
(𝒍 = 𝒓𝒏)
𝑥 = 𝑟[(𝑛 + 1) − (𝑛 𝑐𝑜𝑠 𝛽 + 𝑟 𝑐𝑜𝑠 𝜃)]
𝑐𝑜𝑠 𝛽 =
1
𝑛
√𝑛2 − 𝑠𝑖𝑛2 𝜃
𝑥 = 𝑟[(𝑛 + 1) − (√𝑛2 − 𝑠𝑖𝑛2 𝜃 + 𝑟 𝑐𝑜𝑠 𝜃)]
𝑥 = 𝑟[(1 − 𝑟 𝑐𝑜𝑠 𝜃) + (𝑛 − √𝑛2 − 𝑠𝑖𝑛2 𝜃)]
If (l>>>>n) n2
is very large and (n2
-sin2
  n2
)
𝑥 = 𝑟(1 − 𝑟 𝑐𝑜𝑠 𝜃)
Velocity of the piston
𝑣 =
𝑑𝑥
𝑑𝜃
×
𝑑𝜃
𝑑𝑡
𝑣 = 𝑟𝜔 [𝑠𝑖𝑛 𝜃 +
𝑠𝑖𝑛 2𝜃
2√𝑛2 − 𝑠𝑖𝑛2 𝜃
]
𝑣 = 𝑟𝜔 [𝑠𝑖𝑛 𝜃 +
𝑠𝑖𝑛 2𝜃
2𝑛
]
Acceleration of the piston
𝑓 =
𝑑𝑣
𝑑𝜃
×
𝑑𝜃
𝑑𝑡
𝑓 = 𝑟𝜔2
[𝑐𝑜𝑠 𝜃 +
𝑐𝑜𝑠 2𝜃
𝑛
]
𝑓 = 𝑟𝜔2
𝑐𝑜𝑠 𝜃 (if n>>>cos 2)
Velocity of the piston will be maximum at crank angle , to find 
𝑑𝑣
𝑑𝜃
= 0
(n2
>> sin2
)

34. 33
𝑑
𝑑𝜃
[𝑠𝑖𝑛𝜃 +
𝑠𝑖𝑛2𝜃
2𝑛
] = 0 (𝑛 = 3.5)
𝜃 = 75.5°, 284.5°
𝐴𝑡  = 75.5°, 𝑛 = 3.5
𝑣𝑠 = 1.037𝑟𝜔.
𝐴𝑡  = 284.5°, 𝑛 = 3.5
𝑣𝑠 = −1.037𝑟𝜔.
Acceleration of the piston will be maximum at crank angle , to find 
𝑑𝑎𝑠
𝑑𝜃
= 0
𝑑
𝑑𝜃
[𝑐𝑜𝑠 𝜃 +
𝑐𝑜𝑠 2𝜃
𝑛
] = 0 (n=3.5)
=151°, 209°
𝐴𝑡  = 151°,  = 209°,
𝑎𝑠 = −0.723 · 𝑟 · 𝜔2
.
𝐴𝑡  = 0°,
𝑎𝑠 = −1.3 · 𝑟 · 𝜔.
Angular velocity and angular acceleration of connecting rod
𝑠𝑖𝑛 𝛽 =
𝑠𝑖𝑛 𝜃
𝑛
Differentiating with respecting to time, we get
𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑖𝑛𝑔 𝑟𝑜𝑑 = 𝜔𝑐 =
𝑑𝛽
𝑑𝑡
𝜔𝑐 = 𝜔
𝑐𝑜𝑠 𝜃
√𝑛2 − 𝑠𝑖𝑛2 𝜃
(ω-angular velocity of rod)
𝛼𝑐 = 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑖𝑛𝑔 𝑟𝑜𝑑 =
𝑑𝜔
𝑑𝑡
𝛼𝑐 = −𝜔2
𝑠𝑖𝑛 𝜃 [
𝑛2
− 1
(𝑛2 − 𝑠𝑖𝑛2 𝜃)3 2
⁄
]

35. 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
𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝑓𝑜𝑟𝑐𝑒 = 𝑚𝑓 = 𝑚𝑟𝜔2
(𝑐𝑜𝑠 𝜃 +
𝑐𝑜𝑠 2𝜃
𝑛
)
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
𝐹𝑐 × 𝑐𝑜𝑠 𝛽 = 𝐹
𝐹𝑐 =
𝐹
𝑐𝑜𝑠 𝛽
Thrust on the sides of cylinder
It is the normal reaction on the cylinder walls
𝐹𝑛 = 𝐹𝑐 𝑠𝑖𝑛 𝛽 = 𝑓 𝑡𝑎𝑛 𝛽
Crank Effort
Force is exerted on the crankpin because of the force on the piston.

36. 35
Ft = crank effort
𝐹𝑡 × 𝑟 = 𝐹𝑐𝑟 𝑠𝑖𝑛 𝛽
𝐹𝑡 = 𝐹𝑐 𝑠𝑖𝑛 𝛽 =
𝐹
𝑐𝑜𝑠 𝛽
𝑠𝑖𝑛(𝜃 + 𝛽)
Thrust on the Bearings
𝐹𝑟 = 𝐹𝑐 𝑐𝑜𝑠 𝛽 =
𝐹
𝑐𝑜𝑠 𝛽
𝑐𝑜𝑠(𝜃 + 𝛽)
Turning Moment on Crankshaft
𝑇 = 𝐹𝑡 × 𝑟 =
𝐹
𝑐𝑜𝑠 𝛽
𝑠𝑖𝑛(𝜃 + 𝛽) × 𝑟
𝑇 = 𝐹𝑟 (𝑠𝑖𝑛 𝜃 +
𝑠𝑖𝑛 2𝜃
2√𝑛2 − 𝑠𝑖𝑛2 𝜃
)
𝑇 = 𝐹𝑡 × 𝑟 = 𝐹 × 𝑂𝐷
Dynamically equivalent 2-point mass system (connecting rod)
𝐼𝑐𝑟
𝐺
= 𝑚𝑐𝑟𝐾𝐺
𝑐𝑟
𝐾𝐺 = 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑔𝑦𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑏𝑜𝑢𝑡 𝐺.
𝑏 + 𝑑 = 𝐿
For complete dynamic equivalence b/w actual Connecting Rod & 2-point mass system
1. mb + ms = mcr
2. mb. b = ms. d
3. 𝑚𝑏𝑏2
+ 𝑚𝑠𝑑2
= 𝑚𝑐𝑟𝐾𝑔
2
Solving for 1 & 2 & 3
𝑚𝑏 =
𝑑
𝐿
𝑚𝑐𝑟 𝑚𝑠 =
𝑏
𝐿
𝑚𝑐𝑟
𝐾𝐺
2
= 𝑏. 𝑑
It is the condition on b & d so that 2-point mass system is completely dynamically equivalent.
𝐼 = 𝑚𝑏𝑏2
+ 𝑚𝑠𝑑2
= 𝑚𝑐𝑟. 𝑏. 𝑑
Result can be compared with equivalent length of simple pendulum

