Option 4 : Amplitude

__Explanation:-__

Critical or whirling speed of a shaft

- When the rotational speed of the system coincides with the natural frequency of lateral/transverse vibrations, the shaft tends to bow out with a
**large amplitude.**This speed is termed as critical/whirling speed. - Whirling speed or Critical speed of a shaft is defined as the speed at which a rotating shaft will tend to vibrate violently in the transverse direction if the shaft rotates in the horizontal direction.
- In other words, the whirling or critical speed is the speed at which resonance occurs.
- Hence we can say that Whirling of the shaft occurs when the natural frequency of transverse vibration matches the frequency of a rotating shaft.
- It is 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.

Deflection of the shaft due to transverse vibration of the shaft.

\(y = \frac{e}{{{{\left( {\frac{{{\omega _n}}}{\omega }} \right)}^2} - 1\;}}\)

At critical speed,

\(\omega = {\omega _n} = \sqrt {\frac{k}{m}} = \sqrt {\frac{g}{\delta }} \)

where,

ω = Angular velocity of the shaft, k = Stiffness of shaft, e = initial eccentricity of the center of mass of the rotor

m = mass of rotor, y = additional of rotor due to centrifugal force.

Option 2 : ω_{2} > ω_{1}

__Concept:__

The natural frequency is given by:

\(\omega = \sqrt {\frac{k_{eq}}{m}} =\sqrt {\frac{g}{\delta }} \;\)

__ Calculation__:

__ Given__:

L_{1} = L, \(L_2=\frac{L_1}{2}\)

For cantilever beam:

\(\begin{array}{l} \delta = \frac{{P{L^3}}}{{3EI}}\\ \omega = \sqrt {\frac{g}{\delta }} = \sqrt {\frac{{3EIg}}{{P{L^3}}}} \Rightarrow \omega \propto \sqrt {\frac{1}{{{L^3}}}} \\ \frac{{{\omega _2}}}{{{\omega _1}}} = \sqrt {\frac{{L_1^3}}{{L_2^3}}} = \sqrt {\frac{{{L^3}}}{{{{\left( {\frac{L}{2}} \right)}^3}}}} = 2\sqrt2 \\ \therefore\;{\omega _2} > {\omega _1} \end{array}\)

If two nodes are observed at a frequency of 1800 rpm during whirling of a simply supported long slender rotating shaft, the first critical speed of the shaft in rpm is

Option 1 : 200

**Concept:**

**Whirling frequency** of shaft is,

f = n^{2} f_{c }

where, n = no. of loop (node + 1), f_{c} = critical frequency

Since it is simply supported critical speed and first frequency = f_{n}

**Calculation:**

__Given:__

N = 1800 rpm (for 2 node), n = 3, Nc = ?

1800 = 3^{2} × (N_{c})

**∴ N _{c} = 200 rpm**

Option 2 : \(\frac{{1}}{{f_n^2}}=\frac{{1}}{{f_{n_1}^2}}~+~\frac{{1}}{{f_{n_2}^2}}~+~\frac{{1}}{{f_{n_3}^2}}~+~...+\;\frac{{1}}{{f_{n_s}^2}}\)

**Explanation:**

Dunkerley's method is semi-empirical and gives approximate results but is simple from the point of view of calculation.

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:

\(\frac{1}{{{{\left( {{f_n}} \right)}^2}}} = \frac{1}{{{{\left( {{f_{n1}}} \right)}^2}}} + \frac{1}{{{{\left( {{f_{n2}}} \right)}^2}}} + \frac{1}{{{{\left( {{f_{n3}}} \right)}^2}}} + \ldots + \frac{1}{{{{\left( {{f_{ns}}} \right)}^2}}}\)

Option 2 : Transverse vibration

__Explanation:__

**Whirling speed**or**Critical speed**of a shaft is defined as the speed at which a rotating shaft will tend to vibrate violently in the transverse direction if the shaft rotates in the horizontal direction.- In other words, the whirling or critical speed is the speed at which resonance occurs.
- Hence we can say that
**Whirling of the shaft occurs when the natural frequency of transverse vibration matches the frequency of a rotating shaft.**

Option 3 : diameter and span of the shaft:

__Explanation:-__

Critical or whirling speed of a shaft

- When the rotational speed of the system coincides with the natural frequency of lateral/transverse vibrations, the shaft tends to bow out with a large amplitude. This speed is termed as critical/whirling speed.
- Whirling speed or Critical speed of a shaft is defined as the speed at which a rotating shaft will tend to vibrate violently in the transverse direction if the shaft rotates in the horizontal direction.
- In other words, the whirling or critical speed is the speed at which resonance occurs.
- Hence we can say that Whirling of the shaft occurs when the natural frequency of transverse vibration matches the frequency of a rotating shaft.
- It is 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.

