Mechanical Vibrations Test 1

Concept:

∴ Natural frequency 

\(\omega_n = \sqrt {\frac{{{k_{eq}}}}{m}} \)

\(f= \frac{1}{{2\pi }}\sqrt {\frac{{{k_{eq}}}}{m}} \)

Calculation:

Given: k₁ = k₂ = 20 kN/m, m = 100 kg

This combination is a case of parallel combination.

k_eq = k₁ + k₂ = 40 kN/m = 40000 N/m

\(f= \frac{1}{{2\pi }}\sqrt {\frac{{{k_{eq}}}}{m}} \)

\(= \frac{1}{{2\pi }}\sqrt {\frac{{40 \times {{10}^3}}}{{100}}} = \frac{{20}}{{2\pi }} = \frac{{10}}{\pi }\;Hz\)

Concept:

\( {ω _d} = {ω _n}\sqrt {1 - {ξ ^2}} \)

where, ω_n = natural frequency, ξ = damping ratio

Calculation:

Given: C = 0.2 C_c , mass (m) = 50 kg, stiffness (k) = 30 kN/m

\(ξ = \frac{C}{{{C_c}}} = 0.2\)

\({ω _n} = \sqrt {\frac{k}{m}} = \sqrt {\frac{{30 \times {{10}^3}}}{{50}}} = 24.5\;rad/sec\)

Concept:

\(\xi = \frac{C}{{{C_c}}}\)

Equation of motion

\(m\ddot x + c\dot x + kx = 0\)

Calculation:

Comparing, we get

m = 3 kg, C = 9 Ns/m,  k = 27 N/m

\({C_c} = \sqrt {{km}} = 2\sqrt {27 \times 3} = 18\;\;Ns/m\)

\(\therefore \xi = \frac{C}{{{C_c}}} = \frac{9}{{18}}\)

∴ ξ = 0.5

Concept:

\({\rm{Phase\;difference\;}}\left( \phi \right) = {\tan ^{ - 1}}\left( {\frac{{2\xi \left( {\frac{\omega }{{{\omega _n}}}} \right)}}{{1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}}}} \right)\)

At resonance ω  = ω_n

∴ ϕ = tan^-1 (∞) = 90°

Concept:

Logarithmic decreament \(\left( \delta \right) = \frac{{2\pi \xi }}{{\sqrt {1 - {\xi ^2}} }}\)

Calculation:

ξ = 0.3

\(\therefore \delta = \frac{{2\pi\; \times\; 0.3}}{{\sqrt {1\; - \;{{0.3}^2}} }}\)

∴ δ = 1.976

Ratio of 2 successive amplitude is always constant and given as,

\(\frac{{{X_n}}}{{{X_{n + 1}}}} = {e^\delta } = {e^{1.976}}\)

\(\therefore \frac{{{X_n}}}{{{X_{n + 1}}}} = 7.21\)

Equation of motion will be:

\(0.5\;\ddot x + 300\;x = 0\) 

\(0.5\;\ddot x + 300\;x = 0\) 

Natural frequency:

\(\omega = \sqrt {\frac{k}{m}} = \sqrt {\frac{{300}}{{0.5}}} = 24.5\;rad/s\) 

\(f = \frac{\omega }{{2\pi }} = 3.9\;Hz\)

Note:

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

Concept:

Damping ratio \(= \frac{C}{{{C_c}}} \)

where C_c = 2√ km and c = damping coefficient.

Calculation:

\(16\ddot x + 5\dot x + 4x = 0\). 

By comparing this equation with \(m\ddot x + c\dot x + kx = 0\)

\(\begin{array}{l} m\; = 16,\;c = 5,k = 4\\ {C_c} = 2√ {km} \\ {C_c} = 2√ {4 \times 16} = 16 \end{array}\)

∴ Damping ratio \(= \frac{C}{{{C_c}}} = \frac{5}{{16}}\)

Concept:

\(X = \frac{{\frac{{{m_0}r{\omega ^2}}}{k}}}{{\sqrt {{{\left( {1 - {r^2}} \right)}^2} + {{\left( {2\xi r} \right)}^2}} }}{\rm{\;and\;}}{\omega _n} = \sqrt {\frac{k}{{M\;}}\;} \)

For large ω, the ratio ω/ω_n ≫ 1

\(\therefore X = \frac{{{m_0}r}}{M}\)

Calculation:

\(X = \frac{{{m_0}r}}{M}\)

\(X = \frac{{1\; \times \;1}}{{100}}\)

∴ X = 0.01 cm   

Concept:

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

For cantilever beam:

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

Concept:

\({\rm{Transmissibility\;}} = \frac{{{{\left[ {1 + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]}^{1/2}}}}{{{{\left[ {{{\left( {1 - \frac{{{\omega ^2}}}{{\omega _n^2}}} \right)}^2} + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]}^{1/2}}}}\)

Calculation:

ω  = √2 ω_n

\(\therefore Transmissibility = \frac{{{{\left( {1 + 8{\xi ^2}} \right)}^{\frac{1}{2}}}}}{{{{\left( {1 + 8{\xi ^2}} \right)}^{\frac{1}{2}}}}} = 1.0\)

Concept:

F_T is the force transmitted to the foundation. The disturbing force is F. The ratio of F_T to F is called transmissibility.