37. 36
The equivalent length of simple pendulum is
Generally, 𝑲𝑮
𝟐
≤ 𝒃. 𝒅
(Inertia couple) actual connecting rod < inertia couple of equivalent system (taken)
On the 2-point mass system, a correction couple is applied.
𝑇𝑐 = 𝑚𝑐𝑟𝐾𝐺
2
𝛼 − 𝑚𝑐𝑟𝑏𝑑𝛼
𝑇𝑐 = −𝑚𝑐𝑟𝛼(𝑏𝑑−𝐾𝐺
2)
Correction couple must be applied in the direction of angular acceleration
𝐹
𝑐 =
𝑇𝑐
𝐿

38. 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.
𝐹 = 𝑚1𝒓𝟏𝜔2
+ 𝑚2𝒓𝟐𝜔2
+ 𝑚3𝒓𝟑𝜔2
𝑚1𝒓𝟏𝜔2
+ 𝑚2𝒓𝟐𝜔2
+ 𝑚3𝒓𝟑𝜔2
+ 𝑚𝑐𝒓𝒄𝜔2
= 0
𝑚1𝒓𝟏 + 𝑚2𝒓𝟐 + 𝑚3𝒓𝟑 + 𝑚𝑐𝒓𝒄 = 0
∑ 𝑚𝑟 𝑐𝑜𝑠 𝜃 + 𝑚𝑐𝑟𝑐 𝑐𝑜𝑠 𝜃𝑐 = 0
∑ 𝑚𝑟 𝑠𝑖𝑛 𝜃 + 𝑚𝑐𝑟𝑐 𝑠𝑖𝑛 𝜃𝑐 = 0
𝑚𝑐𝑟𝑐 = √(𝛴𝑚𝑟 𝑐𝑜𝑠 𝜃)2 + (𝛴𝑚𝑟 𝑠𝑖𝑛 𝜃)2
𝑡𝑎𝑛 𝜃𝑐 =
𝛴𝑚𝑟 𝑐𝑜𝑠 𝜃
𝛴𝑚𝑟 𝑠𝑖𝑛 𝜃

39. 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.
𝑚1𝒓1𝜔2
+ 𝑚2𝒓2𝜔2
+ 𝑚3𝒓3𝜔2
+ 𝑚𝑐1𝒓𝑐1𝜔2
+ 𝑚𝑐2𝒓𝑐2𝜔2
= 0
𝑚1𝒓1 + 𝑚2𝒓2 + 𝑚3𝒓3 + 𝑚𝑐1𝒓𝑐1 + 𝑚𝑐2𝒓𝑐2 = 0
𝛴𝑚𝒓 + 𝑚𝑐1𝒓𝑐1 + 𝑚𝑐2𝒓𝑐2 = 0
Let the counter masses be placed in transverse planes at axial locations at O & Q.
Taking moments about O,
𝑚1𝒓1𝑙1𝜔2
+ 𝑚2𝒓2𝑙2𝜔2
+ 𝑚3𝒓3𝑙3𝜔2
+ 𝑚𝑐2𝒓𝑐2𝑙𝑐2𝜔2
= 0
𝑚1𝒓1𝑙1 + 𝑚2𝒓2𝑙2 + 𝑚3𝒓3𝑙3 + 𝑚𝑐2𝒓𝑐2𝑙𝑐2 = 0
𝛴𝑚𝒓𝑙1 + 𝑚𝑐2𝒓𝑐2𝑙𝑐2 = 0

40. 39
This can be also solved analytically,
𝛴𝑚𝒓𝑙1 𝑐𝑜𝑠 𝜃 + 𝑚𝑐2𝒓𝑐2𝑙𝑐2 𝑐𝑜𝑠 𝜃𝐶2 = 0
𝛴𝑚𝒓𝑙1 𝑠𝑖𝑛 𝜃 + 𝑚𝑐2𝒓𝑐2𝑙𝑐2 𝑠𝑖𝑛 𝜃𝐶2 = 0
𝛴𝑚𝒓𝑙 𝑐𝑜𝑠 𝜃 = −𝑚𝑐2𝒓𝑐2𝑙𝑐2 𝑐𝑜𝑠 𝜃𝐶2
𝛴𝑚𝒓𝑙 𝑠𝑖𝑛 𝜃 = −𝑚𝑐2𝒓𝑐2𝑙𝑐2 𝑠𝑖𝑛 𝜃𝐶2
𝑚𝑐2𝒓𝑐2𝑙𝑐2 = √(𝛴𝑚𝒓𝑙 𝑐𝑜𝑠 𝜃)2 + (𝛴𝑚𝒓𝑙 𝑠𝑖𝑛 𝜃)2
𝑡𝑎𝑛 𝜃𝐶2 =
−𝛴𝑚𝒓𝑙1 𝑠𝑖𝑛 𝜃
−𝛴𝑚𝒓𝑙1 𝑐𝑜𝑠 𝜃
Substituting the value of m2 & C2 in above equations, we get
𝑚𝑐1𝒓𝑐1 = √(𝛴𝑚𝒓 𝑐𝑜𝑠 𝜃 + 𝑚𝑐2𝒓𝑐2 𝑐𝑜𝑠 𝜃𝐶2)2 + (𝛴𝑚𝒓 𝑠𝑖𝑛 𝜃 + 𝑚𝑐2𝒓𝑐2 𝑠𝑖𝑛 𝜃𝐶2)2
𝑡𝑎𝑛 𝜃𝐶1 =
−(𝛴𝑚𝒓 𝑠𝑖𝑛 𝜃 + 𝑚𝑐2𝒓𝑐2 𝑠𝑖𝑛 𝜃𝐶2)
−(𝛴𝑚𝒓 𝑐𝑜𝑠 𝜃 + 𝑚𝑐2𝒓𝑐2 𝑐𝑜𝑠 𝜃𝐶2)

41. 40
BALANCING OF RECIPROCATING MASS
𝑓 = 𝑟𝜔2
(𝑐𝑜𝑠 𝜃 +
𝑐𝑜𝑠 2𝜃
𝑛
)
𝐹𝑅 = 𝑚𝑓 = 𝑚𝑟𝜔2
(𝑐𝑜𝑠 𝜃 +
𝑐𝑜𝑠 2𝜃
𝑛
)
𝐹𝑅 = 𝑚𝑟𝜔2
𝑐𝑜𝑠 𝜃 + 𝑚𝑟𝜔2
𝑐𝑜𝑠 2𝜃
𝑛
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 = 𝐹𝑁 × 𝑥 =
𝐹𝐵𝑉 × 𝑥)
Both FU and unbalanced couple vary in magnitude while rotating and causes serious vibration.
→ 𝒎𝒓𝝎𝟐
𝒄𝒐𝒔 𝜽 is called primary unbalancing force and 𝒎𝒓𝝎𝟐 𝒄𝒐𝒔 𝟐𝜽
𝒏
is called secondary unbalancing force.
Partial Balancing of Unbalanced Primary Force in a Reciprocating Engine
The primary unbalanced force (m⋅ω2
⋅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.
𝑚𝑟𝜔2
𝑐𝑜𝑠 𝜃 = 𝐵𝜔2
𝑏 𝑐𝑜𝑠 𝜃
𝒎𝒓 = 𝑩𝒃
But still vertical force of mass B is not balanced (Bω2
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)