Deflection of the shaft due to transverse vibration of the shaft.

\(y = \frac{e}{{{{\left( {\frac{{{\omega _n}}}{\omega }} \right)}^2} - 1\;}}\)

At critical speed,

\(\omega = {\omega _n} = \sqrt {\frac{k}{m}} = \sqrt {\frac{g}{\delta }} \)

where, ω = Angular velocity of the shaft, k = Stiffness of shaft, e = initial eccentricity of the center of mass of the rotor, m = mass of rotor, y = additional of rotor due to centrifugal force.

As the shaft is loaded centrally, it acts as a simply supported beam with a point load at centre:

Deflection is this case is given by:

\({\delta} = \frac{{P{L^3}}}{{48EI}}\)

if the shaft is circular, I = \(\frac{\pi d^4}{64}\)

therefore putting values in equation 1, we will come to know that critical speed d**epends upon the span and diameter of the shaft.**

Option 2 : 5367 rpm

**Concept:**

Let the Transverse natural frequency of rotor is f_{n},

\(\frac{1}{{{{\left( {f_n} \right)}^2}}} = \frac{1}{{{{\left( {{f_1}} \right)}^2}}} + \frac{1}{{{{\left( {{f_2}} \right)}^2}}} + \ldots \)

And we know that,

ω_{n} = 2πf_{n}

⇒ \(\frac{2\pi N}{60}=2\pi f_n\)

∴ N = 60f_{n}

**Calculation:**

**Given:**

f_{1} = 100 Hz, f_{2} =200 Hz

\(∴ \frac{1}{{{{\left( {{f_n}} \right)}^2}}} = \frac{1}{{{{\left( {100} \right)}^2}}} + \frac{1}{{{{\left( {200} \right)}^2}}}\)

\({\left( {{f_n}} \right)^2} = 8000\)

f_{n} = 89.44 Hz

Angular velocity

∴ N = 60 f_{n}

N = 60 × 89.44

N = 5366.56 rpmOption 3 : 90

**Concept:**

Because of the excitation force, there will be excitation in the system of damped vibration. So there will be a lag in the response of the damped vibrating system. After reaching the steady-state, there will be the only steady-state response as written by equation (1). The phase difference between the response of the system (x) and excitations (F0sin(ωt)) can be derived from the diagram below,

Where,

F = excitation force, KA = Maximum spring force, CAω = Maximum damping force (leading by 90° from spring force response as this proportional to velocity)

MAω2 = Maximum Inertial force (leading by 180° from spring force response as this is proportional to inertial acceleration)

From ΔOab,

\({{tan}}\phi = \frac{{{{CA\omega }}}}{{{{KA}} - {{MA}}{{{\omega }}^2}}}\)

\( ⇒ \phi = \;{\tan ^{ - 1}}(\frac{{{{C\omega }}}}{{{{K}} - {{M}}{{{\omega }}^2}}})\)

\(\Rightarrow\phi ={{\tan }^{-1}}\left( \frac{2\xi \frac{\omega }{{{\omega }_{n}}}}{1-{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}} \right){{\;\;\;\;\;}} \ldots \left( 1 \right)\)

For ω/ωn > 1 the phase will be 180˚ as the value 1- (ω/ωn)2 will become negative which means out of phase.

- Phase angle varies from zero at low frequencies to 180° at very high frequencies. It changes very rapidly near the resonance and is 90° at resonance irrespective of damping.
**The phase angle between 0 to 90° for all the values of ω/ωn = 0 to 1**and between 90° to 180° for all the values of ω/ωn > 1- In absence of any damping, the phase angle suddenly changes from zero to 180° at resonance.
- The phase angle φ is zero for all the values of damping ratio at ω/ωn = 0

**Calculation:**

At resonance, (ω/ωn) = 1

Using equation (1),

\(\Rightarrow\phi ={{\tan }^{-1}}\left( \frac{2\xi \frac{\omega }{{{\omega }_{n}}}}{1-{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}} \right)=\tan^{-1}(\infty)=90^\circ\)

So, the phase angle is 90o.

