Mechanical Vibrations MCQ Quiz - Objective Question with Answer for Mechanical Vibrations - Download Free PDF

Explanation:

Flexible Shaft

• A flexible shaft is a type of mechanical component that is specifically designed to transmit rotary motion and torque efficiently between two points that may not be perfectly aligned. It consists of a flexible core that allows it to bend and twist, making it suitable for applications where rigid shafts would be impractical or impossible to use.
• The flexible shaft is made up of a series of tightly wound helical coils, which provide the necessary flexibility and strength to transmit torque. These coils allow the shaft to bend and adapt to changes in alignment while maintaining its ability to rotate. The design ensures low bending rigidity while maintaining relatively high torsional rigidity, allowing the shaft to twist without significant loss of torque transmission.

Advantages:

• Ability to transmit motion and torque around obstacles or in confined spaces.
• Lightweight and compact design compared to rigid shafts.
• High torsional strength while being flexible in bending.
• Reduces the need for precise alignment of components.

Disadvantages:

• Not suitable for applications requiring extremely high torque or bending loads.
• May experience wear and tear over time due to constant bending and twisting.
• Limited in length and load-carrying capacity compared to rigid shafts.

Applications: Flexible shafts are commonly used in applications such as power tools, automotive speedometers, dental equipment, and various industrial machines where motion needs to be transmitted in a flexible and adaptable manner.

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:

The circular frequency (natural frequency in rad/s) for undamped free vibrations is given by:

\( \omega_n = \sqrt{\frac{k}{m}} \)

Where,

• \( k \) = stiffness of the spring (in N/m)
• \( m \) = mass (in kg)

Given:

Mass, \( m = 200~\text{kg} \)

Stiffness, \( k = 80~\text{N/mm} = 80 \times 10^3~\text{N/m} \)

Calculation:

\( \omega_n = \sqrt{\frac{80 \times 10^3}{200}} = \sqrt{400} = 20~\text{rad/s} \)

Concept:

When a mechanical system is subjected to harmonic excitation, the response of the system is influenced by the excitation frequency and the damping present.

The ratio of the amplitude of the system's response to the static deflection under the same force is called the Dynamic Magnification Factor (DMF).

General Formula:

\( \text{DMF} = \frac{1}{\sqrt{(1 - r^2)^2 + (2\xi r)^2}} \)

Where,
\( r = \frac{\omega}{\omega_n} \) is the frequency ratio
\( \xi \) is the damping ratio.

At Resonance:

At resonance, excitation frequency equals natural frequency, i.e., \( \omega = \omega_n \Rightarrow r = 1 \).

Substitute in the equation:

\( \text{DMF}_{\text{resonance}} = \frac{1}{\sqrt{(1 - 1)^2 + (2\xi)^2}} = \frac{1}{2\xi} \)

Explanation:

Dunkerley's Empirical Formula for Natural Frequency of Transverse Vibration:

• Dunkerley's empirical formula is a crucial method used to determine the natural frequency of transverse vibration for a shaft carrying a number of point loads and uniformly distributed load (UDL). This formula is particularly useful in mechanical and structural engineering applications where predicting the vibration characteristics of shafts is essential for ensuring stability and preventing resonance.
• The natural frequency of a system is the frequency at which it tends to oscillate in the absence of any driving or damping force. Dunkerley's formula provides a simplified approach to estimate the natural frequency of a complex system by considering the contributions of individual components separately.

Given:

Let \( f_n \) = Natural frequency of the complete system

\( f_{n1}, f_{n2}, f_{n3}, \dots, f_{ns} \) = Natural frequencies due to individual point loads and UDL

Calculation:

According to Dunkerley's empirical formula, the natural frequency is given by:

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

Explanation:

At resonance (ω = ω_n), phase angle ϕ is 90°.

When ω/ω_n ≫ 1; the phase angle is very close to 180°. Here inertia force increases very rapidly and its magnitude is very large.

Concept:

In vibration system, transmissibility is given by

\(\epsilon =\frac{{{F}_{T}}}{{{F}_{O}}}\)

Where F_T = maximum force transmitted to the ground

\(\epsilon =\frac{\sqrt{1+{{\left( 2\xi \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}}}{\sqrt{{{\left( 1-{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}} \right)}^{2}}+{{\left( \frac{2\xi \omega }{{{\omega }_{n}}} \right)}^{2}}}}\)

Natural frequency \({{\omega }_{n}}=\sqrt{\frac{K}{m}}\)

K = spring stiffness, m = mass

Calculation:

Given, Harmonic excitation force: F(t) = F₀cos(ωt)

F_T = F_O

\(\epsilon =\frac{{{F}_{T}}}{{{F}_{O}}}\)

∴ Transmissibility, ϵ = 1

Also, \(\epsilon =\frac{\sqrt{1+{{\left( 2\xi \frac{w}{{{w}_{n}}} \right)}^{2}}}}{\sqrt{{{\left( 1-{{\left( \frac{w}{{{w}_{n}}} \right)}^{2}} \right)}^{2}}+{{\left( \frac{2\xi w}{{{w}_{n}}} \right)}^{2}}}}=1\)

Let \(\frac{\omega }{{{\omega }_{n}}}=x\)

So, 1 + (2ξ x)² = (1 – x²)² + (2ξ x)²

(1 – x²)² = 1

1 – x² = ± 1

Taking +ve sign x² = 0

∴ Taking –ve sign

x² = 2

Putting value of x:

(ω/ω_n)² = 2

\(\frac{\omega }{{{\omega }_{n}}}=\sqrt{2}\)

∴ ω = √2ω_n

∵ Natural frequency \({{\omega }_{n}}=\sqrt{\frac{K}{m}}\)

Concept:

