## Wednesday, 28 December 2016

### Longitudinal and Transverse Vibrations: Introduction

Introduction:

When elastic bodies such as a spring, a beam and a shaft are displaced from the equilibrium position by the application of external forces, and then released, they execute a vibratory motion. This is due to the reason that, when a body is displaced, the internal forces in the form of elastic or strain energy are present in the body. At release, these forces bring the body to its original position. When the body reaches the equilibrium position, the whole of the elastic or strain energy is converted into kinetic energy due to which the body continuous to move in the opposite direction. The whole of the kinetic energy is again converted into strain energy due to which the body again returns to the equilibrium position. In this way, the vibratory motion is repeated indefinitely.

### Terms Used in Vibratory Motion

The following terms are commonly used in connection with the vibratory motions:

1. Period of vibration or time period: It is the time interval after which the motion is repeated itself. The period of vibration is usually expressed in seconds.
2. Cycle: It is the motion completed during one time period.
3. Frequency: It is the number of cycles described in one second. In S.I. units, the frequency is expressed in Hertz (briefly written in Hz) which is equal to one cycle per second.

## Tuesday, 27 December 2016

### Types of Vibratory Motion

The following types of vibratory motion are important from the subject point of view:
1. Free or Natural vibrations: When no external force acts on the body, after giving it an initial displacement, then the body is said to be under Free or Natural vibrations. The frequency of the free vibrations is called Free natural frequency.
2. Forced vibrations: When the body vibrates under the influence of external force, then the body is said to be under Forced vibrations. The external force applied to the body is a periodic disturbing force created by unbalance. The vibrations have the same frequency as the applied force.
Note: When the frequency of the external force is same as that of the natural vibrations, resonance takes place.
3. Damped vibrations: When there is a reduction in amplitude over every cycle of vibration, the motion is said to be Damped vibration. This is due to the fact that a certain amount of energy possessed by the vibrating system is always dissipated in overcoming frictional resistance to the motion.

## Monday, 26 December 2016

### Types of Free Vibrations

The following three types of free vibrations are important from the subject point of view:
1. Longitudinal vibrations, 2. Transverse vibrations, and 3. Torsional vibrations.
Consider a weightless constraint (spring or shaft) whose one end is fixed and the other end carrying a heavy disc as shown in Fig., This system may execute one of the three above mentioned types of vibrations.

1. Longitudinal Vibrations: When the particles of the shaft or disc moves parallel to the axis of the shaft, as shown in Fig (a), then the vibrations are known as longitudinal vibrations. In this case, the shaft is elongated and shortened alternatively and thus the tensile and compressive stresses are induced alternatively in the shaft.
2. Transverse Vibrations: When the particles of the shaft or disc move approximately perpendicular to the axis of the shaft, as shown in Fig (b), then the vibrations are known as Transverse Vibrations. In this case the shaft is straightened and bent alternatively and bending stresses are induced in the shaft.
3. Torsional Vibrations: When the particles of the shaft or disc move in a circle about the axis of the shaft, as shown in Fig (c), then the vibrations are known as Torsional vibrations. In this case, the shaft is twisted and untwisted alternatively and the torsional shear stresses are induced in the shaft.

Note: If the limit of proportionality (i.e., stress proportional to strain) is not exceeded in the three types of vibrations, then the restoring force in longitudinal and transverse vibrations or the restoring couple in torsional vibrations which is extended on the disc by the shaft ( due to the stiffness of the shaft) is directly proportional to the displacement or mean position. Hence it follows that the acceleration towards the equilibrium position is directly proportional to the displacement from that position and the vibration is, therefore, simple harmonic.

### Natural Frequency of Free Longitudinal Vibrations

The natural frequency of the free longitudinal vibrations may be determined by the following three methods:

1. Equilibrium Method:

Consider a constraint (i.e spring) of negligable mass in an unstained position, as shown in Fig (a).,
Let       s = Stiffness of the constraint. It is the force required to produce unit displacement in the direction of vibration. It is usually expressed in N/m.
m = Mass of the body suspended from the constraint in kg,
W = Weight of the body in newtons = m.g,
δ = Static deflection of the spring in metres due to weight W newtons, and
x = Displacement given to the body by the external force, in metres.
In the equilibrium position, as shown in Fig (b), the gravitational pull W = m.g, is balanced by a force of spring, such that W = s. δ.

Since the mass is now displaced from its equilibrium position by a distance 'x' , as shown in Fig. (c), and is then released, therefore after time 't',
Where      δ = Static deflection i.e. extension or compression of the constraint,
W = Load attached to the free end of constraint,
l = Length of the constraint,
E = Young’s modulus for the constraint, and
A = Cross-sectional area of the constraint.

