SUBJECT DYNAMICS OF MACHINERY (2161901)

Faculty of Degree Engineering-083 Department of Mechanical Engineering – 19

SUBJECT

DYNAMICS OF MACHINERY

(2161901)

IMPORTANT QUESTIONS

WITH ANSWER

PREPARED BY: PROF. ASHOK BAGDA


1) Explain the balancing of several masses rotating in same plane by

Graphical Method.

Consider four numbers of masses of magnitude m1, m2, m3 and m4 at distances of r1,

r2, r3 and r4 from the axis of the rotating shaft. Let θ1, θ2, θ3 and θ4 be the angles of

these masses with the horizontal line OX, as shown in Figure.

Let these masses rotate about an axis through O and perpendicular to the plane of

paper, with a constant angular velocity of ω rad/s.

Figure: Angular position and force polygon

The magnitude and position of the balancing mass may be found out graphically as

discussed below:

First of all, draw the space diagram with the positions of the several masses, as

shown in Figure.

Find out the centrifugal force (or product of the mass and radius of rotation) exerted

by each mass on the rotating shaft.

Now draw the vector diagram with the obtained centrifugal forces (or the product of

the masses and their radii of rotation), such that ab represents the centrifugal force

exerted by the mass m1 (or m1.r1) in magnitude and direction to some suitable scale.

Similarly, draw bc, cd and de to represent centrifugal forces of other masses m2, m3

and m4 (or m2.r2, m3.r3 and m4.r4).

Now, as per polygon law of forces, the closing side ae represents the resultant force

in magnitude and direction, as shown in Figure.



2) What is swaying couple? Derive an expression for swaying couple.

The unbalanced forces along the line of stroke for the two cylinders constitute a couple

about the centre line YY between the cylinders as shown in Figure. 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.

Let a = Distance between the centre lines of the two cylinders.

The swaying couple is maximum or minimum when (cos, sin) θ+ θ is maximum or

minimum. For (cos sin ) θ+ θ to be maximum or minimum,

In order to reduce the magnitude of the swaying couple, revolving balancing masses

are introduced. But, as discussed in the previous article, the revolving balancing masses

because unbalanced forces to act at right angles to the line of stroke.

Since a swaying couple is more harmful than an oscillating couple, therefore a value of

‘c’ from 2/3 to 3/4, in two-cylinder locomotives with two pairs of coupled wheels, is

usually used. But in large four cylinder locomotives with three or more pairs of coupled

wheels, the value of ‘c’ is taken as 2/5.



3) What is Hammer blow? Derive an expression for limiting speed required

for hammer blow.

The maximum magnitude of the unbalanced force along the perpendicular to the line

of stroke is known as hammer blow.

We know that the unbalanced force along the perpendicular to the line of stroke due

to the balancing mass B, at a radius b, in order to balance reciprocating parts only is

B. ω 2 .b sin θ. This force will be maximum when sin θ is unity, i.e. when θ = 90° or

270°.

∴ Hammer blow =± B.ω2

The effect of hammer blow is to cause the variation in pressure between the wheel

and the rail. This variation is shown in Figure, for one revolution of the wheel.

Let P be the downward pressure on the rails (or static wheel load) and Net pressure

between the wheel and the rail= = 2 PB b

If (P±B.ω 2 .b) is negative, then the wheel will be lifted from the rails. Therefore the

limiting condition in order that the wheel does not lift from the rails is given by 2 PB

b = ω, and the permissible value of the angular speed,



4) Explain the balancing of V-Engine.

Consider a symmetrical two cylinder V-engine as shown in Figure, The common crank

OC is driven by two connecting rods PC and QC. The lines of stroke OP and OQ are

inclined to the vertical OY, at an angle α as shown in Figure.

Let m = Mass of reciprocating parts per cylinder,

l = Length of connecting rod,

r = Radius of crank,

n = Ratio of length of connecting rod to crank radius = l / r

θ = Inclination of crank to the vertical at any instant,

ω = Angular velocity of crank.

