ME 304-DOM
EQUATIONS AND IMPORTANT
POINTS TO REMEMBER
MODULE-1
1.
Conditions of static equilibrium
2.
Conditions of Equilibrium of Two Forces
3.
Conditions of Equilibrium of Three Forces
4.
Torque
5.
Example:
PRINCIPLE OF SUPER POSITION
6.
7.
8.
µFR
Rsinϕ = µR
MODULE-2
D’Alembert’s Principle
DYNAMIC ANALYSIS OF SLIDER CRANK MECHANISM
VERY
IMPORTANT
FOR PROBLEM
SOLVING ON
DYNAMIC
ANALYSIS OF
SLIDER CRANK
MECHANISM
MODULE-3
1. Turning Moment Diagram
Torque on Crank
𝐹𝑝 × 𝑟
𝑠𝑖𝑛2θ
𝑇 = 𝐹𝑇 × 𝑟 =
𝑆𝑖𝑛 θ + ϕ = 𝐹𝑝 × 𝑟 𝑠𝑖𝑛θ +
𝑐𝑜𝑠ϕ
2 𝑛2 − 𝑠𝑖𝑛2 θ
Turning Moment Diagram =
Plot between T and θ
Crank Effort Diagram=
Plot between FT and θ
2. Turning moment diagram for a single cylinder,
double acting steam engine (TWO STROKE ENGINE)
3. Turning Moment Diagram for a Four Stroke Cycle
Internal Combustion Engine –Single Cylinder
4. Turning Moment Diagram for a Multi-cylinder Engine
5. Fluctuation of Energy
 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.
 Maximum fluctuation of energy,
∆ E = Maximum energy – Minimum energy
6.Coefficient of Fluctuation of Energy (CE)
It may be defined as the ratio
of the maximum fluctuation
of energy to the work done
per cycle.
7.Coefficient of Fluctuation of Speed
8.Energy Stored in a Flywheel
(Derivation Important -5 Marks Question)
9.Energy Stored in a Flywheel
(Derivation Important -5 Marks Question)…………
ΔE
10. Maximum Acceleration of Flywheel (αf)
Maximum Torque= Tmax
Mean Torque = Tmean
Fluctuation in Torque, ΔT = Tmax – Tmin
We, Know that,
ΔT
α𝑓 =
𝐼
I = mass moment of Inertia = mk2
where, k= radius of gyration
11. Dimensions of Rim of a Flywheel
Centrifugal Force
Flywheel for Punching Press
MODULE-4
Gyroscopic Couple/Torque
Gyroscopic Couple
Summary of Direction Active and Gyroscopic Couple for Aeroplane
Viewing
Direction
Propellor
Rotation
Precision
of Aeroplane
Active Gyroscopic
Couple
Reactive
Gyroscopic
Couple
REAR
Clockwise
Left
Clockwise
Anti-Clockwise
REAR
Clockwise
Right
Anti-Clockwise
Clockwise
REAR
Anti-Clockwise
Left
Anti-Clockwise
Clockwise
REAR
Anti-Clockwise
Right
Clockwise
Anti-Clockwise
FRONT
Clockwise
Left
Anti-Clockwise
Clockwise
FRONT
Clockwise
Right
Clockwise
Anti-Clockwise
FRONT
Anti-Clockwise
Left
Clockwise
Anti-Clockwise
FRONT
Anti-Clockwise
Right
Anti-Clockwise
Clockwise
Equations
Gyroscopic Effect on Ships
Terms Used in a Naval Ship
 The top and front views of a naval
ship are shown in Fig . The fore end of
the ship is called bow and the rear end
is known as stern or aft. The left hand
and right hand sides of the ship, when
viewed from the stern are called port
and star-board respectively.
 We shall now discuss the effect of
gyroscopic couple on the naval ship in
the following three cases:
1. Steering
2. Pitching, and
3. Rolling
On Ship
Conditions
for Pitching
of Ship
 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.