2.  Energy method:

We know that the kinetic energy is due to the motion of the body and the potencil energy is with respect to a certain datum position which is equal to the amount of work required to move the body from the datam position. In the case of vibrations, the datam position is the mean or equilibrium position at which th potencial energy of the body or the system is zero.
In the free vibrations, no energy is transferred to the system or from the system. Therefore the summation of kinetic energy and potencial energy must be a constant quantity which is same at all the times. In other words,
The time period and the natural frequency may be obtained as discussed in the previous method.

3. Rayleigh’s method:

In this method, the maximum kinetic energy at the mean position is equal to the maximum potencial energy (or strain energy) at the extreme position. Assuming the motion executed by the vibration to be simple harmonic, then

Note:

In all the expressions, 'ωis known as natural circular frequency and is generally denoted by 'ω n' .

## Friday, 23 December 2016

### Natural Frequency of Free Transverse Vibrations

Consider a shaft of negligable mass whose one end is fixed and the other end carries a body of weoght W, as shown in fig.

Let           s = Stiffness of shaft,
δ = Static deflection due to weight of the body,
x = Displacement of body from mean position after time t.
m = Mass of body = W/g As discussed in the previous article,

Restoring force  =     – s.x          . . .                       (i)
and accelerating force  =   m   d2 x               . . . (ii)
dt2
Equating equations (iand (ii), the equation of motion becomes

Hence, the time period and the natural frequntly of the transverse vibrations are same as that of longitudinal vibrations. Therefore

Note :

The shape of the curve, into which the vibrating shaft deflects, is identical with the static deflection curve of a cantiliver beam loaded at the end. It has been proved in the text book on strength of Materials that the static deflection of a cantiliver beam loaded at the free end is

δ  =      Wl3                    (in metres)

3EI

Where        W = Load at the free end, in newtons,
l = Length of the shaft or beam in metres,
E = Young’s modulus for the material of the shaft or beam in N/m2, and
I = Moment of inertia of the shaft or beam in m4.

## Wednesday, 21 December 2016

### Effect of Inertia of the Constraint in Longitudinal and Transverse Vibrations

In deriving the expressions for natural frequency of longitudinal and transverse vibrations, we have neglected the inertia of the constraint i.e. shaft. We shall now discuss the effect of the inertia of the constraint, as below:

1. Longitudinal vibration
Consider the constraint whose one end is fixed and other end is free as shown in Fig.
Let         m1 = Mass of the constraint per unit length,
l = Length of the constraint,
mC = Total mass of the constraint = m1. l, and
v = Longitudinal velocity of the free end.
Consider a small element of the constraint at a distance x from the fixed end and of length δx.
∴     Velocity of the small element
If a mass of mC/3 is placed at the free end and the constraint is assumed to be negligible mass, then
Total kinetic energy possesed by the constraint
Hence the two systems are dynamically same. Therefore, inertia of the constraint may be allowed for by ading one-third of its mass to the disc at the free end.
From the above discussion, we find that when the mass of the constraint mC and the mass of the dosc 'm' at the end is given, then natural frequency of vibration,

2. Transverse vibration

Consider a constraint whose one end is fixed and the other end is free as shown in Fig..

Let             m1 = Mass of the constraint per unit length,
l = Length of the constraint,
mC = Total mass of the constraint = m1. l, and
v = Longitudinal velocity of the free end.
Consider a small element of the constraint at a distance from the fixed end and of length δx . The velocity of this element is given by
If a mass of 33mC/140 is placed at the free end and the constraint is assumed to be negligible mass, then
Total kinetic energy possesed by the constraint
Hence the two systems are dynamically same. Therefore, inertia of the constraint may be allowed for by adding 33/140 of its mass to the disc at the free end.
From the above discussion, we find that when the mass of the constraint mC and the mass of the dosc 'm' at the end is given, then natural frequency of vibration,

Notes:

1. If both the ends of the constraint are fixed, and the disc is situated in the middle of it, then proceeding in the similar way as discussed above, we may prove that the inertia of the constraint may be allowed for by adding 13/35 of its mass to the disc.
2. If the constraint is like a simply supported beam, then 17/35 of its mass may be added to the mass of the disc.