Inertia force due to reciprocating parts of cylinder 1, along the line of stroke,

Inertia force due to reciprocating parts of cylinder 2, along the line of stroke,

The balancing of V-engines is only considered for primary and secondary forces as



5) Explain the concept of direct and reverse crank.

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. Since the plane of

rotation of the various cranks (in radial or V-engines) is same, therefore there is no

unbalanced primary or secondary couple.

Consider a reciprocating engine mechanism as shown in Figure. Let the crank OC

(known as the direct crank) rotates uniformly at ω radians per second in a clockwise

direction.

Let at any instant the crank makes an angle θ with the line of stroke OP. 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.

A little consideration will show that when the direct crank revolves in a clockwise

direction, the reverse crank will revolve in the anticlockwise direction. We shall now

discuss the primary and secondary forces due to the mass (m) of the reciprocating

parts at P.

Considering the primary forces



This force is equal to the component of the centrifugal force along the line of stroke,

produced by a mass (m) placed at the crank pin C.

It is assumed that m / 2 is fixed at the direct crank (termed as primary direct crank) pin

C and m / 2 at the reverse crank (termed as primary reverse crank) pin C′, as shown in

Fig. Centrifugal force acting on the primary direct and reverse crank,

Component of the centrifugal force acting on the primary direct crank,

Component of the centrifugal force acting on the primary reverse crank,

∴ Total component of the centrifugal force along the line of stroke,

Considering secondary forces

The crank OD is the 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 as shown in Fig. In the similar way as discussed above, it

will be seen that for the secondary effects, the mass (m) of the reciprocating parts may

be replaced by two masses (each m/2) placed at D and D′ such that OD = OD′ = r/4n.



6) Describe Dunkerley’s method to find the natural frequency of a shaft

carrying several loads.

The natural frequency of transverse vibration for a shaft carrying a number of point

loads and uniformly distributed load is obtained from Dunkerley’s empirical formula.

According to this,

Where,

fn = Natural frequency of transverse vibration of the shaft carrying point loads and

uniformly distributed load.

fn1, fn2, fn3,, etc. = Natural frequency of transverse vibration of each point load.

fns = Natural frequency of transverse vibration of the uniformly distributed load

(or due to the mass of the shaft).

Now, consider a shaft AB loaded as shown in Fig.

Figure: Shaft carrying a number of point loads and a uniformly distributed load.

Let δ1, δ2, δ3, etc. = Static deflection due to the load W1, W2, W3, when considered

separately.

δS = Static deflection due to the uniformly distributed load or due to the

mass of the shaft.

Natural frequency of transverse vibration due to load W1,





Also natural frequency of transverse vibration due to uniformly distributed load or

weight of the shaft,

Therefore, according to Dunkerley’s empirical formula, the natural frequency of the

whole system,



7) Discuss the effect of damping on vibratory systems. What is meant by

under damping, critical damping and over damping?

The motion of a body is resisted by frictional forces. In vibrating systems, the effect of

friction is referred to as damping. The damping provided by fluid resistance is known as

viscous damping.

In damped vibrations, the amplitude of the resulting vibration gradually diminishes. This

is due to the reason that a certain amount of energy is always dissipated to overcome the

frictional resistance. The resistance to the motion of the body is provided partly by the

medium in which the vibration takes place and partly by the internal friction, and in some

cases partly by a dash pot or other external damping device.

Figure: Damped, under damped and over

damped vibration

The damping of a system can be described as being one of the following:

a) Over damped: The system returns to equilibrium without oscillating.

b) Critically damped: The system returns to equilibrium as quickly as possible

without oscillating.

c) Under damped: The system oscillates (at reduced frequency compared to the

undammed case) with the amplitude gradually decreasing to zero.