 Therefore Road reaction over each wheel, =
W/4 = m.g /4 newtons
The positive sign is used when the wheels and rotating parts of the engine rotate in
the same direction. If the rotating parts of the engine revolves in opposite direction,
then negative sign is used.
 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.
 Let the magnitude of this reaction at the two outer or
inner wheels be P newtons. Then
P × x = C or P = C/x
 Vertical reaction at each of the outer or inner wheels,
P /2 = C/ 2x
 Note: We have discussed above that when rotating
parts of the engine rotate in opposite directions, then
–ve sign is used, i.e. net gyroscopic couple,
C = CW – CE,
When C E > CW, then C will be –ve.
 Thus the reaction will be vertically downwards on
the outer wheels and vertically upwards on the inner
wheels.
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2. Effect of the 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 tends to overturn the vehicle.
 We know that centrifugal force,
 This overturning couple is balanced by vertical reactions, which are vertically
upwards on the outer wheels and vertically downwards on the inner wheels.
 Let the magnitude of this reaction at the two outer or inner wheels be Q.
Then
 A little consideration will show that when the vehicle is running at high speeds, PI
may be zero or even negative.
 This will cause the inner wheels to leave the ground thus tending to overturn the
automobile. In order to have the contact between the inner wheels and the ground,
the sum of P/2 and Q/2 must be less than W/4.
Stability of a Two Wheel Vehicle Taking a Turn
 Consider a two wheel vehicle (say a scooter or motor cycle) taking a right
turn as shown in Fig
m = Mass of the vehicle and its rider in kg
W = Weight of the vehicle and its rider in newtons = m.g
h = Height of the centre of gravity of the vehicle and rider
rw = Radius of the wheels
R = Radius of track or curvature,
68
 A little consideration will show that when the
wheels move over the curved path, the vehicle
is always inclined at an angle θ with the vertical
plane as shown in Fig.
 This angle is known as angle of heel (θ).
 In other words, the axis of spin is inclined to the
horizontal at an angle θ, as shown in Fig.
 Thus the angular momentum vector Iω due to
spin is represented by OA inclined to OX at an
angle θ.
 But the precession axis is vertical. Therefore the
spin vector is resolved along OX.
Notes:
 When the engine is rotating in the same
direction as that of wheels, then the
positive sign is used in the
above
expression and if the engine rotates in
opposite direction, then negative sign is
used.
 The gyroscopic couple will act over the
vehicle outwards i.e. in the anticlockwise
direction when seen from the front of the
vehicle. The tendency of this couple is to
overturn the vehicle in outward direction.
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MODULE-5
Part-I- Free Undamped Vibration
1. General Equation of Vibration:
m = mass; x = displacement; s = Stifness of spring
𝑑2𝑥
= Acceleration= 2
𝑑𝑡
2. Natural Frequency of Longitudinal Vibration:
Axial or Longitudinal Deflection
3. Natural frequency of Transverse Vibration:
If Inertia is considered
4. Energy Method:
5. Newton’s Method:
6. Equivalent Stiffness of Spring:
PART-II DAMPED FREE VIBRATION
Displacement of
underdamped
vibration
PART-III FORCED VIBRATION
4
5. Amplitude at Resonance, A
Force Transmitted at Resonance, Ft
𝐴=
6. Maximum Amplitude of Vibration, Amax =
𝐹𝑡
𝑠 1 + 4ζ2
MODULE-6
1. Natural frequency of Transverse Vibration:
If Inertia is considered
2. Whirling of Shaft:
Also,
M= Bending Moment
y1=Distance from neutral
Axis to outer Fiber of beam
I = Moment of Inertia
3. Dunkerley’s Equation:
Natural frequency of Transverse Vibration
4. Natural frequency of Torsional Vibration:
5. Torsionally Equivalent Shaft:
Length of Torsionally Equivalent Shaft, l,
Values of static deflection (δ) for the various types of
beams and under various load conditions and
under various load conditions
Equations of Maximum Bending Moment of Beams