42. 41
Unbalanced force along the line of stroke = 𝑚𝑟𝜔2
𝑐𝑜𝑠 𝜃 − 𝒄. 𝑚𝑟𝜔2
𝑐𝑜𝑠 𝜃
= (1 − 𝒄)𝑚𝑟𝜔2
𝑐𝑜𝑠 𝜃
Unbalanced force along the perpendicular to the line of stroke=c.mrω2
sin.
Resultant unbalanced force at any instant= √((1 − 𝒄)𝑚𝑟𝜔2 𝑐𝑜𝑠 𝜃)2 + (𝒄. 𝑚𝑟𝜔2 𝑠𝑖𝑛 𝜃 . )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 mrω2
.
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 − 𝑐)𝑚𝑟𝜔2
𝑐𝑜𝑠 𝜃
Unbalanced force along cylinder 2= (1 − 𝑐)𝑚𝑟𝜔2
𝑐𝑜𝑠(90 + 𝜃) = −(1 − 𝑐)𝑚𝑟𝜔2
𝑠𝑖𝑛 𝜃
𝐹𝑇 = 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑓𝑜𝑟𝑐𝑒 𝑎𝑙𝑜𝑛𝑔 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 = (1 − 𝑐)𝑚𝑟𝜔2
𝑐𝑜𝑠 𝜃 − (1 − 𝑐)𝑚𝑟𝜔2
𝑠𝑖𝑛 𝜃 = (𝟏 − 𝒄)𝒎𝒓𝝎𝟐
(𝒄𝒐𝒔  − 𝒔𝒊𝒏 𝜽)
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑎𝑛𝑑 𝑀𝑖𝑛𝑖𝑚𝑢𝑚 𝑡𝑟𝑎𝑐𝑡𝑖𝑣𝑒 𝑓𝑜𝑟𝑐𝑒 = ±√𝟐(𝟏 − 𝒄)𝒎𝒓𝝎𝟐

43. 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.
𝑆𝑤𝑎𝑦𝑖𝑛𝑔 𝐶𝑜𝑢𝑝𝑙𝑒 = (1 − 𝑐)𝑚𝜔2
𝑟 𝑐𝑜𝑠 𝜃 ×
𝑎
2
− (1 − 𝑐)𝑚𝜔2
𝑟 𝑐𝑜𝑠(90 + 𝜃) ×
𝑎
2
𝑆𝑤𝑎𝑦𝑖𝑛𝑔 𝐶𝑜𝑢𝑝𝑙𝑒 = (𝟏 − 𝒄)𝒎𝝎𝟐
𝒓(𝒄𝒐𝒔 𝜽 + 𝒔𝒊𝒏 𝜽) ×
𝒂
𝟐
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑎𝑛𝑑 𝑀𝑖𝑛𝑖𝑚𝑢𝑚 𝑆𝑤𝑎𝑦𝑖𝑛𝑔 𝐶𝑜𝑢𝑝𝑙𝑒 = ±
𝒂
√𝟐
(𝟏 − 𝒄)𝒎𝝎𝟐
Secondary Balancing
𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑓𝑜𝑟𝑐𝑒 = 𝑚𝑟𝜔2
𝑐𝑜𝑠 2𝜃
𝑛
𝐼𝑡 𝑐𝑎𝑛 𝑎𝑙𝑠𝑜 𝑏𝑒 𝑤𝑟𝑖𝑡𝑡𝑒𝑛 𝑎𝑠 = 𝑚 (
𝑟
4𝑛
) (2𝜔)2
𝑐𝑜𝑠 2𝜃
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.

44. 43
𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝑓𝑜𝑟𝑐𝑒 = ∑𝑚𝜔2
𝑟 𝑐𝑜𝑠 𝜃
𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝑐𝑜𝑢𝑝𝑙𝑒 = ∑𝑚𝜔2
𝑟𝑙 𝑐𝑜𝑠 𝜃
𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑓𝑜𝑟𝑐𝑒 = ∑𝑚 (
𝑟
4𝑛
) (2𝜔)2
𝑐𝑜𝑠 2𝜃
𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝐶𝑜𝑢𝑝𝑙𝑒
= ∑𝑚 (
𝑟
4𝑛
) (2𝜔)2
𝑙 𝑐𝑜𝑠 2𝜃
Three-cylinder inline engine with crank offset of 120˚
𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝑓𝑜𝑟𝑐𝑒 = ∑𝑚𝜔2
𝑟 𝑐𝑜𝑠 𝜃 = 𝑚𝜔2
𝑟 𝑐𝑜𝑠 𝜃 + 𝑚𝜔2
𝑟 𝑐𝑜𝑠(120 + 𝜃) + 𝑚𝜔2
𝑟 𝑐𝑜𝑠(240 + 𝜃) = 0
𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑓𝑜𝑟𝑐𝑒 = ∑𝑚 (
𝑟
4𝑛
) (2𝜔)2
𝑐𝑜𝑠 2𝜃 = 𝑚 (
𝑟
4𝑛
) (2𝜔)2
(𝑐𝑜𝑠 2𝜃 + 𝑐𝑜𝑠(240 + 2) + 𝑐𝑜𝑠(480 + 2𝜃)) = 0
𝑀𝑝 = 𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝑐𝑜𝑢𝑝𝑙𝑒 = 𝑚𝑟𝜔2
𝑙 𝑐𝑜𝑠(240 + 𝜃) − 𝑚𝑟𝜔2
𝑙 𝑐𝑜𝑠 𝜃 = 𝑚𝑟𝜔2
𝑙(𝑐𝑜𝑠(240 + 𝜃) − 𝑐𝑜𝑠 𝜃)
𝑀𝑝 𝒎𝒂𝒙 = 𝑚𝑟𝜔2
𝑙(2 𝑐𝑜𝑠 30) = √3𝑚𝑟𝜔2
𝑙
𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝐶𝑜𝑢𝑝𝑙𝑒 = 𝑀𝑠 =
𝑚𝑟𝜔2
𝑙
𝑛
(𝑐𝑜𝑠(480 + 2𝜃) − 𝑐𝑜𝑠 2𝜃)
𝑀𝑠 𝒎𝒂𝒙 =
√3𝑚𝑟𝜔2
𝑙
𝑛
Inline 2-cylinder engine
Cranks are 180⁰ apart and have equal reciprocating masses.
𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝑓𝑜𝑟𝑐𝑒 = 𝑚𝑟𝜔2[𝑐𝑜𝑠 𝜃 + 𝑐𝑜𝑠(180° + 𝜃)] = 0
𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝐶𝑜𝑢𝑝𝑙𝑒 = 𝑀𝑝 = 𝑚𝑟𝜔2
[
𝑙
2
𝑐𝑜𝑠 𝜃 + (−
𝑙
2
) 𝑐𝑜𝑠(1800
+ 𝜃)] = 𝑚𝑟𝜔2
𝑙 𝑐𝑜𝑠 𝜃
𝑀𝑝 𝒎𝒂𝒙 = 𝒎𝒓𝝎𝟐
𝒍 𝑎𝑡 𝜃 = 0° & 180°