8) Explain Vibration isolation and transmissibility

A little consideration will show that when an unbalanced machine is installed on the

foundation, it produces vibration in the foundation. In order to prevent these vibrations

or to minimize the transmission of forces to the foundation, the machines are mounted

on springs and dampers or on some vibration isolating material, as shown in Fig.

The arrangement is assumed to have one degree of freedom, when a periodic (i.e.

simple harmonic) disturbing force is applied to a machine of mass m supported by a

spring of stiffness s, then the force is transmitted by means of the spring and the

damper or dashpot to the fixed support or foundation.



The ratio of the force transmitted (FT) to the force applied (F) is known as the isolation

factor or transmissibility ratio of the spring support.

9) What are the classifications of vibration?

Vibrations are classified as below:

According to the actuating force:

a) Free vibration

b) Forced vibration

According to energy dissipation:

a) Undamped vibration

b) Damped vibration

According to behavior of vibrating system:

a) Linear vibration

b) Non – linear vibration

According to motion of system with respect to axis:

a) Longitudinal vibration

b) Transverse vibration

c) Torsional vibration



10) Define the following terms:

a) Time period: It is the time interval after which the motion is repeated itself. The

period of vibration is usually expressed in seconds.

b) Cycle: It is the motion completed during one time period.

c) Frequency: It is the number of cycles described in one second. In S.I. units, the

frequency is expressed in hertz (briefly written as Hz) which is equal to one cycle per

second.

d) Critical or whirling speed: 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.

e) Natural frequency: When no external force acts on the body, after giving it an

initial displacement, then the body is said to be under free vibrations. The frequency

of the free vibrations is called free or natural frequency.

f) Degree of Freedom: The minimum number of independent coordinates required to

determine completely the position of all parts of a system at any instant of time

defines the degree of freedom of the system.

g) Damping Factor: The ratio of the actual damping coefficient (c) to the critical

damping coefficient (Cc) is known as damping factor or damping ratio.

h) Resonance: When the frequency of the external force is same as that of the natural

vibrations, this phenomenon is known as resonance.

i) Forced vibrations: When the body vibrates under the influence of external force,

then the body is said to be under forced vibrations.

j) Damped vibrations: When there is a reduction in amplitude over every cycle of

vibration, the motion is said to be damped vibration.



11) Derive an expression for Torsion ally Equivalent Shaft System.

In General, shaft comprised of multiple cross-sectional diameters with different

lengths. Sometimes it is necessary to assume the shaft as a uniform diameter and

length for some calculations such as finding the natural frequency.

In such cases, we have to replace the different cross-section diameters with the

equivalent shaft of uniform diameter. This will discuss how we can construct such data

of torsionally equivalent shaft for calculation purposes.

Consider a shaft of varying cross-sections as shown in below fig. Let this shaft is

replaced by an equivalent shaft of uniform diameter d and length l as shown figure.

These two shafts must have the same total angle of a twist when equal opposing

torques T is applied at the opposite ends.

d1, d2 and d3 = Diameters for the lengths l1, l2 and l3 respectively,

θ1, θ2 and θ3 = Angles of twist for the lengths l1, l2 and l3 respectively,

θ = Angle of twist for the diameter d and length l,

J1, J2 and J3 = Polar moment of inertia for the shaft of

diameters d1, d2 and d3 respectively.

Since the total angle of the twist of the shaft is equal to the sum of angle of twists of the

different lengths.

θ = θ1 + θ2 + θ3


https://extrudesign.com/natural-frequency-of-free-torsional-vibrations/


As we want to make the multiple cross-sectional shafts into a uniform diametric shaft, so

we have to assume that diameter d of the equivalent shaft should be equal to the one of

the diameters of the actual shaft. So we can assume d = d1



12) Explain Rayleigh’s method for finding the natural frequency of vibratory

system.

In this method, the maximum kinetic energy at the mean position is equal to the

maximum potential energy (or strain energy) at the extreme position. Assuming the

motion executed by the vibration to be simple harmonic, then



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SUBJECT DYNAMICS OF MACHINERY (2161901)

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