Option 2 : occurs when excitation frequency is less than undamped natural frequency

__Explanation:__

Magnification Factor:

\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right)}^2} + {{\left( {2ξ \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)

\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}} }}\)

where \(r=\dfrac{\omega}{\omega_n}\).

Thus M.F = f(r, ξ)

During resonance, the amplitude of vibration is maximum i.e Magnification factor is maximum and for an underdamped harmonic oscillator i.e. (ξ ≠ 0) it happens just before the frequency ratio reaches unity as can be seen from the graph given below.

Thus for an underdamped oscillator, resonance or rather point of maximum amplitude occurs when the excitation frequency is less than the undamped natural frequency.

Option 1 : 6040

k_{1} = k_{2} = 16 MN/m

k_{3} = k_{4} = 32 MN/m, m = 240 kg

springs are parallel, so \({k_{eq}} = {k_1} + {k_2} + {k_3} + {k_4} = 16 + 16 + 32 + 32 \Rightarrow {K_{eq}} = 96\frac{{MN}}{m}\)

we know that at resonance \(\omega = {\omega _n} = \sqrt {\frac{k}{m}} \Rightarrow \frac{{2\pi N}}{{60}} = \sqrt {\frac{{{k_{eq}}}}{m}}\)

\(\Rightarrow N = \frac{{60}}{{2\pi }}\times \sqrt {\frac{{96 \times {{10}^6}}}{{240}}} = 6039.5 \simeq 6040\ rpm\)

Option 4 : natural frequency of the shaft

__Explanation:__

The bending of a rotating shaft depends upon the eccentricity of the center of mass of the rotor and the speed at which the shaft rotates.

Fig shows a rotor having a mass m attached to a shaft.

Where s = stiffness of the shaft, e = initial eccentricity of the center of mass of the rotor, m = mass of the rotor, y = additional deflection of rotor due to centrifugal force, and ω = angular velocity of the shaft.

Now, equating the centrifugal force m(y+e)\({\omega ^2}\) with the force resisting the deflection y, we get

y = **\(\frac{e}{{{{\left( {\frac{{{\omega _n}}}{\omega }} \right)}^2} - 1}}\)**

Where ω_{n} is the natural angular frequency of the shaft.

Now, when ω = ω_{n}, the deflection y is infinitely large (resonance occurs) and the angular speed ω is called the critical speed. i.e. ω_{c} = ω_{n} = **\(\sqrt {\frac{s}{m}} \)** = **\(\sqrt {\frac{g}{\Delta }} \)**

Hence as we can see the **critical speed is an angular speed not any angle of rotation**, so its unit would be the same as the unit of angular speed.

That’s why options 1, 2 and 3 are not correct. **Option 4 is the correct option**.

Option 4 : 5367 r.p.m

__Concept:__

Using Dunkerley’s method

\(\frac{1}{{ω _n^2}} = \frac{1}{{ω _{n1}^2}} + \frac{1}{{ω _{n2}^2}}\)

**Calculation:**

**Given: **ω_{n1} = 100 rpm, ω_{n2 }= 200 rpm

**\(\frac{1}{{ω _n^2}} = \frac{1}{{ω _{n1}^2}} + \frac{1}{{ω _{n2}^2}}\)**

\(= \frac{1}{{{{\left( {100} \right)}^2}}} + \frac{1}{{{{\left( {200} \right)}^2}}} = \frac{1}{{8000}}\)

⇒ ωn = 89.44 cycle/s

∴ N = 89.44 × 60

**∴ N = 5366.56 rpm**

Option 2 : k_{1}m_{2} = m_{1}k_{2}

**Concept:**

For dynamics vibration absorber system under tuned condition,

Frequency is equal i.e.

**f _{1} = f_{2}**

\(\sqrt {\frac{{{k_1}}}{{{m_1}}} = } \sqrt {\frac{{{k_2}}}{{{m_2}}}} \)

Option 1 : whipping speed

__Concept:__

**Whipping speed: **

- It is the speed at which the shaft runs, so that the deflection of the shaft from the axis of rotation becomes infinite.
- This speed is also known as critical or whirling speed. Its depends on eccentricity of the centre of gravity of rotating mass from the axis of rotation of the shaft

Option 1 : Transverse direction

__Explanation:__

Critical or whirling speed of the shaft

Whirling speed or Critical speed of a shaft is defined as the speed at which a rotating shaft will tend to vibrate violently in the **transverse direction** if the shaft rotates in the horizontal direction. In other words, the whirling or critical speed is the speed at which resonance occurs.