45. 44
𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑓𝑜𝑟𝑐𝑒 = 𝐹𝑠 =
𝑚𝑟𝜔2
𝑛
[𝑐𝑜𝑠 2𝜃 + 𝑐𝑜𝑠(360° + 2𝜃)] =
2𝑚𝑟𝜔2
𝑛
𝑐𝑜𝑠 2𝜃
𝐹𝑠 𝒎𝒂𝒙 =
2𝑚𝑟𝜔2
𝑛
𝑤ℎ𝑒𝑛 𝜃 = 0°, 90°, 180°, 270°.
𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝐶𝑜𝑢𝑝𝑙𝑒 = 𝑀𝑠 = 𝑚𝑟𝜔2
[
𝑙
2
𝑐𝑜𝑠 𝜃 + (−
𝑙
2
) 𝑐𝑜𝑠(3600
+ 𝜃)] = 0
Inline four-cylinder Four Stroke Engine
𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝑓𝑜𝑟𝑐𝑒 = 𝑚𝑟𝜔2[𝑐𝑜𝑠 𝜃 + 𝑐𝑜𝑠(180 + 𝜃) + 𝑐𝑜𝑠(180 + 𝜃) + 𝑐𝑜𝑠 𝜃] = 0
𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝐶𝑜𝑢𝑝𝑙𝑒 = 𝑚𝑟𝜔2
[
3𝑙
2
𝑐𝑜𝑠 𝜃 +
𝑙
2
𝑐𝑜𝑠(180 + 𝜃) + (−
3𝑙
2
𝑐𝑜𝑠 𝜃) + (−
𝑙
2
𝑐𝑜𝑠(180 + 𝜃))] = 0
𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑓𝑜𝑟𝑐𝑒 = 𝐹𝑠 =
𝑚𝑟𝜔2
𝑛
[𝑐𝑜𝑠 2𝜃 + 𝑐𝑜𝑠 2(180 + 𝜃) + 𝑐𝑜𝑠 2(180 + 𝜃) + 𝑐𝑜𝑠 2𝜃] =
4𝑚𝑟𝜔2
𝑛
𝑐𝑜𝑠 2𝜃
𝐹𝑠 𝒎𝒂𝒙 =
4𝑚𝑟𝜔2
𝑛
𝑎𝑡  = 0˚, 90˚, 180˚, 270˚
𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝐶𝑜𝑢𝑝𝑙𝑒 = 𝑚𝑟𝜔2
[
3𝑙
2
𝑐𝑜𝑠 𝜃 +
𝑙
2
𝑐𝑜𝑠 2(180 + 𝜃) + (−
3𝑙
2
𝑐𝑜𝑠 2𝜃) + (−
𝑙
2
𝑐𝑜𝑠 2(180 + 𝜃))] = 0

46. 45
Six-cylinder four stroke Engine

47. 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.
𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝑐𝑒𝑛𝑡𝑟𝑖𝑓𝑢𝑔𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 𝑎𝑙𝑜𝑛𝑔 𝑶𝑷 𝑓𝑜𝑟 𝑝𝑟𝑖𝑚𝑎𝑟𝑦 𝑑𝑖𝑟𝑒𝑐𝑡 𝑐𝑟𝑎𝑛𝑘 =
𝑚
2
𝑟𝜔2
𝑐𝑜𝑠 𝜃
𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝑐𝑒𝑛𝑡𝑟𝑖𝑓𝑢𝑔𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 𝑎𝑙𝑜𝑛𝑔 𝑶𝑷 𝑓𝑜𝑟 𝑝𝑟𝑖𝑚𝑎𝑟𝑦 𝑟𝑒𝑣𝑒𝑟𝑠𝑒 𝑐𝑟𝑎𝑛𝑘 =
𝑚
2
𝑟𝜔2
𝑐𝑜𝑠 𝜃
𝑇𝑜𝑡𝑎𝑙 𝑐𝑒𝑛𝑡𝑟𝑖𝑓𝑢𝑔𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 𝑎𝑙𝑜𝑛𝑔 𝑶𝑷 = 𝑚𝑟𝜔2
𝑐𝑜𝑠 𝜃
Component of centrifugal force perpendicular to OP are balanced.
Secondary forces
𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑓𝑜𝑟𝑐𝑒𝑠 = 𝑚𝑟𝜔2 𝑐𝑜𝑠 2𝜃
𝑛
= 𝑚(2𝜔)2 𝑟
4𝑛
𝑐𝑜𝑠 2𝜃
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