### Natural Frequency of Free Transverse Vibrations Due to a Point Load Acting Over a Simply Supported Shaft

Consider a light shaft AB of length 'l' between supports, carrying a point load W at C which is at a distance of lfrom A and lfrom B, as shown in Fig., A little consideration will show that when the shaft is deflected and suddenly released, it will make (or undergo) transverse vibrations. The deflection of the shaft is proportional to the load W which is acting on it and if the shaft is deflected beyond the static equilibrium position under load W, then the load along with the beam will vibrate with simple harmonic motion (S.H.M) (as by helical spring). If 'δ' is the static deflection due to load 'W', then the natural frequency of the transverse vibration is
ωn = g/δ

## Tuesday, 20 December 2016

### static deflection for the various types of beams under various load conditions

Some of the values of the static deflection for the various types of beams under various load conditions are given in the following table.

Values of static deflection (δ) for the various types of beams and under various load conditions

## Friday, 18 November 2016

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

Consider a shaft AB carrying a uniformly distributed load of 'w' per unit length as shown in Fig.,
Let           y1 = Static deflection at the middle of the shaft,
a= Amplitude of vibration at the middle of the shaft, and
w= Uniformly distributed load per unit static deflection at the middle of the shaft = w/y1 .

Now consider a small section of the shaft at a distance 'x' from A and length 'δ x'.
Let        y = Static deflection at a distance 'x' from A, and length 'δ x'
a = Amplitude of its vibration.
∴ Work done on this small section
Since the maximum potential energy at the extreme position is equal to the amount of work done to move the beam from the mean position to one of its extreme position therefore
Maximum potential energy at the extreme position
Assuming that the shape of the curve of a vibrating shaft is similar to the static deflection curve of a beam, therefore
Substituting those values in equation (i), we have maximum potential energy at the extreme position
Since the maximum velocity at the mean position is 'ω.a1', where 'ω'  is the circular frequency of vibration, therefore
Maximum kinetic energy at the mean position
We know that the maximum potential energy at the extreme position is equal to the maximum kinetic energy at the mean position, therefore equating equations (iiand (iii),
When the shaft is simply supported, then the static deflection at a distance 'x' from A is
Where          W = Uniformly distributed load per unit length,
E = Young's modulus or the material o the shaft, and
I = Moment o inertia of the shaft.
Now integrating the above equation (v) within the limits 0 to l,
Substituting the value in equation (ivfrom equation (vi) and (vii), we get circular frequency due to uniformly distributed load,
Natural frequency due to uniformly distributed load,
We know that the static defection of a simply supported shaft due to uniformly distributed load of  'w' per unit length, is
Equation (ixmay be written as,

## Thursday, 17 November 2016

### Natural Frequency of Free Transverse Vibrations of a Shaft Fixed at Both Ends Carrying a Uniformly Distributed Load

Consider a shaft AB fixed at both ends and carrying a uniformly distributed load 'w' per unit length as shown in Fig.,
We know that the static deflection at a distance 'x' from A is given by
Integrating the above equation on both sides  with limits from 0 to l,
Now integrating  y2  within the limits from 0 to l,
We know that
and natural frequency,
Since the static deflection of a shaft fixed at both ends and carrying a uniformly distributed load is

### Natural Frequency of Free Transverse Vibrations For a Shaft Subjected to a Number of Point Loads

Consider a shaft AB of negligible mass loaded with point loads W1 , W2,  Wand W4 etc., as shown in Fig (a)., Let mmm3 and metc., be the corresponding masses in Kg. The natural frequency of such a shaft may be found out by the following two methods:

1. Energy (or Rayleigh’s) method

Let y, y, y3 and yetc. be total deflection under loads W1 , W2W and W4 etc., as shown in Fig (a).,
We know that maximum potencial energy
and maximum kinetic energy
Equating the maximum potencial energy and maximum kinetic energy we have,
Natural frequency of transverse vibration

2. Dunkerley’s method

The natural frequency of transverse vibration for a shaft carrying a number of point loads and uniformly distributed from Dunkerley's empirical formula. According to this

Where          fn = Natural frequency o 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 off the uniformly distributed load (or due to the mass of the shaft).

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

Let δ1 δδetc., = Static deflection due to the load W1 , W2W etc., when considered seperately
δS = Static deflection due to the uniformly distributed load or due to the mass of the shaft.

We know that natural frequency of transverse vibration due to the load W1 ,
Similarly, natural frequency of transverse vibration due to the load W2 ,
and, natural frequency of transverse vibration due to the load W3 ,
Also natural frequency of transverse vibration due to uniformly distributed load or weight of the shaft,
Therefore, according to Dunkerly's empirical formula, the natural frequency of the whole system,

Note: 1. When there is no uniformly distributed load or mass of the shaft is negligible, then δS  = 0.
2. The value of etc., for a simply supported shaft may be obtained from the relation

Where    δ= Static deflection due to load W,
a and b = Distance of the load from the ends,
E = Young's modulus for the material of the shaft,
I = Moment of inertia of the shaft, and
l = Total length of the shaft.