\(T.R = \frac{{{F_T}}}{F} = \frac{{\sqrt {1 + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}{{\sqrt {{{\left( {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right)}^2} + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)

The steady state amplitude for the system:

\(A = \frac{{\frac{F}{k}}}{{\sqrt {{{\left( {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right)}^2} + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)

Calculation:

Given: m = 10 kg, k = 4000 N/m, C = 40 Ns/m, F = 60 N

ω = 40 rad/sec

\({\omega _n} = \sqrt {\frac{k}{m}} = \sqrt {\frac{{4000}}{{10}}} = 20\;rad/sec\)

Frequency Ratio:

\(r = \frac{\omega }{{{\omega _n}}} = \frac{{40}}{{20}} = 2\)

The critical damping coefficient:

\({c_c} = 2m\;{\omega _n} = 2\sqrt {mk} = 2\sqrt {10 \times 4000} = 400\;N.s/m\)

The damping factor, ξ:

\(\xi = \frac{C}{{{C_c}}} = \frac{{40}}{{400}} = 0.1\)

Transmissibility:

\(T.R = \frac{{{F_T}}}{F} = \frac{{\sqrt {1 + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}{{\sqrt {{{\left( {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right)}^2} + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)

\(T = \frac{{\sqrt {1 + {{\left( {2 \times 0.1 \times 2} \right)}^2}} }}{{\sqrt {{{\left( {1 - {2^2}} \right)}^2} + \left( {2 \times 0.1 \times 2} \right)^2} }} = \frac{{1.07703}}{{3.02654}} = 0.3559\)

The amplitude of the force transmitted:

F_T = (T.R) F = (0.3379) × 60

∴ FT = 21.35 N

Concept:

To calculate lowest critical speed (w)

\(\omega = 2\pi f = 2\pi \frac{N}{{60}}\) 

f = frequency, N = rpm, ∴ f = N/60

⇒ N = 60 f

Dunkerley’s empirical formula

\(\frac{1}{f} = \frac{1}{{{f^2}}} + \frac{1}{{f_2^2}}\)

 Calculation:

\(\frac{1}{{{f^2}}} = \frac{1}{{{{200}^2}}} + \frac{1}{{{{310}^2}}}\)

\(\Rightarrow f = 168.059\) 

∴ N = 60 f = 10083.55 rpm

Concept:

For long bearing: fixed-fixed end condition

\(k = \frac{{192\;EI}}{{{\ell ^3}}}\)

\({\omega _n} = \sqrt {\frac{k}{m}} = \sqrt {\frac{{192\;EI}}{{m{\ell ^3}}}} \)

Calculation:

\(I = \frac{{\pi {d^4}}}{{64}} = \frac{{\pi\; \times \;{{\left( {0.02} \right)}^4}}}{{64}} = 7.854 \times {10^{ - 9}}{m^4}\)

\({\omega _n} = \sqrt {\frac{{192\; \times \;2\; \times \;{{10}^{11}}\; \times \;7.854\; \times \;{{10}^{ - 9}}}}{{19.2\; \times \;{{\left( {0.5} \right)}^3}}}} = 354.5\;rad/s\)

\({f_n} = \frac{{{\omega _n}}}{{2\pi }} = \frac{{354.5}}{{2\pi }} = 56.42\;\;Hz\)

N = 60 f_n = 60 × 56.42

∴ N = 3385.13 rpm

Fluid medium ⇒ viscous damping

m = 4 kg, F₀ = 50 N

Resonance amplitude (X) = 20 mm

\(X = \frac{{{F_0}}}{{2\xi k}},\;T = \frac{1}{f} = \frac{1}{{12}}s\)

\(\omega = \frac{{2\pi }}{T} = 2\pi \times 12\)

∴ ω = 24π rad/s

Now,

\(\xi = \frac{C}{{2\sqrt {mk} }}\)

\(\xi k = \frac{C}{2}\sqrt {\frac{k}{m}} = \frac{C}{2}\omega \)