48. 47
Balancing of V-type engines
𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝑓𝑜𝑟𝑐𝑒𝑠 𝑎𝑙𝑜𝑛𝑔 𝑙𝑖𝑛𝑒 𝑜𝑓 𝑠𝑡𝑟𝑜𝑘𝑒 𝑓𝑜𝑟 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 1 = 𝐹𝑃1 = 𝑚𝑟𝜔2
𝑐𝑜𝑠(𝛼 − 𝜃)
𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝐹𝑃1 𝑎𝑙𝑜𝑛𝑔 𝑶𝒀 = 𝑚𝑟𝜔2
𝑐𝑜𝑠(𝛼 − 𝜃) 𝑐𝑜𝑠 𝛼
𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝐹𝑃1 𝑎𝑙𝑜𝑛𝑔 𝑶𝑿 = 𝑚𝑟𝜔2
𝑐𝑜𝑠(𝛼 − 𝜃) 𝑠𝑖𝑛 𝛼
𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝑓𝑜𝑟𝑐𝑒𝑠 𝑎𝑙𝑜𝑛𝑔 𝑙𝑖𝑛𝑒 𝑜𝑓 𝑠𝑡𝑟𝑜𝑘𝑒 𝑓𝑜𝑟 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 2 = 𝐹𝑃2 = 𝑚𝑟𝜔2
𝑐𝑜𝑠(𝛼 + 𝜃)
𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝐹𝑃2 𝑎𝑙𝑜𝑛𝑔 𝑶𝒀 = 𝑚𝑟𝜔2
𝑐𝑜𝑠(𝛼 + 𝜃) 𝑐𝑜𝑠 𝛼
𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝐹𝑃2 𝑎𝑙𝑜𝑛𝑔 𝑶𝑿′ = 𝑚𝑟𝜔2
𝑐𝑜𝑠(𝛼 + 𝜃) 𝑠𝑖𝑛 𝛼
𝑻𝒐𝒕𝒂𝒍 𝑷𝒓𝒊𝒎𝒂𝒓𝒚 𝒇𝒐𝒓𝒄𝒆𝒔 𝒂𝒍𝒐𝒏𝒈 𝑶𝒀 = 𝑭𝑷𝑽 = 𝑚𝑟𝜔2
𝑐𝑜𝑠(𝛼 − 𝜃) 𝑐𝑜𝑠 𝛼 + 𝑚𝑟𝜔2
𝑐𝑜𝑠(𝛼 + 𝜃) 𝑐𝑜𝑠 𝛼
= 𝑚𝑟𝜔2
𝑐𝑜𝑠 𝛼 (𝑐𝑜𝑠(𝛼 − 𝜃) + 𝑐𝑜𝑠(𝛼 + 𝜃))
= 𝑚𝑟𝜔2
𝑐𝑜𝑠 𝛼 (2 × 𝑐𝑜𝑠 𝛼 × 𝑐𝑜𝑠 𝜃)
= 2𝑚𝑟𝜔2
. 𝑐𝑜𝑠2
𝛼 . 𝑐𝑜𝑠 𝜃
𝑻𝒐𝒕𝒂𝒍 𝑷𝒓𝒊𝒎𝒂𝒓𝒚 𝒇𝒐𝒓𝒄𝒆𝒔 𝒂𝒍𝒐𝒏𝒈 𝑶𝑿 = 𝑭𝑷𝑯 = 𝑚𝑟𝜔2
𝑐𝑜𝑠(𝛼 − 𝜃) 𝑠𝑖𝑛 𝛼 − 𝑚𝑟𝜔2
𝑐𝑜𝑠(𝛼 + 𝜃) 𝑠𝑖𝑛 𝛼
= 𝑚𝑟𝜔2
𝑠𝑖𝑛 𝛼 (𝑐𝑜𝑠(𝛼 − 𝜃) − 𝑐𝑜𝑠(𝛼 + 𝜃))
= 𝑚𝑟𝜔2
𝑠𝑖𝑛 𝛼 (2 × 𝑠𝑖𝑛 𝛼 𝑠𝑖𝑛  )
= 2. 𝑚𝑟𝜔2
. 𝑠𝑖𝑛2
𝛼 . 𝑠𝑖𝑛 𝜃
𝑹𝒆𝒔𝒖𝒍𝒕𝒂𝒏𝒕 𝑷𝒓𝒊𝒎𝒂𝒓𝒚 𝒇𝒐𝒓𝒄𝒆 = √(𝐹𝑃𝑉)2 + (𝐹𝑃𝐻)2 = 2. 𝑚𝑟𝜔2
√(𝑐𝑜𝑠2 𝛼 . 𝑐𝑜𝑠 𝜃)2 + (𝑠𝑖𝑛2 𝛼 . 𝑠𝑖𝑛 𝜃)2
𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑓𝑜𝑟𝑐𝑒𝑠 𝑎𝑙𝑜𝑛𝑔 𝑙𝑖𝑛𝑒 𝑜𝑓 𝑠𝑡𝑟𝑜𝑘𝑒 𝑓𝑜𝑟 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 1 = 𝐹𝑆1 = 𝑚𝑟𝜔2
𝑐𝑜𝑠 2(𝛼 − 𝜃)
𝑛
𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝐹𝑆1 𝑎𝑙𝑜𝑛𝑔 𝑶𝒀 = 𝑚𝑟𝜔2
𝑐𝑜𝑠 2(𝛼 − 𝜃)
𝑛
𝑐𝑜𝑠 𝛼
𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝐹𝑃1 𝑎𝑙𝑜𝑛𝑔 𝑶𝑿 = 𝑚𝑟𝜔2
𝑐𝑜𝑠 2(𝛼 − 𝜃)
𝑛
𝑠𝑖𝑛 𝛼
𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑓𝑜𝑟𝑐𝑒𝑠 𝑎𝑙𝑜𝑛𝑔 𝑙𝑖𝑛𝑒 𝑜𝑓 𝑠𝑡𝑟𝑜𝑘𝑒 𝑓𝑜𝑟 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 2 = 𝐹𝑆2 = 𝑚𝑟𝜔2
𝑐𝑜𝑠 2(𝛼 + 𝜃)
𝑛
𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝐹𝑆2 𝑎𝑙𝑜𝑛𝑔 𝑶𝒀 = 𝑚𝑟𝜔2
𝑐𝑜𝑠 2(𝛼 + 𝜃)
𝑛
𝑐𝑜𝑠 𝛼
𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝐹𝑆2 𝑎𝑙𝑜𝑛𝑔 𝑶𝑿′ = 𝑚𝑟𝜔2
𝑐𝑜𝑠 2(𝛼 + 𝜃)
𝑛
𝑠𝑖𝑛 𝛼
𝑻𝒐𝒕𝒂𝒍 𝑺𝒆𝒄𝒐𝒏𝒅𝒂𝒓𝒚 𝒇𝒐𝒓𝒄𝒆𝒔 𝒂𝒍𝒐𝒏𝒈 𝑶𝒀 = 𝑭𝑺𝑽 = 𝑚𝑟𝜔2
𝑐𝑜𝑠 𝛼 (
𝑐𝑜𝑠 2(𝛼 − 𝜃)
𝑛
+
𝑐𝑜𝑠 2(𝛼 + 𝜃)
𝑛
) =
2𝑚
𝑛
𝑟𝜔2
𝑐𝑜𝑠 𝛼 𝑐𝑜𝑠 2𝛼 𝑐𝑜𝑠 2𝜃
𝑻𝒐𝒕𝒂𝒍 𝑺𝒆𝒄𝒐𝒏𝒅𝒂𝒓𝒚 𝒇𝒐𝒓𝒄𝒆𝒔 𝒂𝒍𝒐𝒏𝒈 𝑶𝑿 = 𝑭𝑺𝑯 = 𝑚𝑟𝜔2
𝑠𝑖𝑛 𝛼 (
𝑐𝑜𝑠 2(𝛼 − 𝜃)
𝑛
−
𝑐𝑜𝑠 2(𝛼 + 𝜃)
𝑛
) =
2𝑚
𝑛
𝑟𝜔2
𝑠𝑖𝑛 𝛼 𝑠𝑖𝑛 2𝛼 𝑠𝑖𝑛 2𝜃
𝑹𝒆𝒔𝒖𝒍𝒕𝒂𝒏𝒕 𝑺𝒆𝒄𝒐𝒏𝒅𝒂𝒓𝒚 𝒇𝒐𝒓𝒄𝒆 = √(𝐹𝑆𝑉)2 + (𝐹𝑆𝐻)2 =
2𝑚
𝑛
𝑟𝜔2
√(𝑐𝑜𝑠 𝛼 𝑐𝑜𝑠 2𝛼 𝑐𝑜𝑠 2𝜃)2 + (𝑠𝑖𝑛 𝛼 𝑠𝑖𝑛 2𝛼 𝑠𝑖𝑛 2𝜃)2

49. 48
TURNING MOMENT DIAGRAMS
During 1 revolution of crank shaft, 𝑇 = 𝐹𝑟 (𝑠𝑖𝑛 𝜃 +
𝑠𝑖𝑛 2𝜃
2√𝑛2−𝑠𝑖𝑛2 𝜃
)
𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡(𝑇) = 𝑓(𝜃) = 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑐𝑟𝑎𝑛𝑘 𝑎𝑛𝑔𝑙𝑒
F is the net piston effort, r is the crank radius,  is the crank angle.
T  constant, but we want constant ω.
𝑇 = 𝐼 ∝
𝑇𝑜𝑡𝑎𝑙 𝑤𝑜𝑟𝑘 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 = ∫ 𝑇
4𝜋 2𝜋
⁄
0
𝑑𝜃
Average work produced =
Tmean×4π(2π)
𝑇𝑚𝑒𝑎𝑛 =
∫ 𝑇
4𝜋 2𝜋
⁄
0
𝑑𝜃
4𝜋(2𝜋)
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.