__Additional Information__

- When the natural frequency of the system coincides with the external forcing frequency, it is called
**resonance**. - The speeds at which resonance occurs are known as the
**critical speeds/ whirling speeds or whipping**. At this speed, the shaft runs so that the deflection of the shaft from the axis of rotation becomes infinite and the shaft tends to vibrate violently in the transverse direction.

The rotor shaft of a large electric motor supported between short bearings at both the ends shows a deflection of 1.8 mm in the middle of the rotor. Assuming the rotor to be perfectly balanced and supported at knife edges at both ends, the likely critical speed (in rpm) of the shaft is

Option 2 : 705

**Concept:**

**Critical speed:**

\(\omega_c=\omega_n=\sqrt{\frac{s}{m}}=\sqrt{\frac{g}{∆}}\)

**Calculation:**

**Given:**

Δ = 1.8 mm

\({\omega_c} = \sqrt {\frac{g}{\delta }} \Rightarrow \frac{{2\pi {N_c}}}{{60}} = \sqrt {\frac{g}{\delta }}\)

\(\Rightarrow {N_c} = \frac{{60}}{{2\pi }}\sqrt {\frac{{9.81}}{{1.8 \times {{10}^{ - 3}}}}} = 704.98 \approx 705\ rpm\)

Option 4 : 2990 rpm

__Concept:__

The critical speed of shaft**:** The theoretical angular velocity which excites the natural frequency of a rotating object, such as a shaft. As the speed of rotation approaches the objects natural frequency, resonance occurs regardless of orientation.

It is given by

\(N_c = \frac 1 {2\pi} \sqrt{\frac {g}{δ }} \)

Where, g = Acceleration due to gravity; δ = Deflection;

__Calculation:__

__Given:__

m = 100 kg; δ = 0.1 mm; g = 9.81 m/s2

\(N_c = \frac 1 {2\pi} \sqrt{\frac {g}{δ }} \) \(= \frac {1}{2\pi } \sqrt {\frac {9.81}{0.1\ \times \ 10^{-3}}} =\) 49.84 rps

\(N_c\) = 2990.92 ~ 2990 rpm

\(N_c = \frac 1 {2\pi} \sqrt{\frac {g}{δ }} \)

In above formula Nc has unit of revolution per second (rps). Do remember to convert it into RPM.

The Smallest and largest natural frequencies of a 'n' degree freedom system are ω_{1} and ω_{n} respectively. The approximate natural frequency estimated by Rayleigh’s and Dankerley’s methods is ω_{r} and ω_{d} respectively. Which of the following statements is true?

Option 4 : ωr > ω1 and ωd < ω1

** Explanation**:

**Degree of freedom**:

- The number of independent coordinates required to describe a vibratory system is known as the degree of freedom.
- A simple spring-mass system or a simple pendulum oscillating in one plane are examples of a single degree of freedom.
- A two-mass, two-spring system, constrained to move in one direction, or a double pendulum belongs to two degrees of freedom.

**Shaft carrying multiple loads (Multiple degrees of freedom)**:

There are two methods to find natural frequencies of the system:

**Dunkerley's method (ω**(Approximate results which are_{d})**less than the actual natural frequency**of the system).**Rayleigh Method (ω**(Gives accurate results that are_{r}) or Energy method**slightly greater than the actual natural frequency**of the system)

Given that the smallest natural frequency of the system is ω_{1}, therefore as per the definition of two methods mentioned, **ω _{r} > ω_{1} and ω_{d} < ω_{1}._{ }**

Which of the following are responsible for occurrence of critical or whirling speed of shaft?

1. Eccentric mounting of the rotor

2. Non-uniform distribution of rotor material

3. Bending of shaft due to the weight of the rotor and the shaft itself

4. Environmental effect such as effect of moisture and temperature

Select the correct answer using the code given below.

Option 2 : 1, 2 and 3 only

A critical speed of a rotating shaft is the speed at which the shaft starts to vibrate violently in the transverse direction. Critical speed of shaft is also called whirling speed or whipping speed of shaft. The whirling of the shaft results from mass unbalance of the rotating system. The environment has no effect on whirling speed of shaft.

Option 3 : The speed which equals the natural frequency of the rotor

Turbine shafts which have critical speed less than the normal operating speed are known as flexible shaft and those turbine shafts whose operating speed is less than the critical speed are known as rigid shafts.

If the rotor speed is more than the critical speed of the shaft then it is passed quickly so that there is no time for deflection to grow.