Springs in series:

\(\frac{1}{{{k_e}}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} \)

Springs in parallel:

ke = k1 + k2

Calculation:

Given:

Springs with spring constant k and k are connected in parallel,

Let their equivalent spring constant be k1

k1 = k + k

k1 = 2k

Now k1, k and k are in series connection, let their equivalent spring constant is keq

\(\frac{1}{{{k_{eq}}}} = \frac{1}{{k_e}} + \frac{1}{{k}} + \frac{1}{{k}} \)

\(\frac{1}{{{k_{eq}}}} = \frac{1}{{2k}} + \frac{1}{{k}} + \frac{1}{{k}} \)

\(\frac{1}{k_{eq}}=\frac{1+2+2}{2k}\)

\(\frac{1}{k_{eq}}=\frac{5}{2k}\)

\(k_{eq}=\frac{2k}{5}\)

k_eq = 0.4 k

Concept:

\(Time~ Period= T =2\pi √{\frac Lg}\)

where L = Length of the pendulum, g = acceleration due to gravity

Calculation:

Given:

at moon 

\(g'=\frac g6\)

where g' = acceleration due to gravity at the moon

\(New~Time~ Period= T' =2\pi √ {\frac L{g'}}=2\pi √ {\frac {6L}g}\)

 \(T'=√ {6}T\)

As time period for oscillation increases √6 times, therefore speed becomes √6 times slower.

Concept:

When spring is connected in parallel as shown, the equivalent stiffness is the sum of all individual stiffness of spring.

k_eq = k₁ + k₂

The natural frequency ω_n of a spring-mass system is given by:

\(\omega_n=\sqrt{\frac{k_{eq}}{m}}\;\;\; and\;\;\;\omega_n=2\pi f\)

k_eq = equivalent stiffness and m = mass of body.

Calculation:

Given:

k₁ = 4000 N/m, k₂ = 1600 N/m and m = 1.4 kg

Springs are in parallel correction

∴ k_eq = k₁ + k₂ = 4000 + 1600 = 5600 N/m

\(ω_n = \sqrt {\frac{k_{eq}}{{m}}}\Rightarrow \sqrt {\frac{{5600}}{{1.4}}} = \sqrt {4000} \)

\(f = \frac{ω_n}{{2\pi }} \Rightarrow \frac{{\sqrt {4000} }}{{2\pi }} \approx 10\;Hz\)

Explanation:

In the vibration isolation system, the ratio of the force transmitted to the force applied is known as the isolation factor or transmissibility ratio.

\(T = \frac{{{F_T}}}{{{F_O}}} = \frac{{\sqrt {1 + {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}{{\sqrt {\left[ {\left\{1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2} \right\}^2+ {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]} }}\)

• When ω/ωn = 0 ⇒ TR = 1, (independent of ζ)
• When ω/ωn = 1 and ξ = 0 ⇒ TR = ∞, (independent of ζ)
• When frequency ratio ω/ωn = √2, then all the curves pass through the point TR = 1 for all values of damping factor ξ.
• When frequency ratio ω/ωn < √2, then TR > 1 for all values of damping factor ξ. This means that the force transmitted to the foundation through elastic support is greater than the force applied.

When frequency ratio ω/ωn > √2, then TR < 1 for all values of damping factor ξ. This shows that the force transmitted through elastic support is less than the applied force. Thus vibration isolation is possible only in the range of ω/ωn > √2. Here the force transmitted to the foundation increases as the damping is increased.

Explanation:

Vibration: A body is said to be vibrating if it has a to and fro motion.

Various types of vibrations are mentioned in the table below.

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.

Concept:

Transmissibility is given by:

\(\varepsilon = \frac{{\sqrt {1 + {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}{{\sqrt {{{\left[ {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]}^2} + {{\left[ {2\zeta \frac{\omega }{{{\omega _n}}}} \right]}^2}} }}\)

\(C = 2\zeta \sqrt {KM} \Rightarrow \zeta = \frac{C}{{2\sqrt {KM} }}\)

Calculation:

After additional damper in parallel \(\zeta ' = \frac{{C + C'}}{{2\sqrt {KM} }}\)

Thus  \(\zeta ' > \zeta \)

Now,  \({\omega _d} = {\omega _n}\sqrt {1 - {\zeta ^2}} \)  

 \(\zeta ' > \zeta \)indicating that \(\omega _d' < {\omega _d}\)

\(time\ period\left( {{T_d}} \right) = \frac{{2\pi }}{{{\omega _d}}}\)

Concept:

In the vibration isolation system, the ratio of the force transmitted to the force applied is known as the isolation factor or transmissibility ratio.

\(T = \frac{{{F_T}}}{{{F_O}}} = \frac{{\sqrt {1 + {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}{{\sqrt {\left[ {\left\{1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2} \right\}^2+ {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]} }}\)

• When ω/ω_n = 0 ⇒ TR = 1, (independent of ζ)
• When ω/ω_n = 1 and ξ = 0 ⇒ TR = ∞, (independent of ζ)
• When frequency ratio ω/ω_n = √2, then all the curves pass through the point TR = 1 for all values of damping factor ξ.
• When frequency ratio ω/ω_n < √2, then TR > 1 for all values of damping factor ξ. This means that the force transmitted to the foundation through elastic support is greater than the force applied.
• When frequency ratio ω/ω_n > √2, then TR < 1 for all values of damping factor ξ. This shows that the force transmitted through elastic support is less than the applied force. Thus vibration isolation is possible only in the range of ω/ω_n > √2. Here the force transmitted to the foundation increases as the damping is increased.
• If ω/ωn > √2 then transmissibility increases with an increase in damping if ω/ωn < √2 then transmissibility decreases with an increase in damping