\(\therefore X = \frac{{2{F_0}}}{{2C\;\omega }} = \frac{{{F_0}}}{{C\omega }}\)

\(C = \frac{{{F_0}}}{{X\omega }}\)

\(C = \frac{{50}}{{24\pi }} \times \frac{1}{{0.020}}\)

∴ C = 33.157 Ns/m

Concept:

\(X = \frac{{{F_0}/k}}{{\sqrt {{{\left( {1 - {r^2}} \right)}^2} + {{\left( {2\xi r} \right)}^2}} }}\)

Calculation:

\({\omega _n} = \sqrt {\frac{{{k_{eq}}}}{M}}\)

k_eq = 4 × k = 4 × 0.2 × 10⁶ = 0.8 × 10⁶ N/m

M = 45 kg

\(\therefore {\omega _n} = \sqrt {\frac{{0.8\; \times \;{{10}^6}}}{{45}}} = 133.33\;rad/s\)

ω = 2πn = 2π × 32 = 201.06 rad/s

\(r = \frac{\omega }{{{\omega _n}}} = \frac{{201.06}}{{133.33}}\)

∴ r = 1.508

Now,

Magnification factor of an undamped system is given as,

\(MF = \frac{1}{{\sqrt {{{\left( {1 - {r^2}} \right)}^2} + {{\left( {2\xi r} \right)}^2}} }}\)

For Undamped, ζ = 0

\(\therefore MF = \frac{1}{{\left| {\left( {1 - {r^2}} \right)} \right|}}\)

\(MF = \frac{1}{{\left| {1 - {{\left( {1.51} \right)}^2}} \right|}} = 0.7812\)

Using equation of amplitude

\(X = \frac{{{F_0}/{k_{eq}}}}{{\sqrt {{{\left( {1 - {r^2}} \right)}^2} + {{\left( {2\xi r} \right)}^2}} }}\)

\(\therefore {F_0} = \frac{{m\omega _n^2X}}{{MF}}\)

\({F_0} = \frac{{45\; \times \;{{\left( {133.3} \right)}^2}\; \times \;0.0015}}{{0.7812}}\)

∴ F­₀ = 1.536 kN ≈ 1.54 kN

Concept:

Natural frequency:

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

2ξω_n = c/m

Damping factor:

\(\xi = \frac{c}{{{c_c}}} = \frac{c}{{2\sqrt {km} }} = \frac{c}{{2m{\omega _n}}}\)

Damped frequency:

\({{\rm{\omega }}_{\rm{d}}} = \sqrt {1 - {\xi ^2}} {\omega _n}\)

Calculation:

Given: m = 0.2 kg, k = 1 kN/m = 1000 N/m, c = 8.5 Ns/m

\({\omega _n} = \sqrt {\frac{k}{m}} = \sqrt {\frac{{1000}}{{0.2}}} = 70.71\;rad/s\)

\(\xi = \frac{c}{{2\sqrt {km} }} = \frac{{8.5}}{{2\sqrt {1000 \times 0.2} }} = 0.3\)

\({{\rm{\omega }}_{\rm{d}}} = \sqrt {1 - {\xi ^2}} {\omega _n} = \sqrt {1 - {{\left( {0.3} \right)}^2}} \times 70.71 = 67.45\;rad/s\)

% change in frequency of oscillation = (ω_d – ω_n)/ω_n

= (67.45 - 70.71)/70.71 = -0.0461 = -4.6%

Viscous Damping

• Viscous damping is a type of damping mechanism in which the damping force acting on a moving body is proportional to its velocity. This type of damping is commonly used in mechanical systems to reduce oscillations and dissipate energy in a controlled manner. It is achieved by using a medium, such as oil or gas, to exert a resistive force on the moving body.

Correct Option Analysis:

The correct option is:

Option 2: Velocity of the body

In viscous damping, the damping force is directly proportional to the velocity of the moving body. Mathematically, the damping force (F_d) can be expressed as:

F_d = -c × v

Where:

• c = Damping coefficient (a constant that depends on the properties of the damping medium and the system's design).
• v = Velocity of the body.
• The negative sign indicates that the damping force acts in the direction opposite to the motion, thereby resisting it.

Viscous damping is widely used because of its simplicity and effectiveness. It is employed in many practical applications such as automotive shock absorbers, vibration isolators, and dynamic systems subjected to oscillatory forces. The relationship between the damping force and velocity ensures that the damping mechanism is more effective at higher velocities, thereby providing stability and control in dynamic systems.

Logarithmic decrement:

Logarithmic decrement is defined as the natural logarithm of the ratio of any two successive amplitudes on the same side of the mean line.