Resonance and Whirling MCQ Quiz - Objective Question with Answer for Resonance and Whirling - Download Free PDF

Explanation:

Whirling Speed of a Shaft

• The whirling speed of a shaft, also known as critical speed, is the speed at which the shaft begins to vibrate excessively due to resonance. This phenomenon occurs when the rotational speed of the shaft coincides with the natural frequency of its transverse vibration. At this speed, the dynamic forces acting on the shaft amplify the vibrations, which can lead to structural failure or significant damage if not addressed.
• A rotating shaft can experience various types of vibrations, such as transverse, torsional, and longitudinal vibrations. Among these, transverse vibration is the most critical in the context of whirling speed. As the shaft rotates, any imbalance in mass or external forces can induce vibration. When the rotational speed matches the natural frequency of the transverse vibration mode, resonance occurs, causing the amplitude of vibration to increase drastically.

Importance of Whirling Speed Analysis:

• Helps in designing shafts and rotors to operate safely below or above the critical speed.
• Prevents excessive vibration, which can lead to fatigue, wear, and failure of components.
• Ensures the reliability and longevity of rotating machinery.

Factors Affecting Whirling Speed:

• Shaft geometry: The length, diameter, and material properties of the shaft influence its natural frequency.
• Support conditions: The type and position of bearings or supports affect the boundary conditions, altering the natural frequency.
• Mass distribution: Uneven mass distribution along the shaft can create imbalances, impacting the vibration characteristics.
• External forces: Forces such as aerodynamic loads or magnetic forces can contribute to vibration behavior.

Concept:

For a simply supported shaft with a rotor (concentrated mass) at the center, the critical (whirling) speed is determined using the relation:

$\omega =\sqrt{\frac{g}{\delta }}$

Where, $\delta$ is the static deflection due to the weight of the rotor.

Static deflection at center of simply supported shaft due to point load is given by:

$\delta =\frac{W{L}^3}{48EI}$

Calculation:

Given:

Mass of rotor, $m=12kg,soweight,W=mg=12\times 10=120N$

Span of shaft, $L=1m,EI=4kN\cdot m^2=4000N\cdot m^2$

Substitute in deflection formula:

$\delta =\frac{120\times 1^3}{48\times 4000}=\frac{120}{192000}=\frac{1}{1600}m$

Now, critical speed:

$\omega =\sqrt{\frac{g}{\delta }}=\sqrt{10\times 1600}=\sqrt{16000}=40\sqrt{10}rad/s$

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_d) (Approximate results which are less than the actual natural frequency of the system).
• Rayleigh Method (ω_r) or Energy method (Gives accurate results that are slightly greater than the actual natural frequency of the system)

Given that the smallest natural frequency of the system is ω₁, therefore as per the definition of two methods mentioned, ω_r > ω₁ and ω_d < ω₁. 

Concept:

The natural frequency is given by:

$\omega =\sqrt{\frac{k_{eq}}{m}}=\sqrt{\frac{g}{\delta }}$

Calculation:

Given:

L1 = L, $L_2=\frac{L_1}{2}$

For cantilever beam:

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

Explanation:

Critical Speed Equation (Nc):

All rotating shaft, even in the absence of external load, deflect during rotation. The combined weight of a shaft and wheel can cause deflection that will create resonant vibration at certain speeds, known as Critical Speed.

The equation illustrated below is the Rayleigh-Ritz equation. Good practice suggests that the maximum operating speed should not exceed 75% of the critical speed.

The critical speed,

$N_c=\frac{60}{2\pi }\sqrt{\frac{g}{{\delta }_{st}}}$

Total maximum static deflection (The maximum static deflection (δst) is obtained by adding both the maximum static deflection of the rotating shaft and the load),

⇒ ${\delta }_{st}=\frac{aW{L}^3}{bEI}$

Where,

g = Acceleration due to gravity, E = Yongs Modulus, I = Moment of Inertia about Neutral axis, L = length of the shaft, W = load on the shaft,

a and b = Constants which depends on types of loading and supports

So, Critical speed depends upon the magnitude or location of the load or load carried by the shaft, the length of the shaft, its diameter, and the kind of bearing support.

Additional Information 

• There are two methods used to calculate critical speed, Rayleigh-Ritz and Dunkerley Equation. Both the Rayleigh-Ritz and Dunkerley equation is an approximation to the first natural frequency of vibration, which is assumed to be nearly equal to the critical speed of rotation.
• In general, the Rayleigh-Ritz equation overestimates and The Dunkerley equation underestimates the natural frequency

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.

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.

Concept:

The natural frequency is given by:

$\omega =\sqrt{\frac{k_{eq}}{m}}=\sqrt{\frac{g}{\delta }}$

Calculation:

Given:

L₁ = L, $L_2=\frac{L_1}{2}$

For cantilever beam:

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

Explanation:

Magnification Factor:

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

$M.F={\displaystyle \frac{A}{X_{st}}}={\displaystyle \frac{1}{\sqrt{{\left(1-r^2\right)}^2+{\left(2\xi r\right)}^2}}}$

where $r={\displaystyle \frac{\omega }{{\omega }_n}}$.

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.

Mistake PointsOption 2 is correct, At multiple places, option 3 has been given as the correct answer, which is wrong, because irrespective of the amount of damping, the maximum amplitude occurs before the ratio $r=\frac{\omega }{{\omega }_n}$ reaches unity, which is evident from the plot of magnification factor against the ratio of frequency $\frac{\omega }{{\omega }_n}$. Option 3 is correct for no damping, but the question is for underdamped harmonic oscillator.

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.

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

$y=\frac{{\omega }^2.e}{({\omega }_n{)}^2-{\omega }^2}$

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_d) (Approximate results which are less than the actual natural frequency of the system).
• Rayleigh Method (ω_r) or Energy method (Gives accurate results that are slightly greater than the actual natural frequency of the system)

Given that the smallest natural frequency of the system is ω₁, therefore as per the definition of two methods mentioned, ω_r > ω₁ and ω_d < ω₁. 

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 depends upon the span and diameter of the shaft.

Concept:

Let the Transverse natural frequency of rotor is f_n,

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

And we know that, 

ω_n = 2πf_n

⇒ $\frac{2\pi N}{60}=2\pi f_n$

∴ N = 60f_n

Calculation:

Given:

f₁ = 100 Hz, f₂ =200 Hz  

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

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

f_n = 89.44 Hz

Angular velocity

∴ N = 60 f_n

N = 60 × 89.44

Concept:

Whirling frequency of shaft is,

f = n² 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 = 3² × (N_c)

∴ N_c = 200 rpm

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.