50. 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.
𝐶𝑒 =
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑓𝑙𝑢𝑐𝑡𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑏𝑦 𝑐𝑦𝑐𝑙𝑒
Work done per cycle = Tmean × 
𝑇𝑚𝑒𝑎𝑛 =
𝑃
𝜔
=
𝑃×60
2𝜋𝑁
N = Speed in r.p.m
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑝𝑒𝑟 𝑐𝑦𝑐𝑙𝑒 =
𝑃×60
𝑛
n= no. of working strokes per minute

51. 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,
𝐾 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑓𝑙𝑢𝑐𝑡𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑝𝑒𝑒𝑑 =
𝜔1−𝜔2
𝜔
𝑒 =
1
2
𝐼𝜔1
2
−
1
2
𝐼𝜔2
2
=
1
2
𝐼(𝜔1
2
− 𝜔2
2) = 𝐼
(𝜔1 + 𝜔2)
2
(𝜔1 − 𝜔2) = 𝐼𝜔(𝜔1 − 𝜔2) = 𝐼𝜔2
(𝜔1 − 𝜔2)
𝜔
= 𝐼𝜔2
𝐾
𝑒 = 𝐼𝜔2
𝐾 ⇒ 𝐾 =
𝑒
𝐼𝜔2
=
𝑒
2 ×
𝐼𝜔2
2
=
𝑒
2𝐸
𝑒 = 2𝐸𝐾
Dimensions of Flywheel Rims
𝐶𝑒𝑛𝑡𝑟𝑖𝑓𝑢𝑔𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 𝑜𝑛 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 / 𝑢𝑛𝑖𝑡 𝑙𝑒𝑛𝑔𝑡ℎ = [𝜌 · (𝑟. 𝑑)𝑡]. 𝑟𝜔2
𝑇𝑜𝑡𝑎𝑙 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑐𝑒/ 𝑢𝑛𝑖𝑡 𝑙𝑒𝑛𝑔𝑡ℎ = 2 · 𝜌. 𝑟2
. 𝑡. 𝜔2
For equilibrium
𝜎(2𝑡). 1 = 2𝜌. 𝑟2
. 𝑡. 𝜔2
𝜎 = 𝜌. 𝑟2
. 𝜔2
= 𝜌𝑣2
𝑣 = 2𝜋𝑁 60
⁄
𝑚 = 𝜌. 𝜋. 𝑏. 𝑑. 𝑡
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) =
𝐸𝜋𝑑𝑡
𝑇
𝑃 = 𝑇𝑜𝑟𝑞𝑢𝑒𝑚𝑒𝑎𝑛 × 𝜔𝑚𝑒𝑎𝑛
Actual punching time = Tp
Energy given by motor during punching=
𝐸𝜋𝑑𝑡
𝑇
× 𝑇𝑝
𝛥𝐸 = 𝐸𝜋𝑑𝑡 −
𝐸𝜋𝑑𝑡
𝑇
× 𝑇𝑝
𝛥𝐸 = 𝐸𝜋𝑑𝑡 [1 −
𝑡
4𝑟
]

52. 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)

53. 52
Follower motion programming
𝑆 = 𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑓𝑜𝑙𝑙𝑜𝑤𝑒𝑟 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 = 𝑓() ( -> cam rotation angle)
𝑣 =
𝑑𝑠
𝑑𝑡
=
𝑑𝑠
𝑑
×
𝑑
𝑑𝑡
𝑑𝑠
𝑑𝑡
= 𝑝ℎ𝑦𝑠𝑖𝑐𝑎𝑙 𝑡𝑖𝑚𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒,
𝑑𝑠
𝑑
= 𝑘𝑖𝑛𝑒𝑚𝑎𝑡𝑖𝑐 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒,
𝑑
𝑑𝑡
= 𝜔 = 𝑎𝑛𝑔. 𝑣𝑒𝑙. 𝑜𝑓 𝑐𝑎𝑚
𝑎 =
𝑑𝑣
𝑑𝑡
=
𝑑𝑣
𝑑
×
𝑑
𝑑𝑡
=
𝑑2
𝑠
𝑑𝜃2
(
𝑑𝜃
𝑑𝑡
)
2
𝑗𝑒𝑟𝑘 (𝒋) = 𝜔3
𝑑3
𝑠
𝑑𝜃3
=
𝑑𝑎
𝑑𝑡
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.
𝑆 =
𝑣𝜃
𝜔
 = 𝜔𝑡 𝑠 = 𝑣. 𝑡
ℎ =
𝑣
𝜔
𝑎
𝑣𝑎 =
ℎ𝜔
𝑎
a=0
j=0
𝑣𝑑 =
ℎ𝜔
𝑑
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.

54. 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
𝐴𝑡 𝑎𝑛𝑦 𝑖𝑛𝑠𝑡𝑎𝑛𝑡, 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝒔 =
ℎ
2
−
ℎ
2
𝑐𝑜𝑠 𝛽
𝛽 = 𝜋
𝜃
𝜑
𝒔 =
ℎ
2
−
ℎ
2
𝑐𝑜𝑠 𝜋
𝜃
𝜑
=
ℎ
2
(1 − 𝑐𝑜𝑠 𝜋
𝜃
𝜑
)
=
ℎ
2
(1 − 𝑐𝑜𝑠
𝜋𝜔𝑡
𝜑
)
𝒗 =
𝑑𝑠
𝑑𝑡
=
ℎ
2
𝜋𝜔
𝜑
𝑠𝑖𝑛
𝜋𝜔𝑡
𝜑
=
ℎ
2
𝜋𝜔
𝜑
𝑠𝑖𝑛
𝜋
𝜑
𝒗𝑚𝑎𝑥 =
ℎ
2
𝜋𝜔
𝜑
𝑎𝑡  =
𝜑
2
𝒇 =
𝑑𝒗
𝑑𝒕
=
ℎ
2
(
𝜋𝜔
𝜑
)
2
𝑐𝑜𝑠
𝜋
𝜑
𝒇𝑚𝑎𝑥 =
ℎ
2
(
𝜋𝜔
𝜑
)
2
𝑎𝑡  = 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.

55. 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.
𝒔 = 𝑣𝑜𝑡 +
1
2
𝑓𝑡2
𝒔 =
1
2
𝑓𝑡2
𝑎𝑠 𝑣𝑜 = 0
𝑓 =
2𝑠
𝑡2
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝒔 =
ℎ
2
& 𝜔𝑡 =
𝜑
2
, 𝑡 =
𝜑
2𝜔
𝒇 =
4ℎ𝜔2
𝜑2
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝒗 = 𝑓𝑡 =
4ℎ𝜔2
𝜑2
×
𝜃
𝜔
=
4ℎ𝜔
𝜑2
𝜃
𝒗𝒎𝒂𝒙 =
4ℎ𝜔
𝜑2
×
𝜑
2
=
2ℎ𝜔
𝜑
Here acceleration is abruptly increasing from maximum to minimum, which results in infinite jerk, vibration
and noise.

56. 55
Constant Velocity
Constant velocity of follower implies the displacement of follower is proportional to cam rotation.
𝑣 = 0 →
𝑑𝒔
𝑑𝒕
= 0 → 𝑠 ∝ 𝜃
𝑠 = ℎ
𝜃
𝜑
= ℎ
𝜔𝑡
𝜑
𝑣 =
𝑑𝒔
𝑑𝒕
=
ℎ𝜔
𝜑
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑓 =
𝑑𝒗
𝑑𝒕
= 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

57. 56
Cycloid
A cycloid is locus of point on a circle rotating on a straight line.
𝑠 =
ℎ
𝜋
(
𝜋𝜃
𝜑
−
1
2
𝑠𝑖𝑛
2𝜋𝜃
𝜑
)
𝑣 =
𝑑𝒔
𝑑𝒕
=
𝑑𝒔
𝑑𝜽
×
𝑑𝜽
𝑑𝒕
= [
ℎ
𝜑
−
ℎ
𝜑
𝑐𝑜𝑠
2𝜋
𝜑
] 𝜔
𝑣 = [
ℎ𝜔
𝜑
−
ℎ𝜔
𝜑
𝑐𝑜𝑠
2𝜋
𝜑
] =
ℎ𝜔
𝜑
(1 − 𝑐𝑜𝑠
2𝜋
𝜑
)
𝑣𝑚𝑎𝑥 =
ℎ𝜔
𝜑
𝑎𝑡 𝜃 =
𝜑
2
𝑓 =
𝑑𝒗
𝑑𝒕
=
𝑑𝒗
𝑑𝜽
×
𝑑𝜽
𝑑𝒕
= [
2ℎ𝜋𝜔2
𝜑2
𝑠𝑖𝑛
2𝜋
𝜑
]
𝑓𝑚𝑎𝑥 =
2ℎ𝜋𝜔2
𝜑2
𝑎𝑡 𝜃 =
𝜋
4
There are no abrupt changes in velocity and acceleration.
So, this is the most ideal one to use.

58. 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
𝑣𝑝 = 𝜔1𝑟1 = 𝜔2𝑟2
𝑣𝑝 = 2𝜋𝑁1𝑟1 = 2𝜋𝑁2𝑟2
𝜔1
𝜔2
=
𝑁1
𝑁2
=
𝑟1
𝑟2
𝜔1
𝜔2
=
𝐼12𝐼23
𝐼13𝐼23
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.

59. 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.

60. 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.

61. 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.
𝐶𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑝𝑖𝑡𝑐ℎ (𝒑) =
𝜋𝑑
𝑇
=
𝜋 × (𝑝𝑖𝑡𝑐ℎ 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟)
𝑁𝑜. 𝑜𝑓 𝑡𝑒𝑒𝑡ℎ
Diametral pitch- The number of teeth per unit length of pitch circle diameter in inches.
𝐷𝑖𝑎𝑚𝑒𝑡𝑟𝑎𝑙 𝑃𝑖𝑡𝑐ℎ(𝑷) =
𝑁𝑜. 𝑜𝑓 𝑡𝑒𝑒𝑡ℎ
𝑝𝑖𝑡𝑐ℎ 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
=
𝑇
𝑑
Module- It is the ratio of pitch diameter (in mm) to the number of teeth.
𝑀𝑜𝑑𝑢𝑙𝑒(𝒎) =
𝑝𝑖𝑡𝑐ℎ 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
𝑁𝑜. 𝑜𝑓 𝑡𝑒𝑒𝑡ℎ
=
𝑑
𝑇
𝒑 = 𝝅𝒎
Gear Ratio- It is the number of teeth on the gear to that of the pinion.
𝐺𝑒𝑎𝑟 𝑅𝑎𝑡𝑖𝑜(𝑮) =
𝑁𝑜. 𝑜𝑓 𝑡𝑒𝑒𝑡ℎ 𝑜𝑛 𝐺𝑒𝑎𝑟
𝑁𝑜. 𝑜𝑓 𝑡𝑒𝑒𝑡ℎ 𝑜𝑛 𝑝𝑖𝑛𝑖𝑜𝑛
=
𝑇
𝑡

62. 61
Velocity Ratio- The ratio of angular velocity of the follower to angular velocity of driving gear.
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑟𝑎𝑡𝑖𝑜 (𝑽𝑹) =
𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑓𝑜𝑙𝑙𝑜𝑤𝑒𝑟(𝟐)
𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑑𝑟𝑖𝑣𝑒𝑟 (𝟏)
=
𝜔2
𝜔1
=
𝑁2
𝑁1
=
𝑑1
𝑑2
=
𝑇1
𝑇2
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.
𝐶𝑙𝑒𝑎𝑟𝑒𝑛𝑐𝑒 = 𝐴𝑑𝑒𝑛𝑑𝑢𝑚 − 𝐷𝑒𝑑𝑒𝑛𝑑𝑢𝑚 = (𝑑 + 2𝑚) − (𝑑 − 2𝜋𝑚) = .157𝑚
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.
𝐶𝑜𝑛𝑡𝑎𝑐𝑡 𝑟𝑎𝑡𝑖𝑜 =
𝛼 + 𝛽
𝛾
𝐶𝑜𝑛𝑡𝑎𝑐𝑡 𝑟𝑎𝑡𝑖𝑜 =
𝐴𝑟𝑐 𝑜𝑓 𝑐𝑜𝑛𝑡𝑎𝑐𝑡
𝐶𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑝𝑖𝑡𝑐ℎ
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)

63. 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 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝑣𝐶 𝑎𝑙𝑜𝑛𝑔 𝑡ℎ𝑒 𝑛 − 𝑛 = 𝑣𝐶 × 𝑐𝑜𝑠 𝛼
𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝑣𝐷 𝑎𝑙𝑜𝑛𝑔 𝑡ℎ𝑒 𝑛 − 𝑛 = 𝑣𝐷 × 𝑐𝑜𝑠 𝛽
𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑚𝑜𝑡𝑖𝑜𝑛 𝑎𝑙𝑜𝑛𝑔 𝑡ℎ𝑒 𝑛 − 𝑛 = 𝑣𝐶 × 𝑐𝑜𝑠 𝛼 − 𝑣𝐷 × 𝑐𝑜𝑠 𝛽 = 0
𝜔1 × 𝐴𝐶 × 𝑐𝑜𝑠 𝛼 = 𝜔2 × 𝐵𝐷 × 𝑐𝑜𝑠 𝛽
𝜔1 × 𝐴𝐶 ×
𝐴𝐸
𝐴𝐶
= 𝜔2 × 𝐵𝐷 ×
𝐵𝐹
𝐵𝐷
𝜔1
𝜔2
=
𝐴𝐸
𝐵𝐹
=
𝐵𝑃
𝐴𝑃
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.
𝜔1
𝜔2
=
𝐹𝑃
𝐸𝑃
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.
𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝑣𝐶 𝑎𝑙𝑜𝑛𝑔 𝑡ℎ𝑒 𝑡 − 𝑡 = 𝑣𝐶 × 𝑠𝑖𝑛 𝛼
𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝑣𝐷 𝑎𝑙𝑜𝑛𝑔 𝑡ℎ𝑒 𝑡 − 𝑡 = 𝑣𝐷 × 𝑠𝑖𝑛 𝛽
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑠𝑙𝑖𝑑𝑖𝑛𝑔 = 𝑣𝐶 × 𝑠𝑖𝑛 𝛼 − 𝑣𝐷 × 𝑠𝑖𝑛 𝛽 = (𝜔1 + 𝜔2)𝑷𝑪
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.

64. 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

65. 64
Pressure angles in this case remain constant throughout the engagement of teeth.
∠𝐴𝐸𝑃 = ∠𝐵𝐹𝑃 = 𝜑
𝐴𝐸 = 𝐴𝑃 𝑐𝑜𝑠 𝜑
𝐵𝐹 = 𝐵𝑃 𝑐𝑜𝑠 𝜑
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑟𝑎𝑡𝑖𝑜 𝑜𝑓 𝑔𝑒𝑎𝑟𝑠 =
𝑣1
𝑣2
=
𝐵𝑃
𝐴𝑃
=
𝐵𝐹
𝐴𝐸
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
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.

66. 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)
𝐶𝐷 = [√𝑅𝑎
2 − 𝑅𝑎
2 𝑐𝑜𝑠2 𝜑 − 𝑅𝑎 𝑠𝑖𝑛 𝜑] + [√𝑟𝑎
2 − 𝑟𝑎
2 𝑐𝑜𝑠2 𝜑 − 𝑟𝑎 𝑠𝑖𝑛 𝜑] = √𝑅𝑎
2 − 𝑅𝑎
2 𝑐𝑜𝑠2 𝜑 + √𝑟𝑎
2 − 𝑟𝑎
2 𝑐𝑜𝑠2 𝜑 − (𝑅 + 𝑟) 𝑠𝑖𝑛 𝜑
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
𝐴𝑟𝑐 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ = 𝜔𝑎𝑟 × 𝑡𝑎 = 𝜔𝑎(𝑟 𝑐𝑜𝑠 𝜑)
𝑡𝑎
𝑐𝑜𝑠 𝜑
= (𝑡𝑎𝑛𝑔. 𝑣𝑒𝑙. 𝑜𝑓 𝐻)𝑡𝑎 ×
1
𝑐𝑜𝑠 𝜑
=
𝐴𝑟𝑐 𝑯𝑲
𝑐𝑜𝑠 𝜑
=
𝐴𝑟𝑐 𝑭𝑲 − 𝐴𝑟𝑐 𝑭𝑯
𝑐𝑜𝑠 𝜑
=
𝑭𝑷 − 𝑭𝑷′
𝑐𝑜𝑠 𝜑
=
𝑪𝑷
𝑐𝑜𝑠 𝜑
Similarly, arc of recess is PP’’ = tang. vel. of P × time of recess

67. 66
𝐴𝑟𝑐 𝑜𝑓 𝑟𝑒𝑐𝑒𝑠𝑠 =
𝑷𝑫
𝑐𝑜𝑠 𝜑
𝐴𝑟𝑐 𝑜𝑓 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 =
𝑪𝑷
𝑐𝑜𝑠 𝜑
+
𝑷𝑫
𝑐𝑜𝑠 𝜑
=
𝑪𝑫
𝑐𝑜𝑠 𝜑
𝐴𝑟𝑐 𝑜𝑓 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 =
𝑃𝑎𝑡ℎ 𝑜𝑓 𝑐𝑜𝑛𝑡𝑎𝑐𝑡
𝑐𝑜𝑠 𝜑
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.
𝑇ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑒𝑡ℎ 𝑖𝑛 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 (𝒏) =
𝐴𝑟𝑐 𝑜𝑓 𝑐𝑜𝑛𝑡𝑎𝑐𝑡
𝐶𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑝𝑖𝑡𝑐ℎ
=
𝐶𝐷
𝑐𝑜𝑠 𝜑
1
𝑝
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.

68. 67
Minimum Number of Teeth
We saw previously that maximum addendum radius of wheel to prevent interference is BE.
𝐵𝐸2
= 𝐵𝐹2
+ 𝐹𝐸2
= 𝐵𝐹2
+ (𝐹𝑃 + 𝑃𝐸)2
𝐵𝐸2
= (𝑅 𝑐𝑜𝑠 𝜑)2
+ (𝑅 𝑠𝑖𝑛 𝜑 + 𝑟 𝑠𝑖𝑛 𝜑)2
𝐵𝐸 = 𝑅√1 +
𝑟
𝑅
(
𝑟
𝑅
+ 2) 𝑠𝑖𝑛2 𝜑
Therefore, maximum value of addendum of the wheel is
aw max = BE – pitch radius
𝑎𝑤 𝑚𝑎𝑥 = 𝑅√1 +
𝑟
𝑅
(
𝑟
𝑅
+ 2) 𝑠𝑖𝑛2 𝜑 − 𝑅
= 𝑅 [√1 +
𝑟
𝑅
(
𝑟
𝑅
+ 2) 𝑠𝑖𝑛2 𝜑 − 1]
𝑅 =
𝑚𝑇
2
, 𝑟 =
𝑚𝑡
2
& 𝐺𝑒𝑎𝑟 𝑟𝑎𝑡𝑖𝑜 (𝑮) =
𝑇
𝑡
=
𝑅
𝑟
𝑎𝑤 𝑚𝑎𝑥 =
𝑚𝑇
2
[√1 +
1
𝐺
(
1
𝐺
+ 2) 𝑠𝑖𝑛2 𝜑 − 1]
Let the adopted value of addendum in some cases be aw × m.
𝑆𝑜, 𝑎𝑤 × 𝑚 ≤
𝑚𝑇
2
[√1 +
1
𝐺
(
1
𝐺
+ 2) 𝑠𝑖𝑛2 𝜑 − 1]
𝑇 ≥
2𝑎𝑤
[√1 +
1
𝐺
(
1
𝐺
+ 2) 𝑠𝑖𝑛2 𝜑 − 1]
𝑇 =
2𝑎𝑤
[√1 +
1
𝐺
(
1
𝐺
+ 2) 𝑠𝑖𝑛2 𝜑 − 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 𝑡 =
𝑇
𝐺

69. 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.
𝐺𝐸 = 𝑃𝐸 × 𝑠𝑖𝑛 𝜑 = 𝑟 𝑠𝑖𝑛 𝜑 × 𝑠𝑖𝑛 𝜑 = 𝑟 𝑠𝑖𝑛2
𝜑 =
𝑚𝑡
2
× 𝑠𝑖𝑛2
𝜑
To avoid interference,
𝐺𝐸 ≥ 𝑎𝑟 × 𝑚 (𝑜𝑟)
𝑚𝑡
2
× 𝑠𝑖𝑛2
𝜑 ≥ 𝑎𝑟 × 𝑚 (𝑜𝑟)𝑡 ≥
2 𝑠𝑖𝑛2
𝜑
𝑚
𝑃𝑎𝑡ℎ 𝑜𝑓 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 = 𝐶𝑃 + 𝑃𝐷 =
𝐴𝑑𝑑𝑒𝑚𝑑𝑢𝑚 𝑜𝑓 𝑟𝑎𝑐𝑘
𝑐𝑜𝑠 𝜑
+ √𝑟𝑎
2 − (𝑟 𝑐𝑜𝑠 𝜑)2 − 𝑟 𝑠𝑖𝑛 𝜑
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑝𝑎𝑡ℎ 𝑜𝑓 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 𝑡𝑜 𝑎𝑣𝑜𝑖𝑑 𝑖𝑛𝑡𝑒𝑟𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 𝐷𝐸 = √𝑟𝑎
2 − (𝑟 𝑐𝑜𝑠 𝜑)2